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\hypersetup{
  pdftitle={Two p-adic logarithms: determinant bounds},
  pdfauthor={GPT-6.1 Sol and GPT-6 Astra (OpenAI), Ultra},
  hidelinks,
  pdfcreator={LaTeX via pandoc}}

\title{Two p-adic logarithms: determinant bounds}
\author{GPT-6.1 Sol and GPT-6 Astra (OpenAI), Ultra}
\date{October 2026}

\providecommand{\Eta}{\mathrm{H}}
\begin{document}
\maketitle

\emph{Public domain (CC0).}

A large valuation of \(2^a-3^b\) means that the two powers agree to many
p-adic digits. The determinant method measures how far this agreement
can continue. It places products of powers in a single residue class,
obtains a nonzero algebraic determinant, and compares a height bound
with the large valuation forced by near agreement.

We use the exponential, logarithm, power laws and cyclotomic distances
proved in \href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/TR-BAKER-09.html}{p-adic analysis for linear forms in
logarithms}. The polynomial zero argument is the method proved in
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/TR-BAKER-06.html\#3-the-zero-lemma}{Two logarithms I:
interpolation determinants}. The weighted product formula is
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/NT-LOC/NT-LOC-05.html\#5-norms-and-the-normalized-product-formula}{Places
of number fields in extensions and the product formula}, Theorem 5.2.
The height convention and its power law are proved in
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-TRANS/TR-TRANS-02.html\#4-height-arithmetic-and-finite-sets}{Heights
of algebraic numbers}. The necessary determinant estimates are written
out below.

Throughout, \(v=v_p\) is normalized by \(v(p)=1\), and \(\ln\) is the
real natural logarithm. All algebraic numbers have a specified embedding
into \(\overline{\mathbb Q}_p\). We distinguish the global degree \(d\),
local ramification index \(e\), and local residue degree \(f\).

\subsection{1. Residues and principal
units}\label{residues-and-principal-units}

Put \(K_0=\mathbb Q(\alpha_1,\alpha_2)\), \(d=[K_0:\mathbb Q]\), and
\(F=\mathbb Q_p(\alpha_1,\alpha_2)\). Its residue field has \(p^f\)
elements and \([F:\mathbb Q_p]=ef\le d\). Assume \(v(\alpha_i)=0\).

The multiplicative group of the residue field is cyclic. Here is an
elementary proof. In a finite abelian group, elements whose orders have
prescribed prime power parts can be multiplied to obtain an element of
order equal to the least common multiple \(m\) of all element orders. To
see this, for each prime choose an element whose order contains the
largest occurring power of that prime, and raise it to its complementary
part; the resulting elements have pairwise coprime orders. Their product
has the product of those orders, since a power that kills the product
must kill each component. In a finite subgroup of a field's
multiplicative group, every element is a root of \(X^m-1\), so the group
has at most \(m\) elements. The element just constructed already has
\(m\) distinct powers. It generates the group.

Every nonzero residue has a unique lift to a \((p^f-1)\)-st root of
unity in \(F\). For completeness, take any unit \(a\) with that residue
and set \(g=p^f-1\). Since \(a^g\in1+\mathfrak m_F\), the powers
\((a^g)^{p^n}\) tend to one: the power laws of Lesson 9 first enter
\(v(x-1)>1/(p-1)\), and then add one at each further p-th power.
Consequently

\[
a^{p^{f(n+1)}}-a^{p^{fn}}
=a^{p^{fn}}\bigl((a^g)^{p^{fn}}-1\bigr)\longrightarrow0.
\]

The differences have valuations tending to infinity, so the sequence is
Cauchy. Its limit \(\eta\) has the original residue and satisfies
\(\eta^{p^f}=\eta\). Thus \(\eta^g=1\). If two such lifts have the same
residue, their quotient \(w\) is a principal unit with \(w^g=1\). The
prime-to-p power law in Lesson 9 gives \(v(w^g-1)=v(w-1)\) when
\(w\ne1\), which rules this out. This proves uniqueness.

Lift a generator of the residue group to \(\eta\). We may write

\[
\alpha_i=\eta^{m_i}\epsilon_i,
\qquad \epsilon_i\in1+\mathfrak m_F,
\qquad g=p^f-1.
\tag{10.1}
\]

We use the full residue group here; no condition \(\gcd(m_1,m_2,g)=1\)
is required. Choose the least integer \(t\ge0\) such that

\[
p^t>\frac e{p-1},\qquad c_0=\frac{p^t}e.
\tag{10.2}
\]

Lesson 9 proves \(v(\epsilon_i^{p^t}-1)\ge c_0>1/(p-1)\). Also

\[
\frac{p^t}e\le\frac p{p-1}\le2.
\tag{10.3}
\]

If \(t>0\), minimality gives the first inequality. If \(t=0\), it
follows from \(e\ge1\).

\subsection{2. Three integral determinant
estimates}\label{three-integral-determinant-estimates}

\textbf{Lemma 10.1 (factorials and alternants).} For \(n\ge0\),

\[
\frac n{p-1}-\frac{\ln(n+1)}{\ln p}
\le v(n!)\le\frac n{p-1}.
\tag{10.4}
\]

For \(z_1,\ldots,z_n\in\mathbb Z_p\),

\[
v\!\left(\prod_{i<j}(z_j-z_i)\right)
\ge\sum_{j=0}^{n-1}v(j!)
\ge\frac{n(n-1)}{2(p-1)}-\frac{n\ln n}{\ln p}.
\tag{10.5}
\]

Use \(0\ln0=0\). If \(0\le\nu_1<\cdots<\nu_n\), the determinant
\(\det(z_i^{\nu_j})\) satisfies the same lower bound.

\textbf{Proof.} The digit formula \(v(n!)=(n-s_p(n))/(p-1)\) was proved
in Lesson 9. At a fixed digit sum \(s=a(p-1)+c\), \(0\le c<p-1\), the
smallest nonnegative integer is \((c+1)p^a-1\): moving one digit unit
from a higher position to an unfilled lower position decreases the
number. Concavity of \(\ln(1+x)\) on \([0,p-1]\) gives
\(\ln(c+1)\ge c\ln p/(p-1)\). Hence \(s_p(n)\le(p-1)\ln(n+1)/\ln p\),
proving (10.4).

The polynomial \(\binom Zj\) is p-adically integral on \(\mathbb Z_p\).
It is integral on nonnegative integers, which are dense in
\(\mathbb Z_p\): every residue modulo \(p^h\) has such a representative.
Continuity gives the assertion on the completion. Replacing the powers
in the Vandermonde determinant by falling factorials gives

\[
\det\!\left(\binom{z_i}{j-1}\right)_{i,j=1}^n
=\frac{\prod_{i<j}(z_j-z_i)}{\prod_{j=0}^{n-1}j!}.
\]

The determinant on the left is integral. Sum (10.4), and use
\(\ln(n!)\le n\ln n\), to obtain (10.5).

Finally \(\det(Z_i^{\nu_j})\) is alternating, so it vanishes on
\(Z_i=Z_j\). Division by the monic polynomial \(Z_j-Z_i\) takes place
over \(\mathbb Z\). Each such factor is prime in
\(\mathbb Z[Z_1,\ldots,Z_n]\), because its quotient ring is a polynomial
ring over \(\mathbb Z\), an integral domain. Distinct pair factors are
not associates. They therefore all divide the alternant, with an integer
polynomial quotient. Evaluating that quotient in \(\mathbb Z_p\) and
applying (10.5) proves the last assertion. \(\square\)

\textbf{Lemma 10.2 (centering without an occupancy hypothesis).} Let
\(N=KL\), and let the row labels \(l_i\) consist of \(K\) copies of each
integer \(0,\ldots,L-1\). For any ordering of integers
\(0\le r_i\le R-1\),

\[
\left|\sum_i l_ir_i-\frac{L-1}{2}\sum_i r_i\right|
\le\frac{NL(R-1)}8.
\tag{10.6}
\]

\textbf{Proof.} The weights \(l_i-(L-1)/2\) sum to zero. Their positive
sum is \(K\lfloor L^2/4\rfloor/2\): sum the arithmetic progression on
either side of the midpoint. A weighted sum with \(r_i\in[0,R-1]\) has
maximum at \(R-1\) on the positive weights and zero on the negative
weights, and minimum at the reverse choice. Both absolute values are at
most \(K(R-1)L^2/8=NL(R-1)/8\). \(\square\)

\textbf{Lemma 10.3 (a polynomial zero lemma).} Let \(a_1,a_2\ne0\) lie
in a field of characteristic zero. Let \(E_1,E_2\subset\mathbb Z^2\) be
finite and suppose

\[
\#\{a_1^ra_2^s:(r,s)\in E_1\}\ge L,
\qquad
\#\{b_2r+b_1s:(r,s)\in E_2\}>(K-1)L.
\tag{10.7}
\]

A nonzero polynomial of partial degrees at most \(K-1,L-1\) cannot
vanish at all points

\[
\bigl(b_2(r+r')+b_1(s+s'),a_1^{r+r'}a_2^{s+s'}\bigr),
\quad (r,s)\in E_1,\ (r',s')\in E_2.
\]

\textbf{Proof.} Write \(P(X,Y)=\sum_{j=1}^q Q_j(X)Y^{e_j}\), with
distinct \(0\le e_j<L\) and nonzero \(Q_j\). Evaluation of these
monomials on the at least \(L\) distinct available values has row rank
\(q\): otherwise a nonzero polynomial of degree less than \(L\) would
have at least \(L\) roots. Choose \(q\) labels \((r_i,s_i)\in E_1\)
giving an invertible matrix \((y_i^{e_j})\), where \(x_i=b_2r_i+b_1s_i\)
and \(y_i=a_1^{r_i}a_2^{s_i}\).

The polynomial \(D(X)=\det(Q_j(X+x_i)y_i^{e_j})_{i,j}\) has nonzero
leading coefficient: it is the product of the leading coefficients of
the \(Q_j\)'s times \(\det(y_i^{e_j})\). Its degree is
\(\sum_j\deg Q_j\le(K-1)L\). Proposed vanishing gives, at each point
\((x,y)\) from \(E_2\), a homogeneous linear system with coefficient
determinant \(D(x)\) and nonzero vector \((y^{e_j})\). Thus \(D(x)=0\).
The second cardinality condition supplies more roots than its degree.
\(\square\)

\subsection{3. A close root and the analytic
determinant}\label{a-close-root-and-the-analytic-determinant}

\textbf{Lemma 10.4 (removing the common p-part).} Suppose
\(\epsilon_1,\epsilon_2\) are principal units in \(F\), \(b_i>0\), and
\(u=v_p(\gcd(b_1,b_2))\). If

\[
v(\epsilon_1^{b_1}-\epsilon_2^{b_2})>u+\frac1{p-1},
\]

there is a root of unity \(\zeta\in F\), of order dividing both \(p^u\)
and \(p^t\), such that, for \(b_i'=b_i/p^u\),

\[
\delta=\epsilon_1^{b_1'}-\zeta\epsilon_2^{b_2'},
\qquad
v(\delta)=v(\epsilon_1^{b_1}-\epsilon_2^{b_2})-u.
\tag{10.8}
\]

\textbf{Proof.} Set \(\sigma=\epsilon_1^{b_1'}/\epsilon_2^{b_2'}\). Then
\(v(\sigma^{p^u}-1)\) equals the valuation in the hypothesis. Choose a
\(p^u\)-th root \(\zeta\) maximizing \(v(\sigma-\zeta)\). For every
other root \(\xi\), maximality and the ultrametric inequality give
\(v(\sigma-\xi)\le v(\zeta-\xi)\). The cyclotomic product identity
proved in Lesson 9 gives

\[
\sum_{\xi^{p^u}=1,\ \xi\ne\zeta}v(\zeta-\xi)=u.
\]

Factoring \(\sigma^{p^u}-1\) therefore yields
\(v(\sigma-\zeta)>1/(p-1)\). Distinct p-power roots have distance
valuation at most \(1/(p-1)\), again by Lesson 9. Thus all the other
factors have exactly their root distance valuation, giving
\(v(\sigma-\zeta)=v(\sigma^{p^u}-1)-u\).

The chosen root is closer than any of its other
\(\mathbb Q_p\)-conjugates, so Krasner's lemma, in
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/NT-LOC/NT-LOC-04.html\#4-when-proximity-forces-containment}{Extensions
of complete valued fields}, Theorem 4.1, gives
\(\zeta\in\mathbb Q_p(\sigma)\subset F\). If its order is \(p^w\) with
\(w>0\), the cyclotomic ramification theorem proved in Lesson 9 gives
\(p^{w-1}(p-1)\le e\). Definition (10.2) implies \(w\le t\). Multiplying
\(\sigma-\zeta\) by the unit \(\epsilon_2^{b_2'}\) proves (10.8).
\(\square\)

Take a matrix with rows \((k,l)\), \(0\le k<K\), \(0\le l<L\), and
distinct selected columns \((r_j,s_j)\), \(0\le r_j<R\), \(0\le s_j<S\),
all in one class \(m_1r_j+m_2s_j\equiv c\pmod g\). Its entries are

\[
\binom{b_2r_j+b_1s_j}{k}\alpha_1^{p^t l r_j}\alpha_2^{p^t l s_j}.
\tag{10.9}
\]

\textbf{Lemma 10.5 (analytic valuation).} Let \(\Delta\) be a
determinant of order \(N=KL\) extracted in this way. Write
\(\Lambda=\alpha_1^{b_1}-\alpha_2^{b_2}\). If

\[
v(\Lambda)\ge c_0\left(N-\frac12\right)+u,
\]

then

\[
v(\Delta)\ge\frac{p^tNK(L-1)}{2e}-\frac{N\ln N}{\ln p}.
\tag{10.10}
\]

\textbf{Proof.} Substitute (10.1). Since all columns have one residue
class, the root-of-unity factor in each row is constant across that row;
removing these factors changes no valuation. The positive valuation of
\(\Lambda\) also forces \(m_1b_1\equiv m_2b_2\pmod g\), so the
principal-unit difference has the same valuation. We may therefore work
with \(\epsilon_1,\epsilon_2\).

Lemma 10.4 applies. Interchange the two bases, coefficients and column
coordinates if necessary so that \(p\nmid b_2'\). Put

\[
\theta=b_1'/b_2'\in\mathbb Z_p,\qquad
z_j=r_j+\theta s_j\in\mathbb Z_p,\qquad
\omega=\log(\epsilon_1^{p^t}),\qquad v(\omega)\ge c_0.
\]

The logarithm here is the p-adic logarithm. The deep-unit power laws of
Lesson 9, together with \(\zeta^{p^t}=1\), give

\[
\epsilon_2^{p^t l s_j}
=\epsilon_1^{p^t\theta l s_j}
\bigl(1-\epsilon_1^{-b_1'}\delta\bigr)^{p^t l s_j/b_2'}
=\epsilon_1^{p^t\theta l s_j}(1+\delta q_{ij}),
\qquad v(q_{ij})\ge0.
\tag{10.11}
\]

Indeed the final power has p-adic integral exponent, and the logarithm
and exponential are isometries on their deep domains. Its difference
from one has valuation at least \(v(\delta)\). All the substitutions
take place in those domains, since \(v(\delta)\ge c_0(N-1/2)>1/(p-1)\).

If \(\delta=0\), the power factor in (10.11) is exactly one; set every
\(q_{ij}=0\). This treats a possible zero difference without dividing by
zero.

The factor \(q_{ij}\) in (10.11) depends on \(l_i,s_j\), not on \(k_i\).
Within each fixed \(l\), the binomial polynomials \(\binom Xk\) may be
replaced by \(X^k/k!\) by a triangular row transformation with diagonal
entries one. Factoring \(b_2^{k_i}\) from each row, define

\[
\phi_i(z)=\frac{z^{k_i}}{k_i!}\exp(l_i\omega z),
\qquad
\Delta'=b_2^{\sum_i k_i}
\det\bigl(\phi_i(z_j)(1+\delta q_{ij})\bigr).
\tag{10.12}
\]

Here \(v(\Delta')=v(\Delta)\), and the extracted scalar has nonnegative
valuation.

Expand the determinant by choosing its ordinary part in a set \(I\) of
\(n\) rows and its error part in the other \(N-n\) rows. Expand that
summand by the columns chosen for those \(n\) rows. Every summand then
contains a factor \(\delta^{N-n}\), a pure \(n\)-row minor \(A\) of
\(\phi_i(z_j)\), and a complementary error minor \(B\). We claim

\[
v(A)\ge c_0\left(\frac{n(n-1)}2-\sum_{i\in I}k_i\right)
-\frac{n\ln n}{\ln p},
\qquad
v(B)\ge-\frac{\sum_{i\notin I}k_i}{p-1}.
\tag{10.13}
\]

The bound for \(B\) follows directly from \(z_j\in\mathbb Z_p\), unit
exponential values, integral \(q_{ij}\), and \(v(k_i!)\le k_i/(p-1)\).

For \(A\), expand

\[
\phi_i(z)=\sum_{\nu\ge k_i}\binom\nu{k_i}
l_i^{\nu-k_i}\omega^{\nu-k_i}\frac{z^\nu}{\nu!}.
\]

Its determinant is a sum over distinct orders
\(0\le\nu_1<\cdots<\nu_n\). The node determinant is
\(\det(z_j^{\nu_h})\); Lemma 10.1 bounds its valuation below by
\(n(n-1)/(2(p-1))-n\ln n/\ln p\). The coefficient determinant is an
integer polynomial in \(\omega\), homogeneous of degree
\(\sum_h\nu_h-\sum_{i\in I}k_i\). If it is nonzero this degree is
nonnegative. The factorial denominators lose at most
\(\sum_h\nu_h/(p-1)\). Since \(c_0>1/(p-1)\) and
\(\sum_h\nu_h\ge n(n-1)/2\), the resulting lower bound is exactly the
bound for \(A\) in (10.13). This argument includes \(\omega=0\); terms
of positive degree then vanish.

These series may be regrouped: \(\phi_i\) is analytic on a disc of
radius greater than one, because \(v(\omega)>1/(p-1)\). In a finite
product of its convergent coefficient series only finitely many terms
exceed any fixed positive absolute size. The summation lemma of Lesson 9
justifies the multilinear expansions and the regrouping by distinct
orders.

Now \(\sum_i k_i=N(K-1)/2\). As \(c_0>1/(p-1)\), each complete summand
has valuation at least

\[
(N-n)v(\delta)
+c_0\left(\frac{n(n-1)}2-\frac{N(K-1)}2\right)
-\frac{N\ln N}{\ln p}.
\]

For \(0\le n\le N\),

\[
(N-n)\left(N-\frac12\right)+\frac{n(n-1)}2
=\frac{N(N-1)}2+\frac{(N-n)^2}2
\ge\frac{N(N-1)}2.
\tag{10.14}
\]

Apply the ultrametric inequality to the finite row expansion, then
subtract \(N(K-1)/2\). This proves (10.10). \(\square\)

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={Determinant valuation bounds, centered row weights and the deep-unit height budget}]{../figures/padic-determinant-orders.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 1. Left: Lemma 10.5 with \(K=3,L=4,N=12,p=3,e=1,t=0,u=0\).
Under its precision hypothesis, a summand using \(n\) ordinary rows has
valuation at least \(54+(12-n)^2/2-12\ln12/\ln3\); the dots are these
lower bounds, not measured determinant valuations. The minimum is
attained in the bound at \(n=12\). Right: Lemma 10.2 with
\(K=2,L=4,N=8,R=3\). The weights are \(l_i-3/2\); selecting \(r_i=2\) on
positive weights and zero on negative weights gives the exact extremum
\(8=NL(R-1)/8\). Equations (10.6) and (10.14) prove these two
mechanisms. Bottom: the family \(\alpha_1=1+5^m,\alpha_2=1+2\cdot5^m\)
from Theorem 10.11, with \(1\le m\le12\). The curves show the calculated
height budgets \(h_1h_2/(\ln5)^2\) and \(h_1h_2/(m(\ln5)^2)\), where
\(h_i=h(\alpha_i)\). They illustrate the precision divisor in (10.43);
the curves are height factors, while the exact difference valuation is
\(m\).}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{4. The arithmetic determinant and a parameter
theorem}\label{the-arithmetic-determinant-and-a-parameter-theorem}

For the selected columns define

\[
W=(R-1)b_2+(S-1)b_1,
\qquad
\beta=\frac W2
\left(\prod_{k=1}^{K-1}k!\right)^{-2/(K(K-1))}.
\tag{10.15}
\]

\textbf{Lemma 10.6 (height bound).} If \(\Delta\ne0\), then

\[
v(\Delta)\le\frac{dN}{2ef\ln p}
\left\{\ln N+(K-1)\ln\beta
+\frac{p^tL}{2}\bigl((R-1)h(\alpha_1)+(S-1)h(\alpha_2)\bigr)\right\}.
\tag{10.16}
\]

\textbf{Proof.} The polynomial

\[
P(X,Y)=\det\!\left(\binom{b_2r_j+b_1s_j}{k_i}X^{l_ir_j}Y^{l_is_j}\right)
\in\mathbb Z[X,Y]
\]

satisfies \(\Delta=P(\alpha_1^{p^t},\alpha_2^{p^t})\). Its coefficients
are integers because each binomial is evaluated at a nonnegative
integer.

On \(|X|=|Y|=1\), replace the binomial basis in each \(l\)-block by
\((Z-W/2)^k/k!\), again without changing the determinant. Each row has
Euclidean norm at most \(\sqrt N(W/2)^k/k!\). Hadamard's inequality
gives

\[
\|P\|_{\mathbb T^2}\le N^{N/2}\beta^{N(K-1)/2}.
\tag{10.17}
\]

One can prove the inequality used here by successively subtracting
orthogonal projections of each row onto the span of earlier rows. These
operations preserve the determinant; the resulting perpendicular row
lengths multiply to its absolute value, and each is no larger than its
original row length. If the rows are dependent the determinant is zero.

Every monomial in the determinant expansion has first exponent
\(\sum_i l_ir_{\pi(i)}\). Lemma 10.2 places all such exponents in an
interval of length at most \(NL(R-1)/4\), and places all second
exponents in an interval of length at most \(NL(S-1)/4\). Divide \(P\)
by its smallest actual powers of \(X,Y\), obtaining an integer
polynomial \(Q\) of these partial degrees. Multiplication by those unit
powers changes neither its torus norm nor the valuation at the selected
embedding.

We spell out the torus form of Liouville's inequality. For a complex
polynomial of partial degrees \(a,b\), maximum modulus applied inside
the unit disc, and to the reciprocal polynomial outside it, gives

\[
|Q(x,y)|\le\|Q\|_{\mathbb T^2}\max(1,|x|)^a\max(1,|y|)^b.
\]

Apply the one-variable argument successively. At finite places an
integer polynomial satisfies the same bound without the torus factor.
The weighted product formula, applied to the nonzero value, and the
definition \(h(x)=d^{-1}\sum_w n_w\ln^+|x|_w\), therefore give

\[
v(Q(x,y))\le\frac d{ef\ln p}
\left(\ln\|Q\|_{\mathbb T^2}+a h(x)+b h(y)\right).
\]

Here the selected local weight is \(ef\). A nonzero integer polynomial
has torus norm at least one: any nonzero coefficient is the
corresponding Fourier coefficient on the torus and has absolute value at
most that norm. Thus the archimedean factor used in the product formula
has nonnegative logarithm. Insert (10.17), the two degree widths, and
\(h(\alpha_i^{p^t})=p^t h(\alpha_i)\). This proves (10.16). \(\square\)

\textbf{Theorem 10.7 (p-adic determinant parameter theorem).} Let
\(K\ge3,L\ge2\), and let \(R_1,R_2,S_1,S_2\) be positive integers. Put
\(R=R_1+R_2-1\), \(S=S_1+S_2-1\), \(N=KL\), and define \(\beta\) by
(10.15). Suppose two residue classes \(c_1,c_2\pmod g\) satisfy

\[
\begin{aligned}
\#\{\alpha_1^{p^tr}\alpha_2^{p^ts}:&\ 0\le r<R_1,\ 0\le s<S_1,
\ m_1r+m_2s\equiv c_1\pmod g\}\ge L,\\
\#\{b_2r+b_1s:&\ 0\le r<R_2,\ 0\le s<S_2,
\ m_1r+m_2s\equiv c_2\pmod g\}>(K-1)L.
\end{aligned}
\tag{10.18}
\]

If

\[
\begin{aligned}
p^tK(L-1)\ln p>{}&(d/f+2e)\ln N+(d/f)(K-1)\ln\beta\\
&+\frac{d p^tL}{2f}\bigl((R-1)h(\alpha_1)+(S-1)h(\alpha_2)\bigr),
\end{aligned}
\tag{10.19}
\]

then

\[
v(\alpha_1^{b_1}-\alpha_2^{b_2})
<\frac{p^t}e\left(N-\frac12\right)+u.
\tag{10.20}
\]

\textbf{Proof.} The two rectangles restricted to their indicated classes
are sets \(E_1,E_2\). Their sum lies in the larger rectangle and class
\(c_1+c_2\). Apply Lemma 10.3 with \(a_i=\alpha_i^{p^t}\). A row
dependence in (10.9) would be a nonzero polynomial of the stated partial
degrees vanishing on the sum set. Hence the matrix restricted to that
class has row rank \(N\), and contains a nonzero minor \(\Delta\).

If (10.20) failed, Lemmas 10.5 and 10.6 would both apply. Multiply their
comparison by \(2e\ln p/N\). The result is the opposite weak inequality
to (10.19). This contradiction proves the theorem. \(\square\)

\subsection{\texorpdfstring{5. A bound with the height floor
\(1/d\)}{5. A bound with the height floor 1/d}}\label{a-bound-with-the-height-floor-1d}

\textbf{Theorem 10.8 (two p-adic logarithms).} Suppose
\(\alpha_1,\alpha_2\) are multiplicatively independent p-adic units, and
\(b_1,b_2\) are nonzero integers. Let
\(d=[\mathbb Q(\alpha_1,\alpha_2):\mathbb Q]\), and choose real
\(A_i>1\) satisfying

\[
\ln A_i\ge\max\{h(\alpha_i),1/d\}.
\]

Define

\[
\begin{gathered}
B'=\frac{|b_1|}{d\ln A_2}+\frac{|b_2|}{d\ln A_1},\\
H=\max\left\{\ln B'+\ln\ln p,\frac{10\ln p}d\right\}.
\end{gathered}
\tag{10.21}
\]

Then

\[
v(\alpha_1^{b_1}-\alpha_2^{b_2})
\le2500\,\frac{p^d-1}{(\ln p)^4}\,
d^4\ln A_1\ln A_2\,H^2.
\tag{10.22}
\]

In particular, the same inequality holds with \(132000\) in place of
\(2500\).

\textbf{Proof.} Inverting a base when its exponent is negative makes
both exponents positive, preserving the difference, heights, degree, and
independence. We assume this has been done. Set

\[
G=p^d-1,\quad a_i=\frac{d\ln A_i}{\ln p},\quad
z=Ga_1a_2,\quad B=\frac{dH}{\ln p},\quad k=256.
\tag{10.23}
\]

We have \(B\ge10\), \(a_i\ge1/\ln p\), and \(z\ge d^2\ge1\). For the
last assertion use \(\mathrm e^x-1\ge x^2\) for \(x\ge0\), with
\(x=d\ln p\). Its derivative is \(\mathrm e^x-2x>0\), since
\(\mathrm e^x\ge1+x+x^2/2>2x\), and it vanishes at \(x=0\).

Choose

\[
\begin{aligned}
L&=\lfloor2B\rfloor+2,& K&=\lfloor k z L\rfloor+1,\\
R_1&=\left\lfloor\sqrt{GL a_2/a_1}\right\rfloor+1,&
S_1&=\left\lfloor\sqrt{GL a_1/a_2}\right\rfloor+1,\\
R_2&=\left\lfloor\sqrt{G(K-1)L a_2/a_1}\right\rfloor+1,&
S_2&=\left\lfloor\sqrt{G(K-1)L a_1/a_2}\right\rfloor+1.
\end{aligned}
\tag{10.24}
\]

Thus \(2B<L\le(11/5)B\), \(K\ge k z L\), and

\[
R_1S_1>GL\ge gL,\qquad
R_2S_2>G(K-1)L\ge g(K-1)L.
\tag{10.25}
\]

Multiplicative independence makes the products in the first rectangle
distinct. Pigeonhole among the \(g\) classes gives the first condition
in (10.18).

We first suppose the map \((r,s)\mapsto b_2r+b_1s\) is injective within
every class of the second rectangle. Pigeonhole gives the second
condition. We verify the remaining parameter inequality.

Let \(Q_K=(\prod_{j=1}^{K-1}j!)^{-2/(K(K-1))}\). Elementary integration
of \(\ln x\) gives \(\ln(j!)\ge j\ln j-j\). Since \(x\ln x\) increases
for \(x\ge1\),

\[
\sum_{j=1}^{K-1}j\ln j
\ge\int_0^{K-1}x\ln x\,dx
=\frac{(K-1)^2}2\ln(K-1)-\frac{(K-1)^2}4.
\]

For the first interval the integral is negative and its endpoint value
is zero; for every later unit interval the increasing-function
comparison applies. Consequently

\[
\ln Q_K\le-\ln(K-1)+\frac32
+\frac{\ln(K-1)-1/2}{K}
\le-\ln(K-1)+2.
\tag{10.26}
\]

The last step uses \(\ln m\le(m+1)/2\) for \(m\ge2\), obtained by
differentiating from \(m=2\). Also \(K-1\ge255zL\). Formula (10.24)
gives

\[
\begin{aligned}
W&\le\sqrt{zL}\bigl(1+\sqrt{K-1}\bigr)
\left(\frac{b_1}{a_2}+\frac{b_2}{a_1}\right),\\
\beta&\le\frac{\mathrm e^2}{2\sqrt{255}}
\left(1+\frac1{\sqrt{K-1}}\right)
\left(\frac{b_1}{a_2}+\frac{b_2}{a_1}\right)
<\frac{b_1}{a_2}+\frac{b_2}{a_1}=B'\ln p.
\end{aligned}
\tag{10.27}
\]

Indeed \(\mathrm e<3\), \(\sqrt{255}>15\), and the parenthesis is at
most two. The elementary bound \(\mathrm e<3\) follows by bounding the
tail of \(\sum1/j!\) by a geometric series. Thus \(\ln\beta\le H\).
Using \(d/f\le d\), \(e\le d\), \(p^t\ge1\), and the height budgets, a
sufficient condition for (10.19) is

\[
K(L-1)>\frac{3d}{\ln p}\ln N
+(K-1)B+L\bigl((R-1)a_1+(S-1)a_2\bigr).
\tag{10.28}
\]

This sufficient comparison is valid even when \(\ln\beta<0\), because
\(H>0\) and \(\ln\beta\le H\).

The left side minus \((K-1)B\) is at least \(k z L B\). For the height
term,

\[
L\bigl((R-1)a_1+(S-1)a_2\bigr)
\le2\sqrt{k}\,zL^2+2\sqrt z\,L^{3/2}
<72zLB.
\tag{10.29}
\]

To check the last strict inequality, divide by \(zLB\). The first term
is at most \(32\cdot11/5=70.4\); the second is at most
\(2\sqrt{(11/5)/10}<1\).

Also \(N\le257zL^2\). Since \(z\ge d^2\), we have \(d/z\le1\) and
\(d\ln z/z\le1\): the latter follows from \(d\le\sqrt z\) and
\(\ln z/\sqrt z\le2/\mathrm e<1\), found by differentiation. Using
\(1/\ln p\le1/\ln2<2\), we obtain

\[
\frac{3d\ln N}{zLB\ln p}
\le\frac{6(\ln257+1+2\ln L)}{LB}
\le\frac{6(\ln257+1+2\ln20)}{200}<1.
\tag{10.30}
\]

The middle comparison uses \(B\ge10,L\ge20\) and the decreasing function
\((\ln257+1+2\ln L)/L\). For the last comparison the elementary bounds
\(\ln257<6\), \(\ln20<3\) already suffice. Hence the right side of
(10.28), after removing \((K-1)B\), is less than \(73zLB\); its left
side is at least \(256zLB\). Theorem 10.7 applies and gives
\(v(\Lambda)<2N+u\).

It remains to treat a collision in the second rectangle. Subtract the
colliding labels. With \(n=\gcd(b_1,b_2)\), \(r_0=b_1/n\),
\(s_0=b_2/n\), the differences are \((\ell r_0,-\ell s_0)\) for a
nonzero integer \(\ell\). Put \(\sigma=\alpha_1^{r_0}\alpha_2^{-s_0}\).
Its residue has an order \(q\mid g\). The class congruence gives
\(q\mid\ell\), so

\[
q r_0\le R_2-1,\qquad q s_0\le S_2-1.
\tag{10.31}
\]

If \(q\nmid n\), then \(v(\Lambda)=0\). Otherwise \(n=q p^u n'\), with
\(p\nmid n'\). The prime-to-p power law, followed by the p-power laws of
Lesson 9, shows

\[
v(\Lambda)=v(\sigma^n-1)
\le u+v(\sigma^{q p^t}-1).
\]

If \(u\ge t\), the deep-domain formula gives equality with \(u-t\) in
place of \(u\). If \(u<t\), raising the principal unit to further p-th
powers never decreases its positive valuation, giving the displayed
inequality. Independence ensures \(\sigma^{q p^t}\ne1\). Applying the
weighted product formula to \(X-1\), as in Lemma 10.6, gives

\[
\begin{aligned}
v(\Lambda)
&\le u+\frac d{ef\ln p}
\left(p^tq(r_0h(\alpha_1)+s_0h(\alpha_2))+\ln2\right)\\
&\le u+2\bigl((R_2-1)a_1+(S_2-1)a_2\bigr)+d\\
&\le u+64zL+z.
\end{aligned}
\tag{10.32}
\]

Here (10.3) bounds \(p^t/e\), (10.31) bounds the exponents, and
\(d\le z\).

Finally, \(b_1\le a_2\mathrm e^H\). Since \(\ln^+a_2\le a_2\) and
\(a_1\ln p\ge1\),

\[
u\le\frac{\ln b_1}{\ln p}
\le\frac Bd+\frac{\ln^+a_2}{\ln p}
\le B+a_1a_2\le B+z
\le\frac{11}{100}zB^2.
\tag{10.33}
\]

In the injective case,

\[
2N+u\le\left(514\left(\frac{11}{5}\right)^2+\frac{11}{100}\right)zB^2
<2500zB^2.
\]

The collision bound (10.32) is at most \(14.2zB^2\), since
\(L\le(11/5)B\) and \(B\ge10\). Thus both cases give (10.22), because

\[
zB^2=\frac{p^d-1}{(\ln p)^4}\,d^4\ln A_1\ln A_2\,H^2.
\]

This completes the proof. \(\square\)

The factor \(p^d-1\) in these parameter choices is an upper budget for
the number of residue classes. It is not being asserted to equal the
local residue-group order. This distinction allows the height floor
\(1/d\) without inserting an additional absolute cutoff in \(H\).

\subsection{6. Coprime powers and a worked
example}\label{coprime-powers-and-a-worked-example}

\textbf{Corollary 10.9.} Let \(a>b\ge1\) be coprime positive integers,
and let \(p\nmid ab\) be prime. Then

\[
v_p(a^{p-1}-b^{p-1})
\le250000\,\frac{p-1}{(\ln p)^2}\,\ln(5a)\ln(5b).
\tag{10.34}
\]

\textbf{Proof.} First suppose \(b>1\). Coprimality makes the bases
\(a,b\) multiplicatively independent: a prime dividing either base
forces its exponent to vanish in a proposed relation. In Theorem 10.8
take \(d=1\), \(b_1=b_2=p-1\), and \(\ln A_1=\ln(5a)\),
\(\ln A_2=\ln(5b)\). These exceed their rational heights and one.

For \(b=1\), choose \(c\in\{2,3,5\}\) such that \(c\ne p\) and \(a,c\)
are multiplicatively independent. Such a prime always exists. Dependence
of \(a\ge2\) and a prime \(c\) would force \(a=c^j\) for an integer
\(j\ge1\): taking a prime valuation outside \(c\) in the relation makes
the exponent of \(a\) zero unless every prime divisor is \(c\). Thus at
most one of the three primes is dependent with \(a\), and excluding
\(p\) removes at most one more.

Set \(\alpha_1=a/c,\alpha_2=1/c\). A relation \(\alpha_1^u\alpha_2^v=1\)
is \(a^u c^{-u-v}=1\), so independence of \(a,c\) forces \(u=v=0\). Both
bases are p-adic units, their common number field is \(\mathbb Q\), and
\[
h(a/c)\le\ln\max\{a,c\}\le\ln(5a),\qquad
h(1/c)=\ln c\le\ln5.
\] Use \(\ln A_1=\ln(5a)\), \(\ln A_2=\ln5\), and again \(b_1=b_2=p-1\).
The exact identity \[
\left(\frac ac\right)^{p-1}
-\left(\frac1c\right)^{p-1}
=c^{-(p-1)}(a^{p-1}-1)
\] preserves the valuation, since \(c\ne p\).

In both cases the height choices are the stated \(\ln(5a),\ln(5b)\), and
\[
B'\ln p\le\frac{2(p-1)\ln p}{\ln5},
\qquad \ln(B'\ln p)\le3\ln p<10\ln p.
\] For the second inequality use \(\ln\ln p\le\ln p\), \(p-1\le p\), and
\(\ln(2/\ln5)<\ln2\le\ln p\). Thus \(H=10\ln p\). Substitute it into the
proved bound (10.22), and use the displayed valuation identity when
\(b=1\). This proves (10.34) for every allowed prime, including \(p=2\).
\(\square\)

For positive integers \(a,b\), the independent bases \(2,3\) are 5-adic
units. Taking \(\ln A_1=1\), \(\ln A_2=\ln3\), Theorem 10.8 gives

\[
v_5(2^a-3^b)
\le\frac{10000\ln3}{(\ln5)^4}
\max\!\left\{\ln\left(\frac a{\ln3}+b\right)+\ln\ln5,10\ln5\right\}^{2}.
\tag{10.35}
\]

If \(M=\max(a,b)\), its maximum is at most \(10\ln5+\ln(3M)\). This is
an explicit constant times \((1+\ln M)^2\). For example \(2^3-3=5\), so
its valuation is one; \(2^7-3=125\), so its valuation is three. These
are exact illustrations of the quantity being bounded, rather than
numerical evidence for the theorem.

\textbf{Corollary 10.10 (all powers).} For coprime \(a>b\ge1\), an odd
prime \(p\nmid ab\), and \(n\ge1\),

\[
v_p(a^n-b^n)
\le250000\ln(5a)\ln(5b)\frac p{(\ln p)^2}+v_p(n).
\tag{10.36}
\]

\textbf{Proof.} Let \(q\mid p-1\) be the residue order of \(a/b\). If
\(q\nmid n\), the valuation is zero. Otherwise write \(n=qm\). The base
\((a/b)^q\) is a principal unit in \(\mathbb Q_p\), whose positive
valuation is at least one and thus strictly exceeds \(1/(p-1)\). The
prime-to-p and deep p-power laws proved in Lesson 9 give

\[
v_p(a^n-b^n)=v_p(a^q-b^q)+v_p(m)
=v_p(a^{p-1}-b^{p-1})+v_p(n).
\]

Here \(p\nmid q\) and \(p\nmid(p-1)/q\). Apply Corollary 10.9 and
\(p-1\le p\). \(\square\)

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={An independent auxiliary prime embeds one base in the two-logarithm bound}]{../figures/one-base-embedding.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.1b. Choose \(c\in\{2,3,5\}\) with \(c\ne p\) and
independent of \(a\). At most one candidate is removed by dependence and
at most one by the prime \(p\). The invertible exponent change
\((u,v)\mapsto(u,-u-v)\) proves that \(a/c,1/c\) are independent. The
lower row gives the exact unit-factor identity and the height choices
\(\ln A_1=\ln(5a),\ln A_2=\ln5\). The example \(a=8,p=3\) removes
\(c=2\) by dependence and \(c=3\) by the unit condition, leaving
\(c=5\); both differences have valuation two. The complete argument is
Corollaries 10.9--10.10, and Solution 30 verifies the examples and the
height cutoff. Human-source context: the determinant estimate from
Bugeaud's freely accessible 1999 paper, cited below; the auxiliary-prime
embedding here is proved explicitly from Theorem 10.8.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/one_base_embedding.py}{Figure program}.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{7. A deep-unit refinement}\label{a-deep-unit-refinement}

The precision of the first base can improve the height factor. We state
the coefficient and convergence conditions explicitly.

\textbf{Theorem 10.11 (an E-parameter estimate).} Let
\(\alpha_1,\alpha_2\) be multiplicatively independent nonzero rationals
with \(v_p(\alpha_i)=0\). Put \(g=p-1\), and suppose

\[
v_p(\alpha_1^g-1)\ge E>\frac1{p-1}.
\tag{10.37}
\]

If \(p=2\), require also \(v_2(\alpha_2-1)\ge2\). Let \(b_1,b_2\) be
nonzero integers, \(u=v_p(\gcd(b_1,b_2))\), and require
\(p\nmid b_2/p^u\). Choose

\[
\begin{gathered}
\ln A_i\ge\max\{h(\alpha_i),E\ln p\},\\
B'=\frac{|b_1|}{\ln A_2}+\frac{|b_2|}{\ln A_1},\qquad
H_E=\max\{\ln B'+\ln(E\ln p),10E\ln p\}.
\end{gathered}
\tag{10.38}
\]

Then

\[
v_p(\alpha_1^{b_1}-\alpha_2^{b_2})
\le1300\,\frac{p-1}{E^3(\ln p)^4}\,
\ln A_1\ln A_2\,H_E^2.
\tag{10.39}
\]

Consequently this inequality also holds with \(88000\) in place of
\(1300\).

\textbf{Proof.} Invert bases with negative exponents as before.
Inversion preserves the precision conditions, because the denominators
introduced are units. Decompose \(\alpha_i=\eta^{m_i}\epsilon_i\) as in
(10.1), now in \(\mathbb Q_p\). Since \(p\nmid g\), the prime-to-p power
law gives \(v(\epsilon_1-1)=v(\alpha_1^g-1)\ge E\). Both principal units
are in the exponential domain. For odd \(p\) their positive integer
valuations are at least one, greater than \(1/(p-1)\); for \(p=2\),
(10.37) and the second-base condition give valuations at least two.

If the original difference has valuation zero, the conclusion is
immediate. Otherwise its residue factors agree, so it has the valuation
of \(\epsilon_1^{b_1}-\epsilon_2^{b_2}\). For \(b_i'=b_i/p^u\), the deep
p-power law gives

\[
\delta=\epsilon_1^{b_1'}-\epsilon_2^{b_2'},
\qquad v(\delta)=v(\alpha_1^{b_1}-\alpha_2^{b_2})-u.
\]

The matrix (10.9) is used with exponent multiplier one. In Lemma 10.5
take \(\omega=\log\epsilon_1\), so \(v(\omega)\ge E\). The assumed unit
denominator \(b_2'\) gives \(z_j=r_j+(b_1'/b_2')s_j\in\mathbb Z_p\).
Formula (10.11) holds with multiplier one and no root of unity in
\(\delta\). The coefficient of its error part remains integral whenever
\(v(\delta)>1/(p-1)\).

We give the resulting comparison, keeping the scale change visible. If
\(v(\delta)\ge E(N-1/2)\), the alternant and factorial estimates in
(10.13) give

\[
\begin{aligned}
v(A)&\ge E\left(\frac{n(n-1)}2-\sum_{i\in I}k_i\right)-\frac{n\ln n}{\ln p},\\
v(B)&\ge-\frac{\sum_{i\notin I}k_i}{p-1}
\ge-E\sum_{i\notin I}k_i.
\end{aligned}
\]

The positive coefficient \(E-1/(p-1)\) is what permits replacement of
the Taylor-order sum by \(n(n-1)/2\). Formula (10.14) then gives

\[
v(\Delta)\ge\frac{ENK(L-1)}2-\frac{N\ln N}{\ln p}.
\]

The polynomial height proof of Lemma 10.6 has \(d=e=f=1\) and exponent
multiplier one. Thus a sufficient parameter inequality is

\[
EK(L-1)\ln p>
3\ln N+(K-1)\ln\beta
+\frac L2\bigl((R-1)h(\alpha_1)+(S-1)h(\alpha_2)\bigr).
\tag{10.40}
\]

Under the two cardinality conditions (10.18), the zero lemma and these
two bounds prove \(v(\Lambda)<E(N-1/2)+u\).

Use (10.24) with

\[
G=p-1,\qquad a_i=\frac{\ln A_i}{E\ln p},\qquad
z=Ga_1a_2,\qquad B=\frac{H_E}{E\ln p},\qquad k=256.
\tag{10.41}
\]

Here \(a_i\ge1\), \(z\ge G\ge1\), and \(B\ge10\). The cardinalities,
factorial estimate and height comparison (10.25)--(10.29) apply
unchanged, with \(B' E\ln p=b_1/a_2+b_2/a_1\). In particular
\(\ln\beta\le H_E\), and the height contribution after division by
\(E\ln p\) is less than \(72zLB\).

We check the remaining logarithmic contribution. From \(E>1/G\) and
\(z\ge G\),

\[
\frac1{Ez\ln p}<\frac1{\ln p}<2,
\qquad
\frac{\ln z}{Ez\ln p}\le1.
\]

For the second bound, at \(p=2,3\) use \(1/(E\ln p)<2\) and
\(\ln z/z\le1/\mathrm e\). For \(p\ge5\), the function \(\ln z/z\)
decreases on \(z\ge G\ge4\), giving a bound \(\ln G/\ln p<1\). With
\(N\le257zL^2\), therefore,

\[
\frac{3\ln N}{EzLB\ln p}
\le\frac{6(\ln257+1+2\ln L)}{LB}<1.
\]

The left side of the sufficient parameter inequality, after removing
\((K-1)B\), is at least \(256zLB\), while its remaining right side is
less than \(73zLB\). This verifies (10.40) in the injective case.

In the collision case, take \(q,r_0,s_0\) as in (10.31). If
\(q\nmid n=\gcd(b_1,b_2)\), the valuation is zero. Otherwise
\(\sigma^q\), with \(\sigma=\alpha_1^{r_0}\alpha_2^{-s_0}\), is a deep
principal unit: its residue root has disappeared, and its two principal
factors were already deep. Hence \(v(\Lambda)=u+v(\sigma^q-1)\). The
one-place polynomial bound for \(X-1\) gives

\[
\begin{aligned}
v(\Lambda)
&\le u+\frac{q(r_0h(\alpha_1)+s_0h(\alpha_2))+\ln2}{\ln p}\\
&\le u+E\bigl((R_2-1)a_1+(S_2-1)a_2\bigr)+1\\
&\le u+32EzL+1\le u+Ez(32L+1).
\end{aligned}
\tag{10.42}
\]

Here \(Ez>EG>1\). Finally \(b_1\le a_2\mathrm e^{H_E}\), so

\[
u\le EB+\frac{\ln^+a_2}{\ln p}
\le EB+2Ez\le\frac{12}{100}EzB^2.
\]

The middle bound uses \(\ln^+a_2\le a_2\), \(EGa_1>1\), and
\(1/\ln p<2\). In the injective case,

\[
EN+u\le
\left(257\left(\frac{11}{5}\right)^2+\frac{12}{100}\right)EzB^2
=1244EzB^2<1300EzB^2.
\]

The collision bound is at most \(7.17EzB^2\). Substituting (10.41) into
\(EzB^2\) proves (10.39). \(\square\)

The coefficient condition singles out the first base whose logarithm
supplies the precision \(E\). Interchanging bases is permissible only if
the precision and second-base convergence hypotheses are checked for the
newly ordered pair. At \(p=2\), a unit such as three has \(v_2(3-1)=1\),
on the exponential boundary; its positivity alone does not provide the
deep-unit substitution used in this proof.

For a concrete family take

\[
\alpha_1=1+5^m,\qquad \alpha_2=1+2\cdot5^m,\qquad E=m,\qquad m\ge1.
\]

The two integers are coprime, because twice the first minus the second
is one. They exceed one, so they are multiplicatively independent. Their
fourth powers have distance valuation \(m\) from one, by the
prime-to-five power law. Both height budgets may be their actual
logarithms. For \(b_1=b_2=1\), \(H_E=10m\ln5\) and the exact difference
valuation is \(m\). The bound (10.39) is

\[
v_5(\alpha_1-\alpha_2)
\le520000\,
\frac{\ln(1+5^m)\ln(1+2\cdot5^m)}{m(\ln5)^2}.
\tag{10.43}
\]

The height factor is linear in \(m\), bounded explicitly by
\((m\ln5+\ln2)(m\ln5+\ln3)/(m(\ln5)^2)\). Without the precision divisor,
the product of these heights is quadratic in \(m\).

\subsection{8. Saturated bases and Kummer
degree}\label{saturated-bases-and-kummer-degree}

We now allow an arbitrary number of bases. A root which already belongs
to the number field should be incorporated into the bases before taking
an auxiliary extension. The following construction does this and proves
the resulting extension degree.

Let \(K\) be a number field of degree \(d\), embedded in \(\mathbb C\).
Let \(q\) be a prime, and assume that \(K\) contains all the q-th roots
of unity. For \(n\ge1\) and multiplicatively independent
\(\alpha_1,\ldots,\alpha_n\in K^*\), define

\[
\begin{aligned}
M_q(\boldsymbol\alpha)=\{\boldsymbol v\in\mathbb Z[1/q]^n:\ &\text{for some }T=q^t,
\ T\boldsymbol v\in\mathbb Z^n,\\
&\alpha_1^{Tv_1}\cdots\alpha_n^{Tv_n}=\beta^T
\text{ for some }\beta\in K^*\}.
\end{aligned}
\tag{10.44}
\]

No choice of complex logarithms is involved in this definition.

The complete finite-height count in
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-TRANS/TR-TRANS-02.html\#theorem-2-6-northcott-s-finite-box}{Heights
of algebraic numbers}, Theorem 2.6, bounds the number of algebraic
numbers of degree at most \(d\) and height at most \(H\ge0\) by

\[
\mathcal N(d,H)=d^2\bigl(2^{d+1}\mathrm e^{dH}+1\bigr)^{d+1}.
\tag{10.45}
\]

We use its polynomial coefficient proof and the height arithmetic of
Proposition 2.5 in that earlier lesson.

\textbf{Lemma 10.12 (finite saturation).} The set
\(M=M_q(\boldsymbol\alpha)\) is an additive subgroup of \(\mathbb Q^n\),
contains \(\mathbb Z^n\), and has finite index over it. More precisely,

\[
 [M:\mathbb Z^n]\le\mathcal N\left(d,\sum_{j=1}^n h(\alpha_j)\right).
\tag{10.46}
\]

This index is a power of \(q\). The group \(M\) has a basis of \(n\)
rational vectors over \(\mathbb Z\).

\textbf{Proof.} Powers \(T\) in (10.44) can be enlarged to a common
power of \(q\). Multiplying or dividing the corresponding \(\beta\)'s
proves closure under addition and subtraction. Integral vectors belong
to \(M\) with \(T=1\).

Every coset modulo \(\mathbb Z^n\) has a unique representative
\(\boldsymbol v\) with \(0\le v_j<1\). Choose one associated \(\beta\)
from (10.44). The product inequality and power identity in
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-TRANS/TR-TRANS-02.html\#proposition-2-5-sums-products-powers-and-conjugates}{Heights
of algebraic numbers}, Proposition 2.5, give

\[
 T h(\beta)=h\left(\prod_j\alpha_j^{Tv_j}\right)
 \le T\sum_j v_jh(\alpha_j)\le T\sum_j h(\alpha_j).
\]

Distinct representatives give distinct chosen \(\beta\)'s. Indeed, if
two choices were equal, raise their defining equations to a common power
\(T\). Their quotient would give \(\prod_j\alpha_j^{T(v_j-w_j)}=1\).
Independence forces every \(v_j=w_j\). The earlier finite-height count
(10.45) therefore bounds the number of cosets as asserted.

There are finitely many representatives, and each has a denominator
which is a power of \(q\). Choose a common denominator \(q^a\). Then
\(\mathbb Z^n\subset M\subset q^{-a}\mathbb Z^n\). The finite group
\(M/\mathbb Z^n\) is a subgroup of \((q^{-a}\mathbb Z/\mathbb Z)^n\), so
its order is a power of \(q\).

For completeness, a subgroup \(L\subset\mathbb Z^n\) containing
\(m\mathbb Z^n\), with \(m>0\), has an integral basis of size \(n\).
Prove this by induction. Its first-coordinate image is \(c\mathbb Z\)
for some \(c>0\): take the least positive first coordinate and apply
integer division. Choose \(\boldsymbol x\in L\) with first coordinate
\(c\). The kernel of that projection contains \(m\mathbb Z^{n-1}\), so
has an \((n-1)\)-element basis by induction. Any element of \(L\)
becomes an element of the kernel after subtracting the appropriate
integer multiple of \(\boldsymbol x\). These \(n\) vectors are
independent and span \(L\). Apply this to \(L=q^aM\) and divide its
basis by \(q^a\). \(\square\)

\textbf{Lemma 10.13 (the maximal q-primary root).} The q-primary roots
of unity in \(K\) form a finite cyclic group. If \(\alpha_0\) generates
it and has order \(q^u\), then \(u\ge1\), and \(\alpha_0^a\) is not a
q-th power in \(K\) when \(q\nmid a\).

\textbf{Proof.} Roots of unity have height zero, by the height power
identity. Finiteness follows from (10.45) with \(H=0\). The
finite-subgroup argument of Section 1 applies over any field: its root
count and the existence of an element of maximal exponent prove
cyclicity. The assumption that q-th roots are in \(K\) gives \(u\ge1\).
If \(\beta^q=\alpha_0^a\) with \(q\nmid a\), then \(\beta^{q^{u+1}}=1\),
while its q-th power has order \(q^u\). Thus \(\beta\) has order
\(q^{u+1}\), contradicting maximality of \(u\). \(\square\)

Fix such a generator \(\alpha_0\).

Choose a basis \(\boldsymbol v_1,\ldots,\boldsymbol v_n\) of \(M\). By
(10.44), after choosing a common power \(T=q^t\), there are
\(\theta_i\in K^*\) with

\[
 \theta_i^T=\prod_{j=1}^n\alpha_j^{T v_{ji}},
 \qquad v_{ji}=(\boldsymbol v_i)_j.
\tag{10.47}
\]

\textbf{Theorem 10.14 (the saturated Kummer condition).} The classes of
\(\alpha_0,\theta_1,\ldots,\theta_n\) in \(K^*/(K^*)^q\) are linearly
independent over \(\mathbb F_q\). Consequently

\[
 [K(\alpha_0^{1/q},\theta_1^{1/q},\ldots,\theta_n^{1/q}):K]
 =q^{n+1}.
\tag{10.48}
\]

\textbf{Proof.} Suppose \(\alpha_0^{a_0}\prod_i\theta_i^{a_i}=\beta^q\),
with integers \(a_i\) and \(\beta\in K^*\). Enlarge \(T\) in (10.47)
until it is divisible by the order of \(\alpha_0\). Raising the relation
to \(T\) shows that

\[
 \boldsymbol w=\frac1q\sum_i a_i\boldsymbol v_i\in M:
 \quad \prod_j\alpha_j^{qT w_j}=\beta^{qT}.
\]

Uniqueness of integral coordinates in the basis of \(M\) forces
\(q\mid a_i\) for \(1\le i\le n\). Dividing out the corresponding q-th
powers, we find that \(\alpha_0^{a_0}\) is a q-th power in \(K\). If
\(q\nmid a_0\), its q-th root would have order \(q^{u+1}\),
contradicting the choice of \(\alpha_0\). Thus \(q\mid a_0\) as well.
This proves independence of all the classes.

We prove the degree consequence directly. More generally, let
\(c_1,\ldots,c_r\in K^*\) have independent classes modulo q-th powers,
and choose roots \(\rho_i^q=c_i\). The field
\(L=K(\rho_1,\ldots,\rho_r)\) splits the separable polynomials
\(X^q-c_i\), since all q-th roots of unity are already in \(K\). Fix a
primitive q-th root \(\zeta\). Every \(K\)-automorphism acts by
\(\rho_i\mapsto\zeta^{t_i}\rho_i\), giving an injective group
homomorphism

\[
 \operatorname{Aut}_K(L)\longrightarrow\mathbb F_q^r.
\]

Its image is a vector subspace. If it were proper, Gaussian elimination
would give a nonzero vector \(\boldsymbol a\in\mathbb F_q^r\)
annihilating it. Choose integer representatives \(0\le a_i<q\). The
element \(\prod_i\rho_i^{a_i}\) would then be fixed by every
automorphism and would belong to \(K\). Its q-th power is
\(\prod_i c_i^{a_i}\), contradicting the independence of the classes.

Here the fixed-element assertion and the automorphism count need only
elementary separability. An embedding of an intermediate field extends
across each algebraic generator by choosing a root of the transported
minimal polynomial. In characteristic zero these roots are distinct, so
the number of embeddings of \(L\) equals the product of the successive
degrees, namely \([L:K]\). Since \(L\) is a splitting field, every such
embedding has image \(L\). If an element outside \(K\) were fixed,
choose a different root of its separable minimal polynomial and extend
that embedding to \(L\); the resulting automorphism would move it. Thus
the fixed elements are precisely \(K\).

The automorphism image is therefore all of \(\mathbb F_q^r\), and
\([L:K]=q^r\). Apply this with \(r=n+1\) and the already proved
independent classes. \(\square\)

\textbf{Lemma 10.15 (the residue index does not depend on the saturated
basis).} Suppose \(q\ne p\), a prime of \(K\) above \(p\) is fixed, and
every \(\alpha_j\) is a unit there. The group
\(\langle\alpha_0,\theta_1,\ldots,\theta_n\rangle\subset K^*\) is
independent of the basis and of the root choices in (10.47). Its
generators are p-adic units. If the residue field has size \(p^f\), the
integer

\[
 \delta_q(\boldsymbol\alpha)=
 \frac{p^f-1}{|\langle\overline{\alpha_0},\overline{\theta_1},
 \ldots,\overline{\theta_n}\rangle|}
\tag{10.49}
\]

is consequently well defined.

\textbf{Proof.} Replacing a root choice in (10.47) multiplies
\(\theta_i\) by a q-primary root of unity in \(K\), hence by a power of
\(\alpha_0\). For a change of basis write
\(\boldsymbol v_i'=\sum_j c_{ji}\boldsymbol v_j\), with the integer
matrix \(C\) invertible over \(\mathbb Z\). Raising to a common power
\(T=q^t\) shows that \(\theta_i'/\prod_j\theta_j^{c_{ji}}\) is again a
q-primary root of unity. The new group is contained in the old one.
Applying the inverse integral matrix gives the reverse inclusion.

Taking the valuation in (10.47) gives \(v(\theta_i)=0\); roots of unity
are units too. Reduction is therefore defined. Its image is a subgroup
of the cyclic group of order \(p^f-1\) proved in Section 1, so its order
divides \(p^f-1\), and the quotient in (10.49) is an integer. Basis
independence proves the final assertion. \(\square\)

For example, take \(K=\mathbb Q\), \(q=2\), \(\alpha_1=4\), and
\(\alpha_2=9\). Prime valuations show that
\(M_2(4,9)=\tfrac12\mathbb Z^2\). A saturated basis gives
\(\theta_1=2,\theta_2=3\), and \(\alpha_0=-1\). Theorem 10.14 proves

\[
 [\mathbb Q(\sqrt{-1},\sqrt2,\sqrt3):\mathbb Q]=8.
\]

Starting with \(4,9\) would instead leave both classes trivial modulo
squares. Saturation exposes the arithmetic independence needed for the
auxiliary extension.

\subsection{9. Counting heights and bounding
saturation}\label{counting-heights-and-bounding-saturation}

The index in (10.46) is finite, but its bound uses the sum of the
heights inside an exponential. A sharper count gives a product of
heights and a small power of the field degree. This is the arithmetic
input for reducing a form with many logarithms to independent bases.

Write \(\mathcal H(\gamma)=\exp h(\gamma)\) for multiplicative height.
Let \(K\) have degree \(d\). The number field containing a scalar
\(\theta\) below can be larger than \(K\).

\textbf{Lemma 10.16 (a count at one archimedean place).} Fix an
archimedean place of \(K\), represented by \(\sigma:K\to\mathbb C\), and
put \(f=1\) for a real place and \(f=2\) for a complex place. Fix
\(\phi\ne0\) algebraic and a number field \(L\supset K(\phi)\), of
degree \(\ell\). Use the product-formula absolute values of \(L\), and
set

\[
\begin{aligned}
\Phi_v&=\prod_{w\mid v}|\phi|_w^{1/\ell},&
\Phi_{\widehat v}&=\prod_{w\nmid v}|\phi|_w^{1/\ell},\\
A_v(a)&=\prod_{w\mid v}\max\{|\phi|_w,|a|_w\}^{1/\ell},&
A_{\widehat v}(a)&=\prod_{w\nmid v}\max\{|\phi|_w,|a|_w\}^{1/\ell}.
\end{aligned}
\]

Thus \(\Phi_v\Phi_{\widehat v}=1\), and
\(A_v(a)A_{\widehat v}(a)=\mathcal H(a/\phi)\). If \(X,Y\ge1\),
\(N\ge2\), and \(u=N^{1/(N-1)}\), then

\[
\#\{a\in K:A_v(a)\le X\Phi_v,
\ A_{\widehat v}(a)\le Y\Phi_{\widehat v}\}
\le (N-1)\left(1+(uXY^2)^{d/f}\right)^f.
\tag{10.50}
\]

\textbf{Proof.} Finiteness follows from the earlier finite-height count:
the numbers \(a/\phi\) lie in \(L\), have degree at most \(\ell\), and
have height at most \(\ln(XY)\). The identity for \(A_vA_{\widehat v}\)
follows by factoring \(|\phi|_w\) from each maximum and applying the
product formula to \(\phi\).

The first local inequality implies \(|\sigma a|\le R\), where
\(R=(X\Phi_v)^{d/f}\). This follows because the sum of archimedean
weights above \(v\) is \(f\ell/d\). Indeed, each embedding of \(K\) has
\([L:K]\) extensions; conjugate complex embeddings contribute weight
two, and real embeddings contribute weight one.

Put \(r=(uY^2\Phi_{\widehat v})^{-d/f}\). Draw around each \(\sigma a\)
a ball of radius \(r\) in \(\mathbb R^f\). These balls lie in the ball
of radius \(R+r\). We show that no point is in the interiors of \(N\) of
them.

Otherwise there are distinct \(a_1,\ldots,a_N\) whose images lie in a
ball of radius strictly less than \(r\), with centre \(z\). Put
\(Q=N(N-1)/2\), and let \(\Delta=\prod_{i<j}(a_j-a_i)\ne0\). Translating
the Vandermonde rows replaces \((\sigma a_i)^k\) by \((\sigma a_i-z)^k\)
without changing the determinant. Factoring the radius from its rows and
using Hadamard's inequality gives \(|\sigma\Delta|<N^{N/2}r^Q=u^Qr^Q\).
Here Hadamard's inequality follows by orthogonalizing the columns: this
preserves the determinant and weakly decreases their norms, while the
determinant of the orthogonal columns has absolute value the product of
those norms.

At a finite place \(w\nmid v\), expand the determinant into its
permutation monomials. Every monomial has total degree \(Q\) and degree
at most \(N-1\) in each \(a_i\). With \(\lambda=|\phi|_w\), its absolute
value is at most

\[
\lambda^{-Q}\prod_i\max\{\lambda,|a_i|_w\}^{N-1}.
\]

The ultrametric inequality gives the same bound for \(\Delta\). At an
archimedean embedding \(\tau\) outside \(v\), apply Hadamard to the
powers of \(\tau a_i/\tau\phi\). Each column has norm at most
\(\sqrt N\max\{1,|\tau a_i/\tau\phi|\}^{N-1}\), so the same bound has
the extra factor \(N^{f_wN/2}\), where \(f_w=1\) or two is its
archimedean weight.

Crucially, the sum of these \emph{unselected} archimedean weights is
exactly \(\ell(d-f)/d\). Multiplying their bounds and the finite-place
bounds gives

\[
\prod_{w\nmid v}|\Delta|_w
\le u^{\ell Q(d-f)/d}
Y^{2\ell Q}\Phi_{\widehat v}^{\ell Q}.
\]

The selected place contributes strictly less than \((ur)^{f\ell Q/d}\).
The full product is therefore strictly less than
\((r^{f/d}uY^2\Phi_{\widehat v})^{\ell Q}=1\), contrary to the product
formula. This proves the asserted multiplicity bound for the balls.

Integrate their indicator functions. Their sum is at most \(N-1\)
outside their boundaries, which have volume zero. Scaling a ball's
volume by its radius gives \(M r^f\le(N-1)(R+r)^f\) for the number \(M\)
of centres. Since \(R/r=(uXY^2)^{d/f}\), this is (10.50). \(\square\)

We need a uniform choice of \(N\). For \(d\ge2\), take

\[
m=\lfloor d\ln d\rfloor,\qquad N=m+1,\qquad
u=N^{1/m},\qquad u^d<8.
\tag{10.51}
\]

Here is a proof of the last inequality for every degree. For \(d=2,3\),
the choices are \(N=2,4\), giving \(u^d=4\) in both cases. For
\(d\ge4\), put \(x=d\ln d>4\). Then \(m\ge3x/4\), \(N\le5x/4\), and

\[
\frac{d\ln N}{m}
\le\frac43\left(1+\frac{\ln\ln d}{\ln d}
+\frac{\ln(5/4)}{\ln d}\right)
<\frac43\left(1+\frac38+\frac16\right)=\frac{37}{18}<\ln8.
\]

The first fraction is at most \(1/\mathrm e<3/8\), by differentiating
\(\ln t/t\). The second is less than \(1/6\), since \((5/4)^6<4\).
Finally \(\ln2=2\sum_{j\ge0}(1/3)^{2j+1}/(2j+1)>56/81\), which proves
\(\ln8>37/18\). This series follows by integrating the geometric series
on \([0,1/3]\). The elementary bounds \(8/3<\mathrm e<3\) follow
directly from its exponential series. They give \(1/2<\ln2<1\) and
\(1<\ln3<4/3\); for the last upper bound, use \((8/3)^4>3^3\). These
also verify the two small choices of \(N\).

\textbf{Theorem 10.17 (uniform twisted height count).} Let
\(\theta\ne0\) be algebraic and \(H\ge1\). The following bounds hold:

\[
\begin{aligned}
\#\{a\in K:\mathcal H(\theta a)\le H\}
&\le68d\ln d\,H^{2d} &&(d\ge2),\\
&\le31d\ln d\,H^{2d} &&(d\ge2,\ \theta\in K),\\
&\le17H^2 &&(d=1).
\end{aligned}
\tag{10.52}
\]

\textbf{Proof.} Apply Lemma 10.16 with \(\phi=1/\theta\). Both ratios
\(A_v/\Phi_v\) and \(A_{\widehat v}/\Phi_{\widehat v}\) are at least
one, and their product is \(\mathcal H(\theta a)\). Divide the first
ratio into bands \(2^{(i-1)/d}\le A_v/\Phi_v<2^{i/d}\). There are at
most \(I=1+\lfloor d\ln H/\ln2\rfloor\) bands. Within band \(i\), set
\(X_i=2^{i/d}\), \(Y_i=2^{-(i-1)/d}H\). Then
\(X_iY_i^2=2^{(2-i)/d}H^2\).

Let \(M(H)\) denote the count. Summing (10.50), using \(I\le H^d\),
\(\sum_{i\ge1}2^{2-i}=4\), and \(\sum_{i\ge1}2^{(2-i)/2}=2/(\sqrt2-1)\),
gives

\[
\begin{aligned}
M(H)&\le m(1+4u^d)H^{2d} &&(f=1),\\
M(H)&\le m\left[
\left(1+\frac{4u^{d/2}}{\sqrt2-1}\right)H^d
+4u^dH^{2d}\right] &&(f=2).
\end{aligned}
\tag{10.53}
\]

To see \(I\le H^d\), put \(t=H^d\); if \(k=\lfloor\log_2t\rfloor\), then
\(k+1\le2^k\le t\). With (10.51), the first coefficient is less than 33,
and the two coefficients in the second line are less than 31 and 32: use
\(\sqrt8<3\) and \(1/(\sqrt2-1)=1+\sqrt2<5/2\). Since \(m\le d\ln d\),
both imply the first assertion of (10.52). For \(d=1\), take \(N=2,u=2\)
instead. The first line yields \(M(H)\le H+8H^2\le9H^2<17H^2\).

If \(\theta\in K\), multiplication by \(\theta\) is a bijection, so
reduce to \(\theta=1\). The set contains \(0,1,-1\). All other elements
pair as \(a,a^{-1}\), with equal heights, and at least one member of
each pair satisfies \(|\sigma a|\le1\). Consequently a subset of
cardinality \((M(H)+3)/2\) satisfies the inequalities of Lemma 10.16
with \(X=1,Y=H,\phi=1\). At a complex place this gives
\(M(H)\le2m(1+u^{d/2}H^d)^2-3\le30d\ln d\,H^{2d}\); at a real place it
gives \(M(H)\le2m(1+u^dH^{2d})-3\le18d\ln d\,H^{2d}\). These prove the
second assertion. \(\square\)

\textbf{Theorem 10.18 (the full saturation index).} For independent
\(\alpha_1,\ldots,\alpha_n\in K^*\), define

\[
M_K(\boldsymbol\alpha)=
\{\boldsymbol v\in\mathbb Q^n:
T\boldsymbol v\in\mathbb Z^n,
\ \prod_j\alpha_j^{Tv_j}=\beta^T
\text{ for some }T\ge1,\ \beta\in K^*\}.
\tag{10.54}
\]

Let \(w_K\) be the number of roots of unity in \(K\). This is a lattice
containing \(\mathbb Z^n\), and

\[
w_K[M_K:\mathbb Z^n]
\le C_d\frac{n!\mathrm e^n}{n^n}\prod_jh(\alpha_j),
\qquad
C_d=\begin{cases}58d^{n+1}\ln d,&d\ge2,\\17,&d=1.
\end{cases}
\tag{10.55}
\]

\textbf{Proof.} Common integer powers prove that \(M_K\) is a subgroup.
The coset argument of Lemma 10.12 applies with arbitrary positive
integers \(T\): distinct representatives in \([0,1)^n\) give distinct
elements of \(K\) of height at most \(\sum h(\alpha_j)\). Thus its index
\(J\) is finite. A common denominator puts it between \(\mathbb Z^n\)
and \(T^{-1}\mathbb Z^n\); the integral basis induction there proves it
is a lattice. Its fundamental volume is \(1/J\), since the unit-volume
fundamental domain of \(\mathbb Z^n\) contains \(J\) fundamental domains
of \(M_K\).

The group of all roots of unity in \(K\) is finite by (10.45) with
height zero, and the finite-subgroup argument of Section 1 makes it
cyclic. Each \(h_j=h(\alpha_j)\) is positive: otherwise all powers of
\(\alpha_j\) have height zero and lie in \(K\), so the same
finite-height count would make two powers equal, contradicting
independence.

For \(H>1\), the open diamond

\[
S=\{\boldsymbol x:\sum_jh_j|x_j|<\ln H\}
\quad\text{has volume}\quad
V=\frac{2^n(\ln H)^n}{n!\prod_jh_j}.
\tag{10.56}
\]

In each orthant, scale the coordinates by \(h_j\) and integrate the
simplex \(\sum y_j<\ln H\); induction by its last coordinate gives
volume \((\ln H)^n/n!\). This proves the volume formula.

Average the count
\(\sum_{\boldsymbol v\in M_K}1_S(\boldsymbol t+\boldsymbol v)\) over a
fundamental domain. Translating its integrals tiles \(\mathbb R^n\), so
the integral equals \(V\) and its average equals \(JV\). Some coset
therefore has at least \(JV\) points in \(S\). There are only finitely
many in the bounded diamond. Since \(S\) is open, perturb the
translating vector to a rational vector while retaining this finite set
of points. Denote the resulting rational points by \(P_0,\ldots,P_s\);
their differences still lie in \(M_K\), and \(s+1\ge JV\).

Choose determinations \(\theta_i=\prod_j\alpha_j^{(P_i)_j}\). They are
algebraic because the coordinates are rational. The definition of
\(M_K\) gives \(\theta_i/\theta_0=\mu_i\beta_i\) for \(\beta_i\in K\)
and roots of unity \(\mu_i\) in an algebraic closure. The height power
and product identities give \(h(\theta_0\beta_i)=h(\theta_i)<\ln H\).
Multiplying by a root of unity preserves height: raise to its order and
use the power identity.

The \(\beta_i\) are distinct modulo roots of unity in \(K\). If a
quotient were a root of unity, then so would \(\theta_i/\theta_j\);
raising to a common denominator and then to its order would contradict
independence of the \(\alpha_j\). Consequently their \(w_K\)
root-of-unity multiples give \(w_K(s+1)\) distinct elements of \(K\)
counted by Theorem 10.17 with twist \(\theta_0\). Hence
\(w_KJV\le M(H)\).

Take \(H=\exp(n/(2d))\). At a complex place, (10.53) divided by
\(\exp(n)\) is less than \(d\ln d(31/\sqrt{\mathrm e}+32)<58d\ln d\),
since \(\sqrt{\mathrm e}>3/2\). At a real place the coefficient is less
than \(33d\ln d\). Substitute (10.56) to obtain (10.55). For \(d=1\),
the bound \(17H^2\) gives exactly its stated constant 17. \(\square\)

In particular, since \(w_K\ge1\) and the index is at least one,

\[
\prod_jh(\alpha_j)\ge
\frac{w_K}{C_d}\frac{n^n}{n!\mathrm e^n}.
\tag{10.57}
\]

\textbf{Corollary 10.19 (the q-primary part).} The lattice of Section 8
is exactly \(M_q=M_K\cap\mathbb Z[1/q]^n\). Therefore (10.55) also
bounds \(w_K[M_q:\mathbb Z^n]\).

\textbf{Proof.} Only the reverse inclusion needs proof. For a vector in
the intersection, write a power realizing (10.54) as \(T=q^a m\) with
\(q\nmid m\). Since its coordinates have q-power denominators,
\(q^a\boldsymbol v\) is integral. Put
\(\gamma=\prod_j\alpha_j^{q^av_j}\). The defining equation says
\(\gamma^m=\beta^{q^am}\), so \(\zeta=\gamma/\beta^{q^a}\in K\) has
order dividing \(m\). Raising to \(q^a\) is a bijection on this finite
cyclic group, by Bezout's identity for \(q^a,m\). Choose
\(\eta\in\langle\zeta\rangle\) with \(\eta^{q^a}=\zeta\). Then
\(\gamma=(\beta\eta)^{q^a}\), which realizes (10.44). The inclusion of
finite quotients gives the index inequality. \(\square\)

\textbf{Corollary 10.20 (a primitive circuit relation).} Suppose
\(\alpha_0,\ldots,\alpha_n\in K^*\) are dependent but every \(n\) of
them are independent. There are nonzero integers \(b_i\) with
\(\prod_i\alpha_i^{b_i}=1\) and

\[
|b_i|\le C_d\frac{n!\mathrm e^n}{n^n}
\prod_{j\ne i}h(\alpha_j)\qquad(0\le i\le n),
\tag{10.58}
\]

where \(C_d\) is the degree factor in (10.55).

\textbf{Proof.} The integer vectors whose products are roots of unity
have rational rank one. Two nonproportional such vectors could eliminate
a coordinate, producing a nonzero torsion relation on only \(n\) bases;
raising to the torsion order contradicts their independence. Divide any
nonzero relation vector by its coordinate gcd to obtain a primitive
vector \(\boldsymbol r\). Its product is still a root of unity, since a
power of that product is one. All \(r_i\ne0\), and every integer vector
in this rational line is an integer multiple of \(\boldsymbol r\), by
Bezout's identity for its coordinate gcd.

The vector \((r_1/r_0,\ldots,r_n/r_0)\) belongs to
\(M_K(\alpha_1,\ldots,\alpha_n)\): take \(T=|r_0|w_K\) and
\(\beta=\alpha_0^{-1}\) in (10.54). The root of unity in the relation
disappears upon raising to \(T/r_0\). Its order modulo \(\mathbb Z^n\)
is exactly \(|r_0|\), again by Bezout's identity for the coordinate gcd.
Thus \(|r_0|\le[M_K:\mathbb Z^n]\). Set \(b_i=w_Kr_i\); their product is
one, and (10.55) proves the bound for \(b_0\). Apply the same argument
after omitting each other coordinate. The vector \(\boldsymbol b\) is
the same throughout, so all the bounds hold simultaneously. \(\square\)

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={A selected archimedean packing and an exact saturated height diamond}]{../figures/height-counting-geometry.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.2. Left: the sample \(0,\pm1,\pm i\) in \(\mathbb Q(i)\)
has multiplicative height one and lies in the unit disc. For
\(N=2,d=f=2,X=Y=1\), Lemma 10.16 uses radius \(r=1/2\); the five small
discs have disjoint interiors and lie in the disc of radius \(R+r=3/2\).
The area bound is nine centres. Right: for the rational bases \(4,9\),
the full lattice is \(\tfrac12\mathbb Z^2\). The open diamond
\((\ln4)|x|+(\ln9)|y|<1\) has volume \(2/(\ln4\ln9)\); its coset average
is \(8/(\ln4\ln9)\), between two and three. The zero coset contains the
three displayed points, giving \(1/2,1,2\). Their two rational
root-of-unity multiples are six distinct numbers of height below
\(\mathrm e\). These are samples of the proved packing and averaging
steps (10.50), (10.56), not numerical substitutes for the uniform
bounds. Source mechanism: Loher and Masser, cited below. Reproducible
source:
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/height_counting_geometry.py}{height\_counting\_geometry.py}.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{10. One logarithm with its residue
order}\label{one-logarithm-with-its-residue-order}

The one-dimensional case needs the order of the original residue class,
rather than the size of a saturated residue group. Keeping that order
and the local residue degree in the height inequality gives a sharper
form of the one-logarithm bound.

\textbf{Theorem 10.21 (the actual residue order).} Let \(K\) have degree
\(d\), and let \(\mathfrak p\) be a prime above \(p\), of ramification
index \(e\) and residue degree \(f\). Normalize \(v_p(p)=1\) and
\(\operatorname{ord}_{\mathfrak p}=e v_p\). Suppose \(\alpha\in K^*\) is
a unit at \(\mathfrak p\), and let \(g=|\langle\bar\alpha\rangle|\) be
the order of its nonzero residue class. If
\(b\in\mathbb Z\setminus\{0\}\) and \(\alpha^b\ne1\), then

\[
\operatorname{ord}_{\mathfrak p}(\alpha^b-1)
\le e v_p(b)+\frac d{f\ln p}
\left(\frac{gep}{p-1}h(\alpha)+\ln2\right).
\tag{10.59}
\]

Consequently,

\[
\operatorname{ord}_{\mathfrak p}(\alpha^b-1)
\le\frac d{f\ln p}
\left(\ln(2|b|)+g\left(1+\frac1{p-1}\right)e h(\alpha)\right).
\tag{10.60}
\]

Both bounds include roots of unity, subject to the nonvanishing
condition.

\textbf{Proof.} Replacing \(b\) by \(|b|\) does not change the
valuation: for a unit \(x\), \(x^{-1}-1=-x^{-1}(x-1)\). Hence take
\(b>0\). The order \(g\) divides \(p^f-1\), so it is prime to \(p\). If
\(g\nmid b\), reduction makes the left side zero.

For every nonzero \(\eta\in K\), the normalized height formula gives

\[
\operatorname{ord}_{\mathfrak p}(\eta)\ln p
\le\frac d f h(\eta).
\]

Indeed, when the valuation is positive, the contribution of
\(\mathfrak p\) to \(d h(\eta^{-1})\) is
\(f\operatorname{ord}_{\mathfrak p}(\eta)\ln p\), and all other
contributions are nonnegative. Use \(h(\eta^{-1})=h(\eta)\). When it is
nonpositive, the inequality is immediate. These are the height and
local-degree identities proved in
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-TRANS/TR-TRANS-02.html\#3-the-arithmetic-normalization-and-the-weil-height}{Heights
of algebraic numbers} and \href{https://kokunoyumeto.github.io/open-math-courses-public/courses/NT-LOC/NT-LOC-05.html}{Places of
number fields in extensions and the product formula}.

First suppose \(\alpha\) is not a root of unity and \(g\mid b\). Put
\(\beta=\alpha^g\). Its positive finite valuation \(v_p(\beta-1)\) is at
least \(1/e\). Let \(t\) be the least nonnegative integer with
\(p^t>e/(p-1)\). The proved entrance-time and power-step statements in
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/TR-BAKER-09.html\#4-taking-powers-until-the-exponential-applies}{p-adic
logarithms: power series, heights and the one-logarithm bound} give

\[
c_t=v_p(\alpha^{gp^t}-1)>\frac1{p-1},
\qquad p^t\le\frac{ep}{p-1}.
\]

The second inequality follows from minimality when \(t>0\). When
\(t=0\), its right side is greater than one since \(e\ge1\).

Write \(b=g p^a m\), with \(p\nmid m\). Then \(a=v_p(b)\).
Prime-to-\(p\) powers preserve positive valuation. After step \(t\),
each \(p\)-power adds one. Earlier positive valuations increase strictly
at each step, including a step on the boundary. Thus, whether \(a\ge t\)
or \(a<t\),

\[
v_p(\alpha^b-1)\le a+c_t.
\]

Apply the local height inequality to \(\eta=\alpha^{gp^t}-1\ne0\). The
proved height power and sum inequalities give
\(h(\eta)\le gp^t h(\alpha)+\ln2\). Multiplying the preceding valuation
bound by \(e\) therefore yields

\[
\operatorname{ord}_{\mathfrak p}(\alpha^b-1)
\le e a+\frac d{f\ln p}
\left(gp^t h(\alpha)+\ln2\right).
\]

Use \(p^t\le ep/(p-1)\) to obtain (10.59).

Now suppose \(\alpha\) is a root of unity. Its height is zero by the
height power identity. If the left side is positive, the nontrivial root
\(\alpha^b\) has \(p\)-power order: write its order as \(p^u s\),
\(p\nmid s\), and use the prime-to-\(p\) power step to show that a root
of order dividing \(s\) congruent to one must be one. The completely
proved cyclotomic distance formula in
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/TR-BAKER-09.html\#5-cyclotomic-ramification-and-distances-between-roots}{the
same lesson, Section 5} gives

\[
\operatorname{ord}_{\mathfrak p}(\alpha^b-1)
\le\frac e{p-1}\le\frac{d\ln2}{f\ln p}.
\]

For the second inequality use \(ef\le d\) and \(\ln p\le(p-1)\ln2\),
which follows by induction from \(p\le2^{p-1}\) for integers \(p\ge2\).
This proves (10.59) for torsion as well.

Finally \(v_p(b)\ln p\le\ln|b|\) and \(ef\le d\), so
\(e v_p(b)\le d\ln|b|/(f\ln p)\). Substitute in (10.59) and use
\(p/(p-1)=1+1/(p-1)\) to obtain (10.60). \(\square\)

For example, take \(K=\mathbb Q\), \(\alpha=4\), \(p=5\). The original
residue has order \(g=2\), while the saturated basis \(2\), together
with \(-1\), generates all four nonzero residues. Thus (10.49) gives
saturated index one, whereas the original one-dimensional residue index
is two. The parameters serve different bounds.

The exact power laws give valuation zero for odd \(b\), and \(1+v_5(b)\)
for even \(b\). For \(b=2\cdot5^k\), (10.60) reads

\[
1+k\le k+\frac{7\ln4}{2\ln5}.
\]

At the torsion boundary, \(K=\mathbb Q\), \(p=2\), \(\alpha=-1\),
\(b=1\), the height term vanishes and (10.60) is the equality
\(1=\ln2/\ln2\).

\textbf{Corollary 10.22 (a rational one-base reduction).} Let
\(\alpha_1,\ldots,\alpha_n\in K^*\) be units at \(\mathfrak p\), and let
\(\zeta\in K\) be a root of unity of order \(w\). Let
\(c_0,c_1,\ldots,c_n\) and \(b_1,\ldots,b_n\) be integers, with
\(\boldsymbol b\ne0\), and suppose

\[
b_j=\frac uv c_j\quad(1\le j\le n),
\qquad u\ne0,\quad v\ge1,\quad \gcd(u,v)=1.
\]

Put \(\gamma=\zeta^{c_0}\prod_j\alpha_j^{c_j}\),
\(\Xi=\prod_j\alpha_j^{b_j}\), and \(B_*=\min_{b_j\ne0}|b_j|\). If
\(\gamma\) is not a root of unity and \(\Xi\ne1\), then, with
\(g_\gamma=|\langle\bar\gamma\rangle|\),

\[
\operatorname{ord}_{\mathfrak p}(\Xi-1)
\le\frac d{f\ln p}
\left(\ln(2wB_*)+
g_\gamma\left(1+\frac1{p-1}\right)e h(\gamma)\right).
\tag{10.61}
\]

\textbf{Proof.} The coefficient identities give
\(\Xi^v=\zeta^{-uc_0}\gamma^u\), and hence \(\Xi^{vw}=\gamma^{uw}\ne1\).
All these numbers are local units. Factoring \(X^{vw}-1\) by \(X-1\) and
evaluating at \(\Xi\) shows that
\(\operatorname{ord}_{\mathfrak p}(\Xi-1)\le
\operatorname{ord}_{\mathfrak p}(\gamma^{uw}-1)\): the remaining finite
sum consists of integral units and is therefore integral. Apply (10.60)
to \(\gamma\) and exponent \(uw\). Since \(v b_j=u c_j\) and \(u,v\) are
coprime, \(u\) divides every \(b_j\). Thus \(|u|\le B_*\), proving
(10.61). \(\square\)

This reduces a form whose coefficient vector is rationally proportional
to one integral vector to a single unit, with an exact exponent budget.
Applying it to a many-logarithm estimate also requires a proved bound
for \(h(\gamma)\) in terms of the original heights; the saturation
bounds of Section 9 supply one of the ingredients for that step.

\subsection{11. Integral derivative bases and a multiplicity
reduction}\label{integral-derivative-bases-and-a-multiplicity-reduction}

The many-logarithm argument changes the torus coordinates and then
differentiates in a rational hyperplane. Two facts matter: the change of
derivative basis must preserve p-adic integrality, and the zero estimate
must allow different allocations of points at its different dimension
steps. We establish both, keeping the coordinate degrees separate.

\subsubsection{A minor of minimum
valuation}\label{a-minor-of-minimum-valuation}

\textbf{Lemma 10.23 (integral derivative coordinates).} Let
\(2\le r\le n\). Suppose \(L_0=z_0,L_1,\ldots,L_r\) are independent
integral linear forms, with \(L_i=a_{i0}z_0+\sum_{j=1}^n a_{ij}z_j\),
and \(\sum_j b_jz_j=B_0L_0+\sum_{i=1}^rB_iL_i\), where
\(b_j\in\mathbb Z\), \(B_i\in\mathbb Q\), \(b_n\ne0\), and \(B_r\ne0\).
Set

\[
C_{ij}=b_n a_{ij}-b_j a_{in}\quad(1\le i\le r,\ 1\le j<n),
\qquad E_i=Y_i\frac{\partial}{\partial Y_i}.
\tag{10.62}
\]

The derivations

\[
\delta_j=\sum_{i=1}^rC_{ij}E_i
=\frac1{B_r}\sum_{i<r}C_{ij}(B_rE_i-B_iE_r)
\quad(1\le j<n)
\tag{10.63}
\]

span the torus hyperplane \(\sum_iB_i x_i=0\), of dimension \(r-1\). Let
\(C'\) be the first \(r-1\) rows of \(C\). Choose \(r-1\) columns \(J\)
whose nonzero determinant has minimum \(v_p\) among all such minors.
Then the corresponding \(\delta_j\), \(j\in J\), are independent, and
every \(\delta_j\) has an expression

\[
\delta_j=\sum_{k\in J}\lambda_{kj}\delta_k,
\qquad \lambda_{kj}\in\mathbb Q\cap\mathbb Z_p.
\tag{10.64}
\]

Vanishing of all jets of total order at most \(T\) for these selected
derivations and \(\partial_{Y_0}\) is therefore equivalent to vanishing
for all the displayed derivations and \(\partial_{Y_0}\). The changes of
jets have p-adically integral coefficients.

\textbf{Proof.} Comparing the coefficients of \(z_j\), \(j\ge1\), gives
\(b_j=\sum_iB_i a_{ij}\). Hence \(\sum_iB_i C_{ij}=0\), proving the
equality in (10.63).

The \(r\) row vectors \(a_i=(a_{i1},\ldots,a_{in})\) are independent.
Otherwise a nonzero combination of the \(L_i\) would be a multiple of
\(L_0\), contrary to the assumed independence. Suppose a combination
\(a=\sum_{i<r}q_i a_i\) annihilates the vectors \(b_ne_j-b_je_n\),
\(j<n\). Then \(b_n a_j-b_j a_n=0\) for every \(j<n\), so
\(a=(a_n/b_n)b\). Since \(b=\sum_iB_i a_i\) and \(B_r\ne0\), row
independence first forces \(a_n/b_n=0\), then all \(q_i=0\). Thus \(C'\)
has rank \(r-1\). Its column images in the full matrix are determined by
their first \(r-1\) coordinates through \(\sum_iB_i C_{ij}=0\). They
span exactly the stated hyperplane.

Solve \(C'_j=C'_J\lambda_j\) by Cramer's rule. Each numerator is zero
or, up to sign, another \((r-1)\)-column minor. Minimality of the chosen
denominator's valuation therefore gives \(v_p(\lambda_{kj})\ge0\). This
proves (10.64).

The Euler derivations commute with one another and with
\(\partial_{Y_0}\). Expanding a product of the linear combinations
(10.64) expresses every jet of order at most \(T\) as a combination of
selected jets of that order. The coefficients are products and sums of
the \(\lambda_{kj}\), multiplied by integer multinomial coefficients, so
they are in \(\mathbb Z_p\). Conversely the selected derivations are
among the original ones. This proves both directions of the vanishing
assertion. \(\square\)

These operators also preserve the coordinate degree bounds and local
integral coefficients. In fact,
\(\delta_j(Y^{\boldsymbol m})=(\sum_iC_{ij}m_i)Y^{\boldsymbol m}\). For
a polynomial of degrees at most \(D_i\ge1\), put
\(M_j=\max\{1,\sum_i|C_{ij}|D_i\}\). On a monomial, the multiplier of
\(\partial_{Y_0}^{t_0}\prod_j\delta_j^{t_j}\) is an integer, and its
ordinary absolute value is at most

\[
D_0^{t_0}\prod_jM_j^{t_j}.
\tag{10.65}
\]

Indeed the additive factor is the falling factorial of its exponent,
bounded by \(D_0^{t_0}\), and each Euler factor has the displayed bound.
Nonzero terms retain their torus exponents and lose exactly \(t_0\) in
the additive exponent, so distinct input monomials do not merge. At
finite places the integral multipliers introduce no new denominators.

\subsubsection{Weighted relations with a sharp degree
constant}\label{weighted-relations-with-a-sharp-degree-constant}

We shall use the complete upper successive-minima inequality and
subgroup-minor calculation in
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/TR-BAKER-08.html\#the-subgroup-polynomial-and-its-integer-minors}{Modern
estimates: Matveev and Waldschmidt}. The following deduction uses the
Euclidean radius \(1/\sqrt r\) in the weighted estimate.

\textbf{Lemma 10.24 (weighted independent relations).} Let
\(\mathcal M\subseteq\mathbb Z^r\) have rank \(1\le\nu\le r\), with
basis matrix \(U\) of size \(r\) by \(\nu\). Let \(A_j>0\). If
\(|\det U_J|\prod_{j\in J}A_j\le B\) for every \(\nu\)-row set \(J\),
there are independent \(u_1,\ldots,u_\nu\in\mathcal M\) with

\[
\prod_{i=1}^{\nu}\sum_jA_j|u_{ij}|
\le K_{r,\nu}B,\qquad
K_{r,\nu}=2^\nu(r/\pi)^{\nu/2}
\Gamma(1+\nu/2)\binom r\nu^{1/2}\le r^\nu.
\tag{10.66}
\]

They span \(\mathcal M_{\mathbb Q}\); an integral basis is not asserted.

\textbf{Proof.} Give its real span the weighted Euclidean metric
\(\sum_jA_j^2x_j^2\). The Gram determinant identity gives covolume
\((\sum_{|J|=\nu}\det(U_J)^2\prod_{j\in J}A_j^2)^{1/2}\le\binom r\nu^{1/2}B\).
The unit ball of \(\sum_jA_j|x_j|\) contains the weighted Euclidean ball
of radius \(1/\sqrt r\), by Cauchy--Schwarz. Its \(\nu\)-volume is
therefore at least \((\pi/r)^{\nu/2}/\Gamma(1+\nu/2)\). The complete
volume and upper successive-minima arguments in the earlier lesson give
independent lattice vectors with product of these norms at most
\(2^\nu\) times the covolume divided by that volume. This proves the
first inequality.

For the second use \(\binom r\nu\le r^\nu/\nu!\). It suffices to show
\(a_\nu=2^\nu\Gamma(1+\nu/2)/(\pi^{\nu/2}\sqrt{\nu!})\le1\). The
Gaussian integral gives \(a_0=a_1=1\), and integration by parts in the
defining gamma integral gives \[
\frac{a_{\nu+2}}{a_\nu}=\frac2\pi
\sqrt{\frac{\nu+2}{\nu+1}}<1\quad(\nu\ge0),
\] since \(\pi>3>2\sqrt2\). For completeness, \(\pi>3\) follows from
\(\pi/4=\int_0^1(1+x^2)^{-1}\,dx\): the first eight terms of the
integrated geometric series give the strict lower bound
\(\sum_{j=0}^7(-1)^j/(2j+1)>3/4\). Its omitted remainder is the integral
of \(x^{16}/(1+x^2)>0\). Thus both parity sequences decrease from one.
This proves \(K_{r,\nu}\le r^\nu\). \(\square\)

\subsubsection{Unequal allocations in the zero
estimate}\label{unequal-allocations-in-the-zero-estimate}

Let \(k\) be algebraically closed of characteristic zero, and put
\(G=\mathbb G_a\times\mathbb G_m^r\). Write
\(g^s=(s,\vartheta_1^s,\ldots,\vartheta_r^s)\), where the
\(\vartheta_i\in k^*\) are multiplicatively independent. For nonzero
\(\boldsymbol B\in\mathbb Q^r\), let \(W\) be the span of
\(\partial_{Y_0}\) and the Euler directions in the hyperplane
\(\boldsymbol B\cdot\boldsymbol x=0\). Its dimension is \(r\).

\textbf{Theorem 10.25 (multiplicity and dimension reduction).} Suppose
\(r\ge2\), and let \(S,T\) be nonnegative integers. Choose nonincreasing
nonnegative integer allocations

\[
S_0\ge\cdots\ge S_r,\quad T_0\ge\cdots\ge T_r,\qquad
\sum_{m=0}^rS_m\le S,\quad\sum_{m=0}^rT_m\le T.
\tag{10.67}
\]

Let the positive coordinate bounds \(D_0,\ldots,D_r\) satisfy
\(D_0\ge\max\{D_1,\ldots,D_r,T_r+r\}\), and assume

\[
\begin{aligned}
(S_m+1)\binom{T_m+m+1}{m+1}
&>(m+1)!D_0
\max_{\substack{J\subseteq\{1,\ldots,r\}\\|J|=m}}
\prod_{j\in J}D_j &&(0\le m<r),\\
(S_r+1)\binom{T_r+r}{r}
&>(r+1)!D_0\prod_{j=1}^rD_j.
\end{aligned}
\tag{10.68}
\]

If a nonzero polynomial of these coordinate bounds vanishes at \(g^s\),
\(0\le s\le S\), with all \(W\)-jets of total order at most \(T\), then
there are \(1\le\nu<r\) and independent integral vectors
\(u_1,\ldots,u_\nu\), whose rational span contains \(\boldsymbol B\).
For any \(A_j>0\), put \(R_i=\sum_jA_j|u_{ij}|\) and
\(C_{r,\nu}=\nu!K_{r,\nu}\le\nu!r^\nu\). At least one of

\[
(S_\nu+1)\binom{T_\nu+\nu-1}{\nu-1}\prod_iR_i
\le C_{r,\nu}\max_{|J|=\nu}\prod_{j\in J}A_jD_j,
\tag{10.69}
\]

\[
(S_\nu+1)\binom{T_\nu+\nu}{\nu}\prod_iR_i
\le(\nu+1)C_{r,\nu}D_0
\max_{|J|=\nu}\prod_{j\in J}A_jD_j
\tag{10.70}
\]

holds.

\textbf{Proof.} We use the fully proved stabilizer, jet and
coordinate-degree arguments of the
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/TR-BAKER-08.html\#intersections-and-the-multiplicity-estimate}{multiplicity
subsection of the earlier lesson}, explaining the allocation step
explicitly. Let \(N=r+1\). For \(1\le j\le N+1\), let \(I_j\) be
generated by the translates of all \(W\)-derivatives of \(P\) of order
at most \(\sum_{m=0}^{j-2}T_m\), at translating points \(g^s\) with
\(0\le s\le\sum_{m=0}^{j-2}S_m\); empty sums are zero. Write
\(X_j=Z(I_j)\). Translations and invariant derivatives preserve the
coordinate bounds. Hence
\(G\supsetneq X_1\supseteq\cdots\supseteq X_{N+1}\), and the last set
contains the identity by (10.67). Every set is nonempty.

Choose the first \(j\) for which \(X_j,X_{j+1}\) have the same
dimension, and a common component \(V\) of that dimension. Such a
\(j\le N\) exists, since \(\dim X_1\le N-1\). Before it each dimension
falls, so \(\dim V\le N-j\). As in the cited stabilizer proof, define
\(E=\{x:x+V\subseteq X_j\}\) and \(H=\{x:x+V=V\}\). The set \(E\) is a
finite union of \(H\)-cosets. Its defining ideal, obtained by
translating the generators of \(I_j\) by \(V\), has dimension \(\dim H\)
and the same coordinate degree bounds. That ideal vanishes through order
\(T_{j-1}\) on \(\{g^s:0\le s\le S_{j-1}\}+H\), because the added
derivatives and translates are among those defining \(X_{j+1}\).

Replace \(H\) by its identity component \(H_0\), and put
\(c=N-\dim H_0\). Since \(\dim H_0\le\dim V\), we have \(c\ge j\).
Monotonicity gives at least \(S_{c-1}+1\) point cosets and at least
\(T_{c-1}\) jets on them. All these cosets are distinct. Indeed, if
\(g^q\in H_0\) for \(q\ne0\), the nonzero additive coordinate forces the
additive part of \(H_0\) to be \(\mathbb G_a\). Properness then supplies
a nonzero torus character \(u\) with \(\boldsymbol\vartheta^{qu}=1\),
contrary to multiplicative independence.

Put \(\sigma=\dim(W/(W\cap T(H_0)))\). The complete positive-filtration
upper and lower estimates (8.72)--(8.76) give, for every
codimension-\(c\) coordinate set \(J\),

\[
(S_{c-1}+1)\binom{T_{c-1}+\sigma}{\sigma}\delta(H_0;J)
\le c!\prod_{j\in J}D_j.
\tag{10.71}
\]

Here \(\delta(H_0;J)\) is the nonnegative integral coefficient of the
complementary monomial in the subgroup Hilbert polynomial. To extract a
coefficient from the lower polynomial inequality, send exactly the
coordinates in that complementary monomial to a common infinity, keeping
the others one. The polynomials are squarefree of the same degree, so
division by that power leaves exactly the desired coefficient. This is
the coefficient argument already proved in the earlier lesson.

Since \(W\) is a hyperplane, \(\sigma=c-1\) when \(T(H_0)\subseteq W\),
and \(\sigma=c\) otherwise. In the second case \(c\le r\), and some
\(\delta\) is at least one. The maximum product of \(c\) coordinate
bounds is \(D_0\max_{|J|=c-1}\prod_{j\in J}D_j\), because \(D_0\) is
largest. Inequality (10.71) contradicts (10.68) with \(m=c-1\). Hence
\(T(H_0)\subseteq W\).

Write \(H_0=H_a\times H_m\), and let \(\mathcal M\) be the primitive
character lattice of \(H_m\), of rank \(\nu\). The subgroup and tangent
classification in the earlier lesson shows that
\(\boldsymbol B\in\mathcal M_{\mathbb Q}\). Thus \(\nu\ge1\). If
\(\nu=r\) and \(H_a=\{0\}\), then \(c=r+1,\sigma=r\), and (10.71)
contradicts the last line of (10.68). If \(\nu=r\) and
\(H_a=\mathbb G_a\), then \(c=r,\sigma=r-1\). We may decrease the
available budgets to \(S_r,T_r\); its unique torus coefficient is one.
Thus \((S_r+1)\binom{T_r+r-1}{r-1}\le r!\prod_{j=1}^rD_j\). The last
line of (10.68), and
\(\binom{T_r+r}{r}/\binom{T_r+r-1}{r-1}=(T_r+r)/r\), contradict this
since \(D_0\ge T_r+r\). Therefore \(1\le\nu<r\).

Let \(U\) be a basis matrix of \(\mathcal M\). If \(H_a=\mathbb G_a\),
the subgroup coefficients are \(|\det U_J|\), \(c=\nu,\sigma=\nu-1\).
Decrease its budgets to \(S_\nu,T_\nu\). Multiply (10.71) by
\(\prod_{j\in J}A_j\), and apply Lemma 10.24 to the resulting bounds on
every weighted minor. The independent vectors supplied by that lemma
span \(\mathcal M_{\mathbb Q}\), so their span contains
\(\boldsymbol B\). Its factor \(K_{r,\nu}\) times \(\nu!\) gives
(10.69). If \(H_a=\{0\}\), then \(c=\nu+1,\sigma=\nu\); its coefficients
on \(\{0\}\cup J\) are \(|\det U_J|\). The same argument uses
\((\nu+1)!D_0\) in (10.71), proving (10.70). This accounts for both
additive subgroups and completes the proof. \(\square\)

For the allocations \(S_0=\lfloor S/3\rfloor\),
\(S_i=\lfloor2S/(3r)\rfloor\) for \(i\ge1\), and
\(T_i=\lfloor T/(r+1)\rfloor\), (10.67) holds: \(r\ge2\) makes the point
allocations nonincreasing, and their sums are at most \(S,T\). Since the
\(T_i\) are equal and \(D_0\ge T_r+r\), (10.69) also implies (10.70).
Multiply (10.69) by \((T_\nu+\nu)/\nu\le D_0\); its enlarged right side
is at most the right side of (10.70). Thus the second bound alone
suffices for this choice of allocations.

\textbf{Corollary 10.26 (new algebraic bases).} If the \(\vartheta_j\)
are independent algebraic units in a number field and \(u_i\) are the
vectors in Theorem 10.25, put \(\gamma_i=\prod_j\vartheta_j^{u_{ij}}\).
They are independent units and

\[
h(\gamma_i)\le\sum_j|u_{ij}|h(\vartheta_j)\le R_i
\quad\text{if }A_j\ge h(\vartheta_j).
\tag{10.72}
\]

If \(\boldsymbol b\in\mathbb Z^r\) is in their rational span, there are
integers \(m\ge1,c_i\) with \(m\boldsymbol b=\sum_i c_i u_i\), and
\((\prod_j\vartheta_j^{b_j})^m=\prod_i\gamma_i^{c_i}\).

\textbf{Proof.} Integer products preserve local units, and the proved
height power and product inequalities give (10.72). A multiplicative
relation among the \(\gamma_i\) gives a zero integral combination of the
\(u_i\), by independence of the \(\vartheta_j\); linear independence of
the vectors makes every coefficient zero. Clearing the denominators of a
rational span expression proves the last identity. \(\square\)

\textbf{Example of a sharp weighted factor.} For \(r=2,\nu=1\), take
\(\mathcal M=\mathbb Z(3,-2)\), weights \(A_1=2,A_2=3\), and minor bound
\(B=6\). Its primitive vector has weighted norm 12, exactly
\(K_{2,1}B=2B\). In the torus put \(H_m=\{Y_1^3=Y_2^2\}\),
\(H=\mathbb G_a\times H_m\), and \(g=(1,2,3)\). For \(s=0,1,2\), its
three distinct \(H\)-cosets have characters \(Y_1^3/Y_2^2=(8/9)^s\). The
polynomial \[
P(Y_1,Y_2)=\prod_{s=0}^2(9^sY_1^3-8^sY_2^2)
\] vanishes on all three. The directions \(\partial_{Y_0}\) and
\(2E_1+3E_2\) are tangent to these cosets, so every such derivative
vanishes there. Its coordinate bounds are \(D_1=9,D_2=6\), and the
corresponding weighted relation inequality is the equality
\(3\cdot12=2\max(2\cdot9,3\cdot6)=36\). There is no transverse
multiplicity factor in this case: \(\sigma=\nu-1=0\).

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={Weighted relation geometry and three distinct positive real torus cosets}]{../figures/yu-multiplicity-geometry.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.3. Left: in weighted coordinates \((2u_1,3u_2)\), the
relation \((3,-2)\) becomes \((6,-6)\). Its diamond norm is 12; the
Euclidean circle of radius \(12/\sqrt2\) lies in the diamond and meets
that relation line at the two marked primitive vectors. This is the
equality case of (10.66) for \(r=2,\nu=1\). Right: a positive real
section of the torus projections of the three cosets above. Their
quotient character values \(1,8/9,64/81\) are distinct, and the points
\((1,1),(2,3),(4,9)\) lie on their respective curves. The additive
coordinates \(0,1,2\) are omitted from this projection. The arrow is
tangent to a coset, in the direction \(2E_1+3E_2\); it illustrates
\(\sigma=0\), not a transverse derivative. The complete subgroup, degree
and lattice arguments are (10.66), (10.69)--(10.71) and the example.
Mathematical background: the earlier complete multiplicity lesson and
Yu's free 2013 article, cited below.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_multiplicity_geometry.py}{Figure program}.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{12. Closing the one-base
branch}\label{closing-the-one-base-branch}

The many-logarithm construction may reduce to one new base before the
auxiliary polynomial is needed. A one-logarithm inequality alone does
not finish that step: its constant term must also be compared with the
product of the original heights. We give that comparison with the full
residue-order and root-of-unity factors.

Let \(n\ge2\), let \(K\) have degree \(d\), and let \(\mathfrak p\) lie
above \(p\), with ramification index \(e_{\mathfrak p}\) and residue
degree \(f\). Put \(t=f\ln p\) and \(\ln^*d=\max\{1,\ln d\}\). Take
multiplicatively independent \(\mathfrak p\)-adic units
\(\alpha_1,\ldots,\alpha_n\), integers \(b_j\) not all zero, and write

\[
\Xi=\prod_j\alpha_j^{b_j},\qquad
P_h=\prod_jh(\alpha_j),\qquad
B_* =\min_{b_j\ne0}|b_j|.
\tag{10.73}
\]

Take \(q=2\) for \(p>2\), and \(q=3\) for \(p=2\). In the second case
assume \(\zeta_3\in K\). Let \(w_q=q^u\) be the order of the q-primary
roots of unity in \(K\), and choose their generator \(\alpha_0\). Use
the intrinsic saturated residue index \(\delta(\boldsymbol\alpha)\) from
Lemma 10.15, and define

\[
\begin{aligned}
Q_n&=p^f/\delta(\boldsymbol\alpha),&
M_n&=\max\{Q_n/t^{n+1},\mathrm e^n/n^n\},\\
L&=\max\{\ln(\mathrm e^4(n+1)d),e_{\mathfrak p},t\},&
\varepsilon_n&=10^{-26}/(2n),\\
a&=\begin{cases}
7(p-1)/(p-2),&p\ge5,\ e_{\mathfrak p}=1,\\
14,&p>2\text{ in the other cases},\\
26,&p=2,
\end{cases}&
R_n&=\frac{\mathrm e a}{q}(1+\varepsilon_n)(n+1)d.
\end{aligned}
\tag{10.74}
\]

Here \(\mathrm e=\exp(1)\) is distinct from the ramification index. For
any \(c\ge206\), put

\[
\mathcal C_n=
c a^n\frac{n^n(n+1)^{n+1}d^{n+2}\ln^*d}
{n!w_qt}\,M_nL.
\tag{10.75}
\]

\textbf{Theorem 10.27 (the one-base numerical closure).} Suppose there
are independent integral forms \(L_0=z_0,L_1=a_0z_0+\sum_ja_jz_j\), and
rationals \(B_0,B_1\), with \(B_1\ne0\), such that
\(\sum_jb_jz_j=B_0L_0+B_1L_1\). Set
\(\gamma=\alpha_0^{a_0}\prod_j\alpha_j^{a_j}\), and let \(g_\gamma\) be
its actual residue order at \(\mathfrak p\). Define

\[
\delta_\gamma=\frac{p^f-1}{g_\gamma},\qquad
M_\gamma=\max\{p^f/(\delta_\gamma t^2),\mathrm e\}.
\tag{10.76}
\]

Suppose a positive \(\sigma\) satisfies

\[
h(\gamma)\le\sigma\le
R_n^{n-1}\frac{M_n}{M_\gamma}P_h.
\tag{10.77}
\]

If \(H\ge\max\{\ln B_*,(n+1)t,\ln(4d)\}\), then

\[
\operatorname{ord}_{\mathfrak p}(\Xi-1)
\le\frac1{500}\mathcal C_nP_hH
<\mathcal C_nP_hH.
\tag{10.78}
\]

\textbf{Proof.} Write \(B_1=v_1/v_2\) in lowest terms, with \(v_2>0\).
Coefficient comparison gives \(v_2b_j=v_1a_j\); hence \(v_1\mid b_j\)
and \(|v_1|\le B_*\). Also \(v_2B_0=-v_1a_0\) is integral. Raising the
resulting multiplicative identity to \(w_q\) removes \(\alpha_0\):
\(\Xi^{v_2w_q}=\gamma^{v_1w_q}\).

The vector \((a_1,\ldots,a_n)\) is nonzero, by independence of the two
forms. Thus \(\gamma\) is not a root of unity: a torsion relation would,
after a further power killing \(\alpha_0\), contradict independence of
the \(\alpha_j\). The same argument proves that \(\Xi\) is not torsion.
All powers just used are therefore different from one. The finite
geometric sum and Theorem 10.21 give

\[
\operatorname{ord}_{\mathfrak p}(\Xi-1)
\le\frac d t\left(
\ln(2w_q B_*)+
g_\gamma\frac p{p-1}e_{\mathfrak p}h(\gamma)\right).
\tag{10.79}
\]

Negative \(v_1\) is covered by that theorem. The geometric sum has
integral local-unit terms, so its valuation is nonnegative.

We need two elementary estimates. First,

\[
w_q\le\frac q{q-1}d\le2d.
\tag{10.80}
\]

Indeed
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/TR-BAKER-09.html\#5-cyclotomic-ramification-and-distances-between-roots}{Proposition
9.7}, applied at the prime \(q\), proves that \(\Phi_{q^u}\) is
irreducible over \(\mathbb Q_q\), of degree \(q^{u-1}(q-1)\). It is
consequently irreducible over \(\mathbb Q\) as well. Since its root
\(\alpha_0\) belongs to \(K\), that degree is at most \(d\), proving
(10.80). Thus \(\ln(2w_qB_*)\le2H\).

Second, if \(w_K\) is the total number of roots of unity in \(K\),
Theorem 10.18 and the index lower bound one give the uniform weakened
form

\[
P_h\ge
\frac{w_K n^n}
{58 n!\mathrm e^n d^{n+1}\ln^*d}
\ge\frac{w_q n^n}
{58 n!\mathrm e^n d^{n+1}\ln^*d}.
\tag{10.81}
\]

For \(d=1\), the proved constant 17 is smaller than 58, so the same form
holds. There is no lower bound for a single new height hidden in this
step.

Combining (10.75), (10.81) and \(M_n\ge\mathrm e^n/n^n\) gives

\[
\mathcal C_nP_h\ge
\frac c{58}a^n
\frac{n^n(n+1)^{n+1}}{(n!)^2}\frac d tL
\ge\frac c{58}a^n(n+1)\frac d tL
>2000\frac d t.
\tag{10.82}
\]

Here \(n!\le n^n\), \((n+1)^{n+1}/n^n\ge n+1\), \(a\ge7\), and
\(L\ge4+\ln3>5\). The last numerical comparison is
\(206\cdot49\cdot3\cdot5>2000\cdot58\). The logarithmic term in (10.79)
is therefore at most \(\mathcal C_nP_hH/1000\).

For the height term use \(g_\gamma/M_\gamma<t^2\), directly from
(10.76), and (10.77). Its ratio to \(\mathcal C_nP_hH\) is at most

\[
\frac{n!}{n^n}\left(\frac{\mathrm e}{q}\right)^{n-1}
(1+\varepsilon_n)^{n-1}
\frac{w_q e_{\mathfrak p}t^2}
{c a(n+1)^2d^2\ln^*d\,LH}\frac p{p-1}.
\tag{10.83}
\]

This is an exact cancellation of the \(M_n\), degree and exponential
factors. The local degree inequality gives \(e_{\mathfrak p}\le d\);
also \(L\ge t\), \(H\ge(n+1)t\), \(p/(p-1)\le2\), and (10.80) gives
\(w_q/d\le2\).

For completeness, monotonicity of \(\ln x\) gives
\(\ln(n!)\le\int_1^n\ln x\,dx+\ln n
=n\ln n-n+1+\ln n\). Consequently \(n!\mathrm e^{n-1}/n^n\le n\).
Furthermore \((1+\varepsilon_n)^{n-1}<2\): use \(\ln(1+x)\le x\), which
follows by integrating \((1+x)^{-1}\le1\), and
\(\varepsilon_n(n-1)<1/2\). The exponential series gives
\(\mathrm e<3\), since \(k!\ge2^{k-1}\) for \(k\ge2\), with strict
inequality for \(k\ge3\); hence \(\exp(1/2)<2\). This also proves
\(\ln3>1\) used above.

Thus (10.83) is at most \(8n/(q^{n-1}c a(n+1)^3)\le4n/(c a(n+1)^3)\),
because \(q\ge2,n\ge2\). Since \((n+1)^2\ge4n\) and \(n+1\ge3\), this is
at most \(1/(3ca)\le1/(3\cdot206\cdot7)<1/1000\). The height term is
also at most \(\mathcal C_nP_hH/1000\). Adding the two terms proves
(10.78). \(\square\)

We check the correspondence with the free 2013 preparation. Let
\(\kappa\) be the unique nonnegative integer with
\(p^{\kappa-1}(p-1)\le2e_{\mathfrak p}<p^\kappa(p-1)\). Set \(c_2=7/4\)
for \(p>2\), and \(c_2=13/9\) for \(p=2\). Put \(\vartheta=(p-2)/(p-1)\)
when \(p\ge5,e_{\mathfrak p}=1\), and
\(\vartheta=p^\kappa/(2e_{\mathfrak p})\) otherwise, and let
\(\theta=\vartheta/(1+\varepsilon_n)\). Direct substitution gives

\[
\mathrm e c_2q\frac{p^\kappa}{e_{\mathfrak p}\theta}(n+1)d=R_n.
\tag{10.84}
\]

In the first case \(\kappa=0\); in the other cases the factors are
respectively \(7\) and \(26/3\) before multiplication by
\(\mathrm e(1+\varepsilon_n)(n+1)d\). Thus (10.77) is precisely the
one-base height-product hypothesis of that preparation. Formula (10.75)
is its first numerical coefficient, with any of its listed choices of
\(c\), all at least 206. Its parameter \(h^{(1)}\) is at least
\(\ln B_*\) and \((n+1)t\); its term
\((n+1)(a_0^{(1)}n+a_1^{(1)}+\ln(a_0^{(1)}n+a_2^{(1)})+\ln d)\) is
larger than \(\ln(4d)\), since the listed constants satisfy
\(a_0^{(1)}>3,a_1^{(1)}>2,a_2^{(1)}>2\). Theorem 10.27 therefore proves
the full numerical conclusion in this branch. The proof retains the
original saturated index for the \(n\) bases, and the actual unsaturated
residue order for the one new base.

\subsection{13. Binomial preparation of the auxiliary
polynomial}\label{binomial-preparation-of-the-auxiliary-polynomial}

The auxiliary polynomial uses shifted powers of binomial polynomials and
binomial polynomials in the Euler eigenvalues. We prove their
independence, their exact denominator control and their equivalence to
ordinary jets. These are algebraic statements; no logarithm estimate is
assumed in their proofs.

\subsubsection{Taylor coefficients at integral and fractional
nodes}\label{taylor-coefficients-at-integral-and-fractional-nodes}

For integers \(k\ge1,\ell\ge1,m\ge0\), define

\[
\begin{aligned}
\Delta(z;k)&=\frac{(z+1)\cdots(z+k)}{k!},\quad \Delta(z;0)=1,\\
v(k)&=\operatorname{lcm}(1,\ldots,k),\\
\Theta(z;k,\ell,m)&=\frac1{m!}\frac{d^m}{dz^m}\Delta(z;k)^\ell.
\end{aligned}
\tag{10.85}
\]

\textbf{Lemma 10.28 (prepared Taylor integrality).} For every integer
\(s\),

\[
v(k)^m\Theta(s;k,\ell,m)\in\mathbb Z.
\tag{10.86}
\]

If \(q\) is prime, \(I\in\mathbb Z_{\ge0}\), and \(A,s\in\mathbb Z\),
then

\[
q^{\chi_I\ell(kI+v_q(k!))}v(k)^m
\Theta(s/q^I+A;k,\ell,m)\in\mathbb Z,
\qquad \chi_0=0,\quad\chi_I=1\ (I>0).
\tag{10.87}
\]

In particular \(v(k)^m\Theta(s/q^I+A;k,\ell,m)\) belongs to
\(\mathbb Z[1/q]\), before multiplication by the displayed q-power.

\textbf{Proof.} Generalized binomial coefficients \(\binom Nj\) are
integral for every integer \(N\): for \(N<0\), their product formula is
\((-1)^j\binom{-N+j-1}j\). Vandermonde's polynomial identity gives
\(\Delta(s+Z;k)=\sum_{j=0}^k\binom{s+k}{k-j}\binom Zj\). One proof first
counts subsets for nonnegative integer arguments and then uses
polynomial equality to extend to all arguments. For \(j\ge1\), \[
\binom Zj=\frac{(-1)^{j-1}Z}{j}
\prod_{a=1}^{j-1}(1-Z/a).
\] Each coefficient of \(Z^m\) has denominators consisting of \(j\) and
\(m-1\) numbers among \(1,\ldots,j-1\). Every factor divides \(v(k)\).
Hence multiplication by \(v(k)^m\) makes that coefficient integral. The
constant coefficient is integral too. In the \(\ell\)-th power, a
coefficient of order \(m\) is a sum of products whose orders sum to
\(m\); the same scaling therefore clears all of them. This proves
(10.86), including vanishing coefficients beyond the degree.

Fix \(F(Z)=v(k)^m\Theta(Z;k,\ell,m)\). It is a rational polynomial,
integral at every integer. At \(x=s/q^I+A\), it has no denominator at a
prime \(r\ne q\). To see this without a limiting argument, choose \(B\)
bounding the r-adic denominators of its finitely many coefficients, and
an integer \(N\equiv x\pmod{r^{B+1}}\); the inverse of \(q^I\) modulo
that power supplies \(N\). Both \(x,N\) are r-integral. Factoring
\(x^j-N^j\) shows that \(F(x)-F(N)\) is r-integral, so \(F(x)\) is
r-integral.

The coefficients of \(\Delta(Z;k)^\ell\) have denominator dividing
\((k!)^\ell\). Dividing an ordinary derivative of order \(m\) by \(m!\)
multiplies each monomial coefficient by an integer binomial coefficient
and decreases its degree by \(m\). Thus its value at \(x\) has q-adic
denominator at most \(q^{\ell v_q(k!)+Ik\ell}\); the integer factor
\(v(k)^m\) cannot worsen this bound. Together with the preceding
assertion at every other prime this proves (10.87) for \(I>0\). For
\(I=0\), (10.86) already gives an integer and no q-factor is needed.
\(\square\)

\subsubsection{Independence of the shifted
powers}\label{independence-of-the-shifted-powers}

\textbf{Lemma 10.29 (a binomial polynomial basis).} For \(k,L\ge1\), the
\(kL\) polynomials

\[
F_{a,\ell}(X)=\Delta(X+a;k)^\ell,
\qquad 0\le a<k,\quad1\le\ell\le L,
\tag{10.88}
\]

are a basis of the polynomials of degree at most \(kL\) vanishing at
\(X=-k\). For fixed \(\ell\), their top \(k\) coefficient matrix
satisfies

\[
\det\left([X^{k\ell-j}]F_{a,\ell}\right)_{0\le j,a<k}
=\frac{\prod_{j=0}^{k-1}\binom{k\ell}{j}}
{(k!)^{k\ell}}\prod_{0\le a<b<k}(b-a)\ne0.
\tag{10.89}
\]

\textbf{Proof.} The coefficient in row \(j\), regarded as a polynomial
in \(a\), has degree \(j\) and leading coefficient
\(\binom{k\ell}j/(k!)^\ell\). Indeed expand the degree-\(k\ell\)
polynomial \(\Delta(X;k)^\ell\) after substituting \(X+a\); its highest
term gives that leading coefficient, and lower terms give smaller powers
of \(a\). Changing these row polynomials to monomials is triangular. The
monomial evaluation determinant at \(0,\ldots,k-1\) is the Vandermonde
product in (10.89). It follows either by subtracting columns and
induction, or by observing its zeros at equal columns and its leading
coefficient one. This proves the formula.

In a proposed relation, take the largest \(\ell\) occurring. Lower
levels have degree at most \(k(\ell-1)\). The top \(k\) coefficients and
(10.89) force every coefficient at this largest level to vanish.
Descending proves independence. Every polynomial in (10.88) vanishes at
\(-k\), because the factor with index \(k-a\) is zero there. The space
with this one vanishing condition has dimension \(kL\): evaluation at
\(-k\) is a nonzero linear functional on the \(kL+1\)-dimensional
polynomial space. Thus the independent family is a basis. An invertible
affine substitution preserves its independence. \(\square\)

For \(k=L=2\), Figure 10.4 shows the four basis polynomials and their
exact coefficient matrix. A shared root does not give a relation among
the polynomials.

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={Four binomial basis polynomials with a shared root and their exact coefficient matrix}]{../figures/binomial-auxiliary-basis.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.4. Left: the restrictions to the real interval
\([-3,-1]\) of \(F_{0,1},F_{1,1},F_{0,2},F_{1,2}\) in (10.88) for
\(k=2\). All vanish at \(-2\); the squared polynomials have a double
zero there. Right: their coefficients of \(X,X^2,X^3,X^4\), in the same
column order. The diagonal two-by-two block determinants are
\(-1/2,-1/4\), so the full determinant is \(1/8\). Their constant
coefficients are determined by their common vanishing at \(-2\). This is
the exact independence mechanism in Lemma 10.29 and (10.89), applied to
the auxiliary polynomial below. Human-source background: the binomial
preparation in Yu's free 2013 article, cited below.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/binomial_auxiliary_basis.py}{Figure program}.}
\end{flushleft}
\end{minipage}\par\medskip


\subsubsection{Affine changes of binomial
jets}\label{affine-changes-of-binomial-jets}

\textbf{Lemma 10.30 (Newton reparametrization).} Over a
characteristic-zero field, for \(h\ne0\) and \(c\),

\[
\Delta(hZ+c;t)=\sum_{j=0}^t d_j\Delta(Z;j),\qquad
d_j=\sum_{i=0}^j(-1)^i\binom ji\Delta(c-h(i+1);t),
\qquad d_t=h^t.
\tag{10.90}
\]

If \(h,c\in\mathbb Z[1/q]\), then all \(d_j\) belong to that ring. If
moreover \(h=q^b\), \(b\in\mathbb Z\), the triangular change and its
inverse are p-adically integral for every \(p\ne q\). Products in
several commuting variables preserve the total-order filtration.

\textbf{Proof.} The polynomials \(\Delta(Z;j)\), \(0\le j\le t\), have
degrees \(j\) and nonzero leading coefficients, so form a basis. At
\(Z=-i-1\), their values are \((-1)^j\binom ij\), zero when \(j>i\).
Inverting this triangular evaluation matrix gives (10.90). Explicitly
the inner binomial inversion sum is
\(\binom ab\sum_{i=b}^a(-1)^{i-b}\binom{a-b}{i-b}\), which is zero for
\(a>b\) and one for \(a=b\), by \((1-1)^{a-b}\). Comparing leading
coefficients gives \(d_t=h^t\).

Lemma 10.28 with \(\ell=1,m=0\) makes each displayed value at a
\(\mathbb Z[1/q]\) argument belong to that ring; the sum for \(d_j\)
does too. When \(h=q^b\), the diagonal \(h^t\) is a unit in
\(\mathbb Z[1/q]\). Successive substitution therefore also puts the
inverse matrix in that ring, hence in \(\mathbb Z_p\) for \(p\ne q\). In
products, every new index is coordinatewise at most the old one, so its
total order cannot increase. \(\square\)

The integrality assertion here concerns changes between binomial
families. Conversion between a binomial family and ordinary powers has
leading coefficient \(1/t!\); we use that conversion only for algebraic
equivalence of vanishing.

\subsubsection{From prepared equations to the ordinary zero
estimate}\label{from-prepared-equations-to-the-ordinary-zero-estimate}

\textbf{Proposition 10.31 (the auxiliary polynomial and its jets).} Let
\(\mathcal M\subset\mathbb Z^r\) be finite, with distinct exponent
vectors \(\boldsymbol m\), and let \(A_i\ge0\) bound \(|m_i|\). Put
\(d_i=\lfloor A_i\rfloor\). Fix \(q,I,k,L\) as above, and coefficients
\(c_{\boldsymbol m,a,\ell}\) in a characteristic-zero field, not all
zero. Form

\[
Q(X,Y)=\sum_{\boldsymbol m\in\mathcal M}
\sum_{0\le a<k,\,1\le\ell\le L}
c_{\boldsymbol m,a,\ell}\Delta(q^{-I}X+a;k)^\ell Y^{\boldsymbol m},
\qquad P=Y^{\boldsymbol d}Q.
\tag{10.91}
\]

Then \(P\) is nonzero, is an ordinary polynomial, and has coordinate
bounds \(kL,2A_1,\ldots,2A_r\).

Let \(D_j=\sum_i C_{ij}Y_i\partial_{Y_i}\) be commuting Euler
derivations, and put \(\omega_j(\boldsymbol m)=\sum_iC_{ij}m_i\),
\(1\le j\le r-1\). For a torus point \(\boldsymbol y\) define the
prepared value

\[
\mathcal Q(s;\boldsymbol t)=
\sum_{\boldsymbol m,a,\ell}c_{\boldsymbol m,a,\ell}
v(k)^{t_0}\Theta(q^{-I}s+a;k,\ell,t_0)
\prod_{j=1}^{r-1}\Delta(\omega_j(\boldsymbol m);t_j)
\boldsymbol y^{\boldsymbol m}.
\tag{10.92}
\]

For any \(T\), vanishing of all these values with
\(|\boldsymbol t|\le T\) is equivalent to vanishing of all ordinary jets
of \(Q\), and hence of \(P\), at \((s,\boldsymbol y)\) for
\(\partial_X,D_1,\ldots,D_{r-1}\) through total order \(T\).

\textbf{Proof.} Distinct Laurent monomials are linearly independent. For
each fixed \(\boldsymbol m\), Lemma 10.29, with the invertible
substitution \(X\mapsto q^{-I}X\), proves independence of its
coefficient polynomials. Thus not all coefficients can cancel, and
\(Q\ne0\). Since \(m_i\) is integral and \(m_i\ge-A_i\), it satisfies
\(m_i\ge-\lfloor A_i\rfloor\). Hence \(0\le m_i+d_i\le2A_i\). This
proves the degree and nonvanishing assertions for \(P\).

On its monomial terms, (10.92) is exactly the value of the commuting
operator \((q^Iv(k))^{t_0}\partial_X^{t_0}/t_0!\) times
\(\prod_j\Delta(D_j;t_j)\), applied to \(Q\). The chain rule supplies
\(q^{-It_0}\), canceled by \(q^{It_0}\). Each \(\Delta(D_j;t_j)\) has
leading term \(D_j^{t_j}/t_j!\) and only lower powers. All diagonal
scalars are nonzero. Induction on the torus part of the total order
therefore makes these operators and the ordinary jet operators span
exactly the same filtered space. This proves the first equivalence, in
both directions.

Put \(c_j=\sum_iC_{ij}d_i\). The product rule gives

\[
\partial_X^{t_0}\prod_jD_j^{t_j}P
=Y^{\boldsymbol d}\partial_X^{t_0}
\prod_j(D_j+c_j)^{t_j}Q.
\tag{10.93}
\]

Expanding is triangular with diagonal one and preserves total order. The
inverse uses \(-c_j\) and \(Y^{-\boldsymbol d}\). Since every torus
coordinate is nonzero, evaluation proves the second equivalence.
\(\square\)

When the \(C_{ij}\) are integers, \(s\) is integral and \(q\ne p\),
Lemma 10.28 and integer-valued \(\Delta\) give \[
v(k)^{t_0}\Theta(q^{-I}s+a;k,\ell,t_0)
\prod_j\Delta(\omega_j(\boldsymbol m);t_j)
\in\mathbb Z[1/q]\subseteq\mathbb Z_p.
\] The prepared product \(v(k)^{t_0}\Theta\), rather than \(\Theta\)
alone, needs at most the q-denominator in (10.87). Integrality of the
whole value additionally requires locally integral
\(c_{\boldsymbol m,a,\ell}\) and local-unit torus coordinates.

For the application to Theorem 10.25, take \(D_1,\ldots,D_{r-1}\) to be
the selected integral Euler directions in Lemma 10.23, renumbered. They
span the Euler hyperplane in that theorem's space \(W\). Apply the
proposition at \((s,\vartheta_1^s,\ldots,\vartheta_r^s)\). All prepared
equations then give precisely the ordinary \(W\)-jets required there,
with

\[
D_0=kL,\qquad D_i=2A_i,
\qquad A_i=q^{\nu-I}D_i^{\mathrm{old}},\quad
k=D_{-1}^{\mathrm{old}}+1,\quad L=D_0^{\mathrm{old}}+1.
\tag{10.94}
\]

Here \(\nu\ge0\) is an exponent clearing the denominators of the
q-saturated basis, supplied by Lemma 10.12; it is distinct from the
subgroup rank used in Theorem 10.25. The Euler shift adds
\(\sum_i C_{ij}d_i\) to its eigenvalue; Lemma 10.30 supplies the
corresponding reversible binomial change. Thus the nonzero-polynomial
and prepared-to-ordinary steps require no omitted derivative lemma.
Theorem 10.25 still requires its separate independence, degree and
allocation inequalities.

\subsection{14. Constructing the initial auxiliary
coefficients}\label{constructing-the-initial-auxiliary-coefficients}

The prepared equations are a finite homogeneous linear system. The
weighted Siegel lemma has a complete proof in
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/TR-BAKER-08.html\#small-solutions-over-the-coefficient-field}{the
earlier multiplicity lesson}, equation (8.40), including dependent rows
and integral solutions. We apply it with all weights one and bound every
row explicitly. Figure 10.4 and Lemma 10.29 ensure that a nonzero
coefficient vector produces a nonzero polynomial.

For a number field \(K\) of degree \(d\), use all archimedean
embeddings, counting a complex pair twice. At a finite prime ideal
\(\mathfrak p\), put
\(|x|_{\mathfrak p}=(N\mathfrak p)^{-v_{\mathfrak p}(x)}\). These are
the product-formula conventions of the earlier lesson. For a nonzero
vector \(c\in K^N\), define its Euclidean projective height by \[
h_2(c)=\frac1d\left(
\sum_{\sigma\mid\infty}\log\Bigl(\sum_j|\sigma(c_j)|^2\Bigr)^{1/2}
+\sum_{\mathfrak p}\log\max_j|c_j|_{\mathfrak p}\right).
\] The product formula makes this invariant under multiplication of the
whole vector by any nonzero scalar in \(K\). Replacing the archimedean
Euclidean norm by the maximum defines \(h_\infty(c)\le h_2(c)\).

\textbf{Theorem 10.32 (initial prepared-jet kernel).} Let
\(r,k,L\in\mathbb Z_{\ge1}\), \(S,T\in\mathbb Z_{\ge0}\), and let
\(\mathcal M\subset\mathbb Z^r\) be a finite set of distinct vectors
with \(|m_i|\le A_i\). Take \(\vartheta_i\in K^\times\) and integer
Euler coefficients \(C_{ij}\), \(1\le j<r\). Write
\(\omega_j(\boldsymbol m)=\sum_iC_{ij}m_i\), and put

\[
\begin{aligned}
N&=kL|\mathcal M|,\qquad M=(2S+1)\binom{T+r}{r},\\
\Omega&=\max\bigl(\{0\}\cup\{|\omega_j(\boldsymbol m)|:j<r,\boldsymbol m\in\mathcal M\}\bigr),\\
\mathcal B&=\left(\frac{(S+2k-1+v(k))^k}{k!}\right)^L
\max\{1,\Omega+T\}^{T}.
\end{aligned}
\tag{10.95}
\]

If \(N>M\), there are algebraic integers \(c_{\boldsymbol m,a,\ell}\),
not all zero, for which every prepared value (10.92), with \(I=0\),
vanishes at \((s,\vartheta_1^s,\ldots,\vartheta_r^s)\) for
\(s\in[-S,S]\cap\mathbb Z\) and \(|\boldsymbol t|\le T\). They may be
chosen with

\[
h_\infty(c)\le h_2(c)\le
\frac12\log N+\frac{\log|\Delta_K|}{2d}
+\frac{M}{N-M}
\left(\frac12\log N+\log\mathcal B
+2S\sum_i A_i h(\vartheta_i)\right).
\tag{10.96}
\]

The resulting \(P=Y^{\boldsymbol d}Q\) is nonzero and has the degree and
ordinary-jet properties of Proposition 10.31. When the directions are
the selected basis in Lemma 10.23, these are all the required \(W\)-jets
at the stated points.

\textbf{Proof.} There are \(\binom{T+r}{r}\) nonnegative \(r\)-tuples of
total order at most \(T\). Append a slack coordinate to give total
\(T\); distributing \(T\) objects among \(r+1\) slots amounts to
choosing the \(r\) divider positions among \(T+r\). Thus the prepared
values give at most \(M\) linear equations in the \(N\) coefficients.
Their entries lie in \(K\). Rows that vanish or depend on others cause
no difficulty in (8.40).

We bound a row entry. The coefficient of \(Z^{t_0}\) in
\(\Delta(s+a+v(k)Z;k)^\ell\) is exactly
\(v(k)^{t_0}\Theta(s+a;k,\ell,t_0)\). For \(|s|\le S\), \(0\le a<k\),
the sum of the absolute values of all coefficients of this polynomial is
at most \[
\frac{\prod_{b=1}^k(|s+a+b|+v(k))^\ell}{(k!)^\ell}
\le\left(\frac{(S+2k-1+v(k))^k}{k!}\right)^\ell.
\] The base on the right is at least one, so \(\ell\le L\) gives its
first factor in \(\mathcal B\). For an integer \(\omega\) with
\(|\omega|\le\Omega\), \(|\Delta(\omega;t)|\le(\Omega+t)^t\) when
\(t>0\), directly from its product formula and \(t!\ge1\). The product
over Euler indices is therefore at most \(\max\{1,\Omega+T\}^{T}\), also
when all indices are zero. Every prepared scalar has absolute value at
most \(\mathcal B\) and is an integer, by Lemma 10.28 and integer-valued
\(\Delta\).

At an archimedean embedding, the Euclidean row norm is consequently at
most \[
\sqrt N\,\mathcal B\prod_i
\max\{|\sigma(\vartheta_i)|,|\sigma(\vartheta_i)|^{-1}\}^{SA_i}.
\] At a finite place its maximum norm is at most the analogous product,
since its prepared scalar is an integer. The exponents satisfy
\(|s m_i|\le SA_i\), including negative \(s\) and negative \(m_i\). The
product formula and the proved height formula give \[
\frac1d\sum_v\log\max\{|\vartheta_i|_v,|\vartheta_i|_v^{-1}\}
=2h(\vartheta_i):
\] the sums of the positive and negative parts of
\(\log|\vartheta_i|_v\) are equal. Thus every nonzero row has adelic
norm at most \(\sqrt N\mathcal B\exp(2S\sum_i A_i h(\vartheta_i))\),
which is at least one. Set all weights in (8.40) to one, with
\(J=N,I=M\). Its finite conditions give \(c\in\mathcal O_K^N\); taking
logarithms gives (10.96). Proposition 10.31 proves the
nonzero-polynomial and derivative assertions. \(\square\)

For a fixed finite place \(\mathfrak p\) above \(p\), select a nonzero
coefficient \(c_*\) of minimum p-adic valuation. The scalar
normalization \[
\widetilde c=c/c_*,\qquad
\min_j v_{\mathfrak p}(\widetilde c_j)=0,\qquad
h_2(\widetilde c)=h_2(c)
\tag{10.97}
\] preserves all equations and makes the coefficients locally integral
with one local unit. It need not preserve integrality at other finite
places; the unnormalized vector supplied by the theorem is globally
integral.

This construction supplies an explicit initial coefficient-height
budget. Applying it in a numerical logarithm estimate still requires
comparing (10.96) with that estimate's parameters and supplying the jet
precision required by the following extrapolation bounds.

\subsection{15. P-adic interpolation at integer
nodes}\label{p-adic-interpolation-at-integer-nodes}

We prove the analytic extrapolation estimates directly. The power-series
algebra and its norm are those of
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/TR-BAKER-09.html\#2-convergence-and-legitimate-substitution}{Convergence
and legitimate substitution}, Lemmas 9.1--9.2. We use \(v_p(p)=1\) and
\(v_p(0)=+\infty\). A \emph{normal series} means
\(F(Z)=\sum_{n\ge0}a_n Z^n\in\mathcal A_1\), with \(v_p(a_n)\ge0\) and
\(a_n\to0\). Its divided jet is \(F_j(a)=F^{(j)}(a)/j!\).

\subsubsection{Division and the Hermite
remainder}\label{division-and-the-hermite-remainder}

\textbf{Lemma 10.33 (division inside the unit disc).} If \(v_p(a)>0\)
and \(F\) is normal, then

\[
F(Z)=(Z-a)G(Z)+F(a),\qquad
G(Z)=\sum_{n\ge0}\left(\sum_{j\ge0}a_{n+1+j}a^j\right)Z^n
\tag{10.98}
\]

has a normal quotient \(G\). Consequently, for a finite set of distinct
such points \(a_s\) and an integer \(\mu\ge1\), there are unique \(H,G\)
with \(F=H+WG\), \(W(Z)=\prod_s(Z-a_s)^\mu\), \(\deg H<\mu|E|\), and
\(G\) normal. The polynomial \(H\) has the same jets as \(F\) through
order \(\mu-1\) at every \(a_s\).

\textbf{Proof.} The inner series converges because \(|a|<1\). Its
coefficients have absolute value at most one, and tend to zero, since
their absolute values are at most \(\sup_{m\ge n+1}|a_m|\). The
coefficient recurrence verifies (10.98); the constant coefficient uses
the convergent identity for \(F(a)\). Repeating division by the listed
linear factors gives \(F=H+WG\), with a normal quotient and the stated
remainder degree. Every partial remainder is a polynomial with integral
coefficients: the successive values and the roots are integral, and
products and sums preserve that condition.

Translation of a normal series is legitimate on the unit disc: the
coefficient of \(T^j\) in \(F(a+T)\) is
\(\sum_{n\ge j}\binom nj a_n a^{n-j}\), of absolute value at most
\(\sup_{n\ge j}|a_n|\). The full binomial family has only finitely many
terms of any fixed minimum size, since \(a_n\to0\). Lemma 9.1 therefore
justifies regrouping, and these Taylor coefficients also tend to zero.
Since \(W\) contains \((Z-a_s)^\mu\), the first \(\mu\) Taylor
coefficients of \(F-H\) vanish there. Conversely a polynomial with those
zero coefficients is divisible by that factor. Distinct factors may be
removed successively: a factor nonzero at another point has an
invertible Taylor series there, so removing it preserves the other
vanishing order. Their product therefore divides any polynomial with all
these zero jets. A polynomial of degree below \(\deg W\) with those jets
zero is zero. This proves uniqueness of \(H\). Then \(G\) is unique
because multiplication by the nonzero polynomial \(W\) is injective on
formal power series: the product of the first nonzero coefficients
cannot vanish. \(\square\)

\subsubsection{The exact integer separation
cost}\label{the-exact-integer-separation-cost}

Fix a prime \(p\), an integer \(R\ge1\), and let
\(B=\lfloor\log_p(2R)\rfloor\). We use either the full node set
\(E=[-R,R]\cap\mathbb Z\), or, for a prime \(q\ne p\), its subset of
nodes not divisible by \(q\). Put \(n=|E|\), and

\[
L_s(X)=\prod_{t\in E\setminus\{s\}}\frac{X-t}{s-t},\qquad
\kappa=\begin{cases}0&\text{for the full set},\\1&\text{for the q-deleted set}.\end{cases}
\tag{10.99}
\]

\textbf{Theorem 10.34 (normal-series Hermite extrapolation).} Choose
\(\rho\in\mathbb C_p\) with \(v_p(\rho)=\theta>0\), let
\(\mu\in\mathbb Z_{\ge1}\), and suppose that the normal series \(F\)
satisfies

\[
v_p(F_j(\rho s))+j\theta\ge\Lambda
\qquad(s\in E,\ 0\le j<\mu).
\tag{10.100}
\]

Then every \(x\in\mathbb Z_p\subset\mathbb Q_p\) satisfies

\[
v_p(F(\rho x))\ge
\min\{n\mu\theta,\ \Lambda-((\kappa+1)\mu-1)B\}.
\tag{10.101}
\]

\textbf{Proof.} First \(v_p(L_s(x))\ge-\kappa B\). At a node, its value
is zero or one, so assume \(x\notin E\). For each integer \(a\ge1\),
count the elements of \(E\) in each residue class modulo \(p^a\). For
the full interval these counts differ by at most one. For the q-deleted
interval they differ by at most two: the full counts differ by at most
one, and so do the counts of deleted points \(qt\), because \(q\) is
invertible modulo \(p^a\) and
\(-\lfloor R/q\rfloor\le t\le\lfloor R/q\rfloor\).

Write these counts as \(N_a(u)\) for a residue class \(u\). The
contribution at level \(a\) to the numerator valuation minus the
denominator valuation of \(L_s(x)\) is
\(N_a(x)-N_a(s)+1-\mathbf1_{x\equiv s\bmod p^a}\). If the classes agree
it is zero. Otherwise it is at least \(-\kappa\). For \(a>B\), there is
no denominator contribution, since nonzero differences of integer nodes
have ordinary absolute value at most \(2R<p^a\). The remaining
contributions are nonnegative. Summing proves the bound. Each valuation
is an integer, so the identity
\(v_p(u)=\sum_{a\ge1}\mathbf1_{u\equiv0\bmod p^a}\) applies; numerator
valuations are finite because \(x\notin E\).

The Taylor inverse of \(L_s\) at its own node has the form \[
A_s(U)=L_s(s+U)^{-\mu}
=\prod_{t\ne s}(1+U/(s-t))^{-\mu}.
\] The coefficient of \(U^h\) has valuation at least \(-hB\). Indeed the
coefficients of \((1+V)^{-\mu}\) are the integers
\((-1)^h\binom{\mu+h-1}{h}\), obtained by multiplying \(\mu\) geometric
series; every denominator \((s-t)^h\) has valuation at most \(hB\).
Products preserve the bound.

Let \([A_s]_{\le b}\) mean truncation through degree \(b\). Lemma
10.33's unique Hermite remainder is explicitly \[
H(\rho X)=\sum_{s\in E}L_s(X)^\mu
\sum_{j=0}^{\mu-1}\rho^j F_j(\rho s)(X-s)^j
[A_s(X-s)]_{\le\mu-1-j}.
\] Its degree is at most \(\mu(n-1)+\mu-1=n\mu-1\). At each other node
the factor \(L_s^\mu\) gives order \(\mu\); at its own node,
multiplication by the truncated inverse reproduces the required jets.
This proves the formula by uniqueness, without an assumption about
interpolation matrices.

For \(x\in\mathbb Z_p\), \(x-s\) is integral. Each displayed summand has
valuation at least \(\Lambda-\mu\kappa B-(\mu-1)B\), by (10.100) and the
two bounds just proved. Hence \(H(\rho x)\) has that valuation bound. On
the other hand,
\(v_p(W(\rho x))=\mu\sum_s(\theta+v_p(x-s))\ge n\mu\theta\), and its
normal quotient has integral value. The ultrametric inequality for
\(F=H+WG\) proves (10.101). \(\square\)

\textbf{Corollary 10.35 (the two integer-node steps).} For the full node
set, it is enough to have

\[
\Lambda\ge(2R+1)\mu\theta
+\mu\frac{\log(2R+1)}{\log p}
\quad\Longrightarrow\quad
v_p(F(\rho x))\ge(2R+1)\mu\theta.
\tag{10.102}
\]

For the q-deleted node set, if \(q\mid R\), it is enough to have

\[
\Lambda\ge2(1-1/q)R\mu\theta
+2\mu\frac{\log(2R)}{\log p}
\quad\Longrightarrow\quad
v_p(F(\rho x))\ge2(1-1/q)R\mu\theta.
\tag{10.103}
\]

Both conclusions hold for every \(x\in\mathbb Z_p\), including \(x=s/q\)
for any integer \(s\). In (10.102), the case \(R=0\) is also valid:
Taylor truncation at zero gives
\(v_p(F(\rho x))\ge\min\{\mu\theta,\Lambda\}\).

\textbf{Proof.} The full count is \(2R+1\). The deleted count is
\(2R-2\lfloor R/q\rfloor\), equal to \(2(1-1/q)R\) when \(q\mid R\).
Also \(B\le\log(2R)/\log p\), so the stated costs dominate \((\mu-1)B\)
and \((2\mu-1)B\). Apply the theorem. The separate Taylor proof covers
\(R=0\). Since \(q\ne p\), division of an integer by \(q\) stays in
\(\mathbb Z_p\). \(\square\)

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={Integer nodes grouped by their 3-adic residues and the exact valuations of their differences}]{../figures/integer-node-separation.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.5. The seven nodes \(-3,\ldots,3\), grouped by their
residues modulo three, and the exact matrix \(v_3(s-t)\). The diagonal
is \(+\infty\). Nodes in the same indicated class have difference
valuation one; all other nonzero differences have valuation zero. The
groups split into singletons modulo nine. The drawing is a residue-class
schematic: its line lengths are not p-adic distances. For \(R=3,p=3\),
the exact separation cost in Theorem 10.34 is \(B=1\). The off-diagonal
entries in the row at zero sum to two, matching \(v_3((3!)^2)=2\), but
this factorial total cancels in the integral cardinal factors for the
full node set. The complete mechanism is proved in (10.99)--(10.103).
Human-source context: the normal-series extrapolation in Yu's freely
accessible 1990 and 2013 articles, cited below.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/integer_node_separation.py}{Figure program}.}
\end{flushleft}
\end{minipage}\par\medskip


For an application with jet order \(M\), put \(\mu=M+1\). Replacing
either logarithmic cost above by a larger maximum with a
coefficient-height parameter only strengthens the hypothesis. The
conclusion then follows once normality and the input jet bound (10.100)
hold.

\subsubsection{The error from projecting a logarithmic
curve}\label{the-error-from-projecting-a-logarithmic-curve}

\textbf{Lemma 10.36 (uniform projection error).} Let \(q\ne p\) be a
prime, \(I\ge0\) be an integer, and use the prepared family (10.92) with
integer Euler coefficients and a finite coefficient vector
\(c\in\mathbb C_p^N\setminus\{0\}\). Put \(\delta=\min_j v_p(c_j)\).
Take \(b\in\mathbb Z\setminus\{0\}\), \(a_i\in\mathbb Z_p\), and
\(u_i,\mathcal L\in\mathbb C_p\), satisfying \(v_p(u_i)>1/(p-1)\),
\(v_p(\mathcal L)\ge U\), and \(U-v_p(b)>1/(p-1)\). For
\(y\in\mathbb Z_p\), define

\[
\begin{aligned}
\Phi(y;\boldsymbol t)&=\mathcal Q(y;\boldsymbol t)
\big|_{Y_i=\exp(yu_i)},\\
f(y;\boldsymbol t)&=\mathcal Q(y;\boldsymbol t)
\big|_{Y_i=\exp(y(u_i-a_i\mathcal L/b))}.
\end{aligned}
\tag{10.104}
\]

Then for every nonnegative index \(\boldsymbol t\),

\[
v_p\bigl(f(y;\boldsymbol t)-\Phi(y;\boldsymbol t)\bigr)
\ge\delta+U-v_p(b).
\tag{10.105}
\]

\textbf{Proof.} Here the integer \(v(k)=\operatorname{lcm}(1,\ldots,k)\)
retains its earlier meaning, distinct from the valuation \(v_p\). The
prepared Taylor factor is p-integral at \(q^{-I}y+a\). Here is the
slight extension of Lemma 10.28 needed for \(y\in\mathbb Z_p\): the
rational polynomial \(v(k)^m\Theta(Z;k,\ell,m)\) is integral at
integers. Choose an integer approximating any given p-integral argument
modulo a sufficiently high power of \(p\), exceeding all its coefficient
denominator valuations. Factoring the differences of monomials proves
its value is p-integral, exactly as in the proof of Lemma 10.28. The
Euler binomial factors are integers. Thus every prepared scalar in
(10.92) is p-integral.

All displayed exponentials exist. In each monomial term the ratio of the
projected exponential to the original one is
\(\exp(-y\mathcal L\sum_i m_i a_i/b)\), by Proposition 9.4 in
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/TR-BAKER-09.html\#3-the-exponential-and-the-logarithm}{The
exponential and the logarithm}. Its exponent has valuation at least
\(U-v_p(b)>1/(p-1)\), because \(y\), \(m_i\) and \(a_i\) are p-integral.
The exact first-term exponential valuation in Proposition 9.3 of that
subsection gives that the ratio minus one has valuation at least
\(U-v_p(b)\). The original exponential is a local unit. Multiplying by
the prepared scalar and its coefficient, then summing, proves (10.105).
\(\square\)

In logarithmic applications take \(\mathcal L=p^\kappa\log\Xi\), where
\(\kappa\in\mathbb Z_{\ge0}\). If \(v_p(\Xi-1)\ge U>1/(p-1)\),
Proposition 9.3 gives \(v_p(\mathcal L)\ge U\). The coefficient \(a_i\)
may include a denominator that is a q-power, since it remains
p-integral. The precise domain and coefficient valuation cost of the
projection estimate are thus visible in (10.104)--(10.105).

\subsubsection{Normalizing the prepared
series}\label{normalizing-the-prepared-series}

\textbf{Proposition 10.37 (a normal prepared auxiliary function).}
Retain the integer data of (10.92), the finite nonzero coefficient
vector \(c\), and \(\delta=\min_j v_p(c_j)\). Choose a coefficient
\(c_*\) of that valuation and \(\rho\in\mathbb C_p\) with
\(v_p(\rho)=\theta>0\). Suppose that every logarithmic slope \(w_i\)
satisfies \(v_p(w_i)>\theta+1/(p-1)\). Define \(f(y;\boldsymbol t)\) by
substituting \(Y_i=\exp(yw_i)\) in (10.92), and put

\[
\begin{aligned}
D&=kL,\qquad \gamma=\rho^D(k!)^L/c_*,\\
\Gamma&=v_p(\gamma)=D\theta+L v_p(k!)-\delta,\\
F(Z;\boldsymbol t)&=\gamma f(Z/\rho;\boldsymbol t).
\end{aligned}
\tag{10.106}
\]

For every nonnegative prepared index \(\boldsymbol t\), the function
\(F\) is normal. At an integer node \(s\), its divided jets satisfy

\[
F_j(\rho s;\boldsymbol t)=\gamma\rho^{-j}f_j(s;\boldsymbol t),\qquad
v_p(F_j(\rho s;\boldsymbol t))+j\theta=\Gamma+v_p(f_j(s;\boldsymbol t)).
\tag{10.107}
\]

Consequently, if \(v_p(f_j(s;\boldsymbol t))\ge P\) at the node set of
Theorem 10.34 for \(0\le j<\mu\), then

\[
v_p(f(x;\boldsymbol t))\ge
\min\{n\mu\theta-\Gamma,\ P-((\kappa+1)\mu-1)B\}
\qquad(x\in\mathbb Z_p).
\tag{10.108}
\]

\textbf{Proof.} Consider one prepared polynomial factor after
\(y=Z/\rho\). Its coefficient of \(Z^h\) has valuation at least
\(-h\theta-\ell v_p(k!)\), and its degree is at most \(k\ell-t_0\). To
see this without a coefficient estimate from another source, multiply it
by \((k!)^\ell\): it is the coefficient of \(U^{t_0}\) in \[
\prod_{b=1}^k(q^{-I}Z/\rho+a+b+v(k)U)^\ell.
\] All constants and \(v(k)\) are integers, while \(q\ne p\). A term of
degree \(h\) in \(Z\) therefore has valuation at least \(-h\theta\).
Multiplying by \(\rho^D(k!)^L\) makes every coefficient integral,
because \(h\le k\ell\le D\) and \(\ell\le L\). The Euler binomial
factors are integers, and \(c_j/c_*\) is locally integral. Thus the
normalized polynomial part of every summand has Gauss norm at most one.

The exponential part of that summand is \(\exp(\beta Z/\rho)\), where
\(\beta=\sum_i m_iw_i\). Either \(\beta=0\), or
\(v_p(\beta/\rho)>1/(p-1)\), since the exponents \(m_i\) are integers.
By (9.4), its coefficient of degree \(h\ge1\) has valuation
\(h v_p(\beta/\rho)-v_p(h!)\ge h(v_p(\beta/\rho)-1/(p-1))+1/(p-1)>0\),
tending to infinity. Its constant coefficient is one. It belongs to
\(\mathcal A_1\) with norm one. The product and finite-sum statements of
Lemma 9.2 now prove that \(F\in\mathcal A_1\) with norm at most one,
hence is normal.

The slope hypotheses define \(f\) on the full closed disc
\(|y|\le|\rho|^{-1}\), so the chain rule is valid at every integer node.
It gives (10.107), also for zero jets with infinite valuation. Apply
Theorem 10.34 with \(\Lambda=\Gamma+P\), then subtract \(\Gamma\), to
obtain (10.108). \(\square\)

The larger scale \(D(\theta+1/(p-1))-\delta\) also suffices: (9.4) gives
\(L v_p(k!)\le D/(p-1)\), and a scalar of any additional nonnegative
valuation preserves normality. For the projected slopes in (10.104), it
suffices that \(v_p(u_i)>\theta+1/(p-1)\) and
\(U-v_p(b)>\theta+1/(p-1)\). The ultrametric inequality then supplies
the slope hypotheses above. Input derivative precision remains an
explicit hypothesis in (10.108), distinct from this proof of normality.

\subsection{16. Prepared jets and derivative
precision}\label{prepared-jets-and-derivative-precision}

The interpolation argument can keep the precision of each derivative
separately. We first obtain that precision from the prepared equations,
without changing the selected Euler directions.

\subsubsection{Differentiating along the projected
curve}\label{differentiating-along-the-projected-curve}

\textbf{Lemma 10.38 (the loss in an ordinary divided jet).} Take the
integral forms, matrix \(C\), selected column set \(J\), and derivations
\(\delta_j\) of Lemma 10.23. Use those selected derivations as the
\(D_j\) in (10.92), renumbering them if necessary. Let \(q\ne p\) be a
prime, \(I,\nu\ge0\) be integers, and \(z_j\in\mathbb C_p\),
\(1\le j<n\), satisfy \(v_p(z_j)>1/(p-1)\). Put \(\beta=v_p(b_n)\), and
define

\[
\begin{aligned}
w_i&=\frac1{b_nq^\nu}\sum_{j<n}C_{ij}z_j,\\
\xi_a&=\frac1{b_nq^\nu}\sum_{j<n}\lambda_{aj}z_j\quad(a\in J),\\
C_0&=\max\{v_p(v(k)),\beta\}.
\end{aligned}
\tag{10.109}
\]

Here the coefficients \(\lambda_{aj}\) are those in (10.64), with
\(\lambda_{aj}=\mathbf1_{a=j}\) for selected columns. Suppose that
\(v_p(w_i)>1/(p-1)\) for every \(i\). Define \(f(y;\boldsymbol t)\) by
substituting \(Y_i=\exp(yw_i)\) in the prepared family (10.92). Fix
\(s\in\mathbb Z_p\), an integer \(T\ge0\), and a real \(P_0\). If
\(v_p(f(s;\boldsymbol t'))\ge P_0\) for every \(|\boldsymbol t'|\le T\),
then

\[
v_p\bigl(f_j(s;\boldsymbol t)\bigr)\ge P_0-jC_0
\quad\text{whenever }\ |\boldsymbol t|+j\le T,
\qquad
f_j=\frac1{j!}\frac{d^jf}{dy^j}.
\tag{10.110}
\]

\textbf{Proof.} By (10.64), \(\sum_iw_iE_i=\sum_{a\in J}\xi_a\delta_a\).
Each \(\lambda_{aj}\) is p-integral, and \(q\ne p\), so either
\(\xi_a=0\) or \(v_p(\xi_a)>1/(p-1)-\beta\). Differentiating along the
curve therefore applies the commuting operator
\(V=\partial_X+\sum_{a\in J}\xi_a\delta_a\). This identity follows
directly on a term \(P(X)Y^{\boldsymbol m}\): its restriction is
\(P(y)\exp(y\sum_i m_iw_i)\), whose derivative has the stated two terms.
The exponential laws and convergence in Propositions 9.3--9.4 justify
this differentiation on the finite prepared sum.

Write \(A=q^Iv(k)\). The additive prepared operator is
\(\mathcal B_{t_0}=A^{t_0}\partial_X^{t_0}/t_0!\). For \(h\ge0\), \[
\frac{\partial_X^h}{h!}\mathcal B_{t_0}
=A^{-h}\binom{t_0+h}{h}\mathcal B_{t_0+h}.
\] Its scalar has valuation at least \(-h v_p(v(k))\).

For an indeterminate \(Z\), the defining product for \(\Delta\) gives \[
Z\Delta(Z;t)=(t+1)\Delta(Z;t+1)-(t+1)\Delta(Z;t).
\] Repeated use of this identity expresses \(Z^h\Delta(Z;t)\) as an
integer linear combination of \(\Delta(Z;e)\), with \(t\le e\le t+h\).
Substitution of an Euler derivation is legitimate because all the
operators commute. Multiplication by \(\xi_a^h/h!\) gives coefficients
of valuation at least \(-h\beta\): for \(h>0\), (9.4) gives
\(h v_p(\xi_a)-v_p(h!)\ge-h\beta\); for \(h=0\), the coefficient is one.
A zero \(\xi_a\) causes no problem.

Expand \(V^j/j!\) over \(h_0+\sum_a h_a=j\). The multinomial coefficient
cancels the denominator \(j!\), leaving the product of the divided
powers just considered. Every resulting prepared index has total order
at most \(|\boldsymbol t|+j\), and every scalar has valuation at least
\(-h_0v_p(v(k))-\sum_a h_a\beta\ge-jC_0\). Evaluation at the same curve
point \(s\), followed by the ultrametric inequality, proves (10.110).
This proof retains the factorials; no extra \(v_p(j!)\) loss is needed.
\(\square\)

\subsubsection{Retaining the precision of each
jet}\label{retaining-the-precision-of-each-jet}

\textbf{Theorem 10.39 (interpolation with a linear jet loss).} Retain
the normal function, integer node set, \(\rho\), \(\theta\), \(n\),
\(\mu\), \(B\), and \(\kappa\in\{0,1\}\) of Theorem 10.34. Suppose
\(C\ge0\) and \(v_p(F_j(\rho s))+j\theta\ge\Lambda-jC\) for every node
\(s\) and \(0\le j<\mu\). Then

\[
v_p(F(\rho x))\ge
\min\{n\mu\theta,\ \Lambda-\kappa\mu B-(\mu-1)\max\{B,C\}\}
\qquad(x\in\mathbb Z_p\subset\mathbb Q_p).
\tag{10.111}
\]

\textbf{Proof.} Use the explicit Hermite remainder in the proof of
Theorem 10.34. Its term with jet index \(j\) contains a coefficient of
degree \(h\le\mu-1-j\) in \(A_s\). That proof gives valuations at least
\(-\kappa\mu B\) for \(L_s(x)^\mu\), and at least \(-hB\) for that
coefficient. The remaining power of \(x-s\) is integral. The term thus
has valuation at least \(\Lambda-jC-\kappa\mu B-(\mu-1-j)B\). Since
\(0\le j\le\mu-1\), \[
jC+(\mu-1-j)B\le(\mu-1)\max\{B,C\}.
\] The ultrametric inequality gives the second bound in (10.111) for the
full Hermite remainder. The normal quotient and its root product give
the first bound \(n\mu\theta\), exactly as before. Their sum proves the
claim. \(\square\)

For \(C=0\) this is Theorem 10.34. For \(C>0\), first replacing all jets
by their least precision would instead give the larger loss
\((\mu-1)(C+B)+\kappa\mu B\). The exact Hermite formula avoids that loss
by pairing each jet with only the inverse coefficients it actually uses.

\subsubsection{The input precision for
extrapolation}\label{the-input-precision-for-extrapolation}

\textbf{Corollary 10.40 (prepared vanishing supplies the required
jets).} Retain the selected directions and notation of Lemma 10.38, the
prepared coefficient vector \(c\ne0\), and \(\delta=\min_jv_p(c_j)\).
Suppose the unprojected and projected functions \(\Phi\) and \(f\)
satisfy Lemma 10.36 with \(b=b_n\), projection precision \(U\), and
coefficient minimum \(\delta\). Suppose also that the slopes of \(f\)
satisfy Proposition 10.37 for \(v_p(\rho)=\theta>0\). Let \(\Gamma\) be
its exact scale in (10.106).

At every node of Theorem 10.34, assume \(\Phi(s;\boldsymbol t')=0\) for
all \(|\boldsymbol t'|\le T\). Fix \(\boldsymbol t\) with
\(|\boldsymbol t|+\mu-1\le T\), and write \(M_0=\max\{B,C_0\}\). Then,
for every \(x\in\mathbb Z_p\),

\[
v_p(f(x;\boldsymbol t))\ge
\min\{n\mu\theta-\Gamma,\ U+\delta-\beta
-\kappa\mu B-(\mu-1)M_0\}.
\tag{10.112}
\]

In particular the following explicit budget suffices for
\(v_p(f(x;\boldsymbol t))\ge n\mu\theta-\Gamma\):

\[
U+D\theta+L v_p(k!)\ge n\mu\theta+(\kappa+1)\mu M_0,
\qquad D=kL.
\tag{10.113}
\]

Under that budget the same lower bound holds for
\(\Phi(x;\boldsymbol t)\). For the full interval, \(n=2R+1\) and
\(\kappa=0\); for the q-deleted interval with \(q\mid R\),
\(n=2(1-1/q)R\) and \(\kappa=1\).

These hypotheses have a direct logarithmic model. Take
\(z_1,\ldots,z_n\) with \(v_p(z_j)>\theta+1/(p-1)\), and put \[
u_i=q^{-\nu}\sum_{j=1}^na_{ij}z_j,
\qquad \mathcal L=\sum_{j=1}^nb_jz_j,
\qquad a_i=q^{-\nu}a_{in}.
\] Then \(a_i\in\mathbb Z_p\), and \(w_i=u_i-a_i\mathcal L/b_n\) is
exactly (10.109). If \(v_p(\mathcal L)\ge U\) and
\(U-\beta>\theta+1/(p-1)\), the hypotheses of both the projection and
normality estimates follow from the ultrametric inequality. The
chosen-minor argument supplies the integral coefficients used for the
ordinary input jets.

\textbf{Proof.} Vanishing of \(\Phi\) and (10.105) give
\(v_p(f(s;\boldsymbol t'))\ge U+\delta-\beta\) for every required
prepared index. Lemma 10.38 therefore gives the ordinary input jets with
loss \(jC_0\). By (10.107), the corresponding normal jets have weighted
precision \(\Gamma+U+\delta-\beta-jC_0\). Theorem 10.39 and subtraction
of \(\Gamma\) prove (10.112).

Here \(\beta\le C_0\le M_0\) and \(B\le M_0\). Consequently
\(\beta+\kappa\mu B+(\mu-1)M_0\le(\kappa+1)\mu M_0\). Substitute
\(\Gamma+\delta=D\theta+Lv_p(k!)\) in (10.113) to see that the second
entry of the minimum in (10.112) is at least the first. It also follows
that \(U+\delta-\beta\ge n\mu\theta-\Gamma\). The projection error
(10.105) holds at \(x\), so \(\Phi=f+(\Phi-f)\) has the same lower
bound. The node counts are those proved in Corollary 10.35. \(\square\)

The coefficient minimum cancels from (10.113); it remains in the
valuation conclusion through \(\Gamma\). One may instead choose the
larger normalizing scale \(\Gamma'=D(\theta+1/(p-1))-\delta\) allowed by
Proposition 10.37. Multiplying the normal function by a scalar of
valuation \(\Gamma'-\Gamma\ge0\) repeats the proof with \(\Gamma'\)
throughout. It gives the sufficient budget and its corresponding
conclusion \[
U+D\left(\theta+\frac1{p-1}\right)
\ge n\mu\theta+(\kappa+1)\mu M_0
\quad\Longrightarrow\quad
v_p(f(x;\boldsymbol t)),\ v_p(\Phi(x;\boldsymbol t))\ge n\mu\theta-\Gamma'.
\tag{10.114}
\] Such a scalar exists: its required valuation is the nonnegative
rational \(D/(p-1)-L v_p(k!)\), and a b-th root of \(p^a\) has valuation
\(a/b\) in \(\mathbb C_p\). The conclusion with \(\Gamma'\) does not
assert the stronger conclusion with \(\Gamma\).

Finally, \(v_p(v(k))=\lfloor\log_p k\rfloor\): the highest p-power
dividing a least common multiple is the highest p-power dividing one of
its factors, and the largest such power among \(1,\ldots,k\) is
\(p^{\lfloor\log_pk\rfloor}\). Also \(\beta\le\ln|b_n|/\ln p\). Thus
\(M_0\le\max\{\ln(2R),\ln k,\ln|b_n|\}/\ln p\), a useful real-valued
upper bound in (10.113)--(10.114). To deduce an algebraic zero from the
valuation conclusion one must additionally compare it with the height of
the nonzero prepared value. Those arithmetic and numerical comparisons
are separate from this interpolation argument.

\subsection{17. When extrapolation forces an algebraic
zero}\label{when-extrapolation-forces-an-algebraic-zero}

An analytic lower bound for a valuation forces an algebraic value to
vanish only when it exceeds the upper bound for a nonzero value. We
prove that upper bound with the exact local degree and with the
denominators of the prepared polynomial retained.

\subsubsection{A nonzero prepared value}\label{a-nonzero-prepared-value}

Use the height conventions of Section 14 and the product formula proved
in \href{https://kokunoyumeto.github.io/open-math-courses-public/courses/NT-LOC/NT-LOC-05.html}{Places of number fields in
extensions and the product formula}. The height identities for powers
and inverses are proved in
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-TRANS/TR-TRANS-02.html\#3-the-arithmetic-normalization-and-the-weil-height}{Heights
of algebraic numbers}.

\textbf{Theorem 10.41 (arithmetic precision of a prepared value).} Let
\(K\) have degree \(d\), and let its chosen prime above \(p\) have
ramification index \(e\) and residue degree \(f\). Take the prepared
family (10.92) with a prime \(q\ne p\), integers \(I,J\ge0\),
\(k,L\ge1\), a nonzero vector \(c\in K^N\), and integer Euler arguments
\(\omega_j(\boldsymbol m)\). Let \(|m_i|\le A_i\), put
\(N=kL|\mathcal M|\), \(D=kL\), and take \(x=s/q^J\), \(s\in\mathbb Z\),
with \(|x|\le X\). Suppose \(\eta_i\in K^\times\) are units at the
chosen prime. Evaluate (10.92) at \((x,\boldsymbol\eta)\), and call its
value \(V\). Let \(|\boldsymbol t|\le T\), \(T\in\mathbb Z_{\ge0}\), and
put

\[
\begin{aligned}
\delta&=\min_jv_p(c_j),\qquad
\Omega=\max\bigl(\{0\}\cup\{|\omega_j(\boldsymbol m)|\}\bigr),\\
\mathcal B_I(X,T)&=
\left(\frac{(q^{-I}X+2k-1+v(k))^k}{k!}\right)^L
\max\{1,\Omega+T\}^{T},\\
\Xi_{I,J,t_0}&=
\begin{cases}
0,&I+J=0,\\
\max\{0,Lv_q(k!)+(I+J)(D-t_0)-t_0v_q(v(k))\},&I+J>0.
\end{cases}
\end{aligned}
\tag{10.115}
\]

If \(V\ne0\), then

\[
\begin{aligned}
v_p(V)-\delta\le\frac d{ef\ln p}
\biggl(&\min\{h_\infty(c)+\ln N,\ h_2(c)+\tfrac12\ln N\}\\
&+\ln\mathcal B_I(X,T)+\Xi_{I,J,t_0}\ln q
+2\sum_iA_i h(\eta_i)\biggr).
\end{aligned}
\tag{10.116}
\]

If \(t_0>D\), the value is zero automatically. The same estimate holds
when the Euler arguments are any assigned integers bounded by
\(\Omega\); their linear dependence on \(\boldsymbol m\) is not needed
for this arithmetic assertion.

\textbf{Proof.} Write each row entry as its rational prepared scalar
times \(\boldsymbol\eta^{\boldsymbol m}\). The additive scalar is the
coefficient of \(Z^{t_0}\) in \(\Delta(q^{-I}x+a+v(k)Z;k)^\ell\). The
sum of the absolute values of the coefficients of this polynomial is at
most \[
\frac{\prod_{b=1}^k(q^{-I}|x|+a+b+v(k))^\ell}{(k!)^\ell}
\le\left(\frac{(q^{-I}X+2k-1+v(k))^k}{k!}\right)^\ell.
\] The displayed base is at least one: \(2k-1+v(k)\ge k\), and
\(k^k\ge k!\). Since \(\ell\le L\), its L-th power bounds this
coefficient. The product formula for \(\Delta(\omega;t_j)\) gives the
remaining factor in \(\mathcal B_I\), as in Theorem 10.32. Thus every
rational scalar has ordinary absolute value at most \(\mathcal B_I\).

There is no denominator away from \(q\), by Lemma 10.28 at the argument
\(q^{-I-J}s+a\); the Euler factors are integers. If \(I+J=0\), every
scalar is an integer. Otherwise the coefficient of degree \(h\) of the
ordinary divided derivative has denominator dividing \((k!)^\ell\),
because differentiating a monomial and dividing by \(t_0!\) multiplies
its coefficient by an integer binomial coefficient. Its degree is at
most \(k\ell-t_0\). Evaluation at \(q^{-I-J}s+a\) costs at most
\((I+J)(k\ell-t_0)\) in q-adic valuation. Multiplication by
\(v(k)^{t_0}\) restores \(t_0v_q(v(k))\). Taking \(\ell\le L\) gives the
clearing factor \(q^{\Xi_{I,J,t_0}}\). If \(t_0>D\), every additive
derivative is zero.

Choose a coefficient \(c_*\) of valuation \(\delta\), and put
\(z=V/c_*\). At the chosen prime the maximum norm of the coefficient
vector \(c/c_*\) is exactly one. For \(V\ne0\), the product formula
therefore gives \[
ef\ln p\,(v_p(V)-\delta)
=-\ln|z|_{\mathfrak p}
=\sum_{v\ne\mathfrak p}\ln|z|_v.
\] The factor is \(ef\), since
\(|z|_{\mathfrak p}=p^{-f\operatorname{ord}_{\mathfrak p}(z)}\) and
\(\operatorname{ord}_{\mathfrak p}=ev_p\).

At each archimedean embedding, the triangle inequality bounds the sum by
\(N\) times the coefficient maximum times the entry maximum.
Alternatively Cauchy--Schwarz gives the coefficient Euclidean norm times
\(\sqrt N\) times the entry maximum. This last inequality follows by
applying \((\sum u_jv_j)^2\le(\sum u_j^2)(\sum v_j^2)\) to the absolute
values; its difference is \(\sum_{i<j}(u_iv_j-u_jv_i)^2\ge0\). At finite
places the ultrametric inequality uses maximum norms and introduces no
factor \(N\).

Sum these bounds outside the chosen prime. The normalized coefficient
norms contribute exactly \(d h_\infty(c)\), or \(d h_2(c)\), because
their missing finite norm is one. There are \(d\) archimedean embeddings
with a complex pair counted twice. The rational scalar contributes at
most \(d\ln\mathcal B_I\) there. Its clearing factor contributes at most
\(d\Xi_{I,J,t_0}\ln q\) over the primes above \(q\), by the local-degree
identity for the element \(q\).

At every other place, the torus contribution is bounded by
\(\sum_iA_i\ln\max\{|\eta_i|_v,|\eta_i|_v^{-1}\}\). Its omitted
contribution at the chosen prime is zero, since the \(\eta_i\) are
units. Summing over all places gives \(2d\sum_iA_i h(\eta_i)\), by the
inverse-height identity and product formula. Divide by \(ef\ln p\), and
take the better of the two coefficient norm bounds, to obtain (10.116).
Every step used only integrality and the absolute size of the Euler
factors, proving the final assertion too. \(\square\)

\subsubsection{Fractional nodes and the coefficient
field}\label{fractional-nodes-and-the-coefficient-field}

\textbf{Proposition 10.42 (roots and a common local unit).} Let
\(\vartheta_i\in K^\times\) be units at \(p\), and choose \(q^J\)-th
roots \(\eta_i\) in a fixed algebraic embedding into \(\mathbb C_p\).
Put \(F=K(\eta_1,\ldots,\eta_r)\). For \(x=s/q^J\), consider the
prepared value with torus coordinates \(\eta_i^s\). Theorem 10.41
applies over \(F\). Its torus height term is at most
\(2X\sum_iA_i h(\vartheta_i)\) when \(|x|\le X\), but its local-degree
coefficient is

\[
\frac{[F:\mathbb Q]}{e_Ff_F}
=\frac d{ef}\frac{[F:K]}{[F_{\mathfrak P}:K_{\mathfrak p}]}.
\tag{10.117}
\]

One cannot replace this coefficient by \(d/(ef)\) without further
information.

If every exponent belongs to \(\boldsymbol m_*+q^J\mathbb Z^r\), where
\(\boldsymbol m_*\) is one of the exponent vectors, define
\(\widehat A_i=\max_{\boldsymbol m}|m_i-m_{*,i}|\le2A_i\). Then the
common-factor identity is

\[
\begin{aligned}
V&=\boldsymbol\eta^{s\boldsymbol m_*}V_0,\\
V_0&=\sum_{\boldsymbol m,a,\ell}
c_{\boldsymbol m,a,\ell}\,\mathcal S_{\boldsymbol m,a,\ell}(x;\boldsymbol t)
\boldsymbol\vartheta^{s(\boldsymbol m-\boldsymbol m_*)/q^J}\in K,
\qquad v_p(V)=v_p(V_0).
\end{aligned}
\tag{10.118}
\]

Here \(\mathcal S\) denotes the rational prepared scalar in (10.92). In
this case (10.116) may instead use \(d/(ef)\) and the torus height term
\(2X\sum_i\widehat A_i h(\vartheta_i)\), with the same \(\mathcal B_I\),
clearing factor, and coefficient heights. For a logarithmic curve with
\(\vartheta_i=\exp(u_i)\), choose \(\eta_i=\exp(u_i/q^J)\). The
exponential laws of Proposition 9.4 give \(\eta_i^{q^J}=\vartheta_i\)
and \(\exp(xu_i)=\eta_i^s\), identifying the fractional analytic value
with the stated algebraic value.

\textbf{Proof.} A root of a local unit is a local unit. The height power
identity gives \(h(\eta_i)=h(\vartheta_i)/q^J\), and hence
\(h(\eta_i^s)=|x|h(\vartheta_i)\). Heights of the coefficient vector are
unchanged by extending the field: every archimedean embedding is
repeated \([F:K]\) times, and the corresponding finite-place
contribution is multiplied by that same factor, as proved by the
local-degree identities in the earlier places lesson. Dividing by the
field degree cancels that factor. The relative ramification and residue
degrees multiply the original ones, and their product is
\([F_{\mathfrak P}:K_{\mathfrak p}]\). This proves (10.117) and the
assertion about applying the preceding theorem.

In the coset case, \((\boldsymbol m-\boldsymbol m_*)/q^J\) is an integer
vector. Factoring its common root monomial gives (10.118), term by term.
The common factor is a local unit, so it changes no valuation. Apply the
last assertion of Theorem 10.41 to \(V_0\): the Euler arguments stay
their original integers, and the new exponent bound is
\(\widehat A_i/q^J\). The torus coordinates are \(\vartheta_i^s\), of
height \(|s|h(\vartheta_i)\). Their contribution is at most
\(2|s|\sum_i(\widehat A_i/q^J)h(\vartheta_i)\le2X\sum_i\widehat A_i h(\vartheta_i)\).
The rational scalars and coefficient vector have not changed.
\(\square\)

The two comparisons in Theorem 10.43 exclude every finite valuation in
the case illustrated in Figure 10.6.

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={Disjoint upper and lower intervals for a finite valuation excess}]{../figures/arithmetic-precision-budget.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.6. The prepared scalars are p-integral and the torus
coordinates are local units, so \(\Phi/c_*\) is p-integral and the
valuation excess \(\nu=v_p(\Phi(x;\boldsymbol t))-\delta\) is
nonnegative. If the value is nonzero, the product-formula proof of
Theorem 10.41 gives \(\nu\le\mathcal A_I(X,T')\). The analytic proof
gives \(\nu\ge(n\mu-D)\theta-Lv_p(k!)\). The drawn endpoints
\(\mathcal A=3,G=5\) illustrate these two intervals; they are not
parameters of a specified auxiliary function. Their separation is
exactly the strict second comparison in (10.120), and forces
\(\Phi(x;\boldsymbol t)=0\). The zero value has infinite valuation and
is outside the finite axis. Human-source context: the product-formula
step in Yu's free2013 article, cited below.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/arithmetic_precision_budget.py}{Figure
program}.}
\end{flushleft}
\end{minipage}\par\medskip


\subsubsection{An explicit zero-forcing
step}\label{an-explicit-zero-forcing-step}

\textbf{Theorem 10.43 (extension of the prepared zeros).} Retain the
hypotheses of Corollary 10.40 over the field \(K\), and suppose that at
every integer \(x\) the unprojected function is the algebraic prepared
value at \((x,\vartheta_1^x,\ldots,\vartheta_r^x)\), with
\(\vartheta_i\in K^\times\) local units. Let \(T'=T-\mu+1\ge0\), and let
\(X\ge0\) be real. Put

\[
\begin{aligned}
Q_I&=\begin{cases}0,&I=0,\\L(kI+v_q(k!)),&I>0,\end{cases}\\
\mathcal A_I(X,T')&=\frac d{ef\ln p}\biggl(
\min\{h_\infty(c)+\ln N,\ h_2(c)+\tfrac12\ln N\}\\
&\hspace{28mm}+\ln\mathcal B_I(X,T')+Q_I\ln q
+2X\sum_iA_i h(\vartheta_i)\biggr).
\end{aligned}
\tag{10.119}
\]

The node count \(n\), \(\kappa\), \(\theta\), and \(M_0=\max\{B,C_0\}\)
are those of Corollary 10.40. If

\[
\begin{aligned}
U+D\theta+L v_p(k!)&\ge n\mu\theta+(\kappa+1)\mu M_0,\\
(n\mu-D)\theta-L v_p(k!)&>\mathcal A_I(X,T'),
\end{aligned}
\tag{10.120}
\]

then \(\Phi(x;\boldsymbol t)=0\) for every integer \(|x|\le X\) and
every \(|\boldsymbol t|\le T'\). With the larger normalizing scale of
(10.114), an alternative sufficient pair is

\[
\begin{aligned}
U+D\left(\theta+\frac1{p-1}\right)&\ge n\mu\theta+(\kappa+1)\mu M_0,\\
(n\mu-D)\theta-\frac D{p-1}&>\mathcal A_I(X,T').
\end{aligned}
\tag{10.121}
\]

\textbf{Proof.} The first line of (10.120) supplies the analytic budget
of Corollary 10.40. At every stated index its conclusion gives \[
v_p(\Phi(x;\boldsymbol t))-\delta
\ge(n\mu-D)\theta-L v_p(k!).
\] If the prepared value were nonzero, Theorem 10.41 with \(J=0\) would
give the opposite upper bound \(\mathcal A_I(X,T')\): its torus
coordinates have height \(|x|h(\vartheta_i)\), and its clearing exponent
is at most \(Q_I\). The strict second line of (10.120) contradicts this.
A prepared index with \(t_0>D\) already has zero value. These cases
prove every claimed zero. The same argument using (10.114) instead
proves (10.121). \(\square\)

Both comparisons remove \(\delta\), so they are invariant under scaling
the coefficient vector. The first comparison gives enough analytic
precision; the second makes a nonzero algebraic value impossible. At
fractional nodes one must additionally use the field or common-factor
rule of Proposition 10.42. Applying this step repeatedly requires
checking these inequalities, the slope conditions, and the available
derivative orders at every stage.

\subsection{18. Retaining the derivative orders in the height
budget}\label{retaining-the-derivative-orders-in-the-height-budget}

The uniform estimate (10.115) bounds every coefficient by the sum of all
coefficients. It also discards the factorials in the Euler binomials.
Those bounds suffice for Theorem 10.43, but a numerical induction
benefits from retaining the actual derivative order. We now do so,
including the average over the symmetric initial nodes. The additive
estimate below has the form of Lemma 1.6 in Yu's free 1990 paper; we
give its full proof using only an elementary factorial inequality.

\subsubsection{Positive polynomial
majorants}\label{positive-polynomial-majorants}

Let
\(\Omega_j=\max_{\boldsymbol m\in\mathcal M}|\omega_j(\boldsymbol m)|\),
for \(1\le j<r\), and let \(\Omega_\Sigma=\sum_{j<r}\Omega_j\). For
integers \(\Omega,h\ge0\), define \(E_h(\Omega)=\binom{\Omega+h}{h}\).
Thus \(E_0(\Omega)=E_h(0)=1\). Put \(\Gamma_0=1\); for \(h>0\), put
\(\Gamma_h=0\) if \(r=1\), and
\(\Gamma_h=\binom{\Omega_\Sigma+r-2+h}{h}\) if \(r\ge2\). Use empty
products and sums when \(r=1\). Define the positive polynomial

\[
\begin{aligned}
\mathcal P_{I,X}(Z)
&=\left(\frac1{k!}\prod_{b=1}^k
(q^{-I}X+k-1+b+v(k)Z)\right)^L\\
&=\sum_{u=0}^{D}\mathcal C_{I,u}(X)Z^u,
\qquad D=kL.
\end{aligned}
\tag{10.122}
\]

\textbf{Theorem 10.44 (an order-dependent scalar bound).} For the
rational prepared scalar
\(\mathcal S_{\boldsymbol m,a,\ell}(x;\boldsymbol t)\) in (10.92), with
\(0\le a<k\), \(1\le\ell\le L\), \(|x|\le X\), and \(t_0\le D\), one has

\[
\begin{aligned}
|\mathcal S_{\boldsymbol m,a,\ell}(x;\boldsymbol t)|
&\le\mathcal H_I(X;\boldsymbol t)
:=\mathcal C_{I,t_0}(X)\prod_{j<r}E_{t_j}(\Omega_j),\\
\mathcal B_I^*(X,T)
&:=\max_{|\boldsymbol t|\le T,\ t_0\le D}
\mathcal H_I(X;\boldsymbol t).
\end{aligned}
\tag{10.123}
\]

An index with \(t_0>D\) has zero scalar. If \(\Omega_j=0\), its Euler
factor is \(\Delta(0;t_j)=1\), so changing that index to zero leaves the
scalar unchanged. The maximum in (10.123) is at least one and satisfies

\[
\begin{aligned}
\mathcal B_I^*(X,T)&\le\mathcal B_I(X,T),\\
\mathcal B_I^*(X,T)&\le
\left(\mathrm e\left(2+\frac{q^{-I}X}{k}\right)\right)^D
\max_{\substack{u,h\ge0\text{ integers}\\u\le D,\ u+h\le T}}
v(k)^u\Gamma_h.
\end{aligned}
\tag{10.124}
\]

Here \(\mathcal B_I\) is (10.115); the second bound retains the division
of the total order between the additive and Euler directions. For
\(h>0\), one may further bound \(\Gamma_h\) by
\([\mathrm e(1+(\Omega_\Sigma+r-1)/h)]^h\).

\textbf{Proof.} At a given real or complex \(x\), replace each factor
\(q^{-I}x+a+b\) by its absolute-value upper bound \(q^{-I}X+a+b\).
Expansion of the product bounds the absolute value of each coefficient
by the corresponding coefficient of this polynomial with nonnegative
coefficients. Its constant coefficient after division by \(k!\) is at
least one: each factor \(q^{-I}X+a+b\) is at least \(b\). Consequently
increasing its power from \(\ell\) to \(L\) can only increase every
coefficient. Increasing \(a\) to \(k-1\) has the same property. Taking
the coefficient of \(Z^{t_0}\) gives \(\mathcal C_{I,t_0}(X)\).

For an integer \(\omega\) with \(|\omega|\le\Omega\), the product
defining \(\Delta(\omega;t)\) gives
\(|\Delta(\omega;t)|\le E_t(\Omega)\). Indeed \(|\omega+b|\le\Omega+b\),
for \(1\le b\le t\), and the factorial \(t!\) is retained. When
\(\Omega=0\), the Euler argument is zero and every such binomial is one.
This proves (10.123), the degree-zero case and the duplicate-direction
assertion. Every \(\mathcal C_{I,u}(X)\) is positive. At \(X=0\), it is
the prepared Taylor coefficient at the integer \(k-1\), so it is an
integer by Lemma 10.28 and is at least one. The coefficients increase
with \(X\). In particular the index \(\boldsymbol t=0\) proves
\(\mathcal B_I^*\ge1\).

For the first inequality in (10.124), each \(\mathcal C_{I,u}(X)\) is at
most \(\mathcal P_{I,X}(1)\). Bounding each of its \(k\) factors by
\(q^{-I}X+2k-1+v(k)\) gives the additive part of (10.115). Put
\(\Omega=\max_j\Omega_j\), with \(\Omega=0\) for an empty index set. The
Euler product is at most \(\max\{1,\Omega+T\}^T\), as in Theorem 10.32.
Hence this sharper maximum never exceeds the earlier one.

We prove the additive estimate used for the second inequality. For every
integer \(k\ge1\), monotonicity of \(\log x\) on each interval
\([j-1,j]\) gives \(\log k!\ge\int_1^k\log x\,dx=k\log k-k+1\). Thus
\(k!\ge\mathrm e(k/\mathrm e)^k\), including \(k=1\). At \(y\ge0\),
coefficient comparison with \((Z+k)^{k\ell}/(k!)^\ell\) gives \[
\begin{aligned}
|\Theta(z;k,\ell,u)|
&\le\frac{\binom{k\ell}{u}(|z|+k)^{k\ell-u}}{(k!)^\ell}\\
&\le\frac{\ell^u}{u!}\,
\frac{(|z|+k)^{k\ell}}{(k!)^\ell}
\le\left(\frac{\mathrm e(|z|+k)}k\right)^{k\ell}.
\end{aligned}
\tag{10.125}
\] The first comparison takes absolute values in the differentiated
product. The second uses \(\binom{k\ell}{u}\le(k\ell)^u/u!\) and
\(|z|+k\ge k\). The last uses the factorial bound and
\(\ell^u/u!\le\mathrm e^\ell\), since this is one nonnegative term of
the exponential series. Derivatives above degree \(k\ell\) are zero.
Apply (10.125) with \(z=q^{-I}x+a\), multiply by \(v(k)^u\), and
increase \(\ell\) to \(L\). The base \(\mathrm e(2+q^{-I}X/k)\) is
greater than one. The same bound applies to \(\mathcal C_{I,u}(X)\), by
using \(x=X,a=k-1,\ell=L\).

Finally, \(E_h(\Omega)\) is the coefficient of \(Z^h\) in the product of
\(\Omega+1\) copies of \(1+Z+Z^2+\cdots\). Distributing \(h\) objects
among these slots gives \(\binom{\Omega+h}{h}\). Multiplying these
series over the Euler directions uses \(\Omega_\Sigma+r-1\) slots, so
every fixed Euler product of total order \(h\) is at most \(\Gamma_h\):
it is one of the nonnegative summands of that coefficient. For no Euler
directions only order zero occurs. This proves the second inequality in
(10.124). When \(h>0,r\ge2\), the numerator product for \(\Gamma_h\) is
at most \((\Omega_\Sigma+r-1+h)^h\); the factorial bound gives the
further estimate. When \(r=1\), its left side is zero. \(\square\)

\subsubsection{Averaging the initial
rows}\label{averaging-the-initial-rows}

Keep the data of Theorem 10.32, and write
\(H=\sum_i A_i h(\vartheta_i)\). Let \(r_0\) be the number of Euler
indices with \(\Omega_j>0\). The representative prepared indices after
removing duplicate zero directions, and their row count, are

\[
\begin{aligned}
\mathcal J_T&=\{\boldsymbol t\ge0:|\boldsymbol t|\le T,\ t_0\le D,
\ t_j=0\text{ if }\Omega_j=0\},\\
J_T&=|\mathcal J_T|=
\sum_{u=0}^{\min(D,T)}\binom{T-u+r_0}{r_0},
\qquad M_*=(2S+1)J_T.
\end{aligned}
\tag{10.126}
\]

\textbf{Corollary 10.45 (a sharper initial coefficient budget).} If
\(N>M_*\), the integral kernel in Theorem 10.32 exists with every
originally specified zero and with

\[
\begin{aligned}
h_\infty(c)\le h_2(c)\le&\ \tfrac12\log N+
\frac{\log|\Delta_K|}{2d}\\
&+\frac1{N-M_*}\sum_{s=-S}^S\sum_{\boldsymbol t\in\mathcal J_T}
\left(\tfrac12\log N+
\log\mathcal H_0(|s|;\boldsymbol t)+2|s|H\right).
\end{aligned}
\tag{10.127}
\]

A simpler consequence is

\[
\begin{aligned}
h_2(c)\le&\ \tfrac12\log N+\frac{\log|\Delta_K|}{2d}\\
&+\frac{M_*}{N-M_*}
\left(\tfrac12\log N+\log\mathcal B_0^*(S,T)
+\frac{2S(S+1)}{2S+1}H\right).
\end{aligned}
\tag{10.128}
\]

When \(N>M\) in Theorem 10.32, the right side of (10.128) is no larger
than the right side of (10.96).

\textbf{Proof.} An index above the additive degree has zero scalar. An
index in a zero Euler direction duplicates the row with that index set
to zero, by Theorem 10.44. Removing it therefore preserves every
original vanishing condition. Fixing \(t_0=u\) leaves \(r_0\) Euler
indices of total order at most \(T-u\). The divider argument from
Theorem 10.32 counts them as \(\binom{T-u+r_0}{r_0}\), also when
\(r_0=0\). This proves (10.126).

At each retained row \((s,\boldsymbol t)\), the same adelic row-norm
proof as in Theorem 10.32 now uses \(\mathcal H_0(|s|;\boldsymbol t)\).
At finite places the prepared scalars are still integers. The row norm
is therefore at most
\(\sqrt N\mathcal H_0(|s|;\boldsymbol t)\exp(2|s|H)\). The scalar
majorant is at least one: its positive additive coefficient at integer
\(|s|+k-1\) is an integer by Lemma 10.28, and every retained Euler
factor is a positive integer. Hence the displayed row bound is at least
one, as required by the maximum in (8.40). A row that vanishes for
additional reasons has norm zero and contributes the factor one there.
Apply that proved weighted Siegel lemma with all weights one, retaining
the separate row factors. Its logarithm is exactly (10.127), and its
finite conditions give algebraic-integer coordinates. Dependent rows are
already included in that lemma.

The uniform scalar bound in (10.123) and the identity
\(\sum_{s=-S}^S|s|=S(S+1)\) give (10.128). Finally \(M_*\le M\),
\(\mathcal B_0^*\le\mathcal B\), and \(S(S+1)/(2S+1)\le S\). All
summands are nonnegative. The function \(m/(N-m)\) increases for
\(0\le m<N\), as cross multiplication shows. These facts prove the
comparison with (10.96). The nonzero polynomial and its required
ordinary jets follow from Proposition 10.31, exactly as before.
\(\square\)

\subsubsection{The smaller nonzero-value
budget}\label{the-smaller-nonzero-value-budget}

For the integer target nodes in Theorem 10.43, define, at each
\(\boldsymbol t\in\mathcal J_{T'}\),

\[
\begin{aligned}
\mathcal A_{I,\boldsymbol t}^*(X)=\frac d{ef\ln p}\biggl(&
\min\{h_\infty(c)+\ln N,h_2(c)+\tfrac12\ln N\}\\
&+\ln\mathcal H_I(X;\boldsymbol t)
+\Xi_{I,0,t_0}\ln q+2XH\biggr),\\
\mathcal A_I^*(X,T')&=\max_{\boldsymbol t\in\mathcal J_{T'}}
\mathcal A_{I,\boldsymbol t}^*(X).
\end{aligned}
\tag{10.129}
\]

\textbf{Corollary 10.46 (zero forcing with the retained orders).} Under
the hypotheses of Theorem 10.43, its first analytic comparison and

\[
(n\mu-D)\theta-Lv_p(k!)>\mathcal A_I^*(X,T')
\tag{10.130}
\]

force every zero asserted there. The analogous assertion holds with its
larger normalizing scale and lower bound \((n\mu-D)\theta-D/(p-1)\).
Moreover \(\mathcal A_I^*(X,T')\le\mathcal A_I(X,T')\).

\textbf{Proof.} Repeating the product-formula proof of Theorem 10.41
uses the individual scalar bound \(\mathcal H_I(X;\boldsymbol t)\)
instead of \(\mathcal B_I\), with its same exact clearing exponent. Thus
a nonzero value at a retained index has valuation excess at most
\(\mathcal A_{I,\boldsymbol t}^*\). The analytic lower bound contradicts
(10.130). Every removed index either has zero scalar by additive degree
or duplicates a retained index by setting its zero Euler directions to
zero. The larger normalizing scale gives the stated second conclusion by
the same argument. Finally
\(\mathcal H_I(X;\boldsymbol t)\le\mathcal B_I(X,T')\) and
\(\Xi_{I,0,t_0}\le Q_I\). Taking the maximum proves the comparison with
(10.119). Fractional targets still require the field or common-factor
conditions of Proposition 10.42. \(\square\)

These refinements retain the exact factorials, lcm, derivative orders
and mean node size. To obtain a numerical logarithm estimate, their
bounds and the analytic slope and order conditions must all be checked
for its chosen induction parameters.

\subsection{19. Passing to one exponent
coset}\label{passing-to-one-exponent-coset}

The next auxiliary function is obtained by retaining one coset of the
exponent set and dividing its exponent differences by \(q\). Two facts
justify this step: independent root monomials separate the cosets, and
an invertible affine change preserves the filtered Euler jets. The
saturated Kummer theorem and the Newton reparametrization have already
been proved in Theorem 10.14 and Lemma 10.30. We now combine them,
retaining the coefficient field and the condition on the integer node.
This is the algebraic mechanism of the coset passage in Yu's free 2013
paper, Lemma 5.4.

\subsubsection{Separating the root
monomials}\label{separating-the-root-monomials}

\textbf{Theorem 10.47 (exact coset extraction).} Let \(F\) be a
characteristic-zero field, \(q\) a prime, and
\(\vartheta_i\in F^\times\), \(1\le i\le r\). Choose
\(\eta_i^q=\vartheta_i\), and suppose
\([F(\eta_1,\ldots,\eta_r):F]=q^r\). Take a finite set
\(\mathcal M\subset\mathbb Z^r\), coefficients
\(A_{\boldsymbol m}\in F\), and an integer \(s\) coprime to \(q\). If
\(\sum_{\boldsymbol m}A_{\boldsymbol m}\boldsymbol\eta^{s\boldsymbol m}=0\),
then for each residue vector \(\boldsymbol\rho\in\{0,\ldots,q-1\}^r\),

\[
\sum_{\boldsymbol m\equiv\boldsymbol\rho\ (q)}
A_{\boldsymbol m}\boldsymbol\vartheta^{s(\boldsymbol m-\boldsymbol\rho)/q}=0.
\tag{10.131}
\]

\textbf{Proof.} The \(q^r\) monomials
\(\boldsymbol\eta^{\boldsymbol\rho}\) span the root field over \(F\):
reduce every exponent modulo \(q\), using \(\eta_i^q=\vartheta_i\ne0\),
also for negative exponents. The span is closed under multiplication.
Every nonzero element of this finite-dimensional span has an inverse in
it: multiplication by that element is injective in the surrounding field
and thus surjective on the span. Hence it is the whole root field. Its
degree is \(q^r\), so these spanning monomials are a basis.

Group the given sum by \(\boldsymbol\rho\). Its coefficient after
factoring \(\boldsymbol\eta^{s\boldsymbol\rho}\) is exactly the
expression in (10.131), an element of \(F\). Let \(\boldsymbol\rho'\) be
the least nonnegative residues of \(s\boldsymbol\rho\), and put
\(\boldsymbol b=(s\boldsymbol\rho-\boldsymbol\rho')/q\). Then
\(\boldsymbol\eta^{s\boldsymbol\rho}=\boldsymbol\vartheta^{\boldsymbol b}\boldsymbol\eta^{\boldsymbol\rho'}\).
Multiplication by \(s\) permutes the residue vectors because
\(q\nmid s\). Thus grouping gives a relation among distinct basis
monomials, with each grouped coefficient multiplied by the nonzero
scalar \(\boldsymbol\vartheta^{\boldsymbol b}\). Every coefficient
vanishes, proving (10.131). \(\square\)

This argument extracts exact zeros. A large finite valuation of the sum
does not by itself supply that valuation for each coset.

\subsubsection{The next prepared family}\label{the-next-prepared-family}

\textbf{Theorem 10.48 (coset restriction and affine jet transfer).} Take
a number field \(K\), a prime \(q\ne p\), local units
\(\vartheta_i\in K^\times\), and the integer Euler forms
\(\omega_j(\boldsymbol m)=\sum_iC_{ij}m_i\) in (10.92). Let
\(c\in K^N\setminus\{0\}\). Suppose that a field \(F\supset K\) and
chosen roots \(\eta_i^q=\vartheta_i\) satisfy the degree condition of
Theorem 10.47. Let \(\mathcal E\) be any set of integers coprime to
\(q\). Assume that all original prepared values at
\((s/q,\eta_1^s,\ldots,\eta_r^s)\), \(s\in\mathcal E\), vanish through
total order \(T\).

Choose \(\boldsymbol m_*\) at which some coefficient attains
\(\delta=\min v_p(c_j)\). Retain its exponent coset, and put

\[
\begin{aligned}
\mathcal M'&=\{(\boldsymbol m-\boldsymbol m_*)/q:
\boldsymbol m\in\mathcal M,\ \boldsymbol m\equiv\boldsymbol m_*\ (q)\},\\
c'_{\boldsymbol n,a,\ell}&=c_{\boldsymbol m_*+q\boldsymbol n,a,\ell},\\
Q'(X,Y)&=\sum_{\boldsymbol n,a,\ell}c'_{\boldsymbol n,a,\ell}
\Delta(q^{-(I+1)}X+a;k)^\ell Y^{\boldsymbol n}.
\end{aligned}
\tag{10.132}
\]

Then \(Q'\ne0\). Its new prepared family, with stage \(I+1\) and the
same Euler matrix \(C\), vanishes at \((s,\boldsymbol\vartheta^s)\),
\(s\in\mathcal E\), through the same total order \(T\).

The original prepared values restricted to this coset and the new
prepared values are related by the affine changes

\[
\begin{aligned}
\Delta(qZ+\omega_j(\boldsymbol m_*);t_j)
&=\sum_{u_j=0}^{t_j}d_{t_j,u_j}^{(j)}\Delta(Z;u_j),
\qquad d_{t_j,t_j}^{(j)}=q^{t_j},\\
\min_{|\boldsymbol t|\le T}v_p(\text{original coset value at }s/q)
&=\min_{|\boldsymbol t|\le T}v_p(\text{new prepared value at }s).
\end{aligned}
\tag{10.133}
\]

The minimum identity includes \(\infty\) when all values are zero. The
coefficients satisfy

\[
\min v_p(c'_j)=\delta,\qquad
h_\infty(c')\le h_\infty(c),\qquad h_2(c')\le h_2(c).
\tag{10.134}
\]

Global integrality of \(c\) is retained by \(c'\). For any convex body
\(\mathcal C\) and vector \(x_0\) with
\(\mathcal M\subset q^{-I}\mathcal C-x_0\), one has

\[
\begin{aligned}
\mathcal M'&\subset q^{-(I+1)}\mathcal C-(x_0+\boldsymbol m_*)/q,\\
\deg_{Y_i}(\text{ordinary monomial shift of }Q')
&\le\left\lfloor
\frac{\max_{\boldsymbol m\in\mathcal M}m_i-
\min_{\boldsymbol m\in\mathcal M}m_i}{q}\right\rfloor.
\end{aligned}
\tag{10.135}
\]

Its additive degree is at most \(D=kL\).

\textbf{Proof.} Fix \(s\in\mathcal E\) and a prepared index. The
rational additive and Euler factors, multiplied by the coefficients
\(c\), belong to \(K\subset F\). Theorem 10.47 therefore extracts a zero
separately in every exponent coset. For the chosen one, factoring
\(\boldsymbol\eta^{s\boldsymbol m_*}\) gives a zero with torus monomials
\(\boldsymbol\vartheta^{s\boldsymbol n}\).

The additive scalar at \(s/q\) is exactly
\(v(k)^{t_0}\Theta(q^{-(I+1)}s+a;k,\ell,t_0)\). The old Euler argument
is \(\omega_j(\boldsymbol m_*+q\boldsymbol n)=
q\omega_j(\boldsymbol n)+\omega_j(\boldsymbol m_*)\). Lemma 10.30 gives
the first line of (10.133), and both its matrix and inverse have
coefficients in \(\mathbb Z[1/q]\). The product changes only to indices
with \(u_j\le t_j\), keeping \(t_0\) fixed. On the finite space of
indices of total order at most \(T\), it is triangular with nonzero
diagonal \(q^{\sum_{j<r}t_j}\); its inverse preserves the same
filtration. Every new jet is therefore a linear combination of the
extracted old coset jets, proving all the claimed zeros without reducing
\(T\).

The factored root monomial is a local unit, since every \(\vartheta_i\)
is one. The affine matrix and inverse are p-integral because \(q\ne p\).
Applying the ultrametric inequality to each change shows in both
directions that the minimum valuation cannot decrease. This proves the
second line of (10.133). It compares the jets of an individual coset;
extraction from the whole sum used exact vanishing.

The selected coefficient vector is nonzero and contains a coefficient of
the original minimum valuation. This proves the first assertion of
(10.134). At every place its maximum norm is at most the old one; at
each archimedean embedding the same is true for its Euclidean norm.
Summing logarithms in the projective height definitions proves the
remaining two inequalities. Coordinates of an integral vector remain
integral. Distinct exponents give distinct \(\boldsymbol n\), so Lemma
10.29 proves \(Q'\ne0\), as in Proposition 10.31.

For the geometric inclusion, divide
\(\boldsymbol m+x_0\in q^{-I}\mathcal C\) by \(q\) and subtract
\((x_0+\boldsymbol m_*)/q\). Each coordinate width of the retained
exponent set is divided by \(q\). Multiplying \(Q'\) by the inverse of
its coordinatewise least Laurent monomial makes its torus exponents
nonnegative, with exactly those widths. They are integers, proving
(10.135). The additive degree is unchanged. \(\square\)

Figure 10.7 shows the exponent change in Theorem 10.48 for a square
support at \(q=2\).

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={A parity coset retained and recentered into a smaller integer support}]{../figures/coset-descent-geometry.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.7. The exact support is
\(\mathcal M=\{-2,-1,0,1,2,3\}^2\). The four colors mark its four
classes modulo two. Selecting \(\boldsymbol m_*=(1,0)\) retains the nine
orange points \(m_1\in\{-1,1,3\},m_2\in\{-2,0,2\}\); translating by
\(-\boldsymbol m_*\) and dividing by two gives \(\{-1,0,1\}^2\). The
original coordinate width five becomes retained width four and then new
width two, exactly as in (10.135). For odd \(s\), full root-field degree
separates the four root monomials in Theorem 10.47; their exact zero
sums can then be treated separately. In the selected class the common
local unit is \(\boldsymbol\eta^{s\boldsymbol m_*}\), and the remaining
monomials are \(\boldsymbol\vartheta^{s\boldsymbol n}\), by
(10.132)--(10.133). The displayed support illustrates that algebraic
transformation; it does not assert a particular coefficient vector or
numerical logarithm bound. Human-source context: Yu's free2013 Lemma5.4,
cited below. \href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/coset_descent_geometry.py}{Figure
program}.}
\end{flushleft}
\end{minipage}\par\medskip


\subsubsection{Root-of-unity phases and the coefficient
field}\label{root-of-unity-phases-and-the-coefficient-field}

\textbf{Corollary 10.49 (q-primary phases).} Let
\(\alpha_0,\theta_1,\ldots,\theta_r\) be the saturated generators of
Theorem 10.14. Choose \(\xi^q=\alpha_0\) and \(\rho_i^q=\theta_i\), and
put \(F=K(\xi)\). For an integer \(P\ge1\) coprime to \(q\) and integers
\(d_i\), set

\[
\vartheta_i=\theta_i^P\alpha_0^{d_i},\qquad
\eta_i=\rho_i^P\xi^{d_i},\qquad
[F(\boldsymbol\eta):F]=q^r,\qquad
h(\vartheta_i)=P h(\theta_i).
\tag{10.136}
\]

Thus Theorems 10.47--10.48 apply to these phased torus coordinates. If
the \(\theta_i\) are local units and \(q\ne p\), all the displayed roots
are local units. In particular one may take \(P=p^a\), \(a\ge0\).

\textbf{Proof.} Theorem 10.14 gives total root degree \(q^{r+1}\), and
the same proved Kummer argument for \(\alpha_0\) alone gives
\([F:K]=q\). Hence \([F(\boldsymbol\rho):F]=q^r\). Clearly
\(F(\boldsymbol\eta)\subset F(\boldsymbol\rho)\). Choose integers
\(A,B\) with \(AP+Bq=1\); integer division gives such a pair because
\(P\) and \(q\) are coprime. Then
\(\rho_i=(\eta_i\xi^{-d_i})^A\theta_i^B\), giving the reverse inclusion
and the degree in (10.136). The q-th power identity follows directly. A
root of unity has height zero; applying the product-height inequality in
both directions shows that multiplying by it preserves height. The
proved power identity then gives \(h(\vartheta_i)=P h(\theta_i)\).
Finally taking valuations in the defining power equations proves every
unit assertion. \(\square\)

The coefficient field \(F\) holds the root-of-unity phase, while the
selected new coefficients and torus bases in Theorem 10.48 remain in
\(K\). The next zero set consists of the stated integer nodes coprime to
\(q\); extension to further nodes uses the extrapolation results, with
their slope and precision conditions.

\subsection{20. Retaining the phase congruence in the next
support}\label{retaining-the-phase-congruence-in-the-next-support}

The phase used to make a local unit close to one need not belong to the
global field \(K\). A support congruence controls its powers. After
division by \(q\), the congruence has either one class or \(q\) classes;
in the latter case the q-primary root separates them. We prove both
cases, keeping all the new coefficients in \(K\). This supplies the
phase and support passage of Yu's free 2013 paper, equations
(5.51)--(5.61), before the separate analytic extension to further nodes.

\subsubsection{The congruence and its phase
classes}\label{the-congruence-and-its-phase-classes}

\textbf{Lemma 10.50 (division of a phase congruence).} Let \(q\) be
prime, \(G_0\ge1\) an integer, and \(a\in\mathbb Z\). The congruence
\(qu+a\equiv0\pmod{G_0}\) has the following solutions:

\[
\begin{aligned}
u&\equiv-q^{-1}a\pmod{G_0},
&&\gcd(q,G_0)=1,\\
u&\equiv-a/q+bG_0/q\pmod{G_0},
\quad 0\le b<q,
&&q\mid G_0,\ q\mid a.
\end{aligned}
\tag{10.137}
\]

In the second case there is no solution if \(q\nmid a\), and the
displayed \(q\) classes are distinct. Suppose additionally that
\(\alpha_0\in K^\times\) is not a q-th power, \(\mu_q\subset K\), and
\(\beta^q=\alpha_0\). Then, for every integer \(s\) prime to \(q\), the
\(q\) powers \(\beta^{sb}\), \(0\le b<q\), are linearly independent over
\(K\).

\textbf{Proof.} If \(q\) and \(G_0\) are coprime, integer division gives
integers \(A,B\) with \(Aq+BG_0=1\). Multiplication by \(A\) solves the
congruence, uniquely modulo \(G_0\). If \(q\mid G_0\), reduction modulo
\(q\) first forces \(q\mid a\). Put \(H=G_0/q\). The original congruence
is equivalent to \(u+a/q\equiv0\pmod H\), so \(u=-a/q+bH+G_0v\) for a
unique \(b\) modulo \(q\). This proves the second formula and its
distinctness.

The complete Kummer argument of Theorem 10.14 applied to the one
independent class \(\alpha_0\) proves \([K(\beta):K]=q\). Equivalently
\(1,\beta,\ldots,\beta^{q-1}\) are a basis, since their span is the
field and has that degree, as proved in Theorem 10.47. Multiplication by
\(s\) permutes exponents modulo \(q\); reducing \(\beta^{sb}\) by
\(\beta^q=\alpha_0\) multiplies those basis elements by nonzero scalars
in \(K\). This proves the asserted independence, also for negative
\(s\). \(\square\)

\subsubsection{The phased prepared
values}\label{the-phased-prepared-values}

Take the saturated generators \(\alpha_0,\theta_1,\ldots,\theta_r\) of
Theorem 10.14, with \(\alpha_0\) of order \(q^u\). Let \(G_0\ge1\), put
\(G=q^uG_0\), and choose roots of unity and algebraic roots in a fixed
embedding such that

\[
\begin{aligned}
\zeta^{G_0}&=\alpha_0,\quad \zeta\text{ has order }G,\quad
\xi^q=\zeta,\quad \beta=\xi^{G_0},\quad \beta^q=\alpha_0,\\
\rho_i^q&=\theta_i,\qquad
\gamma_i=\theta_i^P\zeta^{d_i},\qquad
\eta_i=\rho_i^P\xi^{d_i},\qquad \eta_i^q=\gamma_i ,
\end{aligned}
\tag{10.138}
\]

where \(P\ge1\) is prime to \(q\), and \(d_i\in\mathbb Z\). The
existence of \(\zeta\) with the specified order is a hypothesis here;
Lemma 10.53 constructs it for the local residue modulus. Write
\(d\cdot m=\sum_i d_i m_i\), and let \(F=K(\beta)\). Theorem 10.14 gives
\([F(\boldsymbol\rho):F]=q^r\).

Let \(I\ge0,k,L\ge1,T\ge0\) be integers. Let a finite nonempty
\(\Lambda\subset\mathbb Z^r\) satisfy
\(d\cdot\iota\equiv\varepsilon\pmod{G_0}\). Fix \(\iota_0\in\Lambda\),
an integer Euler matrix \(C\), and a nonzero coefficient vector
\(c\in K^N\), with \(N=kL|\Lambda|\). The fractional prepared values are

\[
\begin{aligned}
\Psi(s/q;\boldsymbol t)=
\sum_{\iota,a,\ell}c_{\iota,a,\ell}
v(k)^{t_0}\Theta(q^{-(I+1)}s+a;k,\ell,t_0)
\prod_{j<r}\Delta(C_j(\iota-\iota_0);t_j)
\boldsymbol\eta^{s(\iota-\iota_0)} .
\end{aligned}
\tag{10.139}
\]

Here \(C_j(m)=\sum_i C_{ij}m_i\), the indices \(a,\ell\) retain
\(0\le a<k,1\le\ell\le L\), and the sum ranges over \(\iota\in\Lambda\).

\textbf{Theorem 10.51 (complete phase and support passage).} Suppose
that (10.139) is zero for every \(s\) in a set
\(\mathcal E\subset\mathbb Z\) prime to \(q\), and every
\(|\boldsymbol t|\le T\). Put \(\delta=\min_jv_p(c_j)\), where
\(q\ne p\) and the \(\theta_i\) are units in a fixed embedding into
\(\mathbb C_p\).

Choose \(\iota_{\min}\) at which some coefficient has valuation
\(\delta\), and let \(\lambda_*\in\{0,\ldots,q-1\}^r\) represent
\(\iota_{\min}-\iota_0\) modulo \(q\). First retain the exponents in
this q-coset, and write

\[
\iota-\iota_0=q\lambda+\lambda_*,
\qquad a_*=d\cdot\lambda_*,
\qquad q\,d\cdot\lambda+a_*\equiv0\pmod{G_0}.
\tag{10.140}
\]

If \(\gcd(q,G_0)=1\), retain this whole coset and put
\(\varepsilon'=-q^{-1}a_*\pmod{G_0}\). If \(q\mid G_0\), then
\(q\mid a_*\). Put \(H=G_0/q\), \(a_0=-a_*/q\), and choose the unique
\(b_{\min}\in\{0,\ldots,q-1\}\) for which the \(\lambda\) corresponding
to \(\iota_{\min}\) satisfies
\(d\cdot\lambda\equiv a_0+b_{\min}H\pmod{G_0}\). Retain only that
subclass and put \(\varepsilon'=a_0+b_{\min}H\).

The resulting nonempty set \(\Lambda'\) has
\(d\cdot\lambda\equiv\varepsilon'\pmod{G_0}\). Select
\(\lambda_0\in\Lambda'\) corresponding to \(\iota_{\min}\), retain the
same coefficients \(c'_{\lambda,a,\ell}=c_{\iota,a,\ell}\), and form

\[
\begin{aligned}
Q'(X,Y)&=\sum_{\lambda,a,\ell}c'_{\lambda,a,\ell}
\Delta(q^{-(I+1)}X+a;k)^\ell
Y^{\lambda-\lambda_0},\\
\Psi'(s;\boldsymbol t)&=
\sum_{\lambda,a,\ell}c'_{\lambda,a,\ell}
v(k)^{t_0}\Theta(q^{-(I+1)}s+a;k,\ell,t_0)
\prod_{j<r}\Delta(C_j(\lambda-\lambda_0);t_j)
\boldsymbol\gamma^{s(\lambda-\lambda_0)} .
\end{aligned}
\tag{10.141}
\]

Then \(Q'\ne0\), and \(\Psi'(s;\boldsymbol t)=0\) for every
\(s\in\mathcal E,|\boldsymbol t|\le T\). All those values belong to
\(K\), although the individual \(\gamma_i\) need not belong to \(K\).
The coefficients retain global integrality when present, satisfy
\(\min v_p(c'_j)=\delta\), and have \(h_\infty(c')\le h_\infty(c)\),
\(h_2(c')\le h_2(c)\).

If \(\Lambda\subset q^{-I}\mathcal C-x\), the uncentered support
satisfies

\[
\begin{aligned}
\Lambda'&\subset q^{-(I+1)}\mathcal C-x',
\qquad x'=(x+\iota_0+\lambda_*)/q,\\
\deg_{Y_i}(\text{ordinary monomial shift of }Q')
&\le
\left\lfloor\frac{\max_{\iota\in\Lambda}\iota_i-
\min_{\iota\in\Lambda}\iota_i}{q}\right\rfloor,
\qquad \deg_XQ'\le kL .
\end{aligned}
\tag{10.142}
\]

\textbf{Proof.} Since \(d\cdot(\iota-\iota_0)\) is divisible by \(G_0\),
let \(w_\iota=d\cdot(\iota-\iota_0)/G_0\in\mathbb Z\). The torus part of
an old fractional term is
\(\boldsymbol\rho^{Ps(\iota-\iota_0)}\beta^{s w_\iota}\). The latter
factor belongs to \(F\), while its rational prepared scalar and
coefficient belong to \(K\). Over \(F\), the monomials
\(\boldsymbol\rho^{Ps\lambda_*}\), one for each q-residue vector, are
independent: \([F(\boldsymbol\rho):F]=q^r\) and \(Ps\) is prime to
\(q\), so Theorem 10.47 supplies exactly that basis permutation. Thus
every original zero separates into a zero for each first q-coset.

In the selected coset, substituting (10.140) into the zero and removing
the common nonzero factor \(\boldsymbol\rho^{Ps\lambda_*}\xi^{s a_*}\)
gives \[
\sum_{\iota\text{ in the selected coset},a,\ell}
c_{\iota,a,\ell}\mathcal S_\iota(s;\boldsymbol t)
\boldsymbol\gamma^{s\lambda}=0,
\] where \(\mathcal S_\iota\) contains the additive scalar and the
original Euler arguments. This equality holds for every specified
\(s,\boldsymbol t\). Lemma 10.50 proves that the divided support has
exactly the stated phase congruence when \(q\) and \(G_0\) are coprime.

Suppose \(q\mid G_0\). Lemma 10.50 gives the \(q\) subclasses
\(d\cdot\lambda=a_0+bH+G_0v_\lambda\). Since \(q\mid G_0\), (10.138)
implies \(\zeta^H=\xi^{qH}=\beta\). Multiply the last zero by
\(\zeta^{-s a_0}\). The contribution of subclass \(b\) now belongs to
\(\beta^{sb}K\): each term is its coefficient and rational scalar times
\(\boldsymbol\theta^{Ps\lambda}\alpha_0^{s v_\lambda}\beta^{sb}\). The
\(q\) powers \(\beta^{sb}\) are independent by Lemma 10.50, so every
subclass sum is zero. In particular the subclass containing
\(\iota_{\min}\) is zero for all the specified indices, not just at one
node.

Remove the common local unit \(\boldsymbol\gamma^{s\lambda_0}\). The old
Euler arguments are \[
C_j(\iota-\iota_0)=
qC_j(\lambda-\lambda_0)+C_j(q\lambda_0+\lambda_*).
\] The shift is an integer. Applying the invertible affine binomial
change of Lemma 10.30 in every Euler direction converts the extracted
zeros into the new ones in (10.141). Both changes preserve the
total-order cutoff, as proved in Theorem 10.48. The additive scalar is
already the stage \(I+1\) scalar and is unchanged.

For \(\lambda\in\Lambda'\), the integer
\(v'_\lambda=d\cdot(\lambda-\lambda_0)/G_0\) exists. Each new torus term
is
\(\boldsymbol\theta^{Ps(\lambda-\lambda_0)}\alpha_0^{s v'_\lambda}\in K\).
This proves the coefficient-field assertion. All roots of unity are
local units, since their finite powers are one; roots of the local units
\(\theta_i\) are units too. Thus every removed common torus factor is a
local unit. The coefficient subset contains a coefficient of valuation
\(\delta\); the projective-height and integrality assertions follow from
the proved subset inequalities in Theorem 10.48. Distinct new exponents
and the binomial basis in Lemma 10.29 give \(Q'\ne0\).

Finally divide \(\iota+x\in q^{-I}\mathcal C\) by \(q\) and use
\(\iota=\iota_0+\lambda_*+q\lambda\) to obtain the inclusion in
(10.142). Centering at \(\lambda_0\) changes no coordinate width. Every
retained width is at most the original width divided by \(q\), and is an
integer. The monomial shift and additive bound follow as in Theorem
10.48. \(\square\)

Figure 10.8 gives the exact two-class split in the case \(q\mid G_0\).

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={A divided support with two phase classes separated by the basis one and i}]{../figures/phase-congruence-split.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.8. Here \(q=2,G_0=6,d=(1,1)\), and the first residue
vector is \(\lambda_*=(1,1)\), so \(a_*=2\). The divided support is the
eight points \(\lambda\in\{-2,-1,0,1,2\}^2\) with
\(2(\lambda_1+\lambda_2)+2\equiv0\pmod6\). The original support may be
taken as \(\{(0,0)\}\cup\{2\lambda+(1,1)\}\), with reference
\(\iota_0=(0,0)\); every original point has coordinate sum divisible by
six. Put \(K=\mathbb Q,\alpha_0=-1\), choose a primitive twelfth root
\(\zeta\) with \(\zeta^6=-1\), and write \(\beta=\zeta^3=i\) in a
complex embedding. The normalized phase sum at a fixed odd node is
\(Z_0+i^sZ_1\), with each \(Z_b\in\mathbb Q\). When that sum is zero,
independence in Lemma 10.50 forces both coefficient sums to vanish. Red
points have coordinate sum \(-1\), blue points have sum \(-4\) or \(2\).
The circled reference \(\lambda_0=(-1,0)\) illustrates retaining the
subclass containing a coefficient of minimum valuation; its normalized
phase is one. Choosing a blue reference instead gives phases in
\(\{\pm1\}\), also in \(\mathbb Q\). These are exactly the congruence,
extraction and field-return mechanisms of Theorem 10.51 and
(10.140)--(10.142). The plotted coefficient sums are formal variables.
Human-source context: Yu's free2013 equations5.51--5.59, cited below.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/phase_congruence_split.py}{Figure program}.}
\end{flushleft}
\end{minipage}\par\medskip


\subsubsection{The analytic branch and arithmetic
height}\label{the-analytic-branch-and-arithmetic-height}

\textbf{Corollary 10.52 (the phase does not enlarge the height field).}
If \(v_p(\gamma_i-1)>1/(p-1)\), choose the fractional branch

\[
\eta_i=\exp\bigl(q^{-1}\log\gamma_i\bigr),\qquad
\rho_i^q=\theta_i,\qquad
\rho_i^P\xi^{d_i}=\eta_i .
\tag{10.143}
\]

Such a choice of the \(\rho_i\) exists. It identifies (10.139) with the
fractional values of the corresponding logarithmic curve. At an integer
node \(|s|\le X\), a nonzero new value in (10.141) satisfies

\[
\begin{aligned}
v_p(\Psi'(s;\boldsymbol t))-\delta
\le\frac{[K:\mathbb Q]}{ef\ln p}\biggl(&
\min\{h_\infty(c')+\ln N',h_2(c')+\tfrac12\ln N'\}\\
&+\ln\mathcal H_{I+1}(X;\boldsymbol t)
+\Xi_{I+1,0,t_0}\ln q
+2PX\sum_i A'_i h(\theta_i)\biggr),
\end{aligned}
\tag{10.144}
\]

where \(N'=kL|\Lambda'|\),
\(A'_i=\max_{\lambda\in\Lambda'}|\lambda_i-\lambda_{0,i}|\), and
\(\mathcal H\) uses the new integer Euler arguments in Theorem 10.44. An
index above the additive degree is already zero. Every local degree in
(10.144) is that of \(K\), not the degree of a field containing all
individual phases.

\textbf{Proof.} The exponential laws proved in Proposition 9.4 give the
q-th power of the chosen \(\eta_i\) as \(\gamma_i\), since \(q\ne p\)
changes no slope valuation. Begin with any root \(\rho_i^q=\theta_i\).
The quotient \(\eta_i/(\rho_i^P\xi^{d_i})\) is in \(\mu_q\). Raising to
\(P\) permutes that cyclic group because \(q\nmid P\); multiplication of
\(\rho_i\) by the appropriate q-th root of unity therefore gives
(10.143). This proves the branch identification without a further
root-existence assertion.

Theorem 10.51 writes each new torus term over \(K\) as a power of
\(\alpha_0\) times powers of the \(\theta_i\). At every place, a root of
unity has absolute value one, because its order-th power is one. Thus
these phases contribute zero to every local norm bound, including the
chosen p-adic place. The prepared scalar has the order-dependent
archimedean bound \(\mathcal H_{I+1}\) and the rational clearing factor
of Theorem 10.41. Apply its complete normalized product-formula argument
over \(K\), retaining the coefficient norms and using
\(|Ps(\lambda_i-\lambda_{0,i})|\le PX A'_i\). The signed torus-height
sum is \(2[K:\mathbb Q]PX\sum_i A'_ih(\theta_i)\). This proves (10.144).
\(\square\)

\subsubsection{Constructing the local
phase}\label{constructing-the-local-phase}

\textbf{Lemma 10.53 (lifting the residue phase and its depth).} Let
\(K_{\mathfrak p}\) be the completion at the chosen prime above \(p\),
with ramification index \(e\) and residue field of order \(p^f\).
Suppose \(q\ne p\), and let \(\alpha_0\in K\) have order \(q^u\),
\(u\ge1\). Then \(q^u\mid G=p^f-1\), and there is a primitive G-th root
\(\zeta\in K_{\mathfrak p}\) such that

\[
G_0=G/q^u,\qquad
\zeta^{G_0}=\alpha_0,\qquad
\theta_i\zeta^{-r_i}\equiv1\pmod{\mathfrak p}
\tag{10.145}
\]

for suitable integers \(r_i\), whenever the \(\theta_i\in K^\times\) are
local units. Given a real depth target \(\vartheta\ge0\), let \(t\) be
the least nonnegative integer with \(p^t>e/(p-1)\). Choose an integer
\(j\ge0\) with \(p^t/e+j>\vartheta+1/(p-1)\), put \(h=t+j\), and set

\[
\begin{aligned}
P&=p^h,\qquad d_i=-Pr_i,\qquad
\gamma_i=\theta_i^P\zeta^{d_i},\\
v_p(\gamma_i-1)&\ge p^t/e+j>
\vartheta+\frac1{p-1}.
\end{aligned}
\tag{10.146}
\]

Thus these phases supply the required analytic branch and supernormal
slope depth. This is a sufficient explicit choice of \(P\); using a
smaller prescribed \(P\) in a numerical estimate requires checking its
depth separately.

\textbf{Proof.} We first prove the lifting assertion directly. In a
complete discretely valued field, take a polynomial \(f\) with integral
coefficients and an integral \(x_0\) satisfying \(v_p(f(x_0))>0\),
\(v_p(f'(x_0))=0\). Define \(x_{n+1}=x_n-f(x_n)/f'(x_n)\). A finite
polynomial expansion gives \(f(x+y)=f(x)+f'(x)y+y^2R(x,y)\) with \(R\)
integral when \(x,y\) are integral. The linear terms cancel at the
Newton step, so \(v_p(f(x_{n+1}))\ge2v_p(f(x_n))\). Also
\(f'(x_{n+1})-f'(x_n)\) has positive valuation, leaving the derivative a
unit. Every step remains integral and in the original residue class. The
successive differences have valuations tending to infinity, hence form a
Cauchy sequence. Completeness gives its limit, and the polynomial
difference identity proves that the limit is a root. If \(x,y\) are
roots in that residue class, then
\(0=f(y)-f(x)=(y-x)(f'(x)+(y-x)R(x,y))\). The factor in parentheses is a
unit. Therefore \(x=y\). This proves existence and uniqueness of a
simple-residue lift, including the case of a zero residual error where
the iteration has already stopped.

Apply this to \(X^m-1\), for \(p\nmid m\). Its derivative at a unit is a
unit. In particular a prime-to-\(p\) root of unity reducing to one is
one, by uniqueness in that residue class. Reduction is consequently
injective on all prime-to-\(p\) roots of unity. The reduction of
\(\alpha_0\) has its exact order \(q^u\): a smaller order would make a
nontrivial prime-to-\(p\) root reduce to one. Since the finite
multiplicative residue group has order \(G\), we get \(q^u\mid G\).

The group of G-th roots in \(K_{\mathfrak p}\) maps bijectively to that
residue group: every nonzero residue solves \(X^G-1\) and has its unique
lift. Multiplication commutes with reduction, and the finite-field
cyclicity proved in Section 1 therefore gives a primitive root \(z\).
Write \(\alpha_0=z^{G_0a}\) with \(q\nmid a\), since its order is
\(q^u\). Let \(n=G/q^{v_q(G)}\), so \(\gcd(n,q)=1\). Choose
\(a'\equiv a\pmod{q^u}\) and \(a'\equiv1\pmod n\): Bezout's identity
gives a solution by varying \(a'\) among \(a+q^u b\). Then
\(\gcd(a',G)=1\). The primitive root \(\zeta=z^{a'}\) has
\(\zeta^{G_0}=\alpha_0\). Its reduction generates the residue group, so
every \(\overline{\theta_i}\) is \(\overline\zeta^{\,r_i}\). This proves
(10.145).

Put \(g_i=\theta_i\zeta^{-r_i}\). Its positive valuation distance from
one is at least \(1/e\), unless it is already one. The proved p-power
inequality of Lesson 9 is \[
v_p(g_i^{p^{a+1}}-1)\ge
\min\{p\,v_p(g_i^{p^a}-1),\,v_p(g_i^{p^a}-1)+1\}.
\] The function on the right increases with its argument. For
\(0\le a<t\), the lower bound \(p^a/e\) is at most \(1/(p-1)\), by the
definition of \(t\). Induction therefore gives
\(v_p(g_i^{p^t}-1)\ge p^t/e>1/(p-1)\). Each further p-power adds one by
the exact power law already proved in Lesson 9; if a power is one its
infinite valuation satisfies the same bound. After \(j\) further steps
we obtain (10.146), because \(g_i^P=\gamma_i\). Finally \(P\) is prime
to \(q\), so all the root-field and phase arguments above apply.
Proposition 9.4 gives \(v_p(\log\gamma_i)=v_p(\gamma_i-1)\), including
the infinite value when \(\gamma_i=1\); hence these logarithmic slopes
have the stated depth. \(\square\)

\subsubsection{Filling the missing integer
nodes}\label{filling-the-missing-integer-nodes}

Theorem 10.51 gives the new zeros at nodes prime to \(q\). We can now
prove a precise condition for filling the remaining nodes, while keeping
the height comparison over \(K\).

\textbf{Corollary 10.54 (phase-sensitive q-deleted-node closure).} Take
the new family (10.141), with \(\gamma_i=\exp(u_i)\), and retain the
slope, projection and selected integer Euler-basis hypotheses of
Corollary 10.40 at stage \(I+1\). In particular
\(v_p(w_i)>\theta+1/(p-1)\), \(v_p(\mathcal L)\ge U\), and
\(U-v_p(b_n)>\theta+1/(p-1)\). Let \(R\ge1,\mu\ge1,T\ge0\) be integers
with \(T'=T-\mu+1\ge0\). Suppose that \(\Psi'(s;\boldsymbol t)=0\) for
every \(s\in[-R,R]\cap\mathbb Z\) prime to \(q\) and every
\(|\boldsymbol t|\le T\). Put \(n=2R-2\lfloor R/q\rfloor\),
\(B=\lfloor\log_p(2R)\rfloor\), \(C_0=\max\{v_p(v(k)),v_p(b_n)\}\),
\(M_0=\max\{B,C_0\}\), and \(D=kL\). For \(X\ge0\), define the
nonzero-value budget

\[
\begin{aligned}
\mathcal A_{\mathrm{ph}}(X,T')=
\max_{\substack{|\boldsymbol t|\le T'\\t_0\le D}}
\frac{[K:\mathbb Q]}{ef\ln p}\biggl(&
\min\{h_\infty(c')+\ln N',h_2(c')+\tfrac12\ln N'\}\\
&+\ln\mathcal H_{I+1}(X;\boldsymbol t)
+\Xi_{I+1,0,t_0}\ln q
+2PX\sum_iA'_ih(\theta_i)\biggr).
\end{aligned}
\tag{10.147}
\]

If

\[
\begin{aligned}
U+D\theta+Lv_p(k!)&\ge n\mu\theta+2\mu M_0,\\
(n\mu-D)\theta-Lv_p(k!)&>\mathcal A_{\mathrm{ph}}(X,T'),
\end{aligned}
\tag{10.148}
\]

then \(\Psi'(x;\boldsymbol t)=0\) for every integer \(|x|\le X\) and
every \(|\boldsymbol t|\le T'\). In particular \(X=R\) fills all the
previously omitted integer nodes in the interval. With the larger
normalizing scale, the alternative sufficient pair is

\[
\begin{aligned}
U+D\left(\theta+\frac1{p-1}\right)&\ge n\mu\theta+2\mu M_0,\\
(n\mu-D)\theta-\frac D{p-1}&>\mathcal A_{\mathrm{ph}}(X,T').
\end{aligned}
\tag{10.149}
\]

If \(q\mid R\), the node count is exactly \(n=2(1-1/q)R\). For a desired
order \(0\le O\le T\), the choice \(\mu=T-O+1\) makes the output order
exactly \(T'=O\).

\textbf{Proof.} The q-deleted input set has the stated exact count and
separation cost \(\kappa=1\), by Corollary 10.35. The complete analytic
proof of Corollary 10.40 at the new stage applies to the local torus
coordinates \(\gamma_i\). Its normality, prepared integrality, ordinary
input jets and projection estimates use those local units and their
logarithmic slopes; they do not require each \(\gamma_i\) to lie in
\(K\). The first line of (10.148) is its exact sufficient precision
budget, with \(\kappa+1=2\). Consequently \[
v_p(\Psi'(x;\boldsymbol t))-\delta
\ge(n\mu-D)\theta-Lv_p(k!)
\] at every stated integer and derivative index.

The support congruence in Theorem 10.51 does put each of these integer
values in \(K\). If a value were nonzero, Corollary 10.52 would bound
its valuation excess above by the corresponding summand of (10.147), and
hence by \(\mathcal A_{\mathrm{ph}}\). The second line of (10.148) is a
strict contradiction. An index with \(t_0>D\) is already zero. This
proves the full conclusion. Using the larger scale and its proved lower
bound (10.114) instead proves (10.149); the two scales have their
respective conclusions. Finally \(q\mid R\) gives
\(\lfloor R/q\rfloor=R/q\), and the asserted order choice is the
identity \(T-(T-O+1)+1=O\). \(\square\)

The phase and support passage, the local phase construction and the
conditional closure of the missing nodes now have complete proofs. A
numerical Yu estimate still requires a verified coefficient construction
and a check of the slope, order and strict height comparisons at every
stage with its actual parameters.

\subsection{21. Constructing coefficients for the phased
values}\label{constructing-coefficients-for-the-phased-values}

The initial coefficient construction can also be performed over \(K\)
when the separate phased torus bases lie outside \(K\). What matters is
the field of each normalized row entry. The support congruence
determines that field and removes the root-of-unity phase from every
absolute-value estimate.

\textbf{Theorem 10.55 (an initial kernel over the original field).} Let
\(K\) have degree \(d\) and discriminant \(\Delta_K\). Take nonzero
\(\theta_i\in K\), a root of unity \(\alpha_0\in K\), an integer
\(G_0\ge1\), and a root of unity \(\zeta\) with
\(\zeta^{G_0}=\alpha_0\). Let \(P\in\mathbb Z_{\ge1}\),
\(d_i\in\mathbb Z\), and set \(\gamma_i=\theta_i^P\zeta^{d_i}\). The
separate \(\gamma_i\) need not lie in \(K\).

Take integers \(r,k,L\ge1\), \(S,T\ge0\), and a nonempty finite set
\(\Lambda\subset\mathbb Z^r\) of distinct vectors satisfying
\(\boldsymbol d\cdot\lambda\equiv\varepsilon\pmod{G_0}\), where
\(\boldsymbol d=(d_1,\ldots,d_r)\). Fix \(\lambda_0\in\Lambda\), put
\(A_i=\max_{\lambda\in\Lambda}|\lambda_i-\lambda_{0i}|\), and take
integer Euler forms \(\omega_j(\lambda)=C_j(\lambda-\lambda_0)\),
\(1\le j<r\). Write \(D=kL\), \(N=kL|\Lambda|\), and use
\(\mathcal J_T,J_T,M_*=(2S+1)J_T\) from (10.126), with these new Euler
arguments. Define \(H_{\mathrm{ph}}=P\sum_iA_i h(\theta_i)\).

If \(N>M_*\), there is a nonzero vector \(c\in\mathcal O_K^N\) for which

\[
\begin{aligned}
Q(X,Y)&=\sum_{\lambda,a,\ell}c_{\lambda,a,\ell}
\Delta(X+a;k)^\ell Y^{\lambda-\lambda_0},\\
\Psi_0(s;\boldsymbol t)&=\sum_{\lambda,a,\ell}c_{\lambda,a,\ell}
v(k)^{t_0}\Theta(s+a;k,\ell,t_0)
\prod_{j<r}\Delta(\omega_j(\lambda);t_j)
\gamma^{s(\lambda-\lambda_0)}
=0
\end{aligned}
\tag{10.150}
\]

for every integer \(|s|\le S\) and every \(|\boldsymbol t|\le T\), with
\(0\le a<k\), \(1\le\ell\le L\). The Laurent polynomial \(Q\) is
nonzero. Every normalized monomial in these equations belongs to \(K\),
since

\[
v_\lambda=\frac{\boldsymbol d\cdot(\lambda-\lambda_0)}{G_0}\in\mathbb Z,
\qquad
\gamma^{s(\lambda-\lambda_0)}
=\alpha_0^{s v_\lambda}
\prod_i\theta_i^{Ps(\lambda_i-\lambda_{0i})}\in K.
\tag{10.151}
\]

The vector may be chosen with the separate-row budget

\[
\begin{aligned}
h_\infty(c)\le h_2(c)\le&
\tfrac12\ln N+\frac{\ln|\Delta_K|}{2d}\\
&+\frac1{N-M_*}
\sum_{s=-S}^S\sum_{\boldsymbol t\in\mathcal J_T}
\left(\tfrac12\ln N+
\ln\mathcal H_0(|s|;\boldsymbol t)+2|s|H_{\mathrm{ph}}\right),
\end{aligned}
\tag{10.152}
\]

and hence with the mean-node budget

\[
\begin{aligned}
h_2(c)\le&
\tfrac12\ln N+\frac{\ln|\Delta_K|}{2d}\\
&+\frac{M_*}{N-M_*}
\left(\tfrac12\ln N+\ln\mathcal B_0^*(S,T)
+\frac{2S(S+1)}{2S+1}H_{\mathrm{ph}}\right).
\end{aligned}
\tag{10.153}
\]

These estimates involve \(K\) and the original \(\theta_i\), even when
the individual phases require a larger global field.

\textbf{Proof.} Subtracting the two support congruences gives the
integer \(v_\lambda\). Raising \(\zeta^{G_0}=\alpha_0\) to
\(s v_\lambda\) proves (10.151), for negative \(s\) and negative
exponent differences as well. This identity puts every matrix entry of
(10.150) in \(K\).

The prepared scalar in each entry is an integer: Lemma 10.28 proves the
additive assertion at every integer \(s+a\), and each rising Euler
binomial at an integer is an integer. An index with \(t_0>D\) has zero
row. If \(\omega_j\) vanishes throughout the support, its factor is
\(\Delta(0;t_j)=1\); setting \(t_j=0\) gives the same row and a no
larger total order. Thus imposing the representative rows imposes every
stated equation. Counting them exactly as in (10.126) gives \(M_*\) rows
and \(N\) columns, including any further zero or dependent rows.

We bound each representative row over \(K\). At an archimedean embedding
its Euclidean norm is at most \[
\sqrt N\,\mathcal H_0(|s|;\boldsymbol t)
\prod_i\max\{|\sigma(\theta_i)|,|\sigma(\theta_i)|^{-1}\}^{P|s|A_i}.
\] At a finite place its maximum norm is at most the analogous product
without the first two factors. Indeed the scalar is an integer, while
the root of unity \(\alpha_0^{s v_\lambda}\) in (10.151) has absolute
value one at every place. Theorem 10.44 supplies the scalar bound. The
product formula, with the height conventions of Section 14, gives the
exact identity \[
\frac1d\sum_v\ln\max\{|\theta_i|_v,|\theta_i|_v^{-1}\}
=2h(\theta_i).
\] Consequently the logarithm of the row-norm bound is
\(\tfrac12\ln N+\ln\mathcal H_0(|s|;\boldsymbol t)+2|s|H_{\mathrm{ph}}\).
This bound is nonnegative, since its scalar majorant is at least one by
Theorem 10.44. A zero row has the allowed row factor one.

Apply the complete weighted Siegel lemma (8.40) from
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/TR-BAKER-08.html\#small-solutions-over-the-coefficient-field}{Small
solutions over the coefficient field}, with all weights one and with
these actual \(K\)-rows. Its finite-place conditions give
\(c\in\mathcal O_K^N\setminus\{0\}\). Taking logarithms of its bound
proves (10.152); no discriminant or degree of a field containing the
individual \(\gamma_i\) enters. The monomials \(Y^{\lambda-\lambda_0}\)
are distinct, and Lemma 10.29 gives the independence of the additive
polynomials within each monomial block. Hence the nonzero coefficient
vector gives \(Q\ne0\).

Finally \(\mathcal H_0(|s|;\boldsymbol t)\le\mathcal B_0^*(S,T)\), and
\(\sum_{s=-S}^S|s|=S(S+1)\). Summing the separate bounds proves
(10.153). \(\square\)

Multiplying \(Q\) by a monomial clears its negative exponents. The
complete prepared-to-ordinary jet argument of Proposition 10.31 applies
when the \(\gamma_i\) are nonzero elements of the chosen local field and
the Euler directions are the selected integral basis. Each monomial
factor is nonzero at the evaluation point, so the order of vanishing is
preserved. A coefficient of minimum local valuation can also be divided
out as in (10.97), preserving the heights and equations; this last
normalization is locally integral and need not remain globally integral.

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={Four phased columns give a rational three-row kernel}]{../figures/phase-kernel-matrix.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.9. In \(\mathbb Q_5\), choose the root \(\zeta^2=-1\)
with \(\zeta\equiv2\pmod5\), and put \(P=125\), \(G_0=2\),
\(\alpha_0=-1\), \(\theta=(2,3)\), \(d=(-P,-3P)\). Both
\(\gamma_1=2^P\zeta^{-P}\) and \(\gamma_2=3^P\zeta^{-3P}\) are outside
\(\mathbb Q\), but \(\gamma_1\gamma_2=6^P=b\in\mathbb Q\). For support
\(\Lambda=\{(j,j):0\le j\le3\}\), reference \((0,0)\), \(k=L=1\),
\(S=1\), \(T=0\), the normalized torus column is \(b^{sj}\). The drawn
matrix includes the additive factor \(\Delta(s;1)=s+1\), so its first
row is zero. With \(C=b^2+b+1\), its integer vector \((-b,C,-C,b)\) is a
kernel vector. Equations (10.150)--(10.153) prove the general field and
height mechanism; Solution 28 checks this exact example and its extra
zero. The plotted support points and matrix entries are exact.
Human-source context: the field reduction and initial kernel in Lemma
4.2 of Yu's free 2013 paper, cited below.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/phase_kernel_matrix.py}{Figure program}.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{22. The field and budget at the first fractional
nodes}\label{the-field-and-budget-at-the-first-fractional-nodes}

At a fractional node the value need not be in \(K\), even if its chosen
completion is still \(K_{\mathfrak p}\). We now determine the smaller
field generated by the support, and then prove the arithmetic comparison
needed before the next coset extraction.

\subsubsection{The support-generated root
field}\label{the-support-generated-root-field}

\textbf{Theorem 10.56 (the field of the normalized fractional
monomials).} Retain the saturated Kummer and phase data of Section 20:
in particular \(\mu_q\subset K\), \(q\ne p\), \(q\nmid P\), and
\([K(\beta,\rho_1,\ldots,\rho_r):K]=q^{r+1}\), with
\(\beta^q=\alpha_0\), \(\rho_i^q=\theta_i\). Suppose
\(\gamma_i\in K_{\mathfrak p}\) have \(v_p(\gamma_i-1)>1/(p-1)\), and
use the compatible principal roots
\(\eta_i=\exp(q^{-1}\log\gamma_i)=\rho_i^P\xi^{d_i}\) of Corollary
10.52.

Let \(\Lambda\) be a nonempty finite support with
\(\boldsymbol d\cdot\lambda\equiv\varepsilon\pmod{G_0}\), and fix
\(\lambda_0\in\Lambda\). Define

\[
\begin{aligned}
m_\lambda&=\lambda-\lambda_0,&
w_\lambda&=\frac{\boldsymbol d\cdot m_\lambda}{G_0}\in\mathbb Z,\\
g&=\gcd(G_0,d_1,\ldots,d_r),&
G_*&=G_0/g,\qquad d_{*i}=d_i/g,\\
v_\lambda&=(w_\lambda,Pm_{\lambda1},\ldots,Pm_{\lambda r})
\in\mathbb F_q^{r+1},&
h_\Lambda&=\dim_{\mathbb F_q}\langle v_\lambda:\lambda\in\Lambda\rangle,\\
U_\lambda&=\eta^{m_\lambda},&
E_\Lambda&=K(U_\lambda:\lambda\in\Lambda).
\end{aligned}
\tag{10.154}
\]

Then

\[
P G_*(v_\lambda)_0-\sum_i d_{*i}(v_\lambda)_i=0,\qquad
h_\Lambda\le r,\qquad
[E_\Lambda:K]=q^{h_\Lambda},\qquad
(E_\Lambda)_{\mathfrak P}=K_{\mathfrak p}
\tag{10.155}
\]

for the prime \(\mathfrak P\) selected by these principal roots. The
first relation is a nonzero linear equation over \(\mathbb F_q\). In
particular the global degree can increase while the chosen local degree
stays one.

\textbf{Proof.} Cancelling the positive gcd gives
\(G_*w_\lambda=\sum_i d_{*i}m_{\lambda i}\), and
\(\gcd(G_*,d_{*1},\ldots,d_{*r})=1\). Multiply by \(P\) and reduce
modulo \(q\) to obtain the relation in (10.155). If \(q\nmid G_*\), its
first coefficient is nonzero, since \(q\nmid P\). If \(q\mid G_*\), at
least one \(d_{*i}\) is nonzero modulo \(q\). Thus it defines a
hyperplane of dimension \(r\), proving \(h_\Lambda\le r\). Cancelling
the gcd is essential: the unreduced equation can otherwise be the
identity \(0=0\).

The phase congruence and the compatible roots give \[
U_\lambda=\beta^{w_\lambda}\prod_i\rho_i^{Pm_{\lambda i}},
\qquad
U_\lambda^q=\alpha_0^{w_\lambda}\prod_i\theta_i^{Pm_{\lambda i}}\in K.
\] Choose \(\lambda_1,\ldots,\lambda_h\), \(h=h_\Lambda\), whose residue
vectors form a basis of the support span. Express any \(v_\lambda\) as
\(\sum_j a_jv_{\lambda_j}\), with \(0\le a_j<q\). The corresponding
integer exponent vectors differ by \(q(z_0,z_1,\ldots,z_r)\). The
displayed exact monomial identity therefore gives \[
U_\lambda=\alpha_0^{z_0}\prod_i\theta_i^{z_i}
\prod_{j=1}^h U_{\lambda_j}^{a_j}.
\] Consequently the \(h\) selected roots generate \(E_\Lambda\), and
their q-th powers lie in \(K\), so its degree is at most \(q^h\).

For the reverse bound, the monomials \(\prod_jU_{\lambda_j}^{a_j}\),
\(0\le a_j<q\), have distinct residue vectors
\(\sum_j a_jv_{\lambda_j}\). Reduce their exponents in the full root
field \(K(\beta,\rho_1,\ldots,\rho_r)\). Its degree is \(q^{r+1}\), so
the root monomials with exponents in \(\{0,\ldots,q-1\}^{r+1}\) are a
basis by Theorem 10.47. Our \(q^h\) monomials are nonzero
\(K\)-multiples of distinct elements of that basis, and are therefore
independent. This proves the exact degree, including \(h=0\).

Each \(\eta_i\) lies in \(K_{\mathfrak p}\) by the exponential proof in
Lesson 9, hence \(E_\Lambda\subset K_{\mathfrak p}\). Its closure
contains the closure \(K_{\mathfrak p}\) of \(K\), and is contained in
\(K_{\mathfrak p}\). Thus its chosen completion is precisely
\(K_{\mathfrak p}\). The absolute ramification index and residue degree
at \(\mathfrak P\) are the original \(e,f\). \(\square\)

\subsubsection{A nonzero fractional prepared
value}\label{a-nonzero-fractional-prepared-value}

\textbf{Corollary 10.57 (the phase-sensitive fractional arithmetic
budget).} Take the support of Theorem 10.56, integers \(I\ge0,k,L\ge1\),
a nonzero coefficient vector \(c\in K^N\), \(N=kL|\Lambda|\), and
integer Euler arguments \(C_j(m_\lambda)\). Put \(D=kL\),
\(A_i=\max_\lambda|m_{\lambda i}|\), \(\delta=\min_jv_p(c_j)\), and take
\(x=s/q\), \(s\in\mathbb Z\), with \(|x|\le X\). Let
\(V=\Psi_I(s/q;\boldsymbol t)\) be the prepared value whose torus term
is \(U_\lambda^s\) and whose additive argument is \(q^{-I}x+a\). If
\(t_0>D\), it is zero. Otherwise it belongs to \(E_\Lambda\), and if it
is nonzero,

\[
\begin{aligned}
v_p(V)-\delta\le \mathcal A_{\mathrm{frac},\boldsymbol t}(X)
:=\frac{q^{h_\Lambda}[K:\mathbb Q]}{ef\ln p}\biggl(&
\min\{h_\infty(c)+\ln N,h_2(c)+\tfrac12\ln N\}\\
&+\ln\mathcal H_I(X;\boldsymbol t)
+\Xi_{I,1,t_0}\ln q
+2PX\sum_iA_i h(\theta_i)\biggr).
\end{aligned}
\tag{10.156}
\]

Here \(\mathcal H_I\) and \(\Xi_{I,1,t_0}\) are the already proved
scalar majorant and q-clearing exponent of (10.123) and (10.115).

\textbf{Proof.} Every torus term lies in \(E_\Lambda\), and every
prepared scalar is rational, so \(V\in E_\Lambda\). Lemma 10.28 and
Theorem 10.41 clear each scalar's denominator by \(q^{\Xi_{I,1,t_0}}\),
with no denominator at any other prime; Theorem 10.44 bounds its
ordinary absolute value by \(\mathcal H_I(X;\boldsymbol t)\).

At every place of \(E_\Lambda\), taking the absolute value of
\(U_\lambda^q\) removes the root-of-unity phase. Its signed torus
contribution is bounded by \[
\sum_i \frac{P|s|A_i}{q}
\ln\max\{|\theta_i|_v,|\theta_i|_v^{-1}\}.
\] The torus terms are units at \(\mathfrak P\). Summing this
contribution outside that prime is therefore at most
\(2[E_\Lambda:\mathbb Q]PX\sum_iA_i h(\theta_i)\), by the same
inverse-height identity as in Theorem 10.41. The coefficient heights
retain their values over \(K\): each archimedean embedding is repeated
\([E_\Lambda:K]\) times, and the local degrees over a finite place sum
to that same factor.

Now repeat the full coefficient-normalized product-formula argument of
Theorem 10.41 over \(E_\Lambda\), retaining its Euclidean and maximum
coefficient bounds and the new scalar majorant. Theorem 10.56 gives
\([E_\Lambda:\mathbb Q]=q^{h_\Lambda}[K:\mathbb Q]\) and the unchanged
\(e,f\) at the selected prime. These are exactly the factor and all
terms in (10.156). \(\square\)

The unchanged completion does not remove \(q^{h_\Lambda}\). Nor does a
high valuation at the principal-root prime give that valuation at the
other primes of the global field.

\subsubsection{Closing the first fractional-node
extension}\label{closing-the-first-fractional-node-extension}

\textbf{Corollary 10.58 (a proved first fractional extension).} Retain
the phase family of Corollary 10.57 and all selected-basis, slope,
normality and projection hypotheses of Corollary 10.40 for its local
curve \(\gamma^x=\exp(x\log\gamma)\). Suppose its prepared values vanish
at all integers \(|s|\le R\) through total order \(T\), where
\(R\ge1,T\ge0\) are integers. Choose an integer \(\mu\ge1\) with
\(T'=T-\mu+1\ge0\), and put \(n=2R+1\), \(B=\lfloor\log_p(2R)\rfloor\),
\(C_0=\max\{v_p(v(k)),v_p(b_n)\}\), \(M_0=\max\{B,C_0\}\). Let
\(\mathcal A_{\mathrm{frac}}(X,T')=
\max_{|\boldsymbol t|\le T',\,t_0\le D}
\mathcal A_{\mathrm{frac},\boldsymbol t}(X)\), using (10.156). If

\[
\begin{aligned}
U+D\theta+Lv_p(k!)&\ge n\mu\theta+\mu M_0,\\
(n\mu-D)\theta-Lv_p(k!)&>
\mathcal A_{\mathrm{frac}}(X,T'),
\end{aligned}
\tag{10.157}
\]

then every first fractional prepared value \(\Psi_I(s/q;\boldsymbol t)\)
is zero for integers \(|s|\le qX\) and \(|\boldsymbol t|\le T'\). With
the larger normalizing scale, an alternative sufficient pair is

\[
\begin{aligned}
U+D\left(\theta+\frac1{p-1}\right)&\ge n\mu\theta+\mu M_0,\\
(n\mu-D)\theta-\frac D{p-1}&>
\mathcal A_{\mathrm{frac}}(X,T').
\end{aligned}
\tag{10.158}
\]

\textbf{Proof.} The full integer input interval has separation cost
\(\kappa=0\), as proved in Corollary 10.35. The first inequality of
(10.157) is consequently the sufficient analytic budget (10.113). Its
complete proof, retaining the loss from the selected integer Euler
basis, gives \[
v_p(\Psi_I(x;\boldsymbol t))-\delta
\ge(n\mu-D)\theta-Lv_p(k!)
\] for every \(x\in\mathbb Z_p\) and each output index. The local curve
satisfies the same proof when its separate torus bases lie outside
\(K\), since that argument uses their local logarithms, prepared scalars
and projection data.

For \(x=s/q\), the assumption \(q\ne p\) gives \(x\in\mathbb Z_p\).
Proposition 9.4 identifies the local exponential torus factors with the
principal roots \(\eta^s\), so this analytic value is exactly the
algebraic value of Corollary 10.57. If it were nonzero, (10.156) would
give the opposite upper bound \(\mathcal A_{\mathrm{frac}}(X,T')\),
contradicting the strict second inequality. Indices with \(t_0>D\) are
already zero. This proves every asserted fractional zero. Repeating the
same comparison with the proved larger-scale conclusion (10.114) gives
(10.158). \(\square\)

These hypotheses separate the two tasks in a numerical descent:
constructing the integer input zeros and verifying the precision and
strict height comparisons. Once the fractional zeros are available at
nodes prime to \(q\), Theorem 10.51 performs the exact phase and support
passage.

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={A smaller fractional root field with unequal valuations at its two selected embeddings}]{../figures/fractional-phase-field.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.10. The exact example has
\(K=\mathbb Q,p=5,q=2,\alpha_0=-1,\theta_1=6,P=1,G_0=2,d_1=0,\Lambda=\{0,1\}\).
The cancelled relation is \(w=0\), and the support vectors span
\((0,1)\) over \(\mathbb F_2\), so \(h_\Lambda=1\). The support field
\(E=\mathbb Q(\sqrt6)\) has degree two; the full root field
\(F=\mathbb Q(i,\sqrt6)\) has degree four. Choose the completion sending
\(\sqrt6\) to \(\eta\equiv1\pmod5\). Both these roots and
\(i\equiv2\pmod5\) lie in \(\mathbb Q_5\), so all three chosen
completions in the diagram are \(\mathbb Q_5\). Their global-to-local
degree factors are \(1,2,4\). The value \(V=\tfrac32(1-\sqrt6)\) has
valuation one at this root and zero at the other root \(-\eta\); the two
bars display these exact valuations. Theorem 10.56 and
(10.154)--(10.155) prove the field mechanism, Corollary 10.57 proves its
arithmetic budget, and Solution 29 verifies every number in this
example. Human-source context: the first fractional step and
product-formula field discussion in Yu's free 2013 paper, printed
pp.~352 and 365--366, cited below.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/fractional_phase_field.py}{Figure program}.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{23. Heights of the actual exponent
support}\label{heights-of-the-actual-exponent-support}

The torus contribution can be measured before enclosing the exponents in
a coordinate box. This keeps the geometry of a support through a change
of multiplicative bases and through division by \(q\). Positive and
negative nodes require separate heights unless the support is symmetric.

\subsubsection{A projective height from a support
function}\label{a-projective-height-from-a-support-function}

Let \(K\) have degree \(d\), and use the absolute values and place
multiplicities of Section 14. Thus \(\sum_v\ln|z|_v=0\) for
\(z\in K^\times\); a complex place is counted with multiplicity two.
Take \(\theta=(\theta_1,\ldots,\theta_r)\in(K^\times)^r\), and put
\(\ell_v=(\ln|\theta_1|_v,\ldots,\ln|\theta_r|_v)\). Only finitely many
finite-place vectors \(\ell_v\) are nonzero. For a nonempty compact set
\(C\subset\mathbb R^r\), and a nonempty finite set
\(M\subset\mathbb Z^r\), define

\[
\begin{aligned}
\mathfrak H_\theta(C)&=\frac1d\sum_v\sup_{u\in C}u\cdot\ell_v,\\
\mathfrak H_\theta(M)&=\mathfrak H_\theta(\operatorname{conv}M)
=h_\infty((\theta^m)_{m\in M}).
\end{aligned}
\tag{10.159}
\]

The finite-set equality, and all the properties needed below, are proved
next.

\textbf{Theorem 10.59 (the support height).} The quantities in (10.159)
are finite and nonnegative. For a real vector \(a\), a real \(t\ge0\),
and an integer \(s\),

\[
\begin{aligned}
\mathfrak H_\theta(C+a)&=\mathfrak H_\theta(C),&
\mathfrak H_\theta(tC)&=t\mathfrak H_\theta(C),\\
C\subset C'&\Longrightarrow\mathfrak H_\theta(C)\le\mathfrak H_\theta(C'),&
h_\infty((\theta^{sm})_{m\in M})
&=\begin{cases}
s\mathfrak H_\theta(M),&s\ge0,\\
|s|\mathfrak H_\theta(-M),&s<0.
\end{cases}
\end{aligned}
\tag{10.160}
\]

If \(|m_i|\le A_i\) for every \(m\in M\), then

\[
\mathfrak H_\theta(M),\ \mathfrak H_\theta(-M)
\le2\sum_i A_i h(\theta_i).
\tag{10.161}
\]

These heights are unchanged by extending \(K\). Root-of-unity factors
attached separately to the monomial coordinates change neither height.

\textbf{Proof.} The support function of a finite convex hull is the
maximum over its original points: a convex combination cannot exceed
their maximum, and those points belong to the hull. Since
\(\ln|\theta^m|_v=m\cdot\ell_v\), its normalized sum is exactly the
maximum projective height in (10.159). All sums are finite. For a
compact \(C\), translating by \(a\) adds \(a\cdot\ell_v\) at each place.
The product formula for each \(\theta_i\) makes the sum of these
additions zero. Translating by the negative of any point of \(C\) places
zero in the set; every resulting support function is then nonnegative.
This proves nonnegativity.

Nonnegative scaling multiplies each support function by \(t\), including
\(t=0\); inclusion increases it. Taking \(s\ge0\) in the monomial
identity multiplies the local maximum by \(s\), while \(s<0\) replaces
\(M\) by \(-M\). This proves (10.160). At any place, the box bounds both
support functions by \(\sum_i A_i|\ln|\theta_i|_v|\). The sum of the
positive logarithms for one element is \(d h(\theta_i)\), and the
product formula says the sum of the negative logarithms has the same
absolute value. Their total is \(2d h(\theta_i)\), proving (10.161).

For completeness, an extension \(E/K\) repeats each archimedean
embedding \([E:K]\) times. At a finite place \(v\), the logarithm of the
norm absolute value at a place \(w\) of \(E\) restricting to \(v\) is
multiplied by \([E_w:K_v]\). The local degrees sum to \([E:K]\), by the
places identities proved in \href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/TR-BAKER-02.html}{Absolute values
and places}. The support function is homogeneous for these positive
multipliers. Thus its unnormalized place sum is multiplied by \([E:K]\);
dividing by \([E:\mathbb Q]=[E:K]d\) preserves its value. Every root of
unity has absolute value one at every place, by its finite-order
equation, so its coordinate factors make no contribution. \(\square\)

Symmetry is a sufficient reason for the two signs to agree; it is not
automatic. Over \(\mathbb Q\), for \(\theta=(2,3)\) and
\(M=\{(0,0),(1,0),(0,1)\}\), the monomial vector is \((1,2,3)\), of
height \(\ln3\). The inverse vector \((1,1/2,1/3)\) becomes the
primitive integer vector \((6,3,2)\), of height \(\ln6\). These are
\(\mathfrak H_\theta(M)\) and \(\mathfrak H_\theta(-M)\).

\subsubsection{Contraction and the original multiplicative
bases}\label{contraction-and-the-original-multiplicative-bases}

\textbf{Theorem 10.60 (a height envelope for the divided support).} Let
\(C\subset\mathbb R^r\) be nonempty and compact. If a finite nonempty
support \(M_I\) satisfies \(M_I\subset q^{-I}C-x_I\), then

\[
\mathfrak H_\theta(M_I)\le q^{-I}\mathfrak H_\theta(C),\qquad
\mathfrak H_\theta(-M_I)\le q^{-I}\mathfrak H_\theta(-C).
\tag{10.162}
\]

The support passage
\(M_{I+1}=\{(m-m_*)/q:m\text{ in a retained coset of }M_I\}\) retains
this form of containment. A further restriction of the support and any
choice of reference point retain the bound.

Suppose also that \(\alpha_1,\ldots,\alpha_n\in K^\times\), that
\(B=(b_{ji})\) is a real \(n\)-by-\(r\) matrix of rank \(r\), and that a
positive integer \(\tau\) satisfies \(\tau b_{ji}\in\mathbb Z\) and
\(\theta_i^\tau=\prod_j\alpha_j^{\tau b_{ji}}\). For finite \(D_j\ge0\),
put \(C_B=\{u: |(Bu)_j|\le D_j,\ 1\le j\le n\}\). Then \(C_B\) is
compact and centrally symmetric, and

\[
\mathfrak H_\theta(C_B)=\mathfrak H_\theta(-C_B)
\le2\sum_{j=1}^nD_j h(\alpha_j),\qquad
\mathfrak H_\theta(\pm M_I)
\le2q^{-I}\sum_jD_jh(\alpha_j)
\quad(M_I\subset q^{-I}C_B-x_I).
\tag{10.163}
\]

\textbf{Proof.} Translation, scaling and inclusion in Theorem 10.59 give
(10.162). The next retained support is contained in
\(q^{-(I+1)}C-(x_I+m_*)/q\), so induction preserves exactly this
envelope. Restriction uses inclusion; changing reference uses
translation.

Choose an invertible \(r\)-by-\(r\) row minor of \(B\). Its inverse
bounds each coordinate of \(u\in C_B\) by a fixed finite linear
combination of the \(D_j\). Thus \(C_B\) is bounded; its defining
inequalities make it closed. To see directly that its support maximum is
attained, take a sequence approaching the supremum of a linear function.
Successive bisection of a bounded coordinate interval chooses an
infinite subsequence in nested closed intervals of lengths tending to
zero. Repeating this for the finitely many coordinates gives a
convergent subsequence. Its limit lies in the closed set, and the finite
linear sum tends to its value there. This also proves the required
compactness in finite dimension.

Taking absolute values of the \(\tau\)-power identities and cancelling
\(\tau\) gives \(\ell_v=B^{\mathsf T}(\ln|\alpha_j|_v)_j\). Consequently
\[
u\cdot\ell_v=\sum_j(Bu)_j\ln|\alpha_j|_v
\le\sum_jD_j|\ln|\alpha_j|_v|
\qquad(u\in C_B).
\] Sum this inequality and use the inverse-height identity proved in
Theorem 10.59. The set \(C_B\) is centrally symmetric because its
inequalities use absolute values. Combining the resulting bound with
(10.162) proves (10.163). \(\square\)

The same height bound holds for a nonempty affine box constraint
\(|(Bu)_j-z_j|\le D_j\), with arbitrary real \(z_j\). Indeed its local
support maximum is at most
\(\sum_j z_j\ln|\alpha_j|_v+\sum_jD_j|\ln|\alpha_j|_v|\). The first sum
disappears after summing over places, by the product formula, even if
\(z\) is outside the image of \(B\). Thus the bound uses the half-widths
in the original coordinates and follows the original torus products
through a change of basis. An enclosing box in the new coordinates is
unnecessary.

\subsubsection{The coefficient and nonzero-value
budgets}\label{the-coefficient-and-nonzero-value-budgets}

\textbf{Corollary 10.61 (initial coefficients with the support height).}
Retain every hypothesis and notation of Theorem 10.55, including
\(N>M_*\), and put \(M=\Lambda-\lambda_0\),
\(H_+=\mathfrak H_\theta(M)\), \(H_-=\mathfrak H_\theta(-M)\). Define
\(b(s)=PsH_+\) for \(s\ge0\), and \(b(s)=P|s|H_-\) for \(s<0\). Its
nonzero integral kernel exists with

\[
\begin{aligned}
h_\infty(c)\le h_2(c)\le&
\tfrac12\ln N+\frac{\ln|\Delta_K|}{2d}\\
&+\frac1{N-M_*}\sum_{s=-S}^S\sum_{\boldsymbol t\in\mathcal J_T}
\left(\tfrac12\ln N+\ln\mathcal H_0(|s|;\boldsymbol t)+b(s)\right),
\end{aligned}
\tag{10.164}
\]

and hence with

\[
\begin{aligned}
h_2(c)\le&\ \tfrac12\ln N+\frac{\ln|\Delta_K|}{2d}\\
&+\frac{M_*}{N-M_*}\left(
\tfrac12\ln N+\ln\mathcal B_0^*(S,T)
+\frac{PS(S+1)}{2(2S+1)}(H_++H_-)\right).
\end{aligned}
\tag{10.165}
\]

Both bounds are at least as strong as their respective bounds (10.152)
and (10.153).

\textbf{Proof.} The normalized torus monomials in a row are
\(\alpha_0^{s w_\lambda}\theta^{Ps m_\lambda}\). At an archimedean
embedding, their maximum times
\(\sqrt N\mathcal H_0(|s|;\boldsymbol t)\) bounds the row's Euclidean
norm. At a finite place their maximum bounds its maximum norm, since
every scalar is an integer. The phase has absolute value one. Summing
the logarithms of these maxima gives exactly \(b(s)\), by Theorem 10.59.
The support contains zero, so each local torus maximum is at least one;
the scalar majorant is also at least one, by Theorem 10.44. These are
valid row factors in the completely proved weighted Siegel lemma (8.40),
with a zero row allowed factor one. Its proof therefore gives (10.164)
over \(K\), exactly as in Theorem 10.55. The same row system, integral
vector and polynomial-independence argument impose every original zero.

The positive and negative node sums are both \(S(S+1)/2\). There are
\(J_T\) rows at each node and \(M_*=(2S+1)J_T\). Using the uniform
scalar bound gives (10.165). Finally (10.161) bounds each of \(H_+,H_-\)
by \(2\sum_iA_i h(\theta_i)\); substitution gives the original two torus
terms, without changing any other term. \(\square\)

\textbf{Corollary 10.62 (the support budget for a nonzero value).} In
the phase family of Theorem 10.55, suppose the \(\gamma_i\) are units at
the chosen prime, and consider an integer-node prepared value
\(V=\Psi_I(s;\boldsymbol t)\), with \(t_0\le D\). Put
\(C_c=\min\{h_\infty(c)+\ln N,h_2(c)+\tfrac12\ln N\}\) and
\(\delta=\min_jv_p(c_j)\). If \(V\ne0\), then

\[
v_p(V)-\delta\le\frac d{ef\ln p}
\left(C_c+\ln\mathcal H_I(|s|;\boldsymbol t)
+\Xi_{I,0,t_0}\ln q+b(s)\right).
\tag{10.166}
\]

In the first fractional family of Corollary 10.57, with its support
field \(E_\Lambda\), the nonzero value at \(x=s/q\) satisfies instead

\[
v_p(V)-\delta\le\frac{q^{h_\Lambda}d}{ef\ln p}
\left(C_c+\ln\mathcal H_I(|s|/q;\boldsymbol t)
+\Xi_{I,1,t_0}\ln q+\frac{b(s)}q\right).
\tag{10.167}
\]

Uniformly for \(|s|\le X\) in the integer case and \(|s|\le qX\) in the
fractional case, the torus term may be replaced by

\[
PX\max\{H_+,H_-\}.
\tag{10.168}
\]

The corresponding strict zero-forcing comparisons of Theorem 10.43 and
Corollaries 10.54 and 10.58 remain valid with these smaller budgets,
under all their original analytic and field hypotheses.

\textbf{Proof.} For the integer case the phase congruence puts every
torus monomial in \(K\). Repeat Theorem 10.41's product formula after
dividing by a coefficient of minimum local valuation. Its coefficient
contribution is \(C_c\); Theorem 10.44 supplies the individual scalar
bound, and (10.115) supplies its clearing exponent. At each place the
torus contribution is the maximum of the logarithms of
\(\theta^{Ps m_\lambda}\), with the phase removed. Its sum is
\(d b(s)\). Its omitted contribution at the chosen prime is zero: the
\(\gamma_i\) are units, and taking absolute values of
\(\gamma_i=\theta_i^P\zeta^{d_i}\) shows that the \(\theta_i\) are units
there. These local maxima are nonnegative because \(0\in M\). This
proves (10.166).

In the fractional case,
\(U_\lambda^q=\alpha_0^{w_\lambda}\theta^{Pm_\lambda}\). Taking local
logarithmic absolute values divides the preceding torus maximum by
\(q\), and its field-normalized sum is \(b(s)/q\), by the
field-invariance proof of Theorem 10.59. Theorem 10.56 still gives the
global degree \(q^{h_\Lambda}d\) and the original local \(e,f\); none of
these degree factors has been discarded. The same coefficient-normalized
product formula gives (10.167). The bounds on \(s\) give (10.168).

The analytic conclusions in the cited zero-forcing proofs supply their
unchanged lower bounds for the same valuations. A strict inequality
between that lower bound and the new arithmetic upper bound again
contradicts \(V\ne0\). Taking the maximum over the asserted output
indices proves all the same zeros. This changes only the torus height
estimate; every slope, depth, order and field condition remains a
hypothesis. \(\square\)

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={The exact exponent diamond and its larger coordinate box, with their local height contributions}]{../figures/support-height-envelope.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.11. Over \(\mathbb Q\), with \(\theta=(2,3)\), the
diamond \(C=\{(u_1,u_2):|u_1|+2|u_2|\le2\}\) has seven integer points;
its enclosing box \([-2,2]\times[-1,1]\) has fifteen. At the real,
2-adic and 3-adic places the diamond's support contributions are
\(\ln4,\ln4,\ln3\), attained respectively at \((2,0),(-2,0),(0,-1)\).
The box contributions are \(\ln12,\ln4,\ln3\). Their exact projective
heights are therefore \(\ln48\) and \(\ln144\). The bar lengths are
numerical renderings of these labelled exact logarithms. The polygon,
lattice points, active points and height identities are exact. Equations
(10.159)--(10.168) prove the general support, contraction and budget
mechanism; Solution 31 checks the primitive monomial vectors and a
divided-support example. Human-source context: the original-coordinate
torus identity and support containment (4.16), (5.1), and the
product-formula steps (5.34)--(5.36), (7.13) in Yu's free 2013 paper,
cited below. \href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/support_height_envelope.py}{Figure
program}.}
\end{flushleft}
\end{minipage}\par\medskip


The enclosing-box estimate is valid, but this example loses exactly
\(\ln3\) in its torus height. The support budget avoids that loss, and
(10.163) retains bounds in the original bases after a multiplicative
change of coordinates.

\subsection{24. The field and the numerical scale of the initial
kernel}\label{the-field-and-the-numerical-scale-of-the-initial-kernel}

The initial rows need not generate the entire ambient number field. We
can construct their coefficients over the field they actually generate.
Its discriminant is controlled by the heights of explicit normalized
torus monomials, so an unrelated extension of the ambient field does not
enter the coefficient budget.

\subsubsection{The exact field of the row
entries}\label{the-exact-field-of-the-row-entries}

Retain Theorem 10.55, and write \(M=\Lambda-\lambda_0\),
\(H_+=\mathfrak H_\theta(M)\), \(H_-=\mathfrak H_\theta(-M)\). The
normalized monomials and the field generated by all initial row entries
are

\[
\begin{aligned}
W_\lambda&=\alpha_0^{w_\lambda}\theta^{P(\lambda-\lambda_0)}
\in K,\qquad W_{\lambda_0}=1,\\
E_0&=\begin{cases}
\mathbb Q,&S=0,\\
\mathbb Q(W_\lambda:\lambda\in\Lambda),&S\ge1,
\end{cases}
\qquad d_0=[E_0:\mathbb Q]\le d.
\end{aligned}
\tag{10.169}
\]

\textbf{Theorem 10.63 (initial coefficients over their row field).} The
field in (10.169) is exactly the smallest subfield of \(K\) containing
every coefficient of the initial homogeneous system (10.150). If
\(N>M_*\), that system has a nonzero coefficient vector
\(c\in\mathcal O_{E_0}^N\subset\mathcal O_K^N\) giving the same nonzero
Laurent polynomial and every originally specified zero.

For \(S\ge1\), choose normalized monomials \(z_1,\ldots,z_a\) that
generate \(E_0\). Put \(F_j=\mathbb Q(z_1,\ldots,z_j)\),
\(F_0=\mathbb Q\), so \(F_a=E_0\), and let \(e_j=[F_j:F_{j-1}]\). Omit
steps of degree one, and define \(\Gamma=\sum_j(e_j-1)h(z_j)\). For
\(S=0\), use the empty tower and \(\Gamma=0\). Then

\[
\frac{\ln|\Delta_{E_0}|}{2d_0}
\le\frac12\ln d_0+\Gamma,\qquad
\Gamma\le(d_0-1)P\min\{H_+,H_-\}.
\tag{10.170}
\]

With \(b(s)\) as in Corollary 10.61, the coefficients can be chosen with

\[
\begin{aligned}
h_\infty(c)\le h_2(c)\le&
\frac12\ln(Nd_0)+\Gamma\\
&+\frac1{N-M_*}\sum_{s=-S}^S
\sum_{\boldsymbol t\in\mathcal J_T}
\left(\frac12\ln N+\ln\mathcal H_0(|s|;\boldsymbol t)+b(s)\right),
\end{aligned}
\tag{10.171}
\]

and therefore with

\[
\begin{aligned}
h_2(c)\le&\ \frac12\ln(Nd_0)+\Gamma\\
&+\frac{M_*}{N-M_*}
\left(\frac12\ln N+\ln\mathcal B_0^*(S,T)
+\frac{PS(S+1)}{2(2S+1)}(H_++H_-)\right).
\end{aligned}
\tag{10.172}
\]

The actual discriminant term \(\ln|\Delta_{E_0}|/(2d_0)\) may instead be
retained, replacing \(\frac12\ln d_0+\Gamma\) in either formula.

\textbf{Proof.} Every entry of the row at \(s\) is an integer prepared
scalar times \(W_\lambda^s\). If \(S=0\), all entries are rational
integers, so their field is \(\mathbb Q\). If \(S\ge1\), these
expressions first show that every entry belongs to the monomial field in
(10.169). Conversely take the row at \(s=1\) with derivative index zero.
Its column with \(a=0,\ell=1\) and any \(\lambda\) is
\(\Delta(1;k)W_\lambda=(k+1)W_\lambda\). Every Euler factor there is
\(\Delta(\omega;0)=1\). Division by the nonzero rational integer \(k+1\)
recovers \(W_\lambda\) from that row. Thus every field containing the
entries contains each monomial, proving minimality.

The
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/TR-BAKER-08.html\#the-discriminant-of-the-coefficient-field}{discriminant-and-generators
proof}, equation (8.33), gives the first bound in (10.170) for this
explicit tower, including nonintegral monomials. Each selected \(z_j\)
is a coordinate of the monomial vector containing \(1\). Its ordinary
height is at most that vector's maximum projective height \(PH_+\):
locally, the vector maximum bounds \(\max\{1,|z_j|_v\}\). The inverse
vector also contains \(1\), so it bounds \(h(z_j^{-1})=h(z_j)\) by
\(PH_-\). Consequently \(h(z_j)\le P\min\{H_+,H_-\}\). The integer
degrees have product \(d_0\); induction using \((u-1)+(v-1)\le uv-1\)
gives \(\sum_j(e_j-1)\le d_0-1\). This proves the second bound. The
empty-tower case has \(d_0=1,\Delta_{E_0}=1\).

Apply the completely proved weighted Siegel lemma (8.40) over \(E_0\) to
the actual representative rows. Their number \(M_*\), their integer
scalar bounds and all their dependencies are unchanged. Their normalized
torus row height is \(b(s)\), as in Corollary 10.61. This height can be
computed over \(K\), because the row vector lies in \(E_0\) and
projective heights are invariant under extension. The individual
\(\theta_i\) need not lie in \(E_0\). The Euclidean row bound adds
\(\frac12\ln N+\ln\mathcal H_0\) as before. The finite-place conditions
give \(c\in\mathcal O_{E_0}^N\), and such elements are also algebraic
integers in \(K\).

That lemma supplies \(\frac12\ln N+\ln|\Delta_{E_0}|/(2d_0)\), followed
by the displayed row sum divided by \(N-M_*\). Substitution of (10.170)
proves (10.171). Summing the positive and negative nodes separately
proves (10.172), exactly as in (10.165). A nonzero coefficient vector
still gives a nonzero polynomial by Lemma 10.29, and the same
representative-row argument preserves all jets in (10.150). This proves
every assertion. \(\square\)

\subsubsection{Full widths in the original
coordinates}\label{full-widths-in-the-original-coordinates}

The original torus widths can now be inserted before any numerical
constants are rounded. Let \(\alpha_1,\ldots,\alpha_r\in K^\times\), and
suppose the multiplicative change of basis in Theorem 10.60 satisfies,
for a positive integer \(\tau\),
\(\theta_i^\tau=\omega_i\prod_j\alpha_j^{\tau b_{ji}}\), where
\(\omega_i\) is a root of unity and every \(\tau b_{ji}\) is integral.
The root-of-unity factors contribute zero to every local logarithm, so
the proof of (10.163) still applies. Suppose the original support
coordinates \(B\Lambda\) lie in a translated box of full widths
\(D_1,\ldots,D_r\), and put \(L_\alpha=\sum_j D_j h(\alpha_j)\). After
choosing the reference point \(\lambda_0\), its original coordinates
\(BM\) satisfy affine box constraints with half-widths \(D_j/2\). The
affine-centre proof following (10.163) gives

\[
H_+,H_-\le L_\alpha,\qquad
\Gamma\le(d_0-1)P L_\alpha.
\tag{10.173}
\]

Here the box condition is imposed on \(Bm\); it does not require its
centre to be in the image of \(B\).

\textbf{Corollary 10.64 (an initial budget in the original heights).}
Under these hypotheses and \(N>M_*\), the initial coefficient vector may
be chosen with

\[
\begin{aligned}
h_2(c)\le&\ \frac12\ln(Nd_0)+(d_0-1)P L_\alpha\\
&+\frac{M_*}{N-M_*}
\left(\frac12\ln N+\ln\mathcal B_0^*(S,T)
+\frac{PS(S+1)}{2S+1}L_\alpha\right).
\end{aligned}
\tag{10.174}
\]

Replacing \(d_0\) everywhere by its upper bound \(d\) gives a valid
further bound.

\textbf{Proof.} Substitute both parts of (10.173) into (10.172). The
factor of two from \(H_++H_-\) cancels the two in its denominator. The
replacement by \(d\) increases \(\ln d_0\) and \(d_0-1\), while all
other terms stay unchanged and \(P L_\alpha\ge0\). \(\square\)

In particular a full width \(D_j\) is not a half-width \(D_j\). This
distinction accounts for a factor of two in the torus contribution.

\subsubsection{Substitution at the numerical
scale}\label{substitution-at-the-numerical-scale}

\textbf{Corollary 10.65 (the scaled coefficient inequality).} Suppose
\(S\ge1\), \(N\ge c_*M_*\) with \(c_*>1\), and that positive numbers
\(\mathcal D,c_1,c_2,\sigma_1,\ldots,\sigma_r\) satisfy

\[
h(\alpha_j)\le\sigma_j,\qquad
D_j=\frac{\mathcal D}{c_1c_2rPd\sigma_j},\qquad
Z=\frac{S\mathcal D}{d}.
\tag{10.175}
\]

Then a coefficient vector with every zero in Theorem 10.63 exists with

\[
\begin{aligned}
\frac{h_2(c)}Z\le&
\frac{d\ln(Nd_0)}{2S\mathcal D}
+\frac{d_0-1}{c_1c_2S}\\
&+\frac1{c_*-1}\left(
\frac{d(\frac12\ln N+\ln\mathcal B_0^*(S,T))}{S\mathcal D}
+\frac1{2c_1c_2}\left(1+\frac1{2S+1}\right)\right).
\end{aligned}
\tag{10.176}
\]

An explicit generator tower may improve the second term to \(\Gamma/Z\).
If \(S\ge g>0\), the last factor \(1+1/(2S+1)\) is at most
\(1+1/(2g+1)\).

\textbf{Proof.} Formula (10.175) gives \[
P L_\alpha=P\sum_jD_jh(\alpha_j)
\le P\sum_jD_j\sigma_j
=\frac{\mathcal D}{c_1c_2d}.
\] Also \(M_*/(N-M_*)\le1/(c_*-1)\): divide \(N\ge c_*M_*\) by \(M_*>0\)
and subtract one. Every term in the parentheses of (10.174) is
nonnegative. Divide that bound by \(Z\), insert these inequalities, and
use the exact identity \[
\frac{S+1}{2S+1}=\frac12\left(1+\frac1{2S+1}\right).
\] This proves (10.176), including its coefficient \(\frac1{2c_1c_2}\).
Retaining \(\Gamma\) in (10.172) proves the tower refinement. Finally
\(S\ge g\) increases the positive denominator \(2S+1\), giving the
asserted upper bound. \(\square\)

Equation (10.176) supplies a direct numerical comparison once the column
surplus, scalar majorant and coefficient-field tower have been
determined. It retains the full-width torus term without an
ambient-field discriminant. The later zero-forcing inequalities still
require their own strict comparisons.

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={A rational initial kernel and an algebraic initial kernel use their actual row fields}]{../figures/initial-coefficient-field.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.12. For \(k=L=S=1,T=0,\Lambda=\{0,1,2,3\},P=1\), the
initial rows are \(0\), \((1,1,1,1)\), \(2(1,t,t^2,t^3)\). The vector
\((-t,t^2+t+1,-t^2-t-1,t)\) is their exact integral kernel. Left:
\(t=6\), with ambient \(K=\mathbb Q(\sqrt5)\), gives row field
\(\mathbb Q\), vector \((-6,43,-43,6)\), and no discriminant cost. The
ambient-field cost would be \(\frac14\ln5\). Right: \(t=\sqrt6\)
generates row field \(\mathbb Q(\sqrt6)\), with discriminant \(24\). Its
cost \(\frac14\ln24\) is retained; the generator proof bounds it by
\(\frac12\ln12\). The exact coefficient heights are
\(\ln43,\frac12\ln3770\) on the left and
\(\frac12\ln43,\frac14\ln10180\) on the right. Theorem 10.63 and
equations (10.169)--(10.176) prove the field and budget mechanism;
Solution 32 verifies the rows, discriminants and heights. Human-source
context: the initial coefficient-field reduction in Lemma 4.2 of Yu's
free 2013 paper; the complete discriminant and Siegel inputs are proved
in the earlier lesson linked above, from Matveev's free paper.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/initial_coefficient_field.py}{Figure program}.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{25. Constructing enough columns in one phase
class}\label{constructing-enough-columns-in-one-phase-class}

The coefficient inequalities require a surplus of columns over
equations. We now produce that surplus from the volume of a closed box,
the covolume of the saturated exponent lattice and the number of
possible phase classes. The boundary of the box matters when its
normalized volume is an integer.

\subsubsection{Points in a translated closed
box}\label{points-in-a-translated-closed-box}

\textbf{Theorem 10.66 (the closed-box lattice count).} Let
\(\mathcal L=B\mathbb Z^r\), with \(B\) an invertible real matrix, and
let \(C=\prod_{j=1}^r[0,D_j]\), with every \(D_j>0\). Write

\[
\Delta=|\det B|,\qquad V=\prod_jD_j,\qquad m=\lfloor V/\Delta\rfloor.
\tag{10.177}
\]

There are \(m+1\) distinct points \(y_0,\ldots,y_m\) of \(C\) such that
all \(y_j-y_0\) lie in \(\mathcal L\). Consequently a translate of \(C\)
contains \(m+1\) distinct lattice points, including zero.

If \(\mathcal L\supset\mathbb Z^r\) has finite index
\(J=|\mathcal L/\mathbb Z^r|\), then

\[
\Delta=J^{-1},\qquad m=\lfloor JV\rfloor.
\tag{10.178}
\]

\textbf{Proof.} Take the half-open fundamental cell \(F=B[0,1)^r\). Its
translates by \(\mathcal L\) partition \(\mathbb R^r\), and its volume
is \(\Delta\). For \(x\in F\), let \(n_C(x)\) count the points of
\(x+\mathcal L\) in \(C\). This is a finite sum of indicator functions:
boundedness of \(C-F\) bounds the integer coordinates \(B^{-1}\ell\) of
any participating lattice vector. Translating each integral and using
the partition gives \[
\int_F n_C(x)\,dx
=\sum_{\ell\in\mathcal L}\int_{F+\ell}\mathbf1_C(y)\,dy
=V.
\] The volume and translation rules used here are the same ones proved
in the lattice argument of
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/TR-BAKER-08.html\#small-solutions-over-the-coefficient-field}{Small
solutions over the coefficient field}. If \(V/\Delta>m\), this integral
prevents \(n_C(x)\le m\) everywhere. Some coset therefore supplies
\(m+1\) points.

For the equality case \(V/\Delta=m\), enlarge \(C\) about its centre by
factors \(1+1/n\). Each enlarged closed box has volume strictly greater
than \(m\Delta\), so it contains \(m+1\) points
\(x_n+\ell_{0n},\ldots,x_n+\ell_{mn}\), with \(x_n\in F\). All
participating lattice vectors lie in one fixed bounded set, because the
enlarged boxes lie in the box enlarged by two and \(F\) is bounded.
There are finitely many such lattice vectors. Passing to a subsequence
fixes the distinct vectors \(\ell_0,\ldots,\ell_m\); a further
subsequence makes \(x_n\) converge in the closed bounded cell
\(B[0,1]^r\). Coordinatewise interval bisection gives this convergent
subsequence, as in the proof of Theorem 10.60. The limits \(x+\ell_j\)
lie in \(C\): its closed inequalities are the limits of the enlarged
inequalities. Their differences are the distinct lattice differences
\(\ell_j-\ell_0\). This proves the count also at integer normalized
volume.

For (10.178), since every coordinate unit vector belongs to
\(\mathcal L\), the matrix \(U=B^{-1}\) is integral. Multiplication by
\(B^{-1}\) identifies \(\mathcal L/\mathbb Z^r\) with
\(\mathbb Z^r/U\mathbb Z^r\). Its index is \(|\det U|\). Here is an
integer reduction proof of that identity. Interchanging rows or columns,
changing a sign, and adding an integer multiple of another row or column
preserve index and absolute determinant. The Euclidean algorithm makes
the first column have a positive pivot and zeros below. If an entry in
its first row is not divisible by the pivot, column division and
interchange produce a smaller positive pivot; repeat. This must stop
because positive integers decrease. Column subtraction then clears the
first row and leaves a nonsingular block of size \(r-1\). Induction
diagonalizes the matrix by these operations. The quotient for a diagonal
matrix has the product of the absolute diagonal entries as its index, by
independent residue representatives. This is its absolute determinant.
Thus \(J=|\det U|=1/\Delta\), proving (10.178). \(\square\)

The enlargement argument uses a closed box. For the open interval
\(C=(0,1)\) and lattice \(\mathbb Z\), the normalized volume is one but
no translate contains two lattice points.

\subsubsection{Retaining a single phase}\label{retaining-a-single-phase}

\textbf{Lemma 10.67 (the phase-class count).} Take the \(m+1\) lattice
differences in Theorem 10.66 and write them as \(B\lambda_j\), with
\(\lambda_j\in\mathbb Z^r\). For integers \(G_0\ge1\) and
\(d_1,\ldots,d_r\), put

\[
g=\gcd(G_0,d_1,\ldots,d_r),\qquad H=G_0/g.
\tag{10.179}
\]

There is a set \(\Lambda\) of at least \(\lceil(m+1)/H\rceil\) of these
distinct integer vectors, all satisfying the same congruence
\(\boldsymbol d\cdot\lambda\equiv\varepsilon\pmod{G_0}\). Moreover
\(B\Lambda\subset C-y_0\).

\textbf{Proof.} Each phase value is a multiple of \(g\) modulo \(G_0\),
so there are at most \(H\) possible values. If every class contained
fewer than \(\lceil(m+1)/H\rceil\) of the \(m+1\) points, their total
would be smaller than \(m+1\). Choose a class of at least that size. The
map \(B\) is injective, so no points coalesce. The containment holds
because every original difference was \(y_j-y_0\). This includes the
case \(d_i=0\) for every \(i\), when \(g=G_0,H=1\). \(\square\)

\subsubsection{A certified surplus of prepared
columns}\label{a-certified-surplus-of-prepared-columns}

\textbf{Corollary 10.68 (box volume supplies the initial kernel).}
Suppose the multiplicative and phase hypotheses of Theorem 10.55 are
available for the lattice basis \(B\) and the phase data of Lemma 10.67.
For integers \(k,L\ge1,S,T\ge0\), choose the retained support
\(\Lambda\) there. Its number of coefficient columns satisfies

\[
\begin{aligned}
kL\left\lceil\frac{\lfloor V/\Delta\rfloor+1}{H}\right\rceil
&\le N=kL|\Lambda|\le kL(\lfloor V/\Delta\rfloor+1),\\
N&>\frac{kLV}{\Delta H}.
\end{aligned}
\tag{10.180}
\]

Define the support-independent row upper bound

\[
\overline J_T=\sum_{u=0}^{\min(kL,T)}
\binom{T-u+r-1}{r-1},\qquad
\overline M=(2S+1)\overline J_T.
\tag{10.181}
\]

If \(kLV/(\Delta H)\ge c_*\overline M\), with \(c_*>1\), then
\(N>c_*M_*\). In particular the complete integral initial kernel of
Theorem 10.63 exists and its scaled budget (10.176) applies whenever its
original-width hypotheses hold. The selected support retains that
full-width height bound.

For a saturated lattice of index \(J\) and widths from (10.175), an
explicit sufficient criterion is

\[
\frac{kLJ}{H}\,
\frac{\mathcal D^r}
{(c_1c_2rPd)^r\prod_j\sigma_j}
\ge c_*(2S+1)
\sum_{u=0}^{\min(kL,T)}\binom{T-u+r-1}{r-1}.
\tag{10.182}
\]

\textbf{Proof.} The first inequality in (10.180) is the retained-class
count; the upper bound follows because the retained class is a subset of
the chosen \(m+1\) points. Since \(\lfloor V/\Delta\rfloor+1>V/\Delta\),
and a ceiling is at least its argument, the strict inequality follows.

At additive order \(u\le\min(kL,T)\), allowing all \(r-1\) Euler
directions gives \(\binom{T-u+r-1}{r-1}\) indices of order at most
\(T-u\), by the divider count proved in Theorem 10.32. Every
representative index in (10.126) is among these; zero Euler directions
only remove duplicates. Hence \(M_*\le\overline M\). Combining this fact
with (10.180) proves the strict surplus.

Pick any reference \(\lambda_0\) in the retained class. Its normalized
phase differences are multiples of \(G_0\), and its original coordinates
lie in \(C-y_0-B\lambda_0\). Thus all the hypotheses for the initial
coefficient construction and the full-width estimate are retained. Apply
Theorem 10.63 and, when (10.175) holds, Corollary 10.65. Substitution of
\(\Delta=1/J\) and \(V=\mathcal D^r/[(c_1c_2rPd)^r\prod_j\sigma_j]\)
gives (10.182). \(\square\)

The count certifies the initial coefficient construction. The analytic
slope, derivative precision and strict nonzero-value comparisons used to
extend its zeros remain separate hypotheses.

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={Five lattice points divide into two phase classes, leaving three retained exponent columns}]{../figures/box-phase-support.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.13. The exact lattice has basis columns \((1,0)\),
\((1/2,1/2)\), covolume \(1/2\) and index two over \(\mathbb Z^2\). The
closed box is \(C=[0,2]\times[0,1]\), so \(V/\Delta=4\). Five displayed
points have integral basis coordinates
\((0,0),(0,1),(1,0),(1,1),(2,0)\). For \(G_0=4,d=(2,2)\), there are
\(H=2\) phase classes. The class \(0\) retains three points with
coordinates \((0,0),(1,1),(2,0)\). At \(k=2,L=1,S=1,T=0\), they give
\(N=6\) columns and \(M_*=3\) rows. The lattice, box, points and phase
labels are exact. Theorem 10.66, Lemma 10.67 and Corollary 10.68 prove
(10.177)--(10.182), including the integer-volume boundary case; Solution
33 verifies these data and the derivative-order limitation. Human-source
context: the box and phase selection (4.18)--(4.21) in Yu's free 2013
paper, cited below. \href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/box_phase_support.py}{Figure
program}.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{26. Choosing parameters with a strict coefficient
surplus}\label{choosing-parameters-with-a-strict-coefficient-surplus}

The lattice count becomes useful only when it exceeds the number of
prescribed jets. We give a parameter choice for which that comparison
holds with all integer roundings retained. The node and jet budgets are
real numbers; their floors, rather than the budgets themselves, count
the equations.

\subsubsection{The real parameters}\label{the-real-parameters}

Let \(r\ge2\), and retain the independent \(\mathfrak p\)-adic units
\(\alpha_1,\ldots,\alpha_r\) in a number field of degree \(d\). Write
\(e=e_{\mathfrak p}\), \(f=f_{\mathfrak p}\), and \(t=f\ln p\); thus
\(ef\le d\). Let \(q=2\) for odd \(p\), and \(q=3\) for \(p=2\), with
\(\zeta_3\in K\) in the latter case. Let \(q^u\) be the order of the
q-primary roots of unity, and \(J=q^\nu\) the index of the q-primary
saturation lattice. The roots reduce injectively: a nontrivial q-primary
root in \(1+\mathfrak p\) would retain its positive finite valuation
under its prime-to-\(p\) order by
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/TR-BAKER-09.html\#4-taking-powers-until-the-exponential-applies}{p-adic
logarithms: power series, heights and the one-logarithm bound}, Lemma
9.5, but its power is one. Their reductions form a subgroup of the
finite field's group of \(p^f-1\) nonzero elements. Its distinct cosets
partition that group into sets of size \(q^u\), proving
\(q^u\mid p^f-1\). Use its phase coefficients from Section 20, and put

\[
\begin{aligned}
G&=p^f-1,&G_0&=G/q^u,&
\delta&=\gcd(G_0,d_1,\ldots,d_r),&H&=G_0/\delta,\\
\ell_d&=\max\{1,\ln d\},&
\rho&=\begin{cases}58,&d\ge2,\\17,&d=1,\end{cases}&
g_1&=4+\ln((r+1)d),&A&=\max\{g_1,e,t\}.
\end{aligned}
\tag{10.183}
\]

Take \(\sigma_i\ge h(\alpha_i)>0\), constants \(c_0>1\) and
\(c_1,c_3,c_4>0\), and \(g_0>39\). Choose \(h\ge\max\{g_0,(r+1)t\}\). In
this subsection \(h\) is this real auxiliary parameter, whereas
\(h(\alpha)\) denotes Weil height. Let \(\kappa\ge0\) be the integer
with \(p^{\kappa-1}(p-1)\le2e<p^\kappa(p-1)\), and set \(P=p^\kappa\).
For any \(\tau\ge0\) define

\[
\begin{aligned}
\vartheta&=\begin{cases}(p-2)/(p-1),&p\ge5, e=1,\\P/(2e),&\text{otherwise},\end{cases}
&\theta&=\vartheta/(1+\tau),\\
c_2&=\begin{cases}7/4,&p>2,\\13/9,&p=2,\end{cases}
&a_*&=\begin{cases}7/2,&p\ge5, e=1,\\26/3,&p=2,\\7,&\text{otherwise}.
\end{cases}
\end{aligned}
\tag{10.184}
\]

Choose any \(\widehat\vartheta\ge\vartheta\). The explicit choices
\(3/2\) at \(p=3\), \(5/4\) at \(p=5,e\ge2\), \(1\) at \(p\ge5,e=1\),
\(7/6\) at \(p\ge7,e\ge2\), and \(2\) at \(p=2\) work. Indeed
\(P\le2pe/(p-1)\), so \(P/(2e)\le p/(p-1)\). If \(p\ge5,e=1\), then
\(\kappa=0\), \(P=1\), and \(\vartheta<1\). In particular
\(c_2qP/(e\theta)\ge a_*\) in every case.

Define the positive lower budgets

\[
\begin{aligned}
g_2&=\begin{cases}
c_3q(r+1)^2d,&p\ge5,\ e=1,\text{ or }p\ge7,\\
c_3q(r+1)g_0e/\ln p,&p=2,3,\text{ or }p=5,\ e\ge2,
\end{cases}\\
g_3&=\frac{2c_0c_1c_4}{\rho}
a_*^r\frac{r^r(r+1)^r}{(r!)^2}g_1t,\\
g_4&=\frac{q(r+1)g_3}{c_1\widehat\vartheta t}
\begin{cases}1,&e=1\text{ and }(p\ge5\text{ or }d=1),\\g_1^{-1},&\text{otherwise},\end{cases}\\
g_5&=\frac{c_3q(r+1)g_3}{c_1c_4g_1t},
&1+\epsilon&=\left(1+\frac{r+1}{2g_4}\right)^r.
\end{aligned}
\tag{10.185}
\]

Here the \(d=1\) alternative in \(g_4\) is compatible with \(p=3,e=1\);
\(p=2,d=1\) cannot meet \(\zeta_3\in K\). Set

\[
\begin{aligned}
x&=\nu\ln q,\\
\gamma&=\frac{q^\nu hA}{(h+x)(A+x)},\\
S&=\frac{c_3q(r+1)d(h+x)}t,\\
\mathcal D&=\frac{\gamma}{q^{\nu+u}}
(1+\epsilon)(2+g_2^{-1})c_0c_1c_4
\left(\frac{c_2qPr(r+1)}{e\theta}\right)^r\frac1{r!}\\
&\quad{}\times
\max\left\{\frac{p^f}{\delta t^r},\frac{\mathrm e^r}{r^r}t\right\}
d^{r+1}\ell_d\prod_i\sigma_i\,(A+x),\\
T&=\frac{q(r+1)\mathcal D}{c_1\theta et},\\
k&=\lfloor h+x-1\rfloor+1=\lfloor h+x\rfloor,\\
\widetilde D_0&=\frac{S\mathcal D}{c_1c_4kd(A+x)},\\
L&=\lfloor\widetilde D_0\rfloor+1,\\
D_i&=\frac{\mathcal D}{c_1c_2rPd\sigma_i}.
\end{aligned}
\tag{10.186}
\]

The exponential constant \(\mathrm e=\exp(1)\) in this formula is
different from the ramification index \(e\). The additive factors of the
prepared family have precisely \(k\) choices of degree index and \(L\)
choices of positive power, as in Lemma 10.29.

\textbf{Lemma 10.69 (the saturation factor).} The parameters above
satisfy \(1\le\gamma\le q^\nu\).

\textbf{Proof.} With \(x\ge0\), \(\gamma=\exp(x)hA/((h+x)(A+x))\). The
logarithmic derivative with respect to \(x\) is
\(1-(h+x)^{-1}-(A+x)^{-1}>0\), because \(h>39\) and \(A\ge g_1>5\). At
\(x=0\) the value is one. The upper bound follows from
\((h+x)(A+x)\ge hA\). \(\square\)

\subsubsection{Lower bounds and the rounded equation
count}\label{lower-bounds-and-the-rounded-equation-count}

\textbf{Lemma 10.70 (lower budgets without lost floors).} We have

\[
\begin{gathered}
S\ge g_2,\qquad
\frac{\mathcal D}{A+x}\ge\frac{g_3}{g_1},\qquad
\mathcal D\ge g_3,\qquad T\ge g_4,\\
\widetilde D_0\ge g_5,\qquad
\widetilde D_0<L\le(1+g_5^{-1})\widetilde D_0,\\
\binom{\lfloor T\rfloor+r}{r}
\le(1+\epsilon)\frac{T^r}{r!}.
\end{gathered}
\tag{10.187}
\]

\textbf{Proof.} In the first alternative for \(g_2\), use
\(h+x\ge(r+1)t\) in \(S\). In the second, use \(h+x\ge g_0\) and
\(d/f\ge e\). These give both stated lower bounds, including
\(p=5,e\ge2\).

The q-primary lattice lies in the full saturation lattice of Theorem
10.18. Its index is \(q^\nu\), and \(w_K\ge q^u\). Consequently (10.55)
gives

\[
d^{r+1}\ell_d\prod_i\sigma_i
\ge \frac{q^{\nu+u}}{\rho}\frac{r^r}{r!\mathrm e^r}.
\tag{10.188}
\]

For \(d\ge2\), replacing \(\ln d\) by the larger \(\ell_d\) preserves
this bound; at \(d=1\) the degree and logarithmic factors equal one. The
maximum in (10.186) is at least \((\mathrm e^r/r^r)t\). Combine this
fact with (10.188), \(\gamma\ge1\), \((1+\epsilon)(2+g_2^{-1})\ge2\),
and \(c_2qP/(e\theta)\ge a_*\). This yields
\(\mathcal D/(A+x)\ge g_3/g_1\), including every displayed constant.
Since \(A+x\ge g_1\), it also yields \(\mathcal D\ge g_3\).

If the first alternative in \(g_4\) applies, \(e=1\); use
\(\mathcal D\ge g_3\) and \(\theta\le\widehat\vartheta\). Otherwise use
the stronger bound \(\mathcal D\ge g_3(A+x)/g_1\) and \(A+x\ge e\). Both
give \(T\ge g_4\).

We have \(0<k\le h+x\). Substitute the formula for \(S\) into
\(\widetilde D_0\), and then use the stronger bound for \(\mathcal D\):
\[
\widetilde D_0=
\frac{c_3q(r+1)(h+x)\mathcal D}{c_1c_4kt(A+x)}
\ge\frac{c_3q(r+1)g_3}{c_1c_4g_1t}=g_5.
\] For any positive real \(y\), \(y<\lfloor y\rfloor+1\le y+1\). Taking
\(y=\widetilde D_0\) proves both assertions about \(L\), including the
strict inequality when \(y\) is an integer.

Finally, \[
\binom{\lfloor T\rfloor+r}{r}
\le\frac1{r!}\prod_{j=1}^r(T+j)
\le\frac1{r!}\left(T+\frac{r+1}{2}\right)^r.
\] The last inequality follows from the concavity of \(\ln\): its second
derivative is \(-z^{-2}<0\), so the sum of the logarithms is at most
\(r\) times the logarithm of their average. Divide by \(T^r\), and use
\(T\ge g_4\) and (10.185). This proves the last line of (10.187).
\(\square\)

\textbf{Theorem 10.71 (the parameter choice supplies a kernel).} Use the
closed box with widths \(D_i\) and choose its single phase class as in
Lemma 10.67. In the prepared system impose every node \(|s|\le S\) with
integer \(s\), and every jet of total order at most the real number
\(T\). Write \(s_0=\lfloor S\rfloor\), \(t_0=\lfloor T\rfloor\), let
\(N=kL|\Lambda|\), and let \(M_*\) be the actual representative row
count. Put

\[
c_{01}=c_0\ell_d\frac{p^f}{p^f-1}>1.
\tag{10.189}
\]

Then

\[
\begin{aligned}
N&>\frac{kLq^\nu D_1\cdots D_r}{H}
>c_{01}(2S+1)\binom{t_0+r}{r}\\
&\ge c_{01}(2s_0+1)\binom{t_0+r}{r}
\ge c_{01}M_*.
\end{aligned}
\tag{10.190}
\]

In particular a nonzero global-integral coefficient vector and
polynomial with all these zeros and jets exist, with the field and
height conclusions of Lemma 10.63.

\textbf{Proof.} The first strict inequality is (10.180), with
\(J=q^\nu\), \(\Delta=1/J\), and \(H=G/(q^u\delta)\). For the next
comparison keep all widths in (10.186), and write \(V=\prod_iD_i\).
Since \(L>\widetilde D_0\), \[
\frac{kLq^\nu V}{H}>
\frac{Sq^{\nu+u}\delta\mathcal D^{r+1}}
{c_1^{r+1}c_4c_2^rr^rP^rd^{r+1}(A+x)G\prod_i\sigma_i}.
\] Divide by \((2S+1)(1+\epsilon)T^r/r!\). Substitute (10.186) for
\(\mathcal D\). The degree factors, widths and normalization factors
cancel, leaving \[
\frac{S(2+g_2^{-1})}{2S+1}
\frac{\gamma c_0\ell_d\delta t^r}{G}
\max\left\{\frac{p^f}{\delta t^r},\frac{\mathrm e^r}{r^r}t\right\}
\ge c_{01}.
\] Indeed \(S\ge g_2\) makes the first factor at least one,
\(\gamma\ge1\), and the first member of the maximum gives \(p^f\). The
binomial inequality (10.187) proves the second strict inequality in
(10.190). A derivative order is an integer, so an order at most \(T\) is
exactly an order at most \(t_0\); similarly there are exactly \(2s_0+1\)
nodes. Counting all \(r\) derivative directions gives
\(\binom{t_0+r}{r}\) rows per node by the divider argument of Theorem
10.32. Removing zero additive orders or duplicate Euler directions only
decreases the representative count. This proves the remaining
inequalities.

Since \(N>M_*\), the completely proved integral Siegel argument in Lemma
10.63 applies over the actual row field, with no unproved estimate used
in place of an equation. Its nonzero vector gives a nonzero polynomial
by Lemma 10.29. \(\square\)

\subsubsection{The height term with a real node
budget}\label{the-height-term-with-a-real-node-budget}

\textbf{Corollary 10.72 (the rounded mean torus contribution).} In this
kernel let \(d_0\) be the row-field degree, let \(\mathcal H_0\) be the
scalar row bound from Lemma 10.63 with node budget \(s_0\) and jet
budget \(t_0\), and put \(Z=S\mathcal D/d\) and
\(L_\alpha=\sum_iD_i h(\alpha_i)\). Then

\[
\begin{aligned}
\frac{h_2(\boldsymbol c)}Z
\le{}&\frac{\tfrac12\ln(Nd_0)}Z
 +\frac{d_0-1}{c_1c_2S}\\
&+\frac1{c_{01}-1}\left[
\frac{\tfrac12\ln N+\ln\mathcal H_0}Z
 +\frac{1+1/(2g_2)}{2c_1c_2}
\right].
\end{aligned}
\tag{10.191}
\]

\textbf{Proof.} Apply (10.174) at the integer budget \(s_0\), and use
\(N>c_{01}M_*\) from (10.190). The full original width bound and
\(\sigma_i\ge h(\alpha_i)\) give \(PL_\alpha\le\mathcal D/(c_1c_2d)\),
so the generator term divided by \(Z\) is at most \((d_0-1)/(c_1c_2S)\).
The exact mean of the absolute integer nodes is \(s_0(s_0+1)/(2s_0+1)\).
It satisfies \[
\frac{s_0(s_0+1)}{2s_0+1}
=\frac{s_0}{2}+\frac14-\frac1{4(2s_0+1)}
\le\frac S2+\frac14.
\] Thus its contribution divided by \(Z\) is at most
\((1+1/(2S))/(2c_1c_2)\le(1+1/(2g_2))/(2c_1c_2)\). Substitution proves
(10.191). This leaves the scalar row and row-field generator costs
explicit; each must still be included in a numerical zero-forcing
comparison. \(\square\)

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={The rounded coefficient surplus for rational bases two and five at three}]{../figures/initial-parameter-counts.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.14. Take \(K=\mathbb Q\), \(p=3\),
\((\alpha_1,\alpha_2)=(2,5)\), and the parameters in Solution 34. The
prepared column count and every plotted comparison are exact integer or
rational quantities. There are \(2\lfloor S\rfloor+1\) nodes and
\(\binom{\lfloor T\rfloor+2}{2}\) possible jets per node. The surplus
\(c_{01}=57/20\) is retained through both floors; the strict comparison
follows from the full real parameters, without rounding either budget
upwards. The phase count is one, so no support is lost at that step. See
(10.187)--(10.191), Solution 34, and
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
free paper} for further reading.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{27. The global size and scalar cost of the initial
kernel}\label{the-global-size-and-scalar-cost-of-the-initial-kernel}

We now bound the number of coefficients and the size of each scalar row
entry at the parameters of Section 26. Two factors need separate care. A
prime-to-\(p\) root field is unramified, which strengthens the global
degree bound by the ramification index. The additive prepared derivative
also contains an lcm clearing factor; that factor contributes to its
ordinary absolute value as well as to its integrality.

\subsubsection{The degree available above an unramified root
field}\label{the-degree-available-above-an-unramified-root-field}

\textbf{Lemma 10.73 (the torsion degree divided by ramification).} With
the notation of (10.183), we have

\[
\frac d e\ge q^{u-1}(q-1).
\tag{10.192}
\]

\textbf{Proof.} Put \(F=\mathbb Q(\zeta_{q^u})\subseteq K\). The
cyclotomic Eisenstein proof in
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/TR-BAKER-09.html\#5-cyclotomic-ramification-and-distances-between-roots}{p-adic
logarithms: power series, heights and the one-logarithm bound},
Proposition 9.7 at the prime \(q\), gives
\([F:\mathbb Q]=q^{u-1}(q-1)\): irreducibility over \(\mathbb Q_q\)
implies irreducibility over \(\mathbb Q\). Since \(q\ne p\), the
completion \(F_{\mathfrak q}\) at the prime below \(\mathfrak p\) is
unramified over \(\mathbb Q_p\), by the complete prime-to-\(p\) proof in
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/NT-LOC/NT-LOC-12.html\#1-two-kinds-of-cyclotomic-extension}{Cyclotomic,
quadratic and Kummer extensions of local fields}, Proposition 1.1.

Let \(m=[K:F]\), and choose an \(F\)-basis \(b_1,\ldots,b_m\) of \(K\).
Inside \(K_{\mathfrak p}\), the subspace \(\sum_jF_{\mathfrak q}b_j\)
has dimension at most \(m\) over the complete field \(F_{\mathfrak q}\).
It is complete, hence closed, by the finite-dimensional norm proof in
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/NT-LOC/NT-LOC-04.html\#2-coordinates-control-convergence}{Extensions
of complete valued fields}, Proposition 2.1. It contains \(K\), whose
closure is \(K_{\mathfrak p}\), so it equals that completion. Therefore
\([K_{\mathfrak p}:F_{\mathfrak q}]\le m\).

Ramification indices multiply in a tower: restriction of a normalized
integer valuation multiplies it by the ramification index at each step.
The base root field has ramification index one, so
\(e=e(K_{\mathfrak p}/F_{\mathfrak q})\). The degree formula
\(ef=[L:K]\), fully proved by residue digits in
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/NT-LOC/NT-LOC-04.html\#3-an-integral-basis-from-residue-digits}{Extensions
of complete valued fields}, Theorem 3.1, gives
\(e\le[K_{\mathfrak p}:F_{\mathfrak q}]\le m\). Therefore
\(d/e=[F:\mathbb Q]m/e\ge[F:\mathbb Q]\), as asserted. \(\square\)

\subsubsection{A lower volume and an upper coefficient
count}\label{a-lower-volume-and-an-upper-coefficient-count}

Retain all real parameters in (10.183)--(10.186). Set

\[
\begin{aligned}
b&=\begin{cases}1,&e=1\text{ and }(p\ge5\text{ or }d=1),\\
g_1^{-1},&\text{otherwise},\end{cases}\\
g_{61}&=
2c_0c_1^{1-r}c_4
\left(\frac q{\widehat\vartheta}\right)^r
\frac{(r+1)^r\mathrm e^r}{r!r^r}\,
t\,\frac{q-1}{q}\,g_1\,(bg_3)^{r-1},\\
v&=q^\nu D_1\cdots D_r.
\end{aligned}
\tag{10.193}
\]

\textbf{Lemma 10.74 (a uniform lower box volume).} We have

\[
v\ge g_{61}>0.
\tag{10.194}
\]

\textbf{Proof.} In the maximum defining \(\mathcal D\), use the second
member \((\mathrm e^r/r^r)t\). Substituting one factor of \(\mathcal D\)
into \(v\), and using \(\gamma\ge1\), \(1+\epsilon\ge1\), and
\(2+g_2^{-1}\ge2\), gives \[
v\ge
\frac{2c_0c_1^{1-r}c_4}{q^u}
\left(\frac q\theta\right)^r
\frac{(r+1)^r\mathrm e^r}{r!r^r}\,
t\,\frac d{e^r}\ell_d(A+x)\mathcal D^{r-1}.
\] Lemma 10.73 gives \(d/q^u\ge e(q-1)/q\); also \(\ell_d\ge1\) and
\(\theta\le\widehat\vartheta\). The remaining factor is
\((A+x)(\mathcal D/e)^{r-1}\). In the first case for \(b\), \(e=1\) and
\(\mathcal D\ge g_3\). In the other case, the stronger bound in (10.187)
and \(A+x\ge e\) give \(\mathcal D/e\ge g_3/g_1\). Since \(A+x\ge g_1\),
both cases imply \((A+x)(\mathcal D/e)^{r-1}\ge g_1(bg_3)^{r-1}\). This
proves (10.194). \(\square\)

Let \(w_K\) denote the full root-of-unity order, and put

\[
\begin{aligned}
g_6&=
\rho(1+g_5^{-1})(1+g_{61}^{-1})
\frac{r!\mathrm e^r}
{c_1^{r+1}c_4c_2^rP^rw_Kr^{2r}},\\
g_7&=\frac{c_3q(r+1)g_0g_3}{t},\\
g_8&=\frac{\ln g_7+g_1+
\max\{\ln(g_6d/\mathrm e^{g_1}),0\}}{g_7}
+\frac r{c_3q(r+1)^2}\frac{\ln g_3}{g_3}.
\end{aligned}
\tag{10.195}
\]

\textbf{Theorem 10.75 (the logarithmic coefficient budget).} Suppose
\(g_3\ge\mathrm e\) and \(g_7\ge\mathrm e\). For the selected support of
Theorem 10.71, \(N=kL|\Lambda|\) and \(Z=S\mathcal D/d\) satisfy

\[
N\le g_6S\mathcal D^{r+1},
\qquad Z\ge g_7,\qquad
\ln N\le g_8Z.
\tag{10.196}
\]

\textbf{Proof.} By (10.180), \(|\Lambda|\le v+1\). Also \[
kL\le(1+g_5^{-1})\frac{S\mathcal D}{c_1c_4d(A+x)}.
\] Use \(v\ge g_{61}\) to replace \(v+1\) by at most
\((1+g_{61}^{-1})v\). The full saturation bound (10.55) gives \[
\frac{q^\nu}{d^{r+1}\prod_i\sigma_i}
\le\frac{\rho\ell_dr!\mathrm e^r}{w_Kr^r}.
\] For \(d\ge2\), use the same replacement of \(\ln d\) by \(\ell_d\) as
in (10.188); for \(d=1\) the formula is exact. Substituting the widths
\(D_i\) and \(\ell_d\le g_1\le A+x\) gives the first inequality of
(10.196), with exactly \(g_6\).

The formula for \(S\), together with \(h+x\ge g_0\) and
\(\mathcal D\ge g_3\), gives \(Z\ge g_7\). Taking logarithms in the
first inequality yields \[
\ln N\le\ln(g_6d)+\ln Z+r\ln\mathcal D.
\] Put \(C=g_1+\max\{\ln(g_6d/\mathrm e^{g_1}),0\}>0\); then
\(\ln(g_6d)\le C\), so \(C/Z\le C/g_7\). The function \(\ln y/y\)
decreases for \(y\ge\mathrm e\), because its derivative is
\((1-\ln y)/y^2\). Thus \(\ln Z/Z\le\ln g_7/g_7\).

Finally \(h+x\ge(r+1)t\) gives \(S/d\ge c_3q(r+1)^2\), and therefore \[
\frac{r\ln\mathcal D}{Z}
\le\frac r{c_3q(r+1)^2}\frac{\ln\mathcal D}{\mathcal D}
\le\frac r{c_3q(r+1)^2}\frac{\ln g_3}{g_3}.
\] Combining these three bounds proves \(\ln N\le g_8Z\). \(\square\)

The lower hypotheses here have simple sufficient conditions. If
\(c_0\ge19/10,c_1\ge7/5,c_3\ge47/100,c_4\ge15/4\), then \(g_3/\!t>21\).
Indeed \(a_*\ge7/2\), \(\rho\le58\), \(g_1>5\), and
\(r^r(r+1)^r/(r!)^2\ge1\); hence \[
\frac{g_3}{t}>
\frac2{58}\frac{19}{10}\frac75\frac{15}4
\left(\frac72\right)^2\,5>21.
\] The positive logarithm series in Solution 34 gives \(\ln2>2/3\), so
\(t\ge\ln2>2/3\), and \(g_3>14>\mathrm e\). Also
\(g_7>(47/100)\cdot2\cdot3\cdot39\cdot21>3>\mathrm e\), where
\(\mathrm e<3\) follows by summing its positive series and bounding the
tail geometrically.

\subsubsection{Retaining the clearing factor and the jet
allocation}\label{retaining-the-clearing-factor-and-the-jet-allocation}

Write \(s_0=\lfloor S\rfloor,t_0=\lfloor T\rfloor,D_{\mathrm a}=kL\),
and \(\mathfrak v=v(k)=\operatorname{lcm}(1,\ldots,k)\). For the \(r-1\)
integer Euler forms of the actual prepared system, let \(\Omega_j\) be
the maximum of their absolute arguments on the support, and put \[
m=\sum_{j=1}^{r-1}\Omega_j+r-2,\qquad
U=\min(D_{\mathrm a},t_0).
\tag{10.197}
\]

\textbf{Lemma 10.76 (a complete scalar row bound).} The scalar bound in
Lemma 10.63 may be taken to be

\[
\mathcal H_0=
\left[2\mathrm e\left(\frac{s_0}{k}+2\right)\right]^{D_{\mathrm a}}
\max_{0\le u\le U}
\left\{\mathfrak v^u
\binom{m+t_0-u}{t_0-u}\right\}.
\tag{10.198}
\]

If \(\mathfrak v=1\), a maximizing index is \(u_*=0\). If
\(\mathfrak v>1\), one is \[
u_*=\min\left\{U,\,
\max\left\{0,t_0-\left\lceil\frac m{\mathfrak v-1}\right\rceil+1\right\}\right\}.
\tag{10.199}
\]

More sharply, with \(R=s_0+2k\), the scalar bound may be taken to be \[
\mathcal H_{\rm fine}=
\frac{R^{D_{\mathrm a}}}{(k!)^L}
\max_{0\le u\le U}
\left\{\left(\frac{\mathfrak v}{R}\right)^u
\binom{D_{\mathrm a}}u
\binom{m+t_0-u}{t_0-u}\right\}.
\tag{10.200}
\] A maximizing index \(\widehat u\) is the first index after all the
consecutive increasing steps. A step from \(u\) to \(u+1\), for \(u<U\),
is increasing or constant precisely when \[
\mathfrak v(D_{\mathrm a}-u)(t_0-u)
\ge R(u+1)(m+t_0-u).
\tag{10.201}
\] These steps form an initial interval, so an integer binary search
finds a maximizing index, with no numerical root approximation.

\textbf{Proof.} For \(0\le a<k\), \(1\le\ell\le L\), write \[
F_{a,\ell}(X)=\frac1{(k!)^\ell}
\prod_{j=1}^k(X+a+j)^\ell.
\] It has degree \(k\ell\), and every shift \(a+j\) is at most \(2k\).
In the divided derivative of order \(u\), choose the \(u\)
differentiated linear factors. There are \(\binom{k\ell}{u}\) choices.
For \(|s|\le s_0\), each remaining factor has absolute value at most
\(s_0+2k\). Thus \[
\left|\frac{F_{a,\ell}^{(u)}(s)}{u!}\right|
\le \frac{\binom{k\ell}{u}(s_0+2k)^{k\ell-u}}{(k!)^\ell}
\le\left[2\mathrm e\left(\frac{s_0}{k}+2\right)\right]^{D_{\mathrm a}}.
\] For the last inequality use \(\binom{k\ell}{u}\le2^{k\ell}\), the
positive factor \(s_0+2k\ge1\), and \(\ln(k!)\ge k\ln k-k\). The latter
follows by comparing \(\sum_{j=1}^k\ln j\) with the integral of the
increasing function \(\ln x\) on \([1,k]\). The resulting base is
greater than one, and \(k\ell\le D_{\mathrm a}\). If \(u>k\ell\), the
derivative is zero.

The prepared derivative multiplies this divided derivative by
\(\mathfrak v^u\), by (10.86); discarding that multiplier would give the
wrong ordinary absolute-value bound. Each Euler factor satisfies \[
|\Delta(\omega_j;t_j)|
\le\binom{\Omega_j+t_j}{t_j}.
\] Multiplying the positive generating series for these binomial
coefficients, as in (10.124), shows that their product at total Euler
order \(h\) is at most \(\binom{m+h}{h}\). This increases with \(h\), so
for additive order \(u\) it is at most \(\binom{m+t_0-u}{t_0-u}\). This
proves (10.198).

For \(F(u)=\mathfrak v^u\binom{m+t_0-u}{t_0-u}\), whenever \(u<t_0\), \[
\frac{F(u+1)}{F(u)}
=\mathfrak v\frac{t_0-u}{m+t_0-u}.
\] At \(\mathfrak v=1\) this is at most one. Otherwise it is at least
one exactly when \((\mathfrak v-1)(t_0-u)\ge m\). These ratios decrease
as \(u\) increases. Hence \(F\) increases up to this integer threshold
and decreases afterwards; equality may give two adjacent maximizers.
Clamping the last increasing step to \([0,U]\) gives (10.199), choosing
the larger maximizer in the equality case.

For (10.200), retain the first divided-derivative estimate, rather than
replacing its binomial coefficient by \(2^{k\ell}\). Since
\(\binom{k\ell}{u}\le\binom{D_{\mathrm a}}u\) and \(R^k/k!>1\), it gives
\[
\left|\frac{F_{a,\ell}^{(u)}(s)}{u!}\right|
\le \binom{D_{\mathrm a}}u\,R^{-u}
\left(\frac{R^k}{k!}\right)^\ell
\le\frac{\binom{D_{\mathrm a}}u R^{D_{\mathrm a}-u}}{(k!)^L}.
\] Multiply by the clearing and Euler factors already proved. The ratio
of the resulting expression at \(u+1\) to that at \(u\) is \[
\frac{\mathfrak v}{R}\frac{D_{\mathrm a}-u}{u+1}
\frac{t_0-u}{m+t_0-u}.
\] Every factor depending on \(u\) decreases on \(0\le u<U\); the first
is strictly decreasing. Thus the ratios cross one at most once. Clearing
their positive denominators gives exactly (10.201), which proves the
asserted integer rule and the sharper bound. \(\square\)

\textbf{Corollary 10.77 (the normalized additive cost).} If
\(c_3q/t\le8\), the first factor in (10.198) contributes at most \[
\frac{D_{\mathrm a}}Z
\ln\left[2\mathrm e\left(\frac{s_0}{k}+2\right)\right]
\le\frac{1+g_5^{-1}}{c_1c_4}.
\tag{10.202}
\]

\textbf{Proof.} Since \(h+x>39\), \(k=\lfloor h+x\rfloor\ge39\) and
\((h+x)/k<(k+1)/k\le40/39\). Put \(W=(r+1)d\ge3\). Then \[
\frac{s_0}{k}+2\le\frac Sk+2
<\frac{320}{39}W+2\le\frac{346}{39}W.
\] Since \(\mathrm e^3>18>692/39\), it follows that
\(2\mathrm e(s_0/k+2)<\mathrm e^4W\), and its logarithm is at most
\(g_1\). The inequality \(\mathrm e^3>18\) follows already from the
terms of degrees zero through five in its exponential series at three.
Also \(D_{\mathrm a}/Z\le(1+g_5^{-1})/[c_1c_4(A+x)]\), by (10.187), and
\(g_1\le A+x\). These prove (10.202). For odd \(p\), the hypothesis
holds whenever \(c_3\le4\), since \(\ln p\ge\ln3>1\). At \(p=2\), it
holds whenever \(c_3\le1/2\), since \(\ln2>2/3\). \(\square\)

\textbf{Theorem 10.78 (the initial height with all scalar costs).}
Assume the hypotheses of Theorems 10.71 and 10.75 and Corollary 10.77.
Set \(\Psi=u_*\ln\mathfrak v+\ln\binom{m+t_0-u_*}{t_0-u_*}\). Then the
integral initial kernel can be chosen with \[
\begin{aligned}
\frac{h_2(\boldsymbol c)}Z\le{}&
\frac12g_8+\frac{\ln d_0}{2g_7}
+\frac{d_0-1}{c_1c_2S}\\
&+\frac1{c_{01}-1}
\left[\frac12g_8+\frac{1+g_5^{-1}}{c_1c_4}
+\frac{\Psi}{Z}
+\frac{1+1/(2g_2)}{2c_1c_2}\right].
\end{aligned}
\tag{10.203}
\]

\textbf{Proof.} In (10.191), use \(\ln N/Z\le g_8\), \(Z\ge g_7\), and
the exact scalar bound (10.198) at (10.199). Corollary 10.77 bounds its
additive first factor; its remaining factor has logarithm exactly
\(\Psi\). Substitution proves every term of (10.203). The row-field term
may further be bounded by \(1/[c_1c_2c_3q(r+1)^2]\), since \(d_0\le d\)
and \(S/d\ge c_3q(r+1)^2\), but its exact vanishing at \(d_0=1\) is
preferable there.

More sharply, replace the sum \((1+g_5^{-1})/(c_1c_4)+\Psi/Z\) in
(10.203) by \(\ln\mathcal H_{\rm fine}/Z\), evaluated at the exact
integer maximizer \(\widehat u\) in (10.201). This follows from the same
substitution in (10.191), using (10.200).

This is a height bound for coefficients whose existence has already been
proved. An extension of their zeros still requires the stated analytic
slopes, derivative precision, arithmetic strict comparisons and final
multiplicity contradiction. In particular the lcm and Euler term
\(\Psi\) must be retained in each such comparison. \(\square\)

\subsubsection{Averaging the derivative
orders}\label{averaging-the-derivative-orders}

\textbf{Corollary 10.79 (a mean scalar budget).} Retain every Euler
index, allowing duplicate rows, but omit the identically zero additive
orders above \(D_{\mathrm a}\). Define \[
\begin{aligned}
J_u&=\binom{t_0-u+r-1}{r-1}\quad(0\le u\le U),&
\overline J&=\sum_{u=0}^U J_u,\\
\overline u&=\frac{\sum_{u=0}^UuJ_u}{\overline J},&
\overline h&=\frac{r-1}{r}(t_0-\overline u),&
\overline M&=(2s_0+1)\overline J.
\end{aligned}
\tag{10.204}
\] Let \(H(z)=-z\ln z-(1-z)\ln(1-z)\), with endpoint values zero, and
put \[
\begin{aligned}
A_{\rm mean}&=(D_{\mathrm a}-\overline u)\ln R
-L\ln(k!)+D_{\mathrm a}H(\overline u/D_{\mathrm a}),\\
E_{\rm mean}&=\begin{cases}
\overline h\ln\!\left[\mathrm e(1+m/\overline h)\right],&\overline h>0,\\
0,&\overline h=0,
\end{cases}\\
B_{\rm mean}&=A_{\rm mean}+\overline u\ln\mathfrak v+E_{\rm mean}.
\end{aligned}
\tag{10.205}
\] The mean logarithm of the scalar row bounds is at most
\(B_{\rm mean}\), and an integral initial kernel exists with \[
\begin{aligned}
\frac{h_2(\boldsymbol c)}Z\le{}&
\frac{\tfrac12\ln(Nd_0)}Z+\frac{d_0-1}{c_1c_2S}\\
&+\frac{\overline M}{N-\overline M}
\left[\frac{\tfrac12\ln N+B_{\rm mean}}Z
+\frac{s_0(s_0+1)}
{(2s_0+1)c_1c_2S}\right].
\end{aligned}
\tag{10.206}
\] Moreover \(\overline u\le t_0/(r+1)\).

\textbf{Proof.} At fixed additive order \(u\), the \(r-1\) Euler indices
and their slack coordinate are all the nonnegative integer compositions
of \(t_0-u\) into \(r\) parts. Permuting their coordinates is a
bijection; the sum of their coordinate averages is \(t_0-u\). Thus their
mean total Euler order is \((r-1)(t_0-u)/r\). Weighting by \(J_u\)
proves the formula for \(\overline h\).

Without the additive cutoff, all \(r\) derivative coordinates and their
slack are the compositions of \(t_0\) into \(r+1\) parts, so the mean
additive order is \(t_0/(r+1)\). Removing orders greater than \(U\) can
only decrease that mean: each removed order exceeds every retained one.
This proves the last assertion.

The sharper additive estimate in the proof of Lemma 10.76 has logarithm
at most \[
\ln\binom{D_{\mathrm a}}u+(D_{\mathrm a}-u)\ln R-L\ln(k!).
\] For \(0<u<D_{\mathrm a}\), the binomial theorem at
\(z=u/D_{\mathrm a}\) gives
\(\binom{D_{\mathrm a}}u z^u(1-z)^{D_{\mathrm a}-u}\le1\). Hence
\(\ln\binom{D_{\mathrm a}}u\le D_{\mathrm a}H(u/D_{\mathrm a})\),
including the endpoints by continuity. The second derivative
\(H''(z)=-1/z-1/(1-z)<0\) proves concavity, so averaging gives
\(A_{\rm mean}\).

At total Euler order \(h>0\), the factorial integral bound gives
\(\ln\binom{m+h}{h}\le h\ln[\mathrm e(1+m/h)]\). The right side extends
continuously by zero at \(h=0\). Its second derivative is
\(-m^2/[h(m+h)^2]\le0\), so its average is at most \(E_{\rm mean}\);
when \(m=0\) it is the linear function \(h\). The clearing multiplier
has logarithm \(u\ln\mathfrak v\), whose mean is exactly
\(\overline u\ln\mathfrak v\). This proves (10.205).

By (10.190), \(N>c_{01}(2s_0+1)\binom{t_0+r}{r}\ge c_{01}\overline M\),
so the completely proved weighted integral Siegel argument of Lemma
10.63 applies to these rows. Duplicate or further zero rows do not
increase rank; assigning them the same positive upper scalar bounds
preserves that argument. The row field remains \(E_0\), and the
torus-row mean remains that of (10.174). Use (10.173) for its full-width
and field-generator terms, and substitute the mean scalar bound into the
row sum in (10.171). This is (10.206), with the actual ratio
\(\overline M/(N-\overline M)\). The nonzero polynomial and every
required jet follow as before. \(\square\)

\subsubsection{A second way to clear the global
rows}\label{a-second-way-to-clear-the-global-rows}

\textbf{Corollary 10.80 (row scaling and an alternative mean bound).}
Define \[
B_{\rm proj}=(D_{\mathrm a}-\overline u)\ln R
+D_{\mathrm a}H(\overline u/D_{\mathrm a})+E_{\rm mean}.
\tag{10.207}
\] In (10.206), \(B_{\rm mean}\) may be replaced by \[
\min\{B_{\rm mean},B_{\rm proj}\}.
\tag{10.208}
\] One may also choose the smaller scalar bound separately at each row
before averaging.

\textbf{Proof.} At the additive order \(u\), every original prepared row
contains the common multiplier \(\mathfrak v^u\). Replace it by the
common multiplier \((k!)^L\). This multiplies that entire row by the
nonzero rational number \((k!)^L/\mathfrak v^u\), so it preserves its
kernel and its coefficient field. Every nonzero row has unchanged
projective height; zero rows remain zero. For a nonzero row, the
logarithms of the absolute values of that common scalar sum to zero by
the normalized product formula proved in
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/NT-LOC/NT-LOC-05.html\#5-norms-and-the-normalized-product-formula}{Places
of number fields in extensions and the product formula}, Theorem 5.2.

The divided derivative of \(F_{a,\ell}\) has denominator dividing
\((k!)^\ell\). To verify this at every order, its polynomial
coefficients have that denominator and divided differentiation
multiplies a monomial coefficient by an integer binomial coefficient. At
an integer node its denominator still divides \((k!)^\ell\), hence
divides \((k!)^L\). The Euler binomials are integers at their integer
arguments. Thus the replaced scalar entries are integers, as required
for the same finite-place row estimate.

The sharper derivative bound in Lemma 10.76 now gives the archimedean
scalar bound \[
\binom{D_{\mathrm a}}u R^{D_{\mathrm a}-u}
\binom{m+t_0-u}{t_0-u}.
\] The factorial denominator cancels against the new common multiplier.
Averaging its logarithm with the concavity arguments of Corollary 10.79
gives \(B_{\rm proj}\). Its coefficient matrix has exactly the same
kernel as the original prepared matrix, so the identical weighted Siegel
and full-width argument proves (10.206) with this bound. Both bounds
therefore apply to the same row heights; choosing their smaller value,
globally or row by row, is valid. The original lcm normalization remains
available for every later local jet-precision computation. \(\square\)

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={The complete scalar height budget at the rounded rational parameters}]{../figures/initial-scalar-budget.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.15. The parameters of Solution 34 give the proved
coefficient-height bounds in Solution 35. The sharper allocation and row
average retain the factorial denominator; lcm clearing, Euler orders,
mean torus and logarithmic terms are all included. Left of zero is the
factorial contribution to be subtracted when the highest additive order
is used. The diamonds mark the net sums. Each pair uses either the
actual integer \(\operatorname{lcm}(1,\ldots,50)\) or the completely
proved coarser \(3^{50}\) bound in
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/TR-BAKER-07.html\#an-exponential-bound-for-least-common-multiples}{Two
logarithms II: explicit lower bounds}. The final pair uses the
alternative global row clearing of Corollary 10.80. All bars give proved
upper bounds on the height of some initial kernel. See
(10.198)--(10.208), Solution 35, and
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
free paper} for further reading.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{28. Principal-unit depth and the analytic
torus}\label{principal-unit-depth-and-the-analytic-torus}

The local torus coordinates must define convergent exponential functions
on the enlarged disc used for interpolation. We prove the depth with the
same power \(P\) as in Section 26. The ramification index enters this
choice; the small strict reduction from \(\vartheta\) to \(\theta\) also
has a mathematical purpose.

\subsubsection{The power needed for an enlarged
disc}\label{the-power-needed-for-an-enlarged-disc}

Let \(L/\mathbb Q_p\) be a finite extension with ramification index
\(e\), and normalize \(v_p(p)=1\). Write \(a_p=1/(p-1)\), and choose the
nonnegative integer \(\kappa\) by

\[
p^{\kappa-1}(p-1)\le2e<p^\kappa(p-1),\qquad P=p^\kappa,
\qquad
\vartheta=\begin{cases}
(p-2)/(p-1),&p\ge5, e=1,\\
P/(2e),&\text{otherwise}.
\end{cases}
\tag{10.209}
\]

This integer exists: take the least nonnegative \(\kappa\) satisfying
the strict upper inequality. At \(\kappa=0\), its lower inequality still
holds because \((p-1)/p<1\le2e\). Fix \(0<\theta<\vartheta\); in
particular the choice \(\theta=\vartheta/(1+\varepsilon)\) with
\(\varepsilon>0\) is permitted.

\textbf{Lemma 10.81 (the original principal-unit depth).} If
\(\beta\in L\) and \(v_p(\beta-1)>0\), then

\[
v_p(\beta^P-1)\ge\vartheta+a_p>\theta+a_p.
\tag{10.210}
\]

The statement includes \(\beta^P=1\), with valuation infinity.

\textbf{Proof.} The valuation group of \(L\) is \(e^{-1}\mathbb Z\), by
the complete degree-and-residue proof in
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/NT-LOC/NT-LOC-04.html\#3-an-integral-basis-from-residue-digits}{Extensions
of complete valued fields}, Theorem 3.1. Thus \(v_p(\beta-1)\ge1/e\).
The elementary power step in
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/TR-BAKER-09.html\#4-taking-powers-until-the-exponential-applies}{p-adic
logarithms: power series, heights and the one-logarithm bound}, Lemma
9.5, gives \(c_{j+1}\ge F(c_j)\) for \(c_j=v_p(\beta^{p^j}-1)\), where
\(F(c)=\min\{pc,c+1\}\). Below \(a_p\) the step is exactly
multiplication by \(p\); at or above that threshold the lower bound
still holds. The function \(F\) is increasing and \(F(c)\ge c\) for
\(c\ge0\). If any power is one, all subsequent assertions are immediate.

In the exceptional case \(p\ge5,e=1\), we have \(\kappa=0\), so the
initial bound is \(1=\vartheta+a_p\). Consider the other cases. The
inequalities defining \(\kappa\) give \(a_p<\vartheta\le p a_p\). If
\(\kappa=0\), then \(1/e=2\vartheta>\vartheta+a_p\), which proves the
claim.

Suppose \(\kappa\ge1\). We first prove \(c_{\kappa-1}\ge P/(pe)\). If
none of \(c_0,\ldots,c_{\kappa-2}\) reaches \(a_p\), the exact steps
below the threshold give \(c_{\kappa-1}=p^{\kappa-1}c_0\ge P/(pe)\); for
\(\kappa=1\) this uses no step. Otherwise, after the first such crossing
there is at least one further step before index \(\kappa-1\). That step
gives \(c\ge a_p+1\ge2a_p\). Subsequent steps cannot decrease the
valuation, and the lower defining inequality gives \(P/(pe)\le2a_p\).
This proves the same bound in that case.

Applying the increasing function \(F\) once more yields

\[
c_\kappa\ge\min\{P/e, 1+P/(pe)\}
=\min\{2\vartheta, 1+2\vartheta/p\}
\ge\vartheta+a_p.
\tag{10.211}
\]

Indeed \(2\vartheta>\vartheta+a_p\). For the other member, use
\((1-2/p)\vartheta\le(p-2)/(p-1)=1-a_p\). This remains valid at \(p=2\),
where both sides of that last inequality are zero. Finally
\(\theta<\vartheta\) supplies the strict inequality in (10.210).
\(\square\)

\subsubsection{Coefficients and divided jets of the torus
functions}\label{coefficients-and-divided-jets-of-the-torus-functions}

\textbf{Corollary 10.82 (normality with the actual depth).} Let
\(w_1,\ldots,w_s\in L\) satisfy \(v_p(w_i-1)\ge\vartheta+a_p\), and let
\(\lambda_i\in L\) have \(v_p(\lambda_i)\ge0\). Put

\[
\ell=\sum_i\lambda_i\log w_i,\qquad
F(Z)=\exp(Z\ell)=\sum_{n\ge0}A_nZ^n.
\tag{10.212}
\]

Then \(F\in\mathcal A_{p^\theta}\), its Gauss norm is one, and for
\(n\ge1\)

\[
v_p(A_n)\ge n\vartheta+a_p,\qquad
v_p(A_n)-n\theta\ge n(\vartheta-\theta)+a_p.
\tag{10.213}
\]

If \(v_p(z)\ge-\theta\), its divided jets satisfy

\[
\frac{F^{(j)}(z)}{j!}=F(z)\frac{\ell^j}{j!},\qquad
v_p\!\left(\frac{F^{(j)}(z)}{j!}\right)
\ge j\vartheta+a_p\quad(j\ge1).
\tag{10.214}
\]

If \(\ell=0\), the positive-order coefficients and jets are zero. The
constant coefficient and value have valuation zero.

\textbf{Proof.} The logarithm is an isometry at this depth, by
Proposition 9.3 of the linked preceding lesson. Thus each logarithm has
valuation at least \(\vartheta+a_p\), and the integral multipliers and
ultrametric inequality give the same bound for \(\ell\). If \(\ell=0\),
the assertions follow from \(F=1\). Otherwise \(A_n=\ell^n/n!\). The
complete factorial digit formula (9.4) gives
\(v_p(n!)=(n-s_p(n))a_p\le(n-1)a_p\). This proves the first bound in
(10.213); subtracting \(n\theta\) proves the second. It tends to
infinity, so the series belongs to the complete coefficient algebra
\(\mathcal A_{p^\theta}\). The constant coefficient is one and every
other weighted coefficient has positive valuation, proving that the
Gauss norm is exactly one.

For the asserted nodes, \(v_p(z\ell)\ge\vartheta-\theta+a_p>a_p\). The
exponential converges there and is a unit, by Proposition 9.3. The
coefficient algebra and analytic identities of Lemma 9.2 and Proposition
9.4 justify differentiation and translation on this disc; alternatively
the coefficient identity for each divided derivative is verified
directly by the convergent exponential series. It gives (10.214), with
the same factorial estimate. \(\square\)

This applies to the integer exponent vectors of every torus monomial,
including negative coordinates, and to exponents divided by any power of
\(q\ne p\). Such scalars are p-adically integral. Finite sums of these
exponential series with locally integral coefficients have Gauss norm at
most one. An additional additive polynomial still contributes its own
coefficient norm; (10.213) does not discard that contribution.

The strict choice of \(\theta\) cannot generally be removed. If
\(v_p(\ell)=\vartheta+a_p\), then the exact factorial formula gives
\(v_p(A_n)-n\vartheta=s_p(n)a_p\). At \(n=p^j\) this is the constant
\(a_p\). The weighted coefficients therefore fail to tend to zero at
radius \(p^\vartheta\). In this equality case the exponential does not
converge at a point of valuation \(-\vartheta\). Solution 36 supplies an
actual example.

\subsubsection{Residue phases and a coherent root
choice}\label{residue-phases-and-a-coherent-root-choice}

\textbf{Corollary 10.83 (principal roots of the phased coordinates).}
Let \(G=p^f-1\), let \(\zeta\in L\) be a primitive \(G\)-th root of
unity, and suppose that the residue of the local unit \(\alpha\) is the
residue of \(\zeta^a\). Such residue lifts, including their primitivity,
are proved in Lemma 10.53. Put

\[
\beta=\alpha\zeta^{-a},\qquad
V=\beta^P=\alpha^P\zeta^{-aP}.
\tag{10.215}
\]

For a prime \(q\ne p\), the element

\[
\eta=\exp(q^{-1}\log V)\in L,\qquad
\eta^q=V,\qquad
v_p(\eta-1)=v_p(V-1)\ge\vartheta+a_p
\tag{10.216}
\]

is the unique \(q\)-th root of \(V\) with residue one. Fix
\(\xi\in\mathbb C_p\) with \(\xi^q=\zeta\). A \(q\)-th root \(\rho\) of
\(\alpha\) can be chosen so that

\[
\rho^P\xi^{-aP}=\eta.
\tag{10.217}
\]

The choice may be made separately for every member of a finite family,
without changing any of their \(q\)-th power identities.

\textbf{Proof.} The residue condition gives \(v_p(\beta-1)>0\). Lemma
10.81 therefore proves the depth of \(V\). Since \(q\) is a local unit,
\(q^{-1}\log V\) has this same valuation; Proposition 9.4 gives both
\(\eta^q=V\) and the valuation identity. The series defining \(\eta\)
lies in the complete field \(L\). If two such roots had residue one,
their ratio would be a \(q\)-th root of unity with positive valuation of
its difference from one. The prime-to-\(p\) power identity in Lemma 9.5
forces that ratio to be one. This proves uniqueness, also when \(V=1\).

Start with any \(\rho_0^q=\alpha\) in \(\mathbb C_p\). Then
\(w=\rho_0^P\xi^{-aP}\) satisfies \(w^q=V\), so \(\eta/w\) is a \(q\)-th
root of unity. Fix a primitive one \(\omega\), and write
\(\eta/w=\omega^j\). Since \(P=p^\kappa\) is coprime to \(q\), integer
division gives an integer \(b\) with \(Pb\equiv j\pmod q\). Set
\(\rho=\rho_0\omega^b\). Its \(q\)-th power is still \(\alpha\), and its
\(P\)-th power supplies the required correction, proving (10.217). No
equation for another member of the family is affected. \(\square\)

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={Six exact guaranteed valuation paths for the original principal-unit power}]{../figures/principal-unit-depth.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.16. Start with the smallest possible positive valuation
\(1/e\) and iterate the guaranteed step \(c\mapsto\min\{pc,c+1\}\). The
points show lower bounds, not claims that every step is attained by a
unit. The horizontal line is the exact final depth
\(\vartheta+1/(p-1)\); its special unramified branch is retained at
\(p=5,e=1\). The zero-step case \(p=7,e=2\) is included. Lemma 10.81
proves all six bounds and all other ramification cases. Corollary 10.82
then permits every closed radius \(p^\theta\) with \(\theta<\vartheta\).
See Solution 36, \href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/principal_unit_depth.py}{the
figure program}, and
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
free paper}, equations (2.1), (4.7) and (4.11), for further reading.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{29. The uniform clearing and Euler
budget}\label{the-uniform-clearing-and-euler-budget}

The saturated coordinates explain the integer arguments of the Euler
factors. Their full \(q^\nu\) multiplier matters for ordinary absolute
values. Averaging the derivative orders then controls that multiplier
and the additive lcm cost together.

\subsubsection{Euler arguments in a saturated
basis}\label{euler-arguments-in-a-saturated-basis}

Let \(\beta_1,\ldots,\beta_n\in K^\times\), with \(n\ge2\) and positive
heights \(h_j=h(\beta_j)\). Let \(A_0=(a_{ij})\) be an \(r\)-by-\(n\)
integer matrix, \(2\le r\le n\), and suppose that the original
independent bases \(\alpha_i\) of Section 26 have the form
\(\alpha_i=\zeta_i\prod_j\beta_j^{a_{ij}}\), with roots of unity
\(\zeta_i\), and that

\[
\sum_{j=1}^n|a_{ij}|h_j\le\sigma_i,\qquad
R_j=\frac{|b_n|}{h_j}+\frac{|b_j|}{h_n},
\qquad R_{\max}=\max_{1\le j<n}R_j.
\tag{10.218}
\]

Here \(b_j\in\mathbb Z\) and \(b_n\ne0\). The product and inverse height
inequalities proved earlier give \(h(\alpha_i)\le\sigma_i\), as required
in Section 26.

Let \(B\) be a column-basis matrix of the q-primary saturation lattice
\(M\supseteq\mathbb Z^r\), of index \(J=q^\nu\). Suppose \(B\Lambda\)
lies in a translated closed box of full widths \(D_i\). Fix
\(\lambda_0\in\Lambda\), and write \(\mu=B\lambda,\mu_0=B\lambda_0\).
For any selected indices \(j<n\), put

\[
\omega_j(\lambda)
=J\sum_{i=1}^r(b_na_{ij}-b_ja_{in})(\mu_i-\mu_{0i}).
\tag{10.219}
\]

\textbf{Lemma 10.84 (the original integer Euler arguments).} These are
integer linear forms in \(\lambda-\lambda_0\). They are the eigenvalues
of \(\sum_i(b_na_{ij}-b_ja_{in})Y_i\partial/\partial Y_i\) on the
Laurent monomials \(Y^{J(\mu-\mu_0)}\). With the widths of (10.186),

\[
\Omega_j=\max_{\lambda\in\Lambda}|\omega_j(\lambda)|
\le J R_{\max}\sum_iD_i\sigma_i
=\frac{q^\nu\mathcal D R_{\max}}{c_1c_2Pd}.
\tag{10.220}
\]

\textbf{Proof.} The quotient \(M/\mathbb Z^r\) has order \(J\). The
cyclic subgroup generated by any class has order dividing \(J\), because
its distinct cosets partition this finite group into equal-sized sets.
Hence \(J\) annihilates every class, and \(JB\) has integer entries.
This proves integrality in (10.219). Acting on a Laurent monomial
multiplies it by its integer exponent in the differentiated coordinate,
which proves the eigenvalue identity, including negative exponents.

Both \(\mu\) and \(\mu_0\) lie in the same translated box, so
\(|\mu_i-\mu_{0i}|\le D_i\). No extra factor two is needed for its full
widths. Also \[
|b_na_{ij}-b_ja_{in}|
\le |b_n||a_{ij}|+|b_j||a_{in}|
\le R_j\sigma_i.
\] The last inequality uses separately \(|a_{ij}|h_j\le\sigma_i\) and
\(|a_{in}|h_n\le\sigma_i\), which follow from (10.218). Sum over \(i\),
and substitute \(D_i\sigma_i=\mathcal D/(c_1c_2rPd)\). This gives
(10.220). \(\square\)

The same proof divides the bound by \(q^I\) when the full original
widths have contracted by \(q^{-I}\). It keeps the original
denominator-clearing multiplier \(J\).

\subsubsection{The combined mean cost}\label{the-combined-mean-cost}

Use the full composition rows of Corollary 10.79, including duplicate
Euler rows, and let \(t_0=\lfloor T\rfloor\ge1\). Retain its
\(\overline u,\overline h\), and \(m=\sum_{j=1}^{r-1}\Omega_j+r-2\).
Then \(\overline h>0\).

\textbf{Lemma 10.85 (one budget for clearing and Euler factors).} With
\(\mathfrak v=\operatorname{lcm}(1,\ldots,k)\), put

\[
\begin{aligned}
\mathfrak a_{\rm av}
&=\max\left\{\ln\!\left[\mathrm e(1+m/\overline h)\right],
                   \frac12\ln\mathfrak v\right\},\\
E_{\rm mean}+\overline u\ln\mathfrak v
&\le t_0\mathfrak a_{\rm av}.
\end{aligned}
\tag{10.221}
\]

\textbf{Proof.} The exact composition means in (10.204) give \[
\overline h+2\overline u
=\frac{r-1}{r}t_0+\frac{r+1}{r}\overline u
\le t_0,
\] since \(\overline u\le t_0/(r+1)\). Formula (10.205) bounds the mean
Euler logarithm by \(\overline h\ln[\mathrm e(1+m/\overline h)]\); the
mean clearing logarithm is exactly \(\overline u\ln\mathfrak v\). Both
are bounded by their respective multipliers times
\(\mathfrak a_{\rm av}\). Adding proves (10.221). \(\square\)

For the original parameters, we can retain more information than the
maximum in this lemma. Define

\[
W(d)=\begin{cases}
\ln^3(3d),&d\ge2,\\
\dfrac{\ln6}{\ln2\,\ln3},&d=1,
\end{cases}
\qquad
g_{91}=1+\frac{1+\ln W(d)}{g_0}.
\tag{10.222}
\]

Thus \(g_{91}\) is \(1+(1+3\ln\ln(3d))/g_0\) for \(d\ge2\). Assume in
addition to Section 26 that

\[
h\ge\ln\!\left(\frac{R_{\max}}{dW(d)}\right),
\qquad c_0\ge19/10,\quad c_4\ge15/4,\quad
\widehat\vartheta\le2,
\qquad H_1=h+\nu\ln q.
\tag{10.223}
\]

The added height condition is exactly \(R_{\max}\le dW(d)\mathrm e^h\).
It is the coefficient-height condition associated with (10.218).

\textbf{Theorem 10.86 (the original uniform mean Euler budget).} Under
these assumptions,

\[
\begin{aligned}
\frac m{\overline h}
&\le\frac47\,\mathrm e^{H_1}tW(d)+\frac{r+1}{t_0},\\
\ln\!\left[\mathrm e(1+m/\overline h)\right]
&\le\left(g_{91}+\frac1{2(r-1)}\right)H_1,
\end{aligned}
\tag{10.224}
\]

and

\[
E_{\rm mean}+\overline u\ln\mathfrak v
\le g_{91}t_0H_1.
\tag{10.225}
\]

\textbf{Proof.} First \(T\ge g_4>9\). To verify the latter lower bound
directly, the factor defining \(g_4\) is at least \(g_1^{-1}\). Insert
\(g_3\) from (10.185), cancel \(c_1,g_1,t\), and use
\(q\ge2,r+1\ge3,\rho\le58,a_*\ge7/2\) and \(r!\le r^r\). This gives \[
g_4\ge\frac62\frac{2(19/10)(15/4)}{58}
                 \left(\frac72\right)^2
=\frac{8379}{928}>9.
\] Consequently \(t_0\ge1\) and \(T/t_0<2\).

We have \(\overline h\ge(r-1)t_0/(r+1)\). Sum (10.220), use (10.223),
and use \(T=q(r+1)\mathcal D/(c_1\theta et)\). Then \[
\frac m{\overline h}
\le\mathrm e^{H_1}\frac{T}{t_0}
       \frac{\theta e}{P}\frac{tW(d)}{c_2q}+
       \frac{(r+1)(r-2)}{(r-1)t_0}.
\] Indeed the factor \(d\) in \(R_{\max}\le dW(d)\mathrm e^h\) cancels
the global degree in (10.220). Formula (10.184) gives
\(\theta e/P\le1\), including its special unramified case, and
\(c_2q\ge7/2\). These inequalities prove the first line of (10.224).

Both branches have \(W(d)\ge1\). For \(d\ge2\), \(\ln(3d)\ge\ln6>1\).
For \(d=1\), use \(\ln6=\ln2+\ln3\) and \(\ln2<1\), so
\(W(d)=1/\ln2+1/\ln3>1\). Since \(t\ge\ln2>2/3\) and \(H_1\ge(r+1)t\),
we have \(r+1<3H_1/2\). Moreover \(H_1>39\) implies \[
1+\frac{3H_1}{2}<\frac3{14}H_1^2
\le\frac37\mathrm e^{H_1}.
\] The last inequality follows from the quadratic term of the positive
exponential series. This absorbs the \(1+(r+1)/t_0\) term in the first
line of (10.224), yielding \[
1+m/\overline h\le
\mathrm e^{H_1}W(d)\max\{t,1\}.
\] Take logarithms and add one. As \(H_1\ge g_0\), the constant
\(1+\ln W(d)\) is at most \((g_{91}-1)H_1\). The function \(\ln z/z\)
has maximum \(1/\mathrm e\) for \(z>0\), by differentiating. Hence
\(\ln\max\{t,1\}\le t/\mathrm e\le H_1/[\mathrm e(r+1)]\le H_1/[2(r-1)]\),
since \(\mathrm e>2\). This proves the second line of (10.224).

The complete lcm bound in
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/TR-BAKER-07.html\#an-exponential-bound-for-least-common-multiples}{Two
logarithms II: explicit lower bounds} gives
\(\ln\mathfrak v<k\ln3<(3/2)H_1\). Here \(k\le H_1\), and \(\ln3<3/2\)
follows already from \(1+3/2+(3/2)^2/2>3\) in the positive exponential
series.

Put \(a=g_{91}+1/[2(r-1)]\). We have bounded the combined mean by \[
H_1\left[\frac{r-1}{r}(t_0-\overline u)a+
                       \frac32\overline u\right].
\] This is affine in \(\overline u\), so its maximum on
\([0,t_0/(r+1)]\) occurs at an endpoint. At zero its coefficient of
\(t_0H_1\) is \((r-1)g_{91}/r+1/(2r)\le g_{91}\). At the other endpoint
that coefficient is \(((r-1)g_{91}+2)/(r+1)\le g_{91}\). Both
inequalities use \(g_{91}\ge1\). This proves (10.225). \(\square\)

\subsubsection{The full initial coefficient-height
bound}\label{the-full-initial-coefficient-height-bound}

\textbf{Corollary 10.87 (a uniform bound for all initial scalar rows).}
Assume also the hypotheses of Theorem 10.75 and Corollary 10.77. An
integral initial kernel with all required zeros can be chosen with mean
scalar logarithm at most

\[
\ln\mathcal B_{\rm av}
=kL\ln\!\left[2\mathrm e\left(\frac{s_0}{k}+2\right)\right]
                    +g_{91}TH_1.
\tag{10.226}
\]

In particular

\[
\frac{\ln\mathcal B_{\rm av}}Z
\le\frac{1+g_5^{-1}}{c_1c_4}
                    +\frac{g_{91}}{c_1c_3\theta e}.
\tag{10.227}
\]

With \(d_0\) the actual row-field degree, its coefficients satisfy

\[
\begin{aligned}
\frac{h_2(\boldsymbol c)}Z\le{}&
\frac12g_8+\frac{\ln d_0}{2g_7}
                         +\frac{d_0-1}{c_1c_2S}\\
&+\frac1{c_{01}-1}\left[
\frac12g_8+\frac{1+g_5^{-1}}{c_1c_4}
+\frac{g_{91}}{c_1c_3\theta e}
+\frac{1+1/(2g_2)}{2c_1c_2}\right].
\end{aligned}
\tag{10.228}
\]

\textbf{Proof.} At every additive order, the first factor of (10.198)
bounds its divided derivative before the lcm multiplier. Its logarithm
is independent of the derivative allocation, so it also bounds its
average. Theorem 10.86 bounds the mean lcm and Euler contributions
together by \(g_{91}t_0H_1\le g_{91}TH_1\). This proves (10.226).
Corollary 10.77 bounds its first normalized term. The exact parameter
formulas give \[
\frac{TH_1}{Z}=\frac1{c_1c_3\theta e},
\] proving (10.227).

Use the full composition rows and their actual ratio
\(\overline M/(N-\overline M)\) in (10.206). The column surplus (10.190)
bounds that ratio by \(1/(c_{01}-1)\). Insert (10.227),
\(\ln N/Z\le g_8\), \(Z\ge g_7\), and the exact mean-node bound from
Corollary 10.72. This is (10.228), with every scalar, logarithmic,
field-generator and torus cost retained. The row field and the integral
nonzero coefficient conclusion are those of Lemma 10.63. \(\square\)

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={The saturated-coordinate Euler arguments with their full clearing factor}]{../figures/mean-clearing-euler.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.17. Take the rational bases \(4,9\), their q-primary
saturated generators \(2,3\), and \(J=4\). The map \(B=\tfrac12I\) sends
the integer support on the left to its original coordinates in the upper
right panel. Retain the phase class
\(\lambda_1+\lambda_2\equiv0\pmod2\). With \((b_1,b_2)=(2,3)\), the
prepared Euler argument is \(6\lambda_1-4\lambda_2=4(3\mu_1-2\mu_2)\).
Its extrema at the reference \(\lambda_0=0\) are \(48\) and \(-24\),
marked at the corresponding vertices. The lower panel shows the exact
mean relation at \(r=2\): for \(a=\overline u/t_0\le1/3\), the
normalized Euler mean is \((1-a)/2\) and their weighted sum
\((1-a)/2+2a\) is at most one. Lemma 10.84 keeps the full clearing
factor and original widths; Lemma 10.85 and Theorem 10.86 control the
cost after averaging orders. The coordinates and values are
reconstructed in Solution 37.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/mean_clearing_euler.py}{Figure program}.
Human-source context:
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
free paper}, equations (2.8), (4.23) and (5.6).}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{30. The generator field and its uniform discriminant
cost}\label{the-generator-field-and-its-uniform-discriminant-cost}

The initial scalar bound is now uniform. We next control the field in
which its coefficients are chosen. A large entry of a saturation basis
can make its displayed generators have large heights while leaving the
field unchanged. Centering the basis removes that loss. We then use the
height-product theorem separately after each omission; the cancellation
in \(S\mathcal D/d\) keeps its bound independent of the saturation
index.

\subsubsection{A centered basis with small generator
heights}\label{a-centered-basis-with-small-generator-heights}

\textbf{Lemma 10.88 (a centered saturation basis).} Let \(r\ge2\), and
let \(M\supseteq\mathbb Z^r\) be a rational lattice of finite index
\(J\). Suppose its basis \(B\) corresponds to generators
\(\theta_1,\ldots,\theta_r\in K^\times\) in the independent original
bases \(\alpha_1,\ldots,\alpha_r\), up to roots of unity. Put
\(\sigma_{\max}=\max_i\sigma_i\), where \(h(\alpha_i)\le\sigma_i\).

There is a basis \(B'=BU\), with \(U\in\operatorname{GL}_r(\mathbb Z)\),
that is upper triangular and has the following alternatives for each
column:

\[
\begin{gathered}
B'_{ii}=1\quad\Longrightarrow\quad B'_i=e_i,\\
0<B'_{ii}<1\quad\Longrightarrow\quad
 B'_{ii}\le\tfrac12,\qquad
 |B'_{ji}|\le\tfrac12\quad(j<i),\qquad B'_{ji}=0\quad(j>i).
\end{gathered}
\tag{10.229}
\]

Choose \(\theta'_j=\prod_i\theta_i^{U_{ij}}\). Then

\[
\begin{aligned}
h(\theta'_j)&\le\sum_i|B'_{ij}|h(\alpha_i)
\le\frac r2\sigma_{\max},\\
E&=\mathbb Q(\alpha_0,\theta_1,\ldots,\theta_r)
=\mathbb Q(\alpha_0,\theta'_1,\ldots,\theta'_r),
\end{aligned}
\tag{10.230}
\]

for any root of unity \(\alpha_0\). This change preserves every
normalized monomial, phase class, original support width and integer
Euler argument after relabelling \(\lambda'=U^{-1}\lambda\).

\textbf{Proof.} The cyclic-subgroup coset argument in Lemma 10.84 shows
that \(J\) kills \(M/\mathbb Z^r\). Thus \(L=JM\) is an integer lattice
containing \(J\mathbb Z^r\). Let
\(L_i=L\cap\operatorname{span}(e_1,\ldots,e_i)\). Its i-th coordinate
image is a nonzero subgroup \(h_i\mathbb Z\) containing \(J\mathbb Z\).
Hence \(h_i\) is a positive divisor of \(J\).

Inductively choose \(v_i\in L_i\) whose i-th coordinate is \(h_i\). If
\(h_i=J\), choose \(v_i=Je_i\), which belongs to \(L\). Otherwise,
subtract integer multiples of \(v_{i-1},\ldots,v_1\), in that descending
order, to place coordinate \(j<i\) in the interval \([-h_j/2,h_j/2]\).
Ordinary integer division, with either consistent choice at a midpoint,
makes this possible. Subtracting \(v_j\) does not change a coordinate
larger than \(j\).

The vectors \(v_1,\ldots,v_i\) form a basis of \(L_i\). Indeed, for
\(x\in L_i\), subtract the multiple of \(v_i\) specified by its i-th
coordinate to obtain an element of \(L_{i-1}\), and use induction.
Reading the largest coordinate proves independence. Thus
\(B'=(v_1,\ldots,v_r)/J\) is a basis of \(M\). Its diagonal lies in
\((0,1]\); a proper divisor \(h_i<J\) is at most \(J/2\). Every centered
upper entry has absolute value at most \(h_j/(2J)\le1/2\). The special
choice at \(h_i=J\) gives the first alternative in (10.229).

Both \(B\) and \(B'\) are integer bases of the same lattice, so
expressing one in the other gives inverse integer matrices \(U,U^{-1}\).
The chosen new generators and their inverse products therefore generate
exactly the same field. Clear the denominators of \(B'\) and the finite
root-of-unity factors by a common positive power. The height product
inequality and exact power identity, already proved in the earlier
height lesson, give the first inequality of (10.230). A proper-diagonal
column has at most \(r\) entries of absolute value at most \(1/2\). A
unit column gives height at most \(\sigma_{\max}\le r\sigma_{\max}/2\).
This proves the second inequality.

Finally \(B'\lambda'=B\lambda\) and
\((\theta')^{\lambda'}=\theta^\lambda\). With the phase vector
transformed to \(d'=U^{\mathsf T}d\), we have
\(d'\cdot\lambda'=d\cdot\lambda\). Original-coordinate differences,
hence their full-width bounds and the integer Euler forms in (10.219),
are unchanged. Taking the same integer products of chosen q-th roots
also preserves their powers and generated root field; the inverse
products recover the old roots. The uniqueness assertion in Corollary
10.83 preserves the principal root choices. \(\square\)

The unit-column alternative matters. Centering only the off-diagonal
entries of a column with diagonal one would give the weaker
\((r+1)\sigma_{\max}/2\). Choosing the coordinate unit vector avoids
this extra half-height.

\subsubsection{Each omitted height product and the exact
scale}\label{each-omitted-height-product-and-the-exact-scale}

Retain all parameters in (10.183)--(10.186). Write \(w_K\) for the full
root-of-unity order, and define

\[
\begin{aligned}
g_{11}&=\frac{4q\mathrm e}{\rho}\,
c_0c_1c_3c_4a_*^r
\frac{(r+1)^{r+1}(r-1)^{r-1}}{(r!)^2}g_0g_1,\\
g_{12}&=\begin{cases}
g_1/(2g_7)+1/g_{11},&d\ge2,\\
0,&d=1.
\end{cases}
\end{aligned}
\tag{10.231}
\]

These are the original constants in Yu's equation (3.16);
\(\mathrm e=\exp(1)\) is distinct from the ramification index.

\textbf{Lemma 10.89 (the original omitted-product bound).} For
\(Z=S\mathcal D/d\) and every \(i=1,\ldots,r\),

\[
\begin{gathered}
d^r\ell_d\prod_{j\ne i}\sigma_j
\ge\frac{w_K}{\rho}
\frac{(r-1)^{r-1}}{(r-1)!\,\mathrm e^{r-1}},\\
\frac{2Z}{rd\sigma_i}\ge g_{11}.
\end{gathered}
\tag{10.232}
\]

\textbf{Proof.} Every subset of an independent multiplicative family is
independent: a relation in that subset would extend by zero exponents to
a relation in the full family. Apply the completely proved
height-product inequality (10.57) to the \(r-1\) original bases other
than \(\alpha_i\). Since \(\sigma_j\ge h(\alpha_j)\), replacing the
heights preserves its lower bound. At \(d\ge2\), multiply by
\(d^r\ell_d\) and use \(\ell_d\ge\ln d\). At \(d=1\), the proved
constant is \(17=\rho\). This proves the first line of (10.232) for each
omission separately.

For clarity the exact cancellation in the scale is

\[
\begin{aligned}
Z={}&\frac{c_3q(r+1)hA}{q^ut}
(1+\epsilon)(2+g_2^{-1})c_0c_1c_4
\left(\frac{c_2qPr(r+1)}{e\theta}\right)^r\frac1{r!}\\
&\quad{}\times
\max\left\{\frac{p^f}{\delta t^r},
\frac{\mathrm e^r}{r^r}t\right\}
d^{r+1}\ell_d\prod_j\sigma_j.
\end{aligned}
\tag{10.233}
\]

It follows by substituting \(S\) and \(\gamma\) in (10.186): the factors
\(q^\nu\), \(h+x\) and \(A+x\) cancel exactly. Use \(h\ge g_0\),
\(A\ge g_1\), \((1+\epsilon)(2+g_2^{-1})\ge2\),
\(c_2qP/(e\theta)\ge a_*\), and the second member of the maximum. We
obtain

\[
Z\ge
\frac{2q\mathrm e^r}{q^u}
c_0c_1c_3c_4a_*^r
\frac{(r+1)^{r+1}}{r!}g_0g_1
d^{r+1}\ell_d\prod_j\sigma_j.
\]

Divide by \(\sigma_i\), insert the first line of (10.232), and use
\(w_K\ge q^u\). Multiplication by \(2/(rd)\), together with
\(r(r-1)!=r!\), gives exactly \(g_{11}\) in (10.231). No bound on
\(q^\nu\) has been inferred from an omitted product. \(\square\)

\subsubsection{A uniform coefficient-field and initial height
bound}\label{a-uniform-coefficient-field-and-initial-height-bound}

\textbf{Theorem 10.90 (the original field cost).} Let
\(E=\mathbb Q(\alpha_0,\theta_1,\ldots,\theta_r)\subseteq K\) be the
generator field of the normalized initial rows, and let
\(d_E=[E:\mathbb Q]\). Then

\[
\frac{\ln|\Delta_E|}{2d_EZ}\le g_{12}.
\tag{10.234}
\]

Under the remaining hypotheses of Corollary 10.87, the initial rows have
a nonzero integral kernel \(c\in\mathcal O_E^N\subseteq\mathcal O_K^N\)
giving every original zero and the same nonzero-polynomial conclusion,
with

\[
\begin{aligned}
\frac{h_2(c)}Z\le{}&
g_{12}+\frac{g_8}{2}\\
&+\frac1{c_{01}-1}\left[
\frac{g_8}{2}
+\frac{1+g_5^{-1}}{c_1c_4}
+\frac{g_{91}}{c_1c_3\theta e}
+\frac{1+(2g_2)^{-1}}{2c_1c_2}\right].
\end{aligned}
\tag{10.235}
\]

\textbf{Proof.} At \(d=1\), \(E=\mathbb Q\) and its discriminant is one.
Suppose \(d\ge2\), and use the centered generators of Lemma 10.88. Apply
the complete discriminant-and-generators argument (8.33) in
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/TR-BAKER-08.html\#the-discriminant-of-the-coefficient-field}{Linear
forms in many logarithms} to the tower obtained by first adjoining
\(\alpha_0\), then the \(\theta'_j\). The first height is zero. The
product of its nontrivial degrees is \(d_E\); induction on
\((a-1)+(b-1)\le ab-1\) bounds the sum of the degree decrements by
\(d_E-1\). Consequently

\[
\frac{\ln|\Delta_E|}{2d_E}
\le\frac{\ln d_E}{2}
+(d_E-1)\frac r2\sigma_{\max}.
\tag{10.236}
\]

Choose an index attaining \(\sigma_{\max}\). Lemma 10.89 gives
\(r\sigma_{\max}/2\le Z/(dg_{11})\). Since \(d_E\le d\), the second term
in (10.236), divided by \(Z\), is at most
\((d_E-1)/(dg_{11})<1/g_{11}\). Also \(Z\ge g_7\) and
\(\ln d_E\le\ln d<g_1\), so its first term is at most \(g_1/(2g_7)\).
This proves (10.234), with the original \(g_{12}\).

Every initial row entry is its integer prepared scalar times
\(\alpha_0^{sw_\lambda}\theta^{Ps(\lambda-\lambda_0)}\), by (10.151),
hence belongs to \(E\). Apply the complete weighted integral Siegel
proof (8.40) over this field, allowing the full composition rows and
retaining zero or duplicate rows as in Corollary 10.79. Their normalized
heights do not change upon restriction to \(E\), by the field-invariance
argument in Theorem 10.59. The resulting bound is the original row-sum
inequality with discriminant term \(\ln|\Delta_E|/(2d_E)\).

Insert (10.234), \(\ln N/Z\le g_8\), the exact row ratio bounded by
\(1/(c_{01}-1)\), the mean scalar bound (10.227), and the mean-node
torus term in Corollary 10.72. This is (10.235). The vector is integral
over \(E\), hence over \(K\); it is nonzero, and Lemma 10.29 gives the
nonzero Laurent polynomial. The representative-row identities used in
Theorem 10.55 supply every original jet zero. \(\square\)

The field bound uses the same \(g_{12}\) as the original argument.
Equation (10.235) retains the complete mean scalar budget proved with
\(\operatorname{lcm}(1,\ldots,k)<3^k\). A later individual prepared
value still has its own scalar, denominator and field costs; averaging
the initial rows does not change that value.

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={An integer change of saturation basis preserves the field and original support while reducing generator heights}]{../figures/initial-field-budget.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.18. Over \(K=\mathbb Q(\sqrt5)\), the original bases
\(5,7\) have saturation lattice \((\tfrac12\mathbb Z)\times\mathbb Z\)
and index two. The displayed basis
\(B=\left(\begin{smallmatrix}1/2&5\\0&1\end{smallmatrix}\right)\) has
generators \(\sqrt5,5^5\cdot7\). The integer matrix
\(U=\left(\begin{smallmatrix}1&-10\\0&1\end{smallmatrix}\right)\) gives
\(B'=BU=\operatorname{diag}(1/2,1)\) and generators \(\sqrt5,7\). The
sampled lattice points, blue for an even first coordinate and brown for
an odd one, are labelled by \(\lambda'\), with original coordinates
\(B'\lambda'=(\lambda'_1/2,\lambda'_2)\). In the old basis they have
labels \(U\lambda'\), producing exactly the same points and monomials.
The height bars are numerical renderings of the labelled exact
logarithms; the dashed line is the centered bound
\(r\sigma_{\max}/2=\ln7\). Both lists generate the same quadratic field
of discriminant five. Lemmas 10.88--10.89 and Theorem 10.90 prove the
basis, omitted-product and field-budget mechanisms; Solution 38 checks
this example and the normalized original constants.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/initial_field_budget.py}{Figure program}.
Human-source context:
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
free paper}, equations (3.16), (3.23) and Lemma 4.2;
\href{https://www.impan.pl/shop/en/publication/transaction/download/product/82907}{Loher
and Masser's free paper}, Theorem 3, whose complete height-product proof
is given in Section 9.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{31. Individual scalar values and a complete integer
extension}\label{individual-scalar-values-and-a-complete-integer-extension}

The initial coefficient estimate averages its rows. An individual value
in the induction has its own additive order, ordinary absolute value and
denominator. We retain these costs together with the actual derivative
precision. At the end of this section, all three integer extensions of
the original rational rank-two parameter example are verified uniformly
over its permitted coefficient family.

\subsubsection{The actual projected slopes and the additive
order}\label{the-actual-projected-slopes-and-the-additive-order}

\textbf{Lemma 10.91 (normality at the original logarithmic slopes).} Let
\(q\ne p\), \(\nu\ge0\), \(A_0=(a_{ij})\) an integer \(r\)-by-\(n\)
matrix, and \(b_1,\ldots,b_n\in\mathbb Z\) with \(b_n\ne0\) and
\(v_p(b_j)\ge v_p(b_n)\). Suppose \(z_j\in\mathbb C_p\) satisfy
\(v_p(z_j)\ge\vartheta+1/(p-1)\), and \(0<\theta<\vartheta\). Put

\[
\begin{aligned}
u_i&=q^{-\nu}\sum_{j=1}^na_{ij}z_j,&
\mathcal L&=\sum_jb_jz_j,\\
w_i&=u_i-\frac{q^{-\nu}a_{in}}{b_n}\mathcal L
=q^{-\nu}\sum_{j<n}
\left(a_{ij}-\frac{b_j}{b_n}a_{in}\right)z_j.
\end{aligned}
\tag{10.237}
\]

Every \(u_i,w_i\) has valuation at least
\(\vartheta+1/(p-1)>\theta+1/(p-1)\), with the convention
\(v_p(0)=\infty\). Thus the normality hypotheses of Proposition 10.37
hold for the original and projected curves.

More precisely, retain its prepared family, \(\rho,c_*,\delta,k,L\), and
let \(D_{\mathrm a}=kL\),
\(\mathfrak v=\operatorname{lcm}(1,\ldots,k)\). For a fixed additive
order \(u=t_0\le D_{\mathrm a}\), the following function is normal:

\[
\begin{aligned}
F_u(Z;\boldsymbol t)
&=\frac{\rho^{D_{\mathrm a}-u}(k!)^L}
{c_*\mathfrak v^u}\,f(Z/\rho;\boldsymbol t),\\
G_u&=(D_{\mathrm a}-u)\theta+Lv_p(k!)
-u v_p(\mathfrak v),\\
v_p\!\left(\frac{\rho^{D_{\mathrm a}-u}(k!)^L}
{c_*\mathfrak v^u}\right)&=G_u-\delta .
\end{aligned}
\tag{10.238}
\]

Its ordinary divided jets obey the chain-rule identity (10.107) with
scale \(G_u-\delta\). For \(u>D_{\mathrm a}\), every prepared value is
zero.

\textbf{Proof.} The fractions \(b_j/b_n\) are p-integral by their
valuation hypothesis; \(q^{-\nu}\) is a local unit. Every displayed
linear combination therefore has the asserted lower valuation, by the
ultrametric inequality. In particular the projected slope does not lose
\(v_p(b_n)\): the factors of \(b_n\) have already canceled in (10.237).
The principal-unit depth proof in Lemma 10.81 and its logarithmic
consequence in Corollary 10.82 supply these \(z_j\) when the original
local units are used.

For the scale, the additive factor of a term is the coefficient of
\(V^u\) in \[
\frac1{(k!)^\ell}
\prod_{a=1}^k(q^{-I}Z/\rho+\lambda_{-1}+a+\mathfrak v V)^\ell,
\qquad 1\le\ell\le L,\quad 0\le\lambda_{-1}<k.
\] Factoring \(\mathfrak v^u\) from that coefficient leaves an
integer-coefficient polynomial in \(q^{-I}Z/\rho\), divided by
\((k!)^\ell\), of degree at most \(k\ell-u\). Multiplication by the
factor in (10.238) makes every coefficient p-integral: \(q\) is a local
unit, \(k\ell-u\le D_{\mathrm a}-u\), and \(\ell\le L\). The ratios
\(c_j/c_*\) are p-integral and the integer Euler factors introduce no
denominator.

Every exponential factor has argument \(\beta Z/\rho\), where \(\beta\)
is an integer linear combination of the \(w_i\). Its positive-degree
coefficients tend to zero and have positive valuation, by Corollary
10.82. Its Gauss norm is one. Multiplication by the normalized
polynomial, followed by the finite sum over the support, proves
normality through Lemma 9.2. The chain rule proves the jet identity. The
degree bound proves the final zero assertion. \(\square\)

The scale in (10.238) is allowed to have negative valuation; the
prepared derivative already has the compensating local divisibility. The
order-independent scale \(G_0-\delta\) remains valid, as proved in
Proposition 10.37. We use that common scale in the numerical extension
below.

\subsubsection{A factorial-preserving pointwise
majorant}\label{a-factorial-preserving-pointwise-majorant}

Take a support at contraction depth \(I\), and let \(\Omega_j\) be
nonnegative integer bounds for its \(r-1\) integer Euler arguments. For
\(|x|\le X\), define

\[
R_I=q^{-I}X+2k-1,\qquad
m_I=\sum_{j=1}^{r-1}\Omega_j+r-2,\qquad D_{\mathrm a}=kL.
\tag{10.239}
\]

\textbf{Theorem 10.92 (the individual scalar and denominator budget).}
For a prepared index with additive order \(u\le D_{\mathrm a}\) and
Euler order \(h=\sum_{j>0}t_j\), its rational scalar has ordinary
absolute value at most

\[
\mathcal H_{I,u,h}(X)=
\frac{R_I^{D_{\mathrm a}-u}}{(k!)^L}
\mathfrak v^u\binom{D_{\mathrm a}}u
\binom{m_I+h}{h}.
\tag{10.240}
\]

Uniformly through total order \(T'\), its scalar bound is

\[
\mathcal H_I^{\rm fine}(X,T')=
\frac{R_I^{D_{\mathrm a}}}{(k!)^L}
\max_{0\le u\le\min(D_{\mathrm a},T')}
\left\{\left(\frac{\mathfrak v}{R_I}\right)^u
\binom{D_{\mathrm a}}u\binom{m_I+T'-u}{T'-u}\right\}.
\tag{10.241}
\]

A maximizing index is the first one after all consecutive increasing or
constant steps. For \(u<\min(D_{\mathrm a},T')\), their exact test is

\[
\mathfrak v(D_{\mathrm a}-u)(T'-u)
\ge R_I(u+1)(m_I+T'-u).
\tag{10.242}
\]

At \(x=s/q^J\), \(s\in\mathbb Z\), all scalar denominators divide
\(q^{\Xi_{I,J,u}}\), where

\[
\Xi_{I,J,u}=
\begin{cases}
0,&I+J=0,\\
\max\{0,(I+J)(D_{\mathrm a}-u)+Lv_q(k!)-u v_q(\mathfrak v)\},
&I+J>0.
\end{cases}
\tag{10.243}
\]

Suppose the normalized torus support has original full widths
\(q^{-I}D_i\), \(h(\alpha_i)\le\sigma_i\), and the phase hypotheses of
Theorem 10.55. Let the actual algebraic value \(V\) belong to a number
field \(F\), with degree \(d_F\) and chosen local degrees \(e_F,f_F\).
Its torus monomials are local units and arise from the consistent roots
at \(x=s/q^J\). If \(V\ne0\), then

\[
\begin{aligned}
v_p(V)-\delta\le\frac{d_F}{e_Ff_F\ln p}\biggl[
&h_2(c)+\tfrac12\ln N+\ln\mathcal H_{I,u,h}(X)\\
&+\Xi_{I,J,u}\ln q
+P Xq^{-I}\sum_iD_i\sigma_i\biggr].
\end{aligned}
\tag{10.244}
\]

The field factor is retained even when these roots lie in the old
completion.

\textbf{Proof.} Differentiate the product of \(k\ell\) linear factors at
\(q^{-I}x+\lambda_{-1}\), dividing by \(u!\). Choosing the \(u\)
differentiated factors gives \(\binom{k\ell}u\) terms. Every remaining
factor has ordinary absolute value at most \(R_I\), since
\(\lambda_{-1}\le k-1\) and \(a\le k\). Multiply by \(\mathfrak v^u\),
retaining the denominator \((k!)^\ell\).

If \(u>k\ell\), the derivative is zero. Otherwise
\(\binom{k\ell}u\le\binom{D_{\mathrm a}}u\) and \[
\frac{R_I^{k\ell-u}}{(k!)^\ell}
=R_I^{-u}\left(\frac{R_I^k}{k!}\right)^\ell
\le\frac{R_I^{D_{\mathrm a}-u}}{(k!)^L},
\] because \(R_I\ge k\) makes the parenthesized base at least one. The
proved integer Euler generating-series estimate in Lemma 10.76 bounds
their product at order \(h\) by \(\binom{m_I+h}h\). This proves
(10.240). It increases with \(h\), so maximizing through \(u+h\le T'\)
gives (10.241).

The ratio of consecutive terms inside this maximum is \[
\frac{\mathfrak v}{R_I}
\frac{D_{\mathrm a}-u}{u+1}
\frac{T'-u}{m_I+T'-u}.
\] Both variable factors are nonincreasing, and the first strictly
decreases. At \(m_I=0\), the last factor is one wherever a step is
taken; the terminal value has Euler binomial one. Thus (10.242)
describes an initial interval of steps, including this endpoint case.
For rational \(R_I\), integer division after clearing its denominator
gives an exact binary search.

The scalar is integral at every prime other than \(q\), by the
fractional-node argument of Lemma 10.28. At \(q\), the polynomial
divided derivative has denominator dividing \((k!)^\ell\) and degree at
most \(k\ell-u\). Evaluation at \(q^{-I-J}s+\lambda_{-1}\) costs at most
\((I+J)(k\ell-u)\); its multiplier \(\mathfrak v^u\) recovers
\(u v_q(\mathfrak v)\). Taking \(\ell\le L\) proves (10.243), with the
globally integral \(I+J=0\) case stated separately.

Apply the completely proved nonzero-value product formula of Theorem
10.41 over \(F\), dividing by a coefficient of minimum valuation
\(\delta\). Cauchy--Schwarz gives \(h_2(c)+\frac12\ln N\). The scalar
bounds just proved give its archimedean and q-denominator terms. The
full-width support proof in Theorem 10.64, with contraction as in
Theorem 10.60, bounds the torus contribution by
\(PXq^{-I}\sum_iD_i\sigma_i\). Taking consistent roots divides heights
by \(q^J\), while the node numerator multiplies them by \(|s|\le q^JX\);
these two factors cancel. The phase factors have height zero. Summing
the place contributions and dividing by \(e_Ff_F\ln p\) gives (10.244).
\(\square\)

\subsubsection{All three first integer extensions in the rational
parameter
family}\label{all-three-first-integer-extensions-in-the-rational-parameter-family}

\textbf{Theorem 10.93 (a complete rank-two integer block).} Use the
original rational parameters and support of Solution 34, with bases
\(2,5\) at three. Let \(b_1,b_2\ne0\) be integers satisfying

\[
\begin{gathered}
v_3(b_2)\le v_3(b_1),\qquad
R_{\max}=\frac{|b_2|}{\ln2}+\frac{|b_1|}{\ln5}
\le W\exp(h),\qquad
W=\frac{\ln6}{\ln2\,\ln3},\quad h=g_0,\\
\eta=1-\frac{0.538}{3}=\frac{1231}{1500},
\qquad U=\frac{8Z}{\ln3},\qquad Z=S\mathcal D .
\end{gathered}
\tag{10.245}
\]

Choose the integral initial kernel of Theorem 10.90 for this support and
these Euler arguments. Assume \(v_3(2^{b_1}5^{b_2}-1)\ge U\). Then the
same prepared auxiliary family, without contraction of its support,
satisfies

\[
\Phi_0(s;\boldsymbol t)=0
\quad\text{for }|s|\le\lfloor2^jS\rfloor,\quad
|\boldsymbol t|\le\lfloor\eta^jT\rfloor,\qquad j=1,2,3.
\tag{10.246}
\]

The allowed coefficient family in (10.245), rather than one chosen pair
\(b_1,b_2\), is covered by these comparisons.

\textbf{Proof.} Put \(\beta_1=-2,\beta_2=-5\), both in
\(1+3\mathbb Z_3\), and \(u_i=\log(\beta_i^3)\). Since
\(\beta_1^3-1=-9\) and \(\beta_2^3-1=-126\), the exact logarithm
valuation identity in Proposition 9.3 gives \(v_3(u_i)=2\). Here
\(\nu=0\), \(P=3\), and \(\theta<3/2\). The hypothesis on the
multiplicative difference forces \(b_1+b_2\) even by reduction modulo
three. Hence \(\mathcal L=b_1u_1+b_2u_2=3\log(2^{b_1}5^{b_2})\) has
valuation at least \(U\). The projected slopes are \(w_1=u_1\) and
\(w_2=-(b_1/b_2)u_1\). Lemma 10.91 proves their enlarged-disc normality.
The single selected Euler direction is \(b_2E_1-b_1E_2\), so Lemma 10.38
supplies its actual ordinary jet precision.

The coefficient height (10.235) is uniform over (10.245), since this is
precisely the height condition of Theorem 10.86. It gives
\(h_2(c)/Z\le0.310817025826\). Also
\(k=50,L=35299,D_{\mathrm a}=1764950\), \(v_3(50!)=22\), and
\(v_3(\mathfrak v)=3\). Set \(G_0=D_{\mathrm a}\theta+22L\).

For the induction from \(j\) to \(j+1\), write \[
R_j=\lfloor2^jS\rfloor,\quad T_j=\lfloor\eta^jT\rfloor,\quad
n_j=2R_j+1,\quad \mu_j=T_j-T_{j+1}+1 .
\] Then \(|\boldsymbol t|+\mu_j-1\le T_j\) for every output index. The
initial case has every required zero by the kernel construction.

The height condition gives \[
v_3(b_2)\le\frac{h+\ln W+\ln(\ln2)}{\ln3}<47 .
\] This valuation is an integer, so it is at most46. Since every
separation cost \(B_j=\lfloor\log_3(2R_j)\rfloor\) below is at most7, we
may use \(M_0=46\) in (10.113). In particular \(U-v_3(b_2)>\theta+1/2\).
The exact input budgets in Figure 10.19 satisfy \[
U+G_0>n_j\mu_j\theta+\mu_jM_0,\qquad j=0,1,2.
\] Corollary 10.40 therefore supplies the analytic valuation excess
\((n_j\mu_j-D_{\mathrm a})\theta-22L\) at every output index.

For the arithmetic bound, Lemma 10.84 and (10.245) give the common
integer bound \[
\Omega\le
\left\lceil\frac{\mathcal D}{c_1c_2P}W\exp(h)\right\rceil
\le1313846807806555244846459051 .
\] Use this bound, \(\mathfrak v<3^{50}\), and Theorem 10.92 at
\(I=J=0\), \(X=R_{j+1}\), \(T'=T_{j+1}\). There is no scalar
denominator. The normalized torus values at integer nodes belong to
\(\mathbb Q\), and their full-width height divided by \(Z\) is at most
\(R_{j+1}/(c_1c_2S)\). Retain the coefficient term
\(h_2(c)+\frac12\ln N\), with \(\ln N/Z\le g_8\).

The following are outward certified bounds, all valuation excesses
divided by \(Z\). Their computation and every integer rounding are
detailed in Solution 39.

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}lrr@{}}
\toprule\noalign{}
Extension & Analytic lower bound & Arithmetic upper bound \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
\(j=0\) to \(1\) & 1.339613527059 & 1.269329770046 \\
\(j=1\) to \(2\) & 2.205396661115 & 1.946854665422 \\
\(j=2\) to \(3\) & 3.629068791040 & 3.349870794944 \\
\end{longtable}
}

Each comparison is strict. If a prepared output value were nonzero,
(10.244) would contradict its analytic lower bound. Indices of additive
order exceeding \(D_{\mathrm a}\) are already zero. This proves the
entire output range at each step, and induction proves (10.246).
\(\square\)

This proves the integer block for the stated original rank-two parameter
family. Fractional-node extraction, contracted supports, the other ranks
and primes, the second stopping branch and the final multiplicity
contradiction require their own comparisons.

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={Certified input precision and strict analytic versus arithmetic comparisons for three integer extensions}]{../figures/pointwise-precision-budget.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.19. The constants, floors and permitted coefficient
family are those of (10.245), with \(k=50,L=35299\) and the proved
clearing bound \(3^{50}\). Each upper panel pair compares the available
lower input budget \((U+G_0)/Z\) with the required upper budget
\((n_j\mu_j\theta+\mu_jM_0)/Z\). The lower stacks are certified upper
bounds from the coefficient, individual scalar and full-width torus
terms in (10.244); the black marks are the certified analytic lower
bounds. Their strict gaps force zero values. These are bounds rather
than sampled valuations of an auxiliary function. Lemma 10.91 proves
actual normality, Theorem 10.92 proves the full pointwise budget, and
Theorem 10.93 proves all three integer extensions. Solution 39
reconstructs every bound.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/pointwise_precision_budget.py}{Figure program}.
Human-source context:
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
free paper}, Lemma 5.2 and equations (5.25)--(5.38).}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{32. Nested integer jets and the first fractional
step}\label{nested-integer-jets-and-the-first-fractional-step}

Earlier integer extensions retain more derivatives at their inner nodes.
These derivatives can be used together. The following interpolation
argument keeps the full nested distribution rather than assigning its
smallest multiplicity to every node.

\subsubsection{Hermite interpolation on nested
intervals}\label{hermite-interpolation-on-nested-intervals}

\textbf{Lemma 10.94 (nested normal-series interpolation).} Let
\(0\le R_0<\cdots<R_m\) be integers and let
\(\mu_0\ge\cdots\ge\mu_m\ge1\) be integers. Put

\[
\begin{aligned}
E_j&=[-R_j,R_j]\cap\mathbb Z,&
\mu(s)&=\mu_j\quad(s\in E_j\setminus E_{j-1}),\\
E_{-1}&=\varnothing,&
N_*&=(2R_0+1)\mu_0+
2\sum_{j=1}^m(R_j-R_{j-1})\mu_j .
\end{aligned}
\tag{10.247}
\]

Choose \(v_p(\rho)=\theta>0\), a normal series \(F\), and \(C\ge0\). Set
\(B=\lfloor\log_p(2R_m)\rfloor\) when \(R_m\ge1\), and \(B=0\) when
\(R_m=0\). Suppose \(v_p(F_h(\rho s))+h\theta\ge\Lambda-hC\) for
\(s\in E_m\) and \(0\le h<\mu(s)\). Then

\[
v_p(F(\rho x))\ge
\min\{N_*\theta,\ \Lambda-(\mu_0-1)\max\{B,C\}\}
\qquad(x\in\mathbb Z_p\subset\mathbb Q_p).
\tag{10.248}
\]

\textbf{Proof.} For \(s\in E_m\), form the rational polynomial \[
L_s(X)=\prod_{\substack{t\in E_m\\t\ne s}}
\left(\frac{X-t}{s-t}\right)^{\mu(t)}.
\] It has value one at \(s\) and a zero of order \(\mu(t)\) at every
other node \(t\). We claim \[
v_p(L_s(x))\ge-(\mu_0-\mu(s))B\qquad(x\in\mathbb Z_p).
\] The claim holds at a node by its zero or unit value, so consider a
point that is not a node. Decompose the exponents as \[
\mu(t)=\mu_m+\sum_{j=0}^{m-1}(\mu_j-\mu_{j+1})\mathbf1_{t\in E_j}.
\] For a full integer interval, the counts in residue classes modulo
\(p^a\) differ by at most one. If \(s\in E_j\), the contribution of that
interval, omitting \(s\), at level \(a\) is \[
N_a(x)-N_a(s)+1-\mathbf1_{x\equiv s\pmod{p^a}}\ge0.
\] If \(s\notin E_j\), no node is omitted and the contribution is
\(N_a(x)-N_a(s)\ge-1\). For \(a>B\), every denominator difference
\(s-t\) has valuation less than \(a\), since \(0<|s-t|\le2R_m<p^a\);
those remaining contributions are nonnegative. The base interval \(E_m\)
always contains \(s\). Only increments belonging to intervals not
containing \(s\) can lose valuation, and their sum is \(\mu_0-\mu(s)\).
This proves the claim.

The formal Taylor inverse \[
A_s(U)=L_s(s+U)^{-1}
=\prod_{t\ne s}(1+U/(s-t))^{-\mu(t)}
\] has its degree-\(h\) coefficient of valuation at least \(-hB\). Each
factor has integer binomial coefficients, and each denominator
difference has valuation at most \(B\). Consequently the Hermite
polynomial is \[
H(\rho X)=\sum_{s\in E_m}L_s(X)
\sum_{h=0}^{\mu(s)-1}\rho^h F_h(\rho s)(X-s)^h
[A_s(X-s)]_{\le\mu(s)-1-h}.
\] Its degree is at most \(N_*-1\), and its divided jets at each node
agree with those of \(F\) through order \(\mu(s)-1\). A summand with
inverse coefficient of degree \(a\le\mu(s)-1-h\) has valuation at least
\[
\Lambda-(\mu_0-\mu(s))B-hC-aB
\ge\Lambda-(\mu_0-1)\max\{B,C\}.
\]

For completeness, the normal quotient argument of Lemma 10.33 applies to
the nonuniform root polynomial \(W(Z)=\prod_s(Z-\rho s)^{\mu(s)}\).
Indeed \(W=Z^{N_*}-A\), where every coefficient of \(A\) has positive
valuation. Write \(Q\) and \(R\) for quotient and remainder on division
by \(Z^{N_*}\). Both have Gauss operator norm at most one. Let \(M_A\)
mean multiplication by \(A\). The convergent series \[
G=\sum_{j\ge0}(Q\circ M_A)^jQF
\] is normal, because multiplication by \(A\) strictly decreases Gauss
norm and the normal-series algebra is complete by Lemma 9.2. With
\(H_0=R(F+AG)\), one has \(F=H_0+WG\), \(\deg H_0<N_*\), and \(H_0\) is
normal. Its jets are the required jets. Uniqueness of the polynomial
with those jets follows because a polynomial of degree less than \(N_*\)
divisible by every \((Z-\rho s)^{\mu(s)}\) is zero. Thus \(H_0=H\).

Finally \(v_p(W(\rho x))\ge N_*\theta\), and \(G(\rho x)\) is integral.
Combining this with the bound on \(H\) proves (10.248). \(\square\)

In the prepared-family setting, suppose the values at a node in the
\(j\)-th interval are zero through total order \(T_j\). For a desired
output order \(T'\), choose nonincreasing multiplicities
\(\mu_j\le T_j-T'+1\). Lemma 10.38 and the common normality scale of
Proposition 10.37 give the hypotheses of Lemma 10.94 with
\(\Lambda=U-\beta+G_0\) and \(C=C_0\). If \(M_0=\max\{B,C_0\}\), then
\(\beta\le M_0\), so the sufficient input inequality is

\[
U+G_0\ge N_*\theta+\mu_0M_0.
\tag{10.249}
\]

Under this inequality the prepared analytic valuation excess is at least
\(N_*\theta-G_0\) at every \(x\in\mathbb Z_p\), for every total prepared
order at most \(T'\). The comparison with the unprojected family uses
the same projection estimate as Corollary 10.40; no new precision
hypothesis is introduced.

\subsubsection{The original first fractional
range}\label{the-original-first-fractional-range}

\textbf{Theorem 10.95 (the first fractional step for the rational
family).} Under the hypotheses and with the initial family of Theorem
10.93, put \(S_1=\eta^{-3}S\), \(T_1^*=\eta^3T\). Then

\[
\Phi_0(s/2;\boldsymbol t)=0
\quad\text{for }\ |s|\le2(\lfloor S_1\rfloor+1)=1374,\qquad
|\boldsymbol t|\le\lfloor T_1^*\rfloor=993090.
\tag{10.250}
\]

This range uses the original unfloored \(S,T\), and the coefficient
family is the whole family (10.245).

\textbf{Proof.} Theorem 10.93 and the initial construction give these
nested integer data.

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}lrr@{}}
\toprule\noalign{}
\(j\) & Radius \(R_j\) & Available order \(T_j\) \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
0 & 379 & 1796752 \\
1 & 758 & 1474535 \\
2 & 1517 & 1210101 \\
3 & 3034 & 993090 \\
\end{longtable}
}

For output order \(T'=993090\), take \[
(\mu_0,\mu_1,\mu_2,\mu_3)=(780000,481446,217012,1).
\] Each multiplicity is at most the available order minus \(T'\) plus
one. The ring counts are \(759,758,1518,3034\), hence
\(N_*=1286383318\). Here \(B=7\), \(C_0\le46\), and \(M_0=46\). The
certified bounds are \[
\frac{N_*\theta+780000\cdot46}{Z}\le7.244899709935
<7.294535093395\le\frac{U+G_0}{Z}.
\] Thus (10.249) holds. The analytic valuation excess at every
fractional output argument, divided by \(Z\), is at least
\(7.100020502185\). Division by two preserves membership in
\(\mathbb Z_3\).

We must retain the field of these values. Put
\(\gamma_1=-8,\gamma_2=-125\) and \(\eta_i=\exp(\tfrac12\log\gamma_i)\).
Their logarithms have valuation two, so Lesson 9 proves that these are
square roots in \(\mathbb Q_3\), each congruent to one. Their global
field is \[
F=\mathbb Q(\eta_1,\eta_2)
=\mathbb Q(\sqrt{-2},\sqrt{-5}),\qquad [F:\mathbb Q]=4.
\] The two square classes are independent: \(-2,-5,10\) are all
nonsquares, as their prime valuations or signs show. Equivalently, if
\(\sqrt{-5}=a+b\sqrt{-2}\) with rational \(a,b\), squaring forces
\(ab=0\); \(b=0\) would make \(-5\) a rational square, and \(a=0\) would
make \(5/2\) a rational square. Both are impossible. This proves the
degree. The chosen completion of \(F\) is \(\mathbb Q_3\), since
\(F\subset\mathbb Q_3\) and \(\mathbb Q\) is dense there. Hence its
chosen \(e,f\) are one. Every normalized monomial at \(s/2\) is a
product of integer powers of \(\eta_1^s,\eta_2^s\), so every prepared
value belongs to this ambient field. Individual values may belong to a
proper subfield; the bound over \(F\) remains valid.

The outward parameter intervals give \[
686.229279178168\le S_1\le686.229279178169.
\] Thus \(X=687\) covers the entire output range. In Theorem 10.92 take
\(I=0,J=1,R=X+99=786\), the same common Euler bound of Theorem 10.93,
and \(T'=993090\). Its scalar maximum, using the proved upper clearing
bound \(3^{50}\), occurs at additive order \(470396\), leaving Euler
order \(522694\). The ratio tests are exact integer products. The
entropy estimates proved in Solution 39, with the negative factorial
logarithm retained, give
\(\ln\mathcal H_0^{\rm fine}/Z\le0.208379911543\).

Also \(v_2(50!)=25+12+6+3+1=47\). Equation (10.243) therefore gives the
common denominator exponent \[
\Xi_{0,1,u}\le D_{\mathrm a}+47L=3424003
\] for every additive order. This deliberately retains a valid common
upper bound; the order recovery can only improve it.

The four terms inside the arithmetic bracket of (10.244), divided by
\(Z\), have the following certified upper bounds.

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}lr@{}}
\toprule\noalign{}
Term & Upper bound \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
Coefficient height and \(\tfrac12\ln N\) & 0.310842088027 \\
Individual scalar logarithm & 0.208379911543 \\
q-denominator logarithm & 0.008748404912 \\
Original full-width torus height & 0.714102914710 \\
\end{longtable}
}

Multiplying their sum by the actual ambient global-to-local factor
\(4/\ln3\), using the unrounded rational intervals before the final
rounding, gives an arithmetic valuation excess at most
\(4.522335429896Z\). It is strictly smaller than the analytic lower
bound. Their gap exceeds \(2.577685072290Z\). Thus every nonzero
prepared value would contradict (10.244); orders above \(D_{\mathrm a}\)
are already zero. This proves (10.250). \(\square\)

This establishes the first fractional step for the stated original
rank-two family. The following section completes its finite contraction
chain and terminal contradiction; other parameter cases require separate
arguments.

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={Nested integer jets and certified first fractional precision budgets}]{../figures/nested-integer-jets.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.20. Each unit cell in the upper panel represents one
integer node, with constant height equal to its available multiplicity
\(\mu(s)\); cell edges are half-integers. The exact interval radii are
\(379,758,1517,3034\), and the four ring counts and multiplicities give
\(N_*=1286383318\). The lower panel shows certified bounds divided by
\(Z\): the input required is below the input available, and the
arithmetic upper bound is below the analytic lower bound. The arithmetic
cost includes the quartic ambient field and the full common
q-denominator exponent \(3424003\). These are proved bounds, not sampled
values of an auxiliary function. Lemma 10.94 proves the nested
interpolation; Theorem 10.95 proves the exact original first fractional
range; Solution 40 checks the Hermite basis and calculations.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/nested_integer_jets.py}{Figure program}.
Human-source context:
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
free paper}, Lemma 5.3.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{33. Seventeen contractions and a polynomial
contradiction}\label{seventeen-contractions-and-a-polynomial-contradiction}

The first fractional step can be repeated after restriction to an
exponent coset. Three features make a complete finite descent possible
for the family (10.245): the coefficients remain a subvector of the
initial vector, the actual Euler arguments shrink with the support, and
integer values return to the rational field. We first prove the
interpolation estimate that uses the odd nodes together with two full
intervals.

\subsubsection{Odd nodes inside nested full
intervals}\label{odd-nodes-inside-nested-full-intervals}

\textbf{Lemma 10.96 (mixed nested interpolation).} Let \(p\ne q\) be
primes and \(1\le A\le R\) integers. Put

\[
\begin{aligned}
E_0&=\{s\in\mathbb Z:|s|\le A,\ q\nmid s\},&
E_1&=[-A,A]\cap\mathbb Z,&E_2&=[-R,R]\cap\mathbb Z,\\
\mu(s)&=\mu_0\quad(s\in E_0),&
\mu(s)&=\mu_1\quad(s\in E_1\setminus E_0),&
\mu(s)&=\mu_2\quad(s\in E_2\setminus E_1),
\end{aligned}
\tag{10.251}
\]

where \(\mu_0\ge\mu_1\ge\mu_2\ge1\). Define \[
N_*=(2R+1)\mu_2+(2A+1)(\mu_1-\mu_2)
+|E_0|(\mu_0-\mu_1),\qquad
B=\lfloor\log_p(2R)\rfloor,\quad M=\max\{B,C\}.
\] Suppose \(F\) is normal, \(v_p(\rho)=\theta>0\), \(C\ge0\), and
\(v_p(F_h(\rho s))+h\theta\ge\Lambda-hC\) for \(0\le h<\mu(s)\). Then

\[
v_p(F(\rho x))\ge
\min\{N_*\theta,\ \Lambda-B(\mu_0-\mu_1)-(\mu_0-1)M\}
\quad(x\in\mathbb Z_p\subset\mathbb Q_p).
\tag{10.252}
\]

\textbf{Proof.} Write the multiplicity as
\(\mu_2+(\mu_1-\mu_2)\mathbf1_{E_1}+(\mu_0-\mu_1)\mathbf1_{E_0}\). For a
full integer interval, residue-class counts modulo \(p^a\) differ by at
most one. For \(E_0\), they differ by at most two: subtract the counts
of multiples of \(q\) from those of the full interval, and use that
multiplication by \(q\) permutes residue classes modulo \(p^a\).

Use the cardinal polynomial \[
L_s(X)=\prod_{\substack{t\in E_2\\t\ne s}}
\left(\frac{X-t}{s-t}\right)^{\mu(t)}
\] from the proof of Lemma 10.94. At a residue level \(a\le B\), an
interval containing \(s\) contributes
\(N_a(x)-N_a(s)+1-\mathbf1_{x\equiv s\ (p^a)}\). This is nonnegative for
a full interval and at least \(-1\) for the deleted interval \(E_0\). If
\(x\equiv s\), it is zero; otherwise the asserted bounds follow from the
respective count differences. A level not containing \(s\) contributes
at least \(-1\) for a full interval and \(-2\) for \(E_0\). For \(a>B\),
no denominator difference contributes and all remaining contributions
are nonnegative. Therefore \[
v_p(L_s(x))\ge
-B\bigl(\mu_0-\mu(s)+\mu_0-\mu_1\bigr).
\] At a node this inequality also follows from its zero or unit value.

The coefficient of \(U^a\) in \(L_s(s+U)^{-1}\) has valuation at least
\(-aB\), by the integer negative-binomial expansion proved in Lemma
10.94. In its Hermite formula, a term using the divided jet \(h<\mu(s)\)
and inverse coefficient \(a\le\mu(s)-1-h\) consequently has valuation at
least \[
\Lambda-B(\mu_0-\mu(s)+\mu_0-\mu_1)-hC-aB
\ge\Lambda-B(\mu_0-\mu_1)-(\mu_0-1)M.
\] The polynomial has degree less than \(N_*\) and the prescribed jets.
The monic normal quotient proof of Lemma 10.94 applies unchanged to
\(W(Z)=\prod_{s\in E_2}(Z-\rho s)^{\mu(s)}\); it gives \(F=H+WG\) with
\(G\) normal. Thus \(v_p(W(\rho x)G(\rho x))\ge N_*\theta\). This proves
(10.252). \(\square\)

For the prepared family, \(\Lambda=U-\beta+G_0\), \(C=C_0\), and
\(\beta\le M\). Hence a sufficient input condition is

\[
U+G_0\ge N_*\theta+\mu_0M+B(\mu_0-\mu_1).
\tag{10.253}
\]

It gives analytic valuation excess at least \(N_*\theta-G_0\), with the
same projection comparison as Corollary 10.40. The last term in (10.253)
is the cost of the odd-only data. It cannot be omitted by treating those
nodes as a full interval.

\subsubsection{What each contraction
preserves}\label{what-each-contraction-preserves}

\textbf{Proposition 10.97 (rational contraction invariants).} Start with
the support and integral coefficients of Theorem 10.93, and use
\(\gamma_1=-8,\gamma_2=-125\). Whenever the stage \(I\) family has exact
zeros at \(s/2\), for odd \(s\), through total prepared order \(O\), one
can choose a nonzero stage \(I+1\) family with exact integer zeros at
those same odd \(s\) through order \(O\). At every stage choose a
reference exponent carrying a coefficient of minimum three-adic
valuation. Then

\[
\begin{gathered}
\delta_I=\delta_0,\qquad h_2(c_I)\le h_2(c_0),\qquad N_I\le N_0,\\
\operatorname{width}_i(\mathcal M_I)\le D_i/2^I,\qquad
0\in\mathcal M_I,\qquad
|\omega(\boldsymbol n)|\le\Omega_I:=
\left\lfloor\frac{\Omega_0}{2^I}\right\rfloor,\\
\Omega_0=1313846807806555244846459051,\qquad
\omega(\boldsymbol n)=b_2n_1-b_1n_2.
\end{gathered}
\tag{10.254}
\]

Each family is nonzero, has additive degree at most
\(D_{\mathrm a}=1764950\), and has integral coefficients. Its normality
scale remains \(G_0-\delta_0\), with \(C_0\le46\). Integer prepared
values belong to \(\mathbb Q\); fractional values at \(s/2\) belong to
the fixed field \(F=\mathbb Q(\sqrt{-2},\sqrt{-5})\), whose
global-to-local factor at the chosen three-adic place is four.

\textbf{Proof.} Theorem 10.95 proves that
\(1,\eta_1,\eta_2,\eta_1\eta_2\) are independent over \(\mathbb Q\),
where \(\eta_i^2=\gamma_i\). For odd \(s\), reducing powers of
\(\eta_i^s\) modulo two multiplies these four basis elements by nonzero
rational factors. Thus an exact zero splits into four exact
exponent-coset zeros. Choose a coset containing a coefficient of
valuation \(\delta_0\), with exponent \(\boldsymbol m_*\), and set
\(\boldsymbol n=(\boldsymbol m-\boldsymbol m_*)/2\). Theorem 10.48
proves that the new coefficients are precisely the selected old
coefficients, with no mixing.

The old Euler argument is
\(2\omega(\boldsymbol n)+\omega(\boldsymbol m_*)\). Lemma 10.30 expands
its binomial polynomials in the new binomial basis. The triangular
matrix and its inverse lie in \(\mathbb Z[1/2]\), are three-integral,
and preserve total order. Their diagonal entries are powers of two. They
carry all extracted zero jets to all new zero jets without lowering
\(O\). The additive factor at the old argument \(s/2\) is exactly the
stage \(I+1\) factor at \(s\): \(2^I\mathfrak v\,\partial_X\) cancels
the scaling \(2^{-I}\) in each divided derivative. The factored root
monomial is a local unit.

The chosen subvector retains \(\delta_0\), integrality and the
projective-height bounds of Theorem 10.48. It contains a nonzero
coefficient, and the additive-basis independence of Lemma 10.29 proves
the new polynomial nonzero. Its support contains \(\boldsymbol n=0\),
and every coordinate width is divided by two. Induction gives (10.254).
Since zero is a reference inside the support, \[
|b_2n_1-b_1n_2|\le
2^{-I}(|b_2|D_1+|b_1|D_2)\le2^{-I}\Omega_0;
\] the left side is an integer.

The ordinary analytic torus uses the same \(\log\gamma_i\), the same
projected slopes and integer exponents \(\boldsymbol n\). The proof of
Lemma 10.91 therefore supplies the same normality scale. Powers of two
in additive differentiation are three-adic units, so the ordinary-jet
loss in Lemma 10.38 still uses \(C_0\le46\). At integers the torus is
\(\gamma_1^{sn_1}\gamma_2^{sn_2}\), hence rational. At \(s/2\) it is
\(\eta_1^{sn_1}\eta_2^{sn_2}\). No further roots are introduced when a
support is contracted: the torus bases remain \(\gamma_i\). The global
field is always the same quartic field proved in Theorem 10.95, with
chosen completion \(\mathbb Q_3\). This proves all the assertions.
\(\square\)

The inverse Euler change is three-integral, but need not be integral at
two. It does not change the coefficient vector of the new polynomial.
The height estimate below is applied directly to its new integer Euler
arguments; it does not charge that jet-change matrix as an extra
polynomial coefficient.

\subsubsection{All stages with their original radii and
orders}\label{all-stages-with-their-original-radii-and-orders}

\textbf{Theorem 10.98 (complete finite contraction chain).} Under the
contradiction hypothesis of Theorem 10.93, there are nonzero contracted
families satisfying Proposition 10.97 for \(I=1,\ldots,17\). Define from
the original real parameters

\[
\begin{aligned}
S_I&=\eta^{-3I}S,& T_I&=\eta^{3I}T,&
O_j&=\lfloor\eta^jT_I\rfloor\quad(0\le j\le3),\\
A_I&=2(\lfloor S_I\rfloor+1),&
R_I&=\lfloor4S_I\rfloor,&
X_I&=\lfloor S_I/\eta^3\rfloor+1 .
\end{aligned}
\tag{10.255}
\]

The stage \(I\) family vanishes at every odd integer \(|s|\le A_I\)
through order \(O_0\), at every integer \(|s|\le A_I\) through order
\(O_1\), at every integer \(|s|\le R_I\) through order \(O_2\), and at
every \(s/2\) with \(|s|\le2X_I\) through order \(O_3\).

\textbf{Proof.} Theorem 10.95 and Proposition 10.97 give the first
odd-node input. We give all estimates for the induction.

For closure of the odd-node set, let \(n=A_I\), \(\mu=O_0-O_1+1\),
\(B=\lfloor\log_3(2A_I)\rfloor\), and \(M=\max\{46,B\}\). The radius
\(A_I\) is even, so it contains exactly \(A_I\) odd nodes. The
deleted-node interpolation of Lemma 10.39, with its proved residue-count
loss, has sufficient input and analytic gain

\[
\begin{aligned}
\mathcal B_{I,0}&=n\mu\theta+\mu(B+M),&
\mathcal G_{I,0}&=n\mu\theta-G_0.
\end{aligned}
\tag{10.256}
\]

Indeed its Hermite loss is \(\mu B+(\mu-1)M\); adding \(\beta\le M\)
proves the stated sufficient bound. This fills the even nodes as well.
For extension from the full interval \(A_I\) to \(R_I\), take
\(n=2A_I+1\), \(\mu=O_1-O_2+1\), and the same \(B,M\). Full-interval
interpolation gives

\[
\mathcal B_{I,1}=n\mu\theta+\mu M,\qquad
\mathcal G_{I,1}=n\mu\theta-G_0.
\tag{10.257}
\]

For the next fractional step use Lemma 10.96 with \(q=2\), \(A=A_I\),
\(R=R_I\), and

\[
\begin{aligned}
\mu_0&=\lfloor780000\eta^{3I}\rfloor,&
\mu_1&=O_1-O_3,&\mu_2&=O_2-O_3,\\
N_*&=(2R_I+1)\mu_2+(2A_I+1)(\mu_1-\mu_2)
+A_I(\mu_0-\mu_1),\\
\mathcal B_{I,2}&=N_*\theta+\mu_0M+B(\mu_0-\mu_1),&
\mathcal G_{I,2}&=N_*\theta-G_0,
\end{aligned}
\tag{10.258}
\]

where now \(B=\lfloor\log_3(2R_I)\rfloor\). For all seventeen stages,
the exact floors give \(\mu_0\ge\mu_1\ge\mu_2\ge1\) and
\(\mu_0\le O_0-O_3+1\). The smaller choices \(\mu_1=O_1-O_3\),
\(\mu_2=O_2-O_3\) are valid: each is below the available order
difference plus one. They do not lower the output order \(O_3\). All
\(s/2\) in the asserted range belong to \(\mathbb Z_3\).

Here is the arithmetic estimate used for each output comparison. Let its
total order be \(O\), its absolute argument bound be \(X\), and put
\(J=0\) for integers, \(J=1\) for half-integers. Take respectively \[
(O,X,J)=(O_1,A_I,0),\ (O_2,R_I,0),\ (O_3,X_I,1).
\] Set \(r=X/2^I+99\), \(D=D_{\mathrm a}\), and \(V=3^{50}\). The
pointwise scalar bound in Theorem 10.92 is

\[
\mathcal H_I(O,X)=\max_{0\le u\le\min\{D,O\}}
\frac{r^{D-u}}{(50!)^L}\binom Du V^u
\binom{\Omega_I+O-u}{O-u}.
\tag{10.259}
\]

Keep the uniform denominator exponent \(\Xi_{I,J}=(I+J)D+47L\), which
bounds (10.243) at every order. Proposition 10.97 and Theorem 10.92 give
the nonzero-value upper bound

\[
\begin{aligned}
\mathcal A_{I,J}(O,X)
&=\frac{d_J}{\ln3}\Bigl(
0.310817025826Z+\tfrac12g_8Z+\ln\mathcal H_I(O,X)\\
&\hspace{18mm}+\Xi_{I,J}\ln2+
\frac{Z(X/2^I)}{1.4494\cdot1.75S}\Bigr),\\
d_0&=1,\qquad d_1=4.
\end{aligned}
\tag{10.260}
\]

The coefficient term is the uniform bound of Theorem 10.93; shrinking
the vector can only improve it. The last term is the original full-width
torus bound after \(I\) contractions. The common clearing term retains
\(v_2(50!)=47\). Integer values use the actual rational field; every
fractional value is bounded over the fixed quartic field. This explains
every term in (10.260).

The next table gives an upper bound for \(\max_j\mathcal B_{I,j}/Z\) and
a lower bound for the minimum of the three gaps
\((\mathcal G_{I,j}-\mathcal A_{I,j})/Z\), with the respective output
arguments above. The common available input is
\((U+G_0)/Z\ge7.294535093395\).

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}
  >{\raggedright\arraybackslash}p{(\linewidth - 4\tabcolsep) * \real{0.2727}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 4\tabcolsep) * \real{0.3636}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 4\tabcolsep) * \real{0.3636}}@{}}
\toprule\noalign{}
\begin{minipage}[b]{\linewidth}\raggedright
\(I\)
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
Maximum required input, upper bound
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
Minimum zero-forcing gap, lower bound
\end{minipage} \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
1 & 7.192833169904 & 0.240138949661 \\
2 & 7.156267657061 & 0.358632162785 \\
3 & 7.136548280236 & 0.444858960780 \\
4 & 7.125498620658 & 0.510791197596 \\
5 & 7.119172043534 & 0.563520072068 \\
6 & 7.115527327639 & 0.607407734375 \\
7 & 7.113058026427 & 0.644775387448 \\
8 & 7.112111071831 & 0.677281716283 \\
9 & 7.112117675512 & 0.705361309159 \\
10 & 7.109803325816 & 0.730342112918 \\
11 & 7.110367794024 & 0.754317954142 \\
12 & 7.105538214081 & 0.775918682856 \\
13 & 7.095336861199 & 0.798823134852 \\
14 & 7.111140057941 & 0.818899480846 \\
15 & 7.120519275568 & 0.839066755948 \\
16 & 7.021962525897 & 0.912928544835 \\
17 & 7.002483730960 & 0.985921533843 \\
\end{longtable}
}

These are outward rational bounds. Solution 41 gives their calculation
from the convergent series and exact integer ratio tests already proved
in Solutions 34 and 39. In particular the maximum in (10.259) is bounded
without floating-point exponentiation or constructing the enormous Euler
binomial. Both input and zero-forcing comparisons hold at every one of
the fifty-one entries.

Interpolation now gives excess at least \(\mathcal G_{I,j}\). If a
respective prepared value were nonzero, (10.260) would bound its excess
by a strictly smaller quantity. Thus each closure, extension and
fractional output is zero through its stated order. For odd output
numerators, Proposition 10.97 creates the next contracted family. Since
\(S_{I+1}=S_I/\eta^3\), \(T_{I+1}=\eta^3T_I\), its odd radius and order
are exactly \(2X_I=A_{I+1}\) and \(O_3=\lfloor T_{I+1}\rfloor\). This
proves the induction and all asserted ranges. \(\square\)

\subsubsection{The last support has one
exponent}\label{the-last-support-has-one-exponent}

\textbf{Corollary 10.99 (a complete valuation bound for the stated
family).} For every pair of nonzero integers \(b_1,b_2\) satisfying
(10.245),

\[
v_3(2^{b_1}5^{b_2}-1)<\frac{8Z}{\ln3}.
\tag{10.261}
\]

\textbf{Proof.} Suppose the reverse inequality holds. Theorem 10.98
gives a nonzero stage-seventeen family. The original parameter bounds of
Solution 34 are \(D_1<67805\), \(D_2<29202\). Both are smaller than
\(2^{17}=131072\), so (10.254) gives integer coordinate widths zero. The
support is nonempty and contains its reference zero, hence is exactly
\(\{0\}\). The polynomial is therefore

\[
Q_{17}(X)=\sum_{a,\ell}c_{17,a,\ell}
\Delta(2^{-17}X+a;50)^\ell,\qquad
0\ne Q_{17},\qquad \deg Q_{17}\le1764950.
\tag{10.262}
\]

At this stage \(A_{17}=18092274\), \(O_1=61\). The full integer closure
in Theorem 10.98 includes the prepared order \((0,0)\), so
\(Q_{17}(s)=0\) at all \(36184549\) distinct integers in that interval.
A nonzero degree-\(d\) polynomial has at most \(d\) distinct roots: at
each root polynomial division gives a factor \(X-s\), and division by
the preceding distinct factors leaves the next factor nonzero at the new
root. Induction proves the root bound. It contradicts (10.262). Also
\(2^{b_1}5^{b_2}\ne1\), since its two-adic valuation is the nonzero
integer \(b_1\). Thus no infinite-valuation exception occurs, and
(10.261) follows. \(\square\)

This proves a complete finite descent and contradiction for the entire
rational coefficient family (10.245). Other ranks, fields, primes and
parameter families require their own comparisons; this corollary does
not assert the general Yu estimate (2.9) or its variants.

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={Certified precision margins through seventeen contractions and the terminal support}]{../figures/first-contraction-budget.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.21. At stage \(I\), restriction to a minimum-coefficient
exponent coset sends \(\boldsymbol m\) to
\((\boldsymbol m-\boldsymbol m_*)/2\), with the forward and inverse
Euler changes preserving total jet order. The upper panels give
certified lower margins for all three comparisons at each stage; they
are bounds, not sampled valuations. Integer arithmetic uses
global-to-local factor one, fractional arithmetic factor four at every
stage. The lower panel shows the proved coordinate-width upper bounds
\(D_i/2^I\); the horizontal line is width one. At seventeen both integer
widths vanish, leaving the nonzero polynomial of degree at most1764950
with36184549 integer roots. Lemma10.96, Proposition10.97, Theorem10.98
and Corollary10.99 prove these mechanisms; Solution41 reconstructs the
bounds. \href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/first_contraction_budget.py}{Figure
program}. Human-source context:
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
free paper}, Lemmas5.3--5.4 and the support passage in Section5.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{34. Uniform degree comparisons at the geometric
stop}\label{uniform-degree-comparisons-at-the-geometric-stop}

The general descent has a geometric stopping depth, where the
multiplicity theorem produces fewer independent algebraic bases. Its
degree comparisons can be proved uniformly from the height-product bound
of Theorem 10.18. In particular, they require no additional
individual-height theorem.

We use the parameters (10.183)--(10.186), writing \(a=r+1\),
\(x=\nu\ln q\), and \(\ell_d=\max\{1,\ln d\}\). The independent bases
\(\alpha_i\) belong to the same degree-\(d\) field, and
\(\sigma_i\ge h(\alpha_i)>0\). Use the original choices of
\(\widehat\vartheta\) in (10.184), and \(0\le\tau\le1/100\). The fixed
case constants satisfy the following weaker rational bounds:

\[
\begin{gathered}
\frac{19}{10}\le c_0\le3,\qquad c_2\ge1,\qquad
c_3>\frac{47}{100},\qquad \frac{15}{4}<c_4<21,\\
a_*\ge\frac72,\qquad \rho\le58,\qquad
\theta\le\widehat\vartheta\le2,\qquad e\theta>\frac{49}{100},\\
\eta=1-\frac{c_5}{r+1}>\frac{18}{25},\qquad
1<q\eta^{r+1},\qquad q\eta^{2(r+1)}<1 .
\end{gathered}
\tag{10.263}
\]

The bound \(a_*\ge7/2\), rather than seven, includes the special
unramified case \(p\ge5,e=1\). For the geometric comparisons the
relevant ranges of \(c_5\) are \(0.5267\le c_5\le0.56\) at \(q=2\), and
\(0.753\le c_5\le0.827\) at \(q=3\). The other rational bounds are
direct comparisons with the fixed case constants. The last two
inequalities in (10.263), and all estimates used below, are proved
explicitly.

Let \(I\) be any nonnegative integer with

\[
\begin{gathered}
E_I=(q\eta^{r+1})^I>\exp(3(A+x)),\\
\overline S=q\eta^{-(r+1)I}S,\qquad
\overline T=\eta^{(r+1)I+1}T,\qquad
\mathsf D_0=kL,\qquad
\mathsf D_j=2q^{\nu-I}D_j\quad(1\le j\le r).
\end{gathered}
\tag{10.264}
\]

The original geometric stopping choice is
\(I=\lfloor3(A+x)/\ln(q\eta^{r+1})\rfloor+1\). The widths \(D_j\) and
their normalization are exactly those in (10.186). The positive real
\(\mathsf D_j\) are upper coordinate-degree bounds; the actual degrees
are integers. For the ordinary jet budget set
\(S^\circ=\lfloor\overline S\rfloor\),
\(T^\circ=\lfloor\overline T\rfloor\), and allocate

\[
S_0=\left\lfloor\frac{\overline S}{3}\right\rfloor,\qquad
S_j=\left\lfloor\frac{2\overline S}{3r}\right\rfloor\ (1\le j\le r),
\qquad
T_j=\left\lfloor\frac{\overline T}{r+1}\right\rfloor\ (0\le j\le r).
\tag{10.265}
\]

\textbf{Theorem 10.100 (all geometric-stop degree inequalities).} Under
these hypotheses, the allocations satisfy (10.67), and

\[
\mathsf D_0>20\mathsf D_j\quad(1\le j\le r),\qquad
\mathsf D_0>4\overline T,\qquad
\mathsf D_0>4r.
\tag{10.266}
\]

Consequently
\(\mathsf D_0>\max\{\mathsf D_1,\ldots,\mathsf D_r,T_r+r\}\). For every
\(1\le m<r\) and every \(m\)-element subset \(J\), define the ratios

\[
\begin{aligned}
\mathcal R_0&=\frac{(S_0+1)(T_0+1)}{\mathsf D_0},\\
\mathcal R_{m,J}
&=\frac{(S_m+1)\binom{T_m+m+1}{m+1}}
{(m+1)!\mathsf D_0\prod_{j\in J}\mathsf D_j},\\
\mathcal R_r
&=\frac{(S_r+1)\binom{T_r+r}{r}}
{(r+1)!\mathsf D_0\prod_{j=1}^r\mathsf D_j}.
\end{aligned}
\tag{10.267}
\]

They satisfy the uniform strict bounds

\[
\mathcal R_0>\frac{36}{25},\qquad
\mathcal R_{m,J}>8\quad(1\le m<r),\qquad
\mathcal R_r>31.
\tag{10.268}
\]

Thus every allocation and degree hypothesis of Theorem 10.25 holds, for
every original rank, field and prime parameter case.

\textbf{Proof.} We keep the ramification index \(e\) distinct from
\(\mathrm e=\exp(1)\). The power-series proof of the logarithm gives
\(\ln2>2/3\) and \(\ln3>1\). Hence \(t=f\ln p>2/3\), and \(A\ge g_1>5\).
Also \(2<\mathrm e<3\): the exponential series gives the lower bound,
while \(n!\ge2\cdot3^{n-2}\) for \(n\ge2\) bounds its tail by a
geometric series. For \(d\ge1\) one has \(\ell_d\le d\), since
\(\ln d<d\).

First check (10.263). The inequalities for \(c_5\) give \(\eta>18/25\).
For fixed \(0<c<3\), the function \((1-c/y)^y\) increases for \(y\ge3\).
Its logarithmic derivative is \(\ln(1-c/y)+(c/y)/(1-c/y)>0\): writing
\(z=c/y\), the derivative of \(\ln(1-z)+z/(1-z)\) is \(z/(1-z)^2>0\),
and its value at zero is zero. Therefore \(q\eta^{r+1}\) is bounded
below by \(2(61/75)^3>1\) at \(q=2\), or by \(3(2173/3000)^3>1\) at
\(q=3\). On the other hand, \(\ln(1-z)<-z\) gives
\(q\eta^{2(r+1)}<q\exp(-2c_5)<1\). For the last comparison it suffices
that \(\ln2<7/10\), \(\ln3<11/10\); the finite positive exponential sums
at \(7/10\) through degree four and at \(11/10\) through degree six
already exceed two and three, respectively. Finally \(e\vartheta\ge1/2\)
by (10.184), so \(e\theta\ge50/101>49/100\). The stated upper bounds for
\(\widehat\vartheta\) were proved there.

\subsubsection{Small uniform parameter
bounds}\label{small-uniform-parameter-bounds}

The two alternatives in \(g_2\) give \(g_2>1\). For the first use
\(c_3>47/100,q\ge2,r+1\ge3,d\ge1\). For the second, \(g_0>39,e\ge1\),
and \(p\in\{2,3,5\}\), with \(\ln5<2\). The latter follows from the
finite exponential sum at two through degree three.

Let \[
F_r=\frac{r^r(r+1)^r}{(r!)^2}.
\] Since \(r!\le r^r\), one has \(F_r\ge(1+1/r)^r>2\). The definitions
(10.185) consequently imply

\[
\begin{gathered}
g_5>
\frac{893}{7250}(r+1)\left(\frac72\right)^r>2r,\qquad
g_4>\frac{57}{116}(r+1)\left(\frac72\right)^r>r(r+1),\\
1+\epsilon<\exp(1/2)<2,\qquad
1+g_5^{-1}<\frac54.
\end{gathered}
\tag{10.269}
\]

Indeed \(g_5=(2c_0c_3q/\rho)(r+1)a_*^rF_r\). For its second inequality
the ratio \((r+1)(7/2)^r/r\) increases: its consecutive ratio is
\((7/2)r(r+2)/(r+1)^2\ge28/9\). At \(r=2\) the displayed lower bound is
\(131271/29000>4\). For \(g_4\), its extra factor in (10.185) is at
least \(g_1^{-1}\), and \(\widehat\vartheta\le2\). Thus
\(g_4>(qc_0c_4/\rho)(r+1)a_*^rF_r\), giving its displayed lower bound.
The function \((7/2)^r/r\) increases, and at two its comparison with
\(r(r+1)\) reduces to \(2793/928>1\). The bound on \(\epsilon\) follows
from its defining power and \(\ln(1+z)<z\). Since \(\mathrm e<3\), also
\(\exp(1/2)<2\).

Lemma 10.70 and \(k\ge39\) now give \(\mathsf D_0=kL>kg_5>78r\). They
also give the two useful bounds \[
\frac{S\mathcal D}{c_1c_4d(A+x)}
<\mathsf D_0
<\frac54\frac{S\mathcal D}{c_1c_4d(A+x)}.
\]

\subsubsection{Dominance of the additive
degree}\label{dominance-of-the-additive-degree}

For a non-torsion \(\alpha_i\), Theorem 10.18 with one base, its
saturation index at least one, and \(w_K\ge2\), gives \[
\sigma_i\ge\frac{2}{58\mathrm e\,d^2\ell_d},
\qquad d\sigma_i>\frac1{87d^2}.
\] At \(d=1\), weakening its proved constant17 to58 preserves the bound.
Thus no sharper individual-height result is needed.

Because \(q\eta^{2(r+1)}<1\), we have
\(\eta^{-(r+1)I}>E_I>\exp(3(A+x))\). Also \(q^{I-\nu}> \exp(3A)\). From
the definitions of the two degree bounds,

\[
\frac{\mathsf D_0}{\mathsf D_j}
>\frac{c_3c_2qr(r+1)P}{2c_4}\,
d\sigma_j\,\frac{h+x}{A+x}\,\frac{q^{I-\nu}}t.
\]

Here \(c_2P\ge1\), \((h+x)/(A+x)\ge39/(A+39)\), and \(t\le A\). Use
\(\exp(3A)\ge\exp(2g_1)\exp A
=\mathrm e^8(r+1)^2d^2\exp A\). The function \(\exp A/[A(A+39)]\)
increases for \(A\ge5\), by its logarithmic derivative. The ratio is
therefore larger than \[
\frac{47}{350}\,\frac{9\cdot256}{87}\,
\frac{39\cdot32}{5\cdot44}
=\frac{135143424}{6699000}>20.
\]

Similarly the additive degree divided by the real output order satisfies

\[
\frac{\mathsf D_0}{\overline T}
>\frac{c_3e\theta}{c_4}\frac{h+x}{A+x}
\eta^{-1}\eta^{-(r+1)I}
>\frac{47}{100}\frac{49}{100}\frac{39}{21\cdot44}\,2^{15}>4.
\]

The function \(\exp(3A)/(A+39)\) is increasing for \(A\ge5\); this
justifies its endpoint bound in the second line. Together with
\(\mathsf D_0>4r\), these inequalities imply
\(\mathsf D_0>2(\overline T+r)>T_r+r\).

\subsubsection{Every allocation
inequality}\label{every-allocation-inequality}

The floor definitions in (10.265) have integer sums at most
\(\overline S,\overline T\), hence at most \(S^\circ,T^\circ\). They are
nonincreasing because \(r\ge2\). Moreover \[
S_0+1>\overline S/3,\qquad
S_j+1>2\overline S/(3r),\qquad
T_j+1>\overline T/(r+1).
\] The binomial product gives \(\binom{T_j+b}{b}\ge(T_j+1)^b/b!\) for
every positive integer \(b\). These strict floor bounds retain all
integer losses.

Put \(\lambda=1+g_5^{-1}<5/4\). For \(m=0\), substitution and
cancellation give

\[
\mathcal R_0>
B_0:=\frac{q^2\eta c_4d(A+x)}{3\theta et\,\lambda}
>\frac{36}{25}.
\]

The last inequality uses \(d\ge e\), \(A+x\ge t\), \(q\ge2\),
\(\eta>18/25\), \(c_4>15/4\), and \(\theta\le2\).

For \(1\le m<r\), retain every subset \(J\). The same substitutions give
\[
\mathcal R_{m,J}>
\frac{2B_0}{r}\,
\frac{\prod_{j\in J}\sigma_j}{((m+1)!)^2}
\left(\frac{\eta c_2qPrd}{2e\theta t}
\frac{E_I}{q^\nu}\right)^m.
\] All these subfamilies are independent, so Theorem 10.18 applies
separately to each one: \[
\prod_{j\in J}\sigma_j
\ge\frac{m^m}{29m!\mathrm e^m d^{m+1}\ell_d}.
\] Furthermore \(E_I/q^\nu>\exp(3A)\), and

\[
\frac{\exp(3A)}t
>\frac{8192}{5}(r+1)^2d^2.
\]

Use \(c_2qP/(e\theta)\ge a_*\ge7/2\). The expression in parentheses is
larger than \(2048r(r+1)^2d^3\). Since \(\mathrm e<3\), \(m^m\ge m!\),
and \(d^{2m-1}\ge\ell_d\), dropping only positive factors gives \[
\mathcal R_{m,J}>
\frac{72}{725}\left(\frac{2048}{3}\right)^m
\frac{(r+1)^{2m}}{r((m+1)!)^2}
\ge\frac{72}{725}\frac{(2048/3)^m}{(m+1)^3}>8.
\] For the middle inequality use \((m+1)!\le(m+1)^{m+1}\), and the
increase of \((r+1)^{2m}/r\) for \(r\ge m+1\). The last expression
increases with \(m\ge1\): its consecutive ratio exceeds
\((2048/3)(2/3)^3>1\). At \(m=1\) it is \(147456/17400>8\).

For \(m=r\), use the complete formula for \(\mathcal D\) in (10.186).
Write \[
M=\max\{p^f/(\delta t^r),\,\mathrm e^rt/r^r\}.
\] Since \(\delta\ge1\), \(t>2/3\), \(p^f=\exp t\le\exp A\), and
\(r\ge2\), \(M<\exp A(3/2)^r\). The identities \[
q^{rI}\eta^{(r-1)(r+1)I}=q^I E_I^{r-1}>E_I^r,
\qquad q^{\nu+u}/\gamma\ge1
\] retain the saturation factors before they are bounded. Cancellation
of \(\mathcal D\) yields \[
\mathcal R_r>
\frac{8\eta^r\exp((3r-1)A)}
{135r(r+1)!\,[3(r+1)t]^r\ell_d}.
\] Indeed the remaining factor \(3\lambda(1+\epsilon)(2+g_2^{-1})c_0\)
is less than \(135/2\). Now \(t\le A\), \(\exp A/A>32/5\), and
\(\exp((2r-1)A)\ge[\mathrm e^4(r+1)d]^{2r-1}\). Using
\((r+1)!\le(r+1)^{r+1}\) gives \[
\mathcal R_r>
\frac{8(192/125)^r16^{2r-1}}{135r(r+1)^2}>31.
\] At \(r=2\) the last expression exceeds31. Its consecutive ratio is
\((192/125)16^2r(r+1)/(r+2)^2>1\), because \(r(r+1)/(r+2)^2\ge3/8\).
This proves all three parts of (10.268), hence all the hypotheses
asserted. \(\square\)

\subsubsection{The resulting bases remain in the original
field}\label{the-resulting-bases-remain-in-the-original-field}

\textbf{Corollary 10.101 (weighted rank reduction at the geometric
stop).} Suppose the contracted prepared family at the depth in (10.264)
vanishes at every integer \(|s|\le\overline S\) through total order
\(\overline T\). Suppose its independent torus bases and selected Euler
directions give the hyperplane \(W\) of Theorem 10.25, with nonzero
vector \(\boldsymbol B\). Then there are \(1\le m<r\) and independent
integral vectors \(u_1,\ldots,u_m\), whose rational span contains
\(\boldsymbol B\). Put \(\beta_i=\prod_j\alpha_j^{u_{ij}}\) and
\(R_i=\sum_j\sigma_j|u_{ij}|\). These are independent units in the
original field, \(h(\beta_i)\le R_i\), and

\[
(S_m+1)\binom{T_m+m}{m}\prod_{i=1}^m R_i
\le(m+1)!r^m\mathsf D_0
\left(\frac{2q^{\nu-I}\mathcal D}{c_1c_2rPd}\right)^m.
\tag{10.270}
\]

If the original multiplicative form satisfies
\(\Xi^{N_0}=\zeta\prod_j\alpha_j^{N_0B_j}\), with integral exponents and
a root of unity \(\zeta\in K\), then for some positive integers
\(N_1,w\) and integers \(c_i\),

\[
\Xi^{N_0N_1w}=\prod_{i=1}^m\beta_i^{N_0wc_i}.
\tag{10.271}
\]

\textbf{Proof.} Proposition 10.31 converts all prepared zero jets into
ordinary \(W\)-jets, and clears the Laurent monomial without changing
the cutoff or the additive degree. Its coordinate bounds are precisely
\(\mathsf D_0,\mathsf D_j\) in (10.264). Theorem 10.100 verifies every
degree and allocation hypothesis of Theorem 10.25. Apply that theorem
with weights \(A_j=\sigma_j\). If its first alternative (10.69) holds,
multiply by \((T_m+m)/m\le\mathsf D_0\); it implies the second
alternative (10.70). For that alternative use \(C_{r,m}\le m!r^m\) and
\(\sigma_j\mathsf D_j=2q^{\nu-I}\mathcal D/(c_1c_2rPd)\), independent of
\(j\). This proves (10.270).

The integer products \(\beta_i\) lie in \(K\) and are local units. Their
independence and height inequalities are exactly the elementary argument
of Corollary 10.26, applied to the original \(\alpha_j\). Clear
denominators in the rational-span identity to obtain
\(N_1\boldsymbol B=\sum_i c_i u_i\). Raise the assumed identity for
\(\Xi\) to \(N_1\), then to the order \(w\) of \(\zeta\). This proves
(10.271), without adjoining a new root field. \(\square\)

These degree comparisons apply uniformly to the geometric stopping
depth. Reaching that depth requires the integer and fractional
zero-production estimates with their actual global field and
derivative-precision costs. Once those zeros are available, (10.270)
gives the quantitative lower-rank output used in the general estimate.

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={Uniform geometric-stop degree dominance and multiplicity allocation margins}]{../figures/geometric-stop-bounds.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.22. The coordinate-degree, output-order and rank bounds
in (10.266) put the additive degree above every required degree
threshold. The allocation ratios in (10.268) have proved lower bounds
36/25, 8, 31, uniformly over every original rank and field case,
including \(a_*=7/2\). The plotted numbers are certified universal
bounds, not samples of parameter cases. Theorem 10.100 proves the
cancellations, floors and height-product comparisons; Corollary 10.101
keeps the new bases in the original field and gives their quantitative
height product. Solution 42 checks the limiting rank constants and
allocation algebra.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/geometric_stop_bounds.py}{Figure program}.
Human-source context:
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
free paper}, Section 6.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{35. The original height profile survives rank
reduction}\label{the-original-height-profile-survives-rank-reduction}

The new bases from Corollary 10.101 must satisfy the original
height-product allowance at their smaller rank. A small height product
by itself does not prove this: the allowance changes with both the rank
and the residue index. We retain that change exactly.

Keep all hypotheses and parameters of Theorem 10.100. For any
independent family \(\boldsymbol a\) of \(j\ge1\) local units in the
same field, let \(\delta(\boldsymbol a)\) be its residue index. At
\(j\ge2\) use the q-primary saturated residue index in (10.49). At
\(j=1\) use the actual residue order as in (10.76), without adjoining
the torsion generator to that order. Both are positive integers. Set

\[
\begin{aligned}
M_j(\boldsymbol a)
&=\max\left\{\frac{p^f}{\delta(\boldsymbol a)t^j},
                 \frac{\mathrm e^j}{j^j}t\right\},\\
H_*&=\frac{c_2qP}{e\theta},\qquad
\Gamma_n=\mathrm e H_*(n+1)d\quad(n\ge r).
\end{aligned}
\tag{10.272}
\]

Here \(\mathrm e=\exp(1)\), whereas \(e\) is the ramification index. The
field, \(\theta\), \(c_2\), \(P\), and \(n\) stay fixed when the rank
changes. In particular, the maximum \(M_r(\boldsymbol\alpha)\) is
exactly the one in \(\mathcal D\) in (10.186).

\textbf{Theorem 10.102 (uniform preservation of the original rank
profile).} Under the actual zero hypotheses of Corollary 10.101, let
\(m\), \(\boldsymbol\beta\), and \(R_i\) be its output. For every
\(n\ge r\),

\[
\prod_{i=1}^m R_i
<\frac1{700}\,
\Gamma_n^{r-m}
\frac{M_r(\boldsymbol\alpha)}{M_m(\boldsymbol\beta)}
\prod_{j=1}^r\sigma_j.
\tag{10.273}
\]

This preserves the original residue-index-dependent allowance, including
the different definition at rank one.

\textbf{Proof.} First, \(t>1\) in every original parameter case. At odd
primes this follows from \(t\ge\ln3>1\). At two, \(\zeta_3\in K\), and
the injectivity of the prime-to-two roots under reduction, proved before
(10.183), gives \(3\mid2^f-1\). Thus \(f\) is even and at least two, and
\(t\ge\ln4>1\).

For any \(j\ge1\) and any positive residue index,

\[
t^jM_j(\boldsymbol a)\le p^ft.
\]

Indeed \(\delta(\boldsymbol a)t\ge1\) gives the first part of the
maximum. For the second use \(\ln y\le y-1\) at \(y=t/j>0\): it gives
\((\mathrm e t/j)^j\le\exp t=p^f\). This proves the bound at every rank,
with its actual residue index.

Put \(\lambda=1+g_5^{-1}\), \(E=(q\eta^{r+1})^I\), and
\(C=\lambda(1+\epsilon)(2+g_2^{-1})c_0\). Theorem 10.100 gives
\(C<45/2\), \(\lambda<5/4\), and
\(\mathsf D_0<\lambda S\mathcal D/(c_1c_4d(A+x))\). In (10.270), bound
its denominator from below by

\[
\frac{2\overline S}{3r}
\frac{(\overline T/(r+1))^m}{m!}.
\]

Both lower bounds are strict because \(\lfloor y\rfloor+1>y\). Divide
the resulting upper bound for \(\prod_iR_i\) by
\(\Gamma_n^{r-m}(M_r/M_m)\prod_j\sigma_j\). Substitute the complete
\(\mathcal D,S,T,\overline S,\overline T\) formulas. Writing this
normalized upper bound as \(\mathcal Q\), the exact cancellation gives

\[
\begin{aligned}
\mathcal Q={}&\frac{3r}{2q}\,
\frac{(m+1)!m!}{r!}\,C\,
\frac{\gamma}{q^{\nu+u}}\,
\frac{r^r(r+1)^r}{\mathrm e^{r-m}(n+1)^{r-m}}\\
&\quad{}\times
\left(\frac{2t}{\eta}\right)^m
\ell_d M_m(\boldsymbol\beta)q^{m\nu}
\frac{\eta^{(r+1)I}}{E^m}.
\end{aligned}
\tag{10.274}
\]

The factors \(H_*\) and \(d\) cancel completely. The original \(M_r\)
cancels against the profile ratio; the new \(M_m\) remains. Neither
\((m+1)!\) nor \(m!\) has disappeared.

Monotonicity of \(\ln\) gives
\(\ln(r!)\ge\int_1^r\ln y\,dy=r\ln r-r+1\), hence
\(r^r/r!<\mathrm e^r\). Use \(n\ge r\), \(q\ge2\),
\(\gamma/q^{\nu+u}\le1\), and \(C<45/2\) in (10.274). The stopping
inequalities give

\[
q^{m\nu}\frac{\eta^{(r+1)I}}{E^m}
<\exp\bigl(-3(m+1)A-(2m+3)x\bigr)
\le\exp\bigl(-3(m+1)A\bigr).
\]

Combine this with \(t^mM_m\le p^ft\le\exp(A)A\), \(\mathrm e<3\), and
\(\eta>18/25\). It follows that

\[
\mathcal Q<\frac{135}{8}\,r(m+1)!m!
\left(\frac{25}{3}\right)^m(r+1)^m
\ell_d A\exp\bigl(-(3m+2)A\bigr).
\tag{10.275}
\]

This remaining inequality is uniform in the rank, field, prime,
ramification and residue degree. Set \(V=(r+1)d\), so \(g_1=4+\ln V\)
and \(A\ge g_1>5\). The function \(y\exp(-(3m+2)y)\) decreases for
\(y\ge5\). Also \(g_1\le5V\), since \(\ln V\le V\) and \(V\ge1\), and
\(\ell_d\le d\). Thus (10.275) is at most

\[
\frac{675}{8}\left(\frac{25}{3}\right)^m
\exp\bigl(-4(3m+2)\bigr)
\frac{r(m+1)!m!}{(r+1)^{2m+1}d^{3m}}.
\]

For fixed \(m\ge1\), \(r/(r+1)^{2m+1}\) decreases for \(r\ge1\), by its
logarithmic derivative. Since \(r\ge m+1\), and
\((m+1)!m!\le(m+1)^{2m+1}\), its last factorial-rank factor is smaller
than \(m+1\le2^m\). The latter inequality follows by induction from its
equality at one. Finally \(\mathrm e>2\) gives

\[
\mathcal Q<
\frac{675}{2048}\left(\frac{25}{6144}\right)^m
\le\frac{5625}{4194304}<\frac1{700}.
\tag{10.276}
\]

The final rational comparison is \(5625\cdot700<4194304\). All these
bounds hold for every \(1\le m<r\); they are not deductions from a
finite list of ranks. This proves (10.273). \(\square\)

\subsubsection{A genuine minimum-rank
contradiction}\label{a-genuine-minimum-rank-contradiction}

Let \(a_1,\ldots,a_n\) be independent local units in \(K\), and let
\(a_0\) generate its q-primary roots of unity. Fix
\(\mathcal L=\sum_jb_jz_j\ne0\). A rank-\(j\) representation consists of
integral forms \(L_0=z_0,L_1,\ldots,L_j\) independent over
\(\mathbb Q\), rationals \(B_0,\ldots,B_j\) with
\(\mathcal L=\sum_iB_iL_i\), and positive weights \(s_i\). If
\(L_i=\sum_{k=0}^na_{ik}z_k\), require

\[
\begin{gathered}
\alpha_i=a_0^{a_{i0}}\prod_{k=1}^na_k^{a_{ik}},\qquad
s_i\ge\sum_{k=1}^n|a_{ik}|h(a_k),\\
\prod_{i=1}^js_i\le
\Psi_j(\boldsymbol\alpha)\prod_{k=1}^nh(a_k),\qquad
\Psi_j(\boldsymbol\alpha)=
\Gamma_n^{n-j}\frac{M_n(\boldsymbol a)}{M_j(\boldsymbol\alpha)}.
\end{gathered}
\tag{10.277}
\]

The weighted condition implies \(h(\alpha_i)\le s_i\) by the height
product and power laws; the torsion factor has height zero. The bases
\(\alpha_i\) are independent: a torsion multiplicative relation, after
killing the finite torsion factor, would give a linear relation among
the rows of \(L_i\) modulo \(z_0\), contradicting independence with
\(L_0\). At \(j=n\), take \(L_i=z_i,s_i=h(a_i)\). Its profile equals
one, so a representation exists. There is therefore a least positive
admissible rank. After renumbering its forms one may assume \(B_j\ne0\).

\textbf{Corollary 10.103 (the geometric stop contradicts the original
minimum rank).} Suppose \(r\ge2\) is this least admissible rank, and the
parameters and actual prepared zeros satisfy Corollary 10.101 for its
bases and weights. Then those zeros cannot exist.

\textbf{Proof.} Take the integral vectors \(u_i\) of Corollary 10.101
and put \(L_i'=\sum_{j=1}^ru_{ij}L_j\). Together with \(L_0\), these
forms are independent: their projections modulo \(L_0\) are the
independent combinations of the independent projections of
\(L_1,\ldots,L_r\). Because \(\boldsymbol B\) belongs to the rational
span of the \(u_i\), \(\mathcal L=B_0L_0+\sum_ic_iL_i'\) for rationals
\(c_i\), not all zero. The bases associated with \(L_i'\) are precisely
\(\beta_i=\prod_j\alpha_j^{u_{ij}}\), in the same field and units at the
same prime. The triangle inequality for each coefficient gives

\[
\sum_{k=1}^n\left|\sum_{j=1}^ru_{ij}a_{jk}\right|h(a_k)
\le\sum_{j=1}^r|u_{ij}|s_j=R_i.
\]

Thus every required independence, representation and weighted-height
condition survives. Finally Theorem 10.102 and (10.277) give

\[
\prod_iR_i<\frac1{700}\,
\Gamma_n^{r-m}\frac{M_r}{M_m}\,
\Gamma_n^{n-r}\frac{M_n}{M_r}\prod_kh(a_k)
=\frac1{700}\Psi_m(\boldsymbol\beta)\prod_kh(a_k).
\]

So \(m<r\) is admissible with the original allowance, contrary to the
choice of \(r\). No enlargement of the original field occurs in this
argument. \(\square\)

\subsubsection{A rank-two example with a proved
gap}\label{a-rank-two-example-with-a-proved-gap}

Take \(K=\mathbb Q,p=3,(a_1,a_2)=(2,5)\), and
\(\mathcal L=101z_1+103z_2\). Here \(d=e=f=1\), \(P=3\),
\(H_*=7(1+\tau)\), and \(0\le\tau\le1/100\). The rational valuations at
two and five prove independence and show that the q-primary saturation
is already \(\mathbb Z^2\). The residue group generated by the torsion
root and the two bases has order two, giving
\(\delta(\boldsymbol a)=1\). The proved logarithm bounds
\(1<t=\ln3<11/10\), \(2/3<\ln2<7/10\), \(1<\ln5<2\), and
\(2<\mathrm e<3\) imply

\[
\Gamma_2<64,\qquad M_2(\boldsymbol a)<3,\qquad
M_1(\beta)\ge\mathrm e t>2,
\qquad
\Psi_1(\beta)h(2)h(5)<\frac{672}{5}.
\tag{10.278}
\]

Every possible rank-one form has non-torsion coefficients
\((a_{11},a_{12})=k(101,103)\), where \(k\) is a nonzero integer. Indeed
rational proportionality follows from the representation;
\(\gcd(101,103)=1\) makes its proportionality factor integral. Its
required weight is larger than \(101(2/3)+103=511/3>672/5\). Thus rank
one is impossible, whereas the standard rank-two representation is
admissible. This is a concrete instance of the minimum-rank
contradiction once the geometric-stop zeros are produced. It does not
assert those zero hypotheses without their precision proof.

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={The preserved rank-profile allowance and a rank-two weighted-height gap}]{../figures/rank-profile-contraction.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.23. The universal envelope in (10.276) decreases
geometrically for every smaller rank \(m\ge1\). The plotted first eight
values illustrate the proved formula; its validity for all ranks comes
from Theorem 10.102. The rank-two example has an admissible rank-one
weight below \(672/5\), while every rank-one representation needs a
weight above \(511/3\). Corollary 10.103 proves the general minimum-rank
contradiction, retaining the original residue-index profile and original
field. Solution 43 checks the complete cancellation and the example.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/rank_profile_contraction.py}{Figure program}.
Human-source context:
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
free paper}, (2.10) and (6.18)--(6.19).}
\end{flushleft}
\end{minipage}\par\medskip


The geometric branch now has its full degree, multiplicity and
minimum-rank closure under the actual prepared-zero hypotheses. The
general descent still requires those hypotheses to be proved with their
original integer and fractional precision budgets, and requires the
separate stopping branch when its first phase range ends earlier.

\subsection{36. Complete composition averages at the original general
parameters}\label{complete-composition-averages-at-the-original-general-parameters}

We keep the parameters of Section 26, including their full coordinate
widths, the original coefficient field and every field and prime case.
The derivative coordinates together with their slack form a complete
simplex. Retaining its zero rows makes the average of each coordinate
exact. This gives a smaller initial height budget while keeping the
original kernel and all its zeros.

Throughout this section use the following rational constants. In the
last column, \(\widehat\vartheta\) is the upper bound used in (10.185).

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}llrrrr@{}}
\toprule\noalign{}
Case & Conditions & \(c_0\) & \(c_1\) & \(c_3\) & \(c_4\) \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
I.1 & \(p=3,d\ge2\) & 2.66 & 1.449 & 1.4647 & 20.74 \\
I.2 & \(p=3,d=1\) & 1.9 & 1.4494 & 1.3852 & 20.8 \\
II & \(p=5,e\ge2\) & 2.74 & 1.4372 & 0.8412 & 19 \\
III.1 & \(p\ge5,e=1,d\ge2\) & 2.78 & 1.4341 & 2.992 & 18.7 \\
III.2 & \(p\ge5,e=1,d=1\) & 2.6 & 1.432 & 3.26 & 18.2757 \\
IV & \(p\ge7,e\ge2\) & 3 & 1.4441 & 3.849 & 20 \\
V & \(p=2,\zeta_3\in K\) & 2.5 & 2.5347 & 0.4757 & 3.765 \\
\end{longtable}
}

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}lrrr@{}}
\toprule\noalign{}
Case & \(a_*\) & \(q\) & \(\widehat\vartheta\) \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
I.1, I.2 & 7 & 2 & \(3/2\) \\
II & 7 & 2 & \(5/4\) \\
III.1, III.2 & \(7/2\) & 2 & 1 \\
IV & 7 & 2 & \(7/6\) \\
V & \(26/3\) & 3 & 2 \\
\end{longtable}
}

Every terminating decimal in this table denotes an exact rational
number. Put

\[
F_r=\frac{r^r(r+1)^r}{(r!)^2},\qquad
B_r=\frac{2c_0c_1c_4}{\rho}a_*^rF_r,\qquad
b_r=\frac{7200}{29}\,14^{r-2},\qquad Z=\frac{S\mathcal D}{d}.
\tag{10.279}
\]

\subsubsection{Coupled rounding and coefficient-count
bounds}\label{coupled-rounding-and-coefficient-count-bounds}

\textbf{Lemma 10.104 (uniform lower budgets in every original case).}
For every integer \(r\ge2\), the table and (10.183)--(10.186) give

\[
\begin{gathered}
t>1,\qquad B_r>b_r,\qquad
\mathcal D>B_r(A+x)t,\\
S>\frac75(r+1)^2d,\qquad g_2>58,\qquad g_5>188,\\
\lambda:=1+g_5^{-1}<\frac{189}{188},\qquad
1+\epsilon<\frac{251}{250}.
\end{gathered}
\tag{10.280}
\]

\textbf{Proof.} For odd \(p\), \(t=f\ln p\ge\ln3>1\). At \(p=2\), the
primitive cube root reduces to an element of order three, by the
injectivity proof preceding (10.183). Thus \(3\mid2^f-1\), so \(f\) is
even and at least two. Consequently \(t\ge\ln4>1\).

Direct multiplication of the seven rational rows gives

\[
c_0c_1c_4a_*^2>800,\quad
c_0c_3qa_*^2>202,\quad
\frac{c_0c_4qa_*^2}{\widehat\vartheta}>1000,\quad
c_3q>\frac75.
\tag{10.281}
\]

For explicit checks at the possible minima, the first product is
smallest in III.2 and equals \(833.54005644\). The second is smallest in
III.1 and equals \(203.78512\). The third is smallest in V and equals
\(1060.475\); the fourth is also smallest in V and equals \(1.4271\).
These are exact products of the displayed rationals.

We have \(F_2=9\) and

\[
\frac{F_{r+1}}{F_r}
=\frac{r+2}{r+1}\left(1+\frac2r\right)^r>4.
\tag{10.282}
\]

To verify the last inequality for every real \(r\ge2\), the derivative
of \(r\ln(1+2/r)\) is \(\ln(1+z)-z/(1+z)>0\), where \(z=2/r>0\). The
latter function vanishes at zero and has derivative \(z/(1+z)^2>0\). Its
value at \(r=2\) gives \((1+2/r)^r\ge4\). Since \(a_*\ge7/2\) and
\(\rho\le58\), (10.281)--(10.282) give

\[
B_r>\frac{800}{29}\,9\,14^{r-2}=b_r.
\tag{10.283}
\]

The stronger lower bound in Lemma 10.70 is \(\mathcal D\ge B_r(A+x)t\);
its proof uses the strict inequality \(2+g_2^{-1}>2\), so this
inequality is strict. The formula for \(S\), with \(h\ge(r+1)t\), proves
its bound in (10.280).

In cases III and IV, the smallest possible value of \(g_2=c_3q(r+1)^2d\)
is \(3.26\cdot2\cdot9=58.68\), in III.2. In I, II and V use the other
formula for \(g_2\), \(g_0>39\), and respectively \(\ln3<11/10\),
\(\ln5<2\), and \(\ln2<1\). In II also \(e\ge2\). Each lower bound is
greater than 58.

The formulas for \(g_5\) and \(g_4\), including the smaller alternative
\(g_1^{-1}\) for the latter, give

\[
g_5>\frac{404}{58}(r+1)F_r(7/2)^{r-2},\qquad
g_4>\frac{2000}{58}(r+1)F_r(7/2)^{r-2}.
\tag{10.284}
\]

The first bound is at least \(5454/29>188\). In the second, (10.282)
implies \((r+1)F_r(7/2)^{r-2}> (87/20)r(r+1)\): at \(r=2\), \(9>87/10\),
and each next ratio of \(F_r(7/2)^{r-2}\) is greater than 14 while
\((r+1)/r\le3/2\). Hence \(r(r+1)/(2g_4)<1/300\). Since
\((1+z)^r\le\exp(rz)\) for \(z\ge0\),

\[
1+\epsilon<\exp(1/300)
\le\sum_{j\ge0}(1/300)^j
=\frac{300}{299}<\frac{251}{250}.
\]

This also proves the remaining rounding bounds. The elementary logarithm
inequalities used above follow from the positive exponential series: its
terms through degree five at \(11/10\) already exceed three; its terms
through degree four at two exceed five. \(\square\)

\textbf{Lemma 10.105 (a general logarithmic coefficient budget).} Let
\(\Lambda\) be the selected initial support of Theorem 10.71,
\(N=kL|\Lambda|\), and \(1\le d_E\le d\). Then

\[
\frac{\ln(Nd_E)}Z
<\varepsilon_r:=\frac{3+14/(r+1)}{7b_r}
\le\frac{667}{151200}\,14^{2-r}
<\frac1{225}\,14^{2-r}.
\tag{10.285}
\]

\textbf{Proof.} The height-product bound of Theorem 10.18, applied
separately to each singleton, gives \(d\sigma_i>1/(87d^2)\). Indeed for
\(d\ge2\), its lower bound is
\(d^2\ell_d\sigma_i\ge w_K/(58\mathrm e)>1/87\), using \(w_K\ge2\),
\(\mathrm e<3\), and \(\ell_d\le d\). At \(d=1\) its denominator is 17
and the same weaker bound follows.

Every coordinate of the lattice lies in \(q^{-\nu}\mathbb Z\), by the
annihilator argument in Lemma 10.84. A translated interval of full width
\(D_i\) therefore contains at most \(\lfloor q^\nu D_i\rfloor+1\)
possible coordinates. The phase restriction only decreases the count. As
\(c_1c_2rP>1\), (10.186) gives \(q^\nu D_i<87q^\nu d^2\mathcal D\).
Moreover \(kL\le\lambda S\mathcal D/[c_1c_4d(A+x)]<S\mathcal D\), and
\(\mathcal D>1\). Consequently

\[
N<S\mathcal D(88q^\nu d^2\mathcal D)^r,
\qquad
\ln(Nd_E)<\ln S+(r+1)\ln\mathcal D
 +(2r+1)\ln d+rx+r\ln88.
\tag{10.286}
\]

Set \(S_*=\tfrac75(r+1)^2d\). Both \(S_*\) and \(5b_r\) exceed
\(\mathrm e\); the function \(\ln z/z\) decreases there. Also
\(\ln S_*<2A\), since \(\ln(7/5)<1\) and \(A\ge4+\ln((r+1)d)\). Thus the
contribution of \(\ln S\), after division by \(Z\), is less than
\(10/[7(r+1)^2b_r]\).

Since \(\mathcal D>b_r(A+x)t>5b_r\),

\[
\frac{\ln\mathcal D}{\mathcal D}
<\frac{\ln(5b_r)}{5b_r}
<\frac{3r+2}{5b_r}.
\]

Here \(\ln(36000/29)<8\) and \(\ln14<3\); the positive exponential
series proves both, using respectively terms through degree eight at
eight and through degree four at three. The contribution of
\((r+1)\ln\mathcal D\) is therefore less than \((3r+2)/[7(r+1)b_r]\).

Finally \(\ln88<5<A\), by the positive exponential series through degree
six at five, and \(\ln d<A\). The remaining terms in (10.286) are at
most \((3r+1)(A+x)\). Their contribution is less than
\(5(3r+1)/[7(r+1)^2b_r]\). Adding these three bounds gives

\[
\frac{3r^2+20r+17}{7(r+1)^2b_r}
=\frac{3+14/(r+1)}{7b_r}.
\]

This factor decreases apart from the displayed \(14^{2-r}\). At \(r=2\)
it is \(667/151200<1/225\). All divisions used the original scale
\(Z=S\mathcal D/d\). \(\square\)

\subsubsection{The original height
parameter}\label{the-original-height-parameter}

For the next average take the original

\[
g_0=(r+1)\bigl(a_0r+a_1+\ln(a_0r+a_2)+\ln d\bigr),
\tag{10.287}
\]

with the following choices. In III, \(a_0=2+\ln7\); in I, II and IV,
\(a_0=2+\ln14\); in V, \(a_0=2+\ln26\). The values of \(a_1\), in the
order of the seven rows above, are
\(4.03,4.79,3.44,4.71,5.84,5.12,2.52\). Set \(a_2=a_1\) in I.2 and
III.2, and \(a_2=a_1+\ln2\) otherwise. These choices are compatible with
the larger \(h\) in Section 26.

\textbf{Lemma 10.106 (the general Euler logarithm constant).} For these
original choices, \(g_{91}\) of (10.222) satisfies

\[
g_0>(r+1)(39r/10+36/5+\ln d),
\qquad
g_{91}<1+\frac1{3(r+1)}.
\tag{10.288}
\]

\textbf{Proof.} The elementary positive logarithm series
\(\ln z=2\sum_{j\ge0}y^{2j+1}/(2j+1)\), \(y=(z-1)/(z+1)\), proves
\(\ln7>19/10\), \(\ln14>13/5\), and \(\ln13>5/2\) by retaining
respectively six, twelve and twelve terms. This series and its explicit
geometric tail were proved in Solution 34. In every case \(a_0>39/10\).
Directly at \(r=2\), each of \(2a_0+a_1+\ln(2a_0+a_2)\) exceeds 15. For
the smallest checks in II and III.1 use \(2a_0+a_1>12.64\) and
\(>12.51\), respectively, together with \(2a_0+a_2>13\) and
\(\ln13>5/2\). In V, \(a_0>2+2/3+5/2=31/6\), and \(2a_0+a_2>13\). The
other four checks have larger lower bounds. Increasing \(r\) increases
the first linear term by at least \((39/10)(r-2)\) and the logarithm
cannot decrease. This proves the first bound.

For \(d\ge2\), set \(y=\ln d\). Concavity gives the tangent inequality
\(\ln z\le\ln9+(z-9)/9\). At \(z=y+\ln3\),

\[
1+3\ln(y+\ln3)
\le\frac{19}{3}\ln3-2+\frac y3
<5+\frac y3.
\]

The last inequality uses \(\ln3<11/10\). The first bound in (10.288)
gives \(g_0>(r+1)(15+y)\). Substituting in
\(g_{91}-1=[1+3\ln\ln(3d)]/g_0\) proves the second bound. At \(d=1\),
use \(\ln6<9/5\), \(\ln2>2/3\), and \(\ln3>1\), so
\(W(1)<27/10<\mathrm e\). The last comparison follows from the
exponential series through degree five at one. Hence \(1+\ln W(1)<2\)
and \(g_0>15(r+1)\), which is stronger than required. The exponential
series through degree six at \(9/5\) proves \(\ln6<9/5\). \(\square\)

\subsubsection{Keeping the zero rows in the
simplex}\label{keeping-the-zero-rows-in-the-simplex}

Put \(t_0=\lfloor T\rfloor\), and now retain \textbf{all}
\(\binom{t_0+r}{r}\) composition indices at every node, including those
whose additive order exceeds \(D_{\mathrm a}\). Such rows are zero.
Assign them the same positive scalar majorants as below. Let \(u\) be
the additive coordinate and \(h_{\mathrm E}\) the sum of the \(r-1\)
Euler coordinates. The slack completes the \(r+1\) coordinates to total
\(t_0\).

\textbf{Theorem 10.107 (the complete-simplex mean bound).} Assume the
coefficient-height condition (10.223) and the original choices (10.287).
Retaining these full rows preserves every initial zero and the
coefficient field. Their mean scalar logarithm is at most

\[
D_{\mathrm a}\ln\!\left[2\mathrm e\left(\frac{s_0}{k}+2\right)\right]
 + C_rTH_1,
\qquad
C_r=\frac{r+1/10}{r+1}
 +\frac{17(r-1)}{24(r+1)^2}<1,
\tag{10.289}
\]

where \(s_0=\lfloor S\rfloor\), \(H_1=h+x\). In fact

\[
1-C_r=\frac{23r+193}{120(r+1)^2}>0.
\tag{10.290}
\]

\textbf{Proof.} Permutation of the \(r+1\) composition coordinates is a
bijection. Their means are therefore exactly \(t_0/(r+1)\). Consequently

\[
\overline u=\frac{t_0}{r+1},\qquad
\overline h_{\mathrm E}=\frac{(r-1)t_0}{r+1}.
\tag{10.291}
\]

For \(u\le D_{\mathrm a}\), Lemma 10.76 gives its additive majorant in
(10.198). For \(u>D_{\mathrm a}\) the derivative is zero, so the same
positive expression is still a valid majorant; no division by a zero
entry is used. At Euler total \(h_{\mathrm E}\), retain the majorant
\(\binom{m+h_{\mathrm E}}{h_{\mathrm E}}\). Its logarithm is bounded by
\(h_{\mathrm E}\ln[\mathrm e(1+m/h_{\mathrm E})]\); concavity, as proved
in Corollary 10.79, bounds its mean by its value at
\(\overline h_{\mathrm E}\).

The first part of the proof of Theorem 10.86 used only
\(\overline h\ge(r-1)t_0/(r+1)\). Repeating its substitution with
equality (10.291) gives

\[
\frac m{\overline h_{\mathrm E}}
\le\frac47\mathrm e^{H_1}tW(d)+\frac{r+1}{t_0},
\qquad
1+\frac m{\overline h_{\mathrm E}}
\le\mathrm e^{H_1}W(d)t.
\tag{10.292}
\]

For the second inequality, use exactly its quadratic-series absorption
of \(1+(r+1)/t_0\), with \(H_1>39\), \(H_1\ge(r+1)t\), \(W(d)\ge1\) and
now \(t>1\). Taking logarithms and adding one bounds the mean Euler
contribution by \(\overline h_{\mathrm E}(g_{91}H_1+\ln t)\).
Differentiating \(\ln t/t\) gives its maximum \(1/\mathrm e\); hence
\(\ln t\le H_1/[\mathrm e(r+1)]\).

The complete lcm proof in lesson seven gives
\(\ln\mathfrak v<k\ln3\le H_1\ln3\). Combining this with (10.291), the
sum of the mean clearing and Euler logarithms is less than

\[
\frac{t_0H_1}{r+1}
\left[(r-1)g_{91}+\ln3+
             \frac{r-1}{\mathrm e(r+1)}\right].
\tag{10.293}
\]

Lemma 10.106, \(\ln3<11/10\), and \(1/\mathrm e<3/8\) bound its
coefficient by \(C_r\): the latter exponential inequality follows from
the first four terms at one, whose sum is \(8/3\), and the remaining
terms are positive. The identity (10.290) follows by subtraction. Adding
the allocation-independent additive majorant proves (10.289), since
\(t_0\le T\).

Finally Theorem 10.71 supplies \(N>c_{01}(2s_0+1)\binom{t_0+r}{r}\), so
the complete weighted integral Siegel argument applies to this full row
matrix. The additional zero rows change no kernel equation, and repeated
Euler rows preserve the same equations. Every entry still lies in
\(E=\mathbb Q(\alpha_0,\theta_1,\ldots,\theta_r)\). The row field,
nonzero vector, independent polynomial basis and all prescribed initial
jets are therefore retained. \(\square\)

\textbf{Corollary 10.108 (a sharper general initial coefficient bound).}
In the original generator field \(E\subseteq K\), an integral initial
coefficient vector with every prescribed zero and a nonzero polynomial
may be chosen with

\[
\begin{aligned}
\frac{h_2(c)}Z<{}&g_{12}+\frac{\varepsilon_r}{2}\\
&+\frac1{c_{01}-1}\left[
\frac{\varepsilon_r}{2}
 +\frac{\lambda}{c_1c_4}
 +\frac{C_r}{c_1c_3\theta e}
 +\frac{1+(2g_2)^{-1}}{2c_1c_2}\right].
\end{aligned}
\tag{10.294}
\]

Here \(g_{12}=0\) when \(d=1\); in every case one may further use

\[
g_{12}\le\frac{1+5/(r+1)}{546b_r}.
\tag{10.295}
\]

\textbf{Proof.} Apply the weighted field argument of Theorem 10.90 to
the full rows just proved, retaining its actual discriminant bound.
Lemma 10.105 bounds \(\ln N/Z\). Corollary 10.77 bounds the
allocation-independent additive term; its hypothesis holds for every
displayed case. The identity \(TH_1/Z=1/(c_1c_3\theta e)\), with Theorem
10.107, bounds the combined mean term. The exact row ratio is less than
\(1/(c_{01}-1)\), and Corollary 10.72 bounds the mean torus
contribution. These substitutions give (10.294).

For (10.295), write \(g_3=B_rg_1t\). Then
\(g_1/(2g_7)=1/[2c_3q(r+1)g_0B_r]
 <5/[546(r+1)b_r]\), by \(c_3q>7/5\) and \(g_0>39\). The formula
(10.231) for \(g_{11}\) can be written as

\[
g_{11}=2\mathrm e\,c_3qB_rg_0g_1
\frac{(r+1)(r-1)^{r-1}}{r^r}.
\]

The last ratio exceeds \(1/\mathrm e\), because
\((1+1/(r-1))^{r-1}<\mathrm e\). Thus \(g_{11}>2c_3qB_rg_0g_1>546b_r\),
using \(g_1>5\). Adding its reciprocal proves (10.295); at \(d=1\) its
exact discriminant contribution remains zero. \(\square\)

\subsubsection{An individual value in its normal
coordinates}\label{an-individual-value-in-its-normal-coordinates}

The exact normalizing factor in Lemma 10.91 also gives a second
arithmetic estimate. Here its factorial and lcm valuations cancel
against a common rational rescaling of the actual algebraic value. The
global field degree is still required.

\textbf{Theorem 10.109 (a second arithmetic bound for the normalized
value).} Retain all hypotheses and notation of Theorem 10.92. Let
\(I,J\ge0\), \(x=s/q^J\), \(u\le D_{\mathrm a}\), and \(h_{\mathrm E}\)
the total Euler order. Put

\[
G_u=(D_{\mathrm a}-u)\theta+Lv_p(k!)-uv_p(\mathfrak v),
\qquad
\widehat\Xi=(I+J)(D_{\mathrm a}-u).
\tag{10.296}
\]

If the actual value \(V\in F\) is nonzero, then

\[
\begin{aligned}
v_p(V)+G_u-\delta\le{}&
\frac{d_F}{e_Ff_F\ln p}\biggl[
h_2(c)+\tfrac12\ln N
 +(D_{\mathrm a}-u)\ln R_I
 +\ln\binom{D_{\mathrm a}}u\\
&\quad{}+\ln\binom{m_I+h_{\mathrm E}}{h_{\mathrm E}}
 +\widehat\Xi\ln q
 +PXq^{-I}\sum_iD_i\sigma_i\biggr]
 +(D_{\mathrm a}-u)\theta.
\end{aligned}
\tag{10.297}
\]

For this same normal quantity one may take the smaller of (10.297) and
the right side of (10.244) plus \(G_u\). Both estimates use the same
actual \(F,d_F,e_F,f_F\).

\textbf{Proof.} Multiply the whole algebraic value, at this one fixed
additive order, by the common nonzero rational scalar

\[
W=q^{\widehat\Xi}\frac{(k!)^L}{\mathfrak v^u}\,V\in F.
\tag{10.298}
\]

It is essential that this multiplier is independent of the support index
and the binomial-basis column. For a column of power \(\ell\le L\),
divided differentiation of \(F_{a,\ell}\) has denominator dividing
\((k!)^\ell\). Multiplication by \((k!)^L\) leaves integer polynomial
coefficients. At \(q^{-I}x\), its degree is at most
\(k\ell-u\le D_{\mathrm a}-u\), so \(q^{\widehat\Xi}\) clears every
remaining rational denominator. The Euler binomials are integers at
their original integer arguments. Thus the rational scalar multiplying
each torus monomial in \(W\) is an integer.

Its ordinary absolute value is at most

\[
q^{\widehat\Xi}R_I^{D_{\mathrm a}-u}
\binom{D_{\mathrm a}}u
\binom{m_I+h_{\mathrm E}}{h_{\mathrm E}}.
\]

Indeed the column estimate before taking the largest power is
\((k!)^{L-\ell}\binom{k\ell}u R_I^{k\ell-u}\). Since \(R_I\ge k\), we
have \(R_I^k\ge k!\); increasing \(\ell\) to \(L\) and the binomial to
\(\binom{D_{\mathrm a}}u\) therefore gives the stated bound, exactly as
in Theorem 10.92.

Apply its proved product-formula and row-height argument to this nonzero
\(W\), using the integer scalar bound just obtained. The coefficient
vector and full-width torus term are unchanged. At the chosen prime,
\(q\) is a unit, so \(v_p(W)=v_p(V)+Lv_p(k!)-uv_p(\mathfrak v)\). Adding
\((D_{\mathrm a}-u)\theta\) proves (10.297). Applying (10.244) to the
original \(V\) and adding its same \(G_u\) proves the other bound. Their
minimum is valid because both bound the same number. Restriction of the
completion, or the fact that roots lie in an old completion, has not
replaced the actual global field degree. \(\square\)

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={The exact full derivative simplex and the all-rank initial height budgets}]{../figures/general-descent-precision.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.24. Left: every pair of nonnegative additive and Euler
orders with sum at most six, with the remaining slack, at rank two. The
28 equally weighted rows have mean additive and Euler order two; rows
above the illustrative additive degree two are zero and remain in the
average. Middle: the proved constants \(C_r\) for ranks two through
twelve. Equation (10.290) proves their strict upper bound one for every
rank. Right: the logarithmic coefficient-count bounds \(\varepsilon_r\)
of (10.285), with logarithmic vertical scale; the displayed points
illustrate the analytical all-rank formula. The actual scale is
\(Z=S\mathcal D/d\). Lemmas 10.104--10.106 and Theorem 10.107 keep every
original field and prime case, and Corollary 10.108 retains all scalar,
discriminant and mean torus costs.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/general_descent_precision.py}{Figure program}.
Human-source background:
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
free paper}, equations (1.11), (3.12)--(3.18) and the initial row system
in Section 4.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{37. A fully proved elementary bound for least common
multiples}\label{a-fully-proved-elementary-bound-for-least-common-multiples}

The individual prepared derivatives contain \(v(k)^u\), where
\(v(k)=\operatorname{lcm}(1,\ldots,k)\). We will prove

\[
\ln v(k)<\frac{107}{103}k\qquad(k\ge1).
\tag{10.299}
\]

This strengthens the complete elementary bound in lesson seven. It also
keeps the lcm constant used in the general logarithmic-form parameters.
The proof follows the finite-recurrence method of
\href{https://www.impan.pl/shop/en/publication/transaction/download/product/106154}{Costa
Pereira's freely available paper}, pages 321--323. We give the
recurrence, its induction, and the complete exact finite computation
here. The computation regenerates its primes and encloses each logarithm
by rational numbers; no numerical table is an assumed input.

\subsubsection{A factorial combination}\label{a-factorial-combination}

For \(x>0\), put \(\psi(x)=\sum_{p^a\le x}\ln p\), summing over primes
\(p\) and integers \(a\ge1\). It is zero below one and constant on each
interval \([n,n+1)\). Define the integer weights

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}
  >{\raggedright\arraybackslash}p{(\linewidth - 30\tabcolsep) * \real{0.0476}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 30\tabcolsep) * \real{0.0635}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 30\tabcolsep) * \real{0.0635}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 30\tabcolsep) * \real{0.0635}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 30\tabcolsep) * \real{0.0635}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 30\tabcolsep) * \real{0.0635}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 30\tabcolsep) * \real{0.0635}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 30\tabcolsep) * \real{0.0635}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 30\tabcolsep) * \real{0.0635}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 30\tabcolsep) * \real{0.0635}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 30\tabcolsep) * \real{0.0635}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 30\tabcolsep) * \real{0.0635}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 30\tabcolsep) * \real{0.0635}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 30\tabcolsep) * \real{0.0635}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 30\tabcolsep) * \real{0.0635}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 30\tabcolsep) * \real{0.0635}}@{}}
\toprule\noalign{}
\begin{minipage}[b]{\linewidth}\raggedright
\(r\)
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
1
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
2
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
3
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
5
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
6
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
7
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
10
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
11
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
13
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
14
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
15
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
30
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
42
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
110
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
182
\end{minipage} \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
\(v_r\) & 1 & \(-1\) & \(-1\) & \(-1\) & 1 & \(-1\) & 1 & \(-1\) &
\(-1\) & 1 & 1 & \(-1\) & \(-1\) & \(-1\) & 1 \\
\end{longtable}
}

The symbols \(v_r\) are weights; they are distinct from the integer
\(v(k)\). Put

\[
F(y)=\sum_rv_r\lfloor y/r\rfloor,\qquad
\sigma(x)=\sum_rv_r\ln(\lfloor x/r\rfloor!),\qquad
\omega=-\sum_r\frac{v_r\ln r}{r}.
\tag{10.300}
\]

As usual, \(0!=1\).

\textbf{Lemma 10.110 (the factorial combination and its error).} These
weights satisfy \(\sum v_r/r=0\), \(\sum v_r=-3\) and \(\sum|v_r|=15\).
For every \(y\ge0\), \(-1\le F(y)\le4\). Moreover

\[
\ln v(k)=\psi(k),\qquad
\sigma(x)=\sum_{p^a\le x}(\ln p)F(x/p^a),\qquad
|\sigma(x)-\omega x|\le15(\ln x+1)\quad(x\ge182).
\tag{10.301}
\]

\textbf{Proof.} The exponent of \(p\) in the least common multiple is
the largest integer \(a\) for which \(p^a\le k\); summing its logarithm
proves the first identity. In \(n!\), the exponent of \(p\) is
\(\sum_{a\ge1}\lfloor n/p^a\rfloor\), by counting multiples of each
power. Thus
\(\ln(\lfloor x\rfloor!)=\sum_{p^a\le x}(\ln p)\lfloor x/p^a\rfloor\).
Substituting this finite sum in (10.300) proves the second identity.

For the weight sum, add the four rational identities
\(1+1/15-1/2-1/3-1/5-1/30=0\), \(1/6-1/7-1/42=0\),
\(1/10-1/11-1/110=0\), and \(1/14+1/182-1/13=0\). The other two sums
follow directly from the table.

Write \(F=F_0+F_1+F_2+F_3\), using respectively the weights at
\(\{1,2,3,5,15,30\}\), \(\{6,7,42\}\), \(\{10,11,110\}\), and
\(\{13,14,182\}\). The first rational identity makes \(F_0\) periodic
with period 30. Its values at the residues \(0,1,\ldots,29\), obtained
by substituting in its six floors, are
\(0,1,1,1,1,1,0,1,1,1,0,1,0,1,1,1,1,2,1,2,1,1,1,2,1,1,1,1,1,2\). It is
constant between consecutive integers, so \(F_0\le2\) at every real
argument. The identities for \(1/6\) and \(1/10\), together with
\(\lfloor a+b\rfloor-\lfloor a\rfloor-\lfloor b\rfloor\in\{0,1\}\), show
that \(F_1,F_2\in\{0,1\}\). The identity for \(1/13\) gives
\(F_3\in\{-1,0\}\). Consequently \(F\le4\).

For the lower bound, all denominators divide \(\tau=30030\); the zero
inverse-weight sum makes \(F\) periodic with this period. At integral
\(n\), \(\lfloor n/r\rfloor+\lfloor(\tau-1-n)/r\rfloor=\tau/r-1\).
Therefore \(F(n)+F(\tau-1-n)=3\). Since \(F\le4\), it follows that
\(F(n)\ge-1\). Each floor in \(F\) is constant between consecutive
integers, so the same bounds hold at real \(y\ge0\).

Finally let \(t\ge1\), \(n=\lfloor t\rfloor\). Monotonicity of \(\ln z\)
gives \(n\ln n-n+1\le\ln(n!)\le n\ln n-n+1+\ln n\). The change in
\(t\ln t-t\) between \(n\) and \(t\) is between zero and \(\ln t\).
Hence \(|\ln(n!)-(t\ln t-t)|\le\ln t+1\). Apply this at every
\(t=x/r\ge1\). The \(x\ln x-x\) terms cancel because \(\sum v_r/r=0\);
the remaining linear term is \(\omega x\). The fifteen absolute weights
give the stated error. \(\square\)

\subsubsection{The finite certificate}\label{the-finite-certificate}

The selected half-open intervals in the next table have integer
endpoints. Each pair \((b,c)\) denotes \([b,c)\); pairs in a given row
are disjoint.

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}
  >{\raggedright\arraybackslash}p{(\linewidth - 4\tabcolsep) * \real{0.3333}}
  >{\raggedright\arraybackslash}p{(\linewidth - 4\tabcolsep) * \real{0.3333}}
  >{\raggedright\arraybackslash}p{(\linewidth - 4\tabcolsep) * \real{0.3333}}@{}}
\toprule\noalign{}
\begin{minipage}[b]{\linewidth}\raggedright
Level \(s\)
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
Intervals with \(F\ge s\), for the upper recurrence
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
Intervals with \(F<s\), for the lower recurrence
\end{minipage} \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
1 & (67,126), (157,176), (179,220), (223,275), (277,330), (359,429) &
none \\
2 & (17,22), (23,26), (29,35), (47,52), (59,65), (71,78), (79,88),
(191,210) & (26,29), (65,71), (117,139) \\
3 & (19,21) & (21,31), (33,61), (63,73), (84,103), (110,193), (208,229),
(242,271), (294,323), (325,373), (440,493) \\
4 & none & (440,877) \\
\end{longtable}
}

Let \(\mathcal U,\mathcal L\) be the respective lists of pairs,
retaining their level. The initial cutoffs are
\(a_0=877,a_1=66,a_2=17,a_3=19,a_4=439\). Here \(F(y)\ge s\) for
\(1\le y<a_s\), \(s=0,1\), and \(F(y)<s\) for \(1\le y<a_s\),
\(s=2,3,4\).

\textbf{Lemma 10.111 (exact finite certificate).} The cutoffs and every
interval in the table have the asserted property. Set

\[
\begin{aligned}
A&=1-1/a_0-1/a_1-\sum_{(b,c)\in\mathcal U}1/c,\\
B&=\sum_{(b,c)\in\mathcal U}1/b,\\
C&=1/a_2+1/a_3+1/a_4+\sum_{(b,c)\in\mathcal L}1/c,\\
D&=1-\sum_{(b,c)\in\mathcal L}1/b.
\end{aligned}
\tag{10.302}
\]

With \(Q=10^{12}\), \(N=10^6\), \(U=27/26\), \(L=25/26\), and
\(\varepsilon=9/40000\), the following rational inequalities hold:

\[
\begin{gathered}
A\ge.732878,\quad B\ge.297381,\quad
C\le.263738,\quad D\le.802853,\\
1.046524<\omega<1.046525,\\
AU+BL-\omega-\varepsilon>\frac{6731}{26000000},\qquad
\omega-\varepsilon-CU-DL>\frac{11523}{26000000}.
\end{gathered}
\tag{10.303}
\]

Also \(\psi(x)<Ux\) for \(114\le x<N\), and \(\psi(x)>Lx\) for
\(227\le x<N\). For integers \(1\le n<113\),
\(\psi(n)/n<\psi(113)/113\), and \(U<\psi(113)/113<107/103\).

\textbf{Proof and reproducible finite computation.} The full
standard-library program
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/chebyshev_certificate.py}{chebyshev\_certificate.py}
is part of this lesson's editable sources. We describe every finite
check it performs, including its rational logarithm bounds.

For the floor conditions it substitutes \(\sum v_r(n\mathbin{//}r)\) at
every integer in each initial range and every listed half-open interval.
This covers real arguments as well, because \(F\) is constant on
\([n,n+1)\). It checks the within-level disjointness and that each
interval begins at or beyond its level's initial cutoff. The fractions
in (10.302) are then summed with exact integer numerators and
denominators. Their comparison with the displayed terminating decimals
is an exact rational comparison.

To enclose a logarithm, first write an integer \(m=2^at\), \(1\le t<2\),
and put \(z=(t-1)/(t+1)\), so \(0\le z<1/3\). For \(\ln2\), take
\(z=1/3\). Integrating the finite geometric identity for
\((1-z^2)^{-1}\) proves

\[
0\le\ln\frac{1+z}{1-z}
 -2\sum_{j=0}^{15}\frac{z^{2j+1}}{2j+1}
\le\frac{2z^{33}}{33(1-z^2)}
\le\frac9{132\,3^{33}}<\frac1Q.
\tag{10.304}
\]

Round each of the sixteen positive terms down to an integer multiple of
\(1/Q\). Their sum is a lower bound; adding \(17/Q\) gives an upper
bound, since there are sixteen rounding errors and the tail is less than
\(1/Q\). At \(z=0\) use the exact value zero. Adding \(a\) copies of the
interval for \(\ln2\) gives integer endpoints \(l_m,u_m\) with
\(l_m/Q\le\ln m\le u_m/Q\). This uses integer division and powers only.
Apply these enclosures, with the appropriate signs, to the fifteen terms
defining \(\omega\). They give the displayed strict bracket.
Substituting its two endpoints and the four rational coefficient bounds
proves (10.303); the residual fractions are exactly those shown there.

The sieve begins with unmarked integers from two through \(N\). Each
unmarked \(p\) is prime: a composite would already have been marked by
its smaller prime divisor. It marks multiples starting at \(p^2\);
smaller multiples were marked earlier. For every discovered prime, it
assigns the interval for \(\ln p\) to each of \(p,p^2,\ldots\le N\). No
two different primes assign the same integer. Successive cumulative sums
give integer endpoints \(L_n,U_n\) for \(Q\psi(n)\). The program checks
all the following inequalities by integer arithmetic:

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}llr@{}}
\toprule\noalign{}
Integer range & Checked positive integer expression & Smallest value \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
\(114\le n<N\) & \(27Qn-26U_n\) & 13207722459676, at \(n=199\) \\
\(227\le n<N\) & \(26L_n-25Q(n+1)\) & 55077005461136, at \(n=346\) \\
\(1\le n\le113\) & \(107Qn-103U_n\) & 167297005424 \\
\(1\le n<113\) & \(nL_{113}-113U_n\) & 33103591701014 \\
\end{longtable}
}

The same calculation gives
\(117386725268647\le Q\psi(113)\le117386725271792\) and verifies
\(26L_{113}>27Q\cdot113\). The sieve finds 78498 primes and 78734 prime
powers at most \(N\); these counts are outputs, not inputs. Every
prime-power assignment and cumulative sum is integral, and the maximum
cumulative upper endpoint is checked to be below \(2^{63}\), the storage
limit. The finite sums for (10.302) use arbitrary-size integers.

For a real \(x\in[n,n+1)\), the upper comparison follows from
\(\psi(x)=\psi(n)<Un\le Ux\). The lower comparison follows from
\(\psi(x)=\psi(n)>L(n+1)>Lx\); the extra \(n+1\) in its certificate is
essential. These arguments prove all assertions of the lemma.
\(\square\)

\subsubsection{Continuing the bounds to every real
argument}\label{continuing-the-bounds-to-every-real-argument}

\textbf{Theorem 10.112 (a sharp elementary lcm bound).} For every real
\(x>0\),

\[
\psi(x)\le\frac{\psi(113)}{113}x<\frac{107}{103}x.
\tag{10.305}
\]

Consequently (10.299) holds at every positive integer \(k\).

\textbf{Proof.} We first derive the two recurrences from the actual
factorial combination. For an integer \(j\in[-1,4]\),
\(j-1=-\sum_{s=0}^1\mathbf1_{j<s}+\sum_{s=2}^4\mathbf1_{j\ge s}\). Apply
this at \(j=F(x/p^a)\) in (10.301). All weights \(\ln p\) are positive.
The condition \(x/p^a\in[b,c)\) contributes exactly
\(\psi(x/b)-\psi(x/c)\), with the indicated half-open convention.

For \(s=0,1\), occurrences of \(F<s\) begin only at \(a_s\). Their total
mass is at most \(\psi(x/a_s)\), minus the masses of any disjoint
selected intervals where \(F\ge s\). For \(s=2,3,4\), the mass where
\(F\ge s\) is at least that of its selected intervals. This gives the
upper recurrence. Reverse the two roles to obtain the lower recurrence:

\[
\begin{aligned}
\psi(x)&\le\sigma(x)+\psi(x/a_0)+\psi(x/a_1)
 -\sum_{(b,c)\in\mathcal U}[\psi(x/b)-\psi(x/c)],\\
\psi(x)&\ge\sigma(x)-\sum_{s=2}^4\psi(x/a_s)
 +\sum_{(b,c)\in\mathcal L}[\psi(x/b)-\psi(x/c)].
\end{aligned}
\tag{10.306}
\]

There is no subtraction of intersecting intervals within a level: Lemma
10.111 checked disjointness. Intervals at different levels represent
different indicator terms and must retain their multiplicity.

For \(x\ge N\), (10.301) gives \(|\sigma(x)/x-\omega|<\varepsilon\).
Indeed \((\ln x+1)/x\) decreases for \(x>1\), and \(\ln N<14\). The
latter follows from \(\ln10<7/3\): the exponential series at \(7/3\)
through degree six is already greater than ten.

Now prove simultaneously \(Lx<\psi(x)<Ux\), starting at \(x=N\), by
induction on the interval \([n,n+1)\). The finite base is Lemma 10.111.
Every argument \(x/d\) on the right of (10.306) has \(d\ge17\), so it is
less than \(n\). Every argument requiring an upper estimate is at least
\(N/877>114\). Every argument requiring a lower estimate is at least
\(N/440>227\). Thus either the finite base or the earlier induction
intervals apply to each term. Division of the upper recurrence by \(x\)
gives \(\psi(x)/x<\omega+\varepsilon+(1-A)U-BL<U\). The lower recurrence
gives \(\psi(x)/x>\omega-\varepsilon-CU+(1-D)L>L\). The two strict final
comparisons are precisely (10.303). This proves the bounds at every
\(x\ge N\), and hence the upper bound \(\psi(x)<Ux\) at every
\(x\ge114\).

At \(0<x<1\) the assertion is immediate. At \(1\le x<114\), let
\(n=\lfloor x\rfloor\le113\). Lemma 10.111 gives
\(\psi(x)/x\le\psi(n)/n\le\psi(113)/113\), with equality permitted at
\(x=113\). At \(x\ge114\), use \(\psi(x)/x<U<\psi(113)/113\). The last
strict bound in (10.305) is the certificate at 113. Finally use
\(\ln v(k)=\psi(k)\). \(\square\)

\textbf{Corollary 10.113 (the sharper full-simplex coefficient budget).}
Every field, prime and rank case of Theorem 10.107 and Corollary 10.108
remains valid with \(C_r\) replaced by

\[
C_r^{\sharp}=\frac{r+4/103}{r+1}
 +\frac{17(r-1)}{24(r+1)^2}
=C_r-\frac{63}{1030(r+1)}<C_r<1.
\tag{10.307}
\]

The individual clearing factor also satisfies \(u\ln v(k)<(107/103)uk\)
whenever \(u>0\), and is zero when \(u=0\).

\textbf{Proof.} In (10.293), substitute
\(\ln v(k)<(107/103)k\le(107/103)H_1\), using Theorem 10.112. All Euler,
additive and field arguments are unchanged. The rational calculation of
its mean coefficient now gives (10.307). The full row composition, its
zero rows and its actual coefficient field are the same. In (10.294),
only the scalar mean term containing \(C_r\) changes, so the same
substitution gives the asserted coefficient-height bound. Multiplication
of (10.299) by \(u\) proves the last statement. \(\square\)

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={The factorial recurrence, its prime-power mass bound and its positive induction margins}]{../figures/chebyshev-recurrence.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.25. Left: the actual integer floor combination \(F(n)\)
and the selected intervals in the upper recurrence, colored by their
level; the vertical marks are its initial cutoffs. Middle: \(\psi(n)/n\)
for integers through 1200, with the maximum at 113 and the proved
constants \(27/26\) and \(107/103\). These discrete points illustrate
(10.305); the simultaneous induction (10.306) proves the infinite
statement. Right: the exact positive lower bounds for the two induction
margins in (10.303). The factorial combination is estimated at
\(x\ge10^6\); the finite base is generated by the integer certificate of
Lemma 10.111. \href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/chebyshev_recurrence.py}{Figure
program}. Human method and interval choices:
\href{https://www.impan.pl/shop/en/publication/transaction/download/product/106154}{Costa
Pereira's free paper}, pages 321--323; all required identities, finite
checks, error bounds and induction are given above.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{38. An individual clearing and Euler bound at every original
parameter}\label{an-individual-clearing-and-euler-bound-at-every-original-parameter}

The full-simplex average controls the initial coefficient height. A
later nonzero-value argument needs a bound for each individual row. We
obtain one from a positive generating series. It preserves the factorial
denominator in the additive polynomial and makes the remainder uniform
in the field, prime and rank.

Retain Sections 26 and 29, with \(H_1=h+x\), \(Z=S\mathcal D/d\), the
coefficient-height condition (10.223), and the seven original constant
cases in (10.281). Define \(g_9=\max\{g_{91},107/103\}\).

\subsubsection{The exact sharp regime for the Euler
constant}\label{the-exact-sharp-regime-for-the-euler-constant}

\textbf{Lemma 10.114.} For every original case and every \(r\ge2\),

\[
g_9\le\max\left\{1+\frac1{3(r+1)},\frac{107}{103}\right\}.
\qquad
g_9=\frac{107}{103}\quad\text{if }d=1\text{ or }r\ge8.
\tag{10.308}
\]

\textbf{Proof.} The first statement follows from Lemma 10.106. At
\(r\ge8\), its bound gives \(g_{91}-1<1/27<4/103\); the difference of
the last two fractions is \(5/2781\).

At \(d=1\), the only cases are I.2 and III.2. Four terms of the positive
logarithm series give \(\ln2>693/1000\), and five terms give
\(\ln3>1098/1000\). The already proved \(\ln6<9/5\) therefore gives
\(W(1)<(9/5)/[(693/1000)(1098/1000)]<19/8\). The exponential series
through degree five at \(7/8\) exceeds \(19/8\), so \(\ln W(1)<7/8\).
All these comparisons are between explicit rational numbers; the
positive tails have the required direction.

At rank two in I.2, (10.287) and \(\ln14>13/5,\ln13>5/2\) give
\(g_0>3(9.2+4.79+2.5)=49.47\). In III.2, use \(\ln7>19/10\) to obtain
\(g_0>3(7.8+5.84+2.5)=2421/50\). Every term defining \(g_0\) increases
with \(r\), so the latter lower bound holds throughout both cases. By
(10.222), \(g_{91}-1<(15/8)/(2421/50)=125/3228<4/103\). The last
difference is \(37/332484>0\). Taking the maximum with \(107/103\)
proves (10.308). \(\square\)

\subsubsection{A positive generating-series
majorant}\label{a-positive-generating-series-majorant}

For a support contracted by \(q^{-I}\), \(I\ge0\), take its actual
integer Euler maxima \(\Omega_j\), and set
\(m_I=\sum_{j=1}^{r-1}\Omega_j+r-2\), as in (10.239). Lemma 10.84
permits the same bound with the integer part of its upper estimate. Put
\(a=g_9H_1\), \(z=\exp(-a)\), and \(E_I=(m_I+1)[-\ln(1-z)]\). This
remainder depends on the support, not on the allocation of derivative
orders.

\textbf{Theorem 10.115 (an individual clearing/Euler bound).} For every
pair of nonnegative integers \(u,h_{\mathrm E}\),

\[
u\ln v(k)+\ln\binom{m_I+h_{\mathrm E}}{h_{\mathrm E}}
\le a(u+h_{\mathrm E})+E_I.
\tag{10.309}
\]

In the original parameters the remainder satisfies

\[
\frac{E_I}{Z}<
\frac{1000}{999}\,
\frac{q^{-I}(r-1)}{\mathrm e\,c_1c_2c_3qP(r+1)^2}
+\frac{r-1}{999Z}
<\frac{q^{-I}}{300}+\frac{r-1}{999Z}.
\tag{10.310}
\]

The symbol \(\mathrm e\) in this formula is \(\exp(1)\), and
\(P=p^\kappa\) is the original prime power.

\textbf{Proof.} The positive geometric series and its \(m_I+1\)-fold
product give \((1-z)^{-m_I-1}=\sum_{j\ge0}\binom{m_I+j}{j}z^j\). Indeed
its coefficient counts the \(m_I+1\) nonnegative parts summing to \(j\);
placing \(m_I\) separators among \(m_I+j\) positions gives the displayed
binomial. Positivity bounds any one term by the entire sum. By Theorem
10.112, \(\ln v(k)<(107/103)k\le a\), because \(k\le H_1\) and
\(g_9\ge107/103\). Thus
\(v(k)^u\binom{m_I+h_{\mathrm E}}{h_{\mathrm E}}\le z^{-u-h_{\mathrm E}}(1-z)^{-m_I-1}\).
Taking logarithms proves (10.309), including \(u=0\) or
\(h_{\mathrm E}=0\).

We now bound its actual remainder. The full-width argument (10.220),
with contraction, gives
\(m_I+1\le(r-1)q^{\nu-I}\mathcal D R_{\max}/(c_1c_2Pd)+r-1\). There is
no extra factor \(1/r\): summing the \(r\) widths cancels that factor in
(10.186). Substituting \(R_{\max}\le dW(d)\exp(h)\), \(q^\nu=\exp(x)\),
and the exact formula for \(S\) gives

\[
\frac{m_I+1}{Z}
\le\frac{q^{-I}(r-1)tW(d)\exp(H_1)}
 {c_1c_2c_3qP(r+1)H_1}
+\frac{r-1}{Z}.
\tag{10.311}
\]

For \(0<z<1\), integration gives \(-\ln(1-z)\le z/(1-z)\). Since
\((g_9-1)H_1\ge(g_{91}-1)g_0=1+\ln W(d)\), we have
\(W(d)\exp(H_1)z\le1/\mathrm e\). Also \(H_1>39\) and \(g_9>1\). The
first four exponential terms at 39 exceed 1000, so \(z<1/1000\) and
\((1-z)^{-1}<1000/999\). Finally \(t/H_1\le1/(r+1)\). These
substitutions in (10.311) prove the first inequality of (10.310).

For its uniform last inequality, the definition \(2e<P(p-1)\) gives
\(P\ge3\) in I, \(P\ge5\) in II, and \(P\ge4\) in V; in III and IV use
\(P\ge1\). Direct substitution of the seven rational rows gives the
following lower bounds for \(c_1c_2c_3qP\), all greater than 15:

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}lr@{}}
\toprule\noalign{}
Case & Rational lower bound \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
I.1 & \(445693563/20000000\) \\
I.2 & \(527023581/25000000\) \\
II & \(52892553/2500000\) \\
III.1 & \(18772369/1250000\) \\
III.2 & \(204239/12500\) \\
IV & \(389083863/20000000\) \\
V & \(522494609/25000000\) \\
\end{longtable}
}

For all \(r\ge2\), \((r-1)/(r+1)^2\le1/8\), since \((r-3)^2\ge0\). The
positive exponential series proves \(1/\mathrm e<3/8\), as in Theorem
10.107. The coefficient of \(q^{-I}\) is therefore less than
\((1000/999)(3/8)/(8\cdot15)=25/7992<1/300\). This proves the last
inequality. It holds at every rank and degree; the proof did not test a
finite list of ranks or degrees. \(\square\)

\subsubsection{Retaining the factorial and the actual
field}\label{retaining-the-factorial-and-the-actual-field}

\textbf{Corollary 10.116 (a factorial-preserving general point budget).}
Retain every hypothesis and notation of Theorem 10.92, with its actual
\(R_I=q^{-I}X+2k-1\), number field \(F\), q-denominator exponent
\(\Xi_{I,J,u}\), and \(u\le D_{\mathrm a}\). For an integer total-order
cutoff \(O\ge u+h_{\mathrm E}\), put

\[
\begin{aligned}
B_{I,u}(X,O)&=(D_{\mathrm a}-u)\ln R_I
 +\ln\binom{D_{\mathrm a}}u-L\ln(k!)+aO+E_I,\\
B_I(X,O)&=D_{\mathrm a}\ln(R_I+1)-L\ln(k!)+aO+E_I.
\end{aligned}
\tag{10.312}
\]

Then
\(\ln\mathcal H_{I,u,h_{\mathrm E}}(X)\le B_{I,u}(X,O)\le B_I(X,O)\).
The exact maximization (10.241) remains available, so its logarithm and
\(B_I\) may be replaced by their smaller value when the whole cutoff is
used. For a nonzero value, its same actual normal quantity satisfies

\[
\begin{aligned}
v_p(V)+G_u-\delta\le
\frac{d_F}{e_Ff_F\ln p}\biggl[
&h_2(c)+\tfrac12\ln N+B_{I,u}(X,O)\\
&+\Xi_{I,J,u}\ln q
+PXq^{-I}\sum_iD_i\sigma_i\biggr]+G_u.
\end{aligned}
\tag{10.313}
\]

Here \(G_u=(D_{\mathrm a}-u)\theta+Lv_p(k!)-u v_p(v(k))\), exactly as in
Lemma 10.91. One may take the minimum with the previously proved
normal-coordinate bound (10.297).

\textbf{Proof.} Insert (10.309) into the individual scalar expression
(10.240), and use \(u+h_{\mathrm E}\le O\). This gives its bound by
\(B_{I,u}\). The ordinary binomial expansion gives
\(\binom{D_{\mathrm a}}uR_I^{D_{\mathrm a}-u}\le(R_I+1)^{D_{\mathrm a}}\),
proving \(B_{I,u}\le B_I\). Both bounds keep the negative term
\(-L\ln(k!)\); it has not been canceled by changing the rational
clearing convention.

Apply (10.244) with this scalar bound and then add the same exact
\(G_u\) to both sides. No global field has been restricted, no
q-denominator has been removed, and no derivative order has been
averaged. Equation (10.297) bounds this identical normal quantity in the
identical field. Their minimum is therefore legitimate. At
\(u>D_{\mathrm a}\), the value is zero by Lemma 10.91 and no
nonzero-value argument is needed. \(\square\)

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={Individual scalar peaks with their factorial and the uniform geometric remainder}]{../figures/individual-clearing-euler.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.26. Left: all additive allocations \(0\le u\le8\) with
Euler order \(10-u\), at \(k=4,L=2,D_{\mathrm a}=8,R_I=15,m_I=17\). The
lcm is exactly 12. The plotted scalar is (10.240), including its
denominator \((4!)^2\); its largest value occurs at \(u=1\). The
generating comparison uses \(z=1/13\), so \(a=\ln13\), and displays both
the fixed-allocation and uniform bounds (10.312). This is an
illustrative instance of the positive-series proof, not a substitution
for the original general parameters. Right: the coefficient of \(Z\) in
the first term of the analytical bound (10.310), using
\(1/\mathrm e<3/8\) and \(c_1c_2c_3qP>15\), for \(q=2\) and depths zero,
one and two. Its bound by \(1/(300q^I)\) is proved for every rank; the
separate additive remainder \((r-1)/999\) is retained in (10.310). Proof
locators: Theorem 10.115, Corollary 10.116 and Solution 46.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/individual_clearing_euler.py}{Figure program}.
Human background:
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
free paper}, its definitions (3.16) and the individual scalar estimate
(5.38); the positive-series bound and its uniform remainder are proved
here.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{39. Keeping additive divisibility in the ordinary
jets}\label{keeping-additive-divisibility-in-the-ordinary-jets}

The prepared equations vanish at their input nodes. Their projection
error already contains a useful factor from the additive polynomial.
Keeping that factor through differentiation gives a more precise input
to interpolation, at each fixed additive order. The argument applies to
the original fields and to every contraction depth; it does not average
derivative orders.

\subsubsection{Divisibility of a prepared
scalar}\label{divisibility-of-a-prepared-scalar}

Use the prepared family (10.92), its original and projected curves
(10.104), and the logarithmic slopes of Lemma 10.91. Write

\[
D_{\mathrm a}=kL,\qquad
\ell_p=v_p(\mathfrak v),\qquad
\chi_p=L v_p(k!),\qquad
\gamma_u=\max\{0,u\ell_p-\chi_p\},
\qquad \mathfrak v=\operatorname{lcm}(1,\ldots,k).
\tag{10.314}
\]

The symbol \(\ell_p\) is a valuation, not the exponent of an additive
basis column. The selected Euler arguments are integers, as in Lemma
10.38.

\textbf{Lemma 10.117 (order-dependent projection precision).} Suppose
the hypotheses of Lemma 10.36 hold, with \(b=b_n\), \(\beta=v_p(b_n)\),
and coefficient minimum \(\delta\). At any \(y\in\mathbb Z_p\), every
prepared scalar of additive order \(u\le D_{\mathrm a}\) has valuation
at least \(\gamma_u\). Consequently

\[
v_p\bigl(f(y;\boldsymbol t)-\Phi(y;\boldsymbol t)\bigr)
\ge \delta+U-\beta+\gamma_u,
\qquad u=t_0.
\tag{10.315}
\]

If \(u>D_{\mathrm a}\), both prepared functions vanish identically.

\textbf{Proof.} Consider a column with exponent \(1\le l\le L\). Its
additive prepared scalar is \(\mathfrak v^u\) times the coefficient of
\(V^u\) in

\[
\frac1{(k!)^l}
\prod_{a=1}^k(q^{-I}y+\lambda_{-1}+a+V)^l.
\]

Every coefficient of the numerator is p-integral, since \(q\ne p\) and
\(y\in\mathbb Z_p\). Thus its valuation is at least
\(u\ell_p-lv_p(k!)\ge u\ell_p-\chi_p\). Independently, prepared Taylor
integrality, extended to \(\mathbb Z_p\) in the proof of Lemma 10.36,
gives valuation at least zero. These are simultaneous bounds, giving
their maximum \(\gamma_u\). Each integer Euler binomial is p-integral,
so multiplication by the Euler factors preserves that maximum.

In the proof of Lemma 10.36, the ratio of the two exponential monomials
minus one has valuation at least \(U-\beta\); the original monomial is a
local unit. Multiply this by the scalar bound just proved and by a
coefficient of valuation at least \(\delta\). The ultrametric inequality
proves (10.315). Beyond degree \(D_{\mathrm a}\), every additive
derivative is zero. \(\square\)

This proof also applies at \(y=s/q^J\). A q-denominator has p-adic
valuation zero. It still incurs its full arithmetic cost at primes above
q in Theorem 10.92.

\subsubsection{The precision at one additive
order}\label{the-precision-at-one-additive-order}

Retain the selected derivations and coefficients \(\lambda_{aj}\) of
Lemma 10.38. Assume the full original depth condition
\(v_p(z_j)\ge\vartheta+1/(p-1)\), the p-integrality of \(b_j/b_n\), and
\(0<\theta<\vartheta\). For an integer \(j\ge0\), define

\[
\begin{aligned}
H_{u,j}&=\min\{j,D_{\mathrm a}-u\},\\
M_u(j)&=\min_{0\le a\le H_{u,j}}
\left\{\max\{-a\ell_p,u\ell_p-\chi_p\}
+(j-a)(\vartheta-\beta)\right\}.
\end{aligned}
\tag{10.316}
\]

The minimum is over integers. It computes a lower-bound envelope for the
jets; it does not assert that a jet attains that valuation.

\textbf{Theorem 10.118 (ordinary jets with the additive divisibility
retained).} Fix a prepared index \(\boldsymbol t\) of additive order
\(u\le D_{\mathrm a}\). Suppose the original prepared values at a node
\(s\in\mathbb Z_p\) vanish through total order \(T\). For
\(|\boldsymbol t|+j\le T\),

\[
\begin{aligned}
v_p(f_j(s;\boldsymbol t))
&\ge\delta+U-\beta+M_u(j),\\
v_p((F_u)_j(\rho s;\boldsymbol t))+j\theta
&\ge U-\beta+G_u+M_u(j).
\end{aligned}
\tag{10.317}
\]

Here \(F_u,G_u\) are exactly (10.238), and the subscript \(j\) is
ordinary divided differentiation. In particular, with

\[
C_{\mathrm{sharp}}=\max\{\ell_p,\beta-\vartheta,0\},
\qquad M_u(j)\ge-j C_{\mathrm{sharp}},
\tag{10.318}
\]

one may use the linear loss \(C_{\mathrm{sharp}}\) instead of \(C_0\).
Formula (10.316) can be stronger than this linear consequence.

\textbf{Proof.} Differentiate along the projected curve with the
commuting operator \(\partial_X+\sum_{a\in J}\xi_a\delta_a\) of Lemma
10.38. Its coefficients satisfy
\(v_p(\xi_a)\ge\vartheta+1/(p-1)-\beta\), unless they are zero. For
every positive integer \(h\), the factorial bound (9.4) gives

\[
v_p(\xi_a^h/h!)\ge h(\vartheta-\beta).
\]

The bound is also valid at \(h=0\). Multiplication of an Euler binomial
by a power of its argument has integer expansion coefficients in the
same binomial basis, by the identity used in Lemma 10.38. Thus all Euler
operations of combined order \(j-a\) have scalar valuation at least
\((j-a)(\vartheta-\beta)\), without an additional factorial loss.

The additive part of order \(a\) has multiplier
\((q^I\mathfrak v)^{-a}\binom{u+a}{a}\), whose valuation is at least
\(-a\ell_p\). Its resulting prepared additive order is \(u+a\). The
resulting total prepared order is at most \(|\boldsymbol t|+j\), so its
original value is zero at \(s\). Lemma 10.117 bounds its projected value
by \(\delta+U-\beta+\gamma_{u+a}\). If \(u+a>D_{\mathrm a}\), the term
is identically zero and can be omitted. Every remaining term therefore
has valuation at least

\[
\delta+U-\beta+\gamma_{u+a}-a\ell_p
+(j-a)(\vartheta-\beta).
\]

Since \(\gamma_{u+a}-a\ell_p=\max\{-a\ell_p,u\ell_p-\chi_p\}\),
minimization and the ultrametric inequality prove the first line of
(10.317). The exact chain rule (10.238) proves its second line in the
same normal coordinates.

For the simpler bound, \(\gamma_{u+a}\ge0\) and
\(\vartheta-\beta\ge-C_{\mathrm{sharp}}\), while
\(\ell_p\le C_{\mathrm{sharp}}\). Each term is at least
\(-jC_{\mathrm{sharp}}\), proving (10.318). All q-powers used here are
local units; this assertion does not remove a q-cost from the separate
height estimate. \(\square\)

\subsubsection{Full, nested and q-deleted
interpolation}\label{full-nested-and-q-deleted-interpolation}

For an integer maximum multiplicity \(M\ge1\) and separation bound
\(B\ge0\), put

\[
\begin{aligned}
K_u(M,B)&=\min_{0\le j<M}\{M_u(j)+jB\}\\
&=\min_{0\le a\le H}
\bigl\{\max\{-a\ell_p,u\ell_p-\chi_p\}+aB\\
&\hspace{37mm}+(M-1-a)\min\{0,\vartheta-\beta+B\}\bigr\},\\
H&=\min\{M-1,D_{\mathrm a}-u\}.
\end{aligned}
\tag{10.319}
\]

These are finite integer minima. To evaluate the second one when
\(\ell_p>0\), it suffices to check \(0,H\) and the floor and ceiling of
\((\chi_p-u\ell_p)/\ell_p\), clipped to \([0,H]\). Indeed its expression
is affine on either side of that one breakpoint, and a linear function
on an integer interval takes its minimum at an endpoint. When
\(\ell_p=0\), it equals \((M-1)\min\{0,\vartheta-\beta+B\}\).

To prove the equality in (10.319), write \(j=a+h\). At each fixed \(a\),
the admissible Euler order is \(0\le h\le M-1-a\). Its coefficient in
the expression to minimize is \(\vartheta-\beta+B\). Choosing the
appropriate endpoint gives exactly the second line. No real relaxation
of an integer order is being made.

\textbf{Theorem 10.119 (interpolation at the retained jet precision).}
Use any one of the following proved node systems:

\begin{itemize}
\tightlist
\item
  The full nested intervals of Lemma 10.94. Set \(N_*\) as in (10.247),
  \(M=\mu_0\), and \(E=0\).
\item
  The mixed q-deleted and full intervals of Lemma 10.96. Set \(N_*\) as
  in (10.251), \(M=\mu_0\), and \(E=\mu_0-\mu_1\).
\item
  A q-deleted interval \(|s|\le R\), \(q\nmid s\), with \(q\mid R\) and
  constant multiplicity \(M\). Set \(N_*=2(1-1/q)RM\) and \(E=M\).
\end{itemize}

In the first two cases use their proved separation bound \(B\); in the
last use \(B=\lfloor\log_p(2R)\rfloor\). At every node \(s\), suppose
the original prepared values vanish through order
\(|\boldsymbol t|+\mu(s)-1\), where \(\boldsymbol t\) has fixed additive
order \(u\le D_{\mathrm a}\). Put \(\Lambda_u=U-\beta+G_u\). Then, for
every \(x\in\mathbb Z_p\subset\mathbb Q_p\),

\[
\begin{aligned}
v_p(f(x;\boldsymbol t))+G_u-\delta&\ge\mathcal P_u,\\
v_p(\Phi(x;\boldsymbol t))+G_u-\delta&\ge\mathcal P_u,\\
\mathcal P_u&=\min\{N_*\theta,\Lambda_u-EB-(M-1)B+K_u(M,B)\}.
\end{aligned}
\tag{10.320}
\]

\textbf{Proof.} Lemma 10.91 makes \(F_u\) normal. Theorem 10.118 gives
its weighted jets with baseline \(\Lambda_u=U-\beta+G_u\) and increment
\(M_u(j)\). Use the complete cardinal polynomial, Taylor inverse and
normal quotient constructions of Lemmas 10.94 and 10.96. At a node of
multiplicity \(\mu(s)\), its cardinal factor has valuation at least
\(-B(M-\mu(s)+E)\). The inverse coefficient of degree \(a\le\mu(s)-1-j\)
has valuation at least \(-aB\). These assertions also hold for the
constant q-deleted system, with \(E=M\), by the cardinal and inverse
proof of Theorem 10.39.

A Hermite summand with divided jet \(j\) thus has valuation at least

\[
\Lambda_u+M_u(j)-B(M-\mu(s)+E)-aB
\ge\Lambda_u-EB-(M-1)B+M_u(j)+jB.
\]

Every such \(j\) is less than \(M\), so (10.319) gives the second entry
of (10.320). The same monic normal quotient has valuation at least
\(N_*\theta\) at \(\rho x\), exactly as in those proved interpolation
lemmas. Their sum proves the asserted lower bound for \(f\).

At \(j=0\), (10.316) gives \(M_u(0)=\gamma_u\). Hence
\(K_u(M,B)\le\gamma_u\), and the second entry of (10.320) is at most
\(\Lambda_u+\gamma_u\). Lemma 10.117 gives at least this last precision
for the normalized projection error at \(x\). Adding that error to \(f\)
proves the same assertion for \(\Phi\). \(\square\)

The q-deleted extra term \(EB\) survives. The lower bound also retains
the exact factorial/lcm terms in \(G_u\). From (10.318), its second
entry is at least \(U-\beta+G_u-EB-(M-1)\max\{B,C_{\mathrm{sharp}}\}\).
The finite envelope (10.319) can give additional precision.

\subsubsection{The comparison with an actual algebraic
value}\label{the-comparison-with-an-actual-algebraic-value}

\textbf{Corollary 10.120 (a fixed-order zero criterion).} Retain the
hypotheses and precision \(\mathcal P_u\) of Theorem 10.119. At
\(x=s/q^J\), suppose the phase and field hypotheses of Theorem 10.92
identify the analytic original value \(\Phi(x;\boldsymbol t)\) with its
actual algebraic value \(V\in F\). Take its same coefficient minimum and
fixed additive/Euler orders, full global degree and local degrees. Let
\(\mathcal A_u(x)\) be the smaller of the right sides of (10.312) and
(10.297), with these exact data. Then

\[
\mathcal P_u>\mathcal A_u(x)
\quad\Longrightarrow\quad V=0.
\tag{10.321}
\]

In particular the sufficient input budget

\[
U-\beta+G_u\ge N_*\theta+EB+(M-1)B-K_u(M,B)
\tag{10.322}
\]

and the strict comparison \(N_*\theta>\mathcal A_u(x)\) imply that zero.

\textbf{Proof.} For a nonzero \(V\), Corollary 10.116 and Theorem 10.109
give \(v_p(V)+G_u-\delta\le\mathcal A_u(x)\) in the same actual field.
Theorem 10.119 gives the opposite strict inequality under (10.321), a
contradiction. Substitution of (10.322) proves the last assertion.
\(\square\)

The criterion keeps the separate remainder in (10.311), the original
root field, q-denominators and derivative order. Its two inequalities
must still be verified at the particular original radii, multiplicities
and arithmetic data.

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={The additive divisibility breakpoint and its effect on fixed-order interpolation}]{../figures/additive-jet-precision.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.27. An exact illustrative scalar calculation at
\(p=3,k=4,L=2,D_{\mathrm a}=8\), so \(\ell_p=1,\chi_p=2\), with
\(\beta=0,\vartheta=1,B=0,M=7\). Left: at \(u=0\), the expression
minimized in (10.319) is \(\max\{-a,-2\}\), for integer \(0\le a\le6\);
its minimum is \(-2\), attained from \(a=2\) onward. The dashed line
displays the bound \(-a\) obtained after discarding additive
divisibility. Right: the exact lower-bound envelope is \(K_u(7,0)=u-2\)
for every \(0\le u\le8\), compared with the common linear-loss
contribution \(-6\) from \(C_0=1\). These are contributions to
\(U-\beta+G_u\), not valuations of actual nonzero auxiliary values. The
displayed integers are checked in Solution 47; Theorems 10.118--10.119
prove the general mechanism.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/additive_jet_precision.py}{Figure program}.
Human-source context: the original ordinary-jet extrapolation in
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
freely accessible paper}, Section 5, equations (5.25)--(5.32); the
order-dependent refinement is proved above.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{40. General integer zero production at the original
parameters}\label{general-integer-zero-production-at-the-original-parameters}

We now perform the strict analytic and arithmetic comparisons for every
original field, prime and rank case. The integer radii and jet orders
are formed from the original real parameters. The q-denominator at a
contracted stage is charged in full; its depth-dependent cost is
controlled by the corresponding decrease in the torus height.

\subsubsection{The original height and descent
ranges}\label{the-original-height-and-descent-ranges}

Retain Sections 26, 29 and 36--39. Choose the original
\(\theta=\vartheta/(1+\varepsilon_n)\), where
\(\varepsilon_n=10^{-26}/(2n)\) and \(n\) is the number of original
logarithms, as in (10.74). The estimates below are also valid for
\(\theta=\vartheta/(1+\tau)\) with \(0<\tau\le10^{-6}\). Let \(b_n\ne0\)
have minimum p-adic valuation among the nonzero coefficients. Put

\[
\begin{aligned}
B_{\dagger}&=\min_{b_j\ne0}|b_j|,&
h&\ge\max\{g_0,(r+1)t,\ln B_{\dagger},\ln(R_{\max}/(dW(d)))\},\\
U&=\frac{q^{r+1}S\mathcal D}{et},&
W_*&=\frac{et}{dZ},&W_*U&=q^{r+1},\\
S_I&=S\eta^{-(r+1)I},&T_I&=T\eta^{(r+1)I},&
\eta&=1-\frac{c_5}{r+1}.
\end{aligned}
\tag{10.323}
\]

Here \(t=f\ln p\), \(Z=S\mathcal D/d\), \(k=\lfloor h+\nu\ln q\rfloor\),
and every other parameter is exactly (10.186). The following values of
\(c_5\) complete the original constant table.

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}lrrr@{}}
\toprule\noalign{}
Case & \(r=2\) & \(3\le r\le7\) & \(r\ge8\) \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
I.1 & 0.5377 & 0.55 & 0.56 \\
I.2 & 0.538 & 0.551 & 0.56 \\
II & 0.53 & 0.54 & 0.55 \\
III.1 & 0.528 & 0.536 & 0.55 \\
III.2 & 0.5267 & 0.534 & 0.55 \\
IV & 0.5345 & 0.543 & 0.56 \\
V & 0.753 & 0.78 & 0.827 \\
\end{longtable}
}

Use a nonnegative integer stage \(I\) satisfying
\(I\ln(q\eta^{r+1})\le3(A+\nu\ln q)\) and \(c_5T_{I+1}/(r+1)\ge1\).
These are the original geometric and multiplicity restrictions before
either stopping point. They include \(I=0\): Lemma 10.104 gives
\(T\ge g_4>6(r+1)\), while \(c_5>1/2\) and \(\eta^{r+1}>1/3\).

An admissible stage family has the same selected Euler derivations, an
integral coefficient subvector of the initial kernel, its reference
coefficient of minimum valuation, the phase condition of Theorem 10.55,
full widths at most \(q^{-I}D_i\), and additive factors evaluated at
\(q^{-I}x\). The original torus slopes and their projection are those of
Lemma 10.91. The phase identity (10.151) puts every integer value in
\(K\); the height and coefficient-minimum conclusions of Theorems 10.48
and 10.60 keep the original coefficient bound. Thus these data describe
the full original induction, without a special rational or fixed-rank
family.

\subsubsection{Coupled field bounds}\label{coupled-field-bounds}

\textbf{Lemma 10.121 (surplus and depth in the same field).} Write

\[
C_{01}=c_0\ell_d\frac{p^f}{p^f-1},\qquad
Y=e\theta,\qquad
\Omega_{\mathrm n}=\frac e d\left(\theta+\frac1{p-1}\right).
\tag{10.324}
\]

The following bounds hold simultaneously. In the third column, the lower
bound for \(Y\) is the displayed number divided by \(1+10^{-6}\).

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}
  >{\raggedright\arraybackslash}p{(\linewidth - 6\tabcolsep) * \real{0.2000}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 6\tabcolsep) * \real{0.2667}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 6\tabcolsep) * \real{0.2667}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 6\tabcolsep) * \real{0.2667}}@{}}
\toprule\noalign{}
\begin{minipage}[b]{\linewidth}\raggedright
Case
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
Lower bound for \(C_{01}\)
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
Numerator of the lower bound for \(Y\)
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
Upper bound for \(\Omega_{\mathrm n}\)
\end{minipage} \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
I.1 & \(9c_0/8\) & \(3/2\) & 2 \\
I.2 & \(57/20\) & \(3/2\) & 2 \\
II & \(5c_0/4\) & \(5/2\) & \(3/2\) \\
III.1 & \(c_0\) & \(3/4\) & \(1/2\) \\
III.2 & \(c_0\) & \(3/4\) & 1 \\
IV, \(P=1\) & \(c_0\) & \(1/2\) & \(1/2\) \\
IV, \(P\ge7\) & \(c_0\) & \(7/2\) & \(4/3\) \\
V & \(10/3\) & 2 & \(3/2\) \\
\end{longtable}
}

\textbf{Proof.} The local degree identity gives \(ef\le d\). In I.1, if
\(d=2\), then \(f\le2\) and \(3^f/(3^f-1)\ge9/8\). At \(d=3\), use
\(f\le3\) and \(\ln3>13/12\), giving
\(\ell_d3^f/(3^f-1)>(13/12)(27/26)=9/8\). At \(d\ge4\),
\(\ell_d\ge\ln4>4/3\) suffices. In I.2, \(e=f=d=1\), so
\(C_{01}=(19/10)(3/2)=57/20\). In II, \(e\ge2\). At \(d=2,3\), we have
\(f=1\), giving the factor \(5/4\). At larger \(d\), \(\ell_d>4/3>5/4\).
The other odd-prime surplus bounds follow directly from \(\ell_d\ge1\)
and \(p^f/(p^f-1)>1\).

In V, \(K\) contains a primitive cube root. Its quadratic field has
degree two, so \(d\) is even. Its reduction has order three, hence \(f\)
is even and at least two, as proved in Lemma 10.104. At \(d=2\), \(f=2\)
and the surplus is \((5/2)(4/3)=10/3\). At \(d\ge4\), use
\(\ell_d>4/3\). The logarithm bounds follow from the positive series in
Solution 34; in particular \(\ln2>2/3\) and \(\ln3>1098/1000>13/12\).

Outside III, \(Y=P/[2(1+\tau)]\). The defining strict inequality
\(2e<P(p-1)\) gives \(P\ge3,5,4\) respectively in I, II and V. In IV,
\(P\) is either one or at least \(p\ge7\). In III, \(e=1\) and
\(\vartheta=(p-2)/(p-1)\ge3/4\). These prove the depth column.

For I, use \(\theta\le3/2\), \(e/d\le1\), giving
\(\Omega_{\mathrm n}\le2\); for II use \(\theta\le5/4\). In III,
\(\theta+1/(p-1)\le1\), and \(d\ge2\) in III.1. If IV has \(P=1\), then
\(2e<p-1\), so
\(\Omega_{\mathrm n}=[1/(2(1+\tau))+e/(p-1)]/d<1/d\le1/2\). For its
other branch, \(\theta\le p/(p-1)\) gives
\(\Omega_{\mathrm n}\le(p+1)/(p-1)\le4/3\). Finally Lemma 10.73 gives
\(e/d\le1/2\) in V, and \(\theta\le2\), giving
\(\Omega_{\mathrm n}\le3/2\). \(\square\)

\subsubsection{The additive polynomial and the contracted
denominator}\label{the-additive-polynomial-and-the-contracted-denominator}

\textbf{Lemma 10.122 (integer-stage scalar costs).} At the target radius
\(X=q^{j+1}S_I\), \(j\ge0\), the additive part of the bound (10.312)
satisfies

\[
\frac{D_{\mathrm a}\ln(R_I+1)-L\ln(k!)}Z
<\frac{\lambda}{c_1c_4}
\left(1+\frac{j\ln q}{A+\nu\ln q}\right),
\qquad \lambda<\frac{189}{188}.
\tag{10.325}
\]

At \(I=0\) there is no q-denominator. At \(I\ge1,j\ge1\), its full
denominator cost and the torus height obey

\[
\frac{\Xi_{I,0,u}\ln q}Z
+\frac{PXq^{-I}\sum_iD_i\sigma_i}Z
\le\frac{q^{j+1}}{c_1c_2}
+\frac{\lambda\ln q}{c_1c_4(q-1)(A+\nu\ln q)}.
\tag{10.326}
\]

\textbf{Proof.} Monotonicity of \(\ln x\) gives
\(\ln(k!)\ge\int_1^k\ln x\,dx=k\ln k-k+1\). Thus the additive numerator
in (10.325) is at most \(D_{\mathrm a}\ln[\mathrm e(R_I+1)/k]\). Put
\(H_1=h+\nu\ln q\). Since \(k\ge39\), \(H_1/k<40/39\). We have
\(R_I+1=2k+q^{j+1}S(q\eta^{r+1})^{-I}\). The table gives
\(c_3q^2<77/5\), and \((r+1)d\ge3,t>1\). Therefore

\[
\frac{R_I+1}{k}
<q^j(r+1)d\left(\frac23+\frac{77}5\frac{40}{39}\right)
=\frac{214}{13}q^j(r+1)d.
\]

The exponential series through degree five at three exceeds \(214/13\),
so \(\ln[\mathrm e(R_I+1)/k]<g_1+j\ln q\le A+j\ln q\). Insert
\(D_{\mathrm a}/Z\le\lambda/[c_1c_4(A+\nu\ln q)]\) from Lemma 10.70.
This proves (10.325), retaining the factorial contribution.

Let \(Q=q\eta^{r+1}>1\), \(D=A+\nu\ln q>5\). The exact full-width term
is \(q^{j+1}Q^{-I}/(c_1c_2)\). Formula (10.243) and the proved factorial
bound give \(\Xi_{I,0,u}\le D_{\mathrm a}(I+1/(q-1))\). Only the part
proportional to \(I\) needs absorption. Set \(a=q^{j+1}/(c_1c_2)\),
\(b=\lambda\ln q/(c_1c_4D)\). The function \(aQ^{-z}+bz\) is convex for
\(z\ge0\). On \(0\le z\le3D/\ln Q\), its maximum is bounded by its two
endpoints. At zero its value is \(a\). At the other endpoint its ratio
to \(a\) is \(\exp(-3D)+3\lambda c_2\ln q/(c_4q^{j+1}\ln Q)\).

For odd primes, \(c_5\le14/25\) and the increasing function
\((1-c/y)^y\), \(y\ge3\), gives \(Q\ge2(61/75)^3>43/40\). Hence
\(\ln Q>6/83\), by the first term of the positive logarithm series. With
\(c_4>18\), \(j\ge1\), and \(\ln2<347/500\), the second ratio term is
less than \(4233747/6016000<3/4\). In V, apply the same increasing
function separately in the three rank bands: their endpoints give
\(3(749/1000)^3,3(161/200)^4,3(8173/9000)^9>5/4\). Thus \(\ln Q>2/9\).
Using \(c_4>15/4\) and \(\ln3<1099/1000\), the ratio term is less than
\(300027/470000<2/3\). The first ratio term is less than \(1/1000\):
\(D>5\), and the exponential series through degree six at 15 exceeds
1000. Both endpoint ratios are less than one. Convexity therefore
absorbs the entire \(I\)-dependent cost. The remaining \(1/(q-1)\) term
is exactly the second term in (10.326).

For the stated monotonicity, the derivative of \(y\ln(1-c/y)\) is
\(\ln(1-z)+z/(1-z)>0\), \(z=c/y\); its latter expression has derivative
\(z/(1-z)^2>0\) and value zero at zero. No denominator at q has been
treated as globally harmless. \(\square\)

\subsubsection{The analytic input at every
order}\label{the-analytic-input-at-every-order}

\textbf{Lemma 10.123 (all integer input budgets).} Put
\(\beta=v_p(b_n)\). For a fixed additive order \(u\le D_{\mathrm a}\),
the exact envelopes of Section 39 satisfy

\[
G_u+K_u(M,B)
\ge(D_{\mathrm a}-u)\theta
 -(M-1)\max\{0,\beta-\vartheta-B\}.
\tag{10.327}
\]

Consequently full-node interpolation has precision at least

\[
\min\{N_*\theta,\ U-\beta+(D_{\mathrm a}-u)\theta
 -(M-1)\max\{B,\beta-\vartheta,0\}\}
\tag{10.328}
\]

in the unchanged normal quantity \(v_p(\Phi)+G_u-\delta\).

At any allowed stage \(I\), and for \(0\le j\le r\), define
\(R=\lfloor q^jS_I\rfloor\),
\(\mu=\lfloor c_5\eta^jT_I/(r+1)\rfloor+1\), \(N_*=(2R+1)\mu\), and
\(B=\lfloor\log_p(2R)\rfloor\). Then, simultaneously for all
\(u\le D_{\mathrm a}\),

\[
U-\beta+(D_{\mathrm a}-u)\theta
>N_*\theta+(\mu-1)\max\{B,\beta-\vartheta,0\}.
\tag{10.329}
\]

The analytic gain obeys

\[
W_*N_*\theta>
\frac{c_5q\eta^j}{c_1}\left(2q^j-\frac1{S_I}\right).
\tag{10.330}
\]

\textbf{Proof.} Write \(\ell=v_p(\mathfrak v),\chi=Lv_p(k!)\). In the
expression defining \(M_u(a)\), the identity
\(\chi-u\ell+\max\{-b\ell,u\ell-\chi\}=\max\{\chi-(u+b)\ell,0\}\) is
nonnegative. The remaining Euler contribution is at least
\(a\min\{\vartheta-\beta,0\}\). Thus
\(G_u+M_u(a)\ge(D_{\mathrm a}-u)\theta+a\min\{\vartheta-\beta,0\}\).
Minimizing after adding \(aB\) proves (10.327), and substitution in
Theorem 10.119 proves (10.328).

For each nonzero \(b_j\), its integer factorization gives
\(v_p(b_j)\ln p\le\ln|b_j|\). The minimum-valuation choice of \(b_n\)
therefore gives \(\beta\ln p\le\ln B_\dagger\le h\). Next,
\(\ln(3q^rS)<g_1+(11/10)r+\ln H_1\), since \(3c_3q<24<\mathrm e^4\) and
\(t>1\). We have \(g_1\le4+r+\ln d\), while (10.288) implies
\(H_1>2(4+21r/10+\ln d)\). Also \(\ln H_1<H_1/2\), by the maximum
\(\ln z/z=1/\mathrm e<1/2\). Hence \(\ln(3q^rS)<H_1\). With
\(Y_I=\eta^{-(r+1)I}\ge1\), the inequality \(\ln Y_I\le Y_I-1\) gives
\(\ln(2R+1)\le\ln(3q^rS_I)<Y_IH_1\). It follows that
\(\max\{B,\beta-\vartheta,0\}\le Y_IH_1/\ln p\).

The identities \(S_IT_I=ST\) and \(W_*ST\theta=q(r+1)/c_1\), with
\(\mu>c_5\eta^jT_I/(r+1)\) and \(2R+1>2q^jS_I-1>0\), prove (10.330). For
the upper input bound, the stage restriction gives
\(\mu\le c_5(\eta^j+\eta^{r+1})T_I/(r+1)\). Moreover
\(W_*\beta<1/[7(r+1)b_r]\), using \(ef\le d\),
\(\mathcal D>b_r(A+\nu\ln q)t\), and \(c_3q>7/5\). The remaining jet
loss satisfies
\(W_*(\mu-1)\max\{B,\beta-\vartheta,0\}\le c_5\eta^j/[c_1c_3Y(r+1)]\).

For completeness, \(2c_5(\eta^r+\eta^{r+1})<27/20\) at odd primes and is
less than \(3/2\) in V. To prove these uniform bounds, set
\(c=c_5,y=r+1,z=c/y\). The logarithmic derivative of
\((1-c/y)^{y-1}(2-c/y)\) is \(\ln(1-z)+z/(1-z)-z/[y(1-z)(2-z)]\). Its
first two terms equal \(\int_0^z t/(1-t)^2\,dt\le z^2/[2(1-z)^2]\). The
full derivative is nonpositive because \(c(2-z)\le2(1-z)\): at
\(z\le c/3\) this follows from \(6-8c+c^2>0\) for \(c\le827/1000\). Thus
the expression is largest at \(r=2\). The product \(2c(1-c/3)^2(2-c/3)\)
increases up to \(c=827/1000\), since its derivative has the sign of
\(18-24c+4c^2>0\) there. At the odd endpoint \(c=14/25\), and at the V
endpoint \(c=827/1000\), direct rational substitution gives the asserted
bounds.

Divide the upper input sum by \(q^{r+1}/c_1\). Since \(q\eta>1\), both
\((q\eta)^j\) and \(q^j\) increase with \(j\); their endpoint \(j=r\)
bounds the principal part by these last constants. The floor-radius
remainder is less than \(1/200\), because \(S_I>58\) and \(q^r\ge4\).
The jet-loss remainder is less than \(1/75\), since \(c_3Y>19/10\) at
odd primes and \(c_3Y>19/20\) in V, by Lemma 10.121. The
coefficient-minimum remainder is less than \(1/10000\), using \(c_1<3\)
and \(b_r\ge7200/29\). Their sum is less than \(1/50\). Hence the input
sum is less than \((137/100)q^{r+1}/c_1\) at odd primes and
\((38/25)q^{r+1}/c_1\) in V. These are less than \(W_*U=q^{r+1}\), since
\(c_1>7/5\) or \(c_1>5/2\), respectively. Dropping the nonnegative
\((D_{\mathrm a}-u)\theta\) proves (10.329).

The projection is defined with the required depth: from (10.186),
\(U/H_1>c_3q^{r+2}(r+1)b_r>5000\), while \(\beta<3H_1/2\). Thus
\(U-\beta>1\ge1/(p-1)\). \(\square\)

\subsubsection{The strict arithmetic
comparison}\label{the-strict-arithmetic-comparison}

Use the simultaneous field bounds of Lemma 10.121, denoting its last
three columns by \(C,Y_-,\Omega_+\); thus \(Y_-=Y_0/(1+10^{-6})\). Set
\(\bar\lambda=189/188\), \(\ell_2=347/500\), \(\ell_3=1099/1000\), and

\[
\begin{aligned}
b_r&=\frac{7200}{29}14^{r-2},&
\epsilon_r&=\frac{3+14/(r+1)}{7b_r},\\
\bar g_{12}&=\begin{cases}0&d=1,\\(1+5/(r+1))/(546b_r)&d>1,\end{cases}&
\bar g_9&=\begin{cases}107/103&d=1\text{ or }r\ge8,\\
\max\{107/103,1+1/(3(r+1))\}&\text{otherwise}.\end{cases}
\end{aligned}
\tag{10.331}
\]

The following lower bounds \(S_-\le S\) will retain the floor-radius
loss.

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}lr@{}}
\toprule\noalign{}
Case & \(S_-\) \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
I.1 and I.2 & \((10/11)c_3q(r+1)^2(39r/10+36/5)\) \\
II & \(c_3q(r+1)^2(39r/10+36/5)\) \\
III.1 and both IV branches & \(2c_3q(r+1)^2\) \\
III.2 & \(c_3q(r+1)^2\) \\
V & \((500/347)c_3q(r+1)^2(39r/10+36/5)\) \\
\end{longtable}
}

Indeed \(S=c_3q(r+1)(d/f)H_1/\ln p\), and \(d/f\ge e\). Use \(h\ge g_0\)
with \(\ln3<11/10\) in I, \(e\ge2\) and \(\ln5<2\) in II, and
\(\ln2<347/500\) in V. These are the first three bounds. For III and IV
use \(H_1\ge(r+1)t\) and their minimum degree, two except in III.2. The
resulting \(S_-\) increase with \(r\); direct substitution at \(r=2\)
gives \(S_->58\) in every row. Thus the same lower bound applies to
\(S_I\).

Define

\[
\begin{aligned}
A_r={}&c_1(\bar g_{12}+\epsilon_r)\\
&+\frac{c_1\epsilon_r/2+\bar\lambda/c_4+
C_r^{\sharp}/(c_3Y_-)+117/(232c_2)}{C-1},\\
\mathcal R_r(j)={}&A_r+
c_1\left(\frac1{300}+\frac1{273\cdot999b_r}\right)
+\frac{\bar g_9\eta^{j+1}}{c_3Y_-}\\
&+\frac{\bar\lambda}{c_4}
\left(1+\Omega_++\frac{j\ell_q}{5}
 +\frac{\chi_j\ell_q}{5(q-1)}\right)
 +\frac{q^{j+1}}{c_2},\\
\mathcal L_r(j)={}&c_5q\eta^j\left(2q^j-\frac1{S_-}\right)
 -\mathcal R_r(j),
\qquad \chi_0=0,\quad\chi_j=1\ (j\ge1).
\end{aligned}
\tag{10.332}
\]

The number \(117/232\) bounds \((1+(2g_2)^{-1})/2\), since \(g_2>58\).
Corollary 10.113 and Lemma 10.105 give
\(c_1[h_2(c)+\tfrac12\ln N_0]/Z<A_r\). These bounds persist for every
contracted coefficient subvector: each local vector norm decreases under
deletion, hence so does the projective height, and \(N_I\le N_0\). Its
chosen minimum reference coefficient remains present.

\textbf{Lemma 10.124 (the comparison margin increases along the integer
steps).} For every original rank and field case,
\(\mathcal L_r(j)>\mathcal L_r(0)\) at every integer \(1\le j\le r-1\).

\textbf{Proof.} Write \(c=c_5\), \(a=\bar\lambda\ell_q/(5c_4)\). First
omit the indicator term \(\chi_ja/(q-1)\) in \(\mathcal R_r(j)\), and
allow a real variable \(s\in[0,r-1]\). Call the resulting difference
\(\widetilde{\mathcal L}(s)\). Differentiation and deletion of its
positive terms give

\[
\widetilde{\mathcal L}'(s)
\ge q^{s+1}\left[2c\eta^s\ln(q\eta)-\frac{\ln q}{c_2}\right]-a.
\tag{10.333}
\]

The terms deleted here are positive because \(\ln\eta<0\): they come
from \(-cq\eta^s/S_-\) and \(-\bar g_9\eta^{s+1}/(c_3Y_-)\).

We establish a uniform positive lower bound for the bracket. Put
\(b=c/(r+1-c)\). The inequality \(-\ln(1-z)\le z/(1-z)\) follows by
integration of \(1/(1-z)\). Hence \(\eta^{r-1}\ge\exp[-c+(2-c)b]\) and
\(\ln(q\eta)\ge\ln q-b>0\). For \(s\le r-1\), the bracket is therefore
at least \(2c\exp(-c)\exp((2-c)b)(\ln q-b)-\ln q/c_2\). The factor
\(2c\exp(-c)\) increases for \(0<c<1\). For fixed positive \(a_0,l\),
the derivative of \(\exp(a_0b)(l-b)\) changes sign at most once, from
positive to negative. Its minimum on a compact interval consequently
occurs at an endpoint.

At odd primes, \(5267/10000\le c\le14/25\), \(b\le14/61\),
\(2-c\ge36/25\), and \(\ln2>693/1000\). The exponential at
\(5267/10000\) is less than \(17/10\). At the two endpoints, the
resulting product has the following lower bounds:

\[
\frac{693}{1000},\qquad
\left(\frac{693}{1000}-\frac{14}{61}\right)
\sum_{m=0}^{5}\frac{(504/1525)^m}{m!}
>\frac{129}{200}.
\]

It follows that the bracket is greater than
\(2(5267/10000)(10/17)(129/200)-(347/500)(4/7)
=36901/11900000>1/400\). In V, use \(753/1000\le c\le827/1000\),
\(b\le827/2173\), \(2-c\ge1173/1000\), \(\ln3>1098/1000\), and
\(\exp(753/1000)<213/100\). The endpoint at zero is \(1098/1000\); the
other endpoint, bounded with the exponential series through degree five,
exceeds that number. The bracket is therefore greater than
\(2(753/1000)(100/213)(1098/1000)-(1099/1000)(9/13)
=71469/4615000>1/100\). All exponential upper bounds just used follow by
summing through degree ten and bounding the remaining positive series by
\(x^{11}/[11!(1-x/12)]\), for \(0\le x<1\); the ratio of each following
term to its predecessor is at most \(x/12\). The logarithm intervals
follow from the positive series of Solution 34, with four terms for
\(\ln2\) and five for \(\ln3\), and its geometric tail estimate. Thus
none is an unproved numerical decimal.

For \(s\ge1\), (10.333) is positive: at odd primes,
\(q^{s+1}/400\ge1/100>a\), since \(a<1/125\); in V,
\(q^{s+1}/100\ge9/100>a\), since \(a<3/50\). We must also pay the jump
of the indicator at the first integer. On \(0\le s\le1\), use
\(\eta^s\ge\eta\). At odd primes, \(\eta\ge61/75\),
\(\ln(2\eta)>12/25\), and therefore the bracket exceeds
\(2(5267/10000)(61/75)(12/25)-(347/500)(4/7)>7/500\). In V,
\(\eta\ge2173/3000\), \(\ln(3\eta)>3/4\), giving a bracket greater than
\(2(753/1000)(2173/3000)(3/4)-(1099/1000)(9/13)>1/20\). The two
logarithm lower bounds follow from the first two positive terms, at
\(2\eta\ge122/75\) and \(3\eta\ge2173/1000\). Integrating (10.333) on
this interval yields
\(\widetilde{\mathcal L}(1)-\widetilde{\mathcal L}(0)>q d_0-a>a/(q-1)\),
using \(d_0=7/500\) or \(1/20\). Indeed \(2(7/500)-1/125>1/125\), and
\(3/20-3/50>3/100\). After the full indicator cost is subtracted,
\(\mathcal L_r(1)>\mathcal L_r(0)\). At later integers that cost is
constant, and the positive derivative proves the assertion. \(\square\)

\textbf{Lemma 10.125 (strict margins in all original cases).} For every
\(r\ge2\) and \(0\le j\le r-1\),

\[
\mathcal L_r(j)>\frac1{200}.
\tag{10.334}
\]

\textbf{Proof.} By Lemma 10.124 it suffices to check \(j=0\). For
\(r=2,3,\ldots,7\), substitute their six integer values into
(10.331)--(10.332), using the original constants of (10.186), the
displayed \(c_5\) table and the coupled field bounds. Every quantity is
rational. The following table gives strict lower bounds for the minimum
of those six results.

For every \(r\ge8\), use the single last-band value of \(c_5\),
\(C_r^{\sharp}<1\), \(\eta\le1\), \(\bar g_9=107/103\), and the values
of \(\epsilon_r,\bar g_{12},b_r,S_-\) at \(r=8\). This is a uniform
bound: \(b_r\) increases, \(\epsilon_r\) and \(\bar g_{12}\) decrease,
and \(S_-\) increases. Thus every positive arithmetic cost is enlarged
and every gain is diminished by these substitutions. The inequality
\(C_r^{\sharp}<1\) follows directly from (10.307), or by subtracting it
from one and using \(99/103>17/24\). The last column is the resulting
strict lower bound for every integer \(r\ge8\).

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}
  >{\raggedright\arraybackslash}p{(\linewidth - 4\tabcolsep) * \real{0.2727}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 4\tabcolsep) * \real{0.3636}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 4\tabcolsep) * \real{0.3636}}@{}}
\toprule\noalign{}
\begin{minipage}[b]{\linewidth}\raggedright
Case
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
Minimum over \(2\le r\le7\), greater than
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
Uniformly for \(r\ge8\), greater than
\end{minipage} \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
I.1 & \(80/1000\) & \(76/1000\) \\
I.2 & \(17/1000\) & \(5/1000\) \\
II & \(91/1000\) & \(89/1000\) \\
III.1 & \(52/1000\) & \(65/1000\) \\
III.2 & \(22/1000\) & \(45/1000\) \\
IV, \(P=1\) & \(28/1000\) & \(47/1000\) \\
IV, \(P\ge7\) & \(592/1000\) & \(690/1000\) \\
V & \(248/1000\) & \(402/1000\) \\
\end{longtable}
}

These are finite rational inequalities, obtainable by multiplying
positive denominators. The complete rational inputs and all 56 exact
differences are generated by the
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/integer_comparison_bounds.py}{editable
arithmetic certificate}. Its calculation uses precisely
(10.331)--(10.332), including \(S_-\), the normality contribution and
the separate remainder. Solution 48 spells out the narrowest uniform row
and its comparison with \(1/200\). The eight uniform rows use proved
inequalities for all larger ranks; they are not an inference from
sampled ranks. Lemma 10.124 then supplies every integer step, including
the first q-denominator cost. \(\square\)

\subsubsection{The integer extension
theorem}\label{the-integer-extension-theorem}

\textbf{Theorem 10.126 (general integer zero production).} Assume the
original contradiction \(v_p(\mathcal L)\ge U\), the integral initial
kernel of Corollary 10.113, and an admissible original stage \(I\) as
above. Let \(j\in\{0,\ldots,r-1\}\) if \(I=0\), or
\(j\in\{1,\ldots,r-1\}\) if \(I\ge1\). Suppose all its prepared values
vanish at the integers \(|s|\le\lfloor q^jS_I\rfloor\) and selected
derivative orders \(|\boldsymbol t|\le\lfloor\eta^jT_I\rfloor\). Then
all its prepared values vanish at

\[
|s|\le\lfloor q^{j+1}S_I\rfloor,\qquad
|\boldsymbol t|\le\lfloor\eta^{j+1}T_I\rfloor.
\tag{10.335}
\]

\textbf{Proof.} Fix one target derivative, with additive order \(u\) and
total selected order at most \(O=\lfloor\eta^{j+1}T_I\rfloor\). If
\(u>D_{\mathrm a}\), it is identically zero by polynomial degree.
Otherwise take \(R,\mu,N_*,B\) from Lemma 10.123. The integer inequality
\(\lfloor A\rfloor-\lfloor B\rfloor\ge\lfloor A-B\rfloor\), for
\(A\ge B\), follows by writing \(A-B=m+a\) with integer \(m\) and
\(0\le a<1\). Apply it to \(A=\eta^jT_I\), \(B=\eta^{j+1}T_I\). It gives
\(O+\mu-1\le\lfloor\eta^jT_I\rfloor\). The ordinary-jet transfer in
Theorem 10.118 consequently provides all \(\mu\) input jets at each of
the \(2R+1\) integer nodes. The projected function is normal on the same
enlarged disc by Lemma 10.91 and Corollary 10.82; multiplication of the
arguments by powers of q does not change their p-adic depths. Formula
(10.329) and Theorem 10.119 give analytic precision at least
\(N_*\theta\) in the original quantity \(v_p(\Phi)+G_u-\delta\).

Suppose the integer value \(V\) were nonzero. The phase identity
(10.151) puts it in the original field \(K\). Apply Corollary 10.116 in
that field, with exactly its degree \(d\) and local degrees \(e,f\). The
coefficient term is bounded by \(A_r\). Lemma 10.122 bounds the additive
polynomial and the entire torus/q-denominator contribution. Its
indicator remainder is needed at \(I\ge1\), where \(j\ge1\); at
\(I=0,j\ge1\) the same term is a harmless upper bound, and at \(I=j=0\)
it is absent. The normal scalar satisfies
\(W_*G_u\le\lambda\Omega_{\mathrm n}/(c_1c_4)\), by
\(u\le D_{\mathrm a}\) and Lemma 10.70. The individual Euler/lcm term is
at most \(\bar g_9\eta^{j+1}/(c_1c_3Y_-)\), by Lemma 10.114 and the
target order. Finally its separate scalar remainder is bounded using
(10.311) and \(Z>273(r+1)b_r\), giving
\(\mathcal E_u/Z<1/300+1/(273\cdot999b_r)\). Together these give
\(c_1W_*[v_p(V)+G_u-\delta]\le\mathcal R_r(j)\). But (10.330),
\(S_I\ge S_-\), and (10.334) give the strict opposite inequality
\(c_1W_*N_*\theta>\mathcal R_r(j)+1/200\). Corollary 10.120 now forces
\(V=0\), a contradiction. This holds for every target order and integer,
proving (10.335). \(\square\)

\textbf{Corollary 10.127 (the full initial integer block).} Under the
same original contradiction, the initial auxiliary polynomial of
Corollary 10.113 has every prepared zero at
\(|s|\le\lfloor q^jS\rfloor\),
\(|\boldsymbol t|\le\lfloor\eta^jT\rfloor\), simultaneously for
\(j=0,1,\ldots,r\), for every original field, prime and rank case.

\textbf{Proof.} Its prescribed initial kernel equations give \(j=0\).
The initial stage satisfies both descent restrictions. Apply Theorem
10.126 successively at \(j=0,1,\ldots,r-1\), using each concluded zero
block as the next input. The coefficient vector, field, logarithmic
slopes, radii and real multiplicities stay the original ones throughout.
\(\square\)

Theorem 10.126 also proves the integer part of each subsequent
admissible contraction, once that stage's first q-expanded zero block
has been obtained. Producing that input uses fractional-node
interpolation and coset extraction, separately from the integer
induction proved here.

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={The strict integer-extension margins and the contraction that pays the q-denominator cost}]{../figures/general-integer-extension.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.28. Left: the strict lower bounds in Lemma 10.125, on a
logarithmic vertical scale. Each first marker bounds all six ranks 2--7;
each second marker bounds every rank at least eight by a single proved
uniform estimate. The horizontal line is \(1/200\); the I.2 uniform
marker equals that line as a displayed lower bound, and its exact margin
is strictly larger, as Solution 48 verifies. Right: the convex
denominator-absorption function of Lemma 10.122, divided by its value at
depth zero, for the illustrative case I.2, \(r=2,j=1,D=4+\ln3\),
\(c_5=269/500\). The continuous curve covers \(0\le z\le3D/\ln Q\); dots
select integer depths in that geometric range. This plot does not assert
that all these depths satisfy the separate multiplicity stopping
condition. Both endpoint values lie at most one, so the upper chord
proves the entire q-denominator budget.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/general_integer_extension.py}{Figure program}
and \href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/integer_comparison_bounds.py}{exact
comparison certificate}. Human-source context: the original constant
cases and integer extrapolation in
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
freely accessible paper}, Table 3.1 and Section 5, equations
(5.23)--(5.41); the coupled bounds, floor estimates and all strict
comparisons are proved here.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{41. The full first fractional block and
contraction}\label{the-full-first-fractional-block-and-contraction}

The preceding integer induction gives a hierarchy of zero blocks. Its
earlier blocks have more derivatives than its last block. Retaining all
those derivatives supplies the first fractional extrapolation in every
original field and rank case, with the full degree of the normalized
root field. We then pass to a coefficient class and fill its missing
multiples of q.

Retain the original parameters and contradiction of Section 40. Write
\(c=c_5\), \(a=r+1\), \(H=\eta^{r+1}\), \(S_1=H^{-1}S\), and \(T_1=HT\).
Here the subscript 1 denotes the first stage; it does not replace any
real parameter by its floor. Use the original phase and saturated Kummer
data of Theorems 10.51 and 10.56.

\subsubsection{The mass of the complete nested
input}\label{the-mass-of-the-complete-nested-input}

\textbf{Lemma 10.128 (all initial integer blocks as one interpolation
system).} For every \(0\le j\le r\), set

\[
\begin{aligned}
R_j&=\lfloor q^jS\rfloor,&T_j&=\lfloor\eta^jT\rfloor,&O&=\lfloor HT\rfloor,\\
\mu_j&=T_j-O+1,&
N_*&=(2R_0+1)\mu_0+
2\sum_{j=1}^r(R_j-R_{j-1})\mu_j.
\end{aligned}
\tag{10.336}
\]

These are valid nested multiplicities for every output derivative of
order at most \(O\). They are positive and nonincreasing, and

\[
\begin{aligned}
N_*&=(2R_r+1)\mu_r+
\sum_{j=0}^{r-1}(2R_j+1)(\mu_j-\mu_{j+1}),\\
\frac{c_1W_*N_*\theta}{q^r}&>\Gamma_r,\qquad
\Gamma_r=\frac{2cq}{q^r}\sum_{j=0}^r(q\eta)^j-F_r,\\
F_r&=\frac{q(r+1)}{q^rS_-}
\left[1-H+2\sum_{j=1}^r(\eta^j-H)\right].
\end{aligned}
\tag{10.337}
\]

Here \(S_-\) is the original-case lower bound in Section 40. For a fixed
additive order \(u\le D_{\mathrm a}\), the resulting normalized analytic
precision at every \(x\in\mathbb Z_p\) is at least

\[
\min\{N_*\theta,\ U-\beta+(D_{\mathrm a}-u)\theta
-(\mu_0-1)\max\{B,\beta-\vartheta,0\}\},
\quad B=\lfloor\log_p(2R_r)\rfloor.
\tag{10.338}
\]

\textbf{Proof.} Corollary 10.127 provides the zeros through order
\(T_j\) in each interval. For an output index of order at most \(O\),
its next \(\mu_j-1=T_j-O\) ordinary jets are consequently available
there, by Theorem 10.118. The inequality
\(\lfloor A\rfloor-\lfloor B\rfloor+1>A-B\) gives \(\mu_j>(\eta^j-H)T\).
The smallest real difference, at \(j=r\), is \(c\eta^rT/(r+1)\). The
original first-stage restriction \(cHT/(r+1)\ge1\) makes this at least
\(1/\eta>1\); hence all multiplicities are positive. Monotonicity
follows from \(T_j\ge T_{j+1}\).

Telescoping proves the first equality in (10.337). For its strict lower
bound, count the annuli instead of rounding the multiplicity
differences. The inner interval has more than \(2S-1\) nodes. The
\(j\)-th annulus has more than \(2(q^j-q^{j-1})S-2\) nodes, all
positive. Multiply these lower counts by \((\eta^j-H)T\). The principal
real sum is

\[
2ST\left[(1-H)+\sum_{j=1}^r(q^j-q^{j-1})(\eta^j-H)\right]
=\frac{2cST}{r+1}\sum_{j=0}^r(q\eta)^j.
\]

The removed radius-floor mass is at most
\(T[1-H+2\sum_{j=1}^r(\eta^j-H)]\). Use \(c_1W_*ST\theta=q(r+1)\) and
\(S\ge S_-\). This proves the second inequality of (10.337), without
replacing a difference of floors by its real value.

Finally apply Theorem 10.119 to these full nested intervals, with
\(M=\mu_0\) and \(E=0\). The cancellation (10.327) applies to its same
fixed additive order. It gives (10.338), retaining the complete maximum
jet order and all factorial/lcm contributions in the original normal
quantity \(v_p(\Phi)+G_u-\delta\). \(\square\)

\subsubsection{A comparison in the actual root
field}\label{a-comparison-in-the-actual-root-field}

Let \(C,Y_-,\Omega_+,\bar g_9,\bar g_{12},b_r,\epsilon_r,A_r\) have the
meanings and simultaneous field bounds of (10.331)--(10.332). Write
\(d_r=5\) for \(2\le r\le7\), and \(d_r=6\) for \(r\ge8\); this number
is a lower bound for \(A+\nu\ln q\), not a field degree. Define

\[
\begin{aligned}
\mathcal A_r^{\mathrm f}={}&A_r+
c_1\left(\frac1{300}+\frac1{273\cdot999b_r}\right)
 +\frac{\bar g_9H}{c_3Y_-}+\frac{\bar\lambda}{c_4}\\
&+\frac{\bar\lambda\ell_q}{d_rc_4}
\left(1+\frac1{q-1}\right)
 +\frac{H^{-1}+S_-^{-1}}{c_2}
 +\frac{\bar\lambda\Omega_+}{c_4q^r},\\
\mathcal I_r={}&c_1q-
\frac{c_1}{7(r+1)b_rq^r}
 -\frac{1-H+1/(6(r+1))}{q^rc_3Y_-}.
\end{aligned}
\tag{10.339}
\]

\textbf{Lemma 10.129 (both original first fractional comparisons).} For
every original field, prime and rank case, the arithmetic upper bound
for a nonzero value at \(x=s/q\), \(|x|\le\lfloor S_1\rfloor+1\),
satisfies

\[
\frac{c_1W_*}{q^r}
\bigl[v_p(V)+G_u-\delta\bigr]\le\mathcal A_r^{\mathrm f}.
\tag{10.340}
\]

The second analytic entry of (10.338), in that same normalization, is
greater than \(\mathcal I_r\). Moreover

\[
\Gamma_r-\mathcal A_r^{\mathrm f}>\frac1{50},\qquad
\mathcal I_r-\mathcal A_r^{\mathrm f}>1.
\tag{10.341}
\]

\textbf{Proof.} Theorem 10.56 gives the actual normalized support field
\(E_\Lambda\), with \([E_\Lambda:K]=q^{h_\Lambda}\), \(h_\Lambda\le r\),
and unchanged chosen local degrees \(e,f\). Its nonzero hyperplane
relation requires cancellation of \(\gcd(G_0,d_1,\ldots,d_r)\); those
degree conclusions are therefore valid also when q divides the original
phase modulus. They concern the full field, not just its selected
completion.

Use the individual scalar bound of Corollary 10.116 over \(E_\Lambda\),
as in Corollary 10.57. Its positive coefficient/scalar/torus envelope is
multiplied by \(q^{h_\Lambda}\le q^r\). The normal term \(G_u\) is added
after that multiplication; it is not multiplied by the degree factor.
The coefficient envelope is \(A_r\). The separate remainder is the
second term of \(\mathcal A_r^{\mathrm f}\). Since the output order is
at most \(HT\), its individual Euler/lcm contribution is at most
\(\bar g_9H/(c_3Y_-)\).

The target radius obeys \(X/S\le H^{-1}+1/S_-<q\). To check the last
inequality uniformly, at odd primes Lemma 10.122 gives
\(H\ge(61/75)^3\), hence \(H^{-1}<15/8\); also \(1/S_-<1/58<1/8\). In V
its three rank bands give \(H>5/12\), so \(H^{-1}<12/5\), again leaving
the required room. The additive term is consequently bounded by Lemma
10.122 at radius \(qS\), namely \(\bar\lambda/c_4\) after multiplication
by \(c_1\). Its factorial contribution stays in that estimate. Formula
(10.243) gives the full rational clearing bound
\(\Xi_{0,1,u}\le D_{\mathrm a}(1+1/(q-1))\). Lemma 10.70 converts its
cost to the fifth term of \(\mathcal A_r^{\mathrm f}\). The torus height
gives the sixth term, including the rounded radius's \(+1\). Finally
\(c_1W_*G_u\le\bar\lambda\Omega_+/c_4\), proving the last term and
(10.340). For \(r\ge8\), \(A+\nu\ln q\ge g_1=4+\ln((r+1)d)>6\), since
\(\ln9>2\). Thus the stated \(d_r\) bounds are valid.

For the analytic input, the coefficient-minimum loss satisfies
\(c_1W_*\beta<c_1/[7(r+1)b_r]\), by Lemma 10.123. Its same node estimate
gives \(\max\{B,\beta-\vartheta,0\}\le H_1/\ln p\). Here
\(\mu_0-1=\lfloor T\rfloor-\lfloor HT\rfloor\le(1-H)T+1\), and

\[
c_1W_*T\frac{H_1}{\ln p}
=\frac f{dc_3\theta}\le\frac1{c_3Y_-}.
\]

Since \(T>6(r+1)\), the input loss is bounded by the last term of
\(\mathcal I_r\). Divide by \(q^r\), use \(W_*U=q^{r+1}\), and discard
only the nonnegative \((D_{\mathrm a}-u)\theta\). This proves the
asserted input lower bound.

For ranks 2--7, every expression in (10.337) and (10.339) is rational.
Substitute each of the six integer ranks in each of the eight coupled
field rows. For all larger ranks use the last constant \(c=c_5\), and
set

\[
\begin{aligned}
H_-&=(1-c/9)^9,&
H_+&=\left(\sum_{j=0}^{10}\frac{c^j}{j!}\right)^{-1},\\
E_+(c)&=\sum_{j=0}^{10}\frac{c^j}{j!}
 +\frac{c^{11}}{11!(1-c/12)},\\
\Gamma_{\ge8}^{-}&=
\frac{2cq}{E_+(c)}\sum_{j=0}^{3}q^{-j}
 -\frac{153q}{q^8S_-(8)}.
\end{aligned}
\tag{10.342}
\]

The positive exponential series and its geometric tail prove
\(\exp(c)<E_+(c)\) and \(\exp(c)>1/H_+\). The increasing function
\((1-c/y)^y\) from Lemma 10.122 gives \(H_-\le H<\exp(-c)<H_+\). Also
\(r\ln(1-c/(r+1))\ge-rc/(r+1-c)>-c\), so \(\eta^r>1/E_+(c)\). In the
reversed geometric sum of (10.337), retain its last four terms and use
\(q\eta\le q\). This gives the principal part of \(\Gamma_{\ge8}^{-}\).
For its floor loss, \(F_r\le q(r+1)(2r+1)/(q^rS_-(r))\). Each original
\(S_-(r)/(r+1)^2\) is nondecreasing. The factor \((2r+1)/((r+1)q^r)\)
decreases: its consecutive ratio is less than \(2/q\le1\). Hence its
maximum on \(r\ge8\) is at eight, giving the last term of (10.342).

In the arithmetic upper bound replace \(C_r^{\sharp}\) by one, \(H\) by
\(H_+\) in the Euler term and by \(H_-\) in its inverse, and every
\(\epsilon_r,\bar g_{12},b_r,S_-,q^r\) by its conservative rank-eight
value. In the input lower bound replace \(H\) by \(H_-\) and the
decreasing losses by their rank-eight values. Every substitution has the
correct direction. There are therefore eight rational uniform
comparisons, in addition to the 48 finite-rank comparisons.

The strict lower bounds for the first gap are as follows. Each displayed
integer is divided by 1000.

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}lrr@{}}
\toprule\noalign{}
Case & Minimum at ranks 2--7 & Uniformly at every rank at least 8 \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
I.1 & 1108 & 647 \\
I.2 & 1075 & 587 \\
II & 1121 & 679 \\
III.1 & 1041 & 589 \\
III.2 & 1023 & 580 \\
IV, \(P=1\) & 1040 & 574 \\
IV, \(P\ge7\) & 1473 & 1061 \\
V & 508 & 25 \\
\end{longtable}
}

Every corresponding input gap is strictly greater than one. These are
exact rational substitutions in the formulas just proved; the
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/first_fractional_bounds.py}{editable comparison
certificate} generates all 56 differences and checks them using integer
arithmetic. In particular the smallest uniform fractional comparison, in
V, exceeds \(25/1000>1/50\). The infinite rank range follows from the
displayed monotonicity and exponential bounds, not from numerical rank
sampling. This proves (10.341). \(\square\)

\subsubsection{First fractional zero
production}\label{first-fractional-zero-production}

\textbf{Theorem 10.130 (the full original first fractional block).}
Under the original contradiction and phase data above, the integral
initial auxiliary family has all its prepared zeros at

\[
\Phi(s/q;\boldsymbol t)=0,
\qquad |s|\le q(\lfloor S_1\rfloor+1),\qquad
|\boldsymbol t|\le\lfloor T_1\rfloor.
\tag{10.343}
\]

This holds for every original field, prime and rank case.

\textbf{Proof.} Fix a target derivative. An additive order greater than
\(D_{\mathrm a}\) gives an identically zero derivative. Otherwise use
the complete nested integer input of Lemma 10.128. The projected
function and its unprojected prepared value have the same lower
precision (10.338), by the order-specific projection theorem. Since q
differs from p, \(s/q\in\mathbb Z_p\); the principal exponential laws of
Lesson 9 identify this analytic value with the normalized algebraic
support value in \(E_\Lambda\). Its global degree factor is exactly
\(q^{h_\Lambda}\), and its selected local degrees are the original ones.

If the value were nonzero, Lemma 10.129 would bound its normalized
valuation by \(\mathcal A_r^{\mathrm f}\). The first analytic entry is
strictly larger, by (10.337) and (10.341). The second entry is also
strictly larger, by the input part of (10.341). Their minimum is
therefore strictly larger than the arithmetic upper bound. Corollary
10.120 forces zero. Both inequalities are proved at the original rounded
radius, including its \(+1\), for all derivative indices. \(\square\)

\subsubsection{Completing the first
contraction}\label{completing-the-first-contraction}

\textbf{Theorem 10.131 (the original first full contracted block).} A
coefficient class containing an initial coefficient of minimum valuation
may be chosen, with a nonzero contracted polynomial, original
coefficient-height bound and original minimum. Its phase family is
defined over \(K\), has torus widths at most \(q^{-1}D_i\) and additive
argument \(q^{-1}x\). Initially it has the stronger zeros

\[
\Phi_1(s;\boldsymbol t)=0,
\qquad |s|\le q(\lfloor S_1\rfloor+1),\quad q\nmid s,
\quad |\boldsymbol t|\le\lfloor T_1\rfloor.
\tag{10.344}
\]

It also has all the zeros, including multiples of q, at

\[
\Phi_1(s;\boldsymbol t)=0,
\qquad |s|\le q(\lfloor S_1\rfloor+1),
\quad |\boldsymbol t|\le\lfloor\eta T_1\rfloor.
\tag{10.345}
\]

\textbf{Proof.} Apply the full phase and coset passage of Theorem 10.51
to (10.343), using its second phase partition when q divides \(G_0\).
The saturated root-monomial basis separates the coefficient classes at
\(q\nmid s\). Choose a class containing an initial coefficient attaining
\(\delta\). Its coefficient subvector preserves that minimum, remains
nonzero, and has no larger projective height. The affine prepared-jet
transfer of Theorem 10.48 proves (10.344) at every selected order; its
triangular leading entries are powers of q, hence are units at p.~The
widths and the additive argument are exactly (10.141)--(10.142); Theorem
10.60 preserves the original height envelope. The phase condition of
Theorem 10.55 puts every integer prepared value of this new family in
\(K\). No global root-degree factor is needed for those integer values.

Set \(R=q(\lfloor S_1\rfloor+1)\), \(M=\lfloor cT_1/(r+1)\rfloor+1\),
\(N_*=2(q-1)(\lfloor S_1\rfloor+1)M\), and
\(B=\lfloor\log_p(2R)\rfloor\). The nodes are exactly the q-deleted
interval of (10.344), whose radius is divisible by q. The
floor-difference inequality gives
\(\lfloor\eta T_1\rfloor+M-1\le\lfloor T_1\rfloor\), so all the required
ordinary jets are present. Apply the q-deleted case of Theorem 10.119.
Its extra term is \(EB=MB\), in addition to the inverse-coefficient and
derivative loss.

Define the following arithmetic bound in the original field:

\[
\begin{aligned}
\mathcal A_r^{\mathrm c}={}&A_r+
c_1\left(\frac1{300}+\frac1{273\cdot999b_r}\right)
 +\frac{\bar g_9\eta H}{c_3Y_-}+\frac{\bar\lambda}{c_4}\\
&+\frac{\bar\lambda\ell_q}{d_rc_4}
\left(1+\frac1{q-1}\right)
 +\frac{H^{-1}+S_-^{-1}}{c_2}+\frac{\bar\lambda\Omega_+}{c_4},\\
\mathcal J_r={}&c_1q^{r+1}
 -\frac{c_1}{7(r+1)b_r}
 -\frac{3c}{(r+1)c_3Y_-}.
\end{aligned}
\tag{10.346}
\]

For a nonzero target integer value,
\(c_1W_*[v_p(V)+G_u-\delta]\le\mathcal A_r^{\mathrm c}\). Indeed its
coefficient/scalar envelope is unchanged, its order is at most
\(\eta HT\), its torus width and argument give
\(Rq^{-1}/(c_2S)\le(H^{-1}+S_-^{-1})/c_2\), including the rounded
radius's \(+1\), and \(\Xi_{1,0,u}\le D_{\mathrm a}(1+1/(q-1))\) gives
the full q-denominator term. Its additive scalar is evaluated at
\(q^{-1}x\), of absolute size at most \(\lfloor S_1\rfloor+1<qS\), by
Lemma 10.129; Lemma 10.122 therefore bounds it by the same additive
term. The normal contribution now has no \(q^{-r}\) factor, because the
value is in \(K\).

For the two analytic entries, we have

\[
\begin{aligned}
c_1W_*N_*\theta&>2cq(q-1),\\
c_1W_*\mathcal P_{u,\mathrm{input}}&>\mathcal J_r.
\end{aligned}
\tag{10.347}
\]

The first uses \(\lfloor S_1\rfloor+1>S_1\), \(M>cT_1/(r+1)\), and
\(S_1T_1=ST\). For the second, the node bound of Lemma 10.123 gives
\(\max\{B,\beta-\vartheta,0\}\le H^{-1}H_1/\ln p\). After the exact
cancellation (10.327), the input loss is at most
\(MB+(M-1)\max\{B,\beta-\vartheta,0\}\). Since \(M-1\le cT_1/(r+1)\) and
\(M\le cT_1/(r+1)+1\), it is at most \([2cT_1/(r+1)+1]H^{-1}H_1/\ln p\).
The identity used in Lemma 10.129 and the restriction \(cT_1/(r+1)\ge1\)
bound its normalized value by \(3c/((r+1)c_3Y_-)\). The same
coefficient-minimum estimate bounds \(\beta\). This proves (10.347),
retaining both q-deleted losses.

The two strict comparisons are

\[
2cq(q-1)-\mathcal A_r^{\mathrm c}>\frac1{50},\qquad
\mathcal J_r-\mathcal A_r^{\mathrm c}>1.
\tag{10.348}
\]

They follow from the same 48 finite and eight uniform rational
substitutions used in Lemma 10.129. For \(r\ge8\) bound \(\eta H\) by
\(H_+\) and \(H^{-1}\) by \(H_-^{-1}\); the other monotone terms have
their same rank-eight bounds. The gain has its constant last-band value.
In \(\mathcal J_r\), use \(q^{r+1}\ge q^9\) and the decreasing losses at
rank eight. Strict lower bounds for the first gap, again divided by
1000, are

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}lrr@{}}
\toprule\noalign{}
Case & Minimum at ranks 2--7 & Uniformly at every rank at least 8 \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
I.1 & 372 & 392 \\
I.2 & 338 & 332 \\
II & 402 & 420 \\
III.1 & 366 & 384 \\
III.2 & 326 & 348 \\
IV, \(P=1\) & 371 & 391 \\
IV, \(P\ge7\) & 725 & 836 \\
V & 5635 & 6334 \\
\end{longtable}
}

Every corresponding input gap exceeds one; the
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/first_fractional_bounds.py}{exact certificate}
checks all four comparisons in each of its 56 rows. Thus both analytic
entries exceed the arithmetic bound for every target integer. Corollary
10.120 forces those values to vanish. This proves (10.345), and
completes the first contraction in all original cases. \(\square\)

The higher-order q-deleted zeros (10.344) remain available alongside the
full block (10.345). They belong to the input of later mixed
extrapolations, rather than being discarded after the missing multiples
of q have been filled.

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={The complete initial nested profile and uniform first fractional and contracted margins}]{../figures/first-fractional-profiles.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.29. Left: the actual certified integer radii and
derivative floors of Solutions 39 and 49, with output order993090. Each
annotation counts the full symmetric annulus, including zero in the
first interval. The shaded excess derivative orders give the variable
multiplicities of (10.336), with cardinal degree1304340501; the
fractional target radius687 is the original rounded radius from
(10.343). This is a precision profile, not a graph of auxiliary-function
values. Right: the strict rational lower bounds for every original field
case at every rank at least eight, under the proved uniform
substitutions of (10.342). Fractional values retain their actual
normalized root-field factor at mostq\^{}r; after the exact phase
passage, integer values belong toK. The horizontal threshold is1/50; the
exact V fractional margin exceeds0.025638. Both input comparisons also
pass, as proved in Lemma10.129 and Theorem10.131.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/first_fractional_profiles.py}{Figure program}
and \href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/first_fractional_bounds.py}{exact rational
certificate}. Human-source context: the original fractional range and
first phase passage in
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
free paper}, Section5, equations(5.42)--(5.60); all field, nested-jet,
floor and strict comparison arguments used here are proved in this
lesson.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{42. Retaining the complete mixed
profile}\label{retaining-the-complete-mixed-profile}

The full contracted interval and the higher-order q-deleted interval
provide different information. Both remain available when the integer
block is extended to larger radii. We now put an arbitrary number of
these full blocks into the same interpolation system. The
cardinal-polynomial loss must account separately for the deleted nodes.

\subsubsection{Mixed interpolation with any number of full
intervals}\label{mixed-interpolation-with-any-number-of-full-intervals}

\textbf{Theorem 10.132 (the complete mixed normal interpolation bound).}
Let \(p\ne q\) be primes. Let \(1\le A\le R_1<\cdots<R_m\) be integers,
where \(m\ge1\), and put

\[
\begin{aligned}
E_j&=[-R_j,R_j]\cap\mathbb Z,&
D&=\{s\in\mathbb Z:|s|\le A,\ q\nmid s\},\\
\nu_1&\ge\cdots\ge\nu_m\ge1,&
E&\ge0,& M&=\nu_1+E,\\
\mu(s)&=\nu_m+
\sum_{j=1}^{m-1}(\nu_j-\nu_{j+1})\mathbf1_{E_j}(s)
+E\mathbf1_D(s).
\end{aligned}
\tag{10.349}
\]

All the multiplicities and \(E\) are integers. Their total degree is

\[
\begin{aligned}
N_*&=(2R_m+1)\nu_m+
\sum_{j=1}^{m-1}(2R_j+1)(\nu_j-\nu_{j+1})
+2\bigl(A-\lfloor A/q\rfloor\bigr)E,\\
B&=\lfloor\log_p(2R_m)\rfloor.
\end{aligned}
\tag{10.350}
\]

Choose \(\rho\) with \(v_p(\rho)=\theta>0\) and a normal series \(F\).
Suppose, for real numbers \(\Lambda,C\) with \(C\ge0\), that its divided
jets satisfy \(v_p(F_h(\rho s))+h\theta\ge\Lambda-hC\) at every
\(s\in E_m\) and \(0\le h<\mu(s)\). Then

\[
v_p(F(\rho x))\ge
\min\{N_*\theta,\ \Lambda-EB-(M-1)\max\{B,C\}\}
\qquad(x\in\mathbb Z_p\subset\mathbb Q_p).
\tag{10.351}
\]

For the prepared family, suppose the original values vanish through
order \(|\boldsymbol t|+\mu(s)-1\) at those nodes, with fixed additive
order \(u\le D_{\mathrm a}\). Use the exact jet quantities
\(G_u,M_u(h),K_u(M,B)\) of Theorems 10.117--10.119. Then the projected
and unprojected values have the common lower bound

\[
\begin{aligned}
v_p(f(x;\boldsymbol t))+G_u-\delta&\ge\mathcal P_u,\\
v_p(\Phi(x;\boldsymbol t))+G_u-\delta&\ge\mathcal P_u,\\
\mathcal P_u&=
\min\{N_*\theta,\ U-\beta+G_u-EB-(M-1)B+K_u(M,B)\}.
\end{aligned}
\tag{10.352}
\]

In particular,

\[
\mathcal P_u\ge
\min\{N_*\theta,\ U-\beta+(D_{\mathrm a}-u)\theta
-EB-(M-1)\max\{B,\beta-\vartheta,0\}\}.
\tag{10.353}
\]

\textbf{Proof.} The intervals are nested, and \(D\subset E_1\). The
representation of \(\mu\) in (10.349) is therefore its exact
decomposition into a full outer layer, nonnegative full-interval
increments and one nonnegative deleted increment. Counting each layer
gives (10.350): there are \(2(A-\lfloor A/q\rfloor)\) nonmultiples of q
in the symmetric interval, because zero is a multiple.

At every residue level \(p^a\), counts in the residue classes of a full
consecutive integer interval differ by at most one. To prove this, write
its length as \(kp^a+b\), \(0\le b<p^a\). Each class has either \(k\) or
\(k+1\) representatives. For the deleted interval, subtract the
representatives that are multiples of q. Their indices form the full
interval \([-\lfloor A/q\rfloor,\lfloor A/q\rfloor]\), and
multiplication by q permutes the classes modulo \(p^a\). Each of the two
full-interval count differences is at most one in absolute value.
Deleted-interval counts consequently differ by at most two. A bound of
one does not hold in general.

For \(s\in E_m\), take the cardinal factor

\[
L_s(X)=\prod_{\substack{t\in E_m\\t\ne s}}
\left(\frac{X-t}{s-t}\right)^{\mu(t)}.
\]

If \(x\) is another node, it is zero; if \(x=s\), it is one. Consider a
nonnode \(x\in\mathbb Z_p\). At a residue level \(a\le B\), an increment
on a set containing s contributes
\(N_a(x)-N_a(s)+1-\mathbf1_{x\equiv s\ (p^a)}\). For a full interval
this is nonnegative: if the residues agree it is zero, and otherwise its
first difference is at least \(-1\). For the deleted interval it is at
least \(-1\). An increment on a set not containing s contributes
\(N_a(x)-N_a(s)\), at least \(-1\) for a full interval and at least
\(-2\) for the deleted interval.

The full outer layer always contains s. If \(s\in D\), all the full
increments contain s, and only the deleted increment can lose valuation,
by at most \(E\) at each level. If \(s\in E_1\setminus D\), all full
increments still contain s, but the missing deleted increment can lose
at most \(2E\). If s belongs to a later annulus, the missing full
increments have total weight \(\nu_1-\mu(s)\), and the deleted increment
can again lose at most \(2E\). All three cases give the same bound

\[
v_p(L_s(x))\ge-B\bigl(M-\mu(s)+E\bigr).
\tag{10.354}
\]

For \(a>B\), every nonzero denominator difference has valuation less
than a, since \(0<|s-t|\le2R_m<p^a\). The remaining contributions are
nonnegative. Thus the sum over the residue levels proves (10.354), also
at the nodes by their zero or unit values.

The coefficient of degree a in \[
L_s(s+Z)^{-1}
=\prod_{t\ne s}(1+Z/(s-t))^{-\mu(t)}
\] has valuation at least \(-aB\). Indeed each negative-binomial
coefficient is an integer, and each denominator difference has valuation
at most B. Use the Hermite polynomial constructed in Lemma 10.94, with
this exact multiplicity function. A term with divided jet h and
Taylor-inverse degree \(a\le\mu(s)-1-h\) has valuation at least \[
\Lambda-hC-B(M-\mu(s)+E)-aB
\ge\Lambda-EB-(M-1)\max\{B,C\}.
\] Its degree is less than \(N_*\), and it has the required jets at
every node.

The full monic normal-quotient construction in Lemma 10.94 applies to
\(W(Z)=\prod_{s\in E_m}(Z-\rho s)^{\mu(s)}\). It produces \(F=H+WG\),
with G normal and H the unique polynomial of degree less than \(N_*\)
having those jets. At \(\rho x\), each factor of W has valuation at
least \(\theta\), and G is integral. The quotient contribution is
therefore at least \(N_*\theta\). The ultrametric inequality proves
(10.351).

For the prepared family, Theorem 10.118 gives the weighted jet baseline
\(U-\beta+G_u\) and increment \(M_u(h)\). The same Hermite term now has
valuation at least \[
U-\beta+G_u-EB-(M-1)B+M_u(h)+hB.
\] Every h is less than M. Taking the minimum in the definition of
\(K_u(M,B)\) proves the asserted lower bound for f.~The projection error
has precision at least \(U-\beta+G_u+\gamma_u\), by Lemma 10.117. Since
\(K_u(M,B)\le M_u(0)=\gamma_u\), it has at least the second entry of
(10.352). Adding it gives the same bound for \(\Phi\). Finally the exact
fixed-order cancellation in Lemma 10.123 gives (10.353). This proves the
theorem with both the deleted-cardinal cost and the complete inverse/jet
cost retained. \(\square\)

\subsubsection{An exact small profile}\label{an-exact-small-profile}

Take \(p=2,q=3,A=R_1=9,R_2=18,\nu_1=3,\nu_2=1,E=2\). The twelve deleted
nodes have multiplicity five, the remaining seven inner nodes have
multiplicity three, and the eighteen outer nodes have multiplicity one.
Thus

\[
N_*=12\cdot5+7\cdot3+18\cdot1
=37+19(3-1)+12\cdot2=99,\qquad M=5,\quad B=5.
\tag{10.355}
\]

The deleted nodes have residue counts \((4,3,2,3)\) modulo four. The
discrepancy of two occurs here, so it is necessary to distinguish the
deleted increment from a full interval. The uniform cardinal bounds in
(10.354) are respectively \(-10,-20,-30\) at a deleted inner node, a
nondeleted inner node and an outer node. With \(C=0\), (10.351) gives
precision at least \(\min\{99\theta,\Lambda-30\}\). These are bounds on
the interpolation calculation; they assert no value or zero for a
particular auxiliary function.

For later course stages, the available full-interval orders and higher
q-deleted orders must be inserted into this theorem without discarding
either. The resulting cardinal count and input precision are exact. A
zero-production conclusion additionally requires the corresponding
strict arithmetic comparisons in the actual coefficient field.

\subsubsection{The exact inherited profile at a later
depth}\label{the-exact-inherited-profile-at-a-later-depth}

\textbf{Lemma 10.133 (the complete inherited mixed mass and input
budget).} Retain every original rank, field and prime parameter of
Section 40, and let \(I\ge1\) be an admissible depth with
\(cT_{I+1}/(r+1)\ge1\). Write \(c=c_5,a=r+1,H=\eta^{r+1}\),
\(S_I=H^{-I}S,T_I=H^IT\), and \(O=\lfloor HT_I\rfloor\). Suppose the
stage-I family has all its q-deleted zeros through order
\(\lfloor T_I\rfloor\) at radius \(q(\lfloor S_I\rfloor+1)\), all its
full integer zeros through order \(\lfloor\eta T_I\rfloor\) at that
radius, and all its full integer zeros through order
\(\lfloor\eta^jT_I\rfloor\) at radius \(\lfloor q^jS_I\rfloor\),
\(2\le j\le r\). Put

\[
\begin{aligned}
R_1&=q(\lfloor S_I\rfloor+1),&
R_j&=\lfloor q^jS_I\rfloor\quad(2\le j\le r),\\
\mu_j&=\lfloor\eta^jT_I\rfloor-O+1\quad(0\le j\le r),&
E&=\mu_0-\mu_1,\\
N_I&=(2R_r+1)\mu_r+
\sum_{j=1}^{r-1}(2R_j+1)(\mu_j-\mu_{j+1})
+2(q-1)(\lfloor S_I\rfloor+1)E.
\end{aligned}
\tag{10.356}
\]

These are valid multiplicities for every output derivative of order at
most O. For a fixed additive order \(u\le D_{\mathrm a}\), Theorem
10.132 applies with \(M=\mu_0\), and its first analytic entry satisfies

\[
\begin{aligned}
\frac{c_1W_*N_I\theta}{q^r}
&>\Gamma_{r,I}^{\mathrm{mix}},\\
\Gamma_{r,I}^{\mathrm{mix}}
&=\frac{2cq}{q^r}
\left[\sum_{j=0}^r(q\eta)^j+q-2\right]\\
&\quad-\frac{q(r+1)}{q^rS_I}
\left[2(q+1)(\eta^2-H)+2\sum_{j=3}^r(\eta^j-H)\right].
\end{aligned}
\tag{10.357}
\]

An empty sum is zero. Its second analytic entry, in the same
normalization, is strictly greater than

\[
\mathcal I_r^{\mathrm{mix}}
=c_1q-\frac{c_1}{7(r+1)b_rq^r}
-\frac{1-H+c(1+2H)/(r+1)}{q^rc_3Y_-}.
\tag{10.358}
\]

Here \(b_r,Y_-\) are the original coupled bounds in Lemma 10.121.

\textbf{Proof.} The original bound \(S_I\ge S>58\) gives
\(\lfloor q^2S_I\rfloor>q(\lfloor S_I\rfloor+1)\), since
\(q(q-1)S_I-q-1>0\). All the outer radii are strictly increasing. The
floor-difference inequality gives \(\mu_j>(\eta^j-H)T_I\). At the last
layer, \((\eta^r-H)T_I=c\eta^rT_I/(r+1)\ge1/\eta>1\). Thus the
multiplicities are positive and nonincreasing. Each node has exactly the
available additional ordinary jets required by Theorem 10.118. The top
increment on deleted nodes is
\(E=\lfloor T_I\rfloor-\lfloor\eta T_I\rfloor\ge0\). Formula (10.350) is
therefore precisely (10.356).

For the lower mass, count the nodes by disjoint types, with their full
available multiplicities. There are exactly
\(2(q-1)(\lfloor S_I\rfloor+1)>2(q-1)S_I\) deleted inner nodes, with
multiplicity \(\mu_0\). There are \(2(\lfloor S_I\rfloor+1)+1>2S_I\)
inner multiples of q, with multiplicity \(\mu_1\). The first outer
annulus has more than \(2(q^2-q)S_I-2(q+1)\) nodes, with multiplicity
\(\mu_2\). Every subsequent annulus j has more than
\(2(q^j-q^{j-1})S_I-2\) nodes, with multiplicity \(\mu_j\). Each lower
node count is positive.

Multiply these counts by their respective strict lower multiplicities
\((\eta^j-H)T_I\). The principal mass is

\[
\begin{aligned}
2S_IT_I\bigg[(q-1)(1-H)+(\eta-H)
+\sum_{j=2}^r(q^j-q^{j-1})(\eta^j-H)\bigg]\\
=\frac{2cS_IT_I}{r+1}
\left[\sum_{j=0}^r(q\eta)^j+q-2\right].
\end{aligned}
\tag{10.359}
\]

For the equality, subtract the full initial real annular sum of Lemma
10.128. The difference is \((q-2)(1-\eta)=(q-2)c/(r+1)\). The remaining
removed floor mass is at most
\(T_I[2(q+1)(\eta^2-H)+2\sum_{j=3}^r(\eta^j-H)]\). Use \(S_IT_I=ST\) and
\(c_1W_*ST\theta=q(r+1)\). This proves (10.357), including the larger
rounded inner radius. It does not round the differences of derivative
floors independently.

For the input entry of (10.353), Lemma 10.123 gives \[
B,\ \max\{B,\beta-\vartheta,0\}\le
H^{-I}H_1/\ln p,\qquad H_1=h+\nu\ln q.
\] Also \(E\le cT_I/(r+1)+1\) and \(\mu_0-1\le(1-H)T_I+1\). The two
separate losses are consequently bounded by \[
\left[(1-H+c/(r+1))T_I+2\right]H^{-I}H_1/\ln p.
\] Since \(cHT_I/(r+1)\ge1\), we have \(2/T_I\le2cH/(r+1)\). The
identity \(c_1W_*T_IH^{-I}H_1/\ln p=f/(dc_3\theta)\le1/(c_3Y_-)\)
therefore gives the last term of (10.358). Finally \(W_*U=q^{r+1}\) and
the strict coefficient-minimum bound of Lemma 10.123 give the first two
terms. Discard only the nonnegative \((D_{\mathrm a}-u)\theta\). This
proves the claimed strict input bound, retaining both \(EB\) and the
maximum inverse/jet loss. \(\square\)

Thus the complete inherited profile supplies two explicitly proved
analytic entries at every admissible later depth. To produce fractional
zeros one must compare both of them with the nonzero-value upper bound
in the actual normalized support field. These analytic bounds alone do
not remove its global degree factor or the q-denominator cost.

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={Complete mixed node multiplicities and the deleted residue-count discrepancy}]{../figures/general-mixed-profiles.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.30. The exact profile of (10.355) and Solution51,
withp=2,q=3,A=R1=9,R2=18. Each bar represents one integer node and shows
its complete multiplicity, decomposed into the full outer layer, full
inner increment and q-deleted increment. The total degree is99. The
lower panel counts the deleted nodes in the four residue classes
modulo4; the counts4,3,2,3 show why discrepancy two is needed.
Theorem10.132 proves the cardinal losses and the separateEB cost;
Lemma10.133 supplies the later original floor profile without discarding
any layer. These diagrams describe the interpolation system, not
auxiliary-function values or a later zero-production comparison.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/general_mixed_profiles.py}{Figure program}.
Human-source context: the original integer and fractional phase ranges
in
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
freely accessible paper}, Section5, equations(5.23) and(5.42)--(5.60);
the arbitrary-layer mixed interpolation, floor-mass and fixed-order
input proofs are given here.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{43. The complete prime-two phase
induction}\label{the-complete-prime-two-phase-induction}

In the original case V, retaining more integer derivatives supplies the
fractional interpolation with enough precision throughout the full
descent. We choose each derivative loss from its actual arithmetic
envelope. The subsequent comparison keeps the complete global root-field
factor. It also treats the last contracted block, which can occur one
step beyond the interval used to produce fractional zeros.

Throughout this section take the original case V, so \(p=2,q=3\). Retain
all the original field, phase, coefficient, height and projection
hypotheses of Section 40. Put \(a=r+1,c=c_5,\eta=1-c/a,H=\eta^a,Q=qH\),
and \(D=A+\nu\ln q\). The additive degree remains \(D_{\mathrm a}=kL\);
it is distinct from D. Let \(S_I=H^{-I}S,T_I=H^IT\). The exact original
stopping indices are

\[
L_D=\frac{3D}{\ln Q},\qquad
I^*=\lfloor L_D\rfloor+1,\qquad
I_1=\max\{I\in\mathbb Z_{\ge0}:cT_I/a\ge1\},\qquad
I_0=\min\{I^*,I_1\}.
\tag{10.360}
\]

These indices exist: \(0<H<1,Q>1\), and the initial multiplicity bounds
give \(cT_1/a>1\), as in Section 41. For \(1\le I<I_0\), both
\(I\le L_D\) and \(cHT_I/a\ge1\) hold. At \(J=I_0\), only \(J\le L_D+1\)
and \(cT_J/a\ge1\) are needed for closing the deleted interval.

\subsubsection{Arithmetic costs and chosen ordinary-jet
losses}\label{arithmetic-costs-and-chosen-ordinary-jet-losses}

Use the exact rational coupled bounds
\(Y_-=2/(1+10^{-6}),\Omega_+=3/2,C=10/3\) from Lemma 10.121, and
\(b_r,\epsilon_r,\bar g_{12},\bar g_9,S_-\) from (10.331) and its table.
The coefficient bound \(A_r\) of (10.339) is

\[
A_r=c_1(\bar g_{12}+\epsilon_r)+
\frac{c_1\epsilon_r/2+\bar\lambda/c_4+
C_r^{\sharp}/(c_3Y_-)+117/(232c_2)}{C-1}.
\tag{10.361}
\]

Here \(c_1=2.5347,c_2=13/9,c_3=0.4757,c_4=3.765\),
\(\bar\lambda=189/188\), and \(C_r^{\sharp}\) is (10.307). Define

\[
\begin{aligned}
b&=A_r+c_1\left(\frac1{300}+\frac1{273\cdot999b_r}\right)+\frac{\bar\lambda}{c_4},&
n&=\frac{\bar\lambda\Omega_+}{c_4},\\
e_0&=\frac{\bar g_9}{c_3Y_-},&
\ell&=\frac{\bar\lambda(1099/1000)}{Dc_4},\\
\mathcal C(I)&=b+n+e_0H^I+\ell(I+1/2),\\
B_0(I)&=\mathcal C(I)+\frac{q(1+S_-^{-1})}{c_2}Q^{-I},\\
B_j(I)&=\mathcal C(I)+\ell j+\frac{q^{j+1}}{c_2}Q^{-I}\quad(1\le j<r).
\end{aligned}
\tag{10.362}
\]

The symbol \(e_0\) is a numerical envelope, not the ramification index.
Choose

\[
\begin{aligned}
\varepsilon&=1/100,&\lambda_{\mathrm{node}}&=232/231,\\
\Delta_0(I)&=\frac{B_0(I)+\varepsilon}{2q(q-1)a},&
\Delta_j(I)&=\frac{\lambda_{\mathrm{node}}(B_j(I)+\varepsilon)}{2q^{j+1}a}\quad(1\le j<r),\\
\tau_0(I)&=1,&
\tau_j(I)&=1-\sum_{k=0}^{j-1}\Delta_k(I)\quad(1\le j\le r).
\end{aligned}
\tag{10.363}
\]

All quantities preceding the depth variable are those of the original
construction. These choices change the retained derivative profile; they
do not change its stopping indices, initial coefficients or height
constants.

\textbf{Lemma 10.134 (the adaptive comparison envelopes).} Suppose
\(1\le I<I_0\). A nonzero integer target at radius
\(q(\lfloor S_I\rfloor+1)\), with total order at most \(T_I\), has
normalized valuation \(c_1W_*[v_p(V)+G_u-\delta]\) at most \(B_0(I)\). A
nonzero integer target at radius \(\lfloor q^{j+1}S_I\rfloor\),
\(1\le j<r\), has the same normalized valuation at most \(B_j(I)\). The
additive order u is fixed in each comparison.

For the fractional targets \(x=s/q\),
\(|s|\le q(\lfloor S_I/H\rfloor+1)\), of total order at most
\(\lfloor HT_I\rfloor\), the arithmetic envelope in normalization
\(c_1W_*/q^r\) is

\[
A_{\mathrm f}(I)=b+e_0H^{I+1}+\ell(I+3/2)
+\frac{H^{-1}+S_-^{-1}}{c_2}Q^{-I}+\frac n{q^r}.
\tag{10.364}
\]

Assuming the necessary integer input zeros, the two analytic entries for
deleted closure are bounded below by \(B_0(I)+\varepsilon\) and

\[
J_0(I)=c_1q^{r+1}-\frac{c_1}{7ab_r}
-\frac{2\Delta_0(I)+cH/a}{c_3Y_-}.
\tag{10.365}
\]

For the full extension from the jth to the \((j+1)\)st radius, with
cutoffs \(\lfloor\tau_jT_I\rfloor,\lfloor\tau_{j+1}T_I\rfloor\), they
are bounded below by \(B_j(I)+\varepsilon\) and

\[
J_j(I)=c_1q^{r+1}-\frac{c_1}{7ab_r}-\frac{\Delta_j(I)}{c_3Y_-}.
\tag{10.366}
\]

If \(1\ge\tau_1\ge\cdots\ge\tau_r>H\), the complete mixed fractional
profile has its two analytic entries strictly greater than

\[
\begin{aligned}
\Gamma(I)&=\frac{2qa}{q^r}
\left[(q-1)(1-H)+(\tau_1-H)
+\sum_{j=2}^r(q^j-q^{j-1})(\tau_j-H)\right]\\
&\quad-\frac{qa}{q^rS_-}
\left[2(q+1)(\tau_2-H)+2\sum_{j=3}^r(\tau_j-H)\right],\\
J_{\mathrm f}(I)&=c_1q-\frac{c_1}{7ab_rq^r}
-\frac{1-H+\Delta_0(I)+2cH/a}{q^rc_3Y_-}.
\end{aligned}
\tag{10.367}
\]

\textbf{Proof.} At an integer target, Theorem 10.55 places the actual
value in K. Corollary 10.116 therefore uses its full global degree d and
local degrees e,f.~The coefficient term is at most \(A_r\), and the
scalar remainder is the separate term already included in b, by Theorem
10.115. An order at most \(T_I\) has Euler/lcm cost at most \(e_0H^I\),
by Lemma 10.114. The normal contribution is at most n.

At the full extension radius, Lemma 10.122 bounds the additive part by
\(\bar\lambda/c_4+\ell j\). Retain the actual torus term
\(q^{j+1}Q^{-I}/c_2\) and clearing term \(\ell(I+1/2)\); no chord
absorption is used here. These give \(B_j\).

For deleted closure, the additive argument has size at most
\(q^{-I}q(\lfloor S_I\rfloor+1)\le qSQ^{-I}+q^{1-I}\le S/H+1<qS\). The
last inequality is Lemma 10.129. The scalar monotonicity thus gives the
j=0 additive cost. Its torus cost is at most
\(q(1+S_-^{-1})Q^{-I}/c_2\), since \(q^{-I}\le Q^{-I}\). This proves
\(B_0\).

At a fractional target the corresponding additive argument is at most
\(q^{-I}(\lfloor S_I/H\rfloor+1)\le(S/H)Q^{-I}+q^{-I}<qS\). The torus
cost is at most \((H^{-1}+S_-^{-1})Q^{-I}/c_2\), and the clearing
exponent obeys \(\Xi_{I,1,u}\le D_{\mathrm a}(I+1+1/(q-1))\). Its order
is at most \(HT_I\), giving \(e_0H^{I+1}\). Theorem 10.56 gives global
degree \(dq^{h_\Lambda}\), \(h_\Lambda\le r\), with the original
selected local degrees. Dividing the product-formula bound by \(q^r\)
bounds every positive algebraic-height term by the terms displayed in
(10.364). The normal contribution is added after that product-formula
comparison; it is consequently \(n/q^r\). This proves (10.364),
retaining the worst global factor even if the selected completion has
local degree one.

For deleted closure take \(M_0=\lfloor\Delta_0T_I\rfloor+1\). The
inequality \(\lfloor A\rfloor-\lfloor B\rfloor\ge\lfloor A-B\rfloor\)
ensures that an output through \(\lfloor\tau_1T_I\rfloor\) has all M
ordinary input jets at every deleted node. Their number is
\(2(q-1)(\lfloor S_I\rfloor+1)>2(q-1)S_I\), and \(M_0>\Delta_0T_I\). The
identity \(c_1W_*ST\theta=qa\) gives first entry strictly greater than
\(2q(q-1)a\Delta_0=B_0+\varepsilon\).

Theorem 10.119 has both the deleted loss \(M_0B\) and the inverse/jet
loss. Its exact cancellation (10.327) bounds their sum by
\([2\Delta_0T_I+1]H^{-I}H_1/\ln p\). Here and below, the rounded inner
radius also satisfies \(2R_1<3q^rS_I\); hence the separation bound of
Lemma 10.123 applies to it. Since \(1/T_I\le cH/a\) and
\(c_1W_*T_IH^{-I}H_1/\ln p\le1/(c_3Y_-)\), this gives (10.365). The
coefficient-minimum loss is at most \(c_1/(7ab_r)\). Only the
nonnegative term \((D_{\mathrm a}-u)\theta\) is discarded.

For a full extension take \(M_j=\lfloor\Delta_jT_I\rfloor+1\). The same
floor-difference inequality provides all those ordinary jets between
successive \(\tau\) cutoffs. The full node count exceeds \(2q^jS_I-1\).
Since \(S_I\ge S_->58\) and \(q^j\ge2\), \(1-1/(2q^jS_I)>231/232\). The
factor \(\lambda_{\mathrm{node}}\) in (10.363) thus gives first entry
strictly greater than \(B_j+\varepsilon\). There is no deleted increment
in this full system; \(M_j-1\le\Delta_jT_I\) gives (10.366) by the same
exact fixed-order cancellation. This proves both integer input estimates
without presuming a zero.

For fractional interpolation take \(O=\lfloor HT_I\rfloor\), full radii
\(R_1=q(\lfloor S_I\rfloor+1),R_j=\lfloor q^jS_I\rfloor\) for
\(2\le j\le r\), and multiplicities
\(\mu_j=\lfloor\tau_jT_I\rfloor-O+1\), with
\(\mu_0=\lfloor T_I\rfloor-O+1\). They are positive and nonincreasing.
Every available input order covers the required additional jets,
including when two derivative floors coincide. Apply Theorem 10.132 with
deleted increment \(E=\mu_0-\mu_1\). The disjoint annular counts used in
Lemma 10.133 now multiply \((\tau_j-H)T_I\), giving (10.367). Replacing
\(S_I\) by its lower bound \(S_-\) enlarges the floor loss. No
differences of floors are rounded separately.

For its second entry, \(E\le\Delta_0T_I+1\) and
\(\mu_0-1\le(1-H)T_I+1\). Thus both deleted and inverse/jet losses
together are at most \([(1-H+\Delta_0)T_I+2]H^{-I}H_1/\ln p\). Use
\(2/T_I\le2cH/a\), then divide by \(q^r\). The identities
\(W_*U=q^{r+1}\) and the same minimum-valuation bound give
\(J_{\mathrm f}\). The comparison throughout is in the unchanged normal
quantity with the same u and \(\delta\). \(\square\)

\subsubsection{Continuous depths and every
rank}\label{continuous-depths-and-every-rank}

\textbf{Lemma 10.135 (strict adaptive bounds at every depth).} For every
original V rank \(r\ge2\) and every \(1\le I\le L_D\),

let \(\mathcal G(I)=\Gamma(I)\) at ranks 2--7, and let
\(\mathcal G(I)=qa(2-1/(q^rS_-))(\tau_r(I)-H)\) at ranks at least eight.
The latter is the lower mass from the last full block alone. Then

\[
\begin{aligned}
\Delta_0(I)<c/a,\qquad
\tau_r(I)>H,\qquad
J_j(I)-B_j(I)&>1\ (0\le j<r),\\
\mathcal G(I)-A_{\mathrm f}(I)>19/10,\qquad
J_{\mathrm f}(I)-A_{\mathrm f}(I)&>12/5.
\end{aligned}
\tag{10.368}
\]

For \(2\le r\le7\), each \(\tau_j(I)>\eta^j\). For every \(r\ge8\),
\(\tau_r(I)>2/3\); the original \(\eta^j\) blocks are retained
independently by Theorem 10.126.

There is also a strict closure comparison on the larger interval
\(1\le J\le L_D+1\):

\[
2cq(q-1)-B_0(J)>3,\qquad
c_1q^{r+1}-\frac{c_1}{7ab_r}
-\frac{3c}{ac_3Y_-}-B_0(J)>1.
\tag{10.369}
\]

\textbf{Proof.} Fix r and the original D. Each \(B_j(I)\) is convex: it
is a sum of positive multiples of \(H^I,Q^{-I}\), a constant and an
affine function of I. Consequently each \(\Delta_j\) is convex and each
\(\tau_j\) is concave. Each coefficient of \(\tau_j\) in \(\Gamma\) is
positive. For j=2 this follows from \(2(q^2-q)>2(q+1)/S_-\), and for j
at least three from \(2(q^j-q^{j-1})>2/S_-\); the first coefficient is
positive directly. The bounds \(S_->58,q=3\) imply both inequalities.
The last-block mass is also a positive affine function of \(\tau_r\).
Thus \(\mathcal G-A_{\mathrm f}\), \(J_{\mathrm f}-A_{\mathrm f}\), and
every \(J_j-B_j\) are concave. It suffices to bound their two endpoints;
likewise, an upper bound on \(\Delta_0\) and lower bounds on \(\tau_j\)
follow from the endpoints. This is a proof for a continuous interval,
not a check of sampled depths.

At I=1 every exponential is rational. At \(I=L_D\), use
\(Q^{-L_D}=\exp(-3D)<1/1000\), \(H^{L_D}\le H\), and
\(\ell L_D=3\bar\lambda(1099/1000)/(c_4\ln Q)\). The exponential
inequality follows from the positive series through degree six at 15 and
\(D>5\), as proved in Lemma 10.122. For a rational x greater than one
put

\[
\mathcal L_{20}(x)=2\sum_{k=0}^{19}
\frac{((x-1)/(x+1))^{2k+1}}{2k+1}<\ln x.
\tag{10.370}
\]

The identity follows by integrating the finite geometric expansion of
\(2/(1-z^2)\) from zero to \((x-1)/(x+1)\); its omitted integral is
positive. This also proves the inequality without a numerical logarithm
oracle. Replace the remaining \(1/D\) terms by \(1/5\) at ranks 2--7. In
(10.362), the two upper endpoint common costs are then

\[
\begin{aligned}
C_{\mathrm E}&=b+n+e_0H+3\ell_-/2,\\
C_{\mathrm L}&=b+n+e_0H+
\frac{3\bar\lambda(1099/1000)}{c_4\mathcal L_{20}(Q)}+\ell_-/2,
\qquad \ell_-=\frac{\bar\lambda(1099/1000)}{5c_4}.
\end{aligned}
\tag{10.371}
\]

Use \(Q^{-1}\) and \(1/1000\), respectively, for the torus terms; use
\(e_0H^2\) for the fractional Euler cost. All resulting
\(B_j,\Delta_j,A_{\mathrm f}\) are upper bounds. The resulting
\(\tau_j\) are lower bounds, and the positive coefficients just proved
make their resulting \(\Gamma\) a lower bound. Formula (10.367)
similarly gives a lower input entry.

The following are strict rational lower bounds, each divided by 1000, on
the minima of the two endpoint differences. They are obtained from
(10.361)--(10.367), (10.370)--(10.371) by multiplying positive
denominators. The editable
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/V_adaptive_comparison.py}{rational certificate}
supplies every integer numerator and denominator, and performs precisely
these finite operations.

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}
  >{\raggedright\arraybackslash}p{(\linewidth - 6\tabcolsep) * \real{0.2000}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 6\tabcolsep) * \real{0.2667}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 6\tabcolsep) * \real{0.2667}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 6\tabcolsep) * \real{0.2667}}@{}}
\toprule\noalign{}
\begin{minipage}[b]{\linewidth}\raggedright
Rank
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
Fractional mass gap
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
Fractional input gap
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
Terminal closure gap
\end{minipage} \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
2 & 1930 & 2456 & 3358 \\
3 & 4129 & 2531 & 3657 \\
4 & 7334 & 2838 & 3934 \\
5 & 9512 & 3000 & 4088 \\
6 & 11191 & 3100 & 4186 \\
7 & 12895 & 3168 & 4254 \\
Every rank at least 8 & 7299 & 2520 & 4171 \\
\end{longtable}
}

In the twelve finite endpoint substitutions, every integer input gap
exceeds one, every \(\Delta_0<c/a\), and every \(\tau_j>\eta^j\). These
are also finite rational inequalities with their original c bands, not
extrapolations in r. Concavity proves their full depth ranges.

Here is the analytic bound that justifies the last row for all ranks.
Take \(c=827/1000\) and define

\[
H_-=\left(1-\frac c9\right)^9,\qquad
H_+=\left(\sum_{k=0}^{10}\frac{c^k}{k!}\right)^{-1},\qquad Q_-=3H_-.
\tag{10.372}
\]

Lemma 10.122 proves \(H\ge H_-\). The inequality \(\ln(1-z)<-z\),
obtained by integrating \(-1/(1-z)<-1\), gives \(H<\exp(-c)<H_+\). Thus
\(Q\ge Q_->1\). The increasing \(b_r,S_-\), decreasing
\(\epsilon_r,\bar g_{12}\), \(C_r^{\sharp}<1\), and \(\bar g_9=107/103\)
bound b and \(e_0\) from above by their rank-eight substitutions with
\(C_r^{\sharp}\) replaced by one. Write these bounds as \(b_+,e_+\), let
\(S_8=S_-(8)\), and use \(\ell_+=\bar\lambda(1099/1000)/(6c_4)\). The
original D exceeds six in this rank range, by the parameter bounds in
Lemma 10.129.

The upper common endpoints are (10.371) with \(b_+,e_+,H_+,\ell_+\), and
\(\mathcal L_{20}(Q_-)\). Write them as
\(C_{\mathrm E}^+,C_{\mathrm L}^+\), and put
\(C_{\max}=\max\{C_{\mathrm E}^+,C_{\mathrm L}^+\}\). To bound the sum
of the adaptive losses, the finite positive sums are at most
\(\sum_{j\ge1}q^{-j-1}=1/[q(q-1)]=1/6\) and
\(\sum_{j\ge1}j q^{-j-1}=1/(q-1)^2=1/4\). The first follows by
multiplying a finite geometric sum by \(1-1/q\) and passing to its
vanishing remainder. For the second, write \(j=\sum_{k=1}^j1\) and sum
the nonnegative double series in either order, obtaining
\((1/q^2)/(1-1/q)^2\). Hence

\[
\sum_{j=0}^{r-1}\Delta_j(I)\le\frac1a
\left[\frac{(1+\lambda_{\mathrm{node}})(\mathcal C(I)+\varepsilon)}{12}
+\frac{\lambda_{\mathrm{node}}\ell}{8}
+\left(\frac{1+S_-^{-1}}{4c_2}
+\frac{\lambda_{\mathrm{node}}(r-1)}{2c_2}\right)Q^{-I}\right].
\tag{10.373}
\]

At the early endpoint replace \(Q^{-I}\) by \(Q_-^{-1}\); at the late
endpoint replace it by \(1/1000\). Use \((r-1)/a\le1\) and
\(1/a\le1/9\). If these two torus bounds are v, the resulting loss bound
is

\[
\frac{\lambda_{\mathrm{node}}v}{2c_2}
+\frac19\left[
\frac{(1+\lambda_{\mathrm{node}})(C^++\varepsilon)}{12}
+\frac{\lambda_{\mathrm{node}}\ell_+}{8}
+\frac{(1+S_8^{-1})v}{4c_2}\right].
\tag{10.374}
\]

The exact rational substitutions give \(1-\text{loss}>671/1000>2/3\) at
the early endpoint and \(1-\text{loss}>892/1000>2/3\) at the late
endpoint. Concavity then gives \(\tau_r>2/3\) throughout. Both
comparisons are included in the two uniform certificate rows.

The last full block alone now has normalized fractional mass strictly
greater than

\[
q\cdot9\left(2-\frac1{q^8S_8}\right)(2/3-H_+).
\tag{10.375}
\]

Indeed it has more than \(2q^rS_I-1\) nodes and multiplicity greater
than \((\tau_r-H)T_I\). Use \(a\ge9,q^rS_I\ge q^8S_8\), and
\(c_1W_*ST\theta=qa\). All the other increments in the mixed system are
nonnegative. This is a lower bound for its actual total mass, without
any requirement that every adaptive \(\tau_j\) exceed \(\eta^j\).

At both endpoints, enlarge (10.364) with
\(b_+,e_+,H_+^2,H_-^{-1},S_8^{-1},\ell_+\), \(q^{-r}\le q^{-8}\), and
the respective common depth terms. Subtracting this upper bound from
(10.375) gives differences greater than \(9753/1000\) and \(7299/1000\).
Since \(\Delta_0<c/a\), the input formula (10.367) is bounded below by
(10.358). In that formula use \(1-H\le1-H_-\), \(2cH/a\le2cH_+/9\) and
\(q^r\ge q^8\). Its two gaps exceed \(4974/1000\) and \(2520/1000\).
These bounds prove the asserted uniform fractional comparisons.

For the remaining uniform integer assertions, the two endpoint bounds
give \(B_0<8\). Consequently \(\Delta_0<(801/1200)/a<c/a\). For
\(j\ge1\), \((C_{\max}+\ell_+j)/q^{j+1}\) decreases with j: cross
multiplication reduces its next-step comparison to
\((q-1)C_{\max}+((q-1)j-1)\ell_+>0\). Thus its largest value is at one,
and exact rational substitution gives

\[
\frac{C_{\max}+\ell_+}{q^2}+\frac1{c_2Q_-}<3/2,
\qquad
\frac{\lambda_{\mathrm{node}}(3/2+\varepsilon/q^2)}2<4/5.
\tag{10.376}
\]

It follows that \(B_j<(3/2)q^{j+1}\), \(\Delta_j<4/(5a)\). The full
input gap is bounded below by
\(q^8(qc_1-3/2)-c_1/(7\cdot9b_8)-(4/5)/(9c_3Y_-)\), which exceeds one.
The closure gap is bounded below by
\(c_1q^9-c_1/(7\cdot9b_8)-[2(801/1200)+cH_+]/(9c_3Y_-)-8\), also greater
than one. This proves every integer comparison at every rank, rather
than testing finitely many larger ranks.

Finally, (10.369) needs the interval \([1,L_D+1]\). The function
\(B_0(J)\) remains convex. Its early endpoint has the same upper bound
as before. At the new late endpoint, \(Q^{-J}<1/1000,H^J\le H\), and
\(\ell J\le3\bar\lambda(1099/1000)/(c_4\mathcal L_{20}(Q))+\ell_-\),
using \(\ell_+\) in the uniform case. Its upper bound is therefore the
previous late \(B_0\) bound plus that one extra \(\ell_-\) or
\(\ell_+\). The third column of the table uses exactly this extra cost.
The associated input bound in (10.369) exceeds its arithmetic bound by
more than one in each of the twelve finite rows and the two uniform
rows. Convexity proves (10.369) on the whole larger interval. All strict
assertions of the lemma follow. \(\square\)

\subsubsection{Producing the actual zeros and passing the
phase}\label{producing-the-actual-zeros-and-passing-the-phase}

\textbf{Theorem 10.136 (the complete original V first induction).} Under
the original contradiction and initial integral auxiliary kernel of
Section 40, there are nonzero stage families \(\Phi_I\) for every
\(0\le I\le I_0\). Each has the original coefficient-height bound,
minimum valuation \(\delta\), selected derivations and phase conditions,
torus widths at most \(q^{-I}D_i\), and additive argument \(q^{-I}x\).
Each has every original zero at

\[
\Phi_I(s;\boldsymbol t)=0,
\qquad |s|\le\lfloor qS_I\rfloor,\quad
|\boldsymbol t|\le\lfloor\eta T_I\rfloor.
\tag{10.377}
\]

For each \(1\le I\le I_0\), the stronger radius
\(q(\lfloor S_I\rfloor+1)\) is available at that same full cutoff,
together with all q-deleted zeros at that radius through
\(\lfloor T_I\rfloor\). For every \(1\le I<I_0\), all full original
outer blocks at radii \(\lfloor q^jS_I\rfloor\), cutoffs
\(\lfloor\eta^jT_I\rfloor\), \(2\le j\le r\), remain available. The same
family has the additional full adaptive blocks at those radii through
\(\lfloor\tau_j(I)T_I\rfloor\), with the rounded first radius and cutoff
\(\tau_1\). Its original fractional output is

\[
\Phi_I(s/q;\boldsymbol t)=0,
\qquad |s|\le q(\lfloor S_{I+1}\rfloor+1),\quad
|\boldsymbol t|\le\lfloor T_{I+1}\rfloor.
\tag{10.378}
\]

\textbf{Proof.} Corollary 10.127 and Theorems 10.130--10.131 supply the
initial family, full first fractional block and stage-one family,
including its q-deleted zeros. The original multiplicity bound ensures
\(I_0\ge1\).

Suppose the stage-I q-deleted zeros have been obtained and
\(1\le I<I_0\). First close them at cutoff \(\lfloor\tau_1T_I\rfloor\),
using \(M_0\) from Lemma 10.134. Both analytic entries exceed \(B_0\),
by (10.368); a nonzero value in K has the opposite arithmetic bound.
Corollary 10.120 therefore forces every missing integer value to be
zero. Since \(\Delta_0<c/a\), this includes the original full
\(\eta T_I\) cutoff. Theorem 10.126 independently extends those original
integer blocks through all \(\eta^jT_I\) cutoffs. In addition, apply
Lemma 10.134 successively at \(j=1,\ldots,r-1\), with the just-proved
adaptive block as the next input. Both entries exceed the corresponding
\(B_j\). This gives every stated adaptive block for the very same
coefficient family. No selection or coefficient change occurs during
these integer extensions.

Use their complete mixed multiplicities for a target in (10.378). Both
normalized analytic entries exceed \(A_{\mathrm f}\), by Lemma 10.135.
The projected and unprojected values have the same proved precision, and
the principal exponential laws identify the analytic original value with
the actual algebraic value in the normalized support field. Corollary
10.120, with its full global degree factor and original selected local
degrees, forces that value to vanish. This proves every fractional zero
in (10.378) at its exact rounded radius and every derivative order.

Apply Theorem 10.51 to its nodes prime to q. Retain a class containing
an original coefficient of valuation \(\delta\); when q divides \(G_0\),
perform the required second phase partition too. The entire class
vanishes at every prescribed node and order. The invertible affine
binomial transfer preserves the total-order cutoff; q is a unit at
p.~The new coefficient subvector is integral, nonzero, of the same
minimum and no larger height. Its widths shrink by q and its additive
argument becomes \(q^{-(I+1)}x\). The actual phase identity puts its
integer prepared values in K. This constructs \(\Phi_{I+1}\) with all
q-deleted zeros at the rounded radius and cutoff
\(\lfloor T_{I+1}\rfloor\).

For clarity, the full closure at the successor \(J=I+1\) is valid even
when \(J=I_0\). Set \(M=\lfloor cT_J/a\rfloor+1\) and take the q-deleted
radius \(q(\lfloor S_J\rfloor+1)\). The floor inequality supplies every
ordinary jet for output \(\lfloor\eta T_J\rfloor\). Its first analytic
entry exceeds \(2cq(q-1)\). The two separate deleted and inverse/jet
losses are at most \([2cT_J/a+1]H^{-J}H_1/\ln p\). Since \(cT_J/a\ge1\),
its second entry exceeds \(c_1q^{r+1}-c_1/(7ab_r)-3c/(ac_3Y_-)\). Its
nonzero integer arithmetic envelope is \(B_0(J)\), by the same scalar,
clearing, width and field argument as Lemma 10.134. Formula (10.369)
forces every remaining integer value to vanish on \([1,L_D+1]\). This
completes the full original final block without applying the
fractional-depth comparison beyond its proved interval.

Induction now constructs the families through \(I_0\), with (10.377) and
all the asserted stronger blocks. Every passage retains a coefficient
attaining the original minimum, so none can vanish identically. The
result proves the original V first induction at both possible stopping
indices; the subsequent stopping contradictions and the other prime
cases are separate arguments. \(\square\)

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={Retained integer derivative orders and exact all-rank fractional comparison margins}]{../figures/V-adaptive-profile.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.31. Left: a conservative rank-two early-endpoint
envelope, with D=5, evaluated by the exact rational formulas
(10.361)--(10.363), at illustrative \(S_I=1000,T_I=10000\). The full
cutoffs are \(\lfloor\tau_1T_I\rfloor,\lfloor\tau_2T_I\rfloor\); the
original cutoffs are \(\lfloor\eta T_I\rfloor,\lfloor\eta^2T_I\rfloor\),
and the fractional output is \(\lfloor HT_I\rfloor\). Deleted inner
nodes additionally retain order10000. The panel describes available
derivative cutoffs, not auxiliary-function values. Right: strict
rational lower bounds for the mass and input gaps in Lemma10.135,
minimized over both continuous-depth endpoints. The last column
represents the proved uniform bound for every rank at least eight. The
full factor \(q^r\) has already been charged in these fractional
comparisons. The extra terminal closure is proved separately in
(10.369). \href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/V_adaptive_profile.py}{Figure
program}, \href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/V_adaptive_comparison.py}{rational
certificate}. Human-source context: the original first induction and
stopping indices in
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
freely accessible paper}, Section5,
equations(5.14)--(5.16),(5.20),(5.42)--(5.62). The adaptive bounds,
complete interpolation comparisons and terminal closure are proved
above.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{44. The complete odd-prime phase
induction}\label{the-complete-odd-prime-phase-induction}

For odd p the original integer derivative cutoffs already give enough
fractional precision. We prove this for all seven coupled field rows,
every rank and the entire original depth interval. The proof uses the
full mixed profile of Section 42 and the global degree of the actual
fractional support field.

Retain every original hypothesis of Section 40, now with odd p and q=2.
The rows are I.1, I.2, II, III.1, III.2 and the two branches of IV. Put
\(a=r+1,c=c_5,\eta=1-c/a,H=\eta^a,Q=qH,D=A+\nu\ln q\), and retain the
stopping indices (10.360). Thus \(S_I=H^{-I}S,T_I=H^IT\). The additive
degree \(D_{\mathrm a}=kL\) is unchanged. The original depth restriction
gives \(1\le I<I_0\Rightarrow I\le L_D\) and \(cHT_I/a\ge1\). At the
final stage only \(cT_{I_0}/a\ge1\) is asserted.

\subsubsection{Arithmetic envelopes and the complete input
profile}\label{arithmetic-envelopes-and-the-complete-input-profile}

Use the constants \(c_1,c_2,c_3,c_4\), their original rank bands for c,
and the simultaneous bounds \(C,Y_-,\Omega_+\) in Lemma 10.121. In
particular \(c_2=7/4\). Let \(b_r,\epsilon_r,\bar g_{12},\bar g_9,S_-\)
be exactly the row-dependent bounds in (10.331). The number
\(\bar g_{12}\) is zero in the degree-one rows I.2 and III.2;
\(\bar g_9=107/103\) in those rows and at every rank at least eight.
Define

\[
\begin{aligned}
A_r&=c_1(\bar g_{12}+\epsilon_r)+
\frac{c_1\epsilon_r/2+\bar\lambda/c_4+
C_r^{\sharp}/(c_3Y_-)+117/(232c_2)}{C-1},\\
b&=A_r+c_1\left(\frac1{300}+\frac1{273\cdot999b_r}\right)
+\frac{\bar\lambda}{c_4},\\
e_0&=\frac{\bar g_9}{c_3Y_-},\qquad
n=\frac{\bar\lambda\Omega_+}{c_4},\qquad
\ell=\frac{\bar\lambda(347/500)}{Dc_4},\qquad
\bar\lambda=189/188.
\end{aligned}
\tag{10.379}
\]

The coefficient bound is precisely (10.339); \(C_r^{\sharp}\) is
(10.307). The symbol \(e_0\) again denotes a numerical cost rather than
a local degree. Our two arithmetic envelopes are

\[
\begin{aligned}
A_{\mathrm f}(I)&=b+e_0H^{I+1}+\ell(I+2)
+\frac{H^{-1}+S_-^{-1}}{c_2}Q^{-I}+\frac n{q^r},\\
B_0(J)&=b+n+e_0H^J+\ell(J+1)
+\frac{q(1+S_-^{-1})}{c_2}Q^{-J}.
\end{aligned}
\tag{10.380}
\]

\textbf{Lemma 10.137 (odd-prime fractional and terminal envelopes).} For
an admissible stage \(1\le I<I_0\), a nonzero fractional target
\(x=s/q\), with \(|s|\le q(\lfloor S_{I+1}\rfloor+1)\) and total
derivative order at most \(\lfloor T_{I+1}\rfloor\), has
\((c_1W_*/q^r)[v_p(V)+G_u-\delta]\le A_{\mathrm f}(I)\). Assuming all
the original stage-I integer blocks, its two analytic entries in that
same normalization are strictly greater than

\[
\begin{aligned}
\Gamma_r&=\frac{2cq}{q^r}\sum_{j=0}^r(q\eta)^j
-\frac{qa}{q^rS_-}
\left[2(q+1)(\eta^2-H)+2\sum_{j=3}^r(\eta^j-H)\right],\\
J_{\mathrm f,r}&=c_1q-\frac{c_1}{7ab_rq^r}
-\frac{1-H+c(1+2H)/a}{q^rc_3Y_-}.
\end{aligned}
\tag{10.381}
\]

For any integer stage \(J\ge1\) with \(cT_J/a\ge1\), a nonzero integer
target at the rounded radius \(q(\lfloor S_J\rfloor+1)\), of total order
at most \(\lfloor\eta T_J\rfloor\), has normalized valuation at most
\(B_0(J)\). If all q-deleted zeros are available at this radius through
\(\lfloor T_J\rfloor\), the two analytic closure entries are strictly
greater than

\[
2cq(q-1),\qquad
J_{\mathrm c,r}=c_1q^{r+1}-\frac{c_1}{7ab_r}
-\frac{3c}{ac_3Y_-}.
\tag{10.382}
\]

All comparisons fix the same additive order \(u\le D_{\mathrm a}\) and
use the original coefficient minimum \(\delta\).

\textbf{Proof.} Theorem 10.115 gives the coefficient bound \(A_r\) and
the separate scalar remainder in b. At the fractional radius, the
additive argument has size at most \[
q^{-I}(\lfloor S_I/H\rfloor+1)
\le (S/H)Q^{-I}+q^{-I}<qS.
\] The last strict inequality is the original scalar bound of Lemma
10.129. It gives the additive cost \(\bar\lambda/c_4\). The actual torus
cost is at most \((H^{-1}+S_-^{-1})Q^{-I}/c_2\), since \(S\ge S_-\) and
\(q^{-I}\le Q^{-I}\). The fractional clearing exponent is
\(\Xi_{I,1,u}\le D_{\mathrm a}(I+1+1/(q-1))=D_{\mathrm a}(I+2)\). The
upper logarithm bound \(\ln2<347/500\), proved in Lemma 10.122,
therefore gives \(\ell(I+2)\). The total output order is at most
\(HT_I\), so Lemma 10.114 gives the Euler/lcm cost \(e_0H^{I+1}\).

The actual normalized support field in Theorem 10.56 has global degree
\(dq^{h_\Lambda}\), \(h_\Lambda\le r\), and the original selected local
degrees e,f.~The product-formula comparison charges this entire factor.
Divide its positive algebraic-height terms by \(q^r\); because
\(q^{h_\Lambda-r}\le1\), each is bounded above by its displayed cost.
The normal contribution is added after that comparison and is at most
\(n/q^r\). This proves the first line of (10.380). It remains valid when
the chosen completion has local degree one.

The original full and deleted blocks have the exact nested
multiplicities (10.356). Lemma 10.133 proves their complete mass and
fixed-order input estimates, including the separate deleted-cardinal and
inverse/jet losses. Since q=2 and \(S_I\ge S_-\), (10.357)--(10.358)
imply (10.381). All terms \(\eta^j-H\) in the subtracted floor loss are
positive. Replacing \(S_I\) by \(S_-\) thus decreases that lower mass.
No derivative floor is changed.

At the integer closure radius, \[
q^{-J}q(\lfloor S_J\rfloor+1)
\le qSQ^{-J}+q^{1-J}\le S/H+1<qS.
\] The additive bound is the same. The integer clearing exponent obeys
\(\Xi_{J,0,u}\le D_{\mathrm a}(J+1)\), since the original binomial
denominator contribution is at most \(D_{\mathrm a}/(q-1)\). The torus
cost is at most \(q(1+S_-^{-1})Q^{-J}/c_2\), the Euler/lcm cost at most
\(e_0H^J\), and the normal term at most n.~The actual prepared integer
value lies in K by Theorem 10.55. Corollary 10.116 therefore uses d, e
and f without a fractional root-field factor. These facts prove
\(B_0(J)\).

For the analytic closure choose \(M=\lfloor cT_J/a\rfloor+1\). The
inequality
\(\lfloor T_J\rfloor-\lfloor\eta T_J\rfloor\ge\lfloor cT_J/a\rfloor\)
supplies every required ordinary input jet at every deleted node, for
every output order under consideration. There are
\(2(q-1)(\lfloor S_J\rfloor+1)>2(q-1)S_J\) such nodes. Since
\(M>cT_J/a\) and \(c_1W_*ST\theta=qa\), their normalized mass is
strictly greater than \(2cq(q-1)\).

The exact cancellation (10.327) retains both the cardinal loss \(MB\)
and the inverse/jet loss \((M-1)\max\{B,\beta-\vartheta,0\}\). Lemma
10.123 applies to this rounded interval because
\(2q(\lfloor S_J\rfloor+1)<3q^rS_J\). Their sum is at most
\([2cT_J/a+1]H^{-J}H_1/\ln p\). Use \(1/T_J\le c/a\), the identity
\(c_1W_*T_JH^{-J}H_1/\ln p\le1/(c_3Y_-)\), \(W_*U=q^{r+1}\), and the
strict coefficient-minimum bound from Lemma 10.123. Discard only the
nonnegative term \((D_{\mathrm a}-u)\theta\). The remaining entry is
strictly greater than \(J_{\mathrm c,r}\). This proves (10.382) using
\(cT_J/a\ge1\), without the stronger successor multiplicity restriction.
Orders \(u>D_{\mathrm a}\) vanish identically and require no comparison.
\(\square\)

\subsubsection{A continuous interval and an infinite rank
range}\label{a-continuous-interval-and-an-infinite-rank-range}

\textbf{Lemma 10.138 (all odd-prime strict comparisons).} At every
original odd-prime rank \(r\ge2\), throughout \(1\le I\le L_D\),

\[
\Gamma_r-A_{\mathrm f}(I)>3/4,\qquad
J_{\mathrm f,r}-A_{\mathrm f}(I)>1.
\tag{10.383}
\]

Throughout the larger interval \(1\le J\le L_D+1\),

\[
2cq(q-1)-B_0(J)>1/4,\qquad
J_{\mathrm c,r}-B_0(J)>1.
\tag{10.384}
\]

\textbf{Proof.} Fix the actual r and D. The functions
\(A_{\mathrm f},B_0\) are convex: they are sums of a constant, an affine
function and positive multiples of \(H^I,Q^{-I}\). For example
\((H^I)''=(\ln H)^2H^I>0\). All the analytic lower bounds in
(10.381)--(10.382) are independent of depth. Their gaps are therefore
concave. A concave function on a closed interval is bounded below by the
lesser of its two endpoint values: its defining chord inequality gives
this directly. We will bound precisely those endpoints.

At ranks 2--7 put \(\ell_-=\bar\lambda(347/500)/(5c_4)\) and
\(d_+=3\bar\lambda(347/500)/(c_4\mathcal L_{20}(Q))\), where
\(\mathcal L_{20}\) is the positive rational logarithm bound (10.370).
Lemmas 10.122 and 10.129 give \(D>5\), \(Q>1\), \(L_D>1\), and
\(Q^{-L_D}=\exp(-3D)<1/1000\). Consequently \(\ell\le\ell_-\),
\(\ell L_D<d_+\), and \(H^{L_D+1}\le H^2\). The following are upper
bounds at the early and late endpoints, respectively:

\[
\begin{aligned}
F_{\mathrm E}&=b+e_0H^2+3\ell_-
+\frac{H^{-1}+S_-^{-1}}{c_2Q}+n/q^r,\\
F_{\mathrm L}&=b+e_0H^2+d_++2\ell_-
+\frac{H^{-1}+S_-^{-1}}{1000c_2}+n/q^r,\\
C_{\mathrm E}&=b+n+e_0H+2\ell_-
+\frac{q(1+S_-^{-1})}{c_2Q},\\
C_{\mathrm L}&=b+n+e_0H+d_++2\ell_-
+\frac{q(1+S_-^{-1})}{1000c_2}.
\end{aligned}
\tag{10.385}
\]

The closure late endpoint is \(J=L_D+1\). Its clearing cost is
\(\ell(L_D+2)\), which explains both copies of \(\ell_-\) in
\(C_{\mathrm L}\). We have bounded \(H^{L_D+1}\) by the weaker H there.
These substitutions only enlarge the arithmetic costs.

Here are strict lower bounds, divided by 1000, for all four endpoint
gaps. The first table takes the minimum over ranks 2--7 and both
endpoints. The second gives uniform bounds for every rank at least
eight, justified below.

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}
  >{\raggedright\arraybackslash}p{(\linewidth - 8\tabcolsep) * \real{0.1579}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 8\tabcolsep) * \real{0.2105}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 8\tabcolsep) * \real{0.2105}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 8\tabcolsep) * \real{0.2105}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 8\tabcolsep) * \real{0.2105}}@{}}
\toprule\noalign{}
\begin{minipage}[b]{\linewidth}\raggedright
Case, ranks 2--7
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
Mass gap
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
Input gap
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
Closure gap
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
Closure input gap
\end{minipage} \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
I.1 & 1266 & 1204 & 320 & 9517 \\
I.2 & 1231 & 1163 & 288 & 9472 \\
II & 1278 & 1220 & 347 & 9473 \\
III.1 & 1200 & 1155 & 313 & 9438 \\
III.2 & 1164 & 1136 & 284 & 9418 \\
IV, \(P=1\) & 1219 & 1151 & 312 & 9449 \\
IV, \(P\ge7\) & 1530 & 1553 & 713 & 10088 \\
\end{longtable}
}

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}
  >{\raggedright\arraybackslash}p{(\linewidth - 8\tabcolsep) * \real{0.1579}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 8\tabcolsep) * \real{0.2105}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 8\tabcolsep) * \real{0.2105}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 8\tabcolsep) * \real{0.2105}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 8\tabcolsep) * \real{0.2105}}@{}}
\toprule\noalign{}
\begin{minipage}[b]{\linewidth}\raggedright
Case, every rank at least 8
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
Mass gap
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
Input gap
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
Closure gap
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
Closure input gap
\end{minipage} \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
I.1 & 868 & 1366 & 392 & 739955 \\
I.2 & 815 & 1314 & 332 & 740095 \\
II & 913 & 1406 & 420 & 733980 \\
III.1 & 815 & 1303 & 383 & 732361 \\
III.2 & 780 & 1266 & 347 & 731256 \\
IV, \(P=1\) & 811 & 1300 & 391 & 737433 \\
IV, \(P\ge7\) & 1185 & 1675 & 836 & 737961 \\
\end{longtable}
}

For each finite row, substitute the original rational c band, coupled
constants, \(S_-\), \(b_r\) and \(C_r^{\sharp}\) in (10.379), (10.381),
(10.382) and (10.385). Multiply the positive denominators. The editable
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/odd_depth_endpoints.py}{exact rational
calculation} supplies all 84 individual finite substitutions and their
integer numerators and denominators. Thus the finite columns are finite
rational inequalities; the continuous depths follow from concavity.

We now justify the uniform columns without sampling ranks. For each row
use its fixed c at ranks at least eight, and put

\[
\begin{aligned}
H_-&=(1-c/9)^9,\qquad
E_-(c)=\sum_{j=0}^{10}\frac{c^j}{j!},\qquad H_+=E_-(c)^{-1},\\
E_+(c)&=E_-(c)+\frac{c^{11}}{11!(1-c/12)},\qquad Q_-=qH_-.
\end{aligned}
\tag{10.386}
\]

The positive exponential series gives \(E_-<\exp(c)<E_+\). Indeed the
omitted terms start at \(c^{11}/11!\); the ratio of each subsequent term
to its predecessor is at most \(c/12<1\), with a strict inequality after
the first ratio. Summing the geometric majorant proves the upper
inequality. Lemma 10.122 proves \(H\ge H_-\); integration of
\(-1/(1-z)<-1\) gives \(H<\exp(-c)<H_+\). Direct rational substitution
in each original c row gives \(Q_->1\).

The same integral also gives, for \(0<z<1\), \(\ln(1-z)>-z/(1-z)\),
since \(1/(1-t)<1/(1-z)\) for \(0\le t<z\). Apply it with \(z=c/a\).
Because \(a-c>r\),

\[
\ln(\eta^r)>-\frac{rc}{a-c}>-c,
\qquad \eta^r>\exp(-c)>E_+(c)^{-1}.
\tag{10.387}
\]

Reverse the principal geometric sum in (10.381):
\(q^{-r}\sum_{j=0}^r(q\eta)^j=\sum_{k=0}^r q^{-k}\eta^{r-k}\). At ranks
at least eight it contains its first four positive terms. Each of them
is at least \(q^{-k}\eta^r\). The negative floor bracket in (10.381) is
less than \(2(q+r-1)\), because every difference \(\eta^j-H\) is less
than one. Hence

\[
\begin{aligned}
\Gamma_r&>
\frac{2cq}{E_+(c)}\sum_{k=0}^3q^{-k}
-\frac{2qa(q+r-1)}{q^rS_-}\\
&\ge
\frac{2cq}{E_+(c)}\sum_{k=0}^3q^{-k}
-\frac{2q\cdot9(q+7)}{q^8S_8}
=:\Gamma_+,
\qquad S_8=S_-(8).
\end{aligned}
\tag{10.388}
\]

For the second inequality q=2 makes the subtracted numerator
\(4(r+1)^2\). The sequence \((r+1)^2/2^r\) decreases for \(r\ge8\): its
consecutive ratio is \(((r+2)/(r+1))^2/2<1\). The original \(S_-\)
increases in each row, as is explicit in (10.331). Their combination
proves the rank-eight upper bound on the loss.

The parameters \(b_r,S_-\) increase, \(\epsilon_r,\bar g_{12}\)
decrease, and \(C_r^{\sharp}<1\). All coefficients in (10.379) are
positive. Consequently b is bounded above by \(b_+\), its rank-eight
value with \(C_r^{\sharp}\) replaced by one; \(e_+=\bar g_9/(c_3Y_-)\)
is constant in this rank range. By Lemma 10.129, D exceeds six. Define
\(\ell_+=\bar\lambda(347/500)/(6c_4)\) and
\(d_+^+=3\bar\lambda(347/500)/(c_4\mathcal L_{20}(Q_-))\).

In the four endpoints (10.385), replace b,
\(e_0H^2,e_0H,H^{-1},S_-^{-1},Q^{-1},\ell_-,d_+,n/q^r\) by the
respective upper bounds
\(b_+,e_+H_+^2,e_+H_+,H_-^{-1},S_8^{-1},Q_-^{-1},\ell_+,d_+^+,n/q^8\).
The late torus factor is still \(1/1000\). Call the resulting rational
upper bounds
\(F_{\mathrm E}^+,F_{\mathrm L}^+,C_{\mathrm E}^+,C_{\mathrm L}^+\).
These are valid at every rank at least eight, in each row. Corresponding
uniform analytic lower bounds are

\[
\begin{aligned}
J_{\mathrm f,r}&\ge c_1q-\frac{c_1}{7\cdot9b_8q^8}
-\frac{1-H_-+c(1+2H_+)/9}{q^8c_3Y_-}=:J_{\mathrm f,+},\\
J_{\mathrm c,r}&\ge c_1q^9-\frac{c_1}{7\cdot9b_8}
-\frac{3c}{9c_3Y_-}=:J_{\mathrm c,+}.
\end{aligned}
\tag{10.389}
\]

For the first use \(1-H\le1-H_-\), \(2cH/a\le2cH_+/9\),
\(a\ge9,q^r\ge q^8,b_r\ge b_8\). For the second use those last three
bounds and \(q^{r+1}\ge q^9\).

There are now only fourteen rational substitutions: seven coupled rows
and two endpoints. Subtract their upper F bounds from
\(\Gamma_+,J_{\mathrm f,+}\), and their upper C bounds from
\(2cq(q-1),J_{\mathrm c,+}\). Multiplying positive denominators gives
the uniform table displayed above. The same exact calculation file
records all fourteen values. Over both tables, the minimum mass gap
exceeds \(780/1000>3/4\), minimum fractional input gap exceeds one,
minimum closure gap exceeds \(284/1000>1/4\), and minimum closure input
gap exceeds one. Concavity extends these endpoint bounds to both entire
depth intervals. This proves every assertion. \(\square\)

\subsubsection{The actual phase
induction}\label{the-actual-phase-induction}

\textbf{Theorem 10.139 (the complete odd-prime first induction).} Under
the original contradiction and the initial integral kernel of Section
40, every odd-prime case has a nonzero family \(\Phi_I\) for each
\(0\le I\le I_0\). Each retains the initial coefficient-height bound and
minimum \(\delta\), selected derivations and phase conditions, widths at
most \(q^{-I}D_i\), and additive argument \(q^{-I}x\). It has every
original full integer zero

\[
\Phi_I(s;\boldsymbol t)=0,
\qquad |s|\le\lfloor qS_I\rfloor,\quad
|\boldsymbol t|\le\lfloor\eta T_I\rfloor.
\tag{10.390}
\]

For \(1\le I\le I_0\), the stronger rounded radius
\(q(\lfloor S_I\rfloor+1)\) is available at this full cutoff, and its
q-deleted nodes have every zero through \(\lfloor T_I\rfloor\). For
\(1\le I<I_0\) the same family has all full integer outer blocks at
radii \(\lfloor q^jS_I\rfloor\), cutoffs \(\lfloor\eta^jT_I\rfloor\),
\(2\le j\le r\). Its full fractional output is

\[
\Phi_I(s/q;\boldsymbol t)=0,
\qquad |s|\le q(\lfloor S_{I+1}\rfloor+1),\quad
|\boldsymbol t|\le\lfloor T_{I+1}\rfloor.
\tag{10.391}
\]

\textbf{Proof.} The initial family and all its integer extensions are
Corollary 10.127. The complete first fractional block is Theorem 10.130;
Theorem 10.131 gives the stage-one family, both its rounded full block
and q-deleted block. The original initial multiplicity bound gives
\(I_0\ge1\). These earlier proofs include every original coupled row and
rank.

Assume the stage-I family has its rounded full inner and deleted blocks,
where \(1\le I<I_0\). Theorem 10.126 extends its original full integer
blocks to every radius \(\lfloor q^jS_I\rfloor\) with cutoff
\(\lfloor\eta^jT_I\rfloor\), \(2\le j\le r\). These comparisons use the
same coefficient family throughout. The mixed multiplicities in (10.356)
therefore apply to every target in (10.391) and every specified
derivative order.

If such a value were nonzero, Lemma 10.137 would bound its normalized
valuation above by \(A_{\mathrm f}(I)\), in its actual support field.
Lemma 10.138 says that both analytic entries exceed this bound. The
principal exponential laws in Lemma 10.92 identify the analytic value
with the prepared algebraic value, and the projected and unprojected
values have the identical proved precision. Corollary 10.120, with the
full global factor \(q^r\), the selected local degrees and exact
clearing, yields a contradiction. This proves (10.391) at every
fractional node and derivative order.

Apply Theorem 10.51 to the nodes prime to q in that fractional block.
Choose a phase class containing a coefficient of valuation \(\delta\);
if q divides \(G_0\), apply its required second phase partition as well.
Each chosen class vanishes at all the prescribed nodes and orders. The
invertible affine binomial transfer preserves the total-order cutoff,
since q is a unit at p.~The selected coefficient subvector is integral,
nonzero, of no larger height, and still has minimum \(\delta\). Its
torus widths shrink by q, its additive argument is \(q^{-(I+1)}x\), and
Theorem 10.55 puts its actual integer prepared values in K. Thus it is
\(\Phi_{I+1}\), with every required q-deleted zero through
\(\lfloor T_{I+1}\rfloor\).

Close that new deleted block with J=I+1. Even when \(J=I_0\), the
original indices give \(J\le L_D+1\) and \(cT_J/a\ge1\). The last part
of Lemma 10.137 supplies both analytic entries; (10.384) makes each
strictly exceed the nonzero integer bound \(B_0(J)\). Corollary 10.120
forces every remaining integer value to vanish. Hence the stronger
rounded full inner block is available through
\(\lfloor\eta T_J\rfloor\), including the terminal stage. This step uses
the larger closure interval, without extending the fractional comparison
beyond \(L_D\).

Induction proves all the asserted families through \(I_0\). Their
reference coefficients attain \(\delta\) at every phase passage, so all
are nonzero. Every original integer and fractional radius and floor
cutoff is retained. \(\square\)

\textbf{Corollary 10.140 (the first induction in every original prime
and field case).} For each original case I--V and every rank \(r\ge2\),
the original parameters (10.186), coefficient bounds and minimum,
selected derivations and both stopping indices (10.360) admit nonzero
stage families through \(I_0\), satisfying

\[
|s|\le\lfloor qS_I\rfloor,\quad
|\boldsymbol t|\le\lfloor\eta T_I\rfloor
\ \Longrightarrow\ \Phi_I(s;\boldsymbol t)=0.
\tag{10.392}
\]

For positive stages their full rounded and deleted blocks, and before
the terminal stage their full original outer and fractional blocks, have
exactly the stronger ranges specified in Theorems 10.136 and 10.139.

\textbf{Proof.} Theorem 10.139 covers all odd primes, including both
degree-one rows and both IV branches with their simultaneous field
bounds. Theorem 10.136 covers p=2 with q=3. These exhaust the original
constant table. Both theorems preserve the identical initial kernel data
and stopping indices, so their conclusions combine directly. \(\square\)

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={Original odd-prime derivative profile and strict comparison gaps}]{../figures/odd-all-depth-profiles.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.32. Left: the original rank-two III.2 profile, with
\(c=0.5267,S_I=100,T_I=1000\). The full inner radius is 202, the outer
radius 400, their derivative cutoffs 824 and 679, and the fractional
output cutoff 560. Deleted inner nodes retain order 1000. The
illustration records interpolation inputs and output orders; it does not
plot auxiliary-function values. Solution 53 computes the exact
multiplicities 441, 265, 120 and total interpolation degree 190397.
Right: strict lower bounds for both fractional gaps and the terminal
mass gap in Lemma 10.138, minimized over all finite-rank and
analytically uniform endpoints for each original coupled row. The
comparisons retain the global \(q^r\) field factor. Terminal closure is
proved on its separate larger interval.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/odd_all_depth_profiles.py}{Figure program},
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/odd_depth_endpoints.py}{exact rational
calculation}. Human-source context: the original first induction and
stopping ranges in
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
freely accessible paper}, Section 5, equations (5.13)--(5.20),
(5.42)--(5.62); the full comparison and phase-induction proofs are given
here.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{45. Closing the geometric stopping
branch}\label{closing-the-geometric-stopping-branch}

The first induction now supplies the actual zeros needed by the
geometric reduction. Its last family retains the initial coefficient
minimum, so its polynomial remains nonzero. The degree and rank-profile
comparisons of Sections 34--35 therefore apply to that family with their
original field and weights.

\textbf{Theorem 10.141 (the original geometric stopping contradiction).}
Suppose the original contradiction parameters of Section 40 arise from
the least admissible rank \(r\) in (10.277), with \(r\ge2\). Then the
alternative \(I^*\le I_1\) is impossible.

\textbf{Proof.} If \(I^*\le I_1\), then \(I_0=I^*\). Set \(I\) equal to
this integer. The exact stopping definition gives

\[
Q^I>\exp(3D),\qquad
\overline S=qS_I,\qquad \overline T=\eta T_I,
\qquad
\mathsf D_0=kL,\quad
\mathsf D_j=2q^{\nu-I}D_j\ (1\le j\le r).
\tag{10.393}
\]

Indeed \(I=\lfloor3D/\ln Q\rfloor+1>3D/\ln Q\), including when the
quotient is an integer. Corollary 10.140 gives the nonzero terminal
family and all of its prepared zeros for integer
\(|s|\le\lfloor\overline S\rfloor\), total order at most
\(\lfloor\overline T\rfloor\).

Its coefficients form a nonzero subvector of the original kernel, with
one coefficient attaining \(\delta\). The additive basis still has k
shifted polynomials and powers 1 through L. Its normalized torus
exponent vectors are integral, distinct, and bounded by
\(q^{\nu-I}D_j\). Proposition 10.31 therefore constructs a nonzero
ordinary polynomial with the coordinate bounds in (10.393). The
triangular binomial change and Laurent monomial shift convert exactly
these prepared zeros into ordinary jets through the same total order.
The selected Euler directions still span the original hyperplane W:
affine translations of their eigenvalues give invertible triangular
changes, and their scalar rescalings are nonzero. Lemmas 10.23 and 10.30
prove these assertions.

All original field, height and parameter data remain unchanged. The
independent torus bases and nonzero vector \(\boldsymbol B\) are those
of the minimum-rank representation. The original
\(\theta=\vartheta/(1+\varepsilon_n)\) satisfies the weaker restriction
\(0\le\tau\le1/100\) used in Theorem 10.100. Equation (10.393) is
exactly its geometric depth hypothesis (10.264). Thus Theorem 10.100
verifies every degree inequality and every integer allocation required
by Theorem 10.25, with the strict bounds \(36/25,8,31\).

Corollary 10.101 yields independent integral vectors of some rank
\(1\le m<r\), representing the same nonzero form and producing
local-unit bases in the original field. Theorem 10.102 preserves their
exact residue-index-dependent rank profile, with the strict factor
\(1/700\). Corollary 10.103 then contradicts the defining least rank
\(r\). This rules out \(I^*\le I_1\), including equality. \(\square\)

Figures 10.22--10.23 display the degree thresholds, allocation margins
and preserved rank-profile factor used in this proof. Human-source
context for the original branch is
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
freely accessible paper}, Section 6; the degree, multiplicity, jet and
minimum-rank proofs used here are in this lesson.

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={The actual terminal zero family, ordinary polynomial and smaller-rank contradiction}]{../figures/geometric-stop-coupling.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.33. The branch \(I^*\le I_1\) reaches its exact terminal
integer, including \(I^*=I_1\). Corollary 10.140 supplies the zeros and
the preserved nonzero coefficient. Proposition 10.31 converts them to
ordinary hyperplane jets without increasing their total order; the
polynomial has additive degree at most \(kL\) and torus degrees at most
\(2q^{\nu-I}D_j\). Theorem 10.100 verifies all three allocation ratios.
Corollary 10.101 produces independent integral vectors and new bases in
the original field, and Theorem 10.102 retains the actual
residue-index-dependent profile with factor \(1/700\). Corollary 10.103
gives a smaller admissible rank, contradicting minimality. Arrows
indicate proved implications, with their exact locators; no sampled
parameter case is used.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/geometric_stop_coupling.py}{Figure program}.
Human-source context:
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
free paper}, Section 6. Solution 54 checks the integer endpoint and the
weighted composition.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{46. The multiplicity stop and the retained integer
zeros}\label{the-multiplicity-stop-and-the-retained-integer-zeros}

Retain the original contradiction and the actual stage families of
Theorem 10.140. The remaining alternative to the geometric stop is
\(I_1<I^*\). Write \(a=r+1,c=c_5,H=\eta^a,Q=qH,D=A+\nu\ln q\). In this
section \(S_1=S_{I_1},T_1=T_{I_1}\): the subscript denotes the stopping
stage, which need not be the first stage. We first determine the exact
last integer derivative level and the subsequent depth. We also prove
what a family with the required final zeros would imply. The terminal
integer zeros themselves are constructed in Theorem 10.144 below.

\textbf{Lemma 10.142 (exact parameters of the fixed-order branch).}
There is a unique integer \(r_1\) with \(0\le r_1\le r\) and

\[
1\le\frac{c\eta^{r_1}T_1}{a}<\frac1\eta.
\tag{10.394}
\]

Set \(T_\flat=\eta^{r_1+1}T_1\), \(R=q\lfloor q^{r_1}S_1\rfloor\), and

\[
I_2=\left\lfloor\frac{3D-I_1\ln Q}{\ln q}\right\rfloor+1,
\qquad I_3=I_1+I_2.
\tag{10.395}
\]

Then \(I_1\ge1,I_2\ge1\), \(\eta\le cT_\flat/a<1\), and
\(Q^{I_1}q^{I_2}>\exp(3D)\), including a possible integer endpoint in
the quotient defining \(I_2\).

\textbf{Proof.} Lemma10.70 gives \(T\ge g_4\). In (10.284), \(F_r\ge9\)
and \((7/2)^{r-2}\ge1\), so Lemma10.104 gives \(g_4>(9000/29)a>6a\).
Thus \(cHT/a>1\), using \(c>1/2,H>1/3\). Thus stage1 is allowed and
\(I_1\ge1\). The definition and maximality of \(I_1\) give
\(cT_1/a\ge1\) and \(cHT_1/a<1\). The positive sequence \(c\eta^jT_1/a\)
decreases strictly to zero because \(0<\eta<1\). Its last term at least
one has an integer index \(r_1\). The next term is less than one,
proving the stated two-sided inequality. Since the term at \(j=a\) is
less than one, \(r_1<a\), hence \(r_1\le r\). Strict decrease proves
uniqueness. Multiplying by \(\eta\) gives the assertion for \(T_\flat\).

From \(I_1<\lfloor3D/\ln Q\rfloor+1\) and integrality,
\(I_1\le\lfloor3D/\ln Q\rfloor\le3D/\ln Q\). Thus the numerator defining
\(I_2\) is nonnegative and \(I_2\ge1\). The floor plus one is strictly
larger than its real argument, even when that argument is an integer.
Multiplication by \(\ln q>0\) and exponentiation prove
\(Q^{I_1}q^{I_2}>\exp(3D)\). \(\square\)

\textbf{Proposition 10.143 (the final fixed-order zeros force an
additive polynomial contradiction).} Suppose an admissible stage family
at \(I_3\) retains a nonzero coefficient subvector of the original
kernel, the selected derivations and integral normalized torus exponents
of Proposition10.31, with widths bounded by \(q^{-I_3}D_j\), and has its
prepared zeros for every integer \(|s|\le R\) through
\(\lfloor T_\flat\rfloor\). Then this family cannot exist. In fact,
every torus coordinate degree of its ordinary polynomial is less than
\(145/14336<1\), and its number of distinct integer roots exceeds
\((171/58)kL\).

\textbf{Proof.} The first inequality in Lemma10.142 gives

\[
q^{I_3}H^{I_1}>\exp(3D).
\tag{10.396}
\]

For the following degree estimate use \(T_1<a/(cH)\), the ratio of the
two original parameters in (10.186), and \(c_2qP/(e\theta)\ge7/2\):

\[
\begin{aligned}
q^{\nu-I_3}D_j
&<q^\nu H^{I_1}D_j\exp(-3D)\\
&=q^\nu T_1\frac{D_j}{T}\exp(-3D)\\
&<\frac{q^\nu e\theta t}{cc_2qrPHd\sigma_j}\exp(-3D)\\
&\le\frac{2}{7crH}\frac{t}{d\sigma_j}\exp(-3A).
\end{aligned}
\tag{10.397}
\]

In the last line \(q^\nu\exp(-3D)=\exp(-3A-2\nu\ln q)\le\exp(-3A)\),
with \(\nu\ge0\). Each original base is non-torsion, so the one-base
case of Theorem10.18 gives \(d\sigma_j>1/(87d^2)\), as proved in
Theorem10.100. Also \(c>1/2,r\ge2,H>1/q\ge1/3,t\le A\). Therefore

\[
q^{\nu-I_3}D_j<\frac{522}{7}d^2 A\exp(-3A).
\tag{10.398}
\]

Now \(A\ge g_1=4+\ln(ad)>5\). Hence \(d^2\exp(-2A)\le\exp(-8)/a^2\). The
function \(y\exp(-y)\) decreases for \(y\ge5\), because its derivative
is \((1-y)\exp(-y)<0\). Using \(a\ge3\) and \(\exp(1)>2\), proved in
Theorem10.100, gives

\[
2q^{\nu-I_3}D_j
<\frac{5220}{7\cdot9\cdot2^{13}}
=\frac{145}{14336}<1.
\tag{10.399}
\]

The normalized torus exponent differences in Proposition10.31 are
integers of absolute value at most \(q^{\nu-I_3}D_j<145/28672<1\). Every
such integer is zero. Thus the nonzero ordinary polynomial constructed
there depends only on the additive coordinate \(X\) and has degree at
most \(D_{\rm a}=kL\). Equivalently, an ordinary polynomial whose every
torus degree is less than one is independent of those variables. The
remaining coefficient polynomials cannot cancel: the distinct shifted
powers in Lemma10.29 are independent, as used in Proposition10.31.

It remains to count its roots. We have \(S_1\ge S>58\), and \(r_1\ge0\),
so

\[
2R+1>2q(S_1-1)+1>\frac{57}{29}qS_1.
\tag{10.400}
\]

Put \(\lambda=1+g_5^{-1}<5/4\). The parameter bounds proved in
Lemma10.70 and Theorem10.100 give
\(D_{\rm a}\le\lambda S\mathcal D/(c_1c_4dD)\). Use \(S_1/S=T/T_1\), the
formula for \(T\), and \(T_1<a/(cH)\):

\[
\begin{aligned}
\frac{2R+1}{D_{\rm a}}
&>\frac{57}{29}\frac{q c_1c_4dD}{\lambda\mathcal D}\frac{T}{T_1}\\
&=\frac{57}{29}\frac{q^2a c_4dD}{\lambda e\theta tT_1}\\
&>\frac{57}{29}\frac{q^2 c c_4H}{\lambda\theta}\frac d e\frac D t\\
&>\frac{57}{29}\frac{2(1/2)(15/4)}{(5/4)\,2}
=\frac{171}{58}>1.
\end{aligned}
\tag{10.401}
\]

Here \(qH>1,c_4>15/4,d\ge e,D\ge t\), and \(\theta\le2\) are the
original proved bounds. Prepared vanishing includes total order zero,
since \(T_\flat>0\); Proposition10.31 therefore gives ordinary zeros at
the \(2R+1\) distinct integers. A nonzero polynomial of degree at most
\(D_{\rm a}\) has at most that many roots: divide successively by
\(X-s\) at distinct roots, or use the root-count proof already in
Theorem10.98. The strict count above is impossible. \(\square\)

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={The final integral support collapses to a nonzero additive polynomial}]{../figures/fixed-order-final-polynomial.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.34. A two-coordinate projection of the strict
integral-support bound in Proposition 10.143, with its exact
enlargement. Every coordinate satisfies \(|m_j|<145/28672\); the dashed
boundary is excluded. Only the origin can be integral. The right-hand
arrows give the resulting nonzero additive polynomial of degree at most
\(kL\), and the excessive number of distinct roots. The hypothesis is
the admissible stage-\(I_3\) family with the stated integer zeros.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/fixed_order_final_polynomial.py}{Figure
program}. Human-source context: the final stopping alternative in
Section 7 of
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
freely accessible paper}. The coordinate bound, additive nonzeroness and
root count are proved above.}
\end{flushleft}
\end{minipage}\par\medskip


\subsubsection{Completing the integer zeros at the multiplicity
stop}\label{completing-the-integer-zeros-at-the-multiplicity-stop}

The previous integer theorem assumed that the next full contraction
still had enough derivatives. At the multiplicity stop that assumption
can fail. We use the derivatives that remain at the last integer level,
and reduce the number of Hermite conditions on a short outer interval.
The output radius and derivative order remain the original ones.

\textbf{Theorem 10.144 (the original terminal integer zero block).}
Assume the original contradiction and initial kernel of Theorem 10.140.
Suppose \(I_1<I^*\), and use \(S_1,T_1,r_1,T_\flat,R\) of Lemma 10.142.
The actual stage-\(I_1\) family produced by Theorem 10.140 has all its
prepared zeros at

\[
|s|\le R=q\lfloor q^{r_1}S_1\rfloor,
\qquad |\boldsymbol t|\le\lfloor T_\flat\rfloor.
\tag{10.402}
\]

This holds in every original prime, field and rank case. The coefficient
minimum, coefficient subvector and original phase condition are
retained.

\textbf{Proof.} Theorem 10.140 supplies the actual stage family and its
zeros at \(|s|\le\lfloor qS_1\rfloor\) through order
\(\lfloor\eta T_1\rfloor\). When \(r_1=0\), these include the stated
zeros, since \(q\lfloor S_1\rfloor\le\lfloor qS_1\rfloor\). Assume
henceforth \(r_1\ge1\). Write \(a=r+1,c=c_5,H=\eta^a\), and retain
\(b_r,S_-,Y_-,\bar\lambda,\mathcal R_r(j)\) from Section 40. The
stopping inequalities are \(c\eta^{r_1}T_1/a\ge1\) and \(cHT_1/a<1\).
Also \(I_1\le3(A+\nu\ln q)/\ln(qH)\), by Lemma 10.142.

First consider an integer level \(1\le j\le\min\{r_1,r-1\}\). Suppose
the zeros at radius \(\lfloor q^jS_1\rfloor\) through order
\(\lfloor\eta^jT_1\rfloor\) have been obtained. For a target order at
most \(\lfloor\eta^{j+1}T_1\rfloor\), choose
\(\mu=\lfloor c\eta^jT_1/a\rfloor+1\) at every input node. The floor
inequality used in Theorem 10.126 supplies all these ordinary jets by
Theorem 10.118. Put \(N_*=(2\lfloor q^jS_1\rfloor+1)\mu\). The weaker
stopping condition suffices for the following upper bound:

\[
\mu\le\frac{c(\eta^j+\eta^{r_1})T_1}{a}
\le\frac{2cT_1}{a}.
\tag{10.403}
\]

Indeed the added one is at most \(c\eta^{r_1}T_1/a\). The coefficient
and separation estimates in the proof of Lemma 10.123 use only
\(S_1=S\eta^{-aI_1}\), \(T_1=T\eta^{aI_1}\), and the original height
bounds. They still give \(\beta=v_p(b_n)\),
\(\max\{B,\beta-\vartheta,0\}\le\eta^{-aI_1}H_1/\ln p\), and the same
required projection depth \(U-\beta>1\). Using \(S_1T_1=ST\) and
\(c_1W_*ST\theta=qa\), the full input sum, multiplied by \(c_1W_*/q^a\),
is bounded above by

\[
\frac{c_1W_*}{q^a}
\bigl[N_*\theta+\beta+(\mu-1)\max\{B,\beta-\vartheta,0\}\bigr]
\le\frac{4c}{q}+\frac{2c}{q^rS_-}
 +\frac{c_1}{7ab_rq^a}+\frac{c}{c_3Y_-aq^a}.
\tag{10.404}
\]

Here the first term uses \(j\le r-1\), and both powers of \(\eta\) are
at most one. At odd primes insert
\(c\le14/25,q=2,S_->58,c_1<3,b_r\ge7200/29,c_3Y_->19/10\). The resulting
rational bound is less than \(23/20<c_1\). In V use
\(c\le827/1000,q=3,c_3Y_->19/20\); the bound is less than \(9/8<c_1\).
These exact substitutions are included in the certificate below.
Dropping the nonnegative term \((D_{\rm a}-u)\theta\) from the left side
of the input criterion therefore proves its strict inequality for every
additive order \(u\le D_{\rm a}\). Larger additive orders vanish
identically.

Theorem 10.119 and the scalar identity (10.327) give analytic precision
at least \(N_*\theta\) in the original quantity
\(v_p(\Phi)+G_u-\delta\). Its strict lower bound (10.330) follows from
\(\mu>c\eta^jT_1/a\); it does not use the stronger next-contraction
condition. All integer target values lie in the original field \(K\) by
the retained phase identity (10.151). Lemma 10.122 applies at the same
depth \(I_1\), with \(j\ge1\), and charges the full q-denominator. Thus
their original arithmetic bound is \(\mathcal R_r(j)\). Lemma 10.125
gives \(c_1W_*N_*\theta>\mathcal R_r(j)+1/200\). Corollary 10.120 forces
each value to zero, with the actual degree and local degrees of \(K\).
Successive applications complete all required levels unless \(r_1=r\).
In that last case they give the full radius \(\lfloor q^rS_1\rfloor\)
through order \(\lfloor\eta^rT_1\rfloor\).

It remains to extend this last level when \(r_1=r\). Put
\(z=c\eta^rT_1/a\), so \(1\le z<1/\eta<2\). The last inequality follows
from \(\eta\ge1-827/3000>1/2\). The same floor inequality gives
\(\lfloor\eta^rT_1\rfloor-\lfloor HT_1\rfloor\ge\lfloor z\rfloor=1\).
Consequently every input node provides a value and its first ordinary
jet for every target selected order at most \(\lfloor HT_1\rfloor\). We
retain the extra jet only in an inner interval. Set

\[
R_b=\lfloor q^rS_1\rfloor,\qquad R_0=\lfloor24R_b/25\rfloor,
\qquad
\mu(s)=\begin{cases}2,&|s|\le R_0,\\1,&R_0<|s|\le R_b.\end{cases}
\tag{10.405}
\]

Since \(R_b>q^rS_1-1>231\), these are two strictly nested nonempty
intervals. Their Hermite mass is exactly

\[
N_*=2R_b+2R_0+2,
\qquad \frac{98}{25}R_b<N_*\le\frac{98}{25}R_b+2.
\tag{10.406}
\]

The inequalities follow respectively from \(R_0>24R_b/25-1\) and
\(R_0\le24R_b/25\). Lemma 10.94 and Theorem 10.119 apply with
\(M=2,E=0\). In particular the outer nodes lose one separation factor in
their cardinal polynomial; that loss is already included in \((M-1)B\).
There is no deletion of nodes and no q-deleted cardinal loss.

Because \(1\le z\), the upper bound on \(N_*\) and the same coefficient
and separation bounds give the following upper bound on the entire input
sum, after multiplication by \(c_1W_*/q^a\):

\[
P_r=\frac{98}{25}c\eta^r
 +\frac{2c\eta^r}{q^rS_-}
 +\frac{c_1}{7ab_rq^a}
 +\frac{c\eta^r}{c_3Y_-aq^a}.
\tag{10.407}
\]

For the last term use \(M-1=1\le z\); this pays the outer cardinal loss
and all retained jet costs, rather than dropping them. On the other side
\(T_1<a/(cH)\) and the strict lower mass inequality yield

\[
\frac{c_1W_*N_*\theta}{q^a}>
G_r:=\frac{98}{25}cH\left(1-\frac1{q^rS_-}\right).
\tag{10.408}
\]

The upper arithmetic bound for a nonzero integer target at
\(|s|\le\lfloor q^{r+1}S_1\rfloor\) and order \(\lfloor HT_1\rfloor\) is
the corrected \(\mathcal R_r(r)\) of (10.332). Its full normalized cost
is

\[
C_r=\frac1{q^a}\left[
A_r+c_1\left(\frac1{300}+\frac1{273\cdot999b_r}\right)
 +\frac{\bar g_9H}{c_3Y_-}
 +\frac{\bar\lambda}{c_4}
 \left(1+\Omega_++\frac{\ell_q}{5}
 \left(r+\frac1{q-1}\right)\right)\right]+\frac1{c_2}.
\tag{10.409}
\]

Here \(\ell_q=347/500\) at \(q=2\) and \(1099/1000\) at \(q=3\) are the
proved logarithm upper bounds. In particular the final \(1/c_2\), the
full depth-denominator indicator, the Euler/lcm term, and the separate
scalar remainder all survive.

We now prove the needed strict inequalities for every rank, using the
coupled field rows of Lemma 10.121. For ranks 2--7 substitute the exact
decimal rationals of that table and the rank band for \(c\) into these
three formulas. For ranks at least eight, \(c\) is constant within each
case. Put \(\eta_8=1-c/9,H_8=\eta_8^9,S_8=S_-(8),b_8=b_r(8)\). The
following bounds hold simultaneously for all such ranks:

\[
\begin{aligned}
P_r&\le\frac{98}{25}c\eta_8^8
 +\frac{2c}{q^8S_8}
 +\frac{c_1}{7\cdot9b_8q^9}
 +\frac{c}{9c_3Y_-q^9},\\
G_r&\ge\frac{98}{25}cH_8
 \left(1-\frac1{q^8S_8}\right),\\
C_r&\le\frac{O_8}{q^9}
 +\frac{\bar\lambda\ell_q(8+1/(q-1))}{5c_4q^9}
 +\frac1{c_2}.
\end{aligned}
\tag{10.410}
\]

In the last line \(O_8\) is the bracket before the linear-in-r term in
the formula for \(C_r\), evaluated at rank eight with \(C_r^\sharp\) and
\(H\) enlarged to one and \(\bar g_9=107/103\). This is exactly the
\texttt{high=True} output of the preceding integer certificate. To
verify the enclosing bounds, \(b_r,S_-\) increase with rank, while
\(\epsilon_r,\bar g_{12}\) decrease; \(C_r^\sharp<1\) by Lemma 10.121
and \(\bar g_9=107/103\) for every rank at least eight. The positive
constant terms divided by \(q^{r+1}\) decrease. So does \(r/q^{r+1}\),
since its successive ratio is \((r+1)/(qr)<1\).

For the remaining powers, \((1-c/y)^y\) increases for \(y\ge3\), as
proved in Lemma 10.122. The function \((1-c/y)^{y-1}\) decreases: with
\(w=c/y\), its logarithmic derivative is

\[
\ln(1-w)+\frac{w}{1-w}-\frac{w}{y(1-w)}
\le\frac{w^2}{2(1-w)^2}-\frac{w}{y(1-w)}<0.
\tag{10.411}
\]

The non-strict bound integrates \(t/(1-t)^2\) from zero to \(w\). The
strict sign uses \(c<2(1-w)\), which follows from
\(c\le827/1000<2(1-c/3)\). Thus \(H\ge H_8\) and \(\eta^r\le\eta_8^8\);
replacing the remaining numerator powers by one only enlarges the two
input remainders. This proves all the uniform bounds, without inferring
them from sampled ranks.

The exact rational certificate gives the following strict lower bounds,
each divided by 1000. Each row includes its six finite ranks and its one
uniform bound for every rank at least eight.

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}lrr@{}}
\toprule\noalign{}
Original case & \(c_1-P_r\) & \(G_r-C_r\) \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
I.1 & 21/1000 & 493/1000 \\
I.2 & 21/1000 & 489/1000 \\
II & 20/1000 & 492/1000 \\
III.1 & 20/1000 & 487/1000 \\
III.2 & 19/1000 & 481/1000 \\
IV, first prime branch & 19/1000 & 490/1000 \\
IV, larger prime branch & 26/1000 & 540/1000 \\
V & 872/1000 & 475/1000 \\
\end{longtable}
}

All entries are exact rational substitutions with downward-rounded lower
bounds; \href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/terminal_integer_endpoints.py}{the
editable certificate} retains their exact numerators and denominators.
In particular \(P_r<c_1\) and \(G_r>C_r\). The first inequality proves
the complete input criterion with \(W_*U=q^a\). The second gives
strictly more analytic precision than the full arithmetic bound in the
original field \(K\), so Corollary 10.120 forces every target value to
zero.

We have therefore obtained the radius \(\lfloor q^{r_1+1}S_1\rfloor\)
and derivative order \(\lfloor\eta^{r_1+1}T_1\rfloor\) in every case,
including \(r_1=r\). Finally \(q\lfloor q^{r_1}S_1\rfloor\le
\lfloor q^{r_1+1}S_1\rfloor\), so these include exactly the required
original terminal integer zeros. \(\square\)

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={Retaining a derivative on an inner interval pays the complete terminal integer comparison}]{../figures/terminal-integer-profile.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.35. Left: an illustrative nested interval with
\(R_b=25\), \(R_0=24\). Its 49 inner nodes carry multiplicity two and
its two outer nodes carry multiplicity one, so \(N_*=100\) instead of
102. This small example illustrates the exact cardinal construction;
actual terminal stages satisfy \(R_b>231\). For every integer radius the
retained mass lies strictly above \((98/25)R_b\) and at most two above
that bound. Right: the proved lower margins in the table, for all
original rank and field cases. The input margins include the coefficient
minimum and the outer cardinal loss; the arithmetic margins include the
full global degree, denominator, Euler/lcm and scalar costs.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/terminal_integer_profile.py}{Figure program}.
Human-source context: the terminal integer range in Lemma 5.5 of
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
freely accessible paper}. The complete nested interpolation and both
strict comparisons are proved above.}
\end{flushleft}
\end{minipage}\par\medskip


\subsubsection{Full-node interpolation at the fixed derivative
order}\label{full-node-interpolation-at-the-fixed-derivative-order}

\textbf{Proposition 10.145 (the full fixed-order analytic input).} Use
the original multiplicity-stop parameters of Lemma10.142, and the
original explicit depth choices in Section26. Put
\(O=\lfloor T_\flat\rfloor\), \(R=q\lfloor q^{r_1}S_1\rfloor\),
\(N=2R+1\). Suppose an admissible stage family at any depth \(I\ge I_1\)
retains the initial coefficient bound, minimum coefficient and phase,
and has all its prepared zeros at the integers \(|s|\le R\) through
selected order \(O\). Under the original contradiction, for every
selected order \(\boldsymbol t\) with \(|\boldsymbol t|\le O\) and
additive order \(u\le D_{\rm a}\),

\[
v_p(\Phi^{(I)}(x;\boldsymbol t))+G_u-\delta\ge N\theta
\qquad(x\in\mathbb Z_p\subset\mathbb Q_p).
\tag{10.412}
\]

In particular this holds at each fractional argument \(x=s/q\), with
\(s\in\mathbb Z\), since \(q\ne p\). No additional ordinary jet beyond
the given selected order is required. Larger additive orders vanish by
polynomial degree. This is a local analytic precision assertion.

\textbf{Proof.} Normality and projection depth are those of Lemma10.91
and Theorems10.117--10.119. Multiplication of the torus slopes and
additive argument by \(q^{-I}\) does not change their p-adic depths,
because \(q\ne p\). The retained coefficient bound and minimum provide
the same \(\beta=v_p(b_n)\), \(U\), and \(\delta\). Thus these results
apply at the given depth without its being inside the earlier geometric
range.

Apply the full-node case of Theorem10.119 with \(M=1,E=0\). Every node
has multiplicity one, so the available prepared zero of the target order
itself is sufficient. The complete integer-interval cardinal polynomials
are integral on \(\mathbb Z_p\), by Lemma10.94. Their inverse Taylor
polynomials here have degree zero. Consequently there is neither a
q-deleted loss nor a higher-jet separation loss. By (10.319),
\(K_u(1,B)=M_u(0)=\gamma_u\), and (10.327) at \(M=1\) gives
\(G_u+\gamma_u\ge(D_{\rm a}-u)\theta\). The full precision therefore is

\[
\mathcal P_u\ge
\min\{N\theta,\ U-\beta+(D_{\rm a}-u)\theta\}.
\tag{10.413}
\]

It remains to prove \(U-\beta>N\theta\) for every original case and
rank. We retain the actual large initial multiplicity instead of using
only the coarser lower bound on \(\mathcal D\). From the formula for
\(S\), \(\beta\ln p\le h\le H_1\), and \(ef\le d\),

\[
c_1W_*\beta
\le\frac{f}{c_3d\theta T}
\le\frac1{c_3YT}
\le\frac1{c_3Y_-g_4},\qquad Y=e\theta.
\tag{10.414}
\]

Here \(T\ge g_4\) is Lemma10.70, and the simultaneous field bounds of
Lemma10.121 give \(Y\ge Y_-\). The endpoint condition
\(c\eta^{r_1}T_1/a\ge1\), together with \(c_1W_*S_1T_1\theta=qa\),
bounds the node input by

\[
\frac{c_1W_*N\theta}{q^a}
\le2c\eta^{r_1}q^{r_1+1-r}
 +\frac{c\eta^{r_1}}{q^rS_1}.
\tag{10.415}
\]

Since \(q\eta>1\) and \(r_1\le r\), the first term is at most
\(2cq\eta^r\). The other stopping inequality gives \(T_1<a/(cH)\). Using
\(S_1/S=T/T_1\), \(T\ge g_4\) and \(H>1/3\) therefore gives
\(S_1>Sg_4cH/a>S_-g_4c/(3a)\). The second term is less than
\(3a/(q^rS_-g_4)\). Thus a sufficient exact input bound is

\[
B_r:=2cq\eta^r
 +\frac{3a}{q^rS_-g_4}
 +\frac1{c_3Y_-g_4q^a}<c_1.
\tag{10.416}
\]

This bound charges the coefficient minimum in full. It follows for all
ranks from the following explicit lower bound on the actual \(g_4\).
Write \(F_r=r^ra^r/(r!)^2\). The formulas (10.185) give

\[
g_4\ge L_r:={2c_0c_4qa\over\rho\widehat\vartheta}
a_*^rF_r v_0,
\tag{10.417}
\]

where the simultaneous original choices are

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}lrrrr@{}}
\toprule\noalign{}
Case & \(\rho\) & \(\widehat\vartheta\) & \(a_*\) & \(v_0\) \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
I.1 & 58 & 3/2 & 7 & 1 \\
I.2 & 17 & 3/2 & 7 & 5 \\
II & 58 & 5/4 & 7 & 1 \\
III.1 & 58 & 1 & 7/2 & 5 \\
III.2 & 17 & 1 & 7/2 & 5 \\
both IV branches & 58 & 7/6 & 7 & 1 \\
V & 58 & 2 & 26/3 & 1 \\
\end{longtable}
}

For I.2 and III the definition of \(g_4\) retains its \(g_1\) factor,
which exceeds five; using \(v_0=5\) is a lower bound. In the other rows
that factor cancels or can be enlarged, so \(v_0=1\) suffices. The
values of \(\rho\) follow from the same degree cases as Lemma10.121; the
depth and \(a_*\) choices are exactly (10.184) and its ensuing explicit
upper depths. Thus this uses the actual field row, not independently
combined extrema.

For ranks 2--7, substitute \(L_r\) for \(g_4\), the original rational
constant row and its exact \(c\) band into \(B_r\). For ranks at least
eight, \(c\) is constant and \(\eta^r\) decreases by the derivative
proved in Theorem10.144. The ratio \(F_{r+1}/F_r>4\) is proved in
Lemma10.104; consequently \(L_{r+1}/L_r>14(r+2)/(r+1)\). Both \(a/L_r\)
and \(1/L_r\) decrease. Since \(S_-\) and \(q^r\) increase, each of the
two positive remainders also decreases. Therefore the rank-eight
substitution bounds every larger rank.

The exact rational certificate
\texttt{fixed\_order\_input\_endpoints.py} gives these strict lower
bounds for \(c_1-B_r\), including each case's six finite rows and one
uniform row:

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}lr@{}}
\toprule\noalign{}
Case & Strict input margin \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
I.1 & 77/1000000 \\
I.2 & 39/1000000 \\
II & 80/1000000 \\
III.1 & 89/1000000 \\
III.2 & 23/1000000 \\
IV, first prime branch & 56/1000000 \\
IV, larger prime branch & 67/1000000 \\
V & 54/1000000 \\
\end{longtable}
}

Thus \(c_1W_*[N\theta+\beta]<c_1q^a=c_1W_*U\), proving the required
strict input. Since \((D_{\rm a}-u)\theta\ge0\), the displayed minimum
is at least \(N\theta\). This proves the proposition. \(\square\)

The argument includes the source's zero-additional-jet convention: one
value condition at each node corresponds to \(M=1\) in our Hermite
notation. This distinction prevents accidentally requesting an extra
derivative or discarding the coefficient cost.

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={Full-node cardinal integrality and an exact normal root-polynomial value}]{../figures/fixed-order-value-nodes.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.36. An exact illustration at \(p=5,q=2,R=2,\theta=1\).
The five integer nodes each carry one value condition, with no
additional ordinary jet. At \(x=1/2\in\mathbb Z_5\) their cardinal
values are \(3/128,-5/32,45/64,15/32,-5/128\), with nonnegative
valuations \(0,1,1,1,1\). Right: the monic normal root polynomial
\(W(Z)=\prod_{s=-2}^2(Z-5s)\) has \(W(5/2)=140625/32\), of valuation
six, exceeding its normal bound \(N\theta=5\). This example illustrates
the full-node interpolation mechanism; it is not a complete choice of
logarithmic contradiction parameters. Proposition 10.145 proves the full
original input bound, and Solution 57 checks the example.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/fixed_order_value_nodes.py}{Figure program} and
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/fixed_order_input_endpoints.py}{all-rank input
certificate}. Human-source context: the zero-additional-jet argument in
Section 7 of
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
freely accessible paper}. The normality, cardinal construction and
simultaneous input bounds are proved above.}
\end{flushleft}
\end{minipage}\par\medskip


\subsubsection{Closing successor integers without lowering the
derivative
order}\label{closing-successor-integers-without-lowering-the-derivative-order}

\textbf{Theorem 10.146 (the full fixed-order successor integer block).}
Retain the original multiplicity-stop parameters of Lemma10.142,
including \(I_1<I^*,R=q\lfloor q^{r_1}S_1\rfloor\) and
\(O=\lfloor T_\flat\rfloor\). Suppose \(I_1+1\le J\le I_3\), and an
admissible stage-J coefficient family retains the initial coefficient
bound and minimum, the phase condition, and all prepared zeros through
total order O at the integers \(|s|\le R\) prime to q. Then that same
family has every prepared zero through order O at all integers
\(|s|\le R\).

The proof uses one value condition at each q-deleted node, without any
additional ordinary jet, and keeps the full q-denominator at depth J.

\textbf{Proof.} Put \(a=r+1,c=c_5,H=\eta^a,Q=qH,D=A+\nu\ln q\), and use
\(C,Y_-,\Omega_+,S_-,b_r,A_r,\bar g_9\) of Section40. Let \(L_r\) be the
explicit lower bound on the initial \(g_4\) in (10.417). Thus
\(T\ge g_4\ge L_r\), while \(cT_1/a\ge1>cHT_1/a\). Let \(m_r\) be the
largest nonnegative integer satisfying

\[
\frac{cL_rH^{m_r}}a\ge1.
\tag{10.418}
\]

Then \(I_1\ge m_r\), since otherwise strict decrease would contradict
\(cTH^{I_1+1}/a<1\). In particular \(Q^{I_1}>q\). Here is a uniform
proof of this last fact. Within each constant band H increases with
rank, as proved in Lemma10.122. At the three band starts, exact
substitution gives \(H>11/20\) for odd p and \(H>5/12\) in V.
Lemma10.142 gives \(T>(9000/29)a\). At odd p, \(c>1/2\), and
\((11/20)^8(4500/29)>1\); hence \(I_1\ge8\), so
\(Q^{I_1}>(11/10)^8>2=q\). In V, \(c>3/4\), and \((5/12)^6(6750/29)>1\);
hence \(I_1\ge6\), so \(Q^{I_1}>(5/4)^6>3=q\). The sharper \(m_r\) below
will retain the actual initial multiplicity rather than only these
coarse depths.

There are exactly \(N_*=2(q-1)\lfloor q^{r_1}S_1\rfloor\) q-deleted
nodes. For each target prepared index of order at most O and additive
order \(u\le D_{\rm a}\), Theorem10.119 with \(M=1,E=1\) gives

\[
\mathcal P_u\ge
\min\{N_*\theta,\ U-\beta+(D_{\rm a}-u)\theta-B\},
\qquad B=\lfloor\log_p(2R+1)\rfloor.
\tag{10.419}
\]

Indeed \(K_u(1,B)=\gamma_u\) and \(G_u+\gamma_u\ge(D_{\rm a}-u)\theta\),
by (10.327). The single deleted-cardinal loss is B; there is no
higher-jet inverse loss. Normality and projection apply at J because q
is a p-adic unit, just as in Proposition10.145. Beyond additive degree
\(D_{\rm a}\), the value is identically zero.

We compare both entries with the arithmetic bound for that same normal
quantity in the original field K. The phase identity places every
integer prepared value in K. Set

\[
\begin{aligned}
E_r&=c_1\left(\frac1{300}+\frac1{273\cdot999b_r}\right),\\
K_r&=A_r+E_r+\frac{\bar g_9a}{cc_3Y_-L_r}
 +\frac{\bar\lambda}{c_4}
 \left[1+\Omega_++\frac{3\ell_q}{\mathcal L(Q)}
  +\frac{(1+1/(q-1))\ell_q}{5}\right],\\
F_r(x)&=K_r+\frac{\bar\lambda x\ell_q}{5c_4}
       +\frac{q^x}{c_2Q^{m_r}}.
\end{aligned}
\tag{10.420}
\]

Here \(\bar\lambda=189/188\), \(\ell_2=347/500,\ell_3=1099/1000\), and
\(\mathcal L(Q)>0\) is the five-term positive-series lower bound

\[
\mathcal L(Q)=2\sum_{j=0}^4\frac{z^{2j+1}}{2j+1},
\qquad z=\frac{Q-1}{Q+1};\qquad \ln Q>\mathcal L(Q).
\tag{10.421}
\]

The logarithm series and its positivity are proved in Solution34. For a
nonzero target V, Corollary10.116 and its same-order normal bound
therefore give

\[
c_1W_*[v_p(V)+G_u-\delta]<F_r(r_1).
\tag{10.422}
\]

To verify every term, the coefficient bound is \(A_r\), as in (10.332).
The complete Euler remainder is at most \(E_r\), by (10.310) and
\((r-1)/Z<1/(273b_r)\). Since \(O<T_\flat<a/c\) and \(T\ge L_r\), the
main Euler term is at most \(\bar g_9a/(cc_3Y_-L_r)\). The additive
factorial estimate in Lemma10.122 applies directly: its physical radius
satisfies

\[
q^{-J}R<q^{r_1+1}S Q^{-I_1}q^{-(J-I_1)}
\le q^{r_1}S Q^{-I_1}<q^{r_1-1}S.
\tag{10.423}
\]

Enlarging this radius to \(q^{r_1+1}S\) in that lemma's factorial proof
bounds the scalar by \(\bar\lambda[1+r_1\ell_q/5]/c_4\); this step uses
no geometric-depth absorption. The normal term \(G_u\) is at most
\(D_{\rm a}(\theta+1/(p-1))\), giving the separate
\(\bar\lambda\Omega_+/c_4\). The complete denominator is
\(\Xi_{J,0,u}\le D_{\rm a}[J+1/(q-1)]\). Lemma10.142 gives
\(I_1\le3D/\ln Q\) and

\[
I_3\ln q\le3D+I_1(\ln q-\ln Q)+\ln q
\le\frac{3D\ln q}{\ln Q}+\ln q.
\tag{10.424}
\]

Together with \(D>5,J\le I_3\), this gives precisely the remaining depth
and constant-denominator terms in \(K_r\). Finally the full torus cost
is at most \(q^{r_1}/(c_2Q^{I_1})\), hence at most
\(q^{r_1}/(c_2Q^{m_r})\). Nothing has been dropped at primes above q, or
from the normal scalar or coefficient minimum.

For the first analytic entry, \(T_1<a/(c\eta^{r_1+1})\),
\(c_1W_*ST\theta=qa\), and \(ST=S_1T_1\) give

\[
c_1W_*N_*\theta>
2c(q-1)(q\eta)^{r_1+1}
-\frac{2q(q-1)a}{S_-L_r}.
\tag{10.425}
\]

The floor loss follows from \(c_1W_*\theta=qa/(ST)\),
\(S\ge S_-,T\ge L_r\). The difference of this lower bound and
\(F_r(r_1)\) increases for every real \(0\le r_1\le r\). In fact
\(Q^{m_r}>q\) and \(\eta^{x+1}\ge H\) give the derivative lower bound

\[
q^{x-1}\left[2c(q-1)q^2H\ln(q\eta)-\frac{\ell_q}{c_2}\right]
-\frac{\bar\lambda\ell_q}{5c_4}>0.
\tag{10.426}
\]

For odd p, use \(c\ge5267/10000,H>11/20\),
\(\ln(q\eta)>12/25,c_4>18,c_2=7/4,q=2\). The bound exceeds \(1/3\),
since \(q^{x-1}\ge1/q\). For V use
\(c>3/4,H>5/12,\ln(q\eta)>3/4,c_4>15/4,c_2=13/9,q=3\); the bound exceeds
two. Indeed \(\eta\ge61/75\) at odd p and \(\eta\ge2173/3000\) in V; the
first two terms of the positive logarithm series give
\(\ln(122/75)>12/25\) and \(\ln(2173/1000)>3/4\). The exact endpoint
substitutions below also check \(Q^{m_r}>q\). Thus it suffices to
compare at \(r_1=0\).

For the second entry, \(R\le q^{r+1}S_1\). The separation argument of
Lemma10.123 gives \(\ln(2R+1)<H_1H^{-I_1}+\ln q\). Since
\(\ln q/H_1<1/3\), and \(T_1\ge a/c\),

\[
c_1W_*B<\frac{4c}{3c_3Y_-a},\qquad
c_1W_*\beta\le\frac1{c_3Y_-L_r}.
\tag{10.427}
\]

Consequently its normal input is greater than or equal to the
conservative bound \(c_1q^a-1/(c_3Y_-L_r)-4c/(3c_3Y_-a)\). For its
opposite arithmetic envelope, \(F_r(x)\) increases, and \(Q^{m_r}>q\);
hence

\[
F_r(r_1)<K_r+\frac{q^{r-1}}{c_2}
                  +\frac{\bar\lambda r\ell_q}{5c_4}.
\tag{10.428}
\]

For ranks2--7, choose the largest integer \(m_r\) for which
\(cL_rH^{m_r}/a\ge1\), and substitute the coupled original row in these
formulas. The exact rational certificate gives the following strict
margins; every entry is divided by1000.

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}lrr@{}}
\toprule\noalign{}
Case & First-entry margin at \(r_1=0\) & Second-entry margin \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
I.1 & 38 & 8808 \\
I.2 & 84 & 8769 \\
II & 123 & 8776 \\
III.1 & 87 & 8723 \\
III.2 & 55 & 8650 \\
IV, first prime branch & 105 & 8801 \\
IV, larger prime branch & 232 & 9035 \\
V & 1405 & 60699 \\
\end{longtable}
}

The same table also includes one analytic enclosure for every rank at
least eight. Here c is constant, H and Q increase, while
\(L_{r+1}/L_r>14(r+2)/(r+1)\), as proved in Proposition10.145. Therefore
\(L_r/a\) and \(S_-\) increase. The same \(m_8\) is available for every
larger rank, and \(Q^{I_1}\ge Q_8^{m_8}>q\). Bound \(A_r\) by its
rank-eight envelope with \(C_r^\sharp\) replaced by one. All its other
positive components decrease; \(\bar g_9=107/103\). Thus
\(K_r\le K_8^+\), using the rank-eight lower Q in \(\mathcal L\), and
its lower \(L_r/a\). The mass at \(r_1=0\) increases, and its floor loss
decreases. The rank-eight mass comparison therefore encloses every
larger rank.

For the remaining input comparison, divide both bounds by \(q^{r+1}\).
The constant torus term becomes \(1/(q^2c_2)\); the remaining positive
constant costs decrease after division, and \(r/q^{r+1}\) decreases. The
losses involving \(1/L_r\) and \(1/a\) also decrease. Hence its
rank-eight divided comparison proves every larger rank, with at least
the unscaled positive margin displayed in the table. These monotonicity
arguments, not a finite rank experiment, certify the infinite range.

Both analytic entries strictly exceed the original-K arithmetic bound
for every target. Corollary10.120 forces each remaining integer value to
zero. This closes the entire original R-block at exactly O, for the same
coefficient family, including \(J=I_3\). \(\square\)

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={A q-deleted cardinal loss and the exact successor integer closure margins}]{../figures/fixed-order-successor-profile.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.37. Left: at \(p=2,q=3,R=3\), the four q-deleted integer
nodes carry one value condition each. Their cardinals at zero are
\(-1/6,2/3,2/3,-1/6\), of valuations \(-1,1,1,-1\). The uniform cardinal
loss is \(B=2\), while the actual loss here is one. The normal root
polynomial \(W(Z)=\prod_s(Z-2s)\) has \(W(0)=64\), of valuation six,
above \(N_*\theta=4\) at \(\theta=1\). These exact values illustrate the
interpolation, rather than a full original logarithmic-contradiction
parameter choice. Right: proved strict first-entry margins for every
original case, minimized over all ranks; every second-entry margin also
exceeds eight in the same normalization. Both entries retain the
original field K, coefficient minimum, full denominator and
deleted-cardinal loss. The actual q-deleted family is the hypothesis of
Theorem10.146; Theorem10.149 and the phase construction supply it in
Corollary10.150. Solution58 checks the example and endpoint.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/fixed_order_successor_profile.py}{Figure
program},
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/successor_fixed_order_endpoints.py}{exact
all-rank certificate}. Human-source context: Lemma7.2 of
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
freely accessible paper}. The full successor integer argument and both
strict comparisons are proved above.}
\end{flushleft}
\end{minipage}\par\medskip


\subsubsection{The exact field of a fractional
value}\label{the-exact-field-of-a-fractional-value}

The field generated by all fractional monomials can be larger than the
field generated by a particular prepared value. We can determine the
latter field from the components which survive in that value. This gives
an exact arithmetic factor without identifying distinct local primes.

\textbf{Theorem 10.147 (the surviving Kummer components).} Retain
Theorem10.56, and write \(h=h_\Lambda\). Choose its independent
principal monomials \(U_1,\ldots,U_h\), with \(U_j^q\in K\). For
\(a\in\{0,\ldots,q-1\}^h\), put \(U^a=\prod_jU_j^{a_j}\). Every
fractional prepared value has a unique expansion

\[
V=\sum_a A_aU^a,\qquad A_a\in K,\qquad
W_V=\langle a:A_a\ne0\rangle_{\mathbb F_q},\qquad t_V=\dim W_V.
\tag{10.429}
\]

If \(V\ne0\), then

\[
[K(V):K]=q^{t_V},\qquad K(V)_{\mathfrak P}=K_{\mathfrak p}.
\tag{10.430}
\]

In particular \(t_V=0\) precisely when \(V\in K\). A nonzero constant
component contributes the zero vector and does not increase \(t_V\). The
arithmetic upper bounds (10.156), (10.244) and (10.313) for this value
may use the exact factor \(q^{t_V}d/(ef\ln p)\), while retaining their
other terms and the same coefficient normalization. All those terms
remain present; the normal summand \(G_u\) in (10.313) is added after
the arithmetic bound, as before.

\textbf{Proof.} Theorem10.56 gives the basis \(U^a\) of
\(E=K(U_1,\ldots,U_h)\), of degree \(q^h\). Its exact monomial
identities collect every prepared term into the stated expansion; the
rational scalars and coefficients lie in K. Fix a primitive q-th root
\(\omega\in K\). For \(b\in\mathbb F_q^h\), define

\[
\sigma_b\!\left(\sum_a A_aU^a\right)
=\sum_a A_a\omega^{b\cdot a}U^a.
\tag{10.431}
\]

These are K-automorphisms. Indeed products of basis monomials are
reduced by \(U_j^q\in K\), and \(\omega^q=1\), so the formula respects
multiplication and addition. Its inverse is \(\sigma_{-b}\). Uniqueness
of the expansion shows that \(\sigma_b(V)=V\) if and only if
\(b\cdot a=0\) for every surviving component. Row reduction over
\(\mathbb F_q\) gives exactly \(q^{h-t_V}\) such b. Thus the orbit of V
has \(q^{t_V}\) distinct elements.

Every orbit element is a root of V's minimal polynomial over K, because
each \(\sigma_b\) fixes K. Conversely, the product of \(X-V'\) over the
distinct orbit elements is fixed by every \(\sigma_b\). Its coefficients
belong to K: in the basis expansion of any fixed element, a nonzero
exponent vector a is detected by some b with \(b\cdot a\ne0\), forcing
its coefficient to vanish. The orbit polynomial therefore belongs to
\(K[X]\). These two degree inequalities prove the exact degree without
an additional Galois-theory input. The principal roots lie in
\(K_{\mathfrak p}\), so \(K\subset K(V)\subset K_{\mathfrak p}\). Its
closure is exactly \(K_{\mathfrak p}\), as in Theorem10.56; hence the
selected local indices are the original e,f.

For the arithmetic assertion, let \(F=K(V)\), and divide V by the same
nonzero reference coefficient of minimum valuation \(\delta\). This
quotient belongs to F. The individual torus monomials may still belong
only to E, so we justify the height comparison over F explicitly. At
each place of F extend its absolute value to E. The triangle or
ultrametric bound for the displayed prepared sum holds in that
extension. The absolute value of every fractional monomial is determined
by its q-th power in K. Changing the extension therefore changes none of
its absolute values; its root-of-unity factors also have absolute value
one. Coefficient norms and rational scalar bounds likewise depend only
on the restricted place of K. Consequently the same local bound used in
Theorems10.41 and10.92 bounds \(V/c_{\min}\) at the place of F.

When these bounds are summed, the local weights over a place of K sum to
\([F:K]\) times its original weight, by the extension and norm formula
proved in the prerequisite lesson on places. The coefficient height,
scalar maximum, q-denominator term and signed torus height hence acquire
the factor \([F:K]=q^{t_V}\). The selected local term has weight \(ef\),
as just proved. The complete product-formula proof therefore gives
exactly the claimed arithmetic bound. Adding the same \(G_u\) proves its
normal version. This argument does not require the other primes to have
the selected valuation. \(\square\)

For \(s\) prime to q, multiplication of every support residue vector by
s is invertible over \(\mathbb F_q\). This observation alone does not
reduce \(t_V\): reductions require exact vanishing of components
\(A_a\), rather than a change of representative or a choice of local
embedding. At different prepared indices the surviving components may be
different, so their fields must be checked for the value in question.

Here is an exact phase example. Set
\(K=\mathbb Q,p=73,q=2,\theta_1=74,\theta_2=147,P=1\). Take the roots
\(U_1=\sqrt{74},U_2=\sqrt{147}\) which are congruent to one at73. They
are the principal roots of Lesson9. The phase data may be
\(G_0=36,d_1=d_2=0,\alpha_0=-1\), so the normalized residue relation is
the genuine relation \(w=0\). The square classes of \(-1,74,147\) are
independent: prime valuations at37 and3 first force the exponents of74
and147 to be even, and then the sign forces the remaining exponent to be
even. Thus the full root field has degree eight, and the two principal
monomials generate a degree-four field. For \(V=2-U_1-U_2\), the
surviving vectors are \((0,0),(1,0),(0,1)\), which span dimension two.
Theorem10.147 gives degree four. Its minimal polynomial and norm are

\[
P_V(X)=X^4-8X^3-418X^2+1736X+3577,
\qquad N_{K(V)/\mathbb Q}(V)=3577=49\cdot73.
\tag{10.432}
\]

The four sign embeddings above73 have valuations \(1,0,0,0\), with the
selected principal-root embedding first. Thus the weighted sum is one,
whereas four times the selected valuation is four. These are explicit
algebraic values with the phase relation and principal roots; they are
an illustration of the field and valuation mechanisms, rather than an
auxiliary family with all the integer zeros of the second induction.
Solution59 proves these computations. Theorem10.149 verifies the strict
arithmetic comparison for the actual fractional targets by strengthening
the integer input.

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={Surviving fractional components determine the field and the four local valuations}]{../figures/fractional-value-components.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure10.38. Left: the nonzero residue components of \(U_1U_2\),
\(1+U_1\) and \(2-U_1-U_2\) have span dimensions one, one and two,
giving exact degrees two, two and four over \(\mathbb Q\). A constant
component contributes the origin. Right: at73, the four sign embeddings
of \(2-\sqrt{74}-\sqrt{147}\) have valuations \(1,0,0,0\). The norm has
valuation one; the selected embedding is marked. The bars are exact
valuations, rather than height upper bounds. Theorem10.147 proves the
component and field assertions; Solution59 verifies the polynomial,
principal roots and norm.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/fractional_value_components.py}{Figure program},
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/fractional_value_certificate.py}{exact
calculations}. Human-source context: the fractional phase and
product-formula passage in
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
freely accessible paper}, Section7. The exact field and local
comparisons used here are proved above.}
\end{flushleft}
\end{minipage}\par\medskip


\subsubsection{Increasing the integer input before the fractional
step}\label{increasing-the-integer-input-before-the-fractional-step}

The prescribed integer interval is retained throughout the second
induction. Before taking a fractional value, we can temporarily prove
more integer zeros at the same derivative order. Two such extensions
give enough analytic precision to keep the full global field degree.

\textbf{Lemma 10.148 (the two-radius arithmetic comparisons).} Retain
the original fixed-order parameters, constants and notation of
Theorems10.145--10.146. Set \(\xi=24/25\), \(a=r+1\), and

\[
\begin{aligned}
\mu_r&=q^r/Q^{m_r},&
f_r&=\max\{1,\xi c_1/(2cqH)\},\\
j_r&=\min\{j\in\mathbb Z_{\ge0}:\mu_rf_r\le q^j\},&
\Gamma&=\bar\lambda\Omega_+/c_4,&
\sigma_r&=qa/(S_-L_r).
\end{aligned}
\tag{10.433}
\]

Use the same full-depth \(K_r\) as in Theorem10.146; for ranks at least
eight use its enclosing coefficient bound with \(C_r^\sharp\le1\). The
following quantities are strictly positive in every original case and at
every rank \(r\ge2\):

\[
\begin{aligned}
P_1={}&2cq^2\eta-2q\sigma_r
 -K_r-\frac{\bar\lambda j_r\ell_q}{5c_4}
 -\frac{\mu_rf_r}{c_2},\\
P_2={}&\xi c_1q^r-2\sigma_r
 -K_r-\frac{\bar\lambda(j_r+1)\ell_q}{5c_4}
 -\frac{q\mu_rf_r}{c_2},\\
P_3={}&\xi c_1q-K_r+\Gamma
 -\frac{\bar\lambda j_r\ell_q}{5c_4}
 -\frac{\mu_r}{c_2}
 -\frac{\Gamma+2\sigma_r}{q^r}.
\end{aligned}
\tag{10.434}
\]

Also \(j_r\le2\), and \((1-\xi)c_1q^a>1/(c_3Y_-L_r)\).

\textbf{Proof.} All terms are rational endpoints of the complete bounds
in Theorem10.146. Its \(m_r\) is the last integer with
\(cL_rH^{m_r}/a\ge1\). We give an enclosure for the unbounded rank range
which also retains the factor \(q^r\).

Put \(E_x=(1+1/x)^x\), \(x>0\). Differentiating its logarithm gives
\(\ln(1+1/x)-1/(x+1)>0\): the first term is the integral of \(1/(1+t)\)
from zero to \(1/x\), strictly greater than the interval length times
its endpoint minimum. Thus \(E_x\) increases. The exact factorial ratio
of \(F_r=r^r(r+1)^r/(r!)^2\) is \(F_{r+1}/F_r=E_rE_{r+1}\). At rank
eight the rational product \((9/8)^8(10/9)^9>13/2\). Therefore
\(F_{r+1}/F_r>13/2\) for every \(r\ge8\).

For \(r\ge32\), the c band is constant, and both H and Q increase by
Lemma10.122. The formula (10.417) gives

\[
\frac{L_{r+2}/(r+3)}{L_r/(r+1)}
>a_*^2(13/2)^2.
\tag{10.435}
\]

The following three rational substitutions at \(H_{32}=(1-c/33)^{33}\),
\(Q_{32}=qH_{32}\), verify \(a_*^2(13/2)^2H_{32}^{d_0}>1\) and
\(Q_{32}^{d_0}>q^2\):

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}lrrrr@{}}
\toprule\noalign{}
Rank band & \(c\) used in the enclosure & \(a_*\) & \(q\) & \(d_0\) \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
odd cases other than III & 14/25 & 7 & 2 & 12 \\
III & 11/20 & 7/2 & 2 & 11 \\
V & 827/1000 & 26/3 & 3 & 9 \\
\end{longtable}
}

In the first row the other actual c values are smaller, so their H and Q
are larger. Multiplying the inequality defining \(m_r\) by the displayed
two-rank ratio and by \(H_{32}^{d_0}\) proves \(m_{r+2}\ge m_r+d_0\).
Consequently

\[
\mu_{r+2}\le\mu_r\,q^2/Q_r^{d_0}<\mu_r
\qquad(r\ge32).
\tag{10.436}
\]

This proves decrease along both parity sequences. It is not an enclosure
obtained by holding \(m_r\) fixed while rank grows. Furthermore \(f_r\)
decreases because H increases. Thus \(j_r\) is nonincreasing along each
sequence. The enclosing \(K_r\) decreases: its coefficient and Euler
remainders decrease, \(a/L_r\) decreases, and the increasing Q enlarges
the positive logarithm lower bound in its denominator term. These are
exactly the monotonicities used in Theorem10.146, now without a growing
\(r_1\) scalar cost. Also \(\sigma_r\) decreases, while \(\eta\) and
\(q^r\) increase. Each displayed \(P_i\) therefore increases along each
parity sequence from32 or33. The input comparison likewise improves.

Substitute ranks2--31 and parity endpoints32,33 in each original row,
using exact integer comparisons for \(m_r,j_r\) and the margins. The
following strict margins are divided by1000000.

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}lrrr@{}}
\toprule\noalign{}
Case & \(P_1\) & \(P_2\) & \(P_3\) \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
I.1 & 1028198 & 2077317 & 437081 \\
I.2 & 1376333 & 2808629 & 735003 \\
II & 1237102 & 2462590 & 678380 \\
III.1 & 1212788 & 2458197 & 615506 \\
III.2 & 1309556 & 2722070 & 711872 \\
IV, first prime branch & 1190595 & 2368885 & 589777 \\
IV, larger prime branch & 1318170 & 2496460 & 748768 \\
V & 2912809 & 10621246 & 82270 \\
\end{longtable}
}

All256 substitutions also give \(j_r\le2\) and the strict input
comparison. Their rational formulas and integer calculations are
provided with Figure10.39. Together with the parity argument, these
checks include every rank. \(\square\)

\textbf{Theorem 10.149 (the actual fixed-order fractional zeros).}
Suppose \(I_1\le I<I_3\) and an admissible stage-I family has all its
prepared zeros at integers \(|s|\le R\) through the original order
\(O=\lfloor T_\flat\rfloor\). Then it has all its prepared zeros at
\(s/q\) for integers \(|s|\le R\), through that same order O. The
original coefficient minimum, support widths and phase are retained.

\textbf{Proof.} Put

\[
R_{\rm s}=\left\lfloor\frac{\xi U/(q\theta)-1}{2}\right\rfloor,
\qquad
R_{\rm l}=\left\lfloor\frac{\xi U/\theta-1}{2}\right\rfloor.
\tag{10.437}
\]

The original parameter lower bounds make both radii positive. Indeed
\(U/\theta=c_1q^rST/a\), by (10.186), while
\(S\ge S_->58,T\ge g_4>(9000/29)a\), \(c_1>7/5\), and \(q^r\ge4\). For
\(x\in\mathbb R\), \(\lfloor x\rfloor>x-1\). Therefore, writing
\(N_{\rm s}=2R_{\rm s}+1\), \(N_{\rm l}=2R_{\rm l}+1\),

\[
\begin{aligned}
N_{\rm s}\theta&>\xi U/q-2\theta,&
N_{\rm s}\theta&\le\xi U/q,\\
N_{\rm l}\theta&>\xi U-2\theta,&
N_{\rm l}\theta&\le\xi U,&
qR_{\rm s}&\le R_{\rm l}.
\end{aligned}
\tag{10.438}
\]

For the last inequality use \(qR_{\rm s}\le(\xi U/\theta-q)/2
\le(\xi U/\theta-1)/2\), and integrality. Lemma10.148 and (10.414) give
\(\beta<(1-\xi)U\). Thus both new intervals satisfy the full-node input
inequality \(N\theta<U-\beta\). The original interval already satisfies
it by Proposition10.145. Taking the larger of it and either new interval
preserves that inequality.

We first show that every integer \(|s|\le R_{\rm s}\) is a zero, without
changing the family. The old interval supplies the local normal
precision \((2R+1)\theta\) at every \(x\in\mathbb Z_p\), by
Proposition10.145. The same calculation as in Theorem10.146, now using
all nodes, gives

\[
c_1W_*(2R+1)\theta
>2cq^2\eta-2q\sigma_r.
\tag{10.439}
\]

Here \(q\eta>1\) bounds the original \(r_1\) mass below by its value at
zero; the floor loss uses \(c_1W_*\theta=qa/(ST)\). If \(u>D_{\rm a}\)
the target is identically zero. Otherwise its integer value belongs to K
by the phase identity, so it has the original-K arithmetic bound.

For precision we bound the physical real radius rather than absorbing
the denominator into a geometric stage restriction. From \(ST=S_1T_1\),
\(T_1<a/(cH)\), and \(I\ge I_1\ge m_r\),

\[
q^{-I}R_{\rm s}
\le q^{-I}R_{\rm l}/q
<\frac{\xi c_1q^{r-1}}{2cH Q^{I_1}}S
\le\mu_rf_rS\le q^{j_r}S.
\tag{10.440}
\]

The factorial proof of (10.325), applied directly with this physical
radius enlarged to \(q^{j_r+1}S\), bounds the scalar by
\(\bar\lambda[1+j_r\ell_q/5]/c_4\) in the normalization \(c_1W_*\). This
uses only that factorial proof, not the separate geometric absorption
(10.326). The full coefficient, Euler and normal terms and the entire
denominator through \(I_3\) remain those in \(K_r\). The torus
contribution is at most \(\mu_rf_r/c_2\). The same-order normal
arithmetic cost is therefore less than
\(K_r+\bar\lambda j_r\ell_q/(5c_4)+\mu_rf_r/c_2\). The strict \(P_1\)
comparison forces every proposed nonzero integer value to zero, by
Corollary10.120. Integers already in the old interval retain their
zeros. We now have the full interval of radius \(\max\{R,R_{\rm s}\}\)
at exactly O.

Apply the full-node normal interpolation theorem10.119 to that larger
interval. Its input inequality was just verified, so its precision is at
least \(N_{\rm s}\theta\). In normalization this is greater than
\(\xi c_1q^r-2\sigma_r\). At an integer target of radius \(R_{\rm l}\),
the physical radius and torus height can be larger by at most q than the
previous bounds. The factorial scalar may conservatively use \(j_r+1\);
all other costs still belong to the same \(K_r\). The normal arithmetic
cost is less than \(K_r+\bar\lambda(j_r+1)\ell_q/(5c_4)+q\mu_rf_r/c_2\).
The strict \(P_2\) comparison forces every such integer target to zero.
Thus the original family has all its zeros through O on the full
interval of radius \(\max\{R,R_{\rm l}\}\).

Its normal analytic precision, by the same full-node theorem and its
verified input, is greater than \(\xi U-2\theta\) at every point of
\(\mathbb Z_p\). In normalization it is greater than
\(\xi c_1q^{r+1}-2\sigma_r\). Consider now only the prescribed
fractional targets \(|s|\le R\), \(x=s/q\). Their physical radius
satisfies \(q^{-I}R/q\le q^rS/Q^{I_1}\le\mu_rS\le q^{j_r}S\). Their
torus cost is consequently at most \(\mu_r/c_2\), and their factorial
scalar uses the already justified \(j_r\) envelope.

For a nonzero fractional value, Theorems10.56 and10.147 permit a global
degree factor at most \(q^r\), with the selected e,f unchanged. Its
complete arithmetic bracket therefore has normal upper bound, in the
scale \(c_1W_*\),

\[
q^r\left(K_r-\Gamma
 +\frac{\bar\lambda j_r\ell_q}{5c_4}+\frac{\mu_r}{c_2}\right)
 +\Gamma.
\tag{10.441}
\]

The \(\Gamma\) term bounds \(G_u\), which is added once after the
product-formula arithmetic bracket. All q-denominators are charged at
depth \(I+1\le I_3\); every coefficient, Euler and torus cost remains.
The difference between the analytic lower bound and this upper bound,
divided by \(q^r\), is exactly \(P_3>0\). Hence the value must be zero.
The local exponential branch is the algebraic principal-root branch of
Corollary10.52. Indices beyond additive degree vanish already. This
proves every required fractional zero through O. Neither the two extra
integer intervals nor either interpolation changes any coefficient,
phase, minimum, support width or selected derivative order. \(\square\)

\textbf{Corollary 10.150 (the original second induction and its
contradiction).} Under the original contradiction parameters at least
admissible rank \(r\ge2\), the alternative \(I_1<I^*\) is impossible.

\textbf{Proof.} The actual terminal family at \(I_1\) and all of its
original integer R-block zeros through O are supplied by Theorem10.144.
Suppose a family with those properties has reached depth
\(I_1\le I<I_3\). Theorem10.149 supplies every original fractional zero.
The full saturated Kummer and phase construction of Theorem10.51 then
extracts a nonzero successor family with its q-deleted integer zeros for
\(|s|\le R\) through the same order O. It retains a coefficient
attaining the original minimum, never increases the coefficient height,
preserves the exact phase and divides the support widths by q. Its
rational scalar is the stage \(I+1\) scalar; the affine binomial changes
in that theorem preserve the total-order cutoff. The selected
derivations and integral normalized torus exponents therefore remain the
ones required by Proposition10.31.

Theorem10.146 applies at \(J=I+1\le I_3\) and closes all the missing
integers on the original R-block at the unchanged O. This is the
induction hypothesis for the next depth. Finite induction reaches the
actual family at \(I_3\), including its endpoint. That family has every
hypothesis of Proposition10.143, which makes its integral torus support
constant and yields more distinct integer roots than the degree of its
nonzero additive polynomial. This is a contradiction. The alternative is
therefore impossible. Together with Theorem10.141, both original
stopping branches, including their integer endpoints, have now been
ruled out. \(\square\)

The full global factor is retained in Theorem10.149. Separate primes are
never assigned the same valuation. The stronger intermediate integer
zeros supply the additional local precision before phase extraction. The
required original radius and order remain the induction invariant.

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={Two larger integer intervals and the certified full-field fractional gaps}]{../figures/fixed-fractional-boost-profile.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure10.39. Left: exact illustrative floor data q=3, theta=1,
U=1001, xi=24/25 give radii159 and479, with319 and959 integer nodes. The
old radius25 is shown only to illustrate the two extensions. Their node
masses lie above xi U/q-2 and xi U-2, and3 times159 is at most479. These
data illustrate the floor mechanism, rather than a full original
contradiction parameter choice. The selected order O is unchanged.
Right: proved strict lower fractional gaps P3 after division by the full
q\^{}r field factor, for every original case and every rank. Lemma10.148
proves the two parity enclosures, Theorem10.149 proves both integer
extensions and all original fractional zeros, and Corollary10.150
performs the actual phase induction and stopping contradiction.
Solution60 checks the floors and recurrence.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/fixed_fractional_boost_profile.py}{Figure
program},
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/fixed_fractional_boost_bounds.py}{exact all-rank
certificate}. Human-source context: Section7 of
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
freely accessible paper}. The stronger intermediate integer intervals
and the complete full-field comparison are proved above.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{47. The first explicit Yu
estimate}\label{the-first-explicit-yu-estimate}

The two stopping contradictions now give an explicit estimate for the
original multiplicative form. The bases used during the proof can have
smaller rank than the original list. We keep their height allowance
until it cancels against the auxiliary bound. This also requires an
elementary argument when the least rank is one.

\subsubsection{The constants are uniform in the
rank}\label{the-constants-are-uniform-in-the-rank}

Use the seven cases and their constants from Section 36. Write
\(a=r+1\), \(F_r=r^r(r+1)^r/(r!)^2\), and let \(a_*\), \(\rho\) and
\(\widehat\vartheta\) have their meanings there. For the two IV branches
use the same constants. Define

\[
L_r=\frac{2c_0c_4q(r+1)a_*^rF_r}{\rho\widehat\vartheta}\,v,
\qquad
v=\begin{cases}
5,&\mathrm{I.2,III.1},\\
509/100,&\mathrm{III.2},\\
1,&\mathrm{I.1,II,IV,V}.
\end{cases}
\tag{10.442}
\]

These are lower bounds for the actual \(g_4\). In the branches where
\(g_1\) remains in (10.185), use \(g_1>5\). For III.2, \(d=1\) gives
\(g_1\ge4+\ln3>509/100\). Indeed the first three positive terms of the
logarithm series at \((3-1)/(3+1)=1/2\) sum to \(1+1/12+1/80>109/100\).
In the other branches the factor \(g_1\) cancels, and \(v=1\) is valid.

Define the following lower bounds for the actual \(g_2\):

\[
G_r=c_3q(r+1)^2
\begin{cases}
(39r/10+36/5)/(11/10),&\mathrm{I.1,I.2},\\
(39r/10+36/5)\,20/17,&\mathrm{II},\\
2,&\mathrm{III.1,IV},\\
1,&\mathrm{III.2},\\
(39r/10+36/5)/(347/500),&\mathrm V.
\end{cases}
\tag{10.443}
\]

For I and V, these follow from (10.185), (10.288), \(e\ge1\),
\(\ln3<11/10\) and \(\ln2<347/500\). The logarithm bounds were proved in
the exact series calculations of Section 40 and Solution 48. For II,
\(e\ge2\) and \(\ln5<17/10\); the positive exponential series through
degree four at \(17/10\) already exceeds five. For III and IV use
\(d\ge2\) except in III.2, where \(d=1\). Thus these inequalities apply
to every field in the corresponding case.

\textbf{Lemma 10.151 (closing every numerical constant).} Put

\[
F_0=(1+\epsilon)(2+g_2^{-1})c_0c_1c_3c_4q^2.
\tag{10.444}
\]

For every rank \(r\ge2\), \((1+10^{-26})F_0<c\), where \(c\) is the
following exact case constant. The last column gives strict lower bounds
for the gap.

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}lrr@{}}
\toprule\noalign{}
Case & \(c\) & Gap greater than \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
I.1 & 939 & \(165185/10^6\) \\
I.2 & 636 & \(262798/10^6\) \\
II & 505 & \(146558/10^6\) \\
III.1 & 1794 & \(305548/10^6\) \\
III.2 & 1790 & \(4059/10^6\) \\
IV & 2680 & \(656871/10^6\) \\
V & 206 & \(724409/10^6\) \\
\end{longtable}
}

\textbf{Proof.} Set \(\Delta_r=(r+1)/(2L_r)\). The exact formula
(10.282) gives \(F_{r+1}/F_r>4\), and \(a_*\ge7/2\), so

\[
0<\Delta_{r+1}<\Delta_r/14,\qquad \Delta_2<1.
\tag{10.445}
\]

The second inequality follows by direct substitution in the seven rows.
For \(0<z<1\), \((1+z/14)^2<1+z\): after subtracting one and dividing by
\(z\), its left side is \(1/7+z/196<1\). Since \(r+1\le2r\), it follows
that \((1+\Delta_{r+1})^{r+1}<(1+\Delta_r)^r\). Each displayed \(G_r\)
increases with \(r\). By (10.185), therefore,

\[
(1+10^{-26})F_0
\le (1+10^{-26})(1+\Delta_r)^r
       (2+G_r^{-1})c_0c_1c_3c_4q^2
\le (1+10^{-26})(1+\Delta_2)^2
       (2+G_2^{-1})c_0c_1c_3c_4q^2.
\tag{10.446}
\]

All entries of the final expression are rational. Subtracting it from
the seven integers \(c\), and multiplying positive denominators, gives
the table. The exact fractions and the two positive-series checks are
recorded in the
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_first_main_bound_certificate.py}{reproducible
calculation}. The decrease just proved supplies the quantifier over
every rank; the table alone would only check rank two. \(\square\)

\subsubsection{The bound for the original independent
bases}\label{the-bound-for-the-original-independent-bases}

Let \(a_1,\ldots,a_n\) be multiplicatively independent
\(\mathfrak p\)-adic units in a number field \(K\), with \(n\ge2\).
Write \(d=[K:\mathbb Q]\), \(e=e_{\mathfrak p}\), \(f=f_{\mathfrak p}\),
\(t=f\ln p\) and \(\ell_d=\max\{1,\ln d\}\). At odd \(p\) put \(q=2\).
At \(p=2\), assume \(\zeta_3\in K\) and put \(q=3\). Let \(q^u\) be the
order of the q-primary roots of unity in \(K\). Use the saturated
residue index \(\delta(\boldsymbol a)\) from (10.49), as in (10.272),
for the original list of bases.

Let \(b_1,\ldots,b_n\in\mathbb Z\), not all zero. Renumber so that
\(b_n\ne0\) and \(v_p(b_n)\le v_p(b_j)\) for every \(j\), with
\(v_p(0)=+\infty\). Put

\[
\begin{aligned}
\Xi&=\prod_{j=1}^na_j^{b_j},&
B_\dagger&=\min_{b_j\ne0}|b_j|,\\
R_{\max}&=\max_{j<n}
\left\{\frac{|b_n|}{h(a_j)}+\frac{|b_j|}{h(a_n)}\right\},&
A_n&=\max\{4+\ln((n+1)d),e,t\}.
\end{aligned}
\tag{10.447}
\]

Independence implies \(\Xi\ne1\) and every \(h(a_j)>0\). Use \(W(d)\)
from (10.222). The constants for the statement are

\[
\mathfrak a=\begin{cases}
14,&\mathrm{I,II,IV},\\
7(p-1)/(p-2),&\mathrm{III},\\
26,&\mathrm V,
\end{cases}
\qquad
a_0=\begin{cases}
2+\ln14,&\mathrm{I,II,IV},\\
2+\ln7,&\mathrm{III},\\
2+\ln26,&\mathrm V.
\end{cases}
\tag{10.448}
\]

Use \(c\) from Lemma 10.151 and the exact \(a_1,a_2\) from (10.287):
\(a_1=4.03,4.79,3.44,4.71,5.84,5.12,2.52\) in the seven cases,
\(a_2=a_1\) in I.2 and III.2, and \(a_2=a_1+\ln2\) otherwise. Define

\[
\begin{aligned}
G_1(n,d)&=(n+1)(a_0n+a_1+\ln(a_0n+a_2)+\ln d),\\
\mathsf h&=\max\left\{
\ln\frac{R_{\max}}{dW(d)},\ \ln B_\dagger,
\ G_1(n,d),\ (n+1)t\right\},\\
C_1&=c\mathfrak a^n
\frac{n^n(n+1)^{n+1}d^{n+2}\ell_d}{n!q^ut}
\max\left\{\frac{p^f}{\delta(\boldsymbol a)t^{n+1}},
                   \frac{\mathrm e^n}{n^n}\right\}A_n.
\end{aligned}
\tag{10.449}
\]

Here \(\mathrm e=\exp(1)\); \(e\) is the ramification index. The cases
are I: \(p=3\), with I.1 for \(d>1\) and I.2 for \(d=1\); II:
\(p=5,e\ge2\); III: \(p\ge5,e=1\), with the same degree subdivisions;
IV: \(p\ge7,e\ge2\); V: \(p=2,\zeta_3\in K\).

\textbf{Theorem 10.152 (Yu's first explicit estimate, with proof).}
Under these hypotheses,

\[
\operatorname{ord}_{\mathfrak p}(\Xi-1)
<C_1\mathsf h\prod_{j=1}^nh(a_j).
\tag{10.450}
\]

\textbf{Proof.} Choose \(\tau=10^{-26}/(2n)\) in (10.184). Keep this
same \(\theta=\vartheta/(1+\tau)\) and the same original \(n\) at every
smaller rank. Direct substitution gives

\[
qH_*=\mathfrak a(1+\tau),\qquad
(1+\tau)^n<1+10^{-26}.
\tag{10.451}
\]

For the second inequality, the binomial expansion and
\(\binom nk\le n^k\) give \((1+\tau)^n\le(1-n\tau)^{-1}\). Here
\(n\tau=10^{-26}/2\), and \((1-x)^{-1}<1+2x\) for \(0<x<1/2\).

Take the least admissible rank \(r\) of (10.277), with integral forms
\(L_0=z_0,L_1,\ldots,L_r\), rational coefficients \(B_i\), weights
\(\sigma_i\) and local unit bases \(\alpha_i\). This rank exists by the
explicit rank-n representation in Section 35. The weighted height
condition there and the original integers \(b_j\) supply precisely
(10.218). The function \(G_1(j,d)\) increases for integers \(j\ge1\),
because all its positive linear and logarithmic factors increase.
Consequently \(\mathsf h\) meets the original height conditions
(10.223), (10.287), and \(\mathsf h\ge(r+1)t\).

Suppose first \(r\ge2\). Form the complete original parameters
(10.183)--(10.186) with \(h=\mathsf h\) and this least-rank
representation. Write \(U_r=q^{r+1}S\mathcal D/(et)\). If
\(v_p(\Xi-1)\ge U_r\), the depth and phase construction of Sections 20
and 28, the minimum-valuation derivation choice of Lemma 10.23, and the
integral kernel of Theorem 10.71 give the original contradiction setting
of Section 40. The evaluated linear form is \(P\log\Xi\), whose
valuation is at least \(U_r\). This evaluation is legitimate on the
principal unit disc: (10.280) gives \(U_r>q^{r+1}Sb_r>1\ge1/(p-1)\),
because \(A_r+x\ge e\). Lemma 9.5 and Proposition 9.4 identify its
valuation with that of \(\Xi-1\). The original minimum-valuation
coefficient hypothesis is retained; no bound on newly chosen rational
\(B_i\) replaces it.

Corollary 10.140 constructs the first induction. If \(I^*\le I_1\),
Theorem 10.141 gives the least-rank contradiction, including equality.
If \(I_1<I^*\), Corollary 10.150 constructs the second induction and
gives its contradiction. These exhaustive alternatives prove
\(v_p(\Xi-1)<U_r\).

Now use \(\gamma(\mathsf h+x)(A_r+x)/q^\nu
=\mathsf hA_r\) in (10.186). Exact cancellation, with \(M_r\) and
\(H_*\) as in (10.272), gives

\[
eU_r=F_0(qH_*)^r
\frac{r^r(r+1)^{r+1}d^{r+2}\ell_d}{r!q^ut^2}
M_r(\boldsymbol\alpha)A_r\mathsf h\prod_i\sigma_i.
\tag{10.452}
\]

The profile (10.277) bounds its last weight product by
\((\mathrm eH_*(n+1)d)^{n-r}
(M_n(\boldsymbol a)/M_r(\boldsymbol\alpha))\prod_jh(a_j)\). Thus the
actual smaller-rank residue factor cancels exactly. The ratio between
the remaining rank factor and the desired rank-n factor is at most

\[
\left(\frac{\mathrm e}{q}\right)^{n-r}
\frac{r^r/r!}{n^n/n!}
\left(\frac{r+1}{n+1}\right)^{r+1}\le1.
\tag{10.453}
\]

Indeed consecutive ratios of \(k^k/k!\) are \((1+1/k)^k\ge2\), by the
binomial expansion or its increasing logarithm. Hence the middle
quotient is at most \(2^{r-n}\). Also \(\mathrm e<3\), \(q\ge2\), and
\(r\le n\). Finally \(A_r\le A_n\), and
\(M_n/t^2=t^{-1}\max\{p^f/(\delta(\boldsymbol a)t^{n+1}),
\mathrm e^n/n^n\}\). Lemma 10.151 and the small-\(\tau\) bound therefore
give the claimed strict inequality.

It remains to prove the result for \(r=1\). Write \(B_1=p_1/q_1\) in
lowest terms, \(q_1>0\). Comparison of the integral coefficients in
\(q_1\mathcal L=q_1B_0z_0+p_1L_1\) gives \(q_1B_0\in\mathbb Z\),
\(p_1\mid b_j\) for every \(j\), and \(|p_1|\le B_\dagger\). Set
\(\beta=\alpha_1\) and let \(g\) be its actual residue order, as defined
in Theorem 10.21. It is non-torsion by the independence in (10.277), and

\[
\Xi^{q_1q^u}=\beta^{p_1q^u}\ne1,\qquad
\operatorname{ord}_{\mathfrak p}(\Xi-1)
\le\frac d t\left(2\mathsf h+2eg\sigma_1\right).
\tag{10.454}
\]

The power identity kills the \(z_0\) torsion factor. A positive integer
power of a local unit never decreases the valuation of its difference
from one, by the factorization of \(X^m-1\). Apply the proved bound
(10.60) to that power. We have \(q^u\le p^f-1<p^f\), \(t>1\),
\(\mathsf h\ge(n+1)t\), and \(\mathsf h\ge\ln B_\dagger\); hence
\(\ln(2q^uB_\dagger)<2\mathsf h\). Also \(p/(p-1)\le2\) and
\(h(\beta)\le\sigma_1\). These facts prove the displayed inequality
without a logarithm estimate for a rank-one form.

Let \(\mathcal T=C_1\mathsf h\prod_jh(a_j)\). The height-product lower
bound (10.55), with \(w_K\ge q^u\), and \(M_n\ge\mathrm e^nt/n^n\),
imply

\[
\mathcal T\ge\frac{d\mathsf h}{t}
\frac{c\mathfrak a^n n^n(n+1)^{n+1}A_n}{\rho(n!)^2}
>800\frac{d\mathsf h}{t}.
\tag{10.455}
\]

For the last inequality use \(c\ge200\), \(\mathfrak a\ge7\), \(n\ge2\),
\(A_n>5\), \(\rho\le58\), and \(n^n(n+1)^{n+1}/(n!)^2\ge1\). Thus the
first summand in the elementary rank-one bound is less than
\(\mathcal T/400\).

At rank one, \(\delta(\beta)=(p^f-1)/g\), so
\(M_1(\beta)\ge p^fg/((p^f-1)t)>g/t\). The profile gives
\(\sigma_1\le\Gamma_n^{n-1}(M_n/M_1)\prod_jh(a_j)\). Consequently its
second summand is less than \(2ed\Gamma_n^{n-1}M_n\prod_jh(a_j)\). By
Lemma 10.73, \(eq^u\le2d\). Dividing this summand by \(\mathcal T\) and
substituting \(H_*=\mathfrak a(1+\tau)/q\) gives an upper bound

\[
\frac{4t^2(\mathrm e/q)^{n-1}(n!/n^n)(1+\tau)^{n-1}}
{c\mathfrak a(n+1)^2d\ell_d A_n\mathsf h}
<\frac8{c\mathfrak a(n+1)^3d\ell_d}
\le\frac8{200\cdot7\cdot27}<\frac1{4000}.
\tag{10.456}
\]

Here \(n!/n^n\le2^{1-n}\), by the same consecutive-ratio argument;
\(\mathrm e/q<3/2\), \((1+\tau)^{n-1}<2\), \(A_n\ge t\), and
\(\mathsf h\ge(n+1)t\). The two summands together are strictly smaller
than \((1/400+1/4000)\mathcal T<\mathcal T\). This completes the
rank-one case and the proof. \(\square\)

This is the first estimate in Theorem I of
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
freely accessible 2013 paper}, with its actual residue index,
ramification, degree-one constants, coefficient choice and height
parameter. The proof above derives it from the complete programme
arguments, including both stopping branches.

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={The least-rank proof routes and strict gaps in all seven Yu constants}]{../figures/yu-first-main-bound.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.40. The least admissible rank divides the proof into the
elementary residue-order bound at rank one and the two exhaustive
stopping contradictions at larger rank. The actual smaller-rank residue
factor cancels through (10.277); the remaining rank factor is at most
one. The bars give strict lower bounds for the seven coefficient gaps in
Lemma 10.151, on a logarithmic scale; every bar applies to all ranks,
including the small III.2 gap. Theorem 10.152 proves the final
original-base estimate and Solution 61 checks the cancellation and
rank-one margins.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_first_main_bound_profile.py}{Figure program},
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_first_main_bound_certificate.py}{exact
rational calculation}. Human-source context: Sections 1.1, 1.3, 2 and
3.2 of
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
free paper}.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{48. An application bound in terms of the largest
coefficient}\label{an-application-bound-in-terms-of-the-largest-coefficient}

Keep the field cases and constants of Theorem 10.152. Let
\(a_1,\ldots,a_n\in K^*\) be independent local units, \(n\ge1\), and let
\(b_1,\ldots,b_n\in\mathbb Z\) be not all zero. Set
\(\Xi=\prod_ja_j^{b_j}\) and put \(\Omega=\prod_jh(a_j)\),
\(B\ge\max\{3,|b_1|,\ldots,|b_n|\}\), \(H=\max\{\ln B,t\}\), and
\(C_1^*=(n+1)C_1\). When \(n=1\), use the same displayed formula for
\(C_1\), with \(\delta(a_1)=(p^f-1)/g\), where \(g\) is the original
residue order. The height parameter can be replaced by \((n+1)H\) after
a simple local-height alternative. This replacement will then be useful
for dependent lists of bases.

\subsubsection{An explicit coefficient-height
threshold}\label{an-explicit-coefficient-height-threshold}

For \(n\ge2\), set

\[
W_B=\frac{7899}{7900}\frac{t}{nd}\,
       C_1^*\frac{\Omega}{\max_jh(a_j)}.
\tag{10.457}
\]

\textbf{Lemma 10.153 (the large-coefficient branch supplies every height
condition).} For every original case and every \(n\ge2\),

\[
\ln W_B>a_0n+a_1+\ln d,
\qquad \ln W_B>a_0n+a_2.
\tag{10.458}
\]

\textbf{Proof.} Remove one base of maximal height from the independent
list, and apply (10.55) to the remaining \(n-1\) bases. Use
\(w_K\ge q^u\), \(M_n\ge\mathrm e^nt/n^n\), and the exact formula for
\(C_1^*\). All powers of \(q^u\), \(t\) and \(\ell_d\) cancel, giving

\[
W_B\ge\frac{7899}{7900}\frac{c\mathrm e}{\rho}
\mathfrak a^n
\frac{(n+1)^{n+2}(n-1)^{n-1}}{(n!)^2}\,dA_n.
\tag{10.459}
\]

In all cases \(\mathfrak a\ge\exp(a_0-2)\). For \(2\le n\le511\), taking
logarithms in this lower bound and subtracting \(a_0n+a_1+\ln d\) leaves
at least

\[
\begin{aligned}
&\ln\frac{7899c}{7900\rho}+1
 +(n+2)\ln(n+1)+(n-1)\ln(n-1)\\
&\quad{}-2\ln(n!)-2n+\ln(4+\ln((n+1)d_0))-a_1,
\end{aligned}
\tag{10.460}
\]

where \(d_0=1\) in I.2 and III.2, and \(d_0=2\) in the other five cases.
All 3570 expressions have strictly positive rigorous lower endpoints.
Their minimum is greater than \(5/1000\), in case IV at \(n=14\). To
reproduce this finite certificate, reduce each positive logarithm
argument to \([1,2]\) by powers of two, then use 32 terms of the
positive series \(2\sum_{j\ge0}z^{2j+1}/(2j+1)\), \(z=(x-1)/(x+1)\). The
omitted tail is at most \(2z^{65}/(65(1-z^2))\). Round every lower
endpoint down and every upper endpoint up to \(10^{-24}\). Form
\(\ln(n!)\) by adding the integer logarithm intervals; use its upper
endpoint in the subtracted term and the lower endpoint of the positive
\(\ln A_n\). These directions preserve the lower bound at every
operation. The
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_application_height_certificate.py}{reproducible
exact calculation} records the seven minima and their rational
endpoints.

For the infinitely many \(n\ge512\), concavity of \(\ln x\) puts its
graph above the chord on each interval \([k,k+1]\). Summing their
trapezoid integrals gives

\[
\ln(n!)-\tfrac12\ln n\le\int_1^n\ln x\,dx
=n\ln n-n+1.
\tag{10.461}
\]

Hence \(n!\le\mathrm e\sqrt n(n/\mathrm e)^n\). The integral identities
\(\ln(1+x)>x/(1+x)\) and \(\ln(1-x)>-x/(1-x)\), for \(x>0\) and
\(0<x<1\) respectively, give \((1+1/n)^{n+2}(1-1/n)^{n-1}>1\). Thus the
preceding lower bound implies

\[
\frac{W_B}{d\exp(a_0n)}
>\frac{7899c}{7900\rho\mathrm e}A_n
>\frac{7899c}{7900\rho(11/4)}\,10>\exp(a_1).
\tag{10.462}
\]

Here \(\mathrm e<11/4\), \(\exp6<513\), and therefore
\(A_n\ge4+\ln(n+1)>10\). The final seven inequalities follow from exact
positive-exponential series upper bounds: sum through degree 64 and
bound the remaining tail by its first term divided by \(1-x/66\), at
each \(x=a_1<6\). Each difference is positive; the smallest exceeds
\(47/100\), in II. The same series checks \(\mathrm e<11/4\) and
\(\exp6<513\). This proves the first assertion for all ranks. In the
degree-one cases \(a_2=a_1\); in the others \(\ln d\ge\ln2\) and
\(a_2=a_1+\ln2\). The second assertion follows. \(\square\)

\subsubsection{Conversion to the maximum
coefficient}\label{conversion-to-the-maximum-coefficient}

\textbf{Theorem 10.154 (independent-base application bound).} For the
independent units just described, if \(\Xi\ne1\), then

\[
\operatorname{ord}_{\mathfrak p}(\Xi-1)<C_1^*\Omega H.
\tag{10.463}
\]

At \(n=1\), its right side may be multiplied by \(1/2100\).

\textbf{Proof.} First let \(n\ge2\), and write
\(\mathcal T=C_1^*\Omega H\). The height-product lower bound (10.55),
and the second term in \(M_n\), give

\[
\mathcal T\ge\frac{dH}{t}
\frac{c\mathfrak a^n n^n(n+1)^{n+2}A_n}{\rho(n!)^2}
>7900\frac d t\ln2.
\tag{10.464}
\]

Indeed consecutive ratios of \(k^k/k!\) are at least two, so
\(n^n/n!\ge2^{n-1}\) and \((n+1)^n/n!\ge2^n\). Thus its rank factor is
at least 72. Use \(c\ge200\), \(\mathfrak a\ge7\), \(A_n>5\),
\(\rho\le58\), \(H>1\) and \(\ln2<1\).

The local height inequality in the proof of Theorem 10.21 gives

\[
\operatorname{ord}_{\mathfrak p}(\Xi-1)
\le\frac d t\bigl(nB\max_jh(a_j)+\ln2\bigr).
\tag{10.465}
\]

If \(B/\ln B\le W_B\), its first term is at most
\((7899/7900)\mathcal T\), and its second is strictly less than
\(\mathcal T/7900\). This proves the desired strict bound.

Otherwise \(B/\ln B>W_B\). Renumber the bases so a nonzero coefficient
of minimum p-adic valuation is last; \(C_1\) and \(\Omega\) are
unchanged. Since \(\ln B>1\), we have \(B>W_B\) and then
\(B>W_B\ln W_B\). Lemma 10.153 yields \((n+1)\ln B>G_1(n,d)\). The
singleton case of (10.55), with \(w_K\ge2\), gives
\(\min_jh(a_j)\ge2/(\rho d^2\ell_d\mathrm e)\). Also \(W(d)>1\) and
\(\ell_d\le d\); thus

\[
\frac{R_{\max}}{dW(d)}\le\rho\mathrm e d^2 B.
\tag{10.466}
\]

We have \(\ln\rho+1<6\), since the positive exponential series at five
already exceeds 58. Lemma 10.153 gives \(n\ln B>2\ln d+6\), using
\(a_0>39/10\), \(a_1>0\) and \(n\ge2\). Therefore
\(\ln(R_{\max}/(dW(d)))<(n+1)\ln B\). The other two entries in
\(\mathsf h\) are bounded by \((n+1)H\) directly. Apply Theorem 10.152
to obtain \(\operatorname{ord}_{\mathfrak p}(\Xi-1)
<C_1(n+1)H\Omega=\mathcal T\). This argument uses the proved programme
singleton height bound in place of a separate unproved lower-height
theorem.

Now take \(n=1\). Let \(g\) be the original residue order,
\(\Omega=h(a_1)\), and \(\mathcal T=C_1^*\Omega H\), with the rank-one
definition of \(M_1\). Formula (10.60) bounds the valuation by
\((d/t)\ln(2B)+2edg\Omega/t\). At two, the roots \(-1\) and the
q-primary root together have order \(2q^u\), so \(w_K/q^u\ge2\); at odd
primes it is at least one. Directly in the seven rows,
\(c\mathfrak a w_K/(\rho q^u)>120\) and \(c\mathfrak a>5000\). Use the
singleton height bound, \(M_1\ge\mathrm e t\), and
\(A_1\ge4+\ln2>14/3\), to get

\[
\mathcal T>4480\frac{dH}{t}.
\tag{10.467}
\]

Since \(2^3<3^2\) and \(B\ge3\), \(\ln(2B)<(5/3)H\). Its summand is
therefore less than \(\mathcal T/2688\). For the other summand, retain
the actual residue term \(M_1>g/t\) and \(eq^u\le2d\) from Lemma 10.73.
Exact division by \(\mathcal T\) gives

\[
\frac{2edg\Omega/t}{\mathcal T}
<\frac{t^2}{2c\mathfrak a d\ell_d A_1H}
\le\frac1{2c\mathfrak a d\ell_d}<\frac1{10000}.
\tag{10.468}
\]

Both \(A_1\) and \(H\) are at least \(t\). Finally
\(1/2688+1/10000<1/2100\), since \(12688\cdot2100<26880000\). This
proves the stronger rank-one assertion. \(\square\)

For example, let \(K=\mathbb Q,p=3,a_1=4,b_1=3^k\), \(k\ge1\), and
choose \(B=3^k\). The original residue order is \(g=1\), so
\(\delta(a_1)=2\); the roots in \(\mathbb Q\) give \(q^u=2\). Here
\(t=\ln3\), \(H=k\ln3\), \(A_1=4+\ln2\), and \(M_1=\mathrm e\ln3\): its
other entry is \(3/(2\ln3)<3/2\), whereas \(\mathrm e\ln3>2\).
Substituting the I.2 constants gives

\[
v_3(4^{3^k}-1)=k+1
<\frac{4\cdot636\cdot14}{2100}
  \mathrm e(4+\ln2)k\ln4.
\tag{10.469}
\]

The valuation equality follows from \(v_3(4-1)=1\) and the proved
deep-unit power law, Lemma 9.5. The right side is greater than \(128k\),
using \(4\cdot636\cdot14/2100>16\), \(\mathrm e>2\), \(4+\ln2>4\), and
\(\ln4>1\); hence it strictly exceeds \(k+1\). The application bound is
deliberately uniform rather than sharp in this simple example. It uses
the original rank-one residue order, which is different from the residue
group of a saturated basis.

This is the independent-list part of Theorem 1 in
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
free 2013 paper}, Section 8. Independence supplies positive heights and
the height-product bounds for every sublist. For a dependent list, a
change of basis must also control the new coefficients, its weighted
height product and torsion. Those additional conditions are distinct
from the coefficient-height conversion proved here.

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={The two coefficient-height branches and exact logarithmic margins}]{../figures/yu-application-height.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.41. For \(n\ge2\), the small-coefficient branch proves
the bound directly by the local height inequality; the large-coefficient
branch supplies every entry of \(\mathsf h\) and applies Theorem 10.152.
Both give
\(\operatorname{ord}_{\mathfrak p}(\Xi-1)<\mathcal T=C_1^*\Omega H\).
The bars are strict lower bounds for \(\ln W_B-(a_0n+a_1+\ln d)\),
minimized over all exact ranks 2--511 in each field case. The infinite
range \(n\ge512\) is proved separately in Lemma 10.153, rather than
inferred from the bars. At rank one the two elementary contributions sum
to less than \(1/2100\) of \(\mathcal T\), as shown below the diagram.
Solution 62 checks the factorial bound, branch conversion and rational
example.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_application_height_profile.py}{Figure
program},
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_application_height_certificate.py}{exact
calculation}. Human-source context: Section 8 of
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
free 2013 paper}; the height conversion here uses the proved programme
singleton bound.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{49. A weighted application bound for dependent
bases}\label{a-weighted-application-bound-for-dependent-bases}

Keep the field and prime hypotheses of Theorem 10.154. Let
\(a_1,\ldots,a_n\in K^*\) be local units of multiplicative rank
\(r\ge1\) modulo roots of unity. They can now be dependent. Let
\(b_i\in\mathbb Z\), \(\Xi=\prod_i a_i^{b_i}\ne1\),
\(B\ge\max\{3,|b_1|,\ldots,|b_n|\}\), and \(H=\max\{\ln B,t\}\). Sort
the indexed list by nondecreasing height, keeping a fixed ordering at
equal heights. Define

\[
\begin{aligned}
\varkappa&=\begin{cases}
18,&\mathrm{I,II,IV},\\
9(p-1)/(p-2),&\mathrm{III},\\
34,&\mathrm V,
\end{cases}\\
X_n&=\max\{n,t\},\qquad
L_n=\frac{X_n}{\varkappa(n+5)d},\qquad
H_n(a)=\max\{h(a),L_n\}.
\end{aligned}
\tag{10.470}
\]

Choose \(\boldsymbol\beta\) by reading this sorted list and retaining a
base exactly when it increases the multiplicative rank. This greedy
basis has r elements. Its residue index \(\delta_\beta\) is saturated
when \(r\ge2\), and is the original rank-one residue index when \(r=1\),
as in (10.272). All products below are over the indicated indices, so
repeated bases are counted correctly in the complementary product. Set

\[
\begin{aligned}
\Omega_n&=\prod_{a\in\boldsymbol\beta}h(a)
              \prod_{a\notin\boldsymbol\beta}H_n(a),\\
C_n(\boldsymbol\beta)&=c\mathfrak a^n
\frac{n^n(n+1)^{n+2}d^{n+2}\ell_d}{n!q^ut}
\max\left\{\frac{p^f}{\delta_\beta t^{n+1}},
                  \frac{\mathrm e^n}{n^n}\right\}A_n,\\
\mathcal T_n&=C_n(\boldsymbol\beta)\Omega_nH,
\qquad \lambda=2099/2100,\\
\mathcal W_n&=\lambda\frac{t}{nd}
 C_n(\boldsymbol\beta)\frac{\Omega_n}{h(a_n)}.
\end{aligned}
\tag{10.471}
\]

The n in the constant is the length of the original list. Its residue
index belongs to the specified r-element basis. We will preserve that
basis and index during the proof.

\textbf{Lemma 10.155 (weighted height and coefficient bounds).} The
greedy basis minimizes \(\Omega_n\) over all independent r-element
sublists. For \(n\ge2\),

\[
\mathcal T_n>2100\frac d t\ln2,
\qquad \mathcal W_n>\mathrm e^nd,
\qquad \mathcal W_n>w_K.
\tag{10.472}
\]

Suppose additionally \(r<n\), \(a_1\) is non-torsion and
\(B/\ln B>\mathcal W_n\). Let x be the first index discarded by the
greedy procedure. Its minimal circuit has an exact integer relation with
positive coefficient at \(a_x\), and every coefficient has absolute
value at most K, where

\[
K<\begin{cases}
B/8,&x<n,\\
BdH_n(a_x)/8,&x=n.
\end{cases}
\tag{10.473}
\]

The rank factor needed after removal satisfies, for every \(n\ge2\),

\[
J_n:=\frac{(n+1)^{n+2}}{(n-1)^{n-1}n^2}
       \left(\frac{n+4}{n+5}\right)^{n-2}
>\mathrm e(n+5).
\tag{10.474}
\]

\textbf{Proof.} Rank is ordinary linear rank in the rational space of
formal exponent vectors modulo rational relations whose products are
roots of unity. Clearing denominators proves that these relations form a
subspace. Independence modulo torsion is the same as multiplicative
independence, because a torsion product can be raised to its finite
order. This supplies the elementary linear algebra used by the greedy
procedure.

Write its selected indices as \(i_1<\cdots<i_r\). Any independent
sublist with indices \(j_1<\cdots<j_r\) has \(j_k\ge i_k\): otherwise it
would give k independent vectors before the greedy list first attained
rank k. The function \(h/\max\{h,L_n\}\) increases with h. Factoring
every candidate product as the common product of all \(H_n(a_i)\), times
these r ratios, proves the minimum assertion. Removing a discarded
dependent base or a torsion base leaves the ordered spans unchanged.
Thus it also leaves the greedy basis unchanged, including when the floor
becomes \(L_{n-1}\).

For an independent sublist of length \(j\ge1\), (10.55) gives

\[
\prod h(a)\ge\frac{w_K}{\rho d^{j+1}\ell_d}
              \frac{j^j}{j!\mathrm e^j}.
\tag{10.475}
\]

We will also use this at j=0, with the empty product and the last ratio
both defined to be one. To justify it, apply the untwisted count of
Theorem 10.17 at \(H_0>1\) to the height-zero roots, then let \(H_0\)
decrease to one. It gives \(w_K\le30d\ln d\) for \(d\ge2\), and
\(w_K\le17\) for \(d=1\). In both cases \(w_K\le\rho d\ell_d\),
precisely the required empty-list bound.

To bound \(\mathcal W_n\), remove \(a_n\). If it is selected, put k=r;
otherwise put k=r+1 and use \(H_n(a_n)/h(a_n)\ge1\). In each case
\(1\le k\le n\). Apply the height bound at length k-1, and the floors to
the n-k remaining complementary elements. The second entry of the
maximum in \(C_n\) yields

\[
\begin{aligned}
\mathcal W_n\ge{}&\lambda\frac c\rho\frac{w_K}{q^u}
\frac{dA_n}{n}\,
\mathfrak a^n\mathrm e^{n-k+1}
\frac{(n+1)^{n+2}}{n!}\\
&{}\times\left(\frac{X_n}{\varkappa(n+5)}\right)^{n-k}
\frac{(k-1)^{k-1}}{(k-1)!}.
\end{aligned}
\tag{10.476}
\]

The case constants give \(\mathfrak a/\varkappa\ge13/17>3/4\), and
\(\mathrm e>8/3\), so \(\mathfrak a\mathrm e/\varkappa>2\). Also
\([n/(n+5)]^{n-k}>\mathrm e^{-5}\), since \(\ln(1+5/n)<5/n\). Use the
factorial bound of Lemma 10.153 and \((n+1)^2/n^{3/2}\ge3\) to obtain

\[
\mathcal W_n>
\frac{\lambda(200/58)\,105}{(11/4)^5}
\frac{w_K}{q^u}d\mathrm e^n
>2\frac{w_K}{q^u}d\mathrm e^n.
\tag{10.477}
\]

For the rank factor just used, the derivative of \((x+1)^2/x^{3/2}\) is
\((x+1)(x-3)/(2x^{5/2})\). Its minimum at three is \(16/(3\sqrt3)>3\),
because \(256>243\). The displayed rational prefactor is greater than
two by exact multiplication. Since \(w_K\ge q^u\) and Lemma 10.73 gives
\(q^u\le2d\), both assertions about \(\mathcal W_n\) follow.

For \(\mathcal T_n\), use the height bound at length r and all n-r
complementary floors. Put \(z_n=2n/(n+5)\). Since \(7>z_n\),
\(\mathfrak a\mathrm e/\varkappa>2\) and \(X_n\ge n\), the resulting
bound is

\[
\frac{\mathcal T_n}{dH/t}
\ge\frac{200}{58}\,5\cdot7\,
\frac{(n+1)^{n+2}}{n!}\,z_n^{n-1}.
\tag{10.478}
\]

Consecutive ratios of \((n+1)^{n+2}/n!\) exceed two, so it is at least
\((81/2)2^{n-2}\). At n=2 the remaining factor \(2^{n-2}z_n^{n-1}\) is
\(4/7\); at three and four it is \(9/8\) and \(2048/729\); at all n at
least five it is at least eight. Thus the preceding coefficient is at
least \(81000/29>2100\). Use \(H>1\) and \(\ln2<1\) to prove the
assertion about \(\mathcal T_n\).

For the circuit assertion, let \(a_1,\ldots,a_m\) be the longest initial
independent block, and set x=m+1. Choose a minimal dependent sublist
containing \(a_x\) and v earlier selected bases, with \(1\le v\le r\).
Let S be these v bases. Theorem 10.20 gives an exact integer relation,
and its sign can make the coefficient at x positive. Every earlier
height is at most \(h(a_x)\), so its coefficients satisfy

\[
K\le\rho d^{v+1}\ell_d\frac{v!\mathrm e^v}{v^v}
\frac{h(a_x)\prod_{a\in S}h(a)}{\min_{a\in S}h(a)}.
\tag{10.479}
\]

Put j=r-v+1 and s=n-r-1. After dividing by \(\Omega_n\), the height
denominator is the smallest-height member of S and the basis elements
outside S, an independent list of length j. Therefore

\[
\frac K{\Omega_n}\le
\frac{\rho^2d^{r+3}\ell_d^2}{w_K}\,
\mathrm e^{r+1}\frac{v!j!}{v^vj^j}
\frac{h(a_x)}{H_n(a_x)}L_n^{-s}.
\tag{10.480}
\]

Normalize using \(B>\mathcal W_n\ln B\). In the terminal case x=n,
divide additionally by \(dH_n(a_x)\). The normalized ratio of 8K to the
claimed coefficient bound is at most

\[
\frac{8n\rho^2}{\lambda c\mathfrak a^{r+1}(n+1)^2}
\left[\frac{\varkappa}{\mathfrak a\mathrm e}
             \left(1+\frac5n\right)\right]^s.
\tag{10.481}
\]

This uses \(q^u/w_K\le1\), \(\ell_d/A_n\le1\), \(\ln B>1\),
\(n!\le n^n\), \(v!j!/(v^vj^j)\le1\) and \(h(a_x)/H_n(a_x)\le1\). Every
degree power cancels, since r+3+s=n+2. If x\textless n and \(a_n\) is
selected, it is outside S. Cancel its height against the \(h(a_n)\) in
\(\mathcal W_n\), and use the height bound at length j-1. This yields
the same envelope multiplied by \(1/\mathrm e\); j=1 is covered by the
empty-list bound. If \(a_n\) is not selected, cancel its height against
its modified height. Use s-1 floors to obtain the envelope with exponent
s-1 and the extra factor \(1/(\mathfrak a\mathrm e)\). These
alternatives exhaust the internal case, since S lies before
x\textless n.

We have \(\varkappa/(\mathfrak a\mathrm e)<1/2\). For \(0\le s\le n-2\),
\([(1+5/n)/2]^s\le4/3\): the exponent is zero at n=2; at n=3 it is at
most one; at n=4 use \((9/8)^2<4/3\); at n at least five the base is at
most one. Also \(n/(n+1)^2\le2/9\). Every normalized ratio is therefore
strictly less than

\[
\frac{64\cdot58^2}{27\lambda\cdot200\cdot7^2}
=\frac{107648}{132237}<1.
\tag{10.482}
\]

This proves both circuit bounds, including the field height and all
relation coefficients. The
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_dependent_application_enclosures.py}{exact
calculation} records these rational endpoints.

Finally, for \(2\le n\le15\), direct rational evaluation gives
\(J_n/(n+5)>11/4>\mathrm e\). For n at least sixteen put y=1/n.
Integrating the geometric-series remainders gives
\(\ln(1+x)\ge x-x^2/2\), \(\ln(1+x)\le x-x^2/2+x^3/3\) for x
nonnegative, and \(-\ln(1-y)\ge y+y^2/2\). Substitute them into

\[
\begin{aligned}
\ln\frac{J_n}{n+5}={}&(n+2)\ln(1+y)-(n-1)\ln(1-y)\\
&{}+(n-2)\ln(1+4y)-(n-1)\ln(1+5y).
\end{aligned}
\tag{10.483}
\]

The result is at least \(1+(5/2)y-(119/3)y^2+(125/3)y^3\). Since
\(y\le1/16\), it is greater than \(1+y/48\). Exponentiation proves the
rank-factor assertion for all remaining n. \(\square\)

\textbf{Theorem 10.156 (the dependent-list main application estimate).}
For the stated local units, coefficients and specified greedy basis,

\[
\operatorname{ord}_{\mathfrak p}(\Xi-1)
<C_n(\boldsymbol\beta)\Omega_n\max\{\ln B,t\}.
\tag{10.484}
\]

\textbf{Proof.} Induct on n.~The full-rank case r=n, including n=1, is
Theorem 10.154. Suppose r\textless n and n at least two. If
\(B/\ln B\le\mathcal W_n\), the local height inequality gives
\(\operatorname{ord}_{\mathfrak p}(\Xi-1)
\le(d/t)(nBh(a_n)+\ln2)<\mathcal T_n\): its two contributions are at
most \(\lambda\mathcal T_n\) and strictly less than
\(\mathcal T_n/2100\), by Lemma 10.155.

Otherwise \(B>\mathcal W_n\ln B>\mathcal W_n\), so \(\ln B>n+\ln d\),
\(B>w_K\) and \(H\ge X_n\). First suppose \(a_1\) is non-torsion. Use
the circuit at the first discarded index x, with positive coefficient
\(k_x\) at x and coefficients bounded by K. Raising \(\Xi\) to \(k_x\)
eliminates \(a_x\). Its new integer coefficients have absolute value at
most 2BK. The greedy basis and its residue index are unchanged. If
\(\Xi^{k_x}=1\), then \(\Xi\) is a root of unity; Theorem 10.21 gives
\(\operatorname{ord}_{\mathfrak p}(\Xi-1)\le(d/t)\ln2<\mathcal T_n\).

Otherwise put \(z=dH_n(a_x)\). The new coefficients are at most
\(B^2/4\) for x\textless n, and \(B^2z/4\) for x=n. For every positive
u, \(\ln u\le u/\mathrm e\): differentiating \(\ln u/u\) locates its
maximum at \(u=\mathrm e\). A valid new bound is therefore
\(B'=B^2\exp(z/(4\mathrm e))\), and its corresponding parameter
satisfies

\[
H'\le H\left(2+\frac z{4\mathrm eX_n}\right).
\tag{10.485}
\]

The two entries in the maximum defining \(C_n\) have ratios \(1/t\) and
\(\mathrm e(n-1)^{n-1}/n^n>1/n\) compared with \(C_{n-1}\). The latter
inequality follows from \((n-1)\ln(1-1/n)>-1\). Their maximum thus has
ratio at least \(1/X_n\), and \(A_n\ge A_{n-1}\). Exact cancellation
gives

\[
\frac{C_n(\boldsymbol\beta)}{C_{n-1}(\boldsymbol\beta)}
\ge\mathfrak a\frac d{X_n}
\frac{(n+1)^{n+2}}{(n-1)^{n-1}n^2}.
\tag{10.486}
\]

Let \(\Omega'\) be the modified product of the remaining list with its
new floor and the same basis. Each surviving modified height has ratio
at least \((n+4)/(n+5)\), because \(X_n\ge X_{n-1}\). Consequently

\[
\frac{\Omega_n}{\Omega'}
\ge H_n(a_x)\left(\frac{n+4}{n+5}\right)^{n-r-1}
\ge H_n(a_x)\left(\frac{n+4}{n+5}\right)^{n-2}.
\tag{10.487}
\]

The product of these two ratios exceeds
\(\mathfrak a\mathrm e(n+5)z/X_n\), by Lemma 10.155. Since
\(z/X_n\ge1/(\varkappa(n+5))\),

\[
\begin{aligned}
\left[\mathfrak a\mathrm e(n+5)-\frac1{4\mathrm e}\right]\frac z{X_n}
&\ge\frac{\mathfrak a\mathrm e}{\varkappa}
       -\frac1{4\mathrm e\varkappa(n+5)}\\
&>\frac{104}{51}-\frac1{672}>2.
\end{aligned}
\tag{10.488}
\]

This uses \(\mathfrak a/\varkappa\ge13/17\), \(\mathrm e>8/3\),
\(\varkappa\ge9\) and n+5 at least seven. Thus the original
\(C_n\Omega_n\) absorbs the complete increase of the coefficient-height
parameter. Apply induction to \(\Xi^{k_x}\), and the factored-power
inequality \(\operatorname{ord}_{\mathfrak p}(\Xi-1)
\le\operatorname{ord}_{\mathfrak p}(\Xi^{k_x}-1)\), to prove the claim.

If \(a_1\) is torsion, raise \(\Xi\) to its order \(w\le w_K\). This
removes \(a_1\) and preserves the basis. If the resulting power is one,
use the torsion bound above. Otherwise its coefficients are bounded by
\(wB<B^2\), so \(H'\le2H\). The product-ratio argument, now with deleted
height \(H_n(a_1)=L_n\), gives a ratio greater than
\(\mathfrak a\mathrm e/\varkappa>104/51>2\). Induction and the same
factored-power inequality conclude the proof. \(\square\)

For an indexed example, take \(K=\mathbb Q,p=5\) and the sorted list
\((2,3,4)\). Prime valuations at two and three give vectors
\((1,0),(0,1),(2,0)\); thus the greedy basis is \((2,3)\) and r=2. Its
q-primary saturation has no extra rational exponent, again by prime
valuations. The residue of two generates all four nonzero classes at
five, so \(\delta_\beta=1\). Here \(\varkappa=12\), \(L_3=1/32\), and
every actual height exceeds this floor. Hence

\[
\Omega_3=(\ln2)(\ln3)(\ln4),\qquad
4\cdot2^{-2}=1,\qquad
2^1 3^1 4^2=2^5 3^1=96.
\tag{10.489}
\]

Both sides have valuation \(v_5(96-1)=1\). With B=3,
\(B/\ln B<3<8<\mathcal W_3\), by Lemma 10.155, so this example is in the
local-height branch. The exact circuit identity also illustrates how the
coefficient transfer keeps the selected basis and its residue index. The
full list still has length three in \(C_3\), while its basis has rank
two. For a list of rank zero, \(\Xi\ne1\) is torsion and Theorem 10.21
directly gives \(\operatorname{ord}_{\mathfrak p}(\Xi-1)\le(d/t)\ln2\).

Theorem 10.156 proves the main dependent-list inequality corresponding
to Theorem 1 of
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
free 2013 paper}, Section 8. Its statement uses the specified greedy
basis, which attains the minimum weighted product. The factor \(1/2100\)
established in Theorem 10.154 applies to an independent list of length
one.

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={A weighted basis transfer and universal circuit coefficient bounds}]{../figures/yu-dependent-application.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.42. Left: exact prime-valuation vectors of the rational
list \((2,3,4)\) give the greedy basis \((2,3)\); the circuit
\(4\cdot2^{-2}=1\) transfers exponents \((1,1,2)\) to \((5,1)\). Its
actual value is 96, of valuation one at five. This B=3 example uses the
local-height branch. Right: proved upper bounds for the normalized
circuit coefficient cost in the large-coefficient branch, for all
original fields, ranks and circuit sizes. Lemma 10.155 treats terminal
removal, internal removal with the largest-height base selected, and
internal removal with that base unselected. Each lies below one. Theorem
10.156 then performs the rank-preserving induction, including torsion.
Solution 63 checks the coefficient, height-floor and scalar-rank
cancellations.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_dependent_application_profile.py}{Figure
program},
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_dependent_application_enclosures.py}{exact
universal envelopes}. Human-source context: Section 8 of
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
freely accessible paper}.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{50. The sharper bound for a dependent list of rank
one}\label{the-sharper-bound-for-a-dependent-list-of-rank-one}

Retain the units, sorted indexed list, greedy basis, heights and
constants of Section 49. Suppose its multiplicative rank is one. Write
\(i_0\) for the selected index, \(\beta=a_{i_0}\) for its first
non-torsion base, and \(g=|\langle\overline\beta\rangle|\). Its residue
index is the original one, \(\delta_\beta=(p^f-1)/g\). Put

\[
R=\prod_{i\ne i_0}\frac{H_n(a_i)}{L_n}\ge1,
\qquad U=\frac{t}{d}\mathcal T_n.
\tag{10.490}
\]

The product is over indices other than the selected index of \(\beta\),
including roots of unity and repetitions. A direct reduction to one base
will also handle coefficients that are too small for the
large-coefficient branch of Theorem 10.156.

\subsubsection{Two uniform budgets}\label{two-uniform-budgets}

For \(n\ge2\) define

\[
\begin{aligned}
F_n&=\frac c\rho\frac{w_K}{q^u}\mathfrak a^n\mathrm e^{n-1}
\frac{(n+1)^{n+2}}{n!}
\left(\frac{n}{\varkappa(n+5)}\right)^{n-1},\\
E_n&=\frac{4n!\varkappa^{n-1}(n+5)^{n-1}}
{c\mathfrak a^n n^n(n+1)^{n+2}},\\
\alpha_n&=\left(\frac1{2100}-\frac1{40000}\right)F_n.
\end{aligned}
\tag{10.491}
\]

\textbf{Lemma 10.157 (rank-one scalar budgets).} These quantities
satisfy

\[
U\ge F_nA_nHR,
\qquad \frac{2eg h(\beta)}U<\frac1{40000},
\qquad F_{n+1}>3F_n.
\tag{10.492}
\]

At \(n=2\), the following bounds hold. In the last column the field
degree and \(\ell_d\) are already included.

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}
  >{\raggedright\arraybackslash}p{(\linewidth - 4\tabcolsep) * \real{0.3333}}
  >{\raggedright\arraybackslash}p{(\linewidth - 4\tabcolsep) * \real{0.3333}}
  >{\raggedright\arraybackslash}p{(\linewidth - 4\tabcolsep) * \real{0.3333}}@{}}
\toprule\noalign{}
\begin{minipage}[b]{\linewidth}\raggedright
Field case
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
Lower bound for \(\alpha_2\)
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
Upper bound for \(E_2/(d\ell_d)\)
\end{minipage} \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
I.1 & \(9/4\) & \(1/118314\) \\
I.2 & \(9/4\) & \(1/40068\) \\
II & \(4/3\) & \(1/63630\) \\
III.1 & \(9/4\) & \(1/113022\) \\
III.2 & \(9/4\) & \(1/56385\) \\
IV & \(9/4\) & \(1/337680\) \\
V & \(19/10\) & \(119/5639868\) \\
\end{longtable}
}

Every lower bound is strict, and every last entry is less than
\(1/40000\). For \(n\ge3\), \(\alpha_n>4n/3\).

\textbf{Proof.} The singleton case of (10.55) gives
\(h(\beta)\ge w_K/(\rho\mathrm e d^2\ell_d)\). Insert this and
\(\Omega_n=h(\beta)L_n^{n-1}R\) into \(U\), using the second entry of
the maximum in \(C_n\) and \(X_n\ge n\). All degree and height-floor
powers cancel to give the first inequality.

For the second inequality use the first entry instead. The original
residue order gives \(p^f/\delta_\beta>g\), while
\(L_n\ge t/[\varkappa(n+5)d]\). Exact division yields

\[
\frac{2eg h(\beta)}U
<\frac{2e q^u n!\varkappa^{n-1}(n+5)^{n-1}t^2}
{c\mathfrak a^n n^n(n+1)^{n+2}d^2\ell_dRA_nH}
\le\frac{E_n}{d\ell_dR}.
\tag{10.493}
\]

The last step uses \(eq^u\le2d\) from Lemma 10.73, and \(t^2\le A_nH\).
At two, \(E_2=14\varkappa/(81c\mathfrak a^2)\). In I.1, II, III.1, IV
and V use \(d\ge2\); in the other two cases use \(d=1\). Also
\(\ell_d\ge1\). Direct substitution of the seven case constants gives
the last column. In III, \(\mathfrak a^2/\varkappa\ge49/9\), including
every prime in that case.

For the middle column, use \(\mathrm e>27/10\), obtained by summing its
positive exponential series through degree five. At odd primes
\(w_K/q^u\ge1\). At two, \(-1\) together with the q-primary roots has
order \(2q^u\), so \(w_K/q^u\ge2\). Thus

\[
F_2>\frac c\rho\frac{w_K}{q^u}
\frac{\mathfrak a^2}{\varkappa}\frac{2187}{70}.
\tag{10.494}
\]

Use \(\rho=17\) in degree one and \(\rho=58\) otherwise. Substitution
gives the stated strict rational bounds on \(\alpha_2\). The
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_rank_one_refinement_enclosures.py}{exact
scalar calculations} include each difference as a positive fraction.

The ratio of consecutive \(F_n\) is

\[
\frac{F_{n+1}}{F_n}=
\frac{\mathfrak a\mathrm e}{\varkappa}
\left(1+\frac1{n+1}\right)^{n+3}
\frac{n+1}{n+6}
\left(\frac{(n+1)(n+5)}{n(n+6)}\right)^{n-1}.
\tag{10.495}
\]

We have \(\mathfrak a\mathrm e/\varkappa>104/51\). For \(n=2,3,4,5\),
the rational product after this factor is, respectively,

\[
\frac{56}{27},\quad\frac{15625}{6561},\quad
\frac{1594323}{625000},\quad\frac{11529602}{4348377}.
\tag{10.496}
\]

Multiplying each by \(104/51\) gives a number greater than three. For
all \(n\ge6\), the first power in the ratio is greater than
\(\mathrm e\), by \(\ln(1+x)>x/(1+x)\). The last power is greater than
one and \((n+1)/(n+6)\ge7/12\). The ratio is therefore greater than
\((104/51)(8/3)(7/12)=1456/459>3\). This proves the claim at every rank.

Direct cancellation also gives
\(E_{n+1}/E_n=\mathrm e nF_n/[(n+1)F_{n+1}]<11/12\). Consequently the
last column bounds every \(E_n/(d\ell_dR)\). Finally \(\alpha_2>4/3\) in
every row, so \(\alpha_n>4\cdot3^{n-3}\ge4n/3\) for \(n\ge3\). The
latter inequality follows by induction, starting at three. \(\square\)

\subsubsection{Simultaneous reduction to the selected
base}\label{simultaneous-reduction-to-the-selected-base}

\textbf{Theorem 10.158 (the rank-one application refinement).} Under the
hypotheses of Theorem 10.156, if the list has rank one, then

\[
\operatorname{ord}_{\mathfrak p}(\Xi-1)
<\frac1{2100}C_n(\beta)\Omega_n\max\{\ln B,t\}.
\tag{10.497}
\]

\textbf{Proof.} At \(n=1\) this is Theorem 10.154. Let \(n\ge2\). Put

\[
D_n=\rho\mathrm e d^2\ell_dL_n,
\qquad D_0=\max\{1,D_n\}.
\tag{10.498}
\]

For each non-torsion base at an index \(i\ne i_0\), Theorem 10.20
applied to the dependent pair \((\beta,a_i)\) supplies an exact relation
\(\beta^{u_i}a_i^{k_i}=1\), with \(k_i\) a positive integer and
\(k_i\le\rho\mathrm e d^2\ell_dh(a_i)\). Here \(h(\beta)\le h(a_i)\), so
the smaller height in that theorem is \(h(\beta)\). Taking heights in
the relation gives \(|u_i|/k_i=h(a_i)/h(\beta)\).

Let m be the number of these non-torsion complementary indices. If the
original list has no torsion bases, put \(P=\prod k_i\). Otherwise put
\(P=w_K\prod k_i\). Every torsion base then disappears, and the
relations give

\[
\Xi^P=\beta^M,
\qquad M\in\mathbb Z,
\qquad |M|\le nBP\frac{h(a_n)}{h(\beta)}.
\tag{10.499}
\]

This is an identity in K, with no change in the residue index. If the
power is one, \(\Xi\) is torsion. Theorem 10.21 and
\(\mathcal T_n>2100(d/t)\ln2\) from Lemma 10.155 give the required
bound. Hence assume M is nonzero.

Each \(k_i\le D_0H_n(a_i)/L_n\), so the product of these coefficients is
at most \(D_0^mR\). If there is a non-torsion complementary base, the
largest height occurs among the complementary indices, or equals the
selected height. The singleton height bound then gives

\[
\frac{h(a_n)}{h(\beta)}
\le\max\{1,D_n/w_K\}R.
\tag{10.500}
\]

When there is no such base, this ratio is one. Since \(\ln B\le H\), the
resulting bounds for \(\ln(2|M|)\) are

\[
\begin{cases}
Z_0+2\ln R,&\text{no torsion bases},\\
Z_1+2\ln R,&\text{torsion and a non-torsion complement},\\
H+\ln(2w_K),&\text{only torsion in the complement},
\end{cases}
\tag{10.501}
\]

where

\[
\begin{aligned}
Z_0&=H+\ln(2n)+(n-1)\ln D_0+\ln\max\{1,D_n/w_K\},\\
Z_1&=H+\ln(2n)+(n-2)\ln D_0+\ln\max\{w_K,D_n\}.
\end{aligned}
\tag{10.502}
\]

We show that all three are less than \(\alpha_nA_nHR\). The empty-list
height bound in Lemma 10.155 gives \(w_K\le\rho d\ell_d\). Since
\(D_0\ge D_n\),

\[
\frac{w_K}{D_0}\le\frac{\varkappa(n+5)}{\mathrm eX_n}<45.
\tag{10.503}
\]

Indeed \(\varkappa\le34\), \(X_n\ge n\), \((n+5)/n\le7/2\) and
\(\mathrm e>8/3\). Every displayed alternative is thus at most
\(Z_n+2\ln R\), where

\[
Z_n=H+\ln(2n)+n\ln D_0+\ln45.
\tag{10.504}
\]

For \(n\ge3\), use \(X_n\le nH\), \(H>1\), and
\(\rho\mathrm e/\varkappa<18\). Concavity of the logarithm at eight
gives \(\ln\ell_d\le(\ln d)/8+11/10\): if \(d\ge3\) use
\(\ln x\le x/8+\ln8-1\), and if \(d=1,2\) the left side is zero. Also
\(\ln H\le H-1\), \(\ln18<3\) and \(\ln45<4\). Consequently

\[
Z_n<(n+1)H+\frac{31n}{10}+\ln(2n)+4
          +\frac{9n}{8}\ln d.
\tag{10.505}
\]

By Lemma 10.157, \(\alpha_n>4n/3\); since \(H>1\), its coefficient of
\(\ln d\) is greater than \(9n/8\). For the remaining terms use
\(A_n\ge4+\ln(n+1)+\ln d\), \(\ln(n+1)\ge\ln4>4/3\), and \(\ln(2n)<n\).
At H=1 their comparison reduces to

\[
\frac{64n}{9}>\frac{51n}{10}+5\qquad(n\ge3).
\tag{10.506}
\]

The difference is \(181n/90-5>0\). Increasing H only increases the
difference, because \(\alpha_n[4+\ln(n+1)]>n+1\). Thus
\(Z_n<\alpha_nA_nH\) throughout this infinite range.

At \(n=2\) there are just two possibilities. If the complementary base
is non-torsion, the first alternative is at most
\(H+\ln4+2\ln D_0+2\ln R\). Here \(D_0\le s d\ell_dH\), where
\(s=\max\{1,2\rho\mathrm e/(7\varkappa)\}\). In II,
\(s<319/126<\mathrm e\); in every case \(s<319/63<\exp(5/3)\). The
logarithmic tangent at sixteen gives \(\ln\ell_d\le(\ln d)/16+9/5\).
Using \(\ln4<7/5\) and \(\ln H\le H-1\) bounds its part independent of R
by

\[
\begin{cases}
3H+5+(17/8)\ln d,&\mathrm{II},\\
3H+19/3+(17/8)\ln d,&\text{all other cases}.
\end{cases}
\tag{10.507}
\]

In II, \(H\ge\ln5>8/5\) and \(\alpha_2>4/3\). In V, \(f\ge2\), hence
\(H\ge2\ln2>4/3\), and \(\alpha_2>19/10\). In all remaining cases
\(H>1\) and \(\alpha_2>9/4\). These give \(\alpha_2H>17/8\). Since
\(4+\ln3>5\), the constant comparisons in II and the other cases follow
from \((20/3-3)(8/5)>5\) and \(19/2-3>19/3\), respectively. Thus this
alternative is less than \(\alpha_2A_2H+2\ln R\).

If the complement is torsion, its bound is \(H+\ln(2w_K)\). The same
tangent, with \(\ln\rho<5\), gives at most \(H+15/2+(17/16)\ln d\). The
coefficient comparison already established is stronger than needed. For
the constant part, in II use \((20/3-1)(8/5)>15/2\), and otherwise use
\(19/2-1>15/2\). This proves the comparison in the second possibility
too. All numerical logarithm comparisons used here follow from the
positive series in Solution 64; they require no approximation to be
accepted as an equality.

In every case the bound before the \(2\ln R\) term is less than
\(\alpha_nA_nH\), which is greater than two. The inequality
\(\ln R\le R-1\), for \(R\ge1\), therefore proves

\[
\ln(2|M|)<\alpha_nA_nHR
\le\left(\frac1{2100}-\frac1{40000}\right)U.
\tag{10.508}
\]

Apply Theorem 10.21 to \(\beta^M\), using its actual residue order g.
Factoring the P-th power difference gives

\[
\begin{aligned}
\operatorname{ord}_{\mathfrak p}(\Xi-1)
&\le\operatorname{ord}_{\mathfrak p}(\beta^M-1)\\
&\le\frac d t\left(\ln(2|M|)+2eg h(\beta)\right)
<\frac{\mathcal T_n}{2100},
\end{aligned}
\tag{10.509}
\]

by the two budgets. This includes every size of B and every torsion
pattern. \(\square\)

For example, at five in \(K=\mathbb Q\), the indexed list \((4,16)\) has
rank one and selected base four, with original residue order two and
\(\delta_\beta=2\). Its floor is \(L_2=1/42\), and
\(\Omega_2=(\ln4)(\ln16)\). For coefficients \((0,5^k)\), \(k\ge0\),
choose \(B=\max\{3,5^k\}\). The exact relation \(16\cdot4^{-2}=1\)
permits P=1 and gives \(\Xi=4^{2\cdot5^k}\). Lemma 9.5 proves
\(v_5(\Xi-1)=1+k\). The bound of Theorem 10.158 uses the length-two
constant and this weighted product, while keeping the raw residue order
of four. It does not replace it by the order of a saturated base.

The rank-one factor is also asserted in Theorem 1 of
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
freely accessible paper}. The proof above uses the exact integer
relations and the proved one-base power bound for the whole list; it
therefore also covers the small-coefficient case directly.

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={One-base reduction and the two strict rank-one budgets}]{../figures/yu-rank-one-refinement.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.43. Left: the dependent indexed list \((4,16)\) at five,
with coefficients \((0,5^k)\), reduces exactly to \(4^{2\cdot5^k}\); the
plotted valuations \(1+k\) are exact values, proved by the deep-unit
power law. Right: the normalized logarithmic exponent cost is strictly
below \(1/2100-1/40000\), and the original-residue height cost is
strictly below \(1/40000\). These are universal proved envelopes for
every rank-one list of length at least two, every field case and every
coefficient size, whose sum is \(1/2100\). Lemma 10.157 proves the
scalar budgets and Theorem 10.158 proves the full reduction, including
torsion.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_rank_one_refinement_profile.py}{Figure
program},
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_rank_one_refinement_enclosures.py}{exact
scalar calculations}. Human-source context: Theorem 1 and Section 8 of
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
freely accessible paper}.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{51. An effective p-adic distance between integral
S-units}\label{an-effective-p-adic-distance-between-integral-s-units}

Let S be a finite set of rational primes. An integral S-unit here is a
nonzero integer whose prime factors all belong to S; either sign is
allowed. Let x and y be distinct integral S-units, set
\(M=\max\{|x|,|y|\}\), and let p be any rational prime. Put

\[
g_0=\gcd(|x|,|y|),\qquad x_0=x/g_0,\qquad y_0=y/g_0,
\qquad S_p=S\setminus\{p\}.
\tag{10.510}
\]

The common factor must be retained. For example, \(x=3^N25,y=3^N7\) have
difference \(3^N18\), of valuation \(N+2\) at three. Their coprime parts
have fixed distance of valuation two, however large N is.

Write \(S_p=\{q_1<\cdots<q_n\}\). For \(n\ge1\) the following explicit
constants will bound the distance of the coprime parts. Use the field

\[
K_p=\begin{cases}\mathbb Q,&p>2,\\
\mathbb Q(\zeta_3),&p=2,\end{cases}
\qquad
(d,e,f,q^u)=\begin{cases}(1,1,1,2),&p>2,\\
(2,1,2,3),&p=2.
\end{cases}
\tag{10.511}
\]

Here \(\zeta_3\) is a primitive cube root of unity, and q is the
auxiliary prime of the earlier logarithm estimates, distinct from the
listed primes \(q_i\). Put \(t=f\ln p\). The rational prime bases are
independent, so set \(\boldsymbol\beta=(q_1,\ldots,q_n)\). Use their
saturated residue index when \(n\ge2\), and the raw index at \(n=1\).

These indices are entirely explicit. At odd p and \(n\ge2\), let \(G\)
be the subgroup of \(\mathbb F_p^*\) generated by \(-1,q_1,\ldots,q_n\);
then \(\delta_\beta=(p-1)/|G|\). At odd p and \(n=1\), use
\(\delta_\beta=(p-1)/\operatorname{ord}_p(q_1)\). At two, the indices
are one for \(n\ge2\) and three for \(n=1\). All these finite orders can
be found by enumerating a finite multiplicative group.

Let \(\epsilon=\operatorname{sgn}(x_0/y_0)\), and define

\[
\begin{aligned}
m&=\begin{cases}n,&\epsilon=1,\\n+1,&\epsilon=-1,\end{cases}\\
\Omega_\epsilon&=\begin{cases}
\displaystyle\prod_{i=1}^n\ln q_i,&\epsilon=1,\\
\displaystyle L_m\prod_{i=1}^n\ln q_i,&\epsilon=-1,
\end{cases}\\
\eta_n&=\begin{cases}1/2100,&n=1,\\1,&n\ge2,\end{cases}\\
K_{S,p}^{\epsilon}&=\eta_n C_m(\boldsymbol\beta)\Omega_\epsilon,\\
B_M&=\max\{3,\ln M/\ln2\},\qquad
H_{S,p}(M)=\max\{\ln B_M,t\}.
\end{aligned}
\tag{10.512}
\]

The \(L_m\), \(C_m\), \(\varkappa\) and case constants are those of
Section 49, computed in the specified field \(K_p\). Thus \(C_m\) uses
the original list length m and the index of its n-element basis. The
constant depends on S, p and the sign, but not on the exponents
occurring in x and y.

\textbf{Theorem 10.159 (integral S-unit distance).} If \(n\ge1\), then

\[
v_p(x-y)<v_p(g_0)+K_{S,p}^{\epsilon}H_{S,p}(M).
\tag{10.513}
\]

If \(n=0\), the exact excess \(v_p(x-y)-v_p(g_0)\) is zero unless
\(x_0=-y_0=\pm1\), in which case it is \(v_p(2)\).

\textbf{Proof.} Coprimality of \(|x_0|\) and \(|y_0|\) implies that at
most one is divisible by p.~If one is divisible by p, their difference
is a p-adic unit, so \(v_p(x-y)=v_p(g_0)\). For \(n\ge1\) the constant
is positive, proving the strict inequality in this case. At \(n=0\), all
prime factors of the coprime parts must be p.~If neither part is
divisible by p, their absolute values are both one; distinctness gives
the stated opposite sign case. This proves the entire \(n=0\) assertion.

Now assume \(n\ge1\) and both coprime parts are p-adic units. Their
ratio is

\[
\Xi=\frac{x_0}{y_0}=\epsilon\prod_{i=1}^n q_i^{b_i}\ne1,
\qquad b_i\in\mathbb Z,\qquad
|b_i|\le\frac{\ln M}{\ln q_i}\le\frac{\ln M}{\ln2}.
\tag{10.514}
\]

Indeed the nonnegative exponents in each of \(|x_0|,|y_0|\) are at most
\(\ln M/\ln q_i\), and their difference has absolute value at most their
maximum. Rational prime valuations show that the listed bases are
independent, even modulo roots of unity: a positive rational number that
is a root of unity is one, and each prime valuation then makes its
exponent zero. Their heights are \(h(q_i)=\ln q_i\).

We verify the field and index data. At odd p these are the rational
field data. Its q-primary saturation of prime bases has no additional
exponents: if a rational q-power root of their product exists, prime
valuations make all the indicated exponents integral. Root choices only
introduce the already available sign. Consequently the saturated residue
group is precisely G. The raw singleton index follows from its actual
residue order, as in (10.272).

At two the polynomial \(X^2+X+1\) has no rational root, so
\([K_p:\mathbb Q]=2\). Its reduction is irreducible over
\(\mathbb F_2\), since it vanishes at neither zero nor one. Thus the
residue degree is at least two. The degree inequality \(ef\le d=2\)
gives \(e=1,f=2\). Its root reduces to an element of order three in
\(\mathbb F_4^*\), so including the q-primary torsion generator makes
every saturated residue group at rank at least two equal to that whole
group, irrespective of a choice of saturated basis. Every odd rational
prime reduces to one, so a singleton's raw order is one and its index is
three. The field contains \(\zeta_3\), and cannot contain \(\zeta_9\),
whose degree six is proved by the cyclotomic argument of Lesson 9. Hence
\(q^u=3\). These also verify the two-adic field hypothesis required by
Theorems 10.154 and 10.156.

If \(\epsilon=1\), apply Theorem 10.154 to the n independent prime bases
and bound \(B=B_M\). Its weighted product is \(\prod_i\ln q_i\). The
factor \(\eta_1=1/2100\) is included by that theorem's singleton
conclusion.

If \(\epsilon=-1\), use the indexed list \((-1,q_1,\ldots,q_n)\), whose
length is m=n+1 and rank is n. Its first base has height zero, so the
greedy basis is exactly \(\boldsymbol\beta\), and its weighted product
is \(L_m\prod_i\ln q_i\). The coefficient of the sign base is one, which
is also bounded by \(B_M\). Apply Theorem 10.156, and apply its
sharpened rank-one form, Theorem 10.158, when n=1. Neither argument
changes the selected basis's residue index.

The fields used here both have ramification index e=1. Restricting their
normalized prime-ideal valuation to the rational numbers is therefore
exactly \(v_p\). Since \(y_0\) is a p-adic unit,

\[
v_p(x-y)-v_p(g_0)=v_p(x_0-y_0)
=v_p(\Xi-1)<K_{S,p}^{\epsilon}H_{S,p}(M).
\tag{10.515}
\]

This proves the claimed effective bound in every case. \(\square\)

\textbf{Corollary 10.160 (distance after removing the common factor).}
With \(|z|_p=p^{-v_p(z)}\) and \(n\ge1\),

\[
\frac{|x-y|_p}{|g_0|_p}
>\min\left\{p^{-K_{S,p}^{\epsilon}t},
B_M^{-K_{S,p}^{\epsilon}\ln p}\right\}.
\tag{10.516}
\]

\textbf{Proof.} Exponentiate the strict valuation inequality of Theorem
10.159. The negative exponential of the maximum of two real numbers is
the minimum of their negative exponentials. This gives exactly the two
terms displayed. \(\square\)

For fixed S and p, this lower bound is a fixed negative power of
\(\ln M\) once \(\ln B_M\ge t\). The normalization by the common factor
is necessary; the family at the start of the section proves that
directly.

For the positive example \(S=\{3,5,7\},p=3\), the prime bases are five
and seven. Together with minus one their residues generate
\(\mathbb F_3^*\), so \(\delta_\beta=1\). The data are
\(d=e=f=1,q^u=2,c=636,\mathfrak a=14\), \(t=\ln3\),
\(\Omega=(\ln5)(\ln7)\), and \(A_2=4+\ln3\). Thus the completely
specified constant is

\[
K_{S,3}^{+}=
\frac{636\cdot14^2\cdot81}{\ln3}
\max\left\{\frac3{(\ln3)^3},\frac{\mathrm e^2}4\right\}
(4+\ln3)(\ln5)(\ln7).
\tag{10.517}
\]

For \(x=3^N25,y=3^N7\), the coprime ratio is \(25/7\), with prime
exponents \((2,-1)\). Its difference from one is \(18/7\), of exact
valuation two. The original difference has valuation N+2. The theorem
bounds the excess two by a constant times \(H_{S,3}(3^N25)\), while
retaining the exact contribution N of the common factor.

For a signed example take \(x=25,y=-7,p=2\) and \(S=\{5,7\}\). The ratio
is \(-25/7\), represented by the list \((-1,5,7)\) with coefficients
\((1,2,-1)\). Here \(d=2,e=1,f=2,q^u=3,\delta_\beta=1\),
\(\varkappa=34,m=3\), and \(L_3=3/544\). The weighted product is
\((3/544)(\ln5)(\ln7)\), and the exact difference \(25+7=32\) has
valuation five at two. The sign requires the dependent-list theorem
rather than being silently discarded from the exponent vector.

The logarithmic-form input is the complete programme proof in Theorems
10.154, 10.156 and 10.158. Its scholarly context is Theorem 1 of
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
freely accessible paper}.

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={The common-factor term and the signed local-unit alternative}]{../figures/integral-s-unit-distance.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.44. Left: for \(x=3^N25,y=3^N7\), the exact valuation is
N+2. Its two components are the common-factor valuation N and the fixed
coprime distance of valuation two. The bars show exact integer values,
while Theorem 10.159 bounds the second component in general. Right: a
coprime nonunit case has zero excess valuation; the unit case is reduced
to \(\epsilon\prod q_i^{b_i}\). The example \(25-(-7)=32\) at two
requires the sign base minus one, the field \(\mathbb Q(\zeta_3)\), and
the length-three weighted height floor \(3/544\). Theorem 10.159 proves
the full sign and field conditions, Corollary 10.160 gives the
normalized distance, and Solution 65 checks both examples.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/integral_s_unit_distance.py}{Reproducible exact
calculation and figure}. Human-source context: Theorem 1 of
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
freely accessible paper}.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{52. The one-base case of Yu's second
constant}\label{the-one-base-case-of-yus-second-constant}

The second constant in Yu's free 2013 paper has a different rank
dependence from the first. Its one-base estimate follows directly from
the power bound already proved here. We give its full numerical constant
and keep the original residue order throughout.

Let \(K\) be a number field, \(\mathfrak p\) a prime above p, and
\(\beta\in K^*\) a non-torsion unit at \(\mathfrak p\). Write
\(d=[K:\mathbb Q]\), \(e=e_{\mathfrak p}\), \(f=f_{\mathfrak p}\),
\(t=f\ln p\), and \(\ell_d=\max\{1,\ln d\}\). At odd p put \(q=2\). At
\(p=2\), assume \(\zeta_3\in K\) and put \(q=3\). Let \(q^u\) be the
order of the q-primary roots of unity in K, and let \(w_K\) be the order
of all its roots of unity. The order g of the reduction of \(\beta\) is
its original residue order, without replacing \(\beta\) by a saturated
base. Put

\[
\delta_\beta=\frac{p^f-1}{g},\qquad
A_1=\max\{4+\ln(2d),e,t\},\qquad
a^{(2)}=\begin{cases}7,&p>2,\\13,&p=2.\end{cases}
\tag{10.518}
\]

Use the following exhaustive field cases. Their constants are the freely
accessible parameters (1.29)--(1.30) in Yu's paper.

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}llr@{}}
\toprule\noalign{}
Case & Field and prime hypotheses & \(c^{(2)}\) \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
I.1 & \(p=3,d>1\) & 1438 \\
I.2 & \(p=3,d=1\) & 648 \\
II & \(p=5,e\ge2\) & 690 \\
III.1 & \(p\ge5,e=1,d>1\) & \(495(p-1)/(p-2)\) \\
III.2 & \(p\ge5,e=1,d=1\) & \(557(p-1)/(p-2)\) \\
IV & \(p\ge7,e\ge2\) & 2418 \\
V & \(p=2,\zeta_3\in K\) & 406 \\
\end{longtable}
}

Take a nonzero integer b and \(B\ge\max\{3,|b|\}\). Set
\(H=\max\{\ln B,t\}\) and

\[
\mathcal T_2=
\frac{8c^{(2)}a^{(2)}\mathrm e d^3\ell_d}{q^ut^3}
\max\left\{\frac{p^f}{\delta_\beta},\mathrm e t^2\right\}
A_1h(\beta)H.
\tag{10.519}
\]

Here \(\mathrm e=\exp(1)\), distinct from the ramification index e. The
notation \(\mathcal T_2\) refers to the second Yu constant at list
length one. Substituting \(n=1\) into his formula (1.10) and multiplying
by \(n+1\) gives this expression: the factor \(P=p^\kappa\) cancels and
\(0!=1\).

\textbf{Theorem 10.161 (the second one-base application estimate).}
Under these hypotheses,

\[
\operatorname{ord}_{\mathfrak p}(\beta^b-1)
<\frac{\mathcal T_2}{4000}.
\tag{10.520}
\]

\textbf{Proof.} Put \(U_2=(t/d)\mathcal T_2\). The singleton height
bound (10.55) is \(h(\beta)\ge w_K/(\rho\mathrm e d^2\ell_d)\), with
\(\rho=17\) for \(d=1\) and \(\rho=58\) for \(d\ge2\). Using the second
entry of the maximum gives

\[
U_2\ge JA_1H,\qquad
J=\frac{8c^{(2)}a^{(2)}\mathrm e}{\rho}
\frac{w_K}{q^u}.
\tag{10.521}
\]

For the other contribution, the first entry gives
\(p^f/\delta_\beta=p^fg/(p^f-1)>g\). Exact division yields

\[
\begin{aligned}
\frac{2eg h(\beta)}{U_2}
&<\frac{eq^ut^2}
{4c^{(2)}a^{(2)}\mathrm e d^2\ell_dA_1H}\\
&\le\frac1{2c^{(2)}a^{(2)}\mathrm e d\ell_d}.
\end{aligned}
\tag{10.522}
\]

The last step uses \(t^2\le A_1H\) and \(eq^u\le2d\). The latter follows
from Lemma 10.73: \(d/e\ge q^{u-1}(q-1)\), while \(q/(q-1)\le2\).

At odd primes \(w_K/q^u\ge1\). At two, the roots \(-1\) and the
q-primary roots generate a group of order \(2q^u\), so \(w_K/q^u\ge2\),
as in Lemma 10.157. The positive exponential series gives
\(\mathrm e>27/10\). Also \(\ln2>2/3\), so \(A_1>14/3\) at degree one
and \(A_1>16/3\) at degree at least two.

Since \(B\ge3\) and \(2^3<3^2\), \(\ln(2B)<(5/3)H\). In III,
\(H\ge\ln5>8/5\) and \(\ln2<7/10\), so the sharper inequality is
\(\ln(2B)<(23/16)H\). These logarithm comparisons follow from the
positive series proved in Section 40 and Solution 48; equivalently, the
first four terms of the series for \(\ln5\), with argument \(2/3\),
already exceed \(8/5\).

The following table supplies all numerical endpoints. The column L is a
strict lower bound for \(U_2/H\). The last column is an upper bound for
the sum of the two contributions after division by \(U_2\).

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}lrrr@{}}
\toprule\noalign{}
Case & L & Logarithm multiplier \(\mu\) & Total upper bound \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
I.1 & \(2899008/145\) & \(5/3\) & \(115/1242432\) \\
I.2 & \(2286144/85\) & \(5/3\) & \(235/2286144\) \\
II & \(278208/29\) & \(5/3\) & \(1/5184\) \\
III.1 & \(199584/29\) & \(23/16\) & \(2257/9580032\) \\
III.2 & \(1965096/85\) & \(23/16\) & \(10345/94324608\) \\
IV & \(4874688/145\) & \(5/3\) & \(115/2089152\) \\
V & \(104832/5\) & \(5/3\) & \(295/3040128\) \\
\end{longtable}
}

To obtain each entry, replace \(\mathrm e\) by \(27/10\), \(\ell_d\) by
one, and d by its lower bound: one in I.2 and III.2, two in every other
case. Cases II and IV have \(d\ge e\ge2\); V has \(d\ge2\) because it
contains a primitive cube root. Use the corresponding value of \(\rho\),
the stated torsion ratio, and the lower bound for \(A_1\). In III,
\((p-1)/(p-2)>1\), so replacing this factor by one covers every prime in
that case. The entry in the last column is exactly

\[
\frac\mu L+
\frac1{2c_{\min}^{(2)}a^{(2)}(27/10)d_{\min}}.
\tag{10.523}
\]

All seven entries are less than \(1/4000\). Their smallest positive
difference from \(1/4000\) occurs in III.1 and is \(17251/1197504000\).
Multiplying positive denominators proves each comparison; the
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_second_singleton_budget.py}{exact
calculation} retains every fraction and every positive difference.

The complete one-base power estimate (10.60) now gives

\[
\begin{aligned}
\operatorname{ord}_{\mathfrak p}(\beta^b-1)
&\le\frac d t\bigl(\ln(2B)+2eg h(\beta)\bigr)\\
&<\frac d t\frac{U_2}{4000}
=\frac{\mathcal T_2}{4000}.
\end{aligned}
\tag{10.524}
\]

Non-torsion ensures \(\beta^b\ne1\), so every quantity in this argument
is defined. \(\square\)

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={Seven exact budgets for the second one-base Yu constant}]{../figures/yu-second-singleton-budget.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.45. The logarithm and original residue-order height
contributions are stacked after normalization by \(U_2/4000\). The bars
are uniform rational upper bounds for the seven exhaustive field cases,
and the red line is the allowance. The displayed decimals summarize the
exact fractions in the table. Proof: Theorem 10.161. Human source for
the target constant: equations (1.10), (1.29)--(1.30) of Yu's free 2013
paper, cited below.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_second_singleton_budget.py}{Reproducible
figure and exact certificate}.}
\end{flushleft}
\end{minipage}\par\medskip


For \(K=\mathbb Q,p=3,\beta=4,b=3^k\), \(k\ge1\), take \(B=3^k\). The
raw residue order is one, \(\delta_\beta=2\), \(q^u=2\), \(A_1=4+\ln2\),
and \(H=k\ln3\). The second entry of the maximum dominates:
\(\mathrm e(\ln3)^2>8/3>3/2=p^f/\delta_\beta\). Theorem 10.161 therefore
reads

\[
v_3(4^{3^k}-1)=k+1
<\frac{4\cdot648\cdot7}{4000}
\mathrm e^2(4+\ln2)k\ln4.
\tag{10.525}
\]

The exact valuation is the proved power identity in Lemma 9.5. The right
side exceeds \(126k\): its successive factors exceed \(9/2\), seven,
four, and one. This theorem proves the singleton case. The second
constant at greater list lengths, with its different modified-height
floors, requires its own proof.

\subsection{53. The second constant for every rank-one
list}\label{the-second-constant-for-every-rank-one-list}

The singleton estimate extends to a dependent list of any length whose
multiplicative rank modulo roots of unity is one. The height floor of
the second constant differs from that in Section 49, so we retain it
explicitly. The proof again reduces the whole list to one exact power in
K.

Use the field and prime conventions of Section 52. Take an indexed list
\(a_1,\ldots,a_n\) of local units, of multiplicative rank one modulo
roots of unity. Order its indices by nondecreasing height, with a fixed
order at equal heights, and let \(\beta=a_{i_0}\) be the first
non-torsion member. Let \(b_i\in\mathbb Z\),
\(\Xi=\prod_i a_i^{b_i}\ne1\), \(B\ge\max\{3,|b_1|,\ldots,|b_n|\}\), and
\(H=\max\{\ln B,t\}\). The residue index \(\delta_\beta\) belongs to
this original base, as defined in Section 52.

Let \(\kappa\ge0\) be the integer for which
\(p^{\kappa-1}(p-1)\le2e<p^\kappa(p-1)\), and put \(P=p^\kappa\). Then
\(1\le P\le2pe/(p-1)\le4d\). Set

\[
\begin{aligned}
\varkappa_2&=\begin{cases}25,&p>2,\\48,&p=2,\end{cases}
&X_n&=\max\{n,t\},\\
L_n^{(2)}&=\frac{X_n}{\varkappa_2Pdt},
&H_n^{(2)}(a)&=\max\{h(a),L_n^{(2)}\},\\
\Omega_n^{(2)}&=h(\beta)\prod_{i\ne i_0}H_n^{(2)}(a_i),
&A_n&=\max\{4+\ln((n+1)d),e,t\}.
\end{aligned}
\tag{10.526}
\]

This is precisely the floor in (1.25) and (1.29) of Yu's free paper.
Among the non-torsion choices of \(\beta\), the specified choice
minimizes \(\Omega_n^{(2)}\): factor out the common product of all
modified heights and use that \(h/\max\{h,L_n^{(2)}\}\) increases with
h. Repeated bases are separate indices in the product. Define the second
application constant and its complete budget by

\[
\begin{aligned}
C_n^{(2)}(\beta)&=
\frac{c^{(2)}(a^{(2)}\mathrm e P)^n}{P}
\frac{(n+1)^{n+2}d^{n+2}\ell_d}
{(n-1)!q^ut^3}\\
&\quad\times
\max\left\{\frac{p^f}{\delta_\beta},
\frac{\mathrm e^n t^{n+1}}{n^n}\right\}A_n,\\
\mathcal T_n^{(2)}&=C_n^{(2)}(\beta)\Omega_n^{(2)}H.
\end{aligned}
\tag{10.527}
\]

The constant is \((n+1)\) times formula (1.10) of that paper. Its
residue index is that of the one-element selected basis, even though n
is the length of the entire indexed list. At \(n=1\),
\(\mathcal T_1^{(2)}=\mathcal T_2\) of Section 52.

\subsubsection{Two scalar budgets}\label{two-scalar-budgets}

Suppose \(n\ge2\), and abbreviate \(c=c^{(2)}\), \(a=a^{(2)}\), and
\(\varkappa=\varkappa_2\). Put

\[
\begin{aligned}
R&=\prod_{i\ne i_0}\frac{H_n^{(2)}(a_i)}{L_n^{(2)}}\ge1,
&U&=\frac t d\mathcal T_n^{(2)},\\
F_n&=\frac c\rho\frac{w_K}{q^u}
\frac{a^n\mathrm e^{2n-1}(n+1)^{n+2}}
{n!\varkappa^{n-1}},\\
E_n&=\frac{4(n-1)!\varkappa^{n-1}}
{ca^n\mathrm e^n(n+1)^{n+2}},
&\alpha_n&=\left(\frac1{4000}-\frac1{100000}\right)F_n.
\end{aligned}
\tag{10.528}
\]

\textbf{Lemma 10.162 (second-constant rank-one scalar bounds).} For
every \(n\ge2\),

\[
U\ge F_nA_nHR,\qquad
\frac{2eg h(\beta)}U<\frac1{100000},\qquad
F_{n+1}>3F_n,\qquad \alpha_n>\frac{3n}{2}.
\tag{10.529}
\]

In particular \(\alpha_2>3\).

\textbf{Proof.} Substitute \(\Omega_n^{(2)}=h(\beta)(L_n^{(2)})^{n-1}R\)
into U. Using the singleton height bound (10.55) and the second entry of
its maximum gives

\[
U\ge\frac c\rho\frac{w_K}{q^u}
\frac{a^n\mathrm e^{2n-1}(n+1)^{n+2}}
{(n-1)!n^n}
\left(\frac{X_n}{\varkappa}\right)^{n-1}A_nHR
\ge F_nA_nHR.
\tag{10.530}
\]

For the other term use \(p^f/\delta_\beta>g\),
\(L_n^{(2)}\ge1/(\varkappa Pd)\), \(eq^u\le2d\), and \(t^2\le A_nH\).
Exact cancellation gives

\[
\frac{2eg h(\beta)}U
<\frac{2eq^u(n-1)!\varkappa^{n-1}t^2}
{ca^n\mathrm e^n(n+1)^{n+2}d^2\ell_dRA_nH}
\le\frac{E_n}{d\ell_dR}.
\tag{10.531}
\]

The next table gives strict lower bounds on \(\alpha_2\) and upper
bounds on \(E_2/(d\ell_d)\). Use \(\mathrm e>27/10\), the degree and
torsion bounds from Section 52, and \(\ell_d\ge1\). In III replace
\((p-1)/(p-2)\) by one.

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}lrr@{}}
\toprule\noalign{}
Case & \(\alpha_2\) greater than & \(E_2/(d\ell_d)\) less than \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
I.1 & 9 & \(2500/2080355319\) \\
I.2 & 14 & \(1250/234365481\) \\
II & 4 & \(500/199644669\) \\
III.1 & 3 & \(1000/286446699\) \\
III.2 & 12 & \(10000/1611624357\) \\
IV & 15 & \(2500/3498121809\) \\
V & 9 & \(1600/675264681\) \\
\end{longtable}
}

Every last entry is less than \(1/100000\), by multiplication of
positive denominators. The
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_second_rank_one_budget.py}{exact calculation}
retains each positive gap. For growth, direct division gives

\[
\frac{F_{n+1}}{F_n}
=\frac{a\mathrm e^2}{\varkappa}
\left(1+\frac1{n+1}\right)^{n+3}>3.
\tag{10.532}
\]

Indeed \(a\mathrm e^2/\varkappa>9477/4800>3/2\), and the positive
binomial terms give the last power greater than two. Thus
\(\alpha_n>3\cdot3^{n-2}\ge3n/2\); the elementary inequality
\(2\cdot3^{n-2}\ge n\) follows by induction starting at two. Finally,

\[
\frac{E_{n+1}}{E_n}
=\frac{\varkappa}{a\mathrm e}\frac n{n+2}
\left(\frac{n+1}{n+2}\right)^{n+2}
<\frac{\varkappa}{a\mathrm e^2}<\frac23.
\tag{10.533}
\]

Here \(\ln(1-y)<-y\) for \(0<y<1\), by integrating \(1/(1-y)>1\). The
last numerical comparison uses \(4800/9477<2/3\). Hence the
height-budget bound at two holds at every larger n.~\(\square\)

\subsubsection{Exact reduction and the exponent
budget}\label{exact-reduction-and-the-exponent-budget}

\textbf{Theorem 10.163 (the full second-constant rank-one refinement).}
For every rank-one list just specified, including all coefficient sizes
and torsion patterns,

\[
\operatorname{ord}_{\mathfrak p}(\Xi-1)
<\frac{\mathcal T_n^{(2)}}{4000}.
\tag{10.534}
\]

\textbf{Proof.} For \(n=1\), use Theorem 10.161. Suppose \(n\ge2\), and
put \(D=\rho\mathrm e d^2\ell_dL_n^{(2)}\), \(D_0=\max\{1,D\}\). As in
the exact pair reduction in Theorem 10.158, Theorem 10.20 gives, for
every non-torsion complementary index i, a relation
\(\beta^{u_i}a_i^{k_i}=1\), with \(k_i\in\mathbb Z_{>0}\),
\(k_i\le\rho\mathrm e d^2\ell_dh(a_i)\), and
\(|u_i|/k_i=h(a_i)/h(\beta)\). Let m count these indices. Set
\(Q=\prod k_i\) if no torsion base occurs, and \(Q=w_K\prod k_i\)
otherwise. Then

\[
\Xi^Q=\beta^M,\qquad M\in\mathbb Z,\qquad
|M|\le nBQ\frac{h(a_n)}{h(\beta)}.
\tag{10.535}
\]

Each \(k_i\le D_0H_n^{(2)}(a_i)/L_n^{(2)}\). If a non-torsion
complementary base occurs, its modified height and the singleton height
bound also give \(h(a_n)/h(\beta)\le\max\{1,D/w_K\}R\). Otherwise that
height ratio is one. Consequently, for \(M\ne0\), the three bounds
(10.501)--(10.502) apply with this D, \(D_0\), and R. This follows by
multiplying the exact relation coefficients; it uses no first-constant
estimate.

The empty-list height bound gives \(w_K\le\rho d\ell_d\). Thus

\[
\frac{w_K}{D_0}\le\frac{\varkappa_2Pt}{\mathrm eX_n}
\le192d.
\tag{10.536}
\]

We used \(\varkappa_2\le48\), \(P\le4d\), \(t\le X_n\), and
\(\mathrm e>1\). The factor d is retained in the exponent budget. Each
of the three alternatives is at most \(Z_n+2\ln R\), where

\[
Z_n=H+\ln(2n)+n\ln D_0+\ln192+\ln d.
\tag{10.537}
\]

For odd primes \(t\ge\ln3>1\). At two, the primitive cube root reduces
injectively, so \(3\mid2^f-1\), \(f\ge2\), and \(t\ge\ln4>1\). Therefore
\(X_n/t\le n\) and \(D_0<7nd\ell_d\), since
\(\rho\mathrm e/\varkappa_2\le58(11/4)/25<7\). For \(n\ge3\), the
logarithmic tangent from Section 50, \(\ln\ell_d\le(\ln d)/8+11/10\),
together with \(\ln7<2\), \(\ln192<6\), and \(\ln(2n)<n\), gives

\[
Z_n<H+n\ln n+\frac{41n}{10}+6
       +\left(\frac{9n}{8}+1\right)\ln d.
\tag{10.538}
\]

Lemma 10.162 gives \(\alpha_n>3n/2\). The coefficient of \(\ln d\) in
\(\alpha_nA_nH\) is larger than \(9n/8+1\), since \(H>1\) and \(n\ge3\).
At H=1 the constant difference is larger than \(77n/30-7>0\): use
\(A_n\ge4+\ln(n+1)+\ln d\), \(\ln(n+1)\ge\ln4>4/3\), and
\(\ln n<\ln(n+1)\). Increasing H increases the difference, because its
slope is greater than \(6n-1>0\). Thus \(Z_n<\alpha_nA_nH\) for every
\(n\ge3\).

For \(n=2\), a non-torsion complement gives the sharper bound
\(H+\ln4+2\ln D_0+2\ln R\). Here \(D_0<13d\ell_d\), since
\(2\rho\mathrm e/\varkappa_2<13\). Using \(\ln13<8/3\), \(\ln4<7/5\),
and the tangent \(\ln\ell_d\le(\ln d)/16+9/5\), its part before
\(2\ln R\) is less than \(H+31/3+(17/8)\ln d\). This is less than
\(\alpha_2A_2H\), because \(\alpha_2>3\), \(A_2>5\), \(H>1\),
\(14>31/3\), and \(3>17/8\). If the complement is torsion, its bound is
\(H+\ln(2w_K)<H+15/2+(17/16)\ln d\), which is absorbed by the same lower
bound. All numerical logarithm comparisons follow from positive
exponential series, or the logarithm series already proved in Solution
64.

In every case the bound before \(2\ln R\) is less than
\(\alpha_nA_nH>2\). Since \(\ln R\le R-1\),

\[
\ln(2|M|)<\alpha_nA_nHR
\le\left(\frac1{4000}-\frac1{100000}\right)U.
\tag{10.539}
\]

Apply the complete one-base estimate (10.60) to \(\beta^M\), retaining
its actual residue order g. Since
\(\Xi^Q-1=(\Xi-1)(1+\Xi+\cdots+\Xi^{Q-1})\) and \(\Xi\) is a local unit,

\[
\begin{aligned}
\operatorname{ord}_{\mathfrak p}(\Xi-1)
&\le\operatorname{ord}_{\mathfrak p}(\beta^M-1)\\
&\le\frac d t\bigl(\ln(2|M|)+2eg h(\beta)\bigr)
<\frac{\mathcal T_n^{(2)}}{4000},
\end{aligned}
\tag{10.540}
\]

by Lemma 10.162. If M=0, then \(\Xi\ne1\) is torsion. Theorem 10.21
bounds its valuation by \((d/t)\ln2\). Meanwhile
\(U/4000>\alpha_nA_nHR>2>\ln2\), proving the same strict inequality.
This exhausts every case. \(\square\)

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={Exact rank-one reduction and the second-constant budgets}]{../figures/yu-second-rank-one-budget.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.46. Left: the exact rank-one identity for the indexed
list \(4,16\) at five gives valuation \(1+k\), proved by the power law.
Right: for every list of length at least two, the exponent cost and
original residue-order height cost are strictly below \(24/25\) and
\(1/25\), respectively, after normalization by \(U/4000\). These exact
allowances sum to one. The second modified-height floor and the degree
term in the exponent bound are retained. Proof: Lemma 10.162 and Theorem
10.163. Human source for the target: Theorem 2 and equations (1.10),
(1.25), (1.29)--(1.30) of Yu's free 2013 paper, cited below.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_second_rank_one_budget.py}{Reproducible
figure and exact calculation}.}
\end{flushleft}
\end{minipage}\par\medskip


For \(K=\mathbb Q,p=5,(a_1,a_2)=(4,16)\), the selected base is four,
\(g=2\), and \(\delta_\beta=2\). Here \(P=1\), \(\varkappa_2=25\), and
\(L_2^{(2)}=2/(25\ln5)\). The two actual heights exceed this floor. With
coefficients \((0,5^k)\), \(k\ge0\), the exact reduction is
\(\Xi=4^{2\cdot5^k}\), and the proved valuation is \(1+k\). Theorem
10.163 uses the length-two second constant and \((\ln4)(\ln16)\), while
preserving the original residue order two.

The numerical target is the rank-one refinement in Theorem 2 of
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
free paper}. The proof here supplies this full rank-one application
independently of the omitted general second-constant proof. Rank at
least two and the general coefficient profile remain separate
requirements.

\subsection{54. Sharp two-logarithm constants with the
E-parameter}\label{sharp-two-logarithm-constants-with-the-e-parameter}

The determinant comparison (10.40) contains considerably more
information than the parameter choice used in Theorem 10.11. We now use
smaller rectangles to obtain the constants \(36.1\) and \(53.8\), while
keeping the coefficient ordering and the exponential domain explicit.
The two constants accompany different cutoffs; a smaller numerical
constant need not give the better bound for a particular pair.

\textbf{Lemma 10.164 (factorials and the common p-part).} For an integer
\(K\ge2\), put \[
Q_K=\left(\prod_{j=1}^{K-1}j!\right)^{-2/(K(K-1))}.
\tag{10.541}
\] Then \[
Q_K\le\frac{\mathrm e^{3/2}}{K-1}.
\tag{10.542}
\] If \(a\ge1\) and \(b\ge1\), then \[
\ln b\le a\max\{\ln(b/a),1\}.
\tag{10.543}
\] Consequently, for \(a_i\ge1\), \(g\ge1\), \(q=E\ln p>0\),
\(z=ga_1a_2\), positive integer coefficients \(b_i\), and \[
\mathcal D=\frac{b_1}{a_2}+\frac{b_2}{a_1},\qquad
qB\ge\max\{\ln\mathcal D+2/5,5\},
\tag{10.544}
\] the integer \(u=v_p(\gcd(b_1,b_2))\) satisfies \[
u\le EzB.
\tag{10.545}
\]

\textbf{Proof.} Write \(n=K-1\). Monotonicity of \(\ln x\) gives
\(\ln(j!)\ge\int_1^j\ln x\,dx=j\ln j-j+1\). The function \(x\ln x\) is
convex on \([1,\infty)\), so its trapezoid sums dominate its integral:
\[
\sum_{j=1}^{n}j\ln j
\ge\int_1^n x\ln x\,dx+\frac{n\ln n}{2}
=\frac{n(n+1)}2\ln n-\frac{n^2}{4}+\frac14.
\tag{10.546}
\] Summing the factorial bounds therefore gives \[
\sum_{j=1}^{n}\ln(j!)
\ge\frac{n(n+1)}2\ln n-\frac{3n^2}{4}+\frac n2+\frac14
\ge\frac{n(n+1)}2(\ln n-3/2).
\tag{10.547}
\] Exponentiation proves the first assertion, including \(n=1\).

If \(b\le\mathrm e a\), then \(\ln b\le1+\ln a\le a\). If
\(b>\mathrm e a\), the inequality \[
(a-1)\ln b-a\ln a
\mathrel{>}(a-1)(1+\ln a)-a\ln a
=a-1-\ln a\ge0
\tag{10.548}
\] proves the second assertion for \(a>1\); at \(a=1\) it is immediate.
Finally \(p^u\mid b_1\), and \(\ln(b_1/a_2)\le\ln\mathcal D<qB\), with
\(qB\ge5\). Thus \[
u\le\frac{\ln b_1}{\ln p}
\le\frac{a_2qB}{\ln p}=Ea_2B\le EzB.
\tag{10.549}
\] \(\square\)

\textbf{Theorem 10.165 (sharp rational E-parameter estimates).} Let
\(\alpha_1,\alpha_2\) be multiplicatively independent nonzero rational
p-adic units. Let \(g\) be the least positive integer such that \[
v_p(\alpha_1^g-1)>0,\qquad v_p(\alpha_2^g-1)>0.
\tag{10.550}
\] For odd \(p\), this integer divides \(p-1\); for \(p=2\) it is one.
Suppose \[
v_p(\alpha_1^g-1)\ge E>\frac1{p-1}.
\tag{10.551}
\] Let \(b_1,b_2\) be nonzero integers, put \(u=v_p(\gcd(b_1,b_2))\),
and suppose \[
p\nmid b_2/p^u.
\tag{10.552}
\] When \(p=2\), also suppose \(v_2(\alpha_2-1)\ge2\). Choose real
numbers \(A_i>1\) with \[
\ln A_i\ge\max\{h(\alpha_i),E\ln p\},\qquad
B'=\frac{|b_1|}{\ln A_2}+\frac{|b_2|}{\ln A_1}.
\tag{10.553}
\] For \(\nu=6,4\), define \[
H_\nu=\max\{\ln B'+\ln(E\ln p)+2/5,\ \nu E\ln p,\ 5\}.
\tag{10.554}
\] Then both estimates hold: \[
\begin{aligned}
v_p(\alpha_1^{b_1}-\alpha_2^{b_2})
&\le36.1\,\frac{g}{E^3(\ln p)^4}\,
\ln A_1\ln A_2\,H_6^2,\\
v_p(\alpha_1^{b_1}-\alpha_2^{b_2})
&\le53.8\,\frac{g}{E^3(\ln p)^4}\,
\ln A_1\ln A_2\,H_4^2.
\end{aligned}
\tag{10.555}
\] The coefficient-ordering requirement may be omitted if \textbf{both}
\(v_p(\alpha_i^g-1)\ge E>1/(p-1)\). In that case the pair can be
reordered to make \(b_2/p^u\) a unit; for \(p=2\) both bases are already
in the required exponential domain.

\textbf{Proof.} Negative coefficients are made positive by inverting the
corresponding bases. This preserves heights, residue orders,
independence, and all stated precision conditions. Use the notation \[
q=E\ln p,\qquad a_i=\frac{\ln A_i}{q}\ge1,\qquad
z=ga_1a_2\ge1,\qquad B=\frac{H_\nu}{q}.
\tag{10.556}
\] Then \(B\ge\nu\), \(qB\ge5\), and \(qB\ge\ln\mathcal D+2/5\), where
\(\mathcal D=B'q\). Lemma 10.164 gives \(u\le EzB\).

The principal-unit and error-term argument leading to (10.40) uses only
\(p\nmid g\), the stated first-base precision, the integral ratio
\(b_1'/b_2'\), and the exponential domain of the second base. It
therefore works with this exact common residue order, rather than the
possibly larger \(p-1\). Its sufficient inequality is still (10.40), and
its conclusion under the two cardinalities is \[
v_p(\alpha_1^{b_1}-\alpha_2^{b_2})<E(N-1/2)+u.
\tag{10.557}
\] In particular no assumption of precision \(E\) for the second
logarithm is silently inserted.

For the two choices use the following parameters: \[
\begin{array}{c|cc}
\nu&k&\ell\\ \hline
6&47/10&12/5\\
4&11/2&5/2
\end{array}
\qquad L=\lfloor\ell B\rfloor+2,\quad K=\lfloor kzL\rfloor+1.
\tag{10.558}
\] Set \[
\begin{aligned}
R_1&=\lfloor\sqrt{gLa_2/a_1}\rfloor+1,&
S_1&=\lfloor\sqrt{gLa_1/a_2}\rfloor+1,\\
R_2&=\lfloor\sqrt{g(K-1)La_2/a_1}\rfloor+1,&
S_2&=\lfloor\sqrt{g(K-1)La_1/a_2}\rfloor+1.
\end{aligned}
\tag{10.559}
\] Put \(R=R_1+R_2-1\), \(S=S_1+S_2-1\), \(N=KL\). Since
\(\lfloor x\rfloor+1>x\), the first rectangle has more than \(gL\)
points, and the second has more than \(g(K-1)L\). Independence makes its
multiplicative labels injective. Pigeonholing the first rectangle by its
\(g\) residue classes provides the first cardinality in (10.18). If the
second rectangle's linear labels are injective in every residue class,
pigeonholing also provides the second cardinality.

We first treat this injective case. The rounded sides satisfy \[
(R-1)a_1,\ (S-1)a_2
\le\sqrt{zL}\bigl(1+\sqrt{K-1}\bigr).
\tag{10.560}
\] Thus Lemma 10.164 and the definition (10.15) imply \[
\beta\le
\frac{\mathrm e^{3/2}}{2\sqrt{k}}\,
\frac{(1+\sqrt{K-1})\sqrt K}{K-1}\,\mathcal D.
\tag{10.561}
\] Here \(L\ge16,K\ge76\) when \(\nu=6\), and \(L\ge12,K\ge67\) when
\(\nu=4\). For \(K\ge67\), \[
\frac{(1+\sqrt{K-1})\sqrt K}{K-1}
=\left(1+\frac1{\sqrt{K-1}}\right)
\sqrt{1+\frac1{K-1}}
<\frac98\,\frac{81}{80}<\frac65.
\tag{10.562}
\] The elementary bounds \(\mathrm e^{11/10}<10/3\) and \(k>4\)
consequently give \[
\ln\beta<\ln\mathcal D+2/5\le qB.
\tag{10.563}
\] These exponential bounds can also be verified directly from their
series: after term \(M\), the tail is at most
\((x^{M+1}/(M+1)!)/(1-x/(M+2))\), when \(0<x<M+2\).

The height contribution in (10.40), divided by \(q\), is at most \[
\frac L2\bigl((R-1)a_1+(S-1)a_2\bigr)
\le \sqrt{k}\,zL^2+L\sqrt{zL}.
\tag{10.564}
\] After removing \((K-1)B\) from the comparison's two sides and
dividing by \(zL^2\), it therefore suffices to prove \[
k\left(1-\frac{B+1}{L}\right)
\mathrel{>}\sqrt{k}+\frac1{\sqrt L}
\mathbin{+}\frac{3B\ln(kL^2+L)}{5L^2}.
\tag{10.565}
\] Indeed \(K>kzL\), \(L-1-B>0\), and \(qB\ge5\). Also
\(\ln(kzL^2+L)/z\) decreases for \(z\ge1\): its derivative has the sign
of \(kzL^2/(kzL^2+L)-\ln(kzL^2+L)<0\). This justifies replacing \(z\) by
one only in this upper envelope.

From \(L>\ell B+1\) and \(B\ge\nu\) we obtain \[
\frac{B+1}{L}<\frac{\nu+1}{\ell\nu+1}.
\tag{10.566}
\] Furthermore \(B<L/\ell\), and \(\ln(kL^2+L)/L\) decreases throughout
either range of \(L\). Its derivative has the sign of
\((2kL+1)/(kL+1)-\ln(kL^2+L)<0\), since the first term is less than two
and the second exceeds two. The following exact bounds finish the
sufficient comparison: \[
\begin{array}{c|c|c|c|c}
\nu&\text{left lower bound}&\sqrt{k}\text{ upper}
&L^{-1/2}\text{ upper}&\text{logarithmic upper}\\ \hline
6&141/55&217/100&1/4&9/80\\
4&3&47/20&3/10&67/500
\end{array}
\tag{10.567}
\] For the logarithmic entries use \(\ln((47/10)16^2+16)<36/5\) and
\(\ln((11/2)12^2+12)<67/10\), respectively. The square comparisons and
the exponential-series bound just given establish every entry. The sum
of the three upper bounds is \(1013/400<141/55\) in the first row, and
\(348/125<3\) in the second. Thus (10.40) holds.

The resulting valuation bound satisfies \[
\frac{EN+u}{EzB^2}
\le k\left(\ell+\frac2\nu\right)^2
\mathbin{+}\frac{\ell+2/\nu}{\nu}+\frac1\nu
\mathrel{=}
\begin{cases}
80407/2250<36.1,&\nu=6,\\
101/2<53.8,&\nu=4.
\end{cases}
\tag{10.568}
\] This covers the injective alternative for every real height scale,
not only a finite parameter sample.

It remains to treat a collision of the linear labels within a residue
class. Put \(n=\gcd(b_1,b_2)\), \(r_0=b_1/n\), \(s_0=b_2/n\), and
\(\sigma=\alpha_1^{r_0}\alpha_2^{-s_0}\). If \(q_0\) is the order of the
residue of \(\sigma\), the two colliding points give \[
q_0\mid g,\qquad q_0r_0\le R_2-1,\qquad
q_0s_0\le S_2-1.
\tag{10.569}
\] To see the latter inequalities, their differences are \(t(r_0,-s_0)\)
for a nonzero integer \(t\). Equality of their residue classes forces
\(q_0\mid t\), while both coordinate differences lie within the
rectangle.

If \(q_0\nmid n\), the difference has valuation zero. Otherwise
\(\sigma^{q_0}\) is a deep principal unit, and the power law of Lesson
9, with \(p\nmid q_0\), gives \[
v_p(\alpha_1^{b_1}-\alpha_2^{b_2})
=u+v_p(\sigma^{q_0}-1).
\tag{10.570}
\] Independence ensures \(\sigma^{q_0}\ne1\). The one-place height bound
applied to \(X-1\) yields \[
v_p(\alpha_1^{b_1}-\alpha_2^{b_2})
\le u+E\bigl((R_2-1)a_1+(S_2-1)a_2\bigr)
\mathbin{+}\frac{\ln2}{\ln p}
\le u+2E\sqrt{k}\,zL+\frac{\ln2}{\ln p}.
\tag{10.571}
\] Consequently its quotient by \(EzB^2\) is at most \[
\frac1\nu+
\frac{2\sqrt{k}}{\nu}\left(\ell+\frac2\nu\right)
\mathbin{+}\frac{\ln2}{5\nu}
<\begin{cases}3,&\nu=6,\\4,&\nu=4.\end{cases}
\tag{10.572}
\] Here we used \(z\ge1\), \(qB\ge5\), and \(\ln2<7/10\). Both desired
constants exceed these bounds.

Finally \[
EzB^2=\frac{g}{E^3(\ln p)^4}\ln A_1\ln A_2\,H_\nu^2.
\tag{10.573}
\] This proves the two stated estimates. If both bases have precision
\(E\), one of the coefficients after removal of the common p-part is a
unit; reordering makes it the second coefficient without losing any
precision or convergence hypothesis. \(\square\)

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={Guaranteed determinant margins and valuation coefficients for the two E-parameter bounds}]{../figures/sharp-E-parameter-budget.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.47. Theorem 10.165 uses \(k=47/10,\ell=12/5\) for
\(\nu=6\), and \(k=11/2,\ell=5/2\) for \(\nu=4\). Left: the proved lower
envelope
\(m_\nu(B)=k(1-(B+1)/L)-\sqrt{k}-L^{-1/2}-3B\ln(kL^2+L)/(5L^2)\), with
\(L=\lfloor\ell B\rfloor+2\), remains positive for all \(B\ge\nu\).
Right: the proved upper envelope \(k(L/B)^2+L/B^2+1/B\) for
\((EN+u)/(EzB^2)\) stays below the displayed constants. The teeth come
from the integer rectangle size. The plots display \(4\le B\le20\); the
monotonicity and rational inequalities in the proof establish the whole
infinite ranges, including every \(z\ge1\) and \(qB\ge5\). The collision
alternative has the separate bounds three and four. Human-source
context: Bugeaud--Laurent's freely accessible 1996 paper and Bugeaud's
freely accessible 1999 paper; the complete parameter proof and
coefficient-ordering correction are given here.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/sharp_E_parameter_budget.py}{Figure program}.}
\end{flushleft}
\end{minipage}\par\medskip


For a concrete family, take \(p=5\), \(\alpha_1=1+5^m\),
\(\alpha_2=1+2\cdot5^m\), \(b_1=b_2=1\), \(m\ge1\). The two positive
integer bases are coprime, since their second minus twice their first is
\(-1\); primes dividing either base then prove independence. Their
common residue order is \(g=1\), both have precision \(E=m\), and the
difference has valuation exactly \(m\). With \(\ln A_i=h(\alpha_i)\),
both height cutoffs are attained by \(\nu E\ln5\). Thus the two bounds
give \[
m\le1299.6\,
\frac{\ln(1+5^m)\ln(1+2\cdot5^m)}{m(\ln5)^2},
\qquad
m\le860.8\,
\frac{\ln(1+5^m)\ln(1+2\cdot5^m)}{m(\ln5)^2}.
\tag{10.574}
\] The bound involving \(53.8\) is stronger in this family because its
height cutoff is four rather than six.

\subsection{55. The rational 2-adic bound across both unit
classes}\label{the-rational-2-adic-bound-across-both-unit-classes}

The 2-adic boundary does not require pretending that every principal
unit belongs to the exponential domain. There are two rational unit
classes modulo four. A sign change moves either class into that domain,
and the sign remaining in the linear form determines whether a
logarithmic estimate is needed.

\textbf{Corollary 10.166 (the rational 2-adic estimate).} Let
\(\alpha_1,\alpha_2\) be multiplicatively independent nonzero rationals
with \(v_2(\alpha_i)=0\). Let \(b_1,b_2\) be nonzero integers, choose
\(A_i>1\) with \[
\ln A_i\ge\max\{h(\alpha_i),\ln2\},
\qquad B'=\frac{|b_1|}{\ln A_2}+\frac{|b_2|}{\ln A_1},
\qquad T=\max\{\ln B'+1/25,10\}.
\tag{10.575}
\] Then \[
v_2(\alpha_1^{b_1}-\alpha_2^{b_2})
\le208\,\ln A_1\ln A_2\,T^2.
\tag{10.576}
\] There is no coefficient-ordering requirement and no additional
restriction on either residue class modulo four.

\textbf{Proof.} A rational 2-adic unit has an odd numerator and odd
denominator, so its reduction modulo four is either one or three. Choose
\(\eta_i\in\{1,-1\}\) such that \[
\beta_i=\eta_i\alpha_i\in1+4\mathbb Z_2,\qquad
\eta=\eta_1^{b_1}\eta_2^{-b_2},\qquad
\frac{\alpha_1^{b_1}}{\alpha_2^{b_2}}
=\eta\,\frac{\beta_1^{b_1}}{\beta_2^{b_2}}.
\tag{10.577}
\] The new bases have the same heights and are independent: squaring a
relation among the \(\beta_i\) would give a relation among the original
bases, with both exponents doubled. Inverses and integer powers preserve
\(1+4\mathbb Z_2\).

If \(\eta=-1\), the ratio on the right is three modulo four. Subtracting
one gives valuation exactly one. This already satisfies the asserted
bound, since \(\ln A_i\ge\ln2>69/100\) and \(T\ge10\).

Suppose \(\eta=1\). First enlarge each prescribed logarithmic height by
the same \(\varepsilon>0\), writing
\(a_{i,\varepsilon}=\ln A_i+\varepsilon\), and define \[
E_\varepsilon=\min\{2,a_{1,\varepsilon}/\ln2,a_{2,\varepsilon}/\ln2\}>1,
\qquad q_\varepsilon=E_\varepsilon\ln2.
\tag{10.578}
\] Both \(\beta_i\) have precision at least two, hence at least
\(E_\varepsilon\). The common residue order is one. The both-deep
alternative of Theorem 10.165 applies to these bases and all coefficient
sizes, after the necessary reordering. In particular its bound already
includes the common p-part of the coefficients.

Let \(B'_\varepsilon=|b_1|/a_{2,\varepsilon}+|b_2|/a_{1,\varepsilon}\),
and \(T_\varepsilon=\max\{\ln B'_\varepsilon+1/25,10\}\). Since
\(q_\varepsilon\le2\ln2<7/5<\mathrm e^{2/5}\), the cutoff for \(\nu=6\)
satisfies \[
\begin{aligned}
H_{6,\varepsilon}
&=\max\{\ln B'_\varepsilon+\ln q_\varepsilon+2/5,\ 6q_\varepsilon,\ 5\}\\
&\le T_\varepsilon+19/25
<\frac{11}{10}T_\varepsilon.
\end{aligned}
\tag{10.579}
\] Indeed the logarithmic term is less than \(T_\varepsilon+19/25\),
whereas \(6q_\varepsilon<42/5<10\le T_\varepsilon\) and
\(5\le T_\varepsilon\).

The first estimate in Theorem 10.165 consequently gives \[
\begin{aligned}
v_2(\alpha_1^{b_1}-\alpha_2^{b_2})
&\le\frac{36.1(11/10)^2}{E_\varepsilon^3(\ln2)^4}
a_{1,\varepsilon}a_{2,\varepsilon}T_\varepsilon^2\\
&<\frac{4368100000}{22667121}
a_{1,\varepsilon}a_{2,\varepsilon}T_\varepsilon^2
<208\,a_{1,\varepsilon}a_{2,\varepsilon}T_\varepsilon^2.
\end{aligned}
\tag{10.580}
\] The rational coefficient uses \(E_\varepsilon>1\) and
\(\ln2>69/100\); its gap below \(208\) is \(346661168/22667121>0\). The
bound on \(\ln2\) follows by comparing the exponential series at
\(69/100\) and \(7/10\) with two, using the tail estimate in Theorem
10.165. Now let \(\varepsilon\) decrease to zero. The reciprocal
heights, logarithm and maximum defining \(T_\varepsilon\) are continuous
at the positive prescribed heights. This yields the desired inequality
even when a prescribed height is exactly \(\ln2\). \(\square\)

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={The two sign cases and exact example valuations for the rational 2-adic bound}]{../figures/two-adic-sign-bound.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.48. Left: the exact sign decomposition in Corollary
10.166 sends both bases into \(1+4\mathbb Z_2\). An odd remaining sign
gives valuation one; an even sign uses the fully proved both-deep
estimate with coefficient below \(208\). Right: exact integer valuations
\(v_2(3^{b_1}-5^{b_2})\), \(1\le b_i\le10\). Here
\(\eta_1=-1,\eta_2=1\), so every odd \(b_1\) gives valuation one, while
the even columns require the deep-unit argument. Every cell is
calculated by dividing the nonzero integer difference by two until it
becomes odd. This finite example illustrates the two cases; the proof
covers all independent rational units, signed nonzero exponents, and
prescribed heights at their lower boundary. Human-source context: the
rational 2-adic constant in Bugeaud's freely accessible 1999 Theorem 2;
the complete sign-normalization proof is given here.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/two_adic_sign_bound.py}{Figure program}.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{56. The constant 283 for Fermat differences and arbitrary
powers}\label{the-constant-283-for-fermat-differences-and-arbitrary-powers}

For the special coefficients \(p-1,p-1\), their common p-part is zero
and their logarithmic size is small enough to use a fixed rectangle
height. This supplies the constant \(283\) directly from the proved
determinant machinery, including the one-base case.

\textbf{Theorem 10.167 (Fermat differences and arbitrary powers).} Let
\(a>b\ge1\) be coprime integers, and let \(p\) be an odd prime not
dividing \(ab\). Then \[
v_p(a^{p-1}-b^{p-1})
\le\frac{1631}{6}\,\frac{p-1}{(\ln p)^2}\ln(5a)\ln(5b)
<283\,\frac p{(\ln p)^2}\ln(5a)\ln(5b).
\tag{10.581}
\] For every integer \(n\ge1\), \[
v_p(a^n-b^n)
\le283\,\frac p{(\ln p)^2}\ln(5a)\ln(5b)+v_p(n).
\tag{10.582}
\]

\textbf{Proof.} We first prove a determinant bound for any independent
rational p-adic units \(\alpha_1,\alpha_2\) whose chosen logarithmic
height bounds \(H_i\) satisfy \(H_1>2\), \(H_2>3/2\). Set \[
g=p-1,\quad q=\ln p>1,\quad
a_i=H_i/q>0,\quad z=ga_1a_2,\quad
b_1=b_2=g,\quad E=1,\quad u=0.
\tag{10.583}
\] Here \(z>3\). Indeed \(\ln^2p<p-1\): for \(x>1\), the function
\(\sqrt x-1/\sqrt x-\ln x\) vanishes at one and has derivative
\((\sqrt x-1)^2/(2x^{3/2})>0\). Hence \(\ln^2x<(x-1)^2/x<x-1\). Now
\(g/(\ln p)^2>1\) and \(H_1H_2>3\).

Fermat's theorem and the principal-unit decomposition of Lesson 9 put
both principal factors in \(1+p\mathbb Z_p\), with precision at least
\(E=1>1/(p-1)\). The coefficient \(b_2=g\) is a p-adic unit. Thus the
analytic and arithmetic determinant comparison (10.40) applies. Its
derivation requires positive chosen height bounds above the actual
heights; the separate restriction \(a_i\ge1\) in Theorem 10.165 was used
to simplify that theorem's parameter envelopes, and is not imposed here.

Take \[
L=7,\qquad k=11/2,\qquad K=\lfloor kzL\rfloor+1,
\tag{10.584}
\] and use the four rounded square-root sides from Theorem 10.165 with
these positive \(a_i\). Its cardinality argument depends only on the
products of these sides, so it still supplies the first residue-class
cardinality and, in the injective alternative, the second one. Its two
side-height bounds and factorial calculation also apply unchanged: \[
(R-1)a_1,\ (S-1)a_2
\le\sqrt{zL}(1+\sqrt{K-1}),\qquad
\ln\beta<\ln\mathcal D+2/5,
\tag{10.585}
\] where \(\mathcal D=gq(1/H_1+1/H_2)\). For the latter inequality,
\(K\ge116>67\) and \(k>4\), so the exact factorial and square-root
comparisons in Lemma 10.164 and Theorem 10.165 apply.

Since \(1/H_1+1/H_2<7/6<6/5\) and \(\ln(6/5)<1/5\), \[
\ln\beta<\ln g+\ln q+3/5
<q+\ln q+3/5<2q.
\tag{10.586}
\] The last inequality uses \(q-\ln q\ge1\), proved by differentiating
from its minimum at \(q=1\). Dividing (10.40) by \(q\), we may therefore
remove \((K-1)B\) with \(B=2\). After division by \(zL^2\), a sufficient
comparison is \[
\frac{22}{7}
\mathrel{>}\sqrt{11/2}+\frac1{\sqrt{zL}}
\mathbin{+}\frac{3\ln(kzL^2+L)}{qzL^2}.
\tag{10.587}
\] The logarithmic quotient decreases with \(z\ge3\), by the derivative
calculation in Theorem 10.165. Using \(q>1\), \(z>3\), \(L=7\),
\(\sqrt{11/2}<47/20\), \(1/\sqrt{21}<1/4\), and
\(\ln((11/2)3\cdot7^2+7)<7\), its right side is less than \[
\frac{47}{20}+\frac14+\frac17
=\frac{96}{35}<\frac{22}{7}.
\tag{10.588}
\] The gap is exactly \(2/5\). The logarithmic comparison follows from
the exponential series at seven, whose finite sum already exceeds
\((11/2)3\cdot7^2+7=1631/2\). Thus the sufficient inequality holds and
\[
v_p(\alpha_1^g-\alpha_2^g)<N
\le kzL^2+L
<\left(\frac{539}{2}+\frac73\right)z
=\frac{1631}{6}z.
\tag{10.589}
\]

In the collision alternative, the residue-order argument in Theorem
10.165 gives instead \[
v_p(\alpha_1^g-\alpha_2^g)
\le2\sqrt{k}\,zL+\frac{\ln2}{q}
<\left(\frac{329}{10}+\frac7{30}\right)z
<34z<\frac{1631}{6}z.
\tag{10.590}
\] It uses \(u=0\), \(z>3\), \(q>1\), and \(\ln2<7/10\), and includes
the possibility of valuation zero. This proves the determinant bound in
both alternatives.

If \(b>1\), the coprime positive integer bases \(a,b\) are independent,
as proved in Corollary 10.9. Choose \(H_1=\ln(5a)\), \(H_2=\ln(5b)\);
these dominate their heights and both exceed two. The bound just proved
gives the first asserted inequality.

If \(b=1\), use the auxiliary prime from Corollary 10.10: some
\(c\in\{2,3,5\}\) is different from \(p\) and independent of \(a\), and
the rational bases \(a/c,1/c\) are independent p-adic units. Choose
\(H_1=\ln(5a)>2\) and \(H_2=\ln5>3/2\). They dominate the two rational
heights. The exact identity \[
(a/c)^{p-1}-(1/c)^{p-1}
=c^{-(p-1)}(a^{p-1}-1)
\tag{10.591}
\] preserves valuation because \(p\nmid c\). This proves the first
inequality with \(b=1\) as well. Its replacement by the constant \(283\)
follows from \(1631/6<283\) and \(p-1<p\).

Finally let \(d_0\) be the order of \(a/b\) in the multiplicative group
of the residue field. If \(d_0\nmid n\), the difference has valuation
zero. Otherwise \(d_0\mid p-1\), and the odd-prime deep power law from
Lesson 9 gives \[
\begin{aligned}
v_p(a^n-b^n)
&=v_p((a/b)^{d_0}-1)+v_p(n),\\
v_p(a^{p-1}-b^{p-1})
&=v_p((a/b)^{d_0}-1).
\end{aligned}
\tag{10.592}
\] The powers of \(b\) are units, and \((p-1)/d_0\) is prime to \(p\),
which justifies the second identity. Apply the first part to conclude.
\(\square\)

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={The fixed rectangle budgets proving the constant 283}]{../figures/fermat-283-bound.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.49. Theorem 10.167 uses \(E=1\), \(g=p-1\), \(L=7\),
\(k=11/2\), \(z=gH_1H_2/(\ln p)^2>3\), and \(q=\ln p>1\). Left: the
sufficient determinant comparison has left lower bound \(22/7\) and
right upper bound \(47/20+1/4+1/7=96/35\), with exact gap \(2/5\). The
three colored pieces bound the height contribution, rounded-side error
and logarithmic term. Right: \(N/z\) is at most
\(539/2+7/3=1631/6<283\); the collision case has the smaller bound
\(34\). These are proved uniform budgets, not measured valuations. The
coprime two-base case and the \(b=1\) auxiliary-prime construction use
the same parameters and both satisfy \(H_1>2,H_2>3/2\). Human-source
context: the free Bugeaud--Laurent and Bugeaud determinant papers and
Yamada's freely accessible note; the complete proof, including the
one-base case, is given here.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/fermat_283_bound.py}{Figure program}.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{57. Residue occupancy and the number-field
estimate}\label{residue-occupancy-and-the-number-field-estimate}

The residue order of a pair of units is often smaller than the order of
the entire residue-field multiplicative group. Using that smaller order
does more than shrink the rectangles: a single residue congruence also
limits how often a coordinate can occur. The centered occupancy theorem
from Lesson 6 then improves the arithmetic height of the determinant.

\textbf{Lemma 10.168 (occupancy in one residue class).} Let
\(K,L,R,S,g\) be positive integers, \(N=KL\), and suppose
\(\gcd(m_1,m_2,g)=1\). Let \((r_j,s_j)\), \(1\le j\le N\), be distinct
integer pairs satisfying \[
0\le r_j<R,\quad 0\le s_j<S,\quad m_1r_j+m_2s_j\equiv c\pmod g.
\tag{10.593}
\] Let \(l_1,\ldots,l_N\) contain each integer \(0,\ldots,L-1\) exactly
\(K\) times. Put \[
\begin{aligned}
M_1&=\frac{L-1}{2}\sum_jr_j,&
G_1&=\frac{NL(R+g-1)}8-\frac{gN^2L}{24(S+g-1)},\\
M_2&=\frac{L-1}{2}\sum_js_j,&
G_2&=\frac{NL(S+g-1)}8-\frac{gN^2L}{24(R+g-1)}.
\end{aligned}
\tag{10.594}
\] Then \(G_1,G_2\ge0\), and for every permutation \(\pi\), \[
\left|\sum_i l_ir_{\pi(i)}-M_1\right|\le G_1,\qquad
\left|\sum_i l_is_{\pi(i)}-M_2\right|\le G_2.
\tag{10.595}
\]

\textbf{Proof.} We use the completely proved
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/TR-BAKER-06.html\#11-centering-the-exponents-the-occupancy-bound}{centered
occupancy lemma}, Lemma 6.4. Put \(g_0=\gcd(m_2,g)\), \(g_1=g/g_0\).
Since \(\gcd(m_1,g_0)=1\), every \(r_j\) is in the same class
\(c_0\pmod{g_0}\), with \(0\le c_0<g_0\). Write \(r_j=c_0+g_0r_j'\).
These nonnegative integers lie in a range of length at most \[
R'=\left\lfloor\frac{R-1}{g_0}\right\rfloor+1.
\tag{10.596}
\] For a fixed \(r_j\), division of the original congruence by \(g_0\)
determines \(s_j\) uniquely modulo \(g_1\), because \(m_2/g_0\) is
invertible modulo \(g_1\). Therefore each \(r_j'\) is repeated at most
\[
S'=\left\lfloor\frac{S-1}{g_1}\right\rfloor+1
\tag{10.597}
\] times. There are \(N\) actual selected pairs, so \(N\le R'S'\). Lemma
6.4 applies to this compressed coordinate and capacity: \[
\left|\sum_i\left(l_i-\frac{L-1}{2}\right)r_{\pi(i)}\right|
\le g_0\left(\frac{NLR'}8-\frac{N^2L}{24S'}\right).
\tag{10.598}
\] The constant \(c_0\) disappears since the centered row weights sum to
zero. Now \[
g_0R'\le R+g_0-1,\qquad
\frac{g_0}{S'}\ge\frac{g}{S+g_1-1}.
\tag{10.599}
\] Inserting these two inequalities, and then \(g_0,g_1\le g\), gives
the asserted \(G_1\). The compressed occupancy bound is nonnegative:
\(N\le R'S'\) implies its value is at least \(NLR'/12>0\). Its
successive upper bounds are nonnegative as well. Interchanging the two
coordinates proves the assertion for \(G_2\). \(\square\)

For the rest of this section, let \(g\) be the order of the subgroup
generated by the residues of \(\alpha_1,\alpha_2\), rather than the full
\(p^f-1\) of (10.1). This subgroup is cyclic by the proof in Section 1.
Lift its generator to a root of unity \(\eta\) of order \(g\), and write
\(\alpha_i=\eta^{m_i}\epsilon_i\). The minimality of this subgroup gives
\[
g\mid p^f-1,\qquad \gcd(m_1,m_2,g)=1.
\tag{10.600}
\] The principal factors and the integer \(t\) from (10.2) are
unchanged. All analytic estimates and residue-class zero arguments
proved earlier apply to this subgroup.

\textbf{Theorem 10.169 (the refined determinant parameter theorem).}
Retain the local and global data \(d,e,f,t\) of Section 1, now with the
subgroup order \(g\) just defined. Set \[
D=d/f,\qquad
\gamma_1=\frac{R+g-1}{2R}-\frac{gN}{6R(S+g-1)},\qquad
\gamma_2=\frac{S+g-1}{2S}-\frac{gN}{6S(R+g-1)}.
\tag{10.601}
\] Let \(K\ge3,L\ge2\), \(R=R_1+R_2-1\), \(S=S_1+S_2-1\), \(N=KL\), and
let \(\beta\) be (10.15). Suppose both residue-class cardinalities
(10.18) hold, and \[
\begin{aligned}
p^tK(L-1)\ln p>{}&(D+2e)\ln N+D(K-1)\ln\beta\\
&+Dp^tL\bigl(\gamma_1Rh(\alpha_1)+\gamma_2Sh(\alpha_2)\bigr).
\end{aligned}
\tag{10.602}
\] Then \[
v_p(\alpha_1^{b_1}-\alpha_2^{b_2})
<\frac{p^t}{e}(N-1/2)+u,
\qquad u=v_p(\gcd(b_1,b_2)).
\tag{10.603}
\]

\textbf{Proof.} As in Theorem 10.7, the zero lemma supplies a nonzero
determinant \(\Delta\) with all selected columns in one class. In the
polynomial \(P(X,Y)\) used in Lemma 10.6, every first exponent lies
between \(M_1-G_1\) and \(M_1+G_1\), and every second exponent lies
between \(M_2-G_2\) and \(M_2+G_2\), by Lemma 10.168. Removing the
smallest actual powers gives an integer polynomial of partial degrees at
most \(2G_1,2G_2\). Its torus norm is still bounded by (10.17). The
complete one-place height proof in Lemma 10.6 therefore gives \[
v_p(\Delta)\le\frac{DN}{2e\ln p}
\left(\ln N+(K-1)\ln\beta+
p^tL\bigl(\gamma_1Rh(\alpha_1)+\gamma_2Sh(\alpha_2)\bigr)\right),
\tag{10.604}
\] because \(4G_1/N=\gamma_1LR\) and \(4G_2/N=\gamma_2LS\). The
constants \(\gamma_i\) are nonnegative by Lemma 10.168. The analytic
estimate (10.10) is unchanged. If the asserted valuation inequality
failed, comparing these two bounds and multiplying by \(2e\ln p/N\)
would give the opposite weak inequality to the stated strict parameter
condition. \(\square\)

\textbf{Theorem 10.170 (the explicit number-field bound).} Let
\(\alpha_1,\alpha_2\) be multiplicatively independent nonzero algebraic
numbers, with a fixed p-adic embedding and \(v_p(\alpha_i)=0\). Put \[
d=[\mathbb Q(\alpha_1,\alpha_2):\mathbb Q],\qquad
F=\mathbb Q_p(\alpha_1,\alpha_2),\qquad
[F:\mathbb Q_p]=ef,\qquad D=d/f.
\tag{10.605}
\] Let \(g\) be the order of the subgroup generated by their nonzero
residues. For nonzero integer coefficients \(b_1,b_2\), choose \(A_i>1\)
with \[
\ln A_i\ge\max\{h(\alpha_i),\ln p/D\},
\qquad B'=\frac{|b_1|}{D\ln A_2}+\frac{|b_2|}{D\ln A_1}.
\tag{10.606}
\] Define \[
H=\max\{\ln B'+\ln\ln p+2/5,\ 10\ln p/D,\ 10\}.
\tag{10.607}
\] Then \[
v_p(\alpha_1^{b_1}-\alpha_2^{b_2})
\le24\,\frac{pg}{(p-1)(\ln p)^4}\,
D^4\ln A_1\ln A_2\,H^2.
\tag{10.608}
\] In particular \(g\le p^f-1\) gives the same bound with \(p(p^f-1)\)
in place of \(pg\). Neither unramifiedness nor an ordering of the
coefficient valuations is assumed.

\textbf{Proof.} Inverting bases handles negative coefficients,
preserving all relevant data; multiplying the resulting difference by a
local unit preserves its valuation. We may assume positive coefficients.
Write \[
D_0=D/\ln p,\qquad a_i=D\ln A_i/\ln p\ge1,\qquad
z=ga_1a_2\ge1,\qquad B=D_0H.
\tag{10.609}
\] Then \(D\ge e\ge1\), \(B\ge10\), and \(B\ge10D_0\). Use \[
k=3,\qquad \ell=13/5,\qquad
L=\lfloor\ell B\rfloor+2,\qquad K=\lfloor kzL\rfloor+1
\tag{10.610}
\] and the four rounded square-root sides from Theorem 10.165, with this
\(g\) and these \(a_i\). Their products exceed \(gL\) and \(g(K-1)L\).
Independence gives injective multiplicative labels in the first
rectangle, so pigeonholing its residue classes supplies the first
cardinality.

First assume the linear labels in the second rectangle are injective
within each residue class. Pigeonholing supplies the second cardinality.
Theorem 10.169 applies once its parameter inequality is checked.

Let \(\mathcal D=b_1/a_2+b_2/a_1=B'\ln p\). The factorial proof in Lemma
10.164 and the side bounds in Theorem 10.165 give \[
\beta\le\frac{\mathrm e^{3/2}}{2\sqrt3}\,
\frac{(1+\sqrt{K-1})\sqrt K}{K-1}\,\mathcal D.
\tag{10.611}
\] Here \(L\ge28\) and \(K\ge85\). Since \(K-1\ge84>81\), \[
\frac{(1+\sqrt{K-1})\sqrt K}{K-1}
<\frac{10}{9}\frac{81}{80}=\frac98.
\tag{10.612}
\] The exponential-series bound \(\mathrm e^{11/10}<61/20\), together
with \((9\cdot61)^2<3(16\cdot20)^2\), implies \[
\ln\beta<\ln\mathcal D+2/5\le H.
\tag{10.613}
\] The square comparison is exact: \(301401<307200\).

We next bound the refined height terms. Put \[
X=\sqrt{g(K-1)La_1a_2},\qquad Y=\sqrt{gLa_1a_2}.
\tag{10.614}
\] The rounded sides give \((R+g-1)a_1\le X+Y+ga_1\). Also \[
\frac{gKLa_1}{S+g-1}
\ge\frac{X^2}{X+(S_1+g-1)a_2}
\ge X-(S_1+g-1)a_2
\ge X-Y-ga_2.
\tag{10.615}
\] The middle inequality is \(x^2/(x+y)\ge x-y\) for \(x,y\ge0\).
Inserting these estimates in \(\gamma_1Ra_1\), and then interchanging
the two coordinates, yields \[
\begin{aligned}
\gamma_1Ra_1&\le X/3+2Y/3+ga_1/2+ga_2/6,\\
\gamma_2Sa_2&\le X/3+2Y/3+ga_2/2+ga_1/6.
\end{aligned}
\tag{10.616}
\] Since \(a_i\ge1\), their sum is at most \[
\frac23\sqrt{k}\,zL+\frac43\sqrt{zL}+\frac43z.
\tag{10.617}
\] The nonnegativity of \(\gamma_i\) justifies replacing the actual
heights by their prescribed bounds.

Divide the parameter condition by \(p^t\ln p\). Because
\((D+2e)/p^t\le3D\), \(\ln\beta\le H\), and \(B=D_0H\), it suffices to
prove \[
K(L-1)>3D_0\ln N+(K-1)B
\mathbin{+}L(\gamma_1Ra_1+\gamma_2Sa_2).
\tag{10.618}
\] This replacement remains valid if \(\ln\beta<0\), since \(H>0\). Set
\(\lambda=\ell+2/10=14/5\); then \(\ell B<L-1<L\le\lambda B\). After
dividing by \(zB^2\), the left side minus the coefficient term is larger
than \[
k\ell^2-k\ell-\frac{2k}{B}
\tag{10.619}
\] and the height terms are bounded by \[
\frac23\sqrt{k}\lambda^2
\mathbin{+}\frac43\frac{\lambda^{3/2}}{\sqrt B}
\mathbin{+}\frac43\frac{\lambda}{B}.
\tag{10.620}
\] Moreover \(N\le kzL^2+L\), \(D_0\le B/10\). The function
\(\ln(kz\lambda^2B^2+\lambda B)/z\) decreases for \(z\ge1\), and
\(\ln(k\lambda^2B^2+\lambda B)/B\) decreases for \(B\ge10\). For the
first assertion differentiate and use \(N> \mathrm e\); for the second,
the logarithmic derivative times \(B\) is less than two, while the
logarithm exceeds two. Thus \[
\frac{3D_0\ln N}{zB^2}
\le\frac{3\ln(k\lambda^2B^2+\lambda B)}{10B}
\le\frac{3\ln2380}{100}<\frac{117}{500}.
\tag{10.621}
\] Using \(\sqrt3<26/15\), \(\lambda^{3/2}/\sqrt{10}<149/100\), and
\(\ln2380<39/5\), the sufficient comparison follows from the exact
rational inequality \[
3(13/5)^2-3(13/5)-\frac23\frac{26}{15}(14/5)^2
\mathrel{>}
\frac35+\frac{149}{75}+\frac{28}{75}+\frac{117}{500}.
\tag{10.622}
\] Its two sides are \(3848/1125\) and \(4791/1500\), with positive
difference \(1019/4500\). The square and logarithmic bounds follow by
squaring and the exponential-series certificate already used above. This
establishes the strict parameter condition.

To include the common coefficient p-part, use Lemma 10.164's logarithmic
inequality with \(a=a_2\), \(b=b_1\). Since
\(\ln(b_1/a_2)\le\ln\mathcal D<H\) and \(H\ge10\), \[
u\le\frac{\ln b_1}{\ln p}
\le\frac{a_2H}{\ln p}
=\frac{a_2B}{D}\le zB.
\tag{10.623}
\] Theorem 10.169, \(p^t/e\le C=p/(p-1)\), and \(L\le(14/5)B\) now give
\[
\begin{aligned}
v_p(\alpha_1^{b_1}-\alpha_2^{b_2})
&<CN+u\\
&\le C\left(3(14/5)^2+\frac{14/5}{10}+\frac1{10}\right)zB^2\\
&=\frac{239}{10}CzB^2<24CzB^2.
\end{aligned}
\tag{10.624}
\]

It remains to handle a collision of linear labels inside a residue
class. Write \(n=\gcd(b_1,b_2)\), \(r_0=b_1/n\), \(s_0=b_2/n\), and
\(\sigma=\alpha_1^{r_0}\alpha_2^{-s_0}\). The residue of \(\sigma\) has
order \(q_0\mid g\). The colliding differences give \[
q_0r_0\le R_2-1,\qquad q_0s_0\le S_2-1.
\tag{10.625}
\] If \(q_0\nmid n\), the valuation is zero. Otherwise the prime-to-p
and p-power laws of Lesson 9 give \[
v_p(\alpha_1^{b_1}-\alpha_2^{b_2})
\le u+v_p(\sigma^{q_0p^t}-1).
\tag{10.626}
\] For \(u\ge t\) the deep-unit law gives the sharper exact subtraction
\(u-t\); for \(u<t\), further p-th powers do not decrease positive
valuation. Thus this inequality covers both sides of the exponential
boundary. Independence ensures \(\sigma^{q_0p^t}\ne1\).

The one-place height bound from Lemma 10.6, applied to \(X-1\), gives \[
\begin{aligned}
v_p(\alpha_1^{b_1}-\alpha_2^{b_2})
&\le u+\frac{p^t}{e}\bigl((R_2-1)a_1+(S_2-1)a_2\bigr)
\mathbin{+}\frac{D_0}{e}\ln2\\
&\le zB+2C\sqrt3\,zL+\frac{B}{10}\ln2
<\frac{28}{25}CzB^2<24CzB^2.
\end{aligned}
\tag{10.627}
\] For the displayed coarse bound, divide by \(CzB^2\), use
\(B\ge10,z\ge1,C\ge1\), \(\sqrt3<7/4\), \(\ln2<7/10\), and
\(L\le(14/5)B\). The sum is at most
\(1/10+49/50+7/1000=1087/1000<28/25\). No root of unity outside the
given local field is required.

Finally \[
zB^2=\frac{gD^4}{(\ln p)^4}\ln A_1\ln A_2\,H^2.
\tag{10.628}
\] This proves the claimed constant \(24\) in both alternatives, for all
ramification indices and coefficient sizes. \(\square\)

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={Residue-class capacity and the exact numerical margin for the number-field estimate}]{../figures/BL-number-field-occupancy.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.50. Left: all 32 selected pairs in the \(8\)-by-\(8\)
rectangle satisfying \(r+s\equiv0\pmod2\). Each first coordinate occurs
four times; the congruence reduces the capacity from eight to four. With
\(K=4,L=8\), arrange first coordinates increasingly against the row
labels to attain the exact centered sum \(168\). Middle: the compressed
Lemma 6.4 bound is \(512/3\); its relaxed residue-class bound in Lemma
10.168 is \(5728/27\); ignoring coordinate capacity gives \(224\). These
are proved bounds for every permutation, while \(168\) is the exact
maximum for this example. Right: Theorem 10.170's uniform sufficient
comparison has lower side \(3848/1125\), upper side \(4791/1500\), and
exact positive margin \(1019/4500\). The infinite ranges \(B\ge10\),
\(z\ge1\), \(D_0\le B/10\) are covered by the proof's monotonicity
arguments. The resulting valuation coefficient is \(23.9<24\), including
the common coefficient p-part; the collision bound is separately
smaller. Human-source context: Bugeaud--Laurent's freely accessible 1996
paper, Lemmas 10--13 and Theorem 3; all used occupancy, height, analytic
and numerical arguments are proved in this programme.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/BL_number_field_occupancy.py}{Figure program}.}
\end{flushleft}
\end{minipage}\par\medskip


\textbf{Corollary 10.171 (the complete cutoff table).} Under the
algebraic, local-unit, independence and height hypotheses of Theorem
10.170, put \[
H_{\nu,\eta}
=\max\{\ln B'+\ln\ln p+2/5,\ \nu\ln p/D,\ \eta\}.
\tag{10.629}
\] For each entry in the table, \[
v_p(\alpha_1^{b_1}-\alpha_2^{b_2})
\le c(\nu,\eta)\,
\frac{pgD^4}{(p-1)(\ln p)^4}
\ln A_1\ln A_2\,H_{\nu,\eta}^2,
\tag{10.630}
\] where the constants are exactly \[
\begin{array}{c|ccccc}
&\nu=4&\nu=6&\nu=8&\nu=10&\nu=15\\ \hline
\eta=5&53.8&36.1&28.1&24&18.6\\
\eta=10&51.7&34.8&27.3&23.2&18
\end{array}.
\tag{10.631}
\]

\textbf{Proof.} Use Theorem 10.170's complete proof with
\(B=D_0H_{\nu,\eta}\). Now \(B\ge\nu\) and \(D_0\le B/\eta\). Use the
following parameters and literal minimum integer
\(K_0=\lfloor k(\lfloor\ell\nu\rfloor+2)\rfloor+1\): \[
\begin{array}{cc|cc|r|c}
\eta&\nu&k&\ell&K_0&\delta\\ \hline
5&4&4.3&3.0&61&0.170\\
5&6&3.6&2.8&65&0.093\\
5&8&3.4&2.6&75&0.004\\
5&10&3.0&2.6&85&0.011\\
5&15&3.1&2.3&112&0.049\\
10&4&3.9&3.1&55&0.119\\
10&6&3.7&2.7&67&0.044\\
10&8&3.3&2.6&73&0.043\\
10&10&2.9&2.6&82&0.002\\
10&15&3.0&2.3&109&0.043
\end{array}.
\tag{10.632}
\] Here \(\delta\) is a strict lower bound for the sufficient-comparison
margin, verified below, rather than a measured valuation. Put
\(L=\lfloor\ell B\rfloor+2\), \(K=\lfloor kzL\rfloor+1\), and
\(\lambda=\ell+2/\nu\). The cardinalities, refined side-height
inequalities and common p-part bound remain exactly those proved above.

The factorial logarithmic cutoff is valid in all ten cases. The factor
\[
f(K)=\left(1+\frac1{\sqrt{K-1}}\right)\sqrt{\frac K{K-1}}
\tag{10.633}
\] decreases with \(K\). For \(K\ge82\), it is less than \(9/8\), by the
same comparison as above with \(K-1\ge81\). All such rows have
\(k\ge2.9\), and \[
\left(\frac{601}{200}\frac98\right)^2<4(2.9).
\tag{10.634}
\] The remaining rows have \(K\ge55\), \(k\ge3.3\), and \[
f(K)<\left(1+\frac7{50}\right)\frac{101}{100}
<\frac{29}{25},\qquad
\left(\frac{601}{200}\frac{29}{25}\right)^2<4(3.3).
\tag{10.635}
\] The square comparisons are exact; \(1/\sqrt{54}<7/50\) and
\(\sqrt{55/54}<101/100\). Since \(\mathrm e^{11/10}<601/200\), every row
therefore has \(\ln\beta<\ln\mathcal D+2/5\le H_{\nu,\eta}\).

The uniform determinant comparison reduces, exactly as in Theorem
10.170, to \[
\begin{aligned}
k\ell^2-k\ell-\frac23\sqrt{k}\lambda^2
\mathrel{>}\frac{2k}{\nu}
\mathbin{+}\frac43\sqrt{\lambda^3/\nu}
\mathbin{+}\frac{4\lambda}{3\nu}
\mathbin{+}\frac{3}{\eta\nu}\ln(k\lambda^2\nu^2+\lambda\nu).
\end{aligned}
\tag{10.636}
\] The same two monotonicity arguments cover the entire ranges
\(z\ge1\), \(B\ge\nu\); their logarithmic arguments exceed
\(\mathrm e^2\) already at the minimum parameters. To verify this
comparison explicitly, the following rational upper endpoints suffice:
\[
\begin{array}{cc|rrr}
\eta&\nu&\sqrt{k}&\sqrt{\lambda^3/\nu}
&\ln(k\lambda^2\nu^2+\lambda\nu)\\ \hline
5&4&2.07365&3.27396&6.75321\\
5&6&1.89737&2.26431&7.16332\\
5&8&1.84391&1.70108&7.49012\\
5&10&1.73206&1.48163&7.77486\\
5&15&1.76069&0.98007&8.33483\\
10&4&1.97485&3.41526&6.71309\\
10&6&1.92354&2.15678&7.12592\\
10&8&1.81660&1.70108&7.46065\\
10&10&1.70294&1.48163&7.74136\\
10&15&1.73206&0.98007&8.30233
\end{array}.
\tag{10.637}
\] Each finite decimal means the exact rational with denominator
\(100000\), and is a strict upper bound. The first two columns are
checked by squaring. For a logarithmic endpoint \(x\), use the positive
partial sum \(\sum_{j=0}^{64}x^j/j!\), which exceeds the displayed
argument \(k\lambda^2\nu^2+\lambda\nu\) in every row. Substituting these
endpoints makes the left side minus the right side strictly larger than
the corresponding positive \(\delta\) in the first table. These are ten
exact rational calculations. Together with the monotonicity proof they
establish the comparison for every allowed real parameter, not just the
tabulated minima.

In the injective alternative, the total valuation coefficient, including
\(u\), is at most \[
k(\ell+2/\nu)^2+\frac{\ell+2/\nu}{\nu}+\frac1\nu.
\tag{10.638}
\] For the ten rows in order, its exact values are \[
\begin{array}{cc|r}
\eta&\nu&\text{coefficient upper bound}\\ \hline
5&4&269/5\\
5&6&40537/1125\\
5&8&112391/4000\\
5&10&239/10\\
5&15&55753/3000\\
10&4&25847/500\\
10&6&104149/3000\\
10&8&54571/2000\\
10&10&5779/250\\
10&15&16193/900
\end{array}.
\tag{10.639}
\] Each is at most the stated \(c(\nu,\eta)\); equality in the first row
is harmless because the determinant valuation comparison is strict. In
the collision alternative, the proof above gives a coefficient at most
\[
\frac1\nu+\frac{2\sqrt{k}\lambda}{\nu}
\mathbin{+}\frac{\ln2}{\eta\nu}
<\frac14+\frac{189}{50}+\frac7{200}
=\frac{813}{200}<5.
\tag{10.640}
\] Here \(\nu\ge4,\eta\ge5,\lambda\le18/5\), \(\sqrt{k}<21/10\) and
\(\ln2<7/10\). Every stated constant is at least \(18\), so this
alternative satisfies each estimate as well. The final identity for
\(zB^2\) proves the asserted table. \(\square\)

\subsection{58. The second constant and coefficient
imbalance}\label{the-second-constant-and-coefficient-imbalance}

Two features of an explicit logarithmic-form estimate can be separated:
the constant multiplying the height product, and the parameter measuring
the coefficients. We first compare Yu's two constants, with their own
coefficient parameters. We then give a two-branch estimate in which the
last coefficient has a separate bound.

\subsubsection{Comparing the two
constants}\label{comparing-the-two-constants}

Retain every hypothesis and definition of Theorem 10.152. In particular,
the bases are independent local units, the last nonzero coefficient has
minimum p-adic valuation, and the original saturated residue index is
used. Write \(C_1\) for (10.449), \(P=p^\kappa\), where

\[
p^{\kappa-1}(p-1)\le2e<P(p-1),
\qquad t=f\ln p.
\tag{10.641}
\]

Here \(\kappa\) can be zero. Put \(a^{(2)}=7\) at odd primes and
\(a^{(2)}=13\) at two. Define

\[
C_2=\frac{c^{(2)}}P(a^{(2)}\mathrm eP)^n
\frac{(n+1)^{n+1}d^{n+2}\ell_d}{(n-1)!q^ut^3}
\max\left\{\frac{p^f}{\delta(\boldsymbol a)},
                  \frac{\mathrm e^nt^{n+1}}{n^n}\right\}A_n.
\tag{10.642}
\]

The constants that specify its coefficient parameter are as follows;
\(x=(p-1)/(p-2)\) in the two III rows.

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}lrrr@{}}
\toprule\noalign{}
Case & \(c^{(2)}\) & \(a_0^{(2)}\) & \(a_1^{(2)}\) \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
I.1 & \(1438\) & \(2+\ln21\) & \(1.94\) \\
I.2 & \(648\) & \(2+\ln21\) & \(2.76\) \\
II & \(690\) & \(2+\ln35\) & \(0.71\) \\
III.1 & \(495x\) & \(2+\ln7\) & \(1.99\) \\
III.2 & \(557x\) & \(2+\ln7\) & \(3.32\) \\
IV & \(2418\) & \(2+\ln7\) & \(3.58\) \\
V & \(406\) & \(2+\ln52\) & \(1.48\) \\
\end{longtable}
}

Set

\[
\begin{aligned}
G_2(n,d)&=(n+1)\bigl(a_0^{(2)}n+a_1^{(2)}+\ln(n+1)+\ln d\bigr),\\
\mathsf h_2&=\max\left\{\ln\frac{R_{\max}}{dW(d)},
                 \ln B_\dagger,G_2(n,d),(n+1)t\right\}.
\end{aligned}
\tag{10.643}
\]

These are equations (1.10), (1.12), (1.13), (1.29) and (1.30) of
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
free paper}. The following comparison uses the programme proof of
Theorem 10.152.

\textbf{Lemma 10.172 (uniform comparison of coefficient parameters).}
For every \(n\ge2,d\ge1\) in each of the seven cases,

\[
G_1(n,d)<\frac32G_2(n,d),\qquad
\mathsf h\le\frac32\mathsf h_2.
\tag{10.644}
\]

\textbf{Proof.} In this proof write \(a_0,a_1,a_2\) for the first
constant's parameters in (10.448)--(10.449). Their displayed values give
\(a_0<7\) and \(a_2<7\), so \(\ln(a_0n+a_2)<\ln(n+1)+\ln7\). For
example, \(\mathrm e>27/10\) gives \(\ln26<4\), and \(\ln2<7/10\) checks
every entry involving \(\ln2\). Consequently \((3G_2/2-G_1)/(n+1)\) is
greater than

\[
\left(\frac32a_0^{(2)}-a_0\right)n
+\frac32a_1^{(2)}-a_1+\frac12\ln(n+1)
+\frac12\ln d-\ln7.
\tag{10.645}
\]

The coefficient of n is positive: in every row \(a_0\le a_0^{(2)}+\ln2\)
and \(a_0^{(2)}>3\). The displayed expression therefore has its least
value at n=2,d=1. Substitution and cancellation of logarithms gives,
respectively,

\[
\begin{array}{c|c}
\mathrm{I.1}&22/25+\ln(27/4)+\tfrac12\ln3\\
\mathrm{I.2}&27/20+\ln(27/4)+\tfrac12\ln3\\
\mathrm{II}&-3/8+\ln(125/4)+\tfrac12\ln3\\
\mathrm{III.1}&11/40+\tfrac12\ln3\\
\mathrm{III.2}&57/50+\tfrac12\ln3\\
\mathrm{IV}&9/4-2\ln2+\tfrac12\ln3\\
\mathrm V&17/10+\ln(208/7)+\tfrac12\ln3.
\end{array}
\tag{10.646}
\]

Every entry is positive. For II use \(125/4>\mathrm e\), which follows
from \(\mathrm e<3\); for IV use \(\ln2<7/10\). The elementary bounds on
\(\mathrm e\) follow from its positive power series: the partial sum
through degree four exceeds \(27/10\), while \(k!\ge2^{k-1}\), with
strict inequality for k at least four, bounds the full sum above by
three. This proves the first assertion. The other entries of the two
maxima agree. Since \(G_2>0\), each of them is bounded by
\(3\mathsf h_2/2\), including entries that are negative. This proves the
second assertion. \(\square\)

\textbf{Theorem 10.173 (the second estimate when \(n\le Pt\)).} Under
the hypotheses of Theorem 10.152, if \(n\le p^\kappa f\ln p\), then

\[
\operatorname{ord}_{\mathfrak p}(\Xi-1)
<C_2\mathsf h_2\prod_{j=1}^nh(a_j).
\tag{10.647}
\]

\textbf{Proof.} Multiplying the maximum in \(C_1\) by \(t^{n+1}\) gives
exactly the maximum in \(C_2\). Cancelling all common factors, including
\(n!/(n-1)!=n\), yields

\[
\frac{C_2}{C_1}
=\frac{c^{(2)}}c
 \left(\frac{a^{(2)}\mathrm e}{\mathfrak a}\right)^n
 \left(\frac{Pt}{n}\right)^{n-1}.
\tag{10.648}
\]

The last factor is at least one. Outside III, the middle base is
\(\mathrm e/2>27/20>1\), so its least power for n at least two is its
square. In III the first two factors are \((c_*/c)\mathrm e^n x^{1-n}\),
with \(1<x\le4/3\). For fixed x their least value is at n=2, and at n=2
their least value over x is bounded below by \((c_*/c)(27/10)^2(3/4)\).
The resulting strict rational lower bounds for \(C_2/C_1\) are

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}lr@{}}
\toprule\noalign{}
Case & Strict lower bound for \(C_2/C_1\) \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
I.1 & \(174717/62600\) \\
I.2 & \(19683/10600\) \\
II & \(50301/20200\) \\
III.1 & \(72171/47840\) \\
III.2 & \(1218159/716000\) \\
IV & \(881361/536000\) \\
V & \(147987/41200\) \\
\end{longtable}
}

All exceed \(3/2\). The smallest gap is the exact positive fraction
\(411/47840\) in III.1. Lemma 10.172 and Theorem 10.152 now give
\(C_1\mathsf h<C_2\mathsf h_2\), and the asserted valuation bound
follows. No height floor, residue index, field extension or coefficient
ordering has been changed. \(\square\)

For example, over \(\mathbb Q\) at three, \(P=3\) and \(Pt=3\ln3>3\).
Thus the theorem applies to two or three independent rational local
units. In contrast, for an unramified prime \(p\ge5\), \(P=1\): the
condition is \(n\le f\ln p\). The stated range is part of the theorem's
hypothesis.

\subsubsection{A separate bound for the last
coefficient}\label{a-separate-bound-for-the-last-coefficient}

We use \(\varepsilon\) for the coefficient-imbalance parameter; it is
unrelated to the saturated residue index \(\delta(\boldsymbol a)\). The
two alternatives below allow a small last coefficient \(B_n\) to be
retained explicitly.

\textbf{Lemma 10.174 (a logarithm versus a linear term).} Let
\(T,H_0>0\), \(X,B_n,r\ge1\), and \(0<\varepsilon\le1\). Put

\[
\eta=\frac{23}{500},\qquad \Phi=(1+\eta)T,
\qquad Z=\max\{8\Phi X,\exp(H_0/(1+\eta))\}.
\tag{10.649}
\]

Then

\[
T\max\{\ln(B_nXr),H_0\}
\le\max\left\{\Phi\ln\frac{ZB_n}{\varepsilon},
                        \varepsilon r\right\}.
\tag{10.650}
\]

\textbf{Proof.} The positive exponential series gives

\[
\mathrm e>\sum_{j=0}^6\frac1{j!}=\frac{1957}{720}
>\frac{125}{46}=\frac1{8\eta};
\qquad \frac{1957}{720}-\frac{125}{46}=\frac{11}{16560}>0.
\tag{10.651}
\]

For any \(s>0\), differentiation of \(u\exp(-su)\) on \(u\ge0\) proves
\(\ln y\le y^s/(\mathrm e s)\) for \(y\ge1\). Let \(Y=B_nXr\) and
\(A=ZB_n/\varepsilon\). Then \(A\ge\exp(H_0/(1+\eta))>1\). If
\(Y<A^{1+\eta}\), both entries on the left are at most \(\Phi\ln A\),
proving the inequality in this alternative. For \(Y\ge A^{1+\eta}\), use
\(s=\eta/(1+\eta)\). Since \(Z/X\ge8\Phi\),

\[
\varepsilon r=\frac Z X\frac Y A
\ge8\Phi Y^{\eta/(1+\eta)}
>T\ln Y.
\tag{10.652}
\]

The last strict comparison follows from \(8\eta\mathrm e>1\). Also
\(T\ln Y\ge\Phi\ln A\ge T H_0\). This proves the weak scalar inequality,
and proves it strictly in the second alternative. \(\square\)

\textbf{Theorem 10.175 (a two-branch coefficient estimate).} Keep the
hypotheses of Theorem 10.152. Let \(B\ge\max_j|b_j|\),
\(1\le B_n\le B\), and \(|b_n|\le B_n\). Define

\[
\begin{aligned}
h_{\min}&=\min_jh(a_j),&
T&=C_1\prod_{j=1}^nh(a_j),\\
H_0&=\max\{G_1(n,d),(n+1)t\},&
X&=\max\left\{1,\frac1{dW(d)}
            \left(\frac1{h_{\min}}+\frac1{h(a_n)}\right)\right\}.
\end{aligned}
\tag{10.653}
\]

With \(\Phi,Z\) as in Lemma 10.174, for every \(0<\varepsilon\le1\),

\[
\operatorname{ord}_{\mathfrak p}(\Xi-1)
<\max\left\{\Phi\ln\frac{ZB_n}{\varepsilon},
                         \varepsilon\frac B{B_n}\right\}.
\tag{10.654}
\]

\textbf{Proof.} Put \(r=B/B_n\). The nonzero integer \(b_n\) implies
\(B_\dagger\le |b_n|\le B_n\). For every \(j<n\),

\[
\frac1{dW(d)}\left(\frac{|b_n|}{h(a_j)}
                     +\frac{|b_j|}{h(a_n)}\right)
\le B_nXr.
\tag{10.655}
\]

It follows that \(\mathsf h\le\max\{\ln(B_nXr),H_0\}\). Theorem 10.152
is strict. Combining it with the weak scalar inequality of Lemma 10.174
therefore proves the strict valuation estimate, including its scalar
equality case. \(\square\)

Every constant in this result is explicit and independent of B and
\(B_n\). The factor \(1.046\) measures the logarithmic slack; the price
for that small slack is the fixed factor Z. The proof applies to
arbitrary signed coefficients and allows zero coefficients other than
the designated last one. Its transcendence input is exactly the proved
first estimate, not a citation standing in for a proof.

For \((a_1,a_2)=(2,5)\) over \(\mathbb Q\) at three, take
\((b_1,b_2)=(3^k,1)\), with \(k\ge1\). The bases are independent by
their prime valuations. Here \(B_n=1,B=3^k\), and X=1: substituting W(1)
reduces its other entry to \((\ln3)(\ln10)/((\ln5)(\ln6))<1\), since
\((\ln2)(\ln3)<(\ln2)(\ln5)\). Theorem 10.175 reads

\[
v_3(5\cdot2^{3^k}-1)
<\max\{\Phi\ln(Z/\varepsilon),\varepsilon3^k\}.
\tag{10.656}
\]

The exact valuation is one: modulo nine, \(3^k\equiv3\pmod6\),
\(2^{3^k}\equiv8\), and \(5\cdot8-1\equiv3\). Thus \(B_n\) has stayed
fixed while the other coefficient grows.

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={The seven constant comparisons and the logarithmic versus linear branches}]{../figures/yu-coefficient-transfer.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.51. Left: exact rational lower bounds for \(C_2/C_1\)
under \(n\le Pt\); the dashed line is \(3/2\), the factor in Lemma
10.172. Theorem 10.173 proves the resulting second estimate in that
range. Right: the scalar model \(T=X=B_n=1,H_0=2,r=\exp20\) of Lemma
10.174. With \(z=\varepsilon r\), the two curves are \(\Phi\ln(Zr/z)\)
and z. Their upper envelope bounds 20, the original scalar estimate. The
curves illustrate the proved inequality; they are not sampled
prime-ideal valuations. Theorem 10.175 transfers this mechanism to the
logarithmic form.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_coefficient_transfer.py}{Reproducible
calculation and figure}. Human source for the two constants: equations
(1.9)--(1.13), (1.28)--(1.30) of
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
freely accessible paper}.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{59. Congruence selection and a larger analytic
disc}\label{congruence-selection-and-a-larger-analytic-disc}

A congruence between the unit parts of two monomials gives extra
precision for their quotient. Selecting many such monomials costs a
finite index. The useful balance is between that index and the enlarged
disc on which the auxiliary function is normal. We keep the polynomial
indices separate from the monomial exponents throughout the selection.

\subsubsection{Selecting the monomials}\label{selecting-the-monomials}

Let L be a finite extension of \(\mathbb Q_p\), with ring of integers
\(\mathcal O_L\), uniformizer \(\pi\), ramification index e and residue
field of cardinality \(p^f\). Use \(v_p(p)=1\), so that
\(v_p(\pi)=1/e\). For an integer \(j\ge1\), write
\(U_j=1+\pi^j\mathcal O_L\).

\textbf{Theorem 10.176 (congruence selection with all polynomial
indices).} Let \(u_1,\ldots,u_r\in U_j\), let \(S\subset\mathbb Z^r\) be
finite and nonempty, and let E be a finite nonempty set of auxiliary
indices. For an integer \(m\ge0\), put

\[
H_m=\operatorname{im}\left(\mathbb Z^r\longrightarrow U_j/U_{j+m},
      \quad\lambda\longmapsto\prod_{i=1}^ru_i^{\lambda_i}\right),
\qquad h_m=|H_m|.
\tag{10.659}
\]

There is a nonempty \(S_0\subset S\) such that

\[
h_m\le p^{fm},\qquad
|S_0|\ge\left\lceil\frac{|S|}{h_m}\right\rceil,
\qquad |E\times S_0|\ge\frac{|E\times S|}{p^{fm}},
\tag{10.660}
\]

and, for every \(\lambda,\mu\in S_0\),

\[
v_p\left(\prod_i u_i^{\lambda_i-\mu_i}-1\right)
\ge\frac{j+m}{e}.
\tag{10.661}
\]

Every \(\lambda^0\in S_0\) supplies the full reference fibre
\(E\times\{\lambda^0\}\). Moreover, let q be an integer prime to p, with
\(q\ge2\). Restrict \(S_0\) to any nonempty class \(a+q\mathbb Z^r\),
and replace \(\lambda=a+q\mu\) by \(\mu\). The displayed congruence
remains true for the differences of the new exponents, with the same
units, j and m. This remains true after every finite sequence of such
restrictions and divisions.

\textbf{Proof.} For \(s\ge1\), the map

\[
U_s/U_{s+1}\longrightarrow\mathcal O_L/\pi\mathcal O_L,
\qquad 1+\pi^s a\longmapsto a\pmod\pi
\tag{10.662}
\]

is an isomorphism of groups, where the target is additive. It is induced
by the surjective map on \(U_s\) with kernel \(U_{s+1}\), and the
product of two representatives introduces only the extra term
\(\pi^{2s}ab\), which lies in \(\pi^{s+1}\mathcal O_L\). Multiplying the
successive indices proves \(|U_j/U_{j+m}|=p^{fm}\), including m=0.
Negative exponents are allowed because every \(U_s\) is a group: the
inverse identity is \((1+x)^{-1}-1=-x/(1+x)\).

The monomial map is therefore a homomorphism with at most \(p^{fm}\)
values. One of its fibres on S has at least \(\lceil|S|/h_m\rceil\)
elements. Choose that fibre as \(S_0\). Equal images mean exactly that
the quotient belongs to \(U_{j+m}\). The selection is independent of E,
so every auxiliary index remains available and the reference fibre has
the asserted form.

For the final assertion, let two retained exponents be
\(a+q\mu,a+q\nu\). Put \(w=\prod_i u_i^{\mu_i-\nu_i}\in U_j\). The
original congruence says \(w^q\in U_{j+m}\). But

\[
w^q-1=(w-1)(1+w+\cdots+w^{q-1}),
\qquad 1+w+\cdots+w^{q-1}\equiv q\pmod\pi.
\tag{10.663}
\]

The second factor is a unit, giving \(v_p(w^q-1)=v_p(w-1)\). Thus
\(w\in U_{j+m}\). Induction proves persistence through every subsequent
restriction and division. \(\square\)

In particular, E can be \(\{0,\ldots,D_{-1}\}\times\{0,\ldots,D_0\}\),
while S already satisfies a prescribed residue-phase condition. The
phase subclasses of \(S_0\) are intersections with the original
subclasses; their number cannot increase. The selected coefficient set
retains both complete polynomial-index intervals.

For a concrete example, take \(L=\mathbb Q_3\),
\(u_1=10,u_2=19,j=2,m=1\), and \(S=\{0,\ldots,5\}^2\). Since

\[
10^{\lambda_1}19^{\lambda_2}
\equiv1+9(\lambda_1+2\lambda_2)\pmod{27},
\tag{10.664}
\]

there are three classes of twelve exponents each. The zero class
contains the three even-coordinate exponents \((0,0),(2,2),(4,4)\).
Division of those coordinates by two gives \((0,0),(1,1),(2,2)\), whose
monomials are still one modulo 27.

\subsubsection{Choosing the congruence
level}\label{choosing-the-congruence-level}

Let \(t=f\ln p\), and let A be a positive real number. For a positive
integer r put \(\rho=t/r\) and

\[
\chi(x)=(A+x)\exp(-\rho x),\qquad x_0=1/\rho-A.
\tag{10.665}
\]

The quantity \(p^{fm}(A+m)^{-r}=\chi(m)^{-r}\) combines the worst-case
index loss with r powers of the analytic gain.

\textbf{Lemma 10.177 (integer optimization with explicit rounding
losses).} Suppose \(A\ge A_0\) and \(x_0>3/5\). Choose m by

\[
\begin{cases}
m=1,&3/5<x_0\le1,\\
m=k,&x_0>1,\quad k=\lfloor x_0\rfloor,
       \quad x_0-k\le 1/\rho-1/(\exp\rho-1),\\
m=k+1,&\text{otherwise}.
\end{cases}
\tag{10.666}
\]

This m maximizes \(\chi\) among the positive integers. For each row in
the following table it also satisfies

\[
A+m<\varsigma_1\frac r t,\qquad
\chi(m)>\frac{\chi(x_0)}{\varsigma_2}.
\tag{10.667}
\]

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}rrr@{}}
\toprule\noalign{}
\(A_0\) & \(\varsigma_1\) & \(\varsigma_2\) \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
\(3/2\) & \(1.2133\) & \(1.0202\) \\
\(3/4\) & \(1.3128\) & \(1.0415\) \\
\(1/2\) & \(1.3701\) & \(1.0568\) \\
\(2\) & \(1.176\) & \(1.014\) \\
\end{longtable}
}

Consequently, throughout the same ranges,

\[
\frac{p^{fm}}{(A+m)^r}
<\varsigma_2^r\left(\frac{\mathrm e t}{r}\right)^r p^{-fA}.
\tag{10.668}
\]

\textbf{Proof.} Differentiation gives
\(\chi'(x)=\exp(-\rho x)(1-\rho(A+x))\). If \(x_0\le1\), the greatest
value at a positive integer is at one. Otherwise it suffices to compare
k and k+1. Their quotient is \(\exp(-\rho)(A+k+1)/(A+k)\), and
comparison with one gives exactly the threshold in the statement. The
threshold is between zero and one. Positivity follows from
\(\exp\rho>1+\rho\). For \(\rho\ge1\) the upper bound is immediate; for
\(0<\rho<1\), the power series gives \(\exp\rho<1/(1-\rho)\), which
gives the upper bound as well.

For \(x_0>1\), write \(x_0=k+v\), and define

\[
z(y)=\frac y{\exp y-1},\qquad
H(y)=y+z(y)=\frac y{1-\exp(-y)},\qquad
F(y)=z(y)\exp(1-z(y)).
\tag{10.669}
\]

The relative values at k and k+1 are \(\exp(\rho v)(1-\rho v)\) and
\(\exp(-\rho(1-v))(1+\rho(1-v))\), respectively. The first decreases in
v and the second increases. Their maximum therefore has its least value
at their intersection, where it is \(F(\rho)\). If m=k, then
\(\rho(A+m)\le1\). If m=k+1, the threshold gives
\(\rho(A+m)\le H(\rho)\). Since \(\rho<1/(A+1)\le1/(A_0+1)\), it follows
that

\[
\rho(A+m)\le H(1/(A_0+1)),\qquad
\frac{\chi(m)}{\chi(x_0)}\ge F(1/(A_0+1)).
\tag{10.670}
\]

Here H increases, z decreases from one to zero, and F decreases. For H
the derivative has the sign of \(\exp y-1-y>0\). For z it has the sign
of \(\exp y(1-y)-1<0\). Finally \(z\exp(1-z)\) increases for \(0<z<1\).
These derivative tests prove all the required monotonicity statements.

When \(3/5<x_0\le1\), m=1 and \(\rho(A+1)<(A_0+1)/(A_0+3/5)\). Also

\[
\frac{\chi(1)}{\chi(x_0)}
=\exp(-c)(1+c),\qquad
c=\rho(1-x_0)<\frac{2/5}{A_0+3/5}.
\tag{10.671}
\]

The function \((1+c)\exp(-c)\) decreases for c nonnegative. Thus both
alternatives are reduced to four fixed endpoints. The following decimals
are rational upper or lower endpoints with denominator \(10^{12}\):

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}
  >{\raggedleft\arraybackslash}p{(\linewidth - 6\tabcolsep) * \real{0.2500}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 6\tabcolsep) * \real{0.2500}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 6\tabcolsep) * \real{0.2500}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 6\tabcolsep) * \real{0.2500}}@{}}
\toprule\noalign{}
\begin{minipage}[b]{\linewidth}\raggedleft
\(A_0\)
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
Upper bound for H
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
Lower bound for F
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
Lower bound for the forced ratio
\end{minipage} \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
\(3/2\) & \(1.213297912688\) & \(0.980242084353\) &
\(0.984006473362\) \\
\(3/4\) & \(1.312778225605\) & \(0.960181882841\) &
\(0.963883250822\) \\
\(1/2\) & \(1.370098893207\) & \(0.946280756540\) &
\(0.947923538725\) \\
\(2\) & \(1.175908824386\) & \(0.986228202757\) & \(0.989312214415\) \\
\end{longtable}
}

To verify these without a numerical exponential, for \(0\le y\le1\) use

\[
E_{64}(y)=\sum_{i=0}^{64}\frac{y^i}{i!}
\le\exp y\le E_{64}(y)+\frac{y^{65}}{65!}\frac1{1-y/66}.
\tag{10.672}
\]

The upper tail follows by bounding every subsequent term ratio by y/66.
These rational bounds enclose z. Evaluate \(z\exp(1-z)\) at its lower z
endpoint, using its proved monotonicity, and round down; evaluate H with
the upper z endpoint and round up. For the forced ratio divide \(1+c\)
by an upper bound for \(\exp c\). This gives exactly the displayed
endpoints. Each H endpoint is strictly below \(\varsigma_1\); each lower
ratio is strictly above \(1/\varsigma_2\), as multiplication of positive
rational denominators verifies. The other height bounds are
\(25/21,35/27,15/11,15/13\), respectively, and are also strictly below
\(\varsigma_1\). The
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_congruence_clusters.py}{exact calculation}
records all rational gaps. The monotonicity argument supplies every
allowed A and \(\rho\), rather than a finite sampling of them.

Finally, \(\chi(x_0)=\rho^{-1}\exp(-1+\rho A)\). Raising its strict
reciprocal comparison to the rth power, and using \(\rho r=f\ln p\),
proves the last assertion. \(\square\)

The four rows apply to the local radii used earlier. For
\(A=e\vartheta\), cases I and II of Theorem 10.152 have \(A\ge3/2\);
case III has \(A\ge3/4\); case IV has \(A\ge1/2\); and case V has
\(A\ge2\). Indeed \(e\vartheta=P/2\) outside the unramified odd-prime
case, while in that case \(\vartheta=(p-2)/(p-1)\). The inequalities
defining P give \(P\ge3\) at three, \(P\ge5\) at a ramified five,
\(P\ge1\) at an odd prime at least seven, and \(P\ge4\) at two.

\subsubsection{Normality and divided
derivatives}\label{normality-and-divided-derivatives}

The next statement includes a perturbation of the logarithmic
frequencies. This permits the coefficient relation to be imposed without
losing the enlarged disc.

\textbf{Theorem 10.178 (normality from a congruence class).} Use a class
\(S_0\) from Theorem 10.176 and a reference \(\lambda^0\in S_0\).
Suppose \(j/e>1/(p-1)\), and put

\[
\gamma_\lambda=\ln\left(\prod_i u_i^{\lambda_i-\lambda_i^0}\right).
\tag{10.673}
\]

Let \(N\ge0\) be rational with \(N<(j+m)/e-1/(p-1)\), and let
\(\epsilon_\lambda\in\mathbb C_p\) satisfy
\(v_p(\epsilon_\lambda)>N+1/(p-1)\). Let \(P_\lambda\in\mathbb C_p[z]\)
have degree at most D and all coefficients of valuation at least \(-C\),
where \(D\in\mathbb Z_{\ge0}\), \(C\in\mathbb Q_{\ge0}\). Take
coefficients \(c_\lambda\in\mathbb C_p\), not all zero, and set
\(\beta=\min_\lambda v_p(c_\lambda)\). Then

\[
f(z)=\sum_{\lambda\in S_0}c_\lambda P_\lambda(z)
                 \exp\bigl(z(\gamma_\lambda+\epsilon_\lambda)\bigr)
\tag{10.674}
\]

converges on \(|z|_p\le p^N\). Choose \(\omega,\xi\in\mathbb C_p\) with
\(v_p(\omega)=-N\) and \(v_p(\xi)=C+DN-\beta\). The function
\(\xi f(\omega z)\) is normal on the unit disc: its power-series
coefficients are integral and tend to zero. Moreover, for every
\(a\in\mathbb C_p\) with \(v_p(a)\ge0\) and every integer \(k\ge0\),

\[
v_p\left(\frac{f^{(k)}(a)}{k!}\right)
\ge\beta-C-DN+kN.
\tag{10.675}
\]

\textbf{Proof.} The required scalar valuations are rational. Indeed, a
nonzero element of \(\mathbb C_p\) has an algebraic approximation close
enough to have the same valuation; that approximation lies in a finite
extension of \(\mathbb Q_p\). Roots of integral powers of p therefore
supply \(\omega\) and \(\xi\). The deep-unit logarithm law of
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/TR-BAKER-09.html\#3-the-exponential-and-the-logarithm}{Proposition
9.3} gives \(v_p(\gamma_\lambda)\ge(j+m)/e\). Therefore
\(v_p(\omega(\gamma_\lambda+\epsilon_\lambda))>1/(p-1)\). For any w of
valuation greater than \(1/(p-1)\), the coefficients of \(\exp(wz)\)
satisfy

\[
v_p(w^s/s!)\ge s\bigl(v_p(w)-1/(p-1)\bigr),
\tag{10.676}
\]

and tend to infinity. Here
\(v_p(s!)=\sum_{i\ge1}\lfloor s/p^i\rfloor\le s/(p-1)\), as proved in
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/TR-BAKER-09.html\#3-the-exponential-and-the-logarithm}{Section
3 of Lesson 9}. Thus each rescaled exponential is normal. Every
coefficient of \(P_\lambda(\omega z)\) has valuation at least \(-C-DN\).
Multiplication by \(\xi c_\lambda\) makes all coefficients integral.
Finite sums and multiplication by polynomials preserve convergence and
this coefficient bound, proving normality and the asserted convergence
of f.

For a normal series \(F(z)=\sum_s A_sz^s\), the kth divided derivative
at an integral point b is \(\sum_{s\ge k}A_s\binom sk b^{s-k}\). This
series converges and every summand is integral. Apply this to
\(F(z)=\xi f(\omega z)\) at \(b=a/\omega\), which is integral because N
is nonnegative. The chain rule gives
\(F^{(k)}(a/\omega)=\xi\omega^k f^{(k)}(a)\). Taking valuations gives
precisely the displayed divided-derivative bound. \(\square\)

\textbf{Corollary 10.179 (prepared polynomial factors).} For each
coefficient index, choose integers
\(k\ge1,\ell\ge1,t_0\ge0,I\ge0,a\in\mathbb Z\) with \(k\ell\le D\), and
any finite list of integers \(\gamma_i,t_i\), with \(t_i\ge0\). With
\(V(k)=\operatorname{lcm}(1,\ldots,k)\), let

\[
P(z)=\frac{V(k)^{t_0}}{t_0!}
 \left(\frac d{dz}\right)^{t_0}
 \bigl(\Delta(q^{-I}z+a;k)^\ell\bigr)
 \prod_i\Delta(\gamma_i;t_i).
\tag{10.677}
\]

These factors satisfy Theorem 10.178 with \(C=D/(p-1)\). Thus its
normalizing scalar can be chosen with valuation \(D(N+1/(p-1))-\beta\).
The same coefficient bound holds when the derivative is taken before the
affine substitution.

\textbf{Proof.} The numerator of \(\Delta(q^{-I}z+a;k)^\ell\) has
integral p-adic coefficients, since q is prime to p.~Its denominator is
\((k!)^\ell\), of valuation at most \(k\ell/(p-1)\le D/(p-1)\). Divided
differentiation sends \(z^s\) to \(\binom{s}{t_0}z^{s-t_0}\), so it
preserves this coefficient bound. The factor \(V(k)^{t_0}\) is an
integer, as is every \(\Delta(\gamma_i;t_i)\). For a negative integer
\(\gamma_i\), the latter assertion follows either from a zero factor or
from \(\Delta(-b;t)=(-1)^t\binom{b-1}{t}\) when \(b>t\). Finally, the
chain-rule difference between the two derivative conventions is a power
of q, which is a p-adic unit. The degree remains at most D, so Theorem
10.178 applies. \(\square\)

In the parameters of Lemma 10.177, take
\(j=\lceil e(\vartheta+1/(p-1))\rceil\) and
\(N=(m+e\vartheta)/(e(1+\tau))\), with any positive rational \(\tau\)
for which N is rational. Then \(N<(j+m)/e-1/(p-1)\). Theorems 10.176 and
10.178 therefore give the enlarged disc and preserve it at every
prime-to-p descent stage. Corollary 10.179 gives the full normalization
of the prepared polynomial factors. Global height and product-formula
estimates continue to use their original coefficient field and degree.

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={Congruence classes of monomials and the integer-depth rounding efficiency}]{../figures/yu-congruence-clusters.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.52. Left: the 36 exponents in \(\{0,\ldots,5\}^2\),
colored by \(\lambda_1+2\lambda_2\pmod3\). Each class has twelve points.
The circled exponents \((0,0),(2,2),(4,4)\) contract to
\((0,0),(1,1),(2,2)\), with the same quotient precision modulo 27.
Theorem 10.176 proves the selection and its persistence, including every
auxiliary polynomial index. Right: the scalar rounding efficiency for
\(A=1/2\). The dashed line is the uniform bound \(1/1.0568\); the curve
illustrates Lemma 10.177 and does not supply its infinite-parameter
proof. Theorems 10.178--10.179 turn the retained congruence into a
normal-function and divided-derivative estimate.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_congruence_clusters.py}{Reproducible exact
calculation and figure}. Human-source context: Sections 2, 10 and 11 of
\href{https://matwbn.icm.edu.pl/ksiazki/aa/aa89/aa8944.pdf}{Yu's free
1999 paper}.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{60. Projecting the selected auxiliary
function}\label{projecting-the-selected-auxiliary-function}

The congruence selection concerns a monomial, rather than each of its
individual factors. We now impose the original coefficient relation on
that monomial and carry the resulting precision through interpolation.
The cost of selecting the monomials must also remain in the global
coefficient count.

\subsubsection{The original coefficient
relation}\label{the-original-coefficient-relation}

Let K be a number field, let its chosen completion at p have
ramification index e, and write \(v_p(p)=1\). Take local units
\(a_1,\ldots,a_n\in K^*\), integers \(b_1,\ldots,b_n\) with \(b_n\ne0\),
and put \(\Xi=\prod_j a_j^{b_j}\). Let \(P=p^\kappa\),
\(\kappa\in\mathbb Z_{\ge0}\), and let \(\zeta\) be a root of unity of
order G prime to p in the completion. Suppose integers \(c_j\) make
\(u_j=a_j^P\zeta^{c_j}\) deep units: \(v_p(u_j-1)>1/(p-1)\). Set
\(z_j=\log u_j\). For integers \(A_{ij}\), let
\(L_i(Z)=\sum_j A_{ij}Z_j\), \(1\le i\le r\). Choose a prime \(q\ne p\)
and \(\nu\in\mathbb Z_{\ge0}\), and set \(w_i=\exp(q^{-\nu}L_i(z))\).
These exponentials lie in the chosen completion, by its completeness and
the logarithmic convergence estimates of Lesson 9. Suppose
\(w_i\in U_j\), with \(j/e>1/(p-1)\), and select \(S_0\) as in Theorem
10.176.

\textbf{Theorem 10.180 (the coefficient relation on a selected class).}
Suppose \(V=v_p(\Xi-1)>1/(p-1)\) is finite. Then

\[
\sum_j c_jb_j\equiv0\pmod G,\qquad
\mathcal L:=\sum_j b_jz_j=P\log\Xi,\qquad
v_p(\mathcal L)=\kappa+V.
\tag{10.681}
\]

Fix \(\lambda^0\in S_0\), write \(d_i=\lambda_i-\lambda_i^0\), and let

\[
\begin{aligned}
M_i(Z)&=L_i(Z)-\frac{A_{in}}{b_n}\sum_jb_jZ_j,\\
\gamma_\lambda&=\sum_i d_iq^{-\nu}L_i(z)
 =\log\left(\prod_iw_i^{d_i}\right),\\
\epsilon_\lambda&=-\frac{\mathcal L}{q^\nu b_n}
                         \sum_i d_iA_{in},\qquad
H=\kappa+V-v_p(b_n).
\end{aligned}
\tag{10.682}
\]

The coefficient of \(Z_n\) in every \(M_i\) is zero, and

\[
q^{-\nu}\sum_i d_iM_i(z)
 =\gamma_\lambda+\epsilon_\lambda,\qquad
v_p(\gamma_\lambda)\ge\frac{j+m}{e},\qquad
v_p(\epsilon_\lambda)\ge H.
\tag{10.683}
\]

Consequently, whenever \(N\ge0\) is rational and

\[
N<\frac{j+m}{e}-\frac1{p-1},\qquad H>N+\frac1{p-1},
\tag{10.684}
\]

the projected frequencies satisfy Theorem 10.178 on the whole closed
disc \(|y|_p\le p^N\). This conclusion holds for every retained
auxiliary polynomial index. It persists through every prime-to-p
restriction and exponent division of Theorem 10.176.

\textbf{Proof.} The identity
\(\prod_j u_j^{b_j}=\Xi^P\zeta^{\sum_jc_jb_j}\) shows that the displayed
root of unity reduces to one. Reduction is injective on roots of unity
of order prime to p: if \(t^G=1\) and \(v_p(t-1)>0\), the factorization
of \(t^G-1\) has second factor congruent to the p-adic unit G, forcing
\(t=1\). This proves the phase congruence. All remaining factors are on
the deep-unit disc. The logarithm product and power laws in Propositions
9.3--9.4 therefore give \(\mathcal L=P\log\Xi\), with the stated exact
valuation.

The definition of \(M_i\) cancels its last coefficient exactly. Summing
its identity with coefficients \(q^{-\nu}d_i\) gives the displayed
perturbation. The exponential and logarithm are inverse on this disc, so
their product law identifies \(\gamma_\lambda\) with the logarithm of
the quotient monomial. Theorem 10.176 gives its valuation at least
\((j+m)/e\). The sum \(\sum_i d_iA_{in}\) is an integer and q is a
p-adic unit. Thus the perturbation has valuation at least H, including a
zero perturbation. The two strict disc conditions are precisely those of
Theorem 10.178. Its proof applies to the finite sum over all auxiliary
indices as well as over the exponent indices.

For a restriction \(\lambda=a+q\mu\) and corresponding reference
\(\lambda^0=a+q\mu^0\), both differences divide by q. Thus the new
\(\gamma\) and \(\epsilon\) are the old ones divided by q; their
valuations are unchanged. Iteration proves the last assertion. No new
global field or replacement coefficient relation has been used.
\(\square\)

In a contradiction argument with \(V\ge U\), a bound \(\log|b_n|\le h\)
gives \(H\ge\kappa+U-h/\log p\). Indeed \(p^{v_p(b_n)}\le|b_n|\). This
makes the strict perturbation condition explicit. In applications to
Theorem 10.152 the original minimum-valuation choice of \(b_n\) is
retained.

\subsubsection{The error and its ordinary
derivatives}\label{the-error-and-its-ordinary-derivatives}

\textbf{Lemma 10.181 (Taylor precision of the projection error).} Use
the frequencies of Theorem 10.180 and a positive rational N satisfying
its two disc conditions. Let \(P_\lambda\) have degree at most D and
coefficients of valuation at least \(-C\), with \(D\ge0\) integral and
\(C\ge0\) rational. For a finite nonzero coefficient vector
\(c=(c_\lambda)\), put \(\beta=\min_\lambda v_p(c_\lambda)\), and define

\[
\Phi(y)=\sum_\lambda c_\lambda P_\lambda(y)
                 \exp(y\gamma_\lambda),\qquad
f(y)=\sum_\lambda c_\lambda P_\lambda(y)
                 \exp(y(\gamma_\lambda+\epsilon_\lambda)).
\tag{10.685}
\]

Both functions have the normality conclusion of Theorem 10.178. For
every integral \(a\in\mathbb C_p\) and every \(t\ge0\),

\[
v_p\left(\frac{(f-\Phi)^{(t)}(a)}{t!}\right)
\ge\beta-C+H+\max\{0,(t-D-1)N\}.
\tag{10.686}
\]

If the \(P_\lambda\) are the prepared factors in (10.92), they are
p-integral at every \(a\in\mathbb Z_p\). For these factors the sharper
order-zero error is

\[
v_p(f(a)-\Phi(a))\ge\beta+H
                       \qquad(a\in\mathbb Z_p).
\tag{10.687}
\]

\textbf{Proof.} Normality follows from Theorem 10.178, first with zero
perturbations and then with the actual perturbations. Write
\(\sigma=(j+m)/e\) and \(L=\min\{\sigma,H\}>N+1/(p-1)\). For one
summand, consider \(E(y)=\exp((\gamma+\epsilon)y)-\exp(\gamma y)\). At
an integral a, \(v_p(E(a))\ge H\). For \(l\ge1\),

\[
\frac{E^{(l)}(a)}{l!}
=\frac{\exp(\gamma a)}{l!}
 \left[\gamma^l(\exp(\epsilon a)-1)
 +\exp(\epsilon a)\sum_{h=1}^l
           \binom lh\gamma^{l-h}\epsilon^h\right].
\tag{10.688}
\]

The bracket has valuation at least \(H+(l-1)L\). The factorial identity
of Lesson 9 gives \(v_p(l!)=(l-s_p(l))/(p-1)\le(l-1)/(p-1)\), where
\(s_p(l)\ge1\) is the base-p digit sum. Therefore the displayed divided
derivative has valuation at least \(H+(l-1)N\). All divided derivatives
of \(P_\lambda\) at an integral point have valuation at least \(-C\),
since their Taylor coefficients are sums of integral binomial multiples
of its coefficients. They vanish above D. In the divided product rule
the remaining exponential order is at least \(t-D\). The bounds for
orders zero and positive orders consequently give the asserted maximum.
Multiplication by the coefficients and summation give the result.

For a prepared factor, the integer-valued Taylor proof in Lemma 10.28,
extended to \(\mathbb Z_p\) in Lemma 10.36, gives
\(v_p(P_\lambda(a))\ge0\). Its exponential error has valuation at least
H, by Proposition 9.3. This proves the last bound. \(\square\)

\subsubsection{Interpolation on the enlarged
disc}\label{interpolation-on-the-enlarged-disc}

\textbf{Theorem 10.182 (selected-support extrapolation with every jet
cost).} Use the full prepared family (10.92) on the selected support,
with \(k,L\ge1\), \(D=kL\), coefficient minimum \(\beta\), and the
actual selected derivations of Lemma 10.23. Let the unprojected and
projected functions be \(\Phi(y;\boldsymbol t)\) and
\(f(y;\boldsymbol t)\) from Theorem 10.180. Suppose N is positive and
rational and satisfies that theorem's two strict disc conditions. Put

\[
\begin{aligned}
C_0&=\max\{v_p(V(k)),v_p(b_n)\},\qquad
\Gamma=D\left(N+\frac1{p-1}\right)-\beta,\\
M_0&=\max\{B,C_0\},\qquad V(k)=\operatorname{lcm}(1,\ldots,k).
\end{aligned}
\tag{10.689}
\]

Take one of the full or q-deleted integer node sets E of Theorem 10.34.
Let \(n_E=|E|\), let B be its separation bound, and let \(\eta_E=0\) for
the full interval and \(\eta_E=1\) for the q-deleted interval. Suppose
\(\mu\ge1\), \(|\boldsymbol t|+\mu-1\le T\), and
\(\Phi(s;\boldsymbol t')=0\) at every node for every
\(|\boldsymbol t'|\le T\). Then, for every
\(x\in\mathbb Z_p\subset\mathbb Q_p\), both f and \(\Phi\) have
valuation at least

\[
\min\left\{n_E\mu N-\Gamma,\quad
\beta+H-\eta_E\mu B-(\mu-1)M_0\right\}.
\tag{10.690}
\]

In particular, under the counterhypothesis \(V\ge U\), the budget

\[
\kappa+U+D\left(N+\frac1{p-1}\right)
\ge n_E\mu N+(\eta_E+1)\mu M_0
\tag{10.691}
\]

gives

\[
v_p(f(x;\boldsymbol t)),\ v_p(\Phi(x;\boldsymbol t))
\ge n_E\mu N-D\left(N+\frac1{p-1}\right)+\beta.
\tag{10.692}
\]

The strict perturbation condition must still hold. A sufficient version
under this counterhypothesis is \(\kappa+U-v_p(b_n)>N+1/(p-1)\).

\textbf{Proof.} At a node, Lemma 10.181's prepared-factor estimate and
the assumed zeros give \(v_p(f(s;\boldsymbol t'))\ge\beta+H\). Each
individual projected slope has valuation greater than \(1/(p-1)\): the
unprojected slope has that property by its definition from the deep
\(z_j\), and its correction has valuation at least H. Thus Lemma 10.38
applies with exactly the selected directions, yielding
\(v_p(f_l(s;\boldsymbol t))\ge\beta+H-lC_0\) for \(0\le l<\mu\).

Theorem 10.178 and Corollary 10.179 supply a scalar \(\xi\) of valuation
\(\Gamma\) for which \(F(Z)=\xi f(\omega Z;\boldsymbol t)\) is normal,
where \(v_p(\omega)=-N\). At the scaled node \(s/\omega\), its weighted
jet precision is \(\Gamma+\beta+H-lC_0\), by the chain rule. Apply
Theorem 10.39 with \(\rho=1/\omega\), \(\theta=N\), and its node
indicator equal to \(\eta_E\); subtracting \(\Gamma\) gives the
displayed minimum for f.~The prepared projection error at x is at least
\(\beta+H\), which is no smaller than the second entry of that minimum.
The same minimum therefore applies to \(\Phi\).

Finally \(H\ge\kappa+U-v_p(b_n)\), while

\[
v_p(b_n)+\eta_E\mu B+(\mu-1)M_0
\le(\eta_E+1)\mu M_0.
\tag{10.693}
\]

The budget makes the second entry of the minimum at least the first.
This proves the last conclusion. In this proof only the combined
monomial frequencies need the enlarged precision; the ordinary jet
identity uses the original individual slopes on their original disc.
\(\square\)

\subsubsection{Columns and heights after the
selection}\label{columns-and-heights-after-the-selection}

Retain the global monomial and height setting of Theorems 10.63--10.65.
Let the exponent lattice have covolume \(\Delta\), let the original
coordinate box have full widths \(D_1,\ldots,D_r>0\), and put
\(\mathcal V=\prod_iD_i\). Let the phase map have \(H_{\mathrm{ph}}\)
distinct possible values, as in Lemma 10.67. Let \(h_m\) be the actual
finite unit-quotient image order in Theorem 10.176.

\textbf{Proposition 10.183 (the selected coefficient surplus and
height).} The successive closed-box, phase and unit-congruence
selections give a support \(S_0\) and \(N_c=kL|S_0|\) columns satisfying

\[
N_c\ge kL\left\lceil
 \frac{\left\lceil(\lfloor\mathcal V/\Delta\rfloor+1)/H_{\mathrm{ph}}\right\rceil}
      {h_m}\right\rceil
>\frac{kL\mathcal V}{\Delta H_{\mathrm{ph}}h_m}.
\tag{10.694}
\]

Let \(S\ge1,T\ge0\), \(s_0=\lfloor S\rfloor\), \(t_0=\lfloor T\rfloor\),
and count the whole prepared system by

\[
M=(2s_0+1)\binom{t_0+r}{r}.
\tag{10.695}
\]

If \(kL\mathcal V/(\Delta H_{\mathrm{ph}}h_m)\ge c_*M\), with \(c_*>1\),
there is a nonzero integral coefficient vector giving all these zeros.
For its actual row field \(E_0\), of degree \(d_0\), put
\(L_\alpha=\sum_iD_i h(\alpha_i)\). With the complete scalar bound
\(\mathcal B_0^*(s_0,t_0)\) of Corollary 10.64 it can be chosen with

\[
\begin{aligned}
h_2(c)\le{}&\frac12\log(N_cd_0)+(d_0-1)P L_\alpha\\
&+\frac1{c_*-1}\left[
 \frac12\log N_c+\log\mathcal B_0^*(s_0,t_0)
 +\frac{Ps_0(s_0+1)}{2s_0+1}L_\alpha\right].
\end{aligned}
\tag{10.696}
\]

An actual generator tower or discriminant may replace the corresponding
upper bound as in Theorem 10.63. Every auxiliary pair
\(0\le a<k,1\le\ell\le L\) remains available at every selected exponent;
the resulting Laurent polynomial is nonzero.

\textbf{Proof.} Theorem 10.66 supplies
\(\lfloor\mathcal V/\Delta\rfloor+1\) lattice points. Lemma 10.67
retains at least the inner ceiling, and Theorem 10.176 retains at least
its quotient ceiling. Since
\(\lfloor\mathcal V/\Delta\rfloor+1>\mathcal V/\Delta\), the displayed
strict comparison follows. The auxiliary pairs are independent of both
selections, so their full factor is kL.

There are \(2s_0+1\) nodes and \(\binom{t_0+r}{r}\) indices of total
degree at most \(t_0\), by the divider count already proved in Theorem
10.32. Keep this entire list of M rows, including redundant or zero
rows, rather than changing its average by a nonuniform deletion. The
assumed inequality gives \(N_c>c_*M>M\). The integral Siegel proof in
Theorem 10.63 therefore applies.

The new support is a subset of the old box. After centering at one of
its own points it has the same full-width bounds, so (10.173) still
bounds the generator and both torus heights. The actual row field is
generated by the new normalized monomials: the zero-derivative row at
\(s=1,a=0,\ell=1\) contains \((k+1)W_\lambda\), and the reference
monomial is one. This is the exact field-recovery argument of Theorem
10.63 and requires the retained full auxiliary product. It uses no
ambient-degree replacement. Applying (10.174) to this full row list, and
using \(M/(N_c-M)<1/(c_*-1)\), proves the height bound. The mean torus
term is unchanged because each integer node has the same full index
list; \(\sum_{s=-s_0}^{s_0}|s|=s_0(s_0+1)\). Finally Lemma 10.29 gives
an independent binomial polynomial family for each of the distinct
Laurent monomials. The nonzero coefficient vector therefore gives a
nonzero Laurent polynomial. \(\square\)

The factor \(h_m\le p^{fm}\) in the column condition is essential. Lemma
10.177 controls how that finite-index cost trades against the larger
analytic radius. A parameter choice for a final logarithmic estimate
must satisfy both this coefficient inequality and the strict analytic
and arithmetic comparisons in the zero induction.

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={Selected monomials, projected logarithmic precision and the global column surplus}]{../figures/yu-cluster-projection.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.53. Take \(p=3\), \(u_1=10\), \(u_2=3281/500\), and
\(\Xi=u_1^3u_2=1+3^8\). Upper left: the selected quotient class modulo
27 consists of the twelve points in \(\{0,3\}\times\{0,\ldots,5\}\).
Each retains all four polynomial indices. The circled points have
quotient \(\Xi\) and equal projected frequency. Upper right: the
individual logarithm valuations are exactly two and three; the selected
quotients have proved minimum three, attained at \((3,0)\). The relation
has exact valuation eight. The dashed threshold \(N+1/(p-1)=5/2\) is
strictly below both the quotient minimum and the perturbation precision,
giving normality at \(N=2\). Below: selection leaves 48 of the original
144 columns, against eighteen full prepared rows. The uniform
normalization cost is ten and the output precision is eight, with
interpolation budget \(18=10+8\). Theorems 10.180--10.182 and
Proposition 10.183 prove the mechanism; Solution 74 verifies the exact
rational example and its integral kernel.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_cluster_projection.py}{Reproducible figure
and exact calculation}. Human-source context: Sections 10--11 of
\href{https://matwbn.icm.edu.pl/ksiazki/aa/aa89/aa8944.pdf}{Yu's freely
accessible 1999 paper}.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{61. Constructing the auxiliary function at the second
scale}\label{constructing-the-auxiliary-function-at-the-second-scale}

We now choose the size of the selected support explicitly. Its defining
inequality contains the actual unit-congruence index. Optimizing that
index against the analytic radius removes the factor involving a large
power of the rank. We keep the global residue index, the saturation
index, the polynomial indices and the integer floors throughout.

\subsubsection{Parameters and the selected
support}\label{parameters-and-the-selected-support}

Use a field case of Theorem 10.152. Write \(t=f\log p\), \(P=p^\kappa\),
\(\ell_d=\max\{1,\log d\}\), and let \(\vartheta\) have its earlier
value. Thus \(t>1\), \(A_0=e\vartheta\ge1/2\), and \(A_0\le P\). Let
\(3\le r\le n\), and suppose

\[
\frac r t>A_0+\frac35.
\tag{10.700}
\]

For independent local units \(\alpha_1,\ldots,\alpha_r\), take weights
\(\sigma_i\ge h(\alpha_i)>0\). Put \(\Omega_r=\prod_i\sigma_i\), let J
be their q-primary saturation index, and let \(\delta_r\) be their
saturated residue index. Thus \(J=q^\nu\) for an integer \(\nu\ge0\),
the exponent lattice has covolume \(1/J\), and its phase map has
\((p^f-1)/(q^u\delta_r)\) possible values. The saturation and phase
constructions are those of Sections 8, 18 and 25.

Choose \(m,\zeta_1,\zeta_2\) by Lemma 10.177, using its row for the
current field case. Set

\[
\begin{aligned}
\tau&=\frac{10^{-26}}{2n},\qquad N=\frac{m+A_0}{e(1+\tau)},\\
\mathcal R_r&=\max\left\{\frac{p^f}{\delta_r},
                  \left(\frac{\mathrm e}{r}\right)^rt^{r+1}\right\},\\
A_r&=\max\{4+\log((r+1)d),e,t\}.
\end{aligned}
\tag{10.701}
\]

Here \(\mathrm e=\exp(1)\). It is distinct from the ramification index
e. Take a real \(h\ge(r+1)t\), and put \(H=h+\log J\). The following
constants will be used:

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}lrrrrr@{}}
\toprule\noalign{}
Field cases & \(c_0\) & \(c_1\) & \(c_3\) & \(c_4\) & a \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
I.1, I.2, II & \(27/10\) & \(7/10\) & \(6/5\) & 14 & 7 \\
III.1, III.2 & \(27/10\) & \(7/10\) & \(8/5\) & 14 & 7 \\
IV & \(27/10\) & \(7/10\) & 3 & 14 & 7 \\
V & \(27/10\) & \(41/50\) & \(3/5\) & 6 & 13 \\
\end{longtable}
}

Define the positive real parameters

\[
\begin{aligned}
c_2&=\frac{a}{q^2\zeta_2},\quad g_S=c_3q(r+1)^2d,\quad
K_S=\frac{101}{100}\left(2+\frac1{g_S}\right),\\
S&=\frac{c_3q(r+1)dH}{t},\\
\mathcal D&=K_Sc_0c_1c_4\frac N\vartheta
 (\zeta_2c_2q\mathrm e P)^r\frac{(r+1)^r}{r!}\\
&\qquad{}\times\frac{d^{r+1}\ell_d\Omega_r}{q^uJ}
 \mathcal R_rA_r(1+\tau)^r,\\
T&=\frac{q(r+1)\mathcal D}{c_1eNt},\qquad
D_i=\frac{\mathcal D}{c_1c_2rPd\sigma_i}\quad(1\le i\le r).
\end{aligned}
\tag{10.702}
\]

The polynomial parameters are integers. More precisely, put

\[
k=\lfloor H\rfloor,\qquad
X=\frac{S\mathcal D}{c_1c_4(N/\vartheta)dA_r k},\qquad
L=\lfloor X\rfloor+1,\qquad D_{\mathrm a}=kL.
\tag{10.703}
\]

The degree \(D_{\mathrm a}\) of a prepared polynomial is not the real
scale \(\mathcal D\).

\textbf{Theorem 10.184 (a selected kernel with explicit second-scale
parameters).} For these parameters,

\[
T>101r^2,\qquad X>58,\qquad S\ge g_S.
\tag{10.704}
\]

The closed-box, phase and unit-congruence selections of Proposition
10.183 give \(N_c=kL|S_0|\) coefficients, with every polynomial pair
retained, such that

\[
N_c>c_0(2\lfloor S\rfloor+1)
                  \binom{\lfloor T\rfloor+r}{r}.
\tag{10.705}
\]

In particular there is a nonzero global-integral coefficient vector
whose unprojected prepared values vanish at every integer \(|s|\le S\)
and every prepared index \(|\boldsymbol t|\le T\). The Laurent
polynomial is nonzero, and its coefficients satisfy the actual row-field
and height conclusions of Proposition 10.183.

\textbf{Proof.} First we establish the quantitative bounds that justify
the floors. The saturation estimate (10.55), applied to the q-primary
index J, gives

\[
\Omega_r\ge
\frac{w_KJr^r}{\rho\,d^{r+1}\ell_d\,r!\mathrm e^r},
\qquad
\rho=\begin{cases}17,&d=1,\\58,&d\ge2.\end{cases}
\tag{10.706}
\]

The weights dominate the individual heights, and the full saturation
index dominates its q-primary part, so every factor has the indicated
direction. Also \(w_K\ge q^u\), \(\mathcal R_r>1\), and \(K_S>2\).
Substitution in \(\mathcal D\) yields

\[
\mathcal D>
\frac{2c_0c_1c_4}{\rho}\frac N\vartheta
(c_2qP)^r\frac{r^r(r+1)^r}{(r!)^2}A_r.
\tag{10.707}
\]

The elementary inequalities \(r!/r^r\le2^{1-r}\) and \((1+1/r)^r\ge2\)
follow from the consecutive-ratio and binomial proofs used in Theorem
10.152. Together they imply

\[
\frac{r^r(r+1)^{r+1}}{(r!)^2}
\ge(r+1)2^{2r-1}.
\tag{10.708}
\]

Each row of the parameter table has
\(c_0\ge2,c_3\ge1/3,c_4\ge6,c_2q>3\). For the last assertion, at odd p
use \(2\zeta_2\le2(1.0568)<7/3\); at two use \(3\zeta_2=3.042<13/3\).
Using \(A_r\ge t\), \(A_0=e\vartheta\le P\), and \(P\ge1\), the
preceding bound on \(\mathcal D\) gives

\[
T>\frac{12}{29}(r+1)12^r>101r^2.
\tag{10.709}
\]

Indeed the first expression divided by \(r^2\) increases for \(r\ge3\):
its consecutive ratio is \(12(r+2)r^2/(r+1)^3>1\). At three the required
comparison is \(82944/29>909\), an exact integer comparison after
multiplication by 29.

Since \(k\le H\), the definitions of S and X give
\(X\ge c_3q(r+1)\mathcal D/(c_1c_4(N/\vartheta)A_rt)\). Use the lower
bound for \(\mathcal D\), the factorial comparison, and \(r/t>A_0+3/5\).
It follows that

\[
X>\frac{c_0c_3q}{\rho}(A_0+3/5)
             \frac{r+1}{r}(4c_2qP)^r
\ge\frac{8448}{145}>58.
\tag{10.710}
\]

For the numerical lower bound use \(A_0\ge1/2\) and the table's lower
bounds just proved. The function \((r+1)12^r/r\) increases for
\(r\ge3\), so its least value is at three. Finally \(h\ge(r+1)t\) gives
\(S\ge g_S\), and k is a positive integer.

For the row count, \(T>101r^2\) implies

\[
\binom{\lfloor T\rfloor+r}{r}
\le\frac{(T+r)^r}{r!}
<\frac{101}{100}\frac{T^r}{r!}.
\tag{10.711}
\]

To see the last inequality without approximating a logarithm, bound
\(\binom rj(r/T)^j\) by \((r^2/T)^j\) and sum the geometric series. Its
sum is less than \(101/100\).

Let \(V=\prod_iD_i\), and let \(h_m\) denote the actual unit-quotient
image order. Proposition 10.183 gives \(N_c>kLJV/(H_{\mathrm{ph}}h_m)\),
where \(H_{\mathrm{ph}}=(p^f-1)/(q^u\delta_r)\). Since \(L>X\),

\[
kL>\frac{S\mathcal D}{c_1c_4(N/\vartheta)dA_r}.
\tag{10.712}
\]

Divide this column lower bound by \((2S+1)(101/100)T^r/r!\), and
substitute T and the widths. The powers of \(\mathcal D\) cancel except
for one. To obtain a ratio at least \(c_0\), it is sufficient that

\[
\mathcal D\ge
\frac{101}{100}(2+1/S)c_0c_1c_4\frac N\vartheta
\frac{(c_2qr(r+1)P)^r}{r!\,t^r}
 d^{r+1}\Omega_r A_r\frac{H_{\mathrm{ph}}}{J}
 \frac{h_m}{(eN)^r}.
\tag{10.713}
\]

Lemma 10.177 and \(eN=(m+A_0)/(1+\tau)\) give

\[
\frac{h_m}{(eN)^r}
\le(1+\tau)^r\frac{p^{fm}}{(m+A_0)^r}
<(1+\tau)^r\zeta_2^r
       \left(\frac{\mathrm e t}{r}\right)^rp^{-fA_0}.
\tag{10.714}
\]

Now \(S\ge g_S\), \(\ell_d\ge1\), and \(\mathcal R_r\ge p^f/\delta_r>
((p^f-1)/\delta_r)p^{-fA_0}\). The asserted sufficient bound therefore
follows from the definition of \(\mathcal D\). This proves the strict
coefficient surplus. Proposition 10.183 provides the integral vector,
its height bound and the nonzero polynomial. Every field and residue
index used in that proposition has been retained. \(\square\)

\textbf{Corollary 10.185 (the exact cost of the polynomial rounding).}
The integer degree of this kernel satisfies

\[
\frac{S\mathcal D}{c_1c_4(N/\vartheta)dA_r}
<D_{\mathrm a}
<\frac{59}{58}\frac{S\mathcal D}
                     {c_1c_4(N/\vartheta)dA_r}.
\tag{10.715}
\]

The complete normalizing valuation in Theorem 10.182 may consequently be
bounded above by

\[
\frac{59}{58}\frac{S\mathcal D}{c_1c_4dA_r}
 \left(\vartheta+\frac{\vartheta}{N(p-1)}\right)-\beta.
\tag{10.716}
\]

\textbf{Proof.} The lower bound is \(kL>kX\). The upper bound follows
from \(L\le X+1\) and \(X>58\). Multiply by \(N+1/(p-1)\) and subtract
the coefficient minimum \(\beta\) to obtain the second assertion. This
is the normalization of the combined monomials, justified by Theorem
10.180; it makes no enlarged disc assumption on an individual basis
logarithm. \(\square\)

\subsubsection{The value of the counterhypothesis
scale}\label{the-value-of-the-counterhypothesis-scale}

\textbf{Proposition 10.186 (comparison with the stated second
constant).} For the above kernel define the positive real number

\[
U_r=\frac{q^{r+1}S\mathcal D}{et}.
\tag{10.717}
\]

Use the exact \(C_2\) formula of Section 58 at rank r, with the same
field and the actual residue index \(\delta_r\). Then

\[
eU_r<C_2(r,d,p,\boldsymbol\alpha)\,h\Omega_r.
\tag{10.718}
\]

This statement compares the parameter sizes. It does not assume, or
assert by itself, the valuation inequality \(v_p(\Xi-1)<U_r\).

\textbf{Proof.} Substitution of S and \(\mathcal D\), followed by
\(\zeta_2c_2q^2=a\), gives

\[
eU_r=K_Sc_0c_1c_3c_4q^2\frac N\vartheta(1+\tau)^r
 (a\mathrm e P)^r\frac{(r+1)^{r+1}}{r!}
 \frac{d^{r+2}\ell_d\Omega_r}{q^uJt^2}
 \mathcal R_r A_r H.
\tag{10.719}
\]

The other rounding bound of Lemma 10.177 is \(m+A_0<\zeta_1r/t\). Hence
\(N/\vartheta<\zeta_1r/(A_0t(1+\tau))\). Also \(H/J\le h\):
\(\log J\le J-1\) and \(h>1\) give \(h+\log J\le hJ\). Thus

\[
eU_r<\frac{F_r}{P}(a\mathrm e P)^r
 \frac{(r+1)^{r+1}}{(r-1)!}
 \frac{d^{r+2}\ell_d}{q^ut^3}
 \mathcal R_r A_r h\Omega_r,
\qquad
F_r=K_S\zeta_1c_0c_1c_3c_4q^2
                  \frac P{A_0}(1+\tau)^{r-1}.
\tag{10.720}
\]

For \(r\le n\), the binomial argument in (10.451) gives
\((1+\tau)^{r-1}<1+10^{-26}\). The lower bounds for \(g_S\) in the seven
field cases are respectively

\[
\frac{384}{5},\quad\frac{192}{5},\quad\frac{384}{5},\quad
\frac{512}{5},\quad\frac{256}{5},\quad192,\quad90.
\tag{10.721}
\]

At two the positive-radius regime has \(r\ge4\), since
\(f\ge2,A_0\ge2,t\ge\log4>4/3\). For the other cases \(r\ge3\) is
already an assumption. Outside case III, \(P/A_0=2\). In case III it
equals \(1/\vartheta=(p-1)/(p-2)\). Substitution of these rational
bounds and the parameter table shows

\[
F_r<c^{(2)}
\tag{10.722}
\]

in each case, with the factor \(1/\vartheta\) retained on both sides in
case III. Here are convenient integer upper bounds, obtained by
substituting the displayed rational constants in the formula for
\(F_r\). In the two case-III rows the common factor \(1/\vartheta\) is
divided out in both columns.

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}lrr@{}}
\toprule\noalign{}
Field case & Strict upper bound for \(F_r\) & Stated coefficient \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
I.1 & 627 & 1438 \\
I.2 & 631 & 648 \\
II & 627 & 690 \\
III.1 & 452 & 495 \\
III.2 & 454 & 557 \\
IV & 1763 & 2418 \\
V & 343 & 406 \\
\end{longtable}
}

For example the I.2 entry is the exact rational comparison

\[
\frac{101}{100}\left(2+\frac5{192}\right)
\frac{12133}{10000}\frac{27}{10}\frac7{10}
\frac65\,14\cdot4\cdot2\,(1+10^{-26})<631<648.
\tag{10.723}
\]

The other rows use the same multiplication with their displayed
constants. The \href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_second_scale.py}{editable
calculation} records the exact rational differences as well as these
upper bounds. This is a finite comparison of rational constants; the
preceding monotonicity and small-\(\tau\) arguments cover all ranks,
degrees, primes and ramification indices. The remaining expression is
exactly the displayed \(C_2\) formula. \(\square\)

\subsubsection{Comparing two ranks}\label{comparing-two-ranks}

\textbf{Lemma 10.187 (the rank factor of the second estimate).} Put

\[
F(j)=\frac{(j+1)^{j+1}}{(j-1)!}\quad(j\ge1).
\tag{10.724}
\]

Then \(F(j+1)/F(j)>\mathrm e\). In particular, define
\(\Gamma_2=a\mathrm e^2Pd/q\). Suppose the weights of a rank-r
representation of an original n-base form satisfy

\[
\Omega_r\le\Gamma_2^{n-r}
 \frac{\mathcal R_n(\boldsymbol a)}
      {\mathcal R_r(\boldsymbol\alpha)}
 \prod_{j=1}^nh(a_j).
\tag{10.725}
\]

For every \(3\le r\le n\) in the positive-radius regime, the scale from
Proposition 10.186 then satisfies

\[
eU_r<C_2(n,d,p,\boldsymbol a)\,h\prod_{j=1}^nh(a_j).
\tag{10.726}
\]

All residue indices in these formulas are those of their indicated
families. They need not be equal.

\textbf{Proof.} Direct division gives

\[
\frac{F(j+1)}{F(j)}
=\frac{j+2}{j}\left(1+\frac1{j+1}\right)^{j+1}.
\tag{10.727}
\]

For \(x>0\), integrating \(1/(1+t)>1/(1+x)\) from zero to x gives
\(\log(1+x)>x/(1+x)\). The logarithm of the displayed ratio is
consequently greater than \(2/(j+2)+(j+1)/(j+2)>1\). This proves the
first assertion without a limiting factorial formula.

Insert the assumed weight product in Proposition 10.186. Its
\(\mathcal R_r\) cancels exactly, and \(A_r\le A_n\). After division by
the desired rank-n expression the remaining rank factor is bounded by

\[
\left(\frac{\mathrm e}{q}\right)^{n-r}
                 \frac{F(r)}{F(n)}\le1.
\tag{10.728}
\]

For \(r<n\) it is strictly smaller than \(q^{r-n}\), by the first
assertion. At \(r=n\) it equals one, and Proposition 10.186 already
supplies the strict comparison. This proves the claim. The assertion is
conditional on the displayed weight allowance; preserving that allowance
in a multiplicity argument is a separate mathematical requirement.
\(\square\)

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={Seven uniform prefactor comparisons and the proved lower bound for the jet parameter divided by the square of the rank.}]{../figures/yu-second-scale.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.54.}
\end{flushleft}
\end{minipage}\par\medskip
 On the left, Proposition 10.186 bounds the
auxiliary scale below the stated coefficient in every field case. The
common case-III factor cancels before forming either ratio. On the
right, Theorem 10.184 bounds \(T/r^2\) above the threshold needed to
count all prepared rows; the consecutive-ratio proof covers every
\(r\ge3\), while the plotted ranks are samples. These comparisons
concern the constructed support and its parameters. They do not assert
the global valuation estimate. The constant being compared is the one in
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
freely accessible 2013 paper, p.~320}.

\subsection{62. Small optimizer cases and the first analytic
step}\label{small-optimizer-cases-and-the-first-analytic-step}

The second-scale construction requires a positive optimizer. We first
prove that the first estimate already handles the complementary regime
for every list of at least three bases. We also settle almost all
two-base cases by exact comparisons. We then verify the first analytic
input for the new construction, retaining the original distinguished
coefficient and all ordinary derivative costs.

\subsubsection{When the first constant
suffices}\label{when-the-first-constant-suffices}

Keep the original independent bases, integer coefficients and field
cases of Theorem 10.152, and the exact constants and coefficient
parameters of Section 58. Write \(A_0=e\vartheta\), with e the
ramification index.

\textbf{Theorem 10.188 (the small-optimizer regime).} Suppose \(n\ge3\).
The second estimate

\[
\operatorname{ord}_{\mathfrak p}(\Xi-1)
<C_2\mathsf h_2\prod_{j=1}^n h(a_j)
\tag{10.736}
\]

holds in each of the following ranges:

\begin{itemize}
\tightlist
\item
  outside case III, \(n\le(6/5)Pt\);
\item
  in case III, \(n\le(\vartheta+3/5)t\).
\end{itemize}

In particular it holds whenever \(n/t-A_0\le3/5\).

\textbf{Proof.} The cancellation (10.648) is valid without the
restriction \(n\le Pt\). Outside III, put \(y=n/(Pt)\). It becomes

\[
\frac{C_2}{C_1}
=\frac{c^{(2)}}c\frac{\mathrm e}{2}
 \left(\frac{\mathrm e}{2y}\right)^{n-1}.
\tag{10.737}
\]

All five relevant rows have \(c^{(2)}/c\ge2418/2680=1209/1340\). The
fractions in Section 58 prove this by cross-multiplication; the other
rows are \(1438/939,648/636,690/505,406/206\). If \(y\le6/5\), then
\(\mathrm e>27/10\) gives \(\mathrm e/(2y)>9/8>1\). Since \(n-1\ge2\),

\[
\frac{C_2}{C_1}
>\frac{1209}{1340}\frac{27}{20}\left(\frac98\right)^2
>\frac32.
\tag{10.738}
\]

In III write \(c^{(2)}=c_*/\vartheta\), where \(c_*=495\) or 557, and
put \(z=n/t\). Here \(\mathfrak a=7/\vartheta\), \(P=1\), and
\(3/4\le\vartheta<1\). The same exact cancellation gives

\[
\frac{C_2}{C_1}
=\frac{c_*}c\,\mathrm e
 \left(\frac{\mathrm e\vartheta}{z}\right)^{n-1}.
\tag{10.739}
\]

The function \(v/(v+3/5)\) increases for positive v: its derivative is
\((3/5)/(v+3/5)^2>0\). Thus the assumed \(z\le\vartheta+3/5\) gives
\(\mathrm e\vartheta/z>(27/10)(3/4)/(27/20)=3/2\). Also
\(c_*/c\ge495/1794\), including the III.2 row \(557/1790\). Consequently

\[
\frac{C_2}{C_1}
>\frac{495}{1794}\frac{27}{10}\left(\frac32\right)^2
=\frac{8019}{4784}>\frac32.
\tag{10.740}
\]

The two final rational comparisons are reproduced in the editable
calculation for Figure 10.55. Lemma 10.172 now gives
\(C_1\mathsf h<C_2\mathsf h_2\), and Theorem 10.152 proves the claim.

Finally \(A_0=P/2\) outside III, and \(P\ge1\). Therefore
\(A_0+3/5\le(6/5)P\), since \(3/5\le(7/10)P\). In III, \(e=P=1\), so
\(A_0=\vartheta\). This proves the last assertion, including equality at
the optimizer threshold. \(\square\)

\textbf{Theorem 10.189 (the remaining two-base comparisons).} For
\(n=2\) the second estimate also follows from Theorem 10.152 in every
field case except possibly

\[
p=5,\qquad e=f=1,\qquad d>1.
\tag{10.741}
\]

Theorem 10.220 proves the second estimate in this remaining family. The
comparison here already gives the conclusion whenever
\(\mathsf h_2\ge G_1(2,d)\).

\textbf{Proof.} Theorem 10.173 applies if \(2\le Pt\). Otherwise
\(P=1\): if \(P>1\), it is an integer prime power at least two and
\(t>1\). At two this also contradicts \(P\ge4\). Thus p is odd, \(f=1\),
and \(p<\exp2<9\). The possible primes are three, five and seven. At
three, \(P\ge3\), so only five and seven remain. The defining inequality
\(2e<P(p-1)\) gives \(e=1\) at five, and \(e=1\) or two at seven.

At seven, \(\log7>19/10\). In case III, \(\vartheta=5/6\), so the
rank-two form of the preceding identity gives

\[
\frac{C_2}{C_1}
>\frac{495}{1794}\left(\frac{27}{10}\right)^2
 \frac56\frac{19}{20}>\frac32.
\tag{10.742}
\]

In IV, necessarily \(e=2\) here, it instead gives

\[
\frac{C_2}{C_1}
>\frac{1209}{1340}\left(\frac{27}{20}\right)^2
 \frac{19}{20}>\frac32.
\tag{10.743}
\]

These strict rational comparisons and Lemma 10.172 settle seven.

At five, consider first \(d=1\), the III.2 row. A slightly sharper
comparison of the coefficient parameters suffices:

\[
\mathsf h<\frac{11}{8}\mathsf h_2,\qquad
\frac{C_2}{C_1}>\frac{603231}{438550}>\frac{11}{8}.
\tag{10.744}
\]

To prove the first inequality, set \(a_0=2+\log7\) and
\(x=2a_0+146/25\). The parameters in Sections 47 and 58 give

\[
\frac{G_1(2,1)}3=x+\log x,
\qquad
\frac{G_2(2,1)}3=2a_0+\frac{83}{25}+\log3.
\tag{10.745}
\]

Using \(19/10<\log7<2\), we have \(x<346/25\). Also
\(\exp(27/10)>(27/10)^2(1+7/10+49/200)>346/25\), so \(\log x<27/10\).
Hence

\[
\frac{G_1(2,1)}3<\frac{827}{50},\qquad
\frac{G_2(2,1)}3>\frac{303}{25},\qquad
\frac{11}{8}\frac{303}{25}-\frac{827}{50}=\frac18>0.
\tag{10.746}
\]

All other entries of the coefficient maxima agree and \(\mathsf h_2>0\),
proving the coefficient inequality. For the second inequality use
\(\mathrm e>19/7\), \(\log5>8/5\), and \(\vartheta=3/4\) in the exact
rank-two quotient. This gives

\[
\frac{C_2}{C_1}
>\frac{557}{1790}\left(\frac{19}{7}\right)^2
 \frac34\frac45=\frac{603231}{438550}.
\tag{10.747}
\]

Here its difference from \(11/8\) is \(899/1754200>0\). Combining the
two comparisons with Theorem 10.152 settles the rational field at five.

We justify all elementary logarithm bounds just used. The exponential
series through degree six gives \(\mathrm e>1957/720>19/7>27/10\). Its
sum through degree five is \(163/60\); the remaining terms have first
term \(1/720\) and subsequent ratios at most \(1/7\), giving

\[
\mathrm e<\frac{163}{60}+\frac7{4320}
=\frac{11743}{4320}<\frac{87}{32}<\frac{11}{4}.
\tag{10.748}
\]

Now \(\exp(19/10)=\mathrm e^2/\exp(1/10)
<(11/4)^2/(11/10)=55/8<7\), proving \(\log7>19/10\); \(\mathrm e^2>7\)
proves its upper bound. Similarly
\(\exp(2/5)>1+2/5+2/25+4/375=559/375\). Therefore

\[
\exp(8/5)
<\frac{(87/32)^2}{559/375}<5,
\tag{10.749}
\]

where the last rational comparison is
\(7569\cdot375<5\cdot1024\cdot559\). Thus \(\log5>8/5\). Finally
\(\log3>1\) follows from \(\mathrm e<3\). These estimates prove every
numerical inequality used above. The case \(p=5,e=f=1,d>1\) has not been
inferred from the rational-field calculation. For its stated height
subcase, all the entries in \(\mathsf h\) are at most \(\mathsf h_2\),
because only the G entry differs. Moreover

\[
\frac{C_2}{C_1}
>\frac{495}{1794}\left(\frac{19}{7}\right)^2
 \frac34\frac45>1.
\tag{10.750}
\]

Theorem 10.152 therefore proves the estimate in that subcase as well.
\(\square\)

\subsubsection{The complete first local input
budget}\label{the-complete-first-local-input-budget}

Return to the parameters and the actual kernel of Theorem 10.184.
Suppose its original integer coefficient vector is nonzero and the
distinguished \(b_n\ne0\) has minimum p-adic valuation. Take the height
parameter to satisfy

\[
h\ge\max\{G_2(r,d),(r+1)t,\log B_\dagger\},
\qquad B_\dagger=\min_{b_j\ne0}|b_j|.
\tag{10.751}
\]

These conditions hold with the original \(\mathsf h_2\) when \(r\le n\),
since \(G_2(j,d)\) increases with j. Put \(s_0=\lfloor S\rfloor\),
\(\mu=\lfloor T/(q(r+1))\rfloor+1\), and

\[
B=\lfloor\log_p(2s_0)\rfloor,\qquad
C_0=\max\{v_p(V(k)),v_p(b_n)\},\qquad M_0=\max\{B,C_0\}.
\tag{10.752}
\]

\textbf{Theorem 10.190 (uniform first-step analytic budget).} With these
parameters,

\[
\begin{aligned}
\mu\bigl((2s_0+1)N+M_0\bigr)
&<\frac{437}{2016}U_r<\frac14U_r,\\
\kappa+U_r-v_p(b_n)&>N+\frac1{p-1}.
\end{aligned}
\tag{10.753}
\]

Consequently the actual initial zeros of Theorem 10.184 meet the full
input budget (10.691). Under the counterhypothesis
\(v_p(\Xi-1)\ge U_r\), every prepared index
\(|\boldsymbol t|\le T-\mu+1\) satisfies, for every \(x\in\mathbb Z_p\),

\[
v_p(\Phi(x;\boldsymbol t))
\ge(2s_0+1)\mu N
 -D_{\mathrm a}\left(N+\frac1{p-1}\right)+\beta.
\tag{10.754}
\]

This is a local analytic precision bound. Concluding that an algebraic
value is zero still requires the strict global arithmetic comparison.

\textbf{Proof.} First we control all jet costs by the original height
parameter. The constants give \(c_3\le3,q\le3,t>1\), so
\(2S<18(r+1)dH\). Also \(a_0^{(2)}>3\), \(a_1^{(2)}+\log(r+1)>1\), and
hence

\[
H\ge h\ge G_2(r,d)
>(r+1)(3r+\log d+1)\ge40.
\tag{10.755}
\]

For \(H\ge2\), \(\log H<H/2\): its difference has positive derivative
after two and is positive at two. Furthermore \(\log18<3\) and
\(\log(r+1)\le r\). Thus

\[
\log(2S)
<\frac H2+3+r+\log d<H.
\tag{10.756}
\]

Indeed \((r+1)(3r+\log d+1)/2>3+r+\log d\): its left minus right is at
least \(5r-1>0\) for \(r\ge3\), after keeping the nonnegative
coefficient of \(\log d\). The logarithm bounds follow respectively from
\((27/10)^3>18\) and \(\log(1+r)\le r\), proved by integrating
\((1+x)^{-1}\le1\).

The minimum-valuation coefficient hypothesis gives
\(v_p(b_n)\le\log B_\dagger/\log p\le h/\log p\). This uses the original
integers, not newly chosen rational coefficients. Also
\(v_p(V(k))=\lfloor\log_p k\rfloor\), by the proved
least-common-multiple formula. Since \(k\le H\) and \(\log H<H\), all
costs obey

\[
M_0\le\frac H{\log p}.
\tag{10.757}
\]

The local degree inequality \(d\ge ef\), \(c_3\ge3/5\), \(q\ge2\),
\(r+1\ge4\), and \(A_0\ge1/2\) give

\[
\frac{SN}{H/\log p}
=c_3q(r+1)\frac d f\,N
\ge c_3q(r+1)\frac{m+A_0}{1+\tau}
>\frac{18}{5}>3.
\tag{10.758}
\]

Here \(m\ge1\) and \(0<\tau<1\). It follows that \(M_0<SN/3\) and
\(SN>12\), the latter because \(H/\log p\ge(r+1)f\ge4\). Also
\(S\ge g_S\ge(3/5)\cdot2\cdot16>16\).

By Theorem 10.184, \(T>101r^2\). For \(r\ge3\), \(q\le3\) gives
\(q(r+1)/T<3(r+1)/(101r^2)\le4/303<1/75\). The monotonic decrease of
\((r+1)/r^2\) follows by differentiation, or multiplication of
consecutive positive integers. Consequently

\[
\mu<\frac{76}{75}\frac T{q(r+1)}.
\tag{10.759}
\]

Put \(W=STN/(q(r+1))\). From \(S>16\) and the jet bound,

\[
\mu\bigl((2s_0+1)N+M_0\bigr)
<\frac{76}{75}\left(2+\frac1{16}+\frac13\right)W
=\frac{437}{180}W.
\tag{10.760}
\]

On the other hand, substitution of T and \(U_r\) gives the exact
identity \(U_r=c_1q^{r+1}W\). Every row has \(c_1\ge7/10\); hence
\(U_r\ge(56/5)W\). Division proves the first bound, since
\(437/2016<1/4\).

For strict normality, the identity \(U_r/(SN)=c_1q^rT/(r+1)>1\) follows
from \(T>101r^2\), \(c_1\ge7/10\), and \(r\ge3\). Meanwhile \(N<SN/16\),
\(1/(p-1)\le1<SN/12\), and \(v_p(b_n)<SN/3\). Their sum is less than
\((23/48)SN<SN<U_r\). Since \(\kappa\ge0\), the asserted strict
inequality follows.

For the other strict disc inequality, use the phase and depth
construction of Theorem 10.180. Its deep units lie in \(U_j\) with
integer \(j\ge A_0+e/(p-1)\). Therefore

\[
N=\frac{m+A_0}{e(1+\tau)}
<\frac{j+m}{e}-\frac1{p-1}.
\tag{10.761}
\]

All hypotheses of Theorem 10.182 now hold for the full integer node set,
with \(\eta_E=0\). The budget just proved is stronger than (10.691),
whose left side also has the nonnegative terms
\(\kappa+D_{\mathrm a}(N+1/(p-1))\). Theorem 10.184 supplies every
required initial prepared zero, and \(|\boldsymbol t|+\mu-1\le T\) by
assumption. Equation (10.692) proves the stated precision. \(\square\)

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={Exact positive gaps for the small-optimizer and two-base comparisons, and the fraction of the counterhypothesis scale needed by the first local input.}]{../figures/yu-second-closure.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.55.}
\end{flushleft}
\end{minipage}\par\medskip
 Left: the rational differences in Theorems 10.188
and 10.189 are strictly positive. The comparison uses height ratio
\(3/2\) in the first four bars and \(11/8\) in the rational five-adic
case; the logarithmic vertical scale makes its small positive gap
visible. Right: Theorem 10.190 bounds the full first local input by
\((437/2016)U_r<U_r/4\), uniformly in all its parameters. Solution 76
computes the smaller cost for the actual four-base support of Solution
75 with a unit distinguished coefficient. Neither bar is a sampled
prime-ideal valuation.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_second_closure.py}{Editable exact calculation
and figure}. Human-source context: the two estimates in Section 1.1 and
the coefficient parameters in Section 1.3 of
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
freely accessible paper}.

\subsection{63. The second-scale multiplicity stop and its height
profile}\label{the-second-scale-multiplicity-stop-and-its-height-profile}

The second-scale construction has its own polynomial degree and analytic
radius. We verify the degree comparisons with these parameters before
using the multiplicity theorem. The resulting height-product estimate
has enough room to preserve the second constant through a reduction in
rank. Its one-base endpoint follows from the elementary valuation
estimate.

\subsubsection{Degree and allocation
inequalities}\label{degree-and-allocation-inequalities}

Retain the parameters of Theorem 10.184, including its independent
\(r\ge3\) bases and \(h\ge(r+1)t\). Write \(A=A_r\), \(x=\log J\), and
\(\lambda=59/58\). Set

\[
c_5=\begin{cases}27/50,&q=2,\\4/5,&q=3,\end{cases}
\qquad \eta=1-\frac{c_5}{r+1},\qquad Q=q\eta^{r+1}.
\tag{10.766}
\]

Choose any nonnegative integer I such that

\[
E=Q^I>\exp(5(A+x)),\qquad
\bar S=q\eta^{-(r+1)I}S,\qquad
\bar T=\eta^{(r+1)I+1}T.
\tag{10.767}
\]

Use the coordinate-degree bounds

\[
\mathsf D_0=D_{\mathrm a}=kL,\qquad
\mathsf D_j=2Jq^{-I}D_j\quad(1\le j\le r),
\tag{10.768}
\]

and the integer allocations

\[
S_0=\lfloor\bar S/3\rfloor,\quad
S_j=\lfloor2\bar S/(3r)\rfloor\ (1\le j\le r),\qquad
T_j=\lfloor\bar T/(r+1)\rfloor\ (0\le j\le r).
\tag{10.769}
\]

\textbf{Theorem 10.191 (degree comparisons at the second-scale stop).}
These allocations satisfy (10.67), with total budgets
\(\lfloor\bar S\rfloor,\lfloor\bar T\rfloor\). Moreover,

\[
\mathsf D_0>20\mathsf D_j\quad(1\le j\le r),\qquad
\mathsf D_0>4\bar T,\qquad \mathsf D_0>4r.
\tag{10.770}
\]

For the allocation ratios \(\mathcal R_0,\mathcal R_{m,J},
\mathcal R_r\) defined in (10.267), with the present parameters,

\[
\mathcal R_0>\frac{32}{11},\qquad
\mathcal R_{m,J}>\frac{2^{20}}{29}\ (1\le m<r),\qquad
\mathcal R_r>31.
\tag{10.771}
\]

In particular every allocation and coordinate-degree hypothesis of
Theorem 10.25 holds.

\textbf{Proof.} We use \(\mathrm e=\exp(1)\), reserving e for
ramification. The established bounds give

\[
\begin{gathered}
A\ge4+\log((r+1)d)>5,\quad t>1,\quad
H=h+x\ge(r+1)t>r+1,\\
c_3\ge3/5,\quad c_2q>3,\quad 6\le c_4\le14,\quad
A_0=e\vartheta\ge1/2,\quad \vartheta\le2,\quad d\ge e,\\
eNt<\frac{\zeta_1r}{1+\tau},\quad
\zeta_1<7/5,\quad \zeta_2<11/10,\quad
K_S<21/10,\quad \lambda<11/10,\quad c_0<3.
\end{gathered}
\tag{10.772}
\]

The bounds for \(\zeta_1,\zeta_2\) are those of Lemma 10.177, and
\(K_S<21/10\) follows from \(g_S>16\). The field-case definition of
\(\vartheta\) gives its displayed upper bound, as in Theorem 10.100.
Also \(2<\mathrm e<3\) and \(\ell_d\le d\).

We have \(\eta\ge4/5\). The derivative argument in Theorem 10.100 shows
that \((1-c/y)^y\) increases for \(y>c\). Consequently

\[
Q\ge\begin{cases}2(173/200)^4>1,&q=2,\\
3(4/5)^4>1,&q=3.\end{cases}
\tag{10.773}
\]

On the other hand, \(q\eta^{2(r+1)}<q\exp(-2c_5)<1\). For the last
comparison, the first three exponential terms give \(\exp(27/25)>2\) and
\(\exp(8/5)>3\). It follows that

\[
\eta^{-(r+1)I}>E,\qquad
q^I/J>E/J>\exp(5A),\qquad E/J>\exp(5A).
\tag{10.774}
\]

In particular a finite I with the required property exists. These facts
concern the size of a stopping depth; they do not assert that the
auxiliary function has acquired its required zeros.

For an individual non-torsion base, Theorem 10.18 gives
\(\sigma_j\ge2/(58\mathrm e d^2\ell_d)\), hence \(d\sigma_j>1/(87d^2)\).
The lower bound in Corollary 10.185 and the definitions of S and
\(\mathsf D_j\) give

\[
\begin{aligned}
\frac{\mathsf D_0}{\mathsf D_j}
&>\frac{c_3c_2q r(r+1)PH\vartheta}{2c_4NtA}
                 d\sigma_j\frac{q^I}{J}\\
&>\frac9{392}(r+1)^2d\sigma_j\frac{\exp(5A)}A.
\end{aligned}
\tag{10.775}
\]

In the second line use \(eNt<\zeta_1r\), \(e\vartheta\ge1/2\),
\(P\ge1\), and \(H>r+1\). Since \(\exp(3A)/A\) increases for \(A\ge5\),

\[
\frac{\exp(5A)}A
>256(r+1)^2d^2\frac{2^{15}}5.
\tag{10.776}
\]

Thus the degree ratio is greater than
\(9\cdot256^2\cdot2^{15}/(392\cdot87\cdot5)>20\), where
\((r+1)^4\ge4^4=256\) was used. Similarly,

\[
\frac{\mathsf D_0}{\bar T}
>\frac{c_3A_0H}{c_4A}\eta^{-(r+1)I-1}
>\frac3{35}\frac{2^{25}}5>4.
\tag{10.777}
\]

Here \(H>4\), and \(\exp(5A)/A\) increases for \(A\ge5\). Finally
\(k=\lfloor H\rfloor\ge r+1\) and \(L>X>58\), so
\(\mathsf D_0>58(r+1)>4r\). These inequalities imply
\(\mathsf D_0>\max_j\mathsf D_j\) and \(\mathsf D_0>\bar T+r\ge T_r+r\).

The integer allocations are nonincreasing and have sums at most their
specified integer budgets. Their floors satisfy

\[
S_0+1>\bar S/3,\quad S_j+1>2\bar S/(3r),\quad
\binom{T_j+b}{b}>\frac{(\bar T/(r+1))^b}{b!}\quad(b\ge1).
\tag{10.778}
\]

For the last inequality, each factor \(T_j+l\), \(l\ge1\), is greater
than \(\bar T/(r+1)\). Put

\[
B_0=\frac{q^2\eta c_4dA}{3\lambda A_0t}.
\tag{10.779}
\]

Direct cancellation using the upper bound for \(\mathsf D_0\) gives
\(\mathcal R_0>B_0\). Since \(d/A_0\ge1/2\), \(A\ge t\), \(q\ge2\),
\(\eta\ge4/5\), \(c_4\ge6\), and \(\lambda<11/10\), we have
\(B_0>32/11\).

For every \(1\le m<r\) and m-element coordinate subset J', the same
cancellation yields

\[
\mathcal R_{m,J'}>
\frac{2B_0}{r}\frac{\prod_{j\in J'}\sigma_j}{((m+1)!)^2}
\left(\frac{\eta c_2qrPd}{2eNt}\frac EJ\right)^m.
\tag{10.780}
\]

The bracket is greater than \((6/7)d\exp(5A)\). Each selected family is
independent, so Theorem 10.18 applies to it:

\[
\prod_{j\in J'}\sigma_j
\ge\frac{m^m}{29m!\mathrm e^m d^{m+1}\ell_d}.
\tag{10.781}
\]

Using \(B_0>2\), \(m^m\ge m!\), \(\mathrm e<3\) and
\((2/7)\exp(5A)>2^{21}(r+1)^2d^2\), we obtain

\[
\mathcal R_{m,J'}>
\frac4{29}(2^{21})^m
\frac{(r+1)^{2m}}{r((m+1)!)^2}
\frac{d^{2m-1}}{\ell_d}
\ge\frac4{29}\frac{(2^{21})^m}{(m+1)^3}
\ge\frac{2^{20}}{29}.
\tag{10.782}
\]

Indeed the exponential estimate follows by splitting its exponent into
\(2A+3A\), using \(A\ge4+\log((r+1)d)\) and \(A>5\). For the middle
inequality, \(d^{2m-1}\ge\ell_d\), \((m+1)!\le(m+1)^{m+1}\), and
\((r+1)^{2m}/r\) increases for \(r\ge m+1\). The final expression
increases with m: its consecutive ratio is at least \(2^{21}(2/3)^3>1\).

It remains to treat the full-rank allocation. Keep the entire formula
for \(\mathcal D\) in Theorem 10.184. Cancellation now gives

\[
\begin{aligned}
\mathcal R_r>{}&
\frac{2q q^uJ^{1-r}\eta^r}
 {3r\lambda(r+1)!K_Sc_0\ell_d\mathcal R_r(\boldsymbol\alpha)}
 q^{rI}\eta^{(r-1)(r+1)I}\\
&\quad{}\times
\left(\frac{r}{2eNt\zeta_2\mathrm e(r+1)(1+\tau)}\right)^r.
\end{aligned}
\tag{10.783}
\]

The \(\mathcal R_r\) on the left is the allocation ratio;
\(\mathcal R_r(\boldsymbol\alpha)\) on the right is the residue maximum
of Theorem 10.184. For that maximum,

\[
\mathcal R_r(\boldsymbol\alpha)\le\exp(A)A.
\tag{10.784}
\]

Its first entry is at most \(p^f=\exp t\); for the second,
\((\mathrm et/r)^r\le\exp t\) follows from \(\log(t/r)\le t/r-1\), and
\(t\le A\). Also \(q^{rI}\eta^{(r-1)(r+1)I}=q^IE^{r-1}>E^r\), so
\(J^{1-r}E^r>\exp(5rA)\). Using the strict bounds already proved,

\[
\frac{2q q^u}{3\lambda K_Sc_0}>\frac16,\qquad
\frac{\eta}{2\zeta_1\zeta_2\mathrm e}>\frac1{12},
\tag{10.785}
\]

we conclude that

\[
\mathcal R_r>
\frac{\exp((5r-1)A)}
 {6r(r+1)![12(r+1)]^r\ell_d A}
>\frac{16^{4r}}{6r12^r}>31.
\tag{10.786}
\]

For the second inequality use \(\exp((r-1)A)>A\),
\(\exp(4rA)>[16(r+1)d]^{4r}\), \((r+1)!\le(r+1)^{r+1}\), and
\(\ell_d\le d\). The last expression increases for \(r\ge3\), since its
consecutive ratio is \((16^4/12)r/(r+1)>1\); its value at three exceeds
31 by an integer comparison. This proves all assertions. \(\square\)

\subsubsection{The height product after a reduction in
rank}\label{the-height-product-after-a-reduction-in-rank}

\textbf{Corollary 10.192 (the second height profile at a multiplicity
stop).} Suppose the contracted prepared family with the parameters of
Theorem 10.191 is nonzero and vanishes at every integer \(|s|\le\bar S\)
through total prepared order \(\bar T\). Suppose its independent bases
and selected Euler directions give the hyperplane W of Theorem 10.25,
with a nonzero vector \(\boldsymbol B\). Then there are \(1\le m<r\) and
independent integral vectors \(u_1,\ldots,u_m\), whose rational span
contains \(\boldsymbol B\). Put \(\beta_i=\prod_j\alpha_j^{u_{ij}}\) and
\(R_i=\sum_j\sigma_j|u_{ij}|\). These are independent local units in the
original field, with \(h(\beta_i)\le R_i\), and

\[
\prod_{i=1}^mR_i
<10^{-6}\Gamma_2^{r-m}
\frac{\mathcal R_r(\boldsymbol\alpha)}
     {\mathcal R_m(\boldsymbol\beta)}\Omega_r,
\qquad \Gamma_2=a\mathrm e^2Pd/q.
\tag{10.787}
\]

Here a is the case constant in Theorem 10.184, and each residue maximum
uses its indicated family and its own saturated residue index. At rank
one use the actual residue order of the base.

\textbf{Proof.} Proposition 10.31 converts the stipulated prepared zeros
to ordinary W-jets, and clears the Laurent monomial with the coordinate
degree bounds of Theorem 10.191. That theorem verifies every allocation
and degree hypothesis of Theorem 10.25. As in the proof of Corollary
10.101, both multiplicity alternatives imply

\[
(S_m+1)\binom{T_m+m}{m}\prod_{i=1}^m R_i
\le(m+1)!r^m D_{\mathrm a}
\left(\frac{2Jq^{-I}\mathcal D}{c_1c_2rPd}\right)^m.
\tag{10.788}
\]

Corollary 10.26 proves the independence, field and height assertions. If
an initial multiplicative relation is expressed in the rational span of
\(\boldsymbol B\), clearing denominators and killing its root-of-unity
factor gives the same identity in the new bases, as in (10.271).

Let \(\mathcal Q\) be the product on the left of the claimed bound
divided by \(\Gamma_2^{r-m}(\mathcal R_r/\mathcal R_m)\Omega_r\).

Bound the denominator in the multiplicity output strictly below by
\((2\bar S/(3r))(\bar T/(r+1))^m/m!\), since each floor plus one
strictly exceeds the unrounded quotient. Insert the upper bound for
\(D_{\mathrm a}\) from Corollary 10.185, and divide by the desired
profile. Exact substitution of \(S,T,\mathcal D,\Gamma_2\) gives the
following upper bound for the normalized quotient:

\[
\begin{aligned}
\mathcal Q <{}&\frac{3r}{2q}\frac{(m+1)!m!}{r!}
 \lambda K_Sc_0\,(r+1)^r\mathrm e^{-r}
 \frac{\ell_d\mathcal R_m}{q^u}(1+\tau)^{r-m}\\
&\quad{}\times
\left(\frac{2\zeta_1\zeta_2r\mathrm e^2}{\eta}\right)^m
 J^{m-1}\frac{\eta^{(r+1)I}}{E^m}.
\end{aligned}
\tag{10.789}
\]

For clarity, the final inequality used the bound from Lemma 10.177
\(eNt<\zeta_1r/(1+\tau)\). Before that bound its bracket is
\(2\zeta_2 eNt\mathrm e^2/\eta\), and the power of \(1+\tau\) is r. This
retains the distinction between ramification e and Euler's number
\(\mathrm e\). Both factorials from the multiplicity output and binomial
count remain visible.

For every integer \(m\ge1\), \(\mathcal R_m\le p^ft\le\exp(A)A\). Indeed
its first entry is at most \(p^f\), while \((\mathrm et/m)^m\le\exp t\)
follows from \(\log(t/m)\le t/m-1\). Multiplication by t treats its
second entry. Also the integral bound for \(\log(r!)\) gives

\[
\frac{(r+1)^r}{r!\mathrm e^r}
\le\exp\left(r\log(1+1/r)-1\right)<1.
\tag{10.790}
\]

Equation (10.451) bounds \((1+\tau)^{r-m}\) by \(101/100\). Also
\(g_S>16\), hence \(K_S<21/10\). The other bounds are \(\lambda<11/10\),
\(c_0<3\), and \(q\ge2\). Thus the remaining constant factor is less
than \((3/4)(11/10)(21/10)3(101/100)<6\). Furthermore
\(\zeta_1<7/5,\zeta_2<11/10,\mathrm e<3\) and \(\eta\ge4/5\) give
\(2\zeta_1\zeta_2\mathrm e^2/\eta<693/20<35\).

From \(q\eta^{2(r+1)}<1\) we have \(\eta^{(r+1)I}<E^{-1}\). The stopping
inequality therefore implies

\[
J^{m-1}\frac{\eta^{(r+1)I}}{E^m}
<\exp(-5(m+1)A)J^{-4m-6}
\le\exp(-5(m+1)A).
\tag{10.791}
\]

Consequently

\[
\mathcal Q
<6r(m+1)!m!\,\ell_d A\,(35r)^m
 \exp(-(5m+4)A).
\tag{10.792}
\]

Set \(V=(r+1)d\). Since \(A\ge4+\log V>5\), \(y\exp(-(5m+4)y)\)
decreases for \(y\ge5\). The elementary inequalities \(4+\log V\le5V\)
and \(\ell_d\le d\) bound the last display above by

\[
30\cdot35^m\exp(-20m-16)
\frac{r^{m+1}(m+1)!m!}{(r+1)^{5m+3}d^{5m+2}}.
\tag{10.793}
\]

Here \(m<r\), so \((m+1)!m!\le(m+1)^{2m+1}\le r^{2m+1}\). The last
fraction is strictly smaller than one. Finally \(\mathrm e>2\),
\(35/2^{20}<1\), and \(m\ge1\) give

\[
\mathcal Q
<\frac{30}{2^{16}}\left(\frac{35}{2^{20}}\right)^m
\le\frac{1050}{2^{36}}<10^{-6}.
\tag{10.794}
\]

The last comparison is the integer inequality \(1050\cdot10^6<2^{36}\).
This proves the asserted product bound. In particular the stronger
allowance of Theorem 10.187 survives this multiplicity output, including
its actual change of residue index. \(\square\)

The conclusion is conditional on the stated prepared zeros. The degree
and height-product arguments establish what those zeros imply; their
arithmetic production remains a separate step.

\subsubsection{The one-base endpoint of the second height
profile}\label{the-one-base-endpoint-of-the-second-height-profile}

\textbf{Proposition 10.193 (one base with the stronger rank allowance).}
Use the original independent n-base setting, \(n\ge2\), and the exact
\(C_2\) constant in (10.642). Write \(a=a^{(2)}\), \(c_*=c^{(2)}\),
\(\Gamma_2=a\mathrm e^2Pd/q\), \(F(n)=(n+1)^{n+1}/(n-1)!\), and
\(\Omega_n=\prod_jh(a_j)\). Suppose a rank-one integral-form
representation satisfies the weighted conditions in (10.277), with its
product allowance replaced by

\[
\sigma\le\Gamma_2^{n-1}
 \frac{\mathcal R_n(\boldsymbol a)}{\mathcal R_1(\beta)}\Omega_n.
\tag{10.795}
\]

Here \(\beta\) is its non-torsion unit base, \(g\) is the actual residue
order of \(\beta\), and
\(\mathcal R_1(\beta)=\max\{p^f/\delta(\beta),\mathrm et^2\}\), with
\(\delta(\beta)=(p^f-1)/g\). In particular \(\mathcal R_1>g\). Let the
original height parameter h obey \(h\ge\max\{\log B_\dagger,(n+1)t\}\).
Then

\[
\operatorname{ord}_{\mathfrak p}(\Xi-1)
<\frac1{3000}C_2h\Omega_n.
\tag{10.796}
\]

\textbf{Proof.} Clear the rational coefficient of the one independent
integral form exactly as in (10.454). Its numerator divides every
original nonzero integer coefficient, so its absolute value is at most
\(B_\dagger\). Kill the q-primary torsion by a positive power; the
difference valuation cannot decrease. The elementary residue-order bound
of Theorem 10.21, including signed powers, therefore yields

\[
\operatorname{ord}_{\mathfrak p}(\Xi-1)
\le\frac d t\bigl(2h+2eg\sigma\bigr).
\tag{10.797}
\]

This is the elementary estimate proved in (10.454); no newly chosen
rational coefficient is substituted for the original height parameter.

Put \(\mathcal T=C_2h\Omega_n\). The height-product estimate (10.55),
with saturation index at least one and \(w_K\ge q^u\), gives

\[
\Omega_n\ge\frac{q^u n^n}
 {\rho d^{n+1}\ell_d n!\mathrm e^n},\qquad \rho\le58.
\tag{10.798}
\]

Also \(\mathcal R_n A_n/t^2\ge1/n\). If \(t\le n\), use
\(\mathcal R_n\ge1,A_n\ge t\). If \(t>n\), the second entry of the
residue maximum gives \(\mathcal R_nA_n/t^2\ge(\mathrm et/n)^n>1/n\).
Substituting the full C2 formula therefore proves

\[
\mathcal T\ge
\frac{c_*a^nP^{n-1}F(n)n^{n-1}}{\rho n!}\frac{dh}{t}
\ge9261\frac{dh}{t}.
\tag{10.799}
\]

Indeed \(c_*\ge406,a\ge7,P\ge1,F(n)\ge F(2)=27\), and \(n!\le n^{n-1}\).
The exact last coefficient is \(406\cdot49\cdot27/58=9261\). Thus the
first summand in the elementary bound is less than \(\mathcal T/4000\),
since \(8000<9261\).

For the second summand use the stipulated profile and
\(g/\mathcal R_1<1\). Exact cancellation against the full target gives
the strict normalized upper bound

\[
\frac{2e q^u\mathrm e^{n-2}t^2}
 {c_*a q^{n-1}d^2F(n)\ell_d A_nh}.
\tag{10.800}
\]

The cyclotomic local-degree argument in Theorem 10.73 gives
\(e q^u\le2d\). Also \(A_n\ge t,h\ge(n+1)t\). The expression is
therefore at most

\[
\frac{4\mathrm e^{n-2}}
 {c_*a q^{n-1}dF(n)\ell_d(n+1)}
\le\frac4{406\cdot7\cdot2\cdot27\cdot3}<\frac1{100000}.
\tag{10.801}
\]

For this last comparison, Theorem 10.187 proves
\(F(n)\ge27\mathrm e^{n-2}\); all other factors have their displayed
elementary lower bounds. The exact final integer comparison is
\(400000<460404\). Adding the two contributions gives
\(1/4000+1/100000=13/50000<1/3000\), proving the assertion. \(\square\)

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={Multiplicity allocation bounds, the conditional reduction in rank, and the one-base valuation budget}]{../figures/yu-second-rank-stop.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.56. Left: proved uniform lower bounds for the three
allocation ratios in Theorem 10.191, on a logarithmic scale. All exceed
the required threshold one. Right: Corollary 10.192 uses prepared zeros
as an explicit hypothesis and produces a smaller independent family with
its actual residue index and the stronger height allowance. Below: the
two rank-one contributions, each divided by \(C_2h\Omega_n\), have sum
less than 260 in units of \(10^{-6}\); the target is \(1000/3\) in the
same units. Proposition 10.193 proves the inequality, and Solution 77
checks the algebra and exact endpoints. These are proved bounds, not
sampled field parameters.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_second_rank_stop.py}{Reproducible figure and
exact endpoint calculation}. Human-source context: the multiplicity
method in Section 6 of
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
freely accessible 2013 paper}.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{64. Coefficient heights and the first second-scale zero
extension}\label{coefficient-heights-and-the-first-second-scale-zero-extension}

The selected auxiliary function already vanishes on its initial integer
interval. We estimate the height of its coefficients by averaging over
the entire set of prepared equations. A positive generating series then
bounds each individual value. The resulting strict comparison extends
the zeros to an interval q times as large.

Retain all parameters of Theorem 10.184 and the original integer
representation (10.218). The distinguished nonzero coefficient has
minimum p-adic valuation. Throughout this section assume

\[
h\ge\max\left\{G_2(r,d),(r+1)t,\log B_\dagger,
                   \log\frac{R_{\max}}{dW(d)}\right\}.
\tag{10.806}
\]

Here \(W(d)\) is (10.222), and the original coefficients define
\(R_{\max}\) and \(B_\dagger\). In particular these hypotheses hold for
the original second height parameter when \(r\le n\). Write
\(Z=S\mathcal D/d\), \(\lambda=59/58\), \(H=h+\log J\), and
\(y=1/(r+1)\).

\subsubsection{An individual scalar and its complete-simplex
average}\label{an-individual-scalar-and-its-complete-simplex-average}

\textbf{Lemma 10.194 (the second-scale generating remainder).} At any
contraction depth \(I\ge0\), let \(\Omega_j\) be the actual nonnegative
integer maxima of the Euler arguments, and set

\[
\begin{aligned}
m_I&=\sum_{j=1}^{r-1}\Omega_j+r-2,\\
a_E&=H+1+\log W(d),\qquad a=\max\{107H/103,a_E\},\\
E_I&=(m_I+1)[-\log(1-\exp(-a_E))].
\end{aligned}
\tag{10.807}
\]

For every nonnegative integer additive order u and Euler order v,

\[
u\log V(k)+\log\binom{m_I+v}{v}\le a(u+v)+E_I,
\qquad
\frac{E_I}{Z}<\frac{q^{-I}}{79}+\frac{r-1}{999Z}.
\tag{10.808}
\]

Consequently, at real radius R and total order at most O, the logarithm
of each prepared scalar is at most

\[
D_{\mathrm a}\log(q^{-I}R+2k)-L\log(k!)+aO+E_I.
\tag{10.809}
\]

At the initial full composition rows, the mean scalar logarithm is at
most

\[
D_{\mathrm a}\log(S+2k)-L\log(k!)+C_rTH+E_0,
\qquad
C_r=\frac{r+4/103}{r+1}+\frac{3(r-1)}{8(r+1)^2}
\le1-\frac{483}{824(r+1)}<1.
\tag{10.810}
\]

\textbf{Proof.} Theorem 10.112 gives \(\log V(k)<107k/103\le107H/103\).
The positive series proved in Theorem 10.115 gives, for \(0<z<1\),

\[
\binom{m_I+v}{v}z^v\le(1-z)^{-m_I-1}.
\tag{10.811}
\]

Use \(z=\exp(-a_E)\). The Euler contribution is at most \(a_Ev+E_I\),
while the lcm contribution is at most \((107H/103)u\). Both coefficients
are at most a, proving the first inequality. For the remainder, Lemma
10.84, including its full-width contraction, gives

\[
\frac{m_I+1}{Z}\le
\frac{q^{-I}(r-1)tW(d)\exp H}
 {c_1c_2c_3qP(r+1)H}+\frac{r-1}{Z}.
\tag{10.812}
\]

There is no extra factor \(1/r\) after the r widths are summed. The
definition of \(a_E\) gives \(W(d)\exp(H-a_E)=1/\mathrm e\). Also
\(H>40\) by (10.755), hence \(z<1/1000\). Integration gives
\(-\log(1-z)\le z/(1-z)\), and \(t/H\le1/(r+1)\). Therefore

\[
\frac{E_I}{Z}<\frac{1000}{999}
 \frac{q^{-I}(r-1)}{\mathrm e c_1c_2c_3qP(r+1)^2}
 +\frac{r-1}{999Z}.
\tag{10.813}
\]

Substitution in the second-scale table gives the following lower bounds
for \(c_1c_2c_3qP\), in cases I/II, III, IV and V respectively:

\[
\frac{44100}{5101},\qquad\frac{7840}{2083},\qquad
\frac{18375}{2642},\qquad\frac{328}{39}.
\tag{10.814}
\]

They all exceed \(15/4\). Here use \(P\ge3\) in I/II, \(P\ge1\) in
III/IV, and \(P\ge4\) in V; these follow from (10.641). Since
\((r-1)/(r+1)^2\le1/8\) and \(1/\mathrm e<3/8\), the first remainder
coefficient is less than \(25/1998<1/79\). This proves the individual
bound at every depth and order.

In Theorem 10.92 the additive radius is \(q^{-I}R+2k-1\). Its binomial
coefficient satisfies
\(\binom{D_{\mathrm a}}u (q^{-I}R+2k-1)^{D_{\mathrm a}-u}
\le(q^{-I}R+2k)^{D_{\mathrm a}}\). This proves the asserted scalar bound
and retains \(-L\log(k!)\).

For the mean, retain the separate lcm and Euler coefficients just
proved. We next show \(a_E/H<1+3/[8(r+1)]\). Every second-table
\(a_0^{(2)}>39/10\) and \(a_1^{(2)}\ge71/100\). Since \(r\ge3\) and
\(\log4>4/3\),

\[
G_2(r,d)>(r+1)(40/3+\log d).
\tag{10.815}
\]

For \(d\ge2\), the tangent bound in Lemma 10.106 proves
\(1+\log W(d)<5+(\log d)/3\). This is at most \((3/8)(40/3+\log d)\).
For \(d=1\), that lemma proves \(1+\log W(1)<2\), which gives the
stronger conclusion. All logarithm estimates used here were proved by
positive series in Sections 36 and 40.

At each node keep all \(\binom{\lfloor T\rfloor+r}{r}\) indices,
including zero rows. Permuting the r derivative coordinates and their
slack shows that their exact coordinate means are
\(\lfloor T\rfloor/(r+1)\). The mean additive order is thus
\(\lfloor T\rfloor/(r+1)\), and the mean Euler order is
\((r-1)\lfloor T\rfloor/(r+1)\). Apply the separate lcm bound and the
Euler series with \(a_E\). The resulting mean derivative contribution is
at most \(C_rTH+E_0\). Zero rows can be given the same positive
majorant. Finally direct subtraction gives

\[
1-C_r=\frac{483r+1101}{824(r+1)^2}
      \ge\frac{483}{824(r+1)}.
\tag{10.816}
\]

This proves the mean estimate. \(\square\)

\subsubsection{The number of columns and the coefficient
vector}\label{the-number-of-columns-and-the-coefficient-vector}

\textbf{Lemma 10.195 (second-scale logarithmic column bounds).} Set

\[
b_r=1500\,12^{r-3},\qquad
\varepsilon_r=\frac{8r+15}{9(r+1)b_r},\qquad
e_0=\frac1{79}+\frac1{71928\cdot1500}.
\tag{10.817}
\]

For the actual selected number of columns \(N_c\) and any
\(1\le d_0\le d\),

\[
\mathcal D>b_r A_r,\quad
\frac{\log(N_cd_0)}Z<\varepsilon_r\le\frac{13}{18000},
\quad \frac{E_0}Z<e_0.
\tag{10.818}
\]

The remainder is the one defined in Lemma 10.194.

\textbf{Proof.} The height-product substitution in Theorem 10.184 gives

\[
\frac{\mathcal D}{A_r}>
\frac{2c_0c_1c_4}{58}\frac N\vartheta
(c_2qP)^r\frac{r^r(r+1)^r}{(r!)^2}.
\tag{10.819}
\]

Here \(N/\vartheta>99/100\), since
\(N/\vartheta=(m+A_0)/(A_0(1+\tau))\), \(m\ge1\) and \(\tau<1/100\). At
\(r=3\) the factorial factor is 48. In I/II, III, IV and V, the
resulting rational lower bounds are respectively

\[
\frac{181946699550000000}{3849142313729},\quad
\frac{431281065600000}{262099267823},\quad
\frac{105293228906250}{66850775669},\quad
\frac{6927360000}{63713}.
\tag{10.820}
\]

Each exceeds 1500. The consecutive factorial-factor ratio exceeds four
by (10.282), and \(c_2qP>3\). Induction proves the first bound for all
\(r\ge3\).

The singleton height estimate used in Lemma 10.105 gives
\(d\sigma_i>1/(87d^2)\). Every lattice coordinate is in
\(J^{-1}\mathbb Z\). Counting coordinates in the original box, before
either selection, and using \(c_1c_2rP>1\), gives
\(|S_0|<(88Jd^2\mathcal D)^r\). Corollary 10.185 implies
\(D_{\mathrm a}<S\mathcal D\): use \(\vartheta/N<101/100\), \(c_1c_4>4\)
and \(A_r>5\). Consequently, writing \(x=\log J\),

\[
\log(N_cd_0)<\log S+(r+1)\log\mathcal D
                  +(2r+1)\log d+rx+r\log88.
\tag{10.821}
\]

Let \(S_*=(9/5)(r+1)^2d\). The table gives \(S\ge S_*\). Both \(S_*\)
and \(5b_r\) exceed \(\mathrm e\), on whose right \(\log z/z\)
decreases. Also \(\log S_*<2A_r\), \(\log88<5<A_r\), \(\log d<A_r\), and
\(\log(5b_r)<3r\). For the last comparison, \((27/10)^9>7500\) proves
\(\log7500<9\), and \((27/10)^3>12\) proves \(\log12<3\). The bound
\(\mathrm e>27/10\) was proved earlier. Dividing the four corresponding
groups of terms by Z gives

\[
\begin{aligned}
\frac{\log S}Z&<\frac{10}{9(r+1)^2b_r},\\
\frac{(r+1)\log\mathcal D}Z&<\frac{r}{3(r+1)b_r},\\
\frac{(2r+1)\log d+r\log88}Z
 &<\frac{5(3r+1)}{9(r+1)^2b_r},\\
\frac{rx}Z&<\frac{5r}{9(r+1)b_r}.
\end{aligned}
\tag{10.822}
\]

For the last line use the exact formula for S, \(x/H\le1\),
\(c_3q\ge9/5\), \(\mathcal D>b_rA_r\), and \(A_r\ge t\). Their sum is
\(\varepsilon_r\). Its prefactor \(8+7/(r+1)\) decreases, while \(b_r\)
increases, so its maximum is \(13/18000\) at three. Finally
\(Z>9(r+1)^2b_r\). Thus \((r-1)/(999Z)<1/(71928b_r)\), and Lemma 10.194
gives the stated remainder. \(\square\)

\textbf{Proposition 10.196 (the initial coefficient height at the second
scale).} The integral kernel in Theorem 10.184 can be chosen so that

\[
\begin{aligned}
\frac{h_2(c)}Z<&\ \frac{\varepsilon_r}{2}
 +\frac1{c_1c_2c_3q(r+1)^2}\\
&+\frac{10}{17}\left[
 \frac{\varepsilon_r}{2}
 +\frac{\lambda\vartheta}{c_1c_4N}
 +\frac{C_r}{c_1c_3eN}+e_0
 +\frac1{2c_1c_2}\left(1+\frac1{2g_S+1}\right)\right].
\end{aligned}
\tag{10.823}
\]

Here e in \(eN\) is the ramification index.

\textbf{Proof.} Apply the weighted integral Siegel argument of
Proposition 10.183 to its full row list, using the mean scalar bound of
Lemma 10.194 in the row sum. Its proof is a sum of individual row
heights, so replacing their common maximum by their proved mean is
valid. The coefficient surplus is \(c_0=27/10\), whence the row ratio is
smaller than \(1/(c_0-1)=10/17\).

For the additive part, integration gives \(\log(k!)\ge k\log k-k+1\).
Also \(k>39H/40\), \(c_3q\le6\) and \(t>1\), so

\[
\frac{S+2k}{k}<7(r+1)d,
\qquad
\log\frac{\mathrm e(S+2k)}k<A_r.
\tag{10.824}
\]

Indeed \(2\le(r+1)d/2\) and \((40/39)6+1/2<7\); then \(1+\log7<4\).
Corollary 10.185 bounds the normalized additive term by
\(\lambda\vartheta/(c_1c_4N)\). Direct substitution of S and T gives
\(TH/Z=1/(c_1c_3eN)\). The coefficient-field generator term from
Corollary 10.65 is at most \((d_0-1)/(c_1c_2S)\), which is smaller than
\(1/[c_1c_2c_3q(r+1)^2]\). The full-width mean torus term is precisely
the one displayed; its factor is \(1/(2c_1c_2)\). Use Lemma 10.195 for
both logarithmic column terms and the remainder. This proves the bound
while retaining every initial zero and the nonzero polynomial.
\(\square\)

\subsubsection{The first extension gives exact algebraic
zeros}\label{the-first-extension-gives-exact-algebraic-zeros}

\textbf{Theorem 10.197 (the complete first integer extension).} Choose
the kernel as in Proposition 10.196. Set \(c_5=27/50\) at \(q=2\) and
\(c_5=4/5\) at \(q=3\), and put \(\eta=1-c_5/(r+1)\). Under the
counterhypothesis \(v_p(\Xi-1)\ge U_r\),

\[
\Phi(s;\boldsymbol t)=0
\quad\text{for every integer }|s|\le qS
\quad\text{and every }|\boldsymbol t|\le\eta T.
\tag{10.825}
\]

\textbf{Proof.} Put \(\mu=\lfloor c_5T/(r+1)\rfloor+1\) and
\(s_0=\lfloor S\rfloor\). Since \(T>101r^2\),

\[
\mu<\frac{101}{100}\frac{c_5T}{r+1}.
\tag{10.826}
\]

For example, the relative rounding error is smaller than
\(4/[(27/50)\,101\cdot9]<1/100\); the ratio \((r+1)/r^2\) decreases. The
proof of Theorem 10.190 gives \(M_0<SN/3\), \(S>16\), and strict
normality on the enlarged disc. Consequently

\[
\frac{\mu[(2s_0+1)N+M_0]}{U_r}
<\frac{101}{100}\frac{115}{48}
                      \frac{c_5}{c_1q^r}<\frac14.
\tag{10.827}
\]

For \(q=2\) and \(q=3\) the respective upper bounds, already using only
\(r\ge3,c_1\ge7/10\), are \(20907/89600\) and \(2323/22680\). The exact
floor inequality \(\lfloor\eta T\rfloor+\mu-1\le\lfloor T\rfloor\)
supplies every required input jet. Theorem 10.182 therefore yields

\[
v_p(\Phi(s;\boldsymbol t))-\beta
\ge(2s_0+1)\mu N-D_{\mathrm a}(N+1/(p-1)).
\tag{10.828}
\]

Write \(\mathcal Z=Zd/(et)\). We compare both terms with this same
scale. Let \(d_*\) be the least field degree in the case, put
\(\gamma=c_3qd_*\), and retain \(y=1/(r+1)\). Since \(S\ge\gamma/y^2\),
the positive analytic term is strictly greater than

\[
\mathcal Z\,\mathcal L(y),\qquad
\mathcal L(y)=\frac{2c_5q}{c_1}
                    \left(1-\frac{y^2}{2\gamma}\right).
\tag{10.829}
\]

Suppose a target value is nonzero. It belongs to the original field K:
these are integer nodes at depth zero, so Theorem 10.55 applies. Theorem
10.92 uses its actual degree d and local degrees e,f, and its
q-denominator exponent is zero. The torus term divided by Z is at most
\(q/(c_1c_2)\). For the scalar at radius \(qS\), Lemma 10.194 and the
factorial estimate above give an additive contribution at most
\(\lambda\vartheta B_q/(c_1c_4N)\), where \(B_2=57/50\), \(B_3=61/50\).
Indeed \(qS+2k\le q(S+2k)\), \(A_r>5\), \(\log2<7/10\), and
\(\log3<11/10\).

Moreover

\[
\frac{a\eta}{H}
\le\frac{107}{103}-(c_5-3/8)y,
\qquad C_r\le1-\frac{483}{824}y.
\tag{10.830}
\]

The first follows from \(a/H\le107/103+3y/8\) and multiplication by
\(1-c_5y\); discarding its negative quadratic term only increases the
bound. All second-scale rows satisfy \(c_1c_2c_3q>2\), as follows by
direct substitution: the four products are
\(14700/5101,7840/2083,18375/2642,82/39\). Combining Proposition 10.196
and the additional \((\log N_c)/2\) in Theorem 10.92 now bounds the
arithmetic bracket, with its normalizing loss added, by
\(\mathcal Z\mathcal A(y,A_0)\), where

\[
\begin{aligned}
K&=\frac{107}{103}+\frac{10}{17},\quad
\Delta=c_5-\frac38+\frac{10}{17}\frac{483}{824},\quad
B=\lambda(B_q+10/17),\\
\mathcal A(y,A_0)&=
\frac{q}{c_1c_2}
+\frac{10}{17}\frac1{2c_1c_2}
                         \left(1+\frac{y^2}{2\gamma}\right)
+\frac{y^2}{2}+\frac{22}{17}\frac{13}{18000}
+\frac{27}{17}e_0+\mathcal N\\
&\quad+\frac{101}{100c_1}
       \frac{BA_0/c_4+(K-\Delta y)/c_3}{1+A_0}.
\end{aligned}
\tag{10.831}
\]

The displayed rational factor follows from
\(eN=(m+A_0)/(1+\tau)>(1+A_0)/(101/100)\). It is not a substitution of
\(\vartheta\) for N.

Here are the remaining case data, including an upper bound
\(\mathcal N\) for the normalizing loss divided by \(\mathcal Z\). Set
\(\xi=101/100\).

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}lrrr@{}}
\toprule\noalign{}
Case & \(d_*\) & Lower bound \(a_*\) for \(A_0\) & \(\mathcal N\) \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
I.1 & 2 & \(3/2\) & \(\lambda(3/2+\xi/2)/(c_1c_4)\) \\
I.2 & 1 & \(3/2\) & \(\lambda(3/2+\xi/2)/(c_1c_4)\) \\
II & 2 & \(5/2\) & \(\lambda(5/4+\xi/4)/(c_1c_4)\) \\
III.1 & 2 & \(3/4\) & \(\lambda\xi/(c_1c_4)\) \\
III.2 & 1 & \(3/4\) & \(\lambda\xi/(c_1c_4)\) \\
IV & 2 & \(1/2\) & \(\lambda(7/6+\xi/6)/(c_1c_4)\) \\
V & 2 & 2 & \(\lambda(1+\xi/2)/(c_1c_4)\) \\
\end{longtable}
}

To prove the normalizing bounds, Corollary 10.185, \(A_r\ge t\),
\(d\ge ef\), and \(eN>A_0/\xi\) give

\[
\frac{D_{\mathrm a}(N+1/(p-1))}{\mathcal Z}
<\frac{\lambda}{c_1c_4}
       \left(\frac\vartheta f+\frac\xi{f(p-1)}\right).
\tag{10.832}
\]

Use the bounds for \(\vartheta\) in Section 36. In III,
\(\vartheta+1/(p-1)=1\); in V, \(f\ge2\). The lower bounds for \(A_0\)
follow from (10.641) and its unramified case. This proves every entry in
the table.

It remains to verify the strict comparison uniformly, rather than by
sampling ranks. For fixed y the last fraction defining \(\mathcal A\) is
fractional linear in \(A_0\). It lies between its value at \(a_*\) and
its limit at infinity. For each of these two bounds,
\(\mathcal L-\mathcal A\) is a concave quadratic in y. Its minimum on
\([0,1/4]\) is therefore at an endpoint: subtracting the line joining
the endpoint values gives a nonnegative multiple of \(y(1/4-y)\). Thus
four exact rational comparisons per case prove the assertion for every
rank and every admissible \(A_0\). Substituting the tables gives these
strict lower bounds for the gap:

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}lr@{}}
\toprule\noalign{}
Case & \(\mathcal L-\mathcal A\) greater than \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
I.1 & \(27/500\) \\
I.2 & \(41/1000\) \\
II & \(3/10\) \\
III.1 & \(93/1000\) \\
III.2 & \(93/1000\) \\
IV & \(91/250\) \\
V & \(4/3\) \\
\end{longtable}
}

All exceed \(1/25\). The
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_second_initial_height.py}{exact rational
calculation} records the four differences in each case. We have
therefore bounded the same hypothetical nonzero value strictly above and
strictly below incompatible quantities. It must be zero. Additive orders
exceeding \(D_{\mathrm a}\) vanish by degree, so the conclusion covers
the whole stated cutoff. \(\square\)

The theorem supplies the first enlarged integer block. Further integer
extensions, fractional values and support contraction retain their own
field and denominator costs.

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={The initial height budget and the proved positive gaps for the first integer extension in all seven field cases.}]{../figures/yu-second-initial-height.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.57.}
\end{flushleft}
\end{minipage}\par\medskip
 The left panel shows the complete-simplex
coefficient \(C_r\) and its proved upper envelope; the displayed ranks
are samples of a bound proved for all ranks. The right panel shows the
smaller of the four exact endpoint gaps in each field case. Fractional
linearity in \(A_0\) and concavity in \(1/(r+1)\) make these endpoint
checks uniform. The dashed line is the common proved margin \(1/25\).
Theorem 10.197 gives exact zeros at integer nodes through \(qS\), at
total order at most \(\eta T\); it does not assert fractional zeros.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_second_initial_height.py}{Reproducible
calculation and figure}. Human-source context: the selected-support
induction in Section 11 of
\href{https://matwbn.icm.edu.pl/ksiazki/aa/aa89/aa8944.pdf}{Yu's freely
accessible 1999 paper} and the second constants in Section 1.1 of his
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{freely
accessible 2013 paper}.

\subsection{65. The complete second-scale integer zero
block}\label{the-complete-second-scale-integer-zero-block}

The first enlargement can be repeated without changing the coefficient
vector or its field. The real interval grows by q at each step, while
the available derivative order is multiplied by \(\eta\). We prove the
comparison for every rank, including the accumulated integer rounding
and ordinary-jet costs.

\textbf{Theorem 10.198 (all integer stages at the second scale).} Retain
all hypotheses and notation of Section 64, and choose the coefficient
vector of Proposition 10.196. Under \(v_p(\Xi-1)\ge U_r\),

\[
\Phi(s;\boldsymbol t)=0
\quad\text{if }s\in\mathbb Z,\quad |s|\le q^jS,
\quad |\boldsymbol t|\le\eta^jT,
\qquad 0\le j\le r.
\tag{10.835}
\]

These are the original prepared values at contraction depth zero. In
particular the conclusion through \(q^rS\) retains the actual global
field K and every original polynomial column.

\textbf{Proof.} Put \(y=1/(r+1)\). The initial stage is Theorem 10.184,
and the step from zero to one is Theorem 10.197. We establish the input
and arithmetic comparisons for the remaining steps \(1\le j\le r-1\).

At step j set

\[
R_j=q^jS,\qquad s_j=\lfloor R_j\rfloor,\qquad
\mu_j=\left\lfloor\frac{c_5\eta^jT}{r+1}\right\rfloor+1.
\tag{10.836}
\]

Bernoulli's inequality, proved by induction after multiplying by
\(1-c_5/(r+1)\), gives \(\eta^j>1-c_5\) for \(j\le r\). Equation
(10.709) therefore gives

\[
\frac{r+1}{c_5\eta^jT}
<\frac{29}{12c_5(1-c_5)12^r}
\le\frac{725}{82944}<\frac1{100},
\qquad
\mu_j<\frac{101}{100}\frac{c_5\eta^jT}{r+1}.
\tag{10.837}
\]

Here both choices of \(c_5\) satisfy \(c_5(1-c_5)\ge4/25\). The exact
floor inequality gives

\[
\lfloor\eta^{j+1}T\rfloor+\mu_j-1
\le\lfloor\eta^jT\rfloor.
\tag{10.838}
\]

Thus the preceding block supplies every ordinary input jet.

We next control its separation cost. The proof of (10.756) gives
\(\log(2S)<H/2+3+r+\log d\). Since \(j\le r\), \(q\le3\), and
\(\log3<11/10\),

\[
\log(2R_j)<H/2+3+(21/10)r+\log d<H.
\tag{10.839}
\]

For the last inequality use (10.755). After discarding the nonnegative
\((r-1)\log d/2\), the difference between its lower bound for \(H/2\)
and \(3+(21/10)r+\log d\) is at least \((15r^2-r-25)/10>0\).
Consequently the full interval separation bound of Theorem 10.34 and the
original coefficient and lcm costs obey \(M_j\le H/\log p<SN/3\),
exactly as in (10.757)--(10.758).

Using \(S>16\), \(j\le r-1\), and \(U_r=c_1q^rSTN/(r+1)\), we obtain

\[
\frac{\mu_j[(2s_j+1)N+M_j]}{U_r}
<\frac{101}{100}\frac{c_5}{c_1q}
       \left(2+\frac1{16}+\frac13\right)<\frac{15}{16}.
\tag{10.840}
\]

The two upper bounds, using only \(c_1\ge7/10\), are \(20907/22400\) for
q = 2 and \(2323/2520\) for q = 3. Strict normality is unchanged from
Theorem 10.190. Theorem 10.182 therefore gives at every target integer
node

\[
v_p(\Phi(s;\boldsymbol t))-\beta
\ge (2s_j+1)\mu_jN-D_{\mathrm a}(N+1/(p-1)).
\tag{10.841}
\]

It remains to force exact zeros by the global comparison. Use the same
scale \(\mathcal Z=Zd/(et)\), the same case data
\(a_*,d_*,\gamma=c_3qd_*\), and the same normalizing upper bound
\(\mathcal N\) as in Theorem 10.197. The positive analytic term is
strictly greater than \(\mathcal Z\mathcal L_j(y)\), where

\[
\mathcal L_j(y)=\frac{2c_5q^{j+1}}{c_1}(1-c_5y)^j
 -\frac{c_5q}{c_1\gamma}y^2(1-c_5y)^j.
\tag{10.842}
\]

This retains \(2\lfloor q^jS\rfloor+1>2q^jS-1\). If a target value is
nonzero, it is in K by Theorem 10.55. There is no q-denominator at these
integer nodes and depth zero. Theorem 10.92, Lemma 10.194 and
Proposition 10.196 then bound its arithmetic height, with the
normalizing loss added, by \(\mathcal Z\mathcal A_j(y,A_0)\), as
follows:

\[
\begin{aligned}
\ell_q&=\begin{cases}7/50,&q=2,\\11/50,&q=3,\end{cases}\\
B_j&=\lambda\bigl(1+(j+1)\ell_q+10/17\bigr),\\
K_j(y)&=\left(\frac{107}{103}+\frac38y\right)(1-c_5y)^{j+1}
           +\frac{10}{17}\left(1-\frac{483}{824}y\right),\\
\mathcal A_j(y,A_0)&=\frac{q^{j+1}}{c_1c_2}
 +\frac{10}{17}\frac1{2c_1c_2}\left(1+\frac{y^2}{2\gamma}\right)
\\
&\quad+\frac{y^2}{2}+\frac{22}{17}\frac{13}{18000}
 +\frac{27}{17}e_0+\mathcal N\\
&\quad+\frac{101}{100c_1}
       \frac{B_jA_0/c_4+K_j(y)/c_3}{1+A_0}.
\end{aligned}
\tag{10.843}
\]

Indeed \(q^{j+1}S+2k\le q^{j+1}(S+2k)\), and \(\log q/A_r<\ell_q\)
because \(A_r>5\). The pointwise derivative term is bounded using
\(a/H\le107/103+3y/8\) at order \(\eta^{j+1}T\). The initial mean term
uses \(C_r\le1-483y/824\). Every other contribution is exactly the
corresponding bound in Theorem 10.197, with its actual d,e,f.~In
particular this comparison does not discard the coefficient-field or
logarithmic column term.

Here is a finite exact verification for steps 1 through 4, followed by a
uniform argument for all later steps. For fixed y, the last fraction is
fractional linear in \(A_0\), so it lies between its value at \(a_*\)
and its limit at infinity. At either endpoint
\(\mathcal L_j-\mathcal A_j\) is a polynomial P of degree at most
\(n=j+2\). Its required interval is

\[
0\le y\le b_j:=\frac1{\max\{4,j+2\}},
\tag{10.844}
\]

since \(r\ge3\) and \(j\le r-1\). Write \(P(y)=\sum_{k=0}^n a_ky^k\) and
define

\[
v_i=\sum_{k=0}^i a_kb_j^k
                \frac{\binom ik}{\binom nk},\qquad
P(y)=\sum_{i=0}^n v_i\binom ni
       (y/b_j)^i(1-y/b_j)^{n-i}.
\tag{10.845}
\]

To prove the identity, substitute the first formula into the second and
use \(\binom ni\binom ik=\binom nk\binom{n-k}{i-k}\). The inner binomial
sum is one after taking out \((y/b_j)^k\). The displayed weights are
nonnegative and sum to one. Thus P is at least the smallest of the
rational numbers \(v_i\). Substitution of the seven case rows gives the
strict lower bounds in the second column of the table below. This checks
56 polynomials with 308 coefficients; the identity proves that the
checks cover whole intervals, rather than a sample of ranks.

For the tail first observe

\[
\eta^j\ge\eta^r>\exp(-c_5)>z_q,
\qquad
z_2=29/50,\quad z_3=4/9.
\tag{10.846}
\]

Integration gives \(\log(1-x)>-x/(1-x)\) for \(0<x<1\). Taking
\(x=c_5/(r+1)\) yields \(r\log\eta>-rc_5/(r+1-c_5)>-c_5\). For the
rational exponential bounds, at c = 27/50 and c = 4/5, respectively, the
positive series gives

\[
\exp(c)\le\sum_{k=0}^6\frac{c^k}{k!}
 +\frac{c^7}{7!}\frac1{1-c/8}
 <\begin{cases}50/29,&c=27/50,\\9/4,&c=4/5.\end{cases}
\tag{10.847}
\]

The tail majorant follows because every ratio after the seventh term is
at most c/8; both final comparisons are rational.

For \(j\ge5\), put \(K=107/103+10/17\), \(\xi=101/100\),
\(\varepsilon=13/18000\), and define

\[
\begin{aligned}
\mathfrak a&=\frac{2c_5z_q-1/c_2}{c_1},
&\mathfrak c&=\frac{\xi\lambda\ell_q}{c_1c_4},\\
\mathfrak b&=\frac{c_5q}{16c_1\gamma}
 +\frac{10}{17}\frac1{2c_1c_2}\left(1+\frac1{32\gamma}\right)
 +\frac1{32}+\frac{22}{17}\varepsilon+\frac{27}{17}e_0+\mathcal N\\
&\quad+\frac\xi{c_1}
  \left[\frac{\lambda(1+10/17)}{c_4}
                  +\frac K{c_3(1+a_*)}\right].
\end{aligned}
\tag{10.848}
\]

The bound in Theorem 10.197 implies
\((107/103+3y/8)\eta^{j+1}\le107/103\), so \(K_j(y)\le K\). Use
\(y\le1/4\), \(A_0/(1+A_0)\le1\), and \(1/(1+A_0)\le1/(1+a_*)\). The
preceding estimates give

\[
\mathcal L_j-\mathcal A_j
>F_j:=\mathfrak a q^{j+1}-\mathfrak b-\mathfrak c(j+1).
\tag{10.849}
\]

In every case \(\mathfrak a>0\), \(\mathfrak a(q-1)q^6-\mathfrak c>0\),
and the third column below is a strict lower bound for \(F_5\).
Consequently \(F_{j+1}-F_j>0\) for every \(j\ge5\).

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}
  >{\raggedright\arraybackslash}p{(\linewidth - 4\tabcolsep) * \real{0.2727}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 4\tabcolsep) * \real{0.3636}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 4\tabcolsep) * \real{0.3636}}@{}}
\toprule\noalign{}
\begin{minipage}[b]{\linewidth}\raggedright
Case
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
Steps 1--4: every coefficient greater than
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
Tail: \(F_5\) greater than
\end{minipage} \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
I.1 & \(7/10\) & \(12/5\) \\
I.2 & \(69/100\) & \(119/50\) \\
II & \(93/100\) & \(67/25\) \\
III.1 & \(73/100\) & \(67/50\) \\
III.2 & \(18/25\) & \(33/25\) \\
IV & \(93/100\) & \(41/50\) \\
V & \(9/2\) & \(57/10\) \\
\end{longtable}
}

The \href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_second_integer_extension.py}{exact
calculation} records all rational coefficients, both exponential bounds,
and each tail increment. Every table entry is greater than 1/2. We have
therefore proved \(\mathcal L_j-\mathcal A_j>1/2\) for every remaining
integer step and every admissible rank and field. A nonzero target value
would violate the two bounds for its valuation, so it is zero. Together
with the exact jet inequality, induction completes all stages through r.
Above-degree additive orders vanish identically, as in Theorem 10.197.
\(\square\)

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={The growing integer intervals and decreasing derivative cutoffs, with exact uniform comparison bounds in all seven field cases.}]{../figures/yu-second-integer-extension.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.58.}
\end{flushleft}
\end{minipage}\par\medskip
 Left: the radii \(q^jS\) and orders \(\eta^jT\) of
Theorem 10.198 at r = 6 and q = 2, for j from zero to six. The points
show the real cutoffs; the proof retains their integer floors. Right:
the smallest Bernstein coefficient over steps 1--4 and the tail lower
bound \(F_5\) in each field case. Both give a strict gap greater than
1/2 in units of \(Zd/(et)\); the first step has the separate bound in
Theorem 10.197. The conclusion is an integer zero block at depth zero.
Fractional values and support contractions require their additional
field and denominator estimates.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_second_integer_extension.py}{Reproducible
calculation and figure}. Human-source context: Section 11 of
\href{https://matwbn.icm.edu.pl/ksiazki/aa/aa89/aa8944.pdf}{Yu's freely
accessible 1999 paper} and the second constants in Section 1.1 of his
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{freely
accessible 2013 paper}.

\subsection{66. The first fractional block at the second
scale}\label{the-first-fractional-block-at-the-second-scale}

The integer intervals contain a nested collection of derivative data.
Their smaller intervals retain higher orders, and all these orders can
be used together. At a fractional node the global root field still has
to be paid for. Two quantitative features of the selected construction
supply the comparison: its actual coefficient surplus and the strict
room between its auxiliary scale and the stated second constant.

Retain Sections 61--65, including the original coefficient conditions
(10.806), the saturated phase data and compatible principal roots of
Theorems 10.51 and 10.56. Write

\[
\mathcal Z=\frac{Zd}{et},\qquad
y=\frac1{r+1},\qquad \eta=1-c_5y,\qquad
\mathfrak h=\eta^{r+1},\qquad
S_1=\mathfrak h^{-1}S,\quad T_1=\mathfrak h T.
\tag{10.853}
\]

Thus H continues to mean \(h+\log J\); it is distinct from the order
fraction \(\mathfrak h\). The constants \(c_5\) are those of Theorem
10.197.

\subsubsection{The coefficient surplus and the scale
comparison}\label{the-coefficient-surplus-and-the-scale-comparison}

\textbf{Lemma 10.199 (retaining the full selected-column surplus).} Let
M be the full number of initial equations in Theorem 10.184. Its
selected columns satisfy

\[
\frac{N_c}{M}>c_0p^{fA_0}
       \frac{\delta_r\mathcal R_r}{p^f-1}>c_0p^{fA_0}.
\tag{10.854}
\]

In particular the kernel may be chosen with the bound (10.823) in which
the factor \(10/17\) is replaced by w from this table:

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}lrr@{}}
\toprule\noalign{}
Case & w & \(\chi\) \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
I.1 & \(2/25\) & \(11/10\) \\
I.2 & \(2/25\) & \(51/50\) \\
II & \(1/134\) & \(21/20\) \\
III.1 & \(10/71\) & \(109/100\) \\
III.2 & \(10/71\) & \(6/5\) \\
IV & \(4/23\) & \(13/10\) \\
V & \(5/211\) & \(7/6\) \\
\end{longtable}
}

Every such choice still satisfies Proposition 10.196 and Theorem 10.198.

\textbf{Proof.} Use the column and row bounds in the proof of Theorem
10.184, retaining the factor \(p^{-fA_0}\) in (10.668). The ratio of the
actual column lower bound to the full row upper bound is strictly
greater than

\[
c_0\ell_d\,
\frac{2+1/g_S}{2+1/S}\,
\frac{\delta_r\mathcal R_r}{p^f-1}\,p^{fA_0}.
\tag{10.855}
\]

This follows by substituting (10.702) into the sufficient column
inequality (10.713): the factors \(\zeta_2^r\), \((1+\tau)^r\), the
powers of t and all factorials cancel. Here \(S\ge g_S\) and
\(\ell_d\ge1\). Also \(\delta_r\mathcal R_r\ge p^f\). These inequalities
give the two asserted strict bounds.

The weighted integral Siegel proof in Proposition 10.183 uses the
row-to-surplus ratio \(M/(N_c-M)\). It is now smaller than
\((c_0p^{fA_0}-1)^{-1}\). In the seven cases, respectively, the
following are lower bounds for \(p^{fA_0}\):

\[
5,\quad5,\quad50,\quad3,\quad3,\quad5/2,\quad16.
\tag{10.856}
\]

For I use \(p=3,A_0\ge3/2\) and \(3^3>5^2\). For II use
\(p=5,A_0\ge5/2\) and \(5^5>50^2\). For III use \(p\ge5,A_0\ge3/4\) and
\(5^3>3^4\). For IV use \(p\ge7,A_0\ge1/2\) and \(7>(5/2)^2\). In V,
\(p^f\ge4,A_0\ge2\), so 16 is allowed with equality. Substitution of
these bounds and \(c_0=27/10\) gives exactly w. The strict column
inequality makes the final ratio strict also in V.

Apply the same complete-simplex mean estimate of Lemma 10.194 to the
same full list of equations. The coefficient-field generator term and
the leading logarithmic column term are unchanged; only the row-height
multiplier decreases to w. This proves the claimed height bound. All its
bracket terms are nonnegative. Since every w is less than \(10/17\), the
former height bound and all its integer-zero consequences remain valid.
\(\square\)

\textbf{Lemma 10.200 (room in the second constant).} With \(\chi\) in
the table, the same parameters satisfy

\[
\chi eU_r<C_2(r,d,p,\boldsymbol\alpha)\,h\Omega_r.
\tag{10.857}
\]

The rank-transfer conclusion of Lemma 10.187 remains valid with
\(\chi U_r\) in place of \(U_r\).

\textbf{Proof.} In the seven rows of Proposition 10.186, multiply its
upper prefactors 627, 631, 627, 452, 454, 1763, 343 by the corresponding
\(\chi\). The results are strictly below 1438, 648, 690, 495, 557, 2418,
406, respectively. All these are rational comparisons; for example
\(631(51/50)=32181/50<648\). In III the common \(1/\vartheta\) remains
cancelled on both sides. Every other factor in that proposition is
unchanged. The rank transfer uses exactly the same cancellation of
\(\mathcal R_r\) and the same strict factorial ratio as Lemma 10.187;
\(\chi\) depends on the field case, not the rank. \(\square\)

\subsubsection{Nested interpolation with the selected
monomials}\label{nested-interpolation-with-the-selected-monomials}

\textbf{Theorem 10.201 (the first second-scale fractional zeros).}
Choose the kernel of Lemma 10.199. Under \(v_p(\Xi-1)\ge\chi U_r\), its
original prepared values satisfy

\[
\Phi(s/q;\boldsymbol t)=0
\quad\text{for all }s\in\mathbb Z,\quad
|s|\le q(\lfloor S_1\rfloor+1),\quad
|\boldsymbol t|\le T_1.
\tag{10.858}
\]

The global root field may have degree as large as \(q^rd\). The
assertion retains that degree and the complete rational denominator at
these fractional nodes.

\textbf{Proof.} Since \(\chi>1\), Theorem 10.198 supplies every integer
block. Set

\[
\begin{aligned}
R_j&=\lfloor q^jS\rfloor,&
\mu_j&=\lfloor\eta^jT\rfloor-\lfloor\mathfrak h T\rfloor+1,
                 &&0\le j\le r,\\
N_*&=(2R_0+1)\mu_0+
              2\sum_{j=1}^r(R_j-R_{j-1})\mu_j,&
B&=\lfloor\log_p(2R_r)\rfloor,\\
M_*&=\max\{B,v_p(V(k)),v_p(b_n)\},&
Q_D&=D_{\mathrm a}(N+1/(p-1)).
\end{aligned}
\tag{10.859}
\]

The radii are strictly increasing, and the multiplicities are positive
and nonincreasing. Every required ordinary jet is available, since an
output index has order at most \(\lfloor\mathfrak h T\rfloor\). The
proofs of (10.839) and (10.757) give \(M_*\le H/\log p\).

We use the normal-series interpolation of Lemma 10.94 with the selected
combined frequencies. Theorem 10.180 normalizes the projected function
by a scalar of valuation \(Q_D-\beta\), where \(\beta\) is the minimum
valuation of its nonzero auxiliary coefficients. Its strict disc
conditions follow from Theorem 10.190, since the counterhypothesis is
stronger. The prepared projection error has valuation at least
\(\beta+\kappa+\chi U_r-v_p(b_n)\). Lemma 10.38 gives the ordinary-jet
loss \(lC_0\), with \(C_0=\max\{v_p(V(k)),v_p(b_n)\}\), exactly as in
Theorem 10.182.

In Lemma 10.94 the complete cardinal polynomial at a node of
multiplicity \(\mu(s)\) loses at most \(B(\mu_0-\mu(s))\); its Taylor
inverse of degree a loses at most aB. In a term using jet l, one has
\(a+l\le\mu(s)-1\). Its total loss is consequently at most
\((\mu_0-1)M_*\). The monic quotient contributes \(N_*N\). Subtracting
the normalization and then adding back the projection error proves, for
every \(x\in\mathbb Z_p\),

\[
\begin{aligned}
v_p(\Phi(x;\boldsymbol t))-\beta
&\ge\min\{N_*N-Q_D,\\
&\qquad\kappa+\chi U_r-v_p(b_n)-(\mu_0-1)M_*\}.
\end{aligned}
\tag{10.860}
\]

This proves the needed selected-support nested bound directly from the
general interpolation lemma. It makes no assertion about the unverified
last uniform-multiplicity input at scale \(U_r\).

\subsubsection{The two analytic lower
bounds}\label{the-two-analytic-lower-bounds}

Let \(d_*\) and \(a_*\) have their case values in Theorem 10.197, and
put

\[
\begin{aligned}
S_-&=c_3qd_*(r+1)^2,&
T_-&=\frac{12}{29}(r+1)12^r,\\
\Gamma_r&=\frac{2c_5q}{q^r}\sum_{j=0}^r(q\eta)^j
-\frac{q(r+1)}{q^rS_-}
 \left[1-\mathfrak h+2\sum_{j=1}^r(\eta^j-\mathfrak h)\right],\\
\mathcal I_r&=c_1q\chi-
\frac{101}{100}\frac{1-\mathfrak h+2/T_-}
                         {q^rc_3(1+a_*)}.
\end{aligned}
\tag{10.861}
\]

Counting the annuli, with \(\mu_j>(\eta^j-\mathfrak h)T\), gives

\[
\frac{c_1N_*N}{q^r\mathcal Z}>\Gamma_r.
\tag{10.862}
\]

For completeness, the inner node count exceeds \(2S-1\), and the j-th
annulus count exceeds \(2(q^j-q^{j-1})S-2>0\). Their principal real mass
is \(2c_5ST\sum_{j=0}^r(q\eta)^j/(r+1)\). Their removed floor mass is at
most \(T[1-\mathfrak h+2\sum_{j=1}^r(\eta^j-\mathfrak h)]\). Now
\(c_1STN/\mathcal Z=q(r+1)\) and \(S\ge S_-\). This proves the bound,
including the differences of radius floors.

For the second entry, retain the original integer coefficient:
\(v_p(b_n)\le H/\log p\). Also \(\mu_0-1\le(1-\mathfrak h)T+1\). Since
\(d\ge ef\),

\[
\frac{c_1TH}{\mathcal Z\log p}
=\frac f{dc_3N}
\le\frac1{c_3eN}
<\frac{101}{100c_3(1+A_0)}.
\tag{10.863}
\]

Discarding only the nonnegative \(\kappa\), and using \(T>T_-\), shows
that the second entry multiplied by \(c_1/(q^r\mathcal Z)\) exceeds
\(\mathcal I_r\). Write \(\mathcal N\) for the casewise normalizing
upper bound in Theorem 10.197. The first entry in this same
normalization exceeds \(\Gamma_r-c_1\mathcal N/q^r\).

\subsubsection{The arithmetic upper bound in the actual
field}\label{the-arithmetic-upper-bound-in-the-actual-field}

Theorem 10.56 applies to the actual selected support, which is still in
its original phase class. Its normalized fractional value lies in
\(E_\Lambda\), of degree \(q^{h_\Lambda}d\), where \(h_\Lambda\le r\).
The chosen local degrees are e,f. The common principal-unit monomial
removed in normalizing the value has valuation zero. Thus it suffices to
compare the normalized algebraic value V with the same local lower
bound.

Use \(\varepsilon_r,e_0\) from Lemma 10.195, taking the sharper
\(e_r=1/79+1/(71928b_r)\) in place of its rank-three upper bound when
convenient. Set \(\lambda=59/58\), \(\xi=101/100\), and
\(\ell_2=7/50,\ell_3=11/50\). Define

\[
\begin{aligned}
J_r={}&\frac{\mathfrak h^{-1}+S_-^{-1}}{c_2}
 +\frac w{2c_2}\left(1+\frac1{2S_-}\right)
 +\frac{y^2}{c_2c_3q}
 +(1+w/2)c_1\varepsilon_r+(1+w)c_1e_r,\\
B_r={}&\lambda\left(1+\ell_q+w+\frac q{q-1}\ell_q\right),\\
K_r={}&\left(\frac{107}{103}+\frac38y\right)\mathfrak h
                         +w\left(1-\frac{483}{824}y\right),\\
\mathcal B_r(A_0)={}&J_r+
       \xi\frac{B_rA_0/c_4+K_r/c_3}{1+A_0}.
\end{aligned}
\tag{10.864}
\]

If V is nonzero, the global product-formula proof in Theorems 10.57 and
10.92, with Lemma 10.199 for its coefficients, gives

\[
\frac{c_1}{q^r\mathcal Z}(v_p(V)-\beta)
\le\mathcal B_r(A_0).
\tag{10.865}
\]

Here is each contribution. The torus radius obeys
\(X/S\le\mathfrak h^{-1}+S_-^{-1}<q\), giving the first term of \(J_r\).
The mean torus term in the initial coefficient height gives its second.
The coefficient-field generator gives its third. The leading coefficient
count, the row count in that height bound, and the further
\((\log N_c)/2\) give \((1+w/2)c_1\varepsilon_r\). The two scalar
remainders give \((1+w)c_1e_r\).

The pointwise additive term at \(X<qS\) is bounded as in Theorem 10.197
by \(\lambda(1+\ell_q)\), and the mean additive term has multiplier
\(w\lambda\). The fractional denominator is bounded by
\(D_{\mathrm a}(1+1/(q-1))\), using (10.243) at depth zero and root
denominator q. Corollary 10.185 and \(\log q/A_r<\ell_q\) therefore add
precisely \(\lambda q\ell_q/(q-1)\) to \(B_r\). The individual
derivative bound at order \(\mathfrak h T\) and the initial
complete-simplex mean give \(K_r\), by Lemma 10.194. Finally
\(eN>(1+A_0)/\xi\) gives the last displayed fraction. In this argument
the entire nonnegative global height majorant is multiplied by
\(q^{h_\Lambda}\le q^r\). No multiplicity of primes or equality of
conjugate valuations is assumed.

To verify the radius inequality just used, the function \((1-c/v)^v\)
increases for \(v>c\): its logarithmic derivative is
\(\log(1-z)+z/(1-z)>0\), where \(z=c/v\). The latter inequality follows
by integrating \(1/(1-u)<1/(1-z)\) from zero to z. It follows that
\(\mathfrak h^{-1}+S_-^{-1}\) decreases with r. At odd primes check r =
3 and the smallest \(c_3d_*\) in its row; in V check r = 4. The seven
resulting rational numbers are less than q. Rank three is excluded in V
by the positive-optimizer argument in Proposition 10.186.

\subsubsection{Uniform strict
comparisons}\label{uniform-strict-comparisons}

The required two inequalities are

\[
\Gamma_r-\frac{c_1\mathcal N}{q^r}-\mathcal B_r(A_0)>
\frac12,\qquad
\mathcal I_r-\mathcal B_r(A_0)>\frac1{1000}.
\tag{10.866}
\]

We give finite rational reductions that cover all ranks. At fixed r the
dependence on \(A_0\) is fractional linear. Its maximum is therefore at
\(A_0=a_*\) or its limit at infinity. For r = 3 through 7, omitting the
impossible r = 3 row in V, all displayed quantities are rational and can
be substituted exactly.

For every r at least eight, put

\[
\begin{aligned}
H_-&=(1-c_5/9)^9,\\
E_-&=\sum_{j=0}^{10}\frac{c_5^j}{j!},&
E_+&=E_-+\frac{c_5^{11}}{11!(1-c_5/12)},&
H_+&=E_-^{-1}.
\end{aligned}
\tag{10.867}
\]

The positive exponential series gives \(E_-<\exp(c_5)<E_+\). The
monotonicity just proved and \(\log(1-z)<-z\) give
\(H_-\le\mathfrak h<\exp(-c_5)<H_+\). Equation (10.846) also gives
\(\eta^r>\exp(-c_5)>1/E_+\). In the reversed geometric sum defining
\(\Gamma_r\), retain its last four terms and use \(q\eta<q\). This
proves

\[
\Gamma_r>
\frac{2c_5q}{E_+}\sum_{j=0}^{3}q^{-j}
-\frac{153q}{q^8S_-(8)}.
\tag{10.868}
\]

Indeed the removed floor term is at most \(q(r+1)(2r+1)/(q^rS_-(r))\).
Since \(S_-(r)=c_3qd_*(r+1)^2\), this bound decreases for \(r\ge8\): its
consecutive ratio is \((2r+3)(r+1)/(q(2r+1)(r+2))<1\).

For the arithmetic upper bound replace y by \(1/9\),
\(S_-,b_r,\varepsilon_r,e_r\) by their conservative rank-eight values,
\(\mathfrak h^{-1}\) by \(H_-^{-1}\), and the factor \(\mathfrak h\) in
\(K_r\) by \(H_+\). Replace \(1-483y/824\) there by one. All these
substitutions increase the bound. For the input lower bound, use
\(H_-\), \(T_-(8)\) and \(q^8\); these substitutions decrease it. Use
\(q^8\) also for the upper normalizing loss. This reduces every
remaining rank to two more rational comparisons per case.

There are 34 finite-rank rows and seven uniform rows, each with two
endpoints, for 82 comparisons. The following are strict lower bounds for
their smaller differences:

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}
  >{\raggedright\arraybackslash}p{(\linewidth - 4\tabcolsep) * \real{0.2727}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 4\tabcolsep) * \real{0.3636}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 4\tabcolsep) * \real{0.3636}}@{}}
\toprule\noalign{}
\begin{minipage}[b]{\linewidth}\raggedright
Case
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
Mass minus normalization and arithmetic
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
Input minus arithmetic
\end{minipage} \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
I.1 & \(99/100\) & \(3/25\) \\
I.2 & \(99/100\) & \(3/2000\) \\
II & \(109/100\) & \(3/20\) \\
III.1 & \(93/100\) & \(1/20\) \\
III.2 & \(93/100\) & \(19/100\) \\
IV & \(51/50\) & \(11/25\) \\
V & \(27/25\) & \(69/100\) \\
\end{longtable}
}

The \href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_second_fractional_profile.py}{exact
calculation} records every rational endpoint and difference, the seven
radius bounds and the seven comparisons with the second constant. The
uniform reductions above prove why these finite computations cover all
allowed ranks and fields. Each column exceeds its asserted common lower
bound. Thus both entries of the analytic minimum strictly exceed the
arithmetic upper bound for a nonzero V. This contradiction proves V = 0,
hence the original prepared value is zero. Indices whose additive order
exceeds \(D_{\mathrm a}\) vanish by degree. This completes the whole
stated fractional block. \(\square\)

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={The improved coefficient-height multipliers and the positive fractional input margins in all seven field cases.}]{../figures/yu-second-fractional-profile.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.59.}
\end{flushleft}
\end{minipage}\par\medskip
 Left: the row-to-surplus upper bounds of Lemma
10.199, compared with the sufficient \(10/17\) bound of Proposition
10.196. Right: the smaller exact input margin among all finite-rank and
uniform-tail endpoint comparisons in Theorem 10.201, in units
\(q^rZd/(c_1et)\). Both the full root-field degree and the q-denominator
are included. The theorem concerns the first fractional block;
subsequent support contractions retain their separate depth costs.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_second_fractional_profile.py}{Reproducible
calculation and figure}. Human-source context: the selected-support
induction in Section 11 of
\href{https://matwbn.icm.edu.pl/ksiazki/aa/aa89/aa8944.pdf}{Yu's freely
accessible 1999 paper} and the second constants in Section 1.1 of his
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{freely
accessible 2013 paper}.

\subsection{67. The first support contraction at the second
scale}\label{the-first-support-contraction-at-the-second-scale}

The fractional zeros separate the exponent classes. This produces a
smaller nonzero polynomial whose coefficients are selected from the
original kernel. The new integer values lie in the original number
field. Their rational denominators still contribute to the height bound,
and filling the missing multiples of q uses both losses in the
interpolation formula for a q-deleted interval.

Retain the parameters, kernel and counterhypothesis of Theorem 10.201.
In particular, \(c=c_5\), \(\mathfrak h=\eta^{r+1}\),
\(S_1=S/\mathfrak h\), \(T_1=\mathfrak h T\), and
\(v_p(\Xi-1)\ge\chi U_r\). Let \(\beta\) be the minimum valuation of the
nonzero initial coefficients. The meanings of H, N and \(\mathcal Z\)
remain unchanged.

\textbf{Theorem 10.202 (the first full contracted block at the second
scale).} There is a nonzero contracted polynomial with an integral
coefficient subvector of the kernel in Lemma 10.199. Its coefficient
minimum is \(\beta\), its coefficient height does not increase, its
torus widths are at most \(D_i/q\), and its additive argument is x/q.
Its phase family \(\Phi_1\) has every integer value in K. Put

\[
R=q(\lfloor S_1\rfloor+1).
\tag{10.872}
\]

The family satisfies both collections of zeros

\[
\begin{aligned}
\Phi_1(s;\boldsymbol t)&=0
 &&(|s|\le R,\ q\nmid s,\ |\boldsymbol t|\le\lfloor T_1\rfloor),\\
\Phi_1(s;\boldsymbol t)&=0
 &&(|s|\le R,\ |\boldsymbol t|\le\lfloor\eta T_1\rfloor).
\end{aligned}
\tag{10.873}
\]

Its selected combined-frequency congruence retains the depth of Theorem
10.176. The stronger first collection remains available for subsequent
interpolation.

\textbf{Proof.} Apply the exact phase and support passage of Theorem
10.51 to the fractional zeros of Theorem 10.201. Include the second
phase partition when q divides \(G_0\). The saturated root-monomial
basis separates the exponent classes at every integer s prime to q.
Choose a class containing a coefficient attaining \(\beta\). The
resulting subvector is nonzero and integral, preserves its minimum
valuation, and has no larger projective or Euclidean coefficient height,
by the place-by-place subset inequalities in Theorem 10.48.

The affine prepared-jet transfer in that theorem preserves the whole
total-order cutoff: its triangular leading factors are powers of q,
which are units at p.~The new widths and additive argument are exactly
those of (10.141)--(10.142). All additive polynomial indices remain in
each selected exponent fibre. Their independence in Lemma 10.29
therefore makes the contracted polynomial nonzero. The phase identity in
Theorem 10.55 puts its integer prepared values in K. These arguments
prove the first collection of zeros.

Restriction to a q-class followed by division by q preserves the
selected congruence, by the final assertion of Theorem 10.176.
Consequently Theorem 10.180 and Corollary 10.179 still give the
normalization with the same N. The new coefficient minimum is still
\(\beta\), so the normalizing valuation is \(Q_D-\beta\), where
\(Q_D=D_{\mathrm a}(N+1/(p-1))\). The projection estimate retains the
original coefficient \(b_n\) and the original Euler directions. The
strict disc conditions follow from Theorem 10.190, as in Theorem 10.201.

\subsubsection{Filling the missing
nodes}\label{filling-the-missing-nodes}

Set

\[
\begin{aligned}
M&=\left\lfloor\frac{cT_1}{r+1}\right\rfloor+1,&
n_E&=2(q-1)(\lfloor S_1\rfloor+1),\\
B&=\lfloor\log_p(2R)\rfloor,&
M_*&=\max\{B,v_p(V(k)),v_p(b_n)\}.
\end{aligned}
\tag{10.874}
\]

The q-deleted interval has exactly \(n_E\) nodes, since q divides R. The
elementary inequality \(\lfloor a\rfloor+\lfloor b\rfloor
\le\lfloor a+b\rfloor\), applied to \(a=\eta T_1\) and \(b=cT_1/(r+1)\),
gives

\[
\lfloor\eta T_1\rfloor+M-1\le\lfloor T_1\rfloor.
\tag{10.875}
\]

Thus all required ordinary jets are present. The radius inequality in
Theorem 10.201 gives \(R<q^2S\le q^rS\). The proof of (10.839),
including the original coefficient and least-common-multiple bounds,
therefore gives \(M_*\le H/\log p\).

Apply the q-deleted form (10.690) of Theorem 10.182, before imposing its
sufficient one-term budget. For every output order in the second
collection and every \(x\in\mathbb Z_p\), it gives

\[
\begin{aligned}
v_p(\Phi_1(x;\boldsymbol t))-\beta\ge
\min\{&n_EMN-Q_D,\\
 &\kappa+\chi U_r-v_p(b_n)-MB-(M-1)M_*\}.
\end{aligned}
\tag{10.876}
\]

Here MB is the additional deleted-node loss; the inverse-polynomial and
ordinary-jet loss is the separate term \((M-1)M_*\). Together with the
original coefficient loss they are at most \(2MM_*\), since both B and
\(v_p(b_n)\) are at most \(M_*\).

Write \(\xi=101/100\) and retain \(a_*\), \(T_-\), \(\mathcal N\), and
\(\mathcal B_r(A_0)\) from Theorem 10.201. Define

\[
\mathcal I_r^{\mathrm c}
=c_1\chi q^{r+1}
-\frac{\xi\bigl(2c\mathfrak h/(r+1)+2/T_-\bigr)}
            {c_3(1+a_*)}.
\tag{10.877}
\]

Since \(M>cT_1/(r+1)\), \(\lfloor S_1\rfloor+1>S_1\), and \(S_1T_1=ST\),
the identity in the proof of (10.862) yields

\[
\frac{c_1n_EMN}{\mathcal Z}>2cq(q-1),\qquad
\frac{c_1Q_D}{\mathcal Z}<c_1\mathcal N.
\tag{10.878}
\]

For the other entry use \(M\le cT_1/(r+1)+1\), \(T>T_-\), and the
complete original-coefficient identity (10.863). Discarding only the
nonnegative \(\kappa\) gives

\[
\frac{c_1}{\mathcal Z}
 \bigl(\kappa+\chi U_r-v_p(b_n)-MB-(M-1)M_*\bigr)
>\mathcal I_r^{\mathrm c}.
\tag{10.879}
\]

\subsubsection{The original-field arithmetic
comparison}\label{the-original-field-arithmetic-comparison}

At a target integer, the value lies in K by the proved phase identity.
If it were nonzero, the same product-formula derivation as (10.865)
would give

\[
\frac{c_1}{\mathcal Z}
       \bigl(v_p(\Phi_1(s;\boldsymbol t))-\beta\bigr)
\le\mathcal B_r(A_0).
\tag{10.880}
\]

We justify each changed argument in that bound. The coefficient height
and minimum have their original bounds, and the number of remaining
columns is no larger. The torus contribution has radius
\(q^{-1}R/S\le\mathfrak h^{-1}+S_-^{-1}\). The additive scalar is
evaluated at an argument of absolute value at most
\(q^{-1}R=\lfloor S_1\rfloor+1<qS\). The derivative order \(\eta T_1\)
is smaller than \(T_1\). For the Euler remainder, the depth-one bound
(10.808) gives

\[
\frac{E_1}{Z}<\frac1{79q}+\frac{r-1}{999Z}<e_r.
\tag{10.881}
\]

Only this common upper envelope is required; no comparison of the exact
numerical remainders at successive depths is assumed. Finally, (10.243)
gives the full rational clearing cost
\(\Xi_{1,0,u}\le D_{\mathrm a}(1+1/(q-1))\). Thus every term in (10.864)
remains an upper bound. The normalization here is \(c_1/\mathcal Z\):
the actual global field is K, of degree d, with its original local
degrees e and f.~The factor \(q^r\) paid at fractional nodes is absent,
while the q-denominator remains included.

\subsubsection{Strict inequalities at every permitted
rank}\label{strict-inequalities-at-every-permitted-rank}

The two comparisons are

\[
2cq(q-1)-c_1\mathcal N-\mathcal B_r(A_0)>\frac35,
\qquad
\mathcal I_r^{\mathrm c}-\mathcal B_r(A_0)>9.
\tag{10.882}
\]

Here is a finite rational reduction covering every rank. Use the 34
finite-rank rows and seven uniform rows of Theorem 10.201, with the two
endpoints \(A_0=a_*\) and \(A_0\to\infty\). The arithmetic expression is
the same (10.864), so its proved fractional-linear endpoint reduction
and uniform upper envelope apply unchanged. The normalizing loss is now
\(c_1\mathcal N=\lambda\nu/c_4\), with the same casewise factor \(\nu\)
used there, and has no dependence on r.

For the input comparison at every \(r\ge8\), use \(q^{r+1}\ge q^9\),
\(\mathfrak h<H_+\), \(1/(r+1)\le1/9\), and \(T_-\ge T_-(8)\). Each
substitution decreases the displayed input lower bound. Thus the same 82
rational endpoint comparisons prove the result for all ranks. Strict
lower bounds for their smallest differences are as follows:

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}
  >{\raggedright\arraybackslash}p{(\linewidth - 4\tabcolsep) * \real{0.2727}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 4\tabcolsep) * \real{0.3636}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 4\tabcolsep) * \real{0.3636}}@{}}
\toprule\noalign{}
\begin{minipage}[b]{\linewidth}\raggedright
Case
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
Mass minus normalization and arithmetic
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
Input minus arithmetic
\end{minipage} \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
I.1 & \(61/100\) & 10 \\
I.2 & \(3/5\) & 9 \\
II & \(37/50\) & 10 \\
III.1 & \(63/100\) & 10 \\
III.2 & \(31/50\) & 11 \\
IV & \(69/100\) & 13 \\
V & \(71/10\) & 230 \\
\end{longtable}
}

The \href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_second_first_contraction.py}{exact
calculation} lists every rational endpoint and verifies these strict
bounds. The finite computations certify the rational inequalities after
the proved reductions; the uniform rows cover the infinite tail. Both
entries of the analytic minimum therefore exceed the arithmetic upper
bound for a nonzero target value. Each such value must be zero. Additive
orders above \(D_{\mathrm a}\) vanish by degree. This proves the full
second collection, while retaining the first. \(\square\)

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={The higher-order odd-node data, the completed integer block, and the interpolation jet budget.}]{../figures/yu-second-first-contraction.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.60.}
\end{flushleft}
\end{minipage}\par\medskip
 The exact counting example of Solution 81 has
\(q=2\), R = 38, input order 671 and output order 581. The 38 odd nodes
retain order 671; interpolation of multiplicity 91 adds the 39 even
nodes at order 581. Its cardinal polynomial has degree 3458. The
displayed orders describe available vanishing conditions, not function
values. The example illustrates the exact floors and node counts,
without claiming that these small numbers are an instance of the
second-scale parameter construction. The field and precision statements
are proved in Theorem 10.202.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_second_first_contraction.py}{Reproducible
calculation and figure}.

\subsection{68. Saturation and the cost of contraction
depth}\label{saturation-and-the-cost-of-contraction-depth}

The q-primary saturation index contributes both a factor to the lattice
count and a logarithmic cost to repeated contraction. These two effects
can be balanced in the polynomial scale. We give the construction, its
geometric stopping bounds and the arithmetic estimate for integer nodes.

\subsubsection{A scale that includes the saturation
index}\label{a-scale-that-includes-the-saturation-index}

Retain the field and height assumptions of Sections 61--67, including
(10.806). Write \(A=A_r\), \(x=\log J\), \(B=A+x\), \(H=h+x\), and
\(g=B/A\). Define

\[
\widehat{\mathcal D}=g\mathcal D,\qquad
\widehat T=gT,\qquad \widehat D_i=gD_i,\qquad
\widehat Z=\frac{S\widehat{\mathcal D}}d,\qquad
\widehat{\mathcal Z}=\frac{S\widehat{\mathcal D}}{et}.
\tag{10.884}
\]

Keep \(S,k,N,c_i,\eta\) and the field and residue data fixed. In the
polynomial parameter (10.703), use

\[
\widehat X=
\frac{S\widehat{\mathcal D}}{c_1c_4(N/\vartheta)dBk}=X,
\qquad \widehat L=L,\qquad
\widehat D_{\mathrm a}=D_{\mathrm a},\qquad
\widehat U_r=q^{r+1}\widehat{\mathcal Z}.
\tag{10.885}
\]

\textbf{Theorem 10.203 (the kernel with a saturation-dependent scale).}
There is an integral kernel for these parameters with the full
selected-column surplus of Lemma 10.199. The coefficient-height bounds
of Lemmas 10.194--10.195 and Proposition 10.196 hold with hats on the
scales and with the same rational envelopes. Under
\(v_p(\Xi-1)\ge\chi\widehat U_r\), its integer, fractional and first
contracted zero blocks satisfy Theorems 10.198, 10.201 and 10.202 with
the hatted parameters. Moreover,

\[
\chi e\widehat U_r
<C_2(r,d,p,\boldsymbol\alpha)\,h\Omega_r.
\tag{10.886}
\]

The coefficients are chosen for the hatted system of equations. The
construction retains its actual row field and a coefficient of minimum
valuation throughout phase extraction.

\textbf{Proof.} The integer polynomial parameters are unchanged because
\(\widehat{\mathcal D}/B=\mathcal D/A\). Both the width volume and the
real derivative count acquire a factor \(g^r\). Equivalently, substitute
\(\widehat{\mathcal D},B\) in (10.713): both sides acquire the factor g.
The lattice, phase and unit-congruence selections of Proposition 10.183
apply to this box. The floor estimates (10.711)--(10.712) hold because
\(\widehat T\ge T>101r^2\) and \(\widehat X=X>58\). Keeping the factors
in (10.855) gives the same strict column-to-row surplus, and hence the
same w in Lemma 10.199. Choose its integral coefficient vector for the
complete hatted initial row list.

We check the estimates used in the stated consequences. Corollary 10.185
gives the degree bound with \(\widehat{\mathcal D},B\). The normalizing
estimate (10.832) uses only \(B\ge A\ge t\). The generating-series proof
of Lemma 10.194 applies at arbitrary widths and orders: (10.812),
divided by \(\widehat Z\), has the same leading term and a smaller
\((r-1)/(999\widehat Z)\). The derivative coefficient still satisfies
\(a/H\le107/103+3/[8(r+1)]\), and the complete-simplex mean (10.810) is
unchanged. Estimate (10.824) uses only \(S,k,H\), so its logarithm is
less than \(A\le B\). Also

\[
\frac{\widehat T H}{\widehat Z}=\frac1{c_1c_3eN},\qquad
P\sum_i\widehat D_i\sigma_i
       =\frac{\widehat{\mathcal D}}{c_1c_2d}.
\tag{10.887}
\]

Thus the normalized mean derivative and coefficient-generator bounds
remain valid. For Lemma 10.195 use
\(\widehat{\mathcal D}/A\ge\mathcal D/A>b_r\). The actual lattice count
still gives (10.821) with \(\widehat{\mathcal D}\). The functions
\(\log z/z\) decrease on the same ranges, and \(rx/\widehat Z\)
decreases. Consequently the same \(\varepsilon_r,e_r\) bound all column
and scalar remainders. This proves the asserted coefficient bound.

The radius S and its lower bound \(S_-\) are unchanged, while
\(\widehat T>T_-\). Identity (10.863) is unchanged with hats. The proof
of (10.839), including the bounds for the original \(v_p(b_n)\) and
\(V(k)\), still gives \(\log(2q^rS)<H\). Strict normality follows from
Theorems 10.180 and 10.190 with \(\widehat U_r\ge U_r\). Every
normalized positive interpolation mass uses the unchanged identity
\(c_1S\widehat TN/\widehat{\mathcal Z}=q(r+1)\). Thus the floor,
ordinary-jet, normalization and arithmetic arguments in Theorems 10.198,
10.201 and 10.202 apply to this kernel. Their denominator terms use
\(\log q/B\le\log q/A\), so all their rational endpoint comparisons
remain valid.

For the final constant, substitute the hatted parameters in
(10.719)--(10.720), retaining the factor \(H/J\). The resulting
saturation factor is \(HB/J\). Since \(h>40,A>5,x\ge0\),

\[
\left(1+\frac xh\right)\left(1+\frac xA\right)
\le1+\frac9{40}x+\frac{x^2}{200}
\le1+x+\frac{x^2}{2}\le\exp x.
\tag{10.888}
\]

The last inequality is the positive exponential series. Hence
\(HB/J\le hA\), with equality only at \(x=0\). After this substitution
the prefactors of Lemma 10.200 are identical, proving the strict claimed
bound. The factorial and residue cancellation in Lemma 10.187 uses this
same final bound, so its rank-transfer conclusion also applies.
\(\square\)

\subsubsection{The geometric stop and its rank
product}\label{the-geometric-stop-and-its-rank-product}

Write \(\mathfrak h=\eta^{r+1}\), \(Q=q\mathfrak h\). Let I be a
nonnegative integer with

\[
E=Q^I>\exp(3B),\qquad
\bar S=q\mathfrak h^{-I}S,\qquad
\bar T=\eta\mathfrak h^I\widehat T.
\tag{10.889}
\]

Use \(\mathsf D_0=D_{\mathrm a}\), \(\mathsf D_j=2Jq^{-I}\widehat D_j\),
and the integer allocations (10.769) with these \(\bar S,\bar T\).

\textbf{Theorem 10.204 (degree and rank bounds at the logarithmic
stop).} All the allocation and degree hypotheses of Theorem 10.25 hold.
More precisely,

\[
\begin{gathered}
\mathsf D_0>20\mathsf D_j\ (1\le j\le r),\qquad
\mathsf D_0>4\bar T,\qquad \mathsf D_0>4r,\\
\mathcal R_0>\frac{32}{11},\qquad
\mathcal R_{m,J'}>\frac{2^{10}}{29}\ (1\le m<r),\qquad
\mathcal R_r>31.
\end{gathered}
\tag{10.890}
\]

If the nonzero contracted family has the full prepared integer zeros
through \(\bar S\) and order \(\bar T\), with the independent bases and
hyperplane hypotheses of Corollary 10.192, its rank conclusion holds
with the strict factor \(10^{-6}\) in (10.787), including the actual
residue maxima of both families.

\textbf{Proof.} As in (10.773), \(Q>1\), while \(q\mathfrak h^2<1\).
Thus \(\mathfrak h^{-I}>E\) and \(q^I/J>E/J>\exp(3A+2x)\). The
derivative of \(\exp(2x)/(A+x)\) is positive for \(x\ge0\), since
\(2-1/(A+x)>0\). Consequently

\[
\frac{q^I}{JB}>\frac{\exp(3A)}A,\qquad
\frac{\exp(3B)}B\ge\frac{\exp(3A)}A.
\tag{10.891}
\]

The cancellation (10.775), with B in the denominator, gives
\(\mathsf D_0/\mathsf D_j>
(9/392)(r+1)^2d\sigma_j\exp(3A)/A\). Use \(d\sigma_j>1/(87d^2)\),
\(\exp(2A)>256(r+1)^2d^2\), \(\exp(A)/A>32/5\), and \(r+1\ge4\). The
resulting rational lower bound is
\(9\cdot256^2\cdot32/(392\cdot87\cdot5)>20\). Similarly, the
cancellation in (10.777) gives \(\mathsf D_0/\bar T>(3/35)\exp(3A)/A>
(3/35)2^{15}/5>4\). Finally \(kL>58(r+1)>4r\). The allocation floors
still obey (10.778). Formula (10.779) has B in place of A, so \(B\ge t\)
gives \(\mathcal R_0>32/11\).

For an m-coordinate subset, (10.780) holds with this value of \(B_0\).
Its bracket exceeds \((6/7)d\exp(3A)\). Apply the height-product
estimate (10.781), \(\mathrm e<3\), and

\[
\frac27\exp(3A)>
\frac27\,2^{13}(r+1)^2d^2>
2^{11}(r+1)^2d^2.
\tag{10.892}
\]

Here the two factors arise from \(2A\) and A separately, using
\(A\ge4+\log((r+1)d)>5\). The factorial and field-degree argument of
(10.782) now yields

\[
\mathcal R_{m,J'}>
\frac4{29}\frac{(2^{11})^m}{(m+1)^3}
\ge\frac{2^{10}}{29}.
\tag{10.893}
\]

The last expression increases with m: its consecutive ratio is at least
\(2^{11}(2/3)^3>1\). In the full-rank cancellation (10.783), B cancels
with the same factor in \(\widehat{\mathcal D}\). Also
\(J^{1-r}E^r>\exp(3rA)J^{2r+1}\ge\exp(3rA)\). Using (10.784)--(10.785)
gives

\[
\begin{aligned}
\mathcal R_r
&>\frac{\exp((3r-1)A)}
 {6r(r+1)![12(r+1)]^r\ell_d A}\\
&>\frac{16^{2r}}{6r(r+1)12^r}>31.
\end{aligned}
\tag{10.894}
\]

For the middle inequality use \(\exp((r-1)A)>A\),
\(\exp(2rA)>[16(r+1)d]^{2r}\), \((r+1)!\le(r+1)^{r+1}\), and
\(d^{2r}\ge\ell_d\). The final expression has consecutive ratio
\((256/12)r/(r+2)>1\) for \(r\ge3\), and its value at three is
\(32768/243>31\). This proves the degree and allocation claims.

It remains to check the rank-product constant. Under the stipulated
zeros, Theorem 10.25 and Corollary 10.26 supply the rank output exactly
as in Corollary 10.192. The cancellation (10.789) is unchanged: the
explicit B in \(\widehat{\mathcal D}\) cancels its polynomial
denominator. Its last depth factor satisfies

\[
J^{m-1}\frac{\mathfrak h^I}{E^m}
<\exp(-3(m+1)A)J^{-2m-4}
\le\exp(-3(m+1)A).
\tag{10.895}
\]

Let \(\mathcal Q\) denote the normalized product quotient in the proof
of Corollary 10.192. Keeping all its factorials and its bound
\(\mathcal R_m\le\exp(A)A\), we obtain

\[
\mathcal Q<6r(m+1)!m!\ell_d A(35r)^m
                         \exp(-(3m+2)A).
\tag{10.896}
\]

Set \(V=(r+1)d\). The function \(z\exp(-(3m+2)z)\) decreases for
\(z\ge5\). Since \(A\ge4+\log V\), \(4+\log V\le5V\), and
\(\ell_d\le d\),

\[
\mathcal Q<30\cdot35^m\exp(-12m-8)F_{r,m,d},\qquad
F_{r,m,d}=\frac{r^{m+1}(m+1)!m!}
                  {(r+1)^{3m+1}d^{3m}}.
\tag{10.897}
\]

For \(m=1,r\ge3\), differentiation gives
\(F_{r,1,d}\le2r^2/(r+1)^4\le9/128\). For \(m\ge2\), the function
\(r^{m+1}/(r+1)^{3m+1}\) decreases for \(r\ge m+1\): the numerator of
its logarithmic derivative is \(m+1-2mr<0\). Use the already proved
factorial inequality \(n!\le n^n/2^{n-1}\) at \(n=m,m+1\), followed by
\(r=m+1\). This gives \(F_{r,m,d}<2(m+1)/4^m\le1\). Now
\(\mathrm e>27/10\) gives the exact rational bounds

\[
\begin{aligned}
m=1:&\quad
\mathcal Q<\frac{9450}{128}\left(\frac{10}{27}\right)^{20}<10^{-6},\\
m\ge2:&\quad
\mathcal Q<30\cdot35^2\left(\frac{10}{27}\right)^{32}<10^{-6}.
\end{aligned}
\tag{10.898}
\]

The second bound decreases as m increases, since \(35(10/27)^{12}<1\).
These are integer comparisons after clearing denominators. They prove
the claimed product estimate, with all field and independence assertions
provided by Corollary 10.26. \(\square\)

\subsubsection{The full denominator at an integer
stage}\label{the-full-denominator-at-an-integer-stage}

\textbf{Lemma 10.205 (scalar costs before the logarithmic stop).} Let a
selected family be at integer depth \(I\ge0\), with widths
\(q^{-I}\widehat D_i\) and additive argument \(q^{-I}x\). Take
\(j\ge1\), \(S_I=\mathfrak h^{-I}S\), and target radius
\(R=q^{j+1}S_I\). Suppose \(I\log Q\le3B\). For every additive order
\(u\le D_{\mathrm a}\),

\[
\begin{aligned}
\frac{c_1}{\widehat Z}
\left[\Xi_{I,0,u}\log q+
 PRq^{-I}\sum_i\widehat D_i\sigma_i\right]
&\le\frac{q^{j+1}}{c_2}
 +\frac{\lambda\vartheta\log q}{c_4NB(q-1)},\\
\frac{D_{\mathrm a}\log(q^{-I}R+2k)-L\log(k!)}{\widehat Z}
&<\frac{\lambda\vartheta}{c_1c_4N}
               \left(1+\frac{(j+1)\log q}{B}\right).
\end{aligned}
\tag{10.899}
\]

Here \(\lambda=59/58\), and \(\Xi_{0,0,u}=0\). The exponent
\(\Xi_{I,0,u}\) is the full denominator exponent from (10.243).

\textbf{Proof.} That formula bounds
\(\Xi_{I,0,u}\le D_{\mathrm a}(I+1/(q-1))\). The hatted degree estimate
gives \(c_1D_{\mathrm a}/\widehat Z<
\lambda\vartheta/(c_4NB)\). The torus contribution is exactly
\(q^{j+1}Q^{-I}/c_2\). Put \(a_j=q^{j+1}/c_2\),
\(b=\lambda\vartheta\log q/(c_4NB)\). The second derivative of
\(a_jQ^{-z}+bz\) is positive. On \(0\le z\le3B/\log Q\), convexity
therefore bounds this function by the larger endpoint value. At zero it
is \(a_j\). At the other endpoint its ratio to \(a_j\) is

\[
\exp(-3B)+
\frac{3\lambda(\vartheta/N)c_2\log q}
                  {c_4q^{j+1}\log Q}.
\tag{10.900}
\]

We have \(\vartheta/N<1+\tau<101/100=\xi\). At \(q=2\), use
\(Q\ge2(173/200)^4\). At \(q=3\), rank three is excluded by the
optimizer condition, so \(Q\ge3(21/25)^5\). These follow from the
increasing function \((1-c/v)^v\) proved in Theorem 10.201. Also
\(\log Q>2(Q-1)/(Q+1)\): integrate \(2/(1-z^2)>2\) from zero to
\((Q-1)/(Q+1)\). With \(\log2<7/10\), \(\log3<11/10\), and the largest
\(c_2\) in each q-case, exact substitution bounds the second term by
\(3/5\) at two and \(2/5\) at three. The
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_second_depth_buffer.py}{rational calculation}
records every field case. The first term is less than \(1/1000\), since
\(B>5\) and \(\exp(15)>2^{15}>1000\). Thus the depth-proportional cost
is absorbed. The remaining \(1/(q-1)\) term gives precisely the
displayed remainder.

For the additive scalar use \(Q^{-I}\le1\) and
\(q^{-I}R+2k\le q^{j+1}(S+2k)\). The factorial argument of (10.824)
bounds \(\log[\mathrm e(S+2k)/k]\) by \(A\le B\). Substitution of the
hatted degree bound proves the second inequality. \(\square\)

The lemma charges all denominator primes in the global height. Later
zero-production comparisons must combine this arithmetic bound with the
selected-series interpolation bound; fractional nodes also require their
actual global root-field degree.

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={The saturation factor and the convex bound for the cost of contraction depth.}]{../figures/yu-second-depth-buffer.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.61.}
\end{flushleft}
\end{minipage}\par\medskip
 Left: the ratio \((h+x)(A+x)/(hA\exp x)\) at
\(h=41,A=6\); Theorem 10.203 proves its upper bound for every admissible
height. The curve uses continuous x; actual saturation indices have
\(x=\nu\log q\). Right: the convex endpoint argument of Lemma 10.205 at
\(B=6,j=1\), with \((q,r)=(2,3),(3,4)\) and \(\vartheta/N\) replaced by
its upper bound \(101/100\). The separate factorial-denominator
remainder remains in the lemma. These parameter illustrations explain
the inequalities; they are not full arithmetic instances.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_second_depth_buffer.py}{Reproducible figure
and exact comparisons}. Human-source context: the selected-support and
contraction methods in Sections 10--11 of
\href{https://matwbn.icm.edu.pl/ksiazki/aa/aa89/aa8944.pdf}{Yu's freely
accessible 1999 paper}; the parameter, depth and product inequalities
used here are proved above.

\subsection{69. Integer zeros at a contracted
depth}\label{integer-zeros-at-a-contracted-depth}

Contraction increases the integer radius while decreasing the available
derivative order. Their product remains fixed. The scalar denominator
grows with depth, but Lemma 10.205 absorbs its depth-dependent part into
the torus-height budget. We now combine that arithmetic estimate with
the selected-function interpolation theorem.

\textbf{Theorem 10.206 (the integer extension after contraction).} Use
the hatted kernel of Theorem 10.203 and assume
\(v_p(\Xi-1)\ge\chi\widehat U_r\). Put \(c=c_5\),
\(\mathfrak h=\eta^{r+1}\), and \(Q=q\mathfrak h\). Let \(I\ge1\) be an
integer with \(I\log Q\le3B\), and set

\[
S_I=\mathfrak h^{-I}S,\qquad T_I=\mathfrak h^I\widehat T.
\tag{10.901}
\]

Suppose a nonzero phase family \(\Phi_I\) is obtained by the prime-to-p
support and phase restrictions of Theorems 10.48, 10.51 and 10.176,
retaining a coefficient of the original minimum valuation \(\beta\). Its
widths are at most \(q^{-I}\widehat D_i\), its additive argument is
\(q^{-I}x\), and its integer values belong to K. If

\[
\Phi_I(s;\boldsymbol t)=0
\quad(s\in\mathbb Z,\ |s|\le qS_I,\
                         |\boldsymbol t|\le\eta T_I),
\tag{10.902}
\]

then, for every \(1\le j\le r\),

\[
\Phi_I(s;\boldsymbol t)=0
\quad(s\in\mathbb Z,\ |s|\le q^jS_I,\
                         |\boldsymbol t|\le\eta^jT_I).
\tag{10.903}
\]

All cutoffs mean their integer floors. The coefficient vector, its
height bound and its minimum valuation remain unchanged in these
extensions. In particular, Theorems 10.202--10.203 provide the required
starting block at \(I=1\); its higher-order q-deleted block remains
available. At later depths this theorem applies once the stated starting
block has been obtained.

\textbf{Proof.} Let \(1\le j\le r-1\), and assume the j-th block. Use
the full node interval \(|s|\le s_j\), where

\[
s_j=\lfloor q^jS_I\rfloor,\qquad
\mu_j=\left\lfloor\frac{c\eta^jT_I}{r+1}\right\rfloor+1,
\qquad
\lfloor\eta^{j+1}T_I\rfloor+\mu_j-1
\le\lfloor\eta^jT_I\rfloor.
\tag{10.904}
\]

The last inequality follows from
\(\eta^{j+1}T_I+c\eta^jT_I/(r+1)=\eta^jT_I\) and the elementary floor
inequality. It holds even when \(\mu_j=1\). Thus all ordinary jets
required by Theorem 10.182 are supplied by the assumed prepared zeros.

\subsubsection{The two interpolation
bounds}\label{the-two-interpolation-bounds}

Write \(M_*\) for the maximum of the full-interval separation cost,
\(v_p(V(k))\), and the original \(v_p(b_n)\). Equation (10.839) and
\(H>40>1\) imply

\[
\begin{aligned}
\log(2q^jS_I)&<H+I\log(\mathfrak h^{-1})
                      \le\mathfrak h^{-I}H,\\
M_*&\le\frac{\mathfrak h^{-I}H}{\log p},\qquad
v_p(b_n)\le\frac H{\log p}.
\end{aligned}
\tag{10.905}
\]

Indeed \(1+z\le\exp z\) for \(z\ge0\), so \(H+z\le H\exp z\). The
separation bound is \(\lfloor\log_p(2q^jS_I)\rfloor\), and the other two
costs have the bounds already proved in (10.757). The last displayed
bound is kept separately; it is the original integer coefficient, not a
coefficient enlarged by contraction.

All phase restrictions retain the selected combined-frequency congruence
by Theorem 10.176. The strict disc conditions of Theorem 10.180
consequently persist, and its perturbation still uses the original
\(b_n\). Theorem 10.190 and \(\chi\widehat U_r\ge U_r\) give the strict
perturbation condition. Multiplication of the additive argument by the
p-adic unit \(q^{-I}\) changes neither its coefficient integrality nor
its degree \(D_{\mathrm a}\). Thus Corollary 10.179 supplies the
normalizing valuation \(Q_D-\beta\), where
\(Q_D=D_{\mathrm a}(N+1/(p-1))\).

The prepared projection error of Lemma 10.181 at every integer node has
valuation at least \(\beta+\kappa+\chi\widehat U_r-v_p(b_n)\). Its
ordinary divided jets are those of the original selected directions, as
in the proof of Theorem 10.182; their l-th cost is at most
\(l\max\{v_p(V(k)),v_p(b_n)\}\). The full-cardinal interpolation bound
(10.690) therefore gives, for every target integer and every permitted
output order,

\[
v_p(\Phi_I(s;\boldsymbol t))-\beta\ge
\min\left\{
 (2s_j+1)\mu_jN-Q_D,\quad
 \kappa+\chi\widehat U_r-v_p(b_n)-(\mu_j-1)M_*
\right\}.
\tag{10.906}
\]

There is no deleted-node loss in this full interval. In particular, the
second entry is used as it stands, without replacing it by the stronger
sufficient one-term budget (10.691).

Put \(y=1/(r+1)\), \(\lambda=59/58\), and \(\xi=101/100\). Retain the
case constants \(a_*,d_*\) of Theorem 10.197, \(\gamma=c_3qd_*\), and
its normalizing factor \(\nu\), so
\(c_1Q_D/\widehat{\mathcal Z}<\lambda\nu/c_4\). The lower bound
\(S\ge\gamma/y^2\), the identity \(S_IT_I=S\widehat T\), and
\(c_1S\widehat TN/\widehat{\mathcal Z}=q(r+1)\) give

\[
\begin{aligned}
\frac{c_1((2s_j+1)\mu_jN-Q_D)}
     {q^j\widehat{\mathcal Z}}
&>\mathcal L_j(y),\\
\mathcal L_j(y)
&=2cq(1-cy)^j\left(1-\frac{y^2}{2q^j\gamma}\right)
                      -\frac{\lambda\nu}{c_4q^j}.
\end{aligned}
\tag{10.907}
\]

Here \(2\lfloor q^jS_I\rfloor+1>2q^jS_I-1\) and
\(\mu_j>c\eta^jT_I/(r+1)\); no multiplicity rounding is discarded. Also
\(\mu_j-1\le c\eta^jT_I/(r+1)\). Applying the hatted identity (10.863),
and discarding only \(\kappa\ge0\), gives the following strict lower
bound for the other normalized entry:

\[
\mathcal I_j
=c_1\chi q^{r+1-j}
-\frac{\xi}{q^jc_3(1+a_*)}
       \left(\frac1{\widehat T}+\frac{c\eta^j}{r+1}\right).
\tag{10.908}
\]

That is, the entry is greater than \(\mathcal I_j\). The term
\(1/\widehat T\) accounts for the original \(b_n\), and the second term
accounts for the ordinary jets and separation.

\subsubsection{The arithmetic upper
bound}\label{the-arithmetic-upper-bound}

The phase identity (10.151), retained at each restriction, puts every
target integer value in K. Its global degree and local degrees are
therefore d,e,f.~Restricting the coefficient vector does not increase
\(h_2(c)\) or its column count, and the retained minimum is \(\beta\).
Theorem 10.92 applies with its full q-denominator.

For an explicit envelope, put \(\varepsilon=13/18000\),
\(e_0=1/79+1/(71928\cdot1500)\), and \(\ell_q=7/50\) at two and
\(11/50\) at three. Let \(E_-(c),E_+(c)\) be the degree-ten rational
exponential bounds in Theorem 10.201, and write \(H_+=1/E_-(c)\). In
particular \(\exp(-c)<H_+\). Define

\[
\begin{aligned}
W(y)&=\frac w{2c_2}\left(1+\frac{y^2}{2\gamma}\right)
 +\frac{y^2}{c_2c_3q}
 +(1+w/2)c_1\varepsilon+(1+w)c_1e_0,\\
b_j&=\lambda\left(1+(j+1)\ell_q+w+
                                  \frac{\ell_q}{q-1}\right),\\
k_j(y)&=\left(\frac{107}{103}+\frac38y\right)
                    (1-cy)^{j+1}H_+
                      +w\left(1-\frac{483}{824}y\right),\\
\mathcal A_j(y,A_0)&=\frac q{c_2}+\frac{W(y)}{q^j}
 +\frac{\xi}{q^j}
       \frac{b_jA_0/c_4+k_j(y)/c_3}{1+A_0}.
\end{aligned}
\tag{10.909}
\]

If a target value is nonzero, we claim

\[
\frac{c_1}{q^j\widehat{\mathcal Z}}
       (v_p(\Phi_I(s;\boldsymbol t))-\beta)
\le\mathcal A_j(y,A_0).
\tag{10.910}
\]

To check every term, apply (10.244) with \(R=q^{j+1}S_I\) and output
order \(O\le\eta^{j+1}T_I\). Since \(t=f\log p\), its field factor
converts the height normalization into \(c_1/(q^j\widehat Z)\). Lemma
10.205 bounds the combined torus and depth-dependent scalar cost by
\(q/c_2\), with the remaining \(\lambda\ell_q/(q-1)\) contribution to
\(b_j\). Its additive-factorial bound gives \(\lambda(1+(j+1)\ell_q)\),
since \(\log q/B<\ell_q\). The initial mean additive cost adds
\(w\lambda\). Use \(\vartheta/N<\xi A_0/(1+A_0)\) for these terms.

The initial mean torus cost, coefficient-field generator, logarithmic
column counts and scalar remainders give the four terms of W,
respectively, exactly as in (10.864). We used
\(\varepsilon_r\le\varepsilon\), \(e_r\le e_0\), and the arbitrary-depth
bound \(E_I/\widehat Z<e_0\) from Lemmas 10.194--10.195. For the
pointwise derivative cost, \(a/H\le107/103+3y/8\) and
\(\mathfrak h^I\le\mathfrak h<\exp(-c)<H_+\): the strict exponential
inequality follows by integrating \(1/(1-z)>1\) to obtain
\(\log(1-z)<-z\). The mean derivative cost has the multiplier
\(wC_r\le w(1-483y/824)\). Finally \(1/(eN)<\xi/(1+A_0)\) gives exactly
\(k_j\). This proves the complete arithmetic bound in the actual field.

\subsubsection{Comparisons for every rank and integer
step}\label{comparisons-for-every-rank-and-integer-step}

We prove that both analytic entries are strictly larger than this
arithmetic bound. For fixed y, its dependence on \(A_0\ge a_*\) is
fractional linear, so its maximum is at \(a_*\) or at its limit at
infinity. For \(1\le j\le4\), at either endpoint
\(P(y)=\mathcal L_j(y)-\mathcal A_j(y,A_0)\) is a polynomial of degree
at most \(j+2\). Its required interval is

\[
0\le y\le\frac1{\max\{r_0+1,j+2\}},\qquad
r_0=\begin{cases}3,&q=2,\\4,&q=3.\end{cases}
\tag{10.911}
\]

The q = 3 rank restriction is the positive-optimizer consequence in
Proposition 10.186. Use the Bernstein identity (10.845) on this
interval. Substitution of the seven case rows gives 56 polynomials, with
308 rational Bernstein coefficients, all greater than 1/4. The
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_second_later_integer.py}{exact calculation}
constructs the coefficients from the displayed formulas and reconstructs
each polynomial exactly. The nonnegative weights in (10.845) sum to one,
so the comparison holds on the whole interval.

For \(j\ge5\), one has \(r\ge j+1\ge6\), hence \(y\le1/7\). The
inequality \(\log(1-z)>-z/(1-z)\) used in (10.846) gives
\(\eta^j\ge\eta^{r-1}>\exp(-c)>1/E_+(c)\). Both \(q^{-j}\) and
\((j+1)q^{-j}\) decrease with j. Replace y by 1/7 in W, replace \(k_j\)
by \((107/103+3/56)H_++w\), and replace j by five in the remaining
decreasing costs. This bounds the mass from below and the arithmetic
expression from above by

\[
\begin{aligned}
L_*&=\frac{2cq}{E_+(c)}
       \left(1-\frac1{98q^5\gamma}\right)
                         -\frac{\lambda\nu}{c_4q^5},\\
A_*(A_0)&=\frac q{c_2}+\frac{W(1/7)}{q^5}
 +\frac\xi{q^5}
 \frac{b_5A_0/c_4+[(107/103+3/56)H_++w]/c_3}{1+A_0}.
\end{aligned}
\tag{10.912}
\]

The same two endpoints reduce this tail to 14 rational comparisons. Each
gives \(L_*-A_*>3/100\).

For the input entry put \(y_0=1/(r_0+1)\) and
\(T_0=(12/29)(r_0+1)12^{r_0}\). Since \(\widehat T>T_0\), \(r+1-j\ge2\),
and \(q^j\ge q\), the input is greater than

\[
I_*=c_1\chi q^2-
       \frac{\xi(1/T_0+cy_0)}{qc_3(1+a_*)}.
\tag{10.913}
\]

Meanwhile \((j+1)/q^j\le2/q\), \(W(y)\le W(y_0)\), and
\(\eta^{j+1}\le1\) give the uniform arithmetic upper bound

\[
U_*=\frac q{c_2}+\frac{W(y_0)}q+
\frac\xi q\left[
 \frac{\lambda(1+w+\ell_q/(q-1)+2\ell_q)}{c_4}
 +\frac{(107/103+3y_0/8)H_++w}{c_3(1+a_*)}
\right].
\tag{10.914}
\]

Each of the seven rational differences \(I_*-U_*\) exceeds 7/5. Thus the
finite-step mass gap exceeds 1/4, the remaining mass gap exceeds 3/100,
and the input gap exceeds 7/5, all in the normalization
\(c_1/(q^j\widehat{\mathcal Z})\). These are exact rational comparisons
after the analytic reductions; no rank or contraction depth is left to
numerical sampling.

A nonzero target value would have its normalized valuation at most
\(\mathcal A_j\) and strictly greater than that quantity by both entries
of the interpolation minimum. It must therefore be zero. Additive orders
greater than \(D_{\mathrm a}\) vanish identically. Induction proves
every asserted block through j = r. The phase family is unchanged during
this induction, so all previously known zeros, including a stronger
q-deleted block, are retained.

At depth one, Theorem 10.202 supplies a full radius
\(q(\lfloor S_1\rfloor+1)>qS_1\) with order \(\lfloor\eta T_1\rfloor\).
Also \(\log Q<\log3<3B\). This verifies the stated unconditional
application at that depth. \(\square\)

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={Floored integer radii and derivative orders at a contracted depth, and strict lower comparison margins.}]{../figures/yu-second-later-integer.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.62.}
\end{flushleft}
\end{minipage}\par\medskip
 Left: the counting example of Solution 83 at
\(q=2,r=3,I=2,S=21/2,\widehat T=1200\). Each full integer interval
carries the displayed derivative order; the arrows give the ordinary
interpolation multiplicities. This is an illustration of exact floor
arithmetic, not a number-field instance of the parameter construction.
Right: exact strict lower margins from Theorem 10.206, in units of
\(c_1/(q^j\widehat{\mathcal Z})\): 1/4 for steps 1--4, 3/100 for the
infinite tail, and 7/5 for the input entry. The proof connects these
comparisons to the actual selected functions.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_second_later_integer.py}{Reproducible
calculation and figure}. Human-source context: the extrapolation and
contraction methods of Section 11 of
\href{https://matwbn.icm.edu.pl/ksiazki/aa/aa89/aa8944.pdf}{Yu's freely
accessible 1999 paper}; the selected-function and height estimates used
here are proved in this lesson.

\subsection{70. Fractional zeros and successive
contractions}\label{fractional-zeros-and-successive-contractions}

At a contracted depth, the inner interval retains two different orders:
a higher order away from multiples of q, and a lower order at every
integer. The outer integer intervals supplied by Theorem 10.206 add
further layers. We use all these layers, while retaining the denominator
of the fractional target and the degree of its actual root field.

Use the hatted kernel of Theorem 10.203 throughout. Put \(a=r+1\),
\(c=c_5\), \(\mathfrak h=\eta^a\), \(Q=q\mathfrak h\), and

\[
S_I=\mathfrak h^{-I}S,\qquad
T_I=\mathfrak h^I\widehat T,\qquad
\widehat{\mathcal Z}=\frac{S\widehat{\mathcal D}}{et}.
\tag{10.917}
\]

The logarithmic scale is \(B=A_r+\log J\); it is distinct from the
integer separation cost used below. Retain the counterhypothesis
\(v_p(\Xi-1)\ge\chi\widehat U_r\), including the casewise \(\chi\) and
coefficient-height multiplier w of Lemma 10.199.

\subsubsection{The complete mixed fractional
step}\label{the-complete-mixed-fractional-step}

\textbf{Theorem 10.207 (fractional zeros after a contraction).} Let
\(I\ge1\) be an integer satisfying

\[
I\log Q\le3B,\qquad \frac{cT_{I+1}}a\ge1.
\tag{10.918}
\]

Suppose a nonzero selected family \(\Phi_I\) has been obtained by the
support and phase restrictions of Theorems 10.48, 10.51 and 10.176. It
retains the original coefficient minimum \(\beta\), an integral
coefficient subvector, the complete polynomial-index fibres, widths
\(q^{-I}\widehat D_i\), and additive argument \(q^{-I}x\). Put
\(R_1=q(\lfloor S_I\rfloor+1)\) and \(R_j=\lfloor q^jS_I\rfloor\) for
\(2\le j\le r\). Suppose its prepared zeros comprise:

\begin{itemize}
\tightlist
\item
  the q-deleted integers through \(R_1\), to order
  \(\lfloor T_I\rfloor\);
\item
  every integer through \(R_1\), to order \(\lfloor\eta T_I\rfloor\);
\item
  every integer through \(R_j\), to order \(\lfloor\eta^jT_I\rfloor\),
  for \(2\le j\le r\).
\end{itemize}

Then

\[
\Phi_I(s/q;\boldsymbol t)=0
\quad\left(s\in\mathbb Z,\quad
 |s|\le q(\lfloor S_{I+1}\rfloor+1),\quad
 |\boldsymbol t|\le\lfloor T_{I+1}\rfloor\right).
\tag{10.919}
\]

\textbf{Proof.} Set \(O=\lfloor T_{I+1}\rfloor\) and define

\[
\begin{aligned}
\mu_j&=\lfloor\eta^jT_I\rfloor-O+1 &&(0\le j\le r),\\
E&=\mu_0-\mu_1,\qquad M=\mu_0,\\
N_I&=(2R_r+1)\mu_r+
 \sum_{j=1}^{r-1}(2R_j+1)(\mu_j-\mu_{j+1})
 +2(q-1)(\lfloor S_I\rfloor+1)E,\\
B_I&=\lfloor\log_p(2R_r)\rfloor,\qquad
M_* =\max\{B_I,v_p(V(k)),v_p(b_n)\},\\
Q_D&=D_{\mathrm a}\bigl(N+1/(p-1)\bigr).
\end{aligned}
\tag{10.920}
\]

These are legitimate nested radii and multiplicities. Indeed
\(S_I\ge S>16\) gives \(q(q-1)S_I-q-1>0\), hence \(R_2>R_1\), and all
subsequent radii increase. The last real order difference is
\((\eta^r-\mathfrak h)T_I=c\eta^rT_I/a\ge1/\eta>1\). Thus every
\(\mu_j\) is positive, and they are nonincreasing. At a node in layer j,
an output order at most O plus its largest ordinary-jet order is at most
\(O+\mu_j-1\), exactly the provided input floor. The deleted increment E
is a nonnegative integer.

Apply the normal-series assertion (10.351) of Theorem 10.132, with the
deleted radius equal to \(R_1\). Theorem 10.180 gives the normalizing
scalar valuation \(Q_D-\beta\). Theorem 10.182 and Lemma 10.38 give the
ordinary-jet loss \(l\max\{v_p(V(k)),v_p(b_n)\}\), with original
baseline \(\beta+\kappa+\chi\widehat U_r-v_p(b_n)\). Theorem 10.176
preserves the combined-frequency congruence under each stipulated
restriction; the factor \(q^{-I}\) is a p-adic unit. The normal-disc
conditions are those of Theorem 10.190 for the hatted scale. Subtract
the normalizing valuation after applying (10.351), and add back the
projection error. This gives

\[
\begin{aligned}
v_p(\Phi_I(x;\boldsymbol t))-\beta\ge
\min\{&N_IN-Q_D,\\
 &\kappa+\chi\widehat U_r-v_p(b_n)
             -EB_I-(\mu_0-1)M_*\}
\qquad(x\in\mathbb Z_p).
\end{aligned}
\tag{10.921}
\]

In particular, both the deleted-cardinal cost and the inverse/jet cost
remain. No prepared-family constants from (10.352) or (10.353) are
substituted for this selected-family normalization.

The proof of (10.839), together with \(H>1\), gives

\[
\log(2q^rS_I)<H+I\log(\mathfrak h^{-1})
 \le\mathfrak h^{-I}H,\qquad
M_*\le\frac{\mathfrak h^{-I}H}{\log p},\qquad
v_p(b_n)\le\frac H{\log p}.
\tag{10.922}
\]

For the middle inequality use \(H+z\le H\exp z\) for \(z\ge0\). Keep the
original coefficient bound separately. The floor estimates are
\(E\le cT_I/a+1\) and \(\mu_0-1\le(1-\mathfrak h)T_I+1\). Using
(10.863), \(T_I=\mathfrak h^I\widehat T\), and
\(2/T_I\le2c\mathfrak h/a\), the second analytic entry, multiplied by
\(c_1/(q^r\widehat{\mathcal Z})\), is strictly greater than

\[
\mathcal I_r=c_1q\chi-
 \frac{\xi}{q^rc_3(1+a_*)}
 \left[1-\mathfrak h+\frac{c(1+2\mathfrak h)}a+\frac1{T_-}\right],
\quad \xi=\frac{101}{100},\quad
T_- =\frac{12}{29}a12^r.
\tag{10.923}
\]

Here \(a_*\) is the casewise lower bound for \(A_0\) in Theorem 10.197.
The term \(1/T_-\) pays for the original \(b_n\); it is additional to
the two rounded interpolation losses.

To count the first entry, separate the inner deleted nodes, the
remaining inner multiples of q, and the outer annuli. Their counts are
\(2(q-1)(\lfloor S_I\rfloor+1)\), \(2(\lfloor S_I\rfloor+1)+1\), and
\(2(R_j-R_{j-1})\). The first two exceed \(2(q-1)S_I\) and \(2S_I\). The
first annulus exceeds \(2(q^2-q)S_I-2(q+1)>0\); later annuli exceed
\(2(q^j-q^{j-1})S_I-2>0\). Also \(\mu_j>(\eta^j-\mathfrak h)T_I\).
Telescoping gives

\[
\begin{aligned}
&(q-1)(1-\mathfrak h)+(\eta-\mathfrak h)
 +\sum_{j=2}^r(q^j-q^{j-1})(\eta^j-\mathfrak h)\\
&\hspace{25mm}=\frac ca
       \left[\sum_{j=0}^r(q\eta)^j+q-2\right].
\end{aligned}
\tag{10.924}
\]

Use \(S_IT_I=S\widehat T\), \(c_1S\widehat TN/\widehat{\mathcal Z}=qa\),
and \(S_I\ge S_-=c_3qd_*a^2\). The normalized mass is strictly greater
than

\[
\begin{aligned}
\Gamma_r={}&\frac{2cq}{q^r}
       \left[\sum_{j=0}^r(q\eta)^j+q-2\right]\\
&-\frac{qa}{q^rS_-}
 \left[2(q+1)(\eta^2-\mathfrak h)
       +2\sum_{j=3}^r(\eta^j-\mathfrak h)\right].
\end{aligned}
\tag{10.925}
\]

The first analytic entry therefore exceeds
\(\Gamma_r-\lambda\nu/(c_4q^r)\), where \(\lambda=59/58\) and \(\nu\) is
the case factor in Theorem 10.197, so that
\(c_1\mathcal N=\lambda\nu/c_4\).

\subsubsection{The full fractional denominator over the permitted
depths}\label{the-full-fractional-denominator-over-the-permitted-depths}

Let \(X=\lfloor S_{I+1}\rfloor+1\), the absolute target radius for
\(x=s/q\). By Theorem 10.56 its normalized value V lies in the actual
root field of degree at most \(q^rd\), with chosen local degrees
e,f.~Removing its common principal-unit monomial does not change the
valuation. The original minimum \(\beta\) and the coefficient-height
bound of Lemma 10.199 remain valid for the subvector.

Put \(y=1/a\), let \(\varepsilon_r,e_r\) be the bounds of Lemma 10.195,
and use \(\ell_2=7/50,\ell_3=11/50\). Define

\[
\begin{aligned}
u_r&=\frac{\mathfrak h^{-1}+S_-^{-1}}{c_2},\\
W_r&=\frac w{2c_2}\left(1+\frac1{2S_-}\right)
 +\frac{y^2}{c_2c_3q}
 +(1+w/2)c_1\varepsilon_r+(1+w)c_1e_r,\\
b_r'&=\lambda\left(1+\ell_q+w+\frac q{q-1}\ell_q\right),\\
k_r'&=\left(\frac{107}{103}+\frac38y\right)\mathfrak h^2
                 +w\left(1-\frac{483}{824}y\right),\\
L_Q&=\frac{2(Q-1)}{Q+1},\qquad
v_r=\frac{3\lambda l_q}{c_4L_Q},\qquad
l_2=\frac7{10},\quad l_3=\frac{11}{10},\\
\mathcal B_{r,0}(A_0)&=u_r+W_r+
 \xi\frac{b_r'A_0/c_4+k_r'/c_3}{1+A_0},\\
\mathcal B_{r,1}(A_0)&=\frac{u_r}{32768}+W_r+
 \xi\frac{(b_r'/c_4+v_r)A_0+k_r'/c_3}{1+A_0}.
\end{aligned}
\tag{10.926}
\]

If V is nonzero, Theorem 10.92 and the same height contributions as in
Theorem 10.201 give

\[
\frac{c_1}{q^r\widehat{\mathcal Z}}(v_p(V)-\beta)
 \le\max\{\mathcal B_{r,0}(A_0),\mathcal B_{r,1}(A_0)\}.
\tag{10.927}
\]

Here is the depth estimate and a check of the other contributions. Since
\(q^{-I}\le Q^{-I}\), the torus cost is at most \(u_rQ^{-I}\). Formula
(10.243) bounds the full denominator by \(D_{\mathrm a}(I+1+1/(q-1))\).
Its part proportional to I has normalized height at most
\(\lambda(\vartheta/N)I\log q/(c_4B)\). The sum of this linear term and
\(u_rQ^{-I}\) is convex as a function of real I on \([0,3B/\log Q]\). At
zero its value is \(u_r\). At the other endpoint it is bounded by

\[
u_r\exp(-3B)+\frac{3\lambda(\vartheta/N)\log q}{c_4\log Q}
 <\frac{u_r}{32768}+\xi\frac{A_0}{1+A_0}v_r.
\tag{10.928}
\]

Indeed \(B>5\), \(\exp(15)>2^{15}\),
\(\vartheta/N=(1+\tau)A_0/(m+A_0)<\xi A_0/(1+A_0)\), \(\log q<l_q\), and
\(\log Q>L_Q>0\), as proved in Lemma 10.205. Convexity bounds every
intermediate integer depth. The remainder \(1+1/(q-1)\) gives exactly
the term \(\lambda q\ell_q/(q-1)\) in \(b_r'\).

The mean torus, coefficient generator, coefficient count and scalar
remainders give \(W_r\), with the same terms as in (10.864). The scalar
radius obeys \(q^{-I}X\le\mathfrak h^{-1}S+1<qS\), by Theorem 10.201.
Thus its additive term is bounded by \(\lambda(1+\ell_q)\), while the
initial mean contributes \(w\lambda\). The target derivative order is at
most \(T_{I+1}\le\mathfrak h^2\widehat T\), because \(I\ge1\). Together
with the unchanged initial complete-simplex mean, Lemma 10.194 gives
\(k_r'\). At depth I its Euler remainder is bounded by
\(1/(q^I79)+(r-1)/(999\widehat Z)<e_r\). Finally use \(eN>(1+A_0)/\xi\),
retaining the full factor \(q^r\) on every nonnegative height
contribution. No equality between valuations at different primes is
assumed.

\subsubsection{Exact comparisons covering every
rank}\label{exact-comparisons-covering-every-rank}

For the finite ranks 3 through 7, or 4 through 7 in case V, use the
optimizer hypothesis from Theorem 10.184:

\[
\frac rt>A_0+\frac35,\qquad t>1,
\qquad a_*\le A_0<r.
\tag{10.929}
\]

Each \(\mathcal B_{r,i}\) is fractional linear in \(A_0\). Its maximum
on this interval is bounded by its values at \(a_*\) and r. This uses
the actual rank restriction even when the upper endpoint r itself is not
attainable.

For all ranks at least eight, use the exponential bounds
\(H_-=(1-c/9)^9\), \(H_+=E_-^{-1}\), and \(E_+\) from Theorem 10.201.
They give \(H_-\le\mathfrak h<\exp(-c)<H_+\) and \(\eta^r>1/E_+\).
Keeping the last four geometric terms and bounding every removed order
fraction by one gives

\[
\Gamma_r>
\frac{2cq}{E_+}\sum_{j=0}^3q^{-j}
-\frac{18q(q+7)}{q^8S_-(8)}.
\tag{10.930}
\]

The discarded floor term is at most \(2q(r+1)(q+r-1)/(q^rS_-(r))\). It
decreases with r: its consecutive ratio is
\((q+r)(r+1)/(q(q+r-1)(r+2))<1\).

For both arithmetic endpoints, replace y and all decreasing size
remainders by their rank-eight bounds, replace \(\mathfrak h^{-1}\) by
\(H_-^{-1}\), the derivative factor \(\mathfrak h^2\) by \(H_+^2\), and
\(1-483y/824\) by one. The function \(2(Q-1)/(Q+1)\) increases with Q,
so using \(Q=qH_-\) increases \(v_r\). All these changes increase the
arithmetic bounds. In the input loss, use
\(1-H_-+c(1+2H_+)/9+1/T_-(8)\), and use \(q^8\) in its denominator and
in the normalizing loss. These changes decrease the two analytic lower
bounds. The resulting arithmetic expressions are again fractional
linear; their extrema on \([a_*,\infty)\) are bounded by the value at
\(a_*\) and the limit at infinity.

There are 34 finite-rank rows and seven uniform rows, with two depth
endpoints and two parameter endpoints each. Exact rational substitution
in these 164 comparisons gives the following strict lower bounds:

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}
  >{\raggedright\arraybackslash}p{(\linewidth - 4\tabcolsep) * \real{0.2727}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 4\tabcolsep) * \real{0.3636}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 4\tabcolsep) * \real{0.3636}}@{}}
\toprule\noalign{}
\begin{minipage}[b]{\linewidth}\raggedright
Case
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
Mass minus normalization and arithmetic
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
Input minus arithmetic
\end{minipage} \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
I.1 & \(107/100\) & \(1/5\) \\
I.2 & \(107/100\) & \(2/25\) \\
II & \(11/10\) & \(21/100\) \\
III.1 & \(103/100\) & \(13/100\) \\
III.2 & \(51/50\) & \(7/25\) \\
IV & \(26/25\) & \(49/100\) \\
V & \(18/25\) & \(39/100\) \\
\end{longtable}
}

The \href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_second_later_fractional.py}{exact
calculation} records every rational difference. The reductions above
prove why these comparisons cover all permitted depths, ranks and field
parameters. In particular both analytic entries strictly exceed the
nonzero-value arithmetic bound, by more than \(7/10\) and \(2/25\),
respectively. Thus V, and hence the original prepared value, is zero.
Additive orders above \(D_{\mathrm a}\) vanish by degree. This proves
the entire stated fractional block. \(\square\)

\subsubsection{The following integer
block}\label{the-following-integer-block}

\textbf{Theorem 10.208 (the next contracted starting profile).} Under
the hypotheses of Theorem 10.207, a support and phase restriction
produces a nonzero family \(\Phi_{I+1}\) of the same kind at depth
\(I+1\), with the same minimum \(\beta\). At
\(R=q(\lfloor S_{I+1}\rfloor+1)\), it has q-deleted zeros through order
\(\lfloor T_{I+1}\rfloor\) and full integer zeros through order
\(\lfloor\eta T_{I+1}\rfloor\).

\textbf{Proof.} Apply the exact phase/coset passage of Theorems 10.48
and 10.51 to the whole fractional block, including the extra phase
partition when q divides \(G_0\). Keep the class containing an original
coefficient of minimum valuation. The Kummer monomial basis separates
the classes for \(q\nmid s\). The affine jet change is triangular with
diagonal powers of q, so it retains every provided index. Theorem 10.55
identifies the new integer values in K itself. Theorems 10.29 and 10.176
preserve nonzero support, the polynomial-index fibres, and the
combined-frequency congruence. Thus the same argument as Theorem 10.202
gives the stipulated q-deleted zeros, widths \(q^{-I-1}\widehat D_i\),
and additive argument \(q^{-I-1}x\).

To fill the multiples of q, put \(T'=T_{I+1}\) and
\(M'=\lfloor cT'/a\rfloor+1\). Then
\(\lfloor\eta T'\rfloor+M'-1\le\lfloor T'\rfloor\). There are exactly
\(n_E=2(q-1)(\lfloor S_{I+1}\rfloor+1)\) input nodes. Their separation
maximum is at most \(\mathfrak h^{-I-1}H/\log p\), since
\(R<q^2S_{I+1}\le q^rS_{I+1}\). The q-deleted form of Theorem 10.182
consequently gives analytic entries

\[
n_EM'N-Q_D,\qquad
\kappa+\chi\widehat U_r-v_p(b_n)
 -M'\lfloor\log_p(2R)\rfloor-(M'-1)M_*.
\tag{10.931}
\]

In normalization \(c_1/\widehat{\mathcal Z}\), their strict lower bounds
are

\[
2cq(q-1)-\frac{\lambda\nu}{c_4},\qquad
\mathcal I_r^{\rm c}=c_1\chi q^{r+1}
 -\frac{\xi}{c_3(1+a_*)}\left(\frac{3c}{a}+\frac1{T_-}\right).
\tag{10.932}
\]

For the first use \(n_E>2(q-1)S_{I+1}\), \(M'>cT'/a\), and
\(S_{I+1}T'=S\widehat T\). For the second, the two separation/jet terms
have total multiplier \(2M'-1\le2cT'/a+1\). Since \(cT'/a\ge1\), their
normalized loss is at most \(3c/a\); the original coefficient again
contributes the separate \(1/T_-\).

The arithmetic upper bound is the same maximum of
\(\mathcal B_{r,0},\mathcal B_{r,1}\), now in this integer
normalization. Indeed the actual field is K; the torus radius is
\(q^{-I-1}R=q^{-I}(\lfloor S_{I+1}\rfloor+1)\), precisely the preceding
fractional radius after scaling. Formula (10.243) has the same total
depth \(I+1\). The derivative order has decreased, and the coefficient
height has not increased. These facts preserve each contribution to the
preceding height bound. The convex estimate still uses the previous I,
so it remains valid even if \((I+1)\log Q>3B\).

For the same 164 endpoint comparisons the two new differences exceed
\(7/10\) and 9, respectively. In the uniform rank-eight row, \(q^{r+1}\)
is replaced by \(q^9\) in the positive term, and the loss uses
\(a=9,T_-(8)\). All substitutions have the required direction. Thus a
nonzero integer value would contradict the product-formula bound. Every
missing multiple of q is filled, and the stronger q-deleted block
remains available. \(\square\)

\textbf{Corollary 10.209 (iteration up to the geometric or order
threshold).} Define

\[
I_G=\min\{I\ge1:Q^I>\exp(3B)\},\qquad
I_M=\max\{I\ge1:cT_I/a\ge1\}.
\tag{10.933}
\]

These are finite integers. At every \(1\le I\le\min(I_G,I_M)\), an
actual nonzero selected family exists with the two rounded starting
blocks of Theorem 10.208. At every such I with \(I<I_G\), it also has
all the outer integer blocks required by Theorem 10.207. If
\(I_G\le I_M\), the actual family at \(I_G\) supplies the zeros in
Theorem 10.204 and hence its rank conclusion whenever the remaining
independence and hyperplane hypotheses there hold.

\textbf{Proof.} The inequalities \(Q>1\) and \(0<\mathfrak h<1\) give
finiteness. The set defining \(I_M\) is nonempty: \(\mathfrak h>1/3\),
checked at the least rank in each q-case using its monotonicity, and
\(cT_1/a>(27/50)(1/3)(12/29)12^3>1\). Theorems 10.202--10.203 construct
the rounded depth-one profile. If \(I<I_G\), Theorem 10.206 supplies its
outer integer blocks, since \(I\log Q\le3B\). If also \(I<I_M\), then
\(cT_{I+1}/a\ge1\), so Theorems 10.207--10.208 construct the next
profile. Induction proves the whole interval. At \(I_G\), its rounded
full interval contains \(|s|\le qS_{I_G}\), with the required order
\(\eta T_{I_G}\); these are exactly the zeros in Theorem 10.204.
\(\square\)

When \(I_M<I_G\), the constructed family has reached the order threshold
first. The corollary does not extend the mixed step past its
multiplicity hypothesis. That endpoint requires a separate argument with
a fixed derivative order.

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={Nested mixed multiplicities at a contracted depth, and uniform positive interpolation margins in all seven field cases.}]{../figures/yu-second-later-fractional.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.63.}
\end{flushleft}
\end{minipage}\par\medskip
 Left: the exact node types and multiplicities for
\(q=2,r=3,I=1,S=21/2,\widehat T=1200,c=27/50\), with output order
\(\lfloor T_2\rfloor\). This is a counting illustration, not a complete
number-field parameter instance. Right: strict rational lower bounds for
the input-minus-arithmetic margins in Theorem 10.207; each includes both
depth endpoints and every rank in its field case. Theorem 10.208
supplies the next starting block, and Corollary 10.209 states the
resulting finite induction.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_second_later_fractional.py}{Reproducible
exact calculation and figure}. Human-source context: Section 11 of
\href{https://matwbn.icm.edu.pl/ksiazki/aa/aa89/aa8944.pdf}{Yu's freely
accessible 1999 paper} and the second constants in Section 1.1 of his
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{freely
accessible 2013 paper}.

\subsection{71. The last integer extension and the fixed-order
endpoint}\label{the-last-integer-extension-and-the-fixed-order-endpoint}

The derivative-order threshold leaves one more full integer extension
available. This extension identifies the actual starting interval for a
continuation at fixed order. We then prove the contradiction that such a
continuation must reach; the intermediate fractional steps are a
separate requirement.

\textbf{Theorem 10.210 (the last integer extension).} Under the
hypotheses of Theorem 10.206, the same selected family satisfies

\[
\Phi_I(s;\boldsymbol t)=0
\quad(s\in\mathbb Z,\ |s|\le q^{r+1}S_I,
                         |\boldsymbol t|\le\mathfrak h T_I).
\tag{10.936}
\]

All cutoffs mean their integer floors. This assertion does not require
\(cT_{I+1}/(r+1)\ge1\).

\textbf{Proof.} Write \(a=r+1\). Theorem 10.206 supplies the full
integer interval of radius \(s_r=\lfloor q^rS_I\rfloor\) through order
\(\lfloor\eta^rT_I\rfloor\). Set

\[
\mu=\left\lfloor\frac{c\eta^rT_I}{a}\right\rfloor+1,
\qquad
\lfloor\mathfrak h T_I\rfloor+\mu-1
\le\lfloor\eta^rT_I\rfloor.
\tag{10.937}
\]

The second inequality follows from
\(\mathfrak h T_I+c\eta^rT_I/a=\eta^rT_I\). It holds when \(\mu=1\) as
well. Thus every ordinary divided jet required by Theorem 10.182 is
available for each asserted target order.

Retain the original minimum coefficient valuation \(\beta\), the
normalizing cost \(Q_D=D_{\mathrm a}(N+1/(p-1))\), and the original
integer \(b_n\). The input nodes form a complete integer interval. The
interpolation bound is therefore

\[
v_p(\Phi_I(s;\boldsymbol t))-\beta\ge
\min\left\{
 (2s_r+1)\mu N-Q_D,\quad
 \kappa+\chi\widehat U_r-v_p(b_n)-(\mu-1)M_*
\right\}.
\tag{10.938}
\]

There is no deleted-node loss. The separation and jet estimate
\(M_*\le\mathfrak h^{-I}H/\log p\) remains valid at \(j=r\): equation
(10.839) bounds \(\log(2q^rS)\), and the proof of (10.905) then applies
unchanged. The original coefficient separately satisfies
\(v_p(b_n)\le H/\log p\). The strict disc, perturbation, phase and
congruence conditions are those verified in Theorem 10.206. The integer
target lies in K, and the family has not changed.

Use \(y=1/a\), \(\lambda=59/58\), \(\xi=101/100\), and the case data
\(a_*,d_*,w,\nu,\chi\) of Theorem 10.197 and Lemmas 10.199--10.200.
Define

\[
S_-=c_3qd_*a^2,\qquad T_-=(12/29)a12^r.
\tag{10.939}
\]

In units \(c_1/(q^r\widehat{\mathcal Z})\), the two interpolation
entries are strictly greater than

\[
\begin{aligned}
L_r&=2cq\eta^r\left(1-\frac1{2q^rS_-}\right)
                          -\frac{\lambda\nu}{c_4q^r},\\
J_r&=c_1\chi q-
 \frac{\xi(1/T_-+c\eta^r/a)}{q^rc_3(1+a_*)}.
\end{aligned}
\tag{10.940}
\]

For the first entry use \(\mu>c\eta^rT_I/a\), \(2s_r+1>2q^rS_I-1\), and
\(c_1S_IT_IN/\widehat{\mathcal Z}=qa\). For the second use
\(\mu-1\le c\eta^rT_I/a\), the separate original coefficient bound, and
(10.863). The strict lower bound \(\widehat T>T_-\) preserves the strict
inequality even at a floor endpoint.

Here is the arithmetic estimate with all terms retained. Put
\(\ell_q=7/50\) for \(q=2\), and \(\ell_q=11/50\) for \(q=3\). Let
\(E_-(c)<\exp c<E_+(c)\) be the rational bounds from Theorem 10.201 and
set \(H_+=1/E_-(c)\). Define

\[
\begin{aligned}
\epsilon&=\frac{13}{18000},\qquad
e_0=\frac1{79}+\frac1{71928\cdot1500},\\
W_r&=\frac{w}{2c_2}\left(1+\frac1{2S_-}\right)
 +\frac{y^2}{c_2c_3q}
 +(1+w/2)c_1\epsilon+(1+w)c_1e_0,\\
b_r&=\lambda\left(1+(r+1)\ell_q+w+
                                      \frac{\ell_q}{q-1}\right),\\
k_r&=\left(\frac{107}{103}+\frac{3y}{8}\right)
                   \eta^{r+1}H_+
                   +w\left(1-\frac{483y}{824}\right),\\
\mathcal A_r(z)&=\frac q{c_2}+\frac{W_r}{q^r}
       +\frac{\xi}{q^r}\frac{b_rz/c_4+k_r/c_3}{1+z}.
\end{aligned}
\tag{10.941}
\]

For a nonzero target, \(c_1(v_p(\Phi_I)-\beta)/(q^r\widehat{\mathcal Z})
\le\mathcal A_r(A_0)\). To verify this endpoint of the arithmetic
argument in Theorem 10.206, first apply Lemma 10.205 at \(j=r\). Its
statement covers every \(j\ge1\), and absorbs the complete
depth-dependent denominator and torus cost. The remaining constant
denominator gives \(\ell_q/(q-1)\). The physical radius is at most
\(q^{r+1}S\), giving the additive factorial term
\(\lambda(1+(r+1)\ell_q)\); the initial mean contributes \(w\lambda\).
These terms are multiplied by \(\vartheta/N<\xi A_0/(1+A_0)\).

The derivative order is at most \(\eta^{r+1}\mathfrak h^I\widehat T\).
Since \(I\ge1\) and \(\mathfrak h<\exp(-c)<H_+\), its pointwise
contribution is bounded by the first part of \(k_r\). The mean
derivative contribution is the second part. The remaining
coefficient-generator, mean torus, column-count and scalar-remainder
terms are exactly \(W_r\), with the uniform remainder bounds used in
Theorem 10.206. Thus the arithmetic argument applies at \(j=r\); its
numerical comparison is supplied below.

For ranks 3 through 7, or 4 through 7 in case V, evaluate
\(\mathcal A_r(z)\) at \(z=a_*\) and at infinity. A function
\((bz+k)/(1+z)\) has derivative \((b-k)/(1+z)^2\), so its maximum on
\([a_*,\infty)\) occurs at one of these endpoints.

For every \(r\ge8\) use \(y_8=1/9\), \(S_8=c_3qd_*9^2\), and
\(T_8=(12/29)9\cdot12^8\). The integral inequality
\(\log(1-u)>-u/(1-u)\) gives \(\eta^r>\exp(-c)>1/E_+(c)\), because
\(r<a-c\). Consequently uniform lower bounds for the two entries are

\[
\begin{aligned}
L_{\ge8}&=\frac{2cq}{E_+(c)}
          \left(1-\frac1{2q^8S_8}\right)
                           -\frac{\lambda\nu}{c_4q^8},\\
J_{\ge8}&=c_1\chi q-
       \frac{\xi(1/T_8+c/9)}{q^8c_3(1+a_*)}.
\end{aligned}
\tag{10.942}
\]

For an upper bound on \(\mathcal A_r\), use \(y_8,S_8\) in \(W_r\),
bound \(\eta^{r+1}H_+\) by \(H_+^2\), and bound \(1-483y/824\) by 1.
Each replacement increases its nonnegative term. Finally use \(r=8\) in
\(q^{-r}\) and \((r+1)q^{-r}\); both decrease for \(r\ge8\). These
operations give a uniform upper bound for all remaining ranks.

The resulting exact rational differences exceed the following values.

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}lrr@{}}
\toprule\noalign{}
Case & Mass minus arithmetic & Input minus arithmetic \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
I.1 & \(91/1000\) & \(33/100\) \\
I.2 & \(91/1000\) & \(11/50\) \\
II & \(91/1000\) & \(27/100\) \\
III.1 & \(66/1000\) & \(29/100\) \\
III.2 & \(66/1000\) & \(11/25\) \\
IV & \(49/1000\) & \(29/50\) \\
V & \(1/20\) & \(3/4\) \\
\end{longtable}
}

There are 34 finite-rank rows and seven uniform tail rows, each with two
parameter endpoints. The displayed formulas therefore reduce the proof
to 82 comparisons of each interpolation entry with its arithmetic bound.
The exact rational calculations, including the tail substitutions, are
given in the reproducible source
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_second_terminal_parameters.py}{\texttt{yu\_second\_terminal\_parameters.py}}.
They yield the strict margins in the table. Hence each entry exceeds the
arithmetic upper bound, contradicting a nonzero target. Additive orders
above \(D_{\mathrm a}\) vanish by degree. \(\square\)

\textbf{Lemma 10.211 (the actual fixed-order starting interval).}
Suppose \(I_M<I_G\) in Corollary 10.209. Put \(I_1=I_M\),
\(S_1=\mathfrak h^{-I_1}S\), and \(T_1=\mathfrak h^{I_1}\widehat T\).
There is a unique integer \(0\le j_0\le r\) for which

\[
1\le\frac{c\eta^{j_0}T_1}{a}<\frac1\eta.
\tag{10.943}
\]

Define

\[
\begin{aligned}
T_{\mathrm f}&=\eta^{j_0+1}T_1,\qquad
O=\lfloor T_{\mathrm f}\rfloor,\qquad
R=q\lfloor q^{j_0}S_1\rfloor,\\
I_2&=\left\lfloor\frac{3B-I_1\log Q}{\log q}\right\rfloor+1,
\qquad I_3=I_1+I_2.
\end{aligned}
\tag{10.944}
\]

Then \(\eta\le cT_{\mathrm f}/a<1\), \(I_2\ge1\), and

\[
q^{I_3}\mathfrak h^{I_1}>\exp(3B).
\tag{10.945}
\]

The actual family at \(I_1\) from Corollary 10.209 has all integer zeros
\(|s|\le R\) through order O. Its original minimum coefficient, phase,
integral coefficient subvector and complete polynomial fibres are
retained.

\textbf{Proof.} The definition of \(I_M\) gives \(cT_1/a\ge1\) and
\(c\mathfrak h T_1/a<1\). The sequence \(c\eta^jT_1/a\) decreases
strictly. Its last index with value at least 1 is therefore a unique
\(j_0\in\{0,\ldots,r\}\). Multiplication by \(\eta\) gives the asserted
interval for \(cT_{\mathrm f}/a\).

Since \(I_1<I_G\), we have \(I_1\log Q\le3B\), hence \(I_2\ge1\). The
floor plus one is strictly larger than its argument, including when that
argument is an integer. Thus \(Q^{I_1}q^{I_2}>\exp(3B)\), as asserted.

If \(j_0=0\), the rounded starting interval from Corollary 10.209
contains \(|s|\le q\lfloor S_1\rfloor\) through order
\(\lfloor\eta T_1\rfloor\). If \(1\le j_0<r\), Theorem 10.206 gives
radius \(\lfloor q^{j_0+1}S_1\rfloor\) and order
\(\lfloor\eta^{j_0+1}T_1\rfloor\). If \(j_0=r\), Theorem 10.210 gives
the final integer level with the same order. In every case
\(q\lfloor q^{j_0}S_1\rfloor
\le\lfloor q^{j_0+1}S_1\rfloor\), so the required interval is present.
None of these extensions changes the family. \(\square\)

\textbf{Proposition 10.212 (the final additive-polynomial
contradiction).} Suppose a stage-\(I_3\) family retains a nonzero
integral coefficient subvector of the hatted kernel of Theorem 10.203,
its complete polynomial fibres, and the integral normalized support of
Proposition 10.31, with widths at most \(q^{-I_3}\widehat D_i\). Suppose
its prepared values vanish at every integer \(|s|\le R\) through order
O, where \(R,O,I_3\) are defined in Lemma 10.211. Then such a family
cannot exist. Each torus degree is strictly less than \(609/163840\),
whereas the number of integer roots exceeds \((1305/236)D_{\mathrm a}\).

\textbf{Proof.} The coordinate-degree bound from Proposition 10.31 and
Theorem 10.191 is \(2Jq^{-I_3}\widehat D_i\). The strict stopping
inequality of Lemma 10.211 and \(T_1<a/(c\mathfrak h)\) give

\[
\begin{aligned}
2Jq^{-I_3}\widehat D_i
&<2JT_1\frac{\widehat D_i}{\widehat T}\exp(-3B)\\
&<\frac{2eNt}{cc_2qrP\mathfrak h d\sigma_i}\exp(-3A).
\end{aligned}
\tag{10.946}
\]

Equation (10.702) and Theorem 10.203 give the ratio
\(\widehat D_i/\widehat T=eNt/(c_2qraPd\sigma_i)\). The remaining
exponential factor satisfies \(J\exp(-3B)=J^{-2}\exp(-3A)\le\exp(-3A)\).
The optimizer gives \(eNt<\zeta_1r<7r/5\). Also \(c>1/2\), \(c_2q>3\),
\(P\ge1\), and \(\mathfrak h>1/3\). The one-base height bound in Theorem
10.18 gives \(1/(d\sigma_i)<87d^2\). Therefore the degree is strictly
less than \((2436/5)d^2\exp(-3A)\).

Since \(A\ge4+\log(ad)>5\), \(a\ge4\), and \(\exp(1)>2\),

\[
d^2\exp(-2A)\le\frac{\exp(-8)}{a^2},\qquad
\frac{2436}{5}d^2\exp(-3A)
 <\frac{2436}{5\cdot16\cdot2^{13}}
 =\frac{609}{163840}<1.
\tag{10.947}
\]

The nonnegative integer torus degree is thus zero. The remaining
ordinary polynomial depends only on the additive coordinate and has
degree at most \(D_{\mathrm a}\). It is nonzero: its retained
coefficient vector is nonzero, and the complete polynomial fibres are
expressed in the independent polynomial basis proved in Lemma 10.29 and
Proposition 10.31.

Finally \(S_1\ge S>16\) and \(j_0\ge0\) give
\(2R+1>2q(S_1-1)>(15/8)qS_1\). The hatted degree bound from Corollary
10.185 is

\[
D_{\mathrm a}<
\frac{\lambda S\widehat D}{c_1c_4(N/\vartheta)dB}.
\tag{10.948}
\]

Using \(S_1/S=\widehat T/T_1\) and \(\widehat T=qa\widehat D/(c_1eNt)\),
we obtain

\[
\begin{aligned}
\frac{2R+1}{D_{\mathrm a}}
&>\frac{15q^2ac_4dB}{8\lambda e\vartheta tT_1}\\
&>\frac{15q^2cc_4\mathfrak h}{8\lambda\vartheta}
                         \frac de\frac Bt
 >\frac{45}{8\lambda}=\frac{1305}{236}>1.
\end{aligned}
\tag{10.949}
\]

For the last step use \(q\mathfrak h>1\), \(q\ge2\), \(c>1/2\),
\(c_4\ge6\), \(d/e\ge1\), \(B/t\ge1\), and \(\vartheta\le2\). Order zero
is available since \(O\ge0\). By Proposition 10.31 these zeros yield
\(2R+1\) distinct additive roots. For each distinct root, division by
its linear factor gives zero remainder. Induction on the number of
distinct roots therefore makes their product divide the polynomial. Its
degree cannot exceed that of the nonzero polynomial, contradicting the
displayed strict inequality. \(\square\)

Proposition 10.212 gives the contradiction once the final family and its
zeros have been constructed. Lemma 10.211 provides the starting family
at \(I_1\); Theorems 10.216--10.217 propagate its fractional zeros and
restore the full integer intervals through the same order O, retaining
the complete root-field and rational-denominator costs.

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={Final integer comparison margins and the conditional additive-polynomial contradiction.}]{../figures/yu-second-terminal-parameters.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.64. Left: the strict rational lower margins proved in
Theorem 10.210, in units \(c_1/(q^r\widehat{\mathcal Z})\). The proof
includes every rank through its finite rows and uniform tail. Right:
Proposition 10.212, conditional on obtaining the final-stage zeros.
Integral support collapses to the additive coordinate; the nonzero
polynomial then has more roots than its degree. Its normalized support
bound is half the coordinate-degree bound. The reproducible figure
source is
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_second_terminal_parameters.py}{\texttt{yu\_second\_terminal\_parameters.py}}.
The stopping alternatives in the freely accessible Yu (2013), Section 7,
provide context; every mathematical input and bound used here is proved
in this programme.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{72. Keeping the derivative order through the final
contractions}\label{keeping-the-derivative-order-through-the-final-contractions}

The interval constructed in Lemma 10.211 has reached the smallest
derivative scale used by the geometric induction. We now keep its order
fixed. Before each fractional step, two integer extensions supply enough
interpolation mass to pay for the entire root-field degree. After
restriction to a phase class, interpolation on the q-deleted interval
restores the missing integers at that same order.

Use the hatted kernel of Theorem 10.203 and the counterhypothesis
\(v_p(\Xi-1)\ge\chi\widehat U_r\). Assume \(I_M<I_G\), and retain
\(I_1,S_1,T_1,T_{\mathrm f},O,R,I_3\) from Lemma 10.211. Put \(a=r+1\),
\(c=c_5\), \(\eta=1-c/a\), \(\mathfrak h=\eta^a\), and
\(Q=q\mathfrak h>1\). The constants \(a_*,d_*,w,\nu,\chi\) have their
case values in Theorem 10.197 and Lemma 10.199. Here
\(c_1Q_D/\widehat{\mathcal Z}<\lambda\nu/c_4\), where
\(Q_D=D_{\mathrm a}(N+1/(p-1))\) and \(\lambda=59/58\). Also set
\(\xi=101/100\) and \(\zeta=99/100\). The symbol \(\zeta\) in this
section is distinct from the optimizer constant \(\zeta_2\).

\subsubsection{The lower bound for the actual stopping
depth}\label{the-lower-bound-for-the-actual-stopping-depth}

\textbf{Lemma 10.213 (size and depth bounds).} Define

\[
\begin{aligned}
P_0&=\max\{1,a_*\},& F_r&=\frac{r^r(r+1)^r}{(r!)^2},\\
L_r&=\frac{2c_0c_4qa}{58}(c_2q)^rF_rP_0^{r-1}
               \max\left\{1,\frac{5(a_*+3/5)}r\right\},\\
m_r&=\max\{m\in\mathbb Z_{\ge0}:cL_r\mathfrak h^m/a\ge1\},\\
\mu_r&=q^rQ^{-m_r},&
j_r&=\min\{j\in\mathbb Z_{\ge0}:\mu_r\le q^j\},\\
S_-&=c_3qd_*a^2,& \sigma_r&=\frac{qa}{S_-L_r}.
\end{aligned}
\tag{10.952}
\]

Then \(\widehat T>L_r\), \(I_1\ge m_r\ge1\), and
\(c_1N/\widehat{\mathcal Z}<\sigma_r\). For \(\alpha=-\log\mathfrak h\)
and every integer \(I_1\le I<I_3\), the pair \((I_1,I)\) belongs to the
triangle

\[
m_r\le x\le\frac{3B}{\log Q},\qquad
x\le z\le\frac{3B+\alpha x}{\log q}.
\tag{10.953}
\]

\textbf{Proof.} Substitute (10.707), with the hatted scales, into
\(\widehat T=qa\widehat{\mathcal D}/(c_1eNt)\). Since
\(A_0=e\vartheta\), this gives

\[
\widehat T>
\frac{2c_0c_4qa}{\rho A_0}(c_2qP)^rF_r\frac Bt.
\tag{10.954}
\]

Use \(\rho\le58\), \(P\ge A_0\), \(P\ge P_0\), and \(B/t\ge1\). The
optimizer inequality \(r/t>A_0+3/5\) and \(B>5\) also give
\(B/t>5(A_0+3/5)/r\ge5(a_*+3/5)/r\). This proves \(\widehat T>L_r\). For
existence without a logarithmic rounding assumption, (10.708),
\(c_0\ge2,c_4\ge6,q\ge2\) and \(c_2q>3\) give \(L_r/a>(12/29)12^r\).
Since \(c>1/2\), \(\mathfrak h>1/3\), and \(r\ge3\), this makes
\(cL_r\mathfrak h/a>1\). Since \(0<\mathfrak h<1\), the maximum defining
\(m_r\) exists. The integer \(I_1=I_M\) is the last depth with
\(c\widehat T\mathfrak h^{I_1}/a\ge1\), so \(I_1\ge m_r\). The identity
\(c_1S\widehat TN/\widehat{\mathcal Z}=qa\) and \(S\ge S_-\) prove the
bound for \(\sigma_r\).

Finally \(I_1\log Q\le3B\). The exact floor in (10.944) gives
\(I\le I_3-1\le I_1+(3B-I_1\log Q)/\log q
=(3B+\alpha I_1)/\log q\). These are exactly the triangle inequalities.
Its vertices are \((m_r,m_r)\), \((m_r,(3B+\alpha m_r)/\log q)\), and
\((3B/\log Q,3B/\log Q)\). \(\square\)

\subsubsection{A joint bound for the radius and the rational
denominator}\label{a-joint-bound-for-the-radius-and-the-rational-denominator}

Write \(\ell_2=7/50\), \(\ell_3=11/50\), and \(l_q=5\ell_q\). Define the
positive rational logarithm bound

\[
v_Q=\frac{Q-1}{Q+1},\qquad
L_Q=2\sum_{k=0}^4\frac{v_Q^{2k+1}}{2k+1}<\log Q.
\tag{10.955}
\]

The strict inequality is the positive logarithm series proved in
Solution 34. In particular no numerical logarithm determines \(m_r\) or
\(j_r\): their definitions use rational powers and integer comparisons.

For \(z\ge a_*\), put

\[
\begin{aligned}
b_r&=1500\,12^{r-3},\qquad
\varepsilon_r=\frac{8r+15}{9ab_r},\qquad
 e_r=\frac1{79}+\frac1{71928b_r},\\
W_r&=\frac w{2c_2}\left(1+\frac1{2S_-}\right)
 +\frac1{a^2c_2c_3q}\\
&\quad +(1+w/2)c_1\varepsilon_r+(1+w)c_1e_r,\\
K_j(z)&=W_r+\frac{\xi}{1+z}
 \left\{\frac{\lambda[1+w+q\ell_q/(q-1)+j\ell_q]z}{c_4}\right.\\
&\qquad\left. +\frac{(107/103+3/(8a))a/(cL_r)
                +w(1-483/(824a))}{c_3}\right\},\\
V_\theta(z)&=\max\left\{\begin{aligned}
 &\frac{\theta\mu_r}{c_2}
 +\frac{\xi z}{1+z}\frac{\lambda l_q}{c_4}
                     \min\left(\frac{m_r}{5},\frac3{L_Q}\right),\\
 &\frac{\theta}{c_2}\min\left(\frac{q^r}{32768},\mu_r\right)
 +\frac{\xi z}{1+z}\frac{3\lambda l_q}{c_4L_Q}
 \end{aligned}\right\},\\
\mathcal H_{\rm s}(z)&=K_{j_r}(z)+V_1(z),\\
\mathcal H_{\rm l}(z)&=K_{j_r+1}(z)+V_q(z).
\end{aligned}
\tag{10.956}
\]

\textbf{Lemma 10.214 (complete fixed-order arithmetic bounds).} Let
\(\theta\in\{1,q\}\). At a depth \(I_1\le I<I_3\), consider a nonzero
prepared value of an admissible selected subfamily, at order at most O.
If its physical radius X obeys

\[
\frac XS\le\theta q^r
               \exp(\alpha I_1-I\log q),
\tag{10.957}
\]

then the following bounds hold, provided \(X\le q^jS\). For an integer
value in K at depth I, its normalized valuation
\(c_1(v_p(V)-\beta)/\widehat{\mathcal Z}\) is at most
\(K_j(A_0)+V_\theta(A_0)\). The same bound holds for an integer value at
depth \(I+1\) with the same physical radius. For a fractional value at
depth I, divide the normalized valuation by \(q^r\); its upper bound is
again \(K_j(A_0)+V_\theta(A_0)\). The fractional assertion retains the
full global field degree and the chosen local degrees e,f.

\textbf{Proof.} The global height argument is Theorem 10.92 with the
coefficient bound of Lemma 10.199 and the hatted estimates of Theorem
10.203. Its pointwise torus contribution is at most \(X/(c_2S)\). The
part of its denominator proportional to I is at most
\(\lambda(\vartheta/N)I\log q/(c_4B)\). To bound their sum, on the
triangle in Lemma 10.213 consider

\[
F(x,z)=\frac{\theta q^r}{c_2}e^{\alpha x-z\log q}
           +\frac{\lambda\vartheta}{c_4NB}z\log q.
\tag{10.958}
\]

The Hessian of the exponential term is its positive value times the
outer product of \((\alpha,-\log q)\) with itself; the other term is
linear. Thus F is convex. Expressing a point in the triangle as a convex
combination of its vertices bounds F by its largest vertex value. At
both vertices on the upper edge, the exponential is \(e^{-3B}\), and the
vertex with \(x=z=3B/\log Q\) has the larger linear term. Only that
vertex and \((m_r,m_r)\) need to be retained.

At the first vertex the torus term is \(\theta\mu_r/c_2\). Use \(B>5\),
\(m_r\log Q\le3B\), \(\log q<l_q\), and \(\log Q>L_Q\) to bound its
denominator part by the first linear term in \(V_\theta\). At the other
vertex, \(e^{-3B}<2^{-15}\) and \(e^{-3B}\le Q^{-m_r}\). Its torus term
is therefore bounded by the minimum in the second entry. The denominator
term is at most \(3\lambda(\vartheta/N)\log q/(c_4\log Q)\). In both
entries use \(\vartheta/N<\xi A_0/(1+A_0)\). This proves exactly the
bound \(V_\theta(A_0)\).

Here are all the remaining contributions. The initial mean torus height,
coefficient-field generator, logarithmic column counts and two scalar
remainders give the four terms of \(W_r\), exactly as in (10.864), with
the sharper \(e_r\) permitted there. The factorial estimate in (10.824)
and \(X\le q^jS\) give

\[
\log\frac{\mathrm e(X+2k)}k
 \le j\log q+\log\frac{\mathrm e(S+2k)}k
 <B+j\log q.
\tag{10.959}
\]

The pointwise additive contribution is consequently bounded by
\(\lambda(1+j\ell_q)\), and its initial mean has multiplier
\(w\lambda\). The fractional denominator in (10.243) is
\(D_{\mathrm a}(I+1+1/(q-1))\). The constant part contributes
\(\lambda q\ell_q/(q-1)\). An integer at depth I has a smaller
denominator; an integer at depth \(I+1\) has the same total denominator
as the fractional target just considered. Thus the same bound covers all
three cases without discarding a power of q.

For the derivative term, \(O\le T_{\mathrm f}<a/c\) and
\(\widehat T>L_r\), so \(O/\widehat T<a/(cL_r)\). Lemma 10.194 supplies
the multiplier \(107/103+3/(8a)\). The unchanged initial mean is bounded
by \(w(1-483/(824a))\). Finally \(1/(eN)<\xi/(1+A_0)\). These terms are
exactly the second fraction in \(K_j\); the additive terms use
\(\vartheta/N<\xi A_0/(1+A_0)\).

An integer value is in K by Theorem 10.55. A fractional normalized value
is in the field of Theorem 10.56, of degree at most \(q^rd\) and with
the specified local degrees e,f.~Its removed common principal-unit
monomial has valuation zero. Multiply the entire nonnegative height
majorant by \(q^r\) in this case. The coefficient minimum is still
\(\beta\), and restriction has not increased the coefficient height.
This proves all the stated bounds without an equality assumption for
valuations at different primes. \(\square\)

\subsubsection{Rational comparisons in every
rank}\label{rational-comparisons-in-every-rank}

Put \(C=\lambda\nu/c_4\) and

\[
D_r=\frac{103\xi c}{100c_3(1+a_*)a}.
\tag{10.960}
\]

\textbf{Lemma 10.215 (the fixed-order comparison margins).} In every
admissible case and rank, \(j_r\le1\), and each of the following
quantities is strictly positive for \(a_*\le z\le r\):

\[
\begin{aligned}
P_1(z)&=2cq^2\eta-2q\sigma_r-C-\mathcal H_{\rm s}(z),\\
P_2(z)&=\zeta\chi c_1q^r-2\sigma_r-C-\mathcal H_{\rm l}(z),\\
P_3(z)&=\zeta\chi c_1q-
                 \frac{C+2\sigma_r}{q^r}-\mathcal H_{\rm s}(z),\\
P_4(z)&=2cq(q-1)\eta-2(q-1)\sigma_r-C-D_r-
                                               \mathcal H_{\rm s}(z),\\
P_5(z)&=c_1\chi q^{r+1}-
            \frac{\xi}{c_3(1+a_*)L_r}-D_r-\mathcal H_{\rm s}(z),\\
P_6&=(1-\zeta)c_1\chi q^{r+1}-
                              \frac{\xi}{c_3(1+a_*)L_r},\\
P_7&=1-\frac{\zeta\chi c_1}{2cq\mathfrak h}.
\end{aligned}
\tag{10.961}
\]

The strict lower bounds for \(P_1,P_2,P_3,P_4\), in units of
\(10^{-6}\), are as follows.

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}lrrrr@{}}
\toprule\noalign{}
Case & \(P_1\) & \(P_2\) & \(P_3\) & \(P_4\) \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
I.1 & 2422817 & 4303236 & 315715 & 507605 \\
I.2 & 2422663 & 3859563 & 204821 & 507452 \\
II & 2506006 & 4370042 & 312956 & 604168 \\
III.1 & 2408114 & 3728458 & 138904 & 489560 \\
III.2 & 2407902 & 4338090 & 291349 & 489352 \\
IV & 2369299 & 4839514 & 416027 & 469693 \\
V & 9589027 & 72053237 & 338353 & 5464556 \\
\end{longtable}
}

Moreover \(P_5>10\), \(P_6>11/100\), and \(P_7>1/4\).

\textbf{Proof.} We give the finite reduction and its unbounded-rank
argument. For each fixed rank the arithmetic bounds are maxima of two
fractional-linear functions of z. Under \(u=1/(1+z)\), each of these is
affine. Its maximum on an interval is attained at an endpoint. It is
therefore sufficient to substitute \(z=a_*\) and \(z=r\), since the
optimizer gives \(a_*\le A_0<r\).

For the tail, the factorial computation in Lemma 10.148 proves
\(F_{r+1}/F_r>13/2\) for \(r\ge8\). The ratio of \(\max(1,K/(r+2))\) to
\(\max(1,K/r)\) is at least \(r/(r+2)\): check separately whether K lies
below r, between r and \(r+2\), or above \(r+2\). Consequently

\[
\frac{L_{r+2}/(r+3)}{L_r/(r+1)}
>(c_2qP_0)^2(13/2)^2\frac r{r+2}=:G_r
\qquad(r\ge32).
\tag{10.962}
\]

Both \(\mathfrak h\) and Q increase with rank by the logarithmic
derivative argument in Theorem 10.201. For each case, direct rational
substitution at ranks 32 and 33 gives

\[
G_r\mathfrak h_r^{d_0}>1,\qquad Q_r^{d_0}>q^2,
\qquad d_0=\begin{cases}10,&q=2,\\9,&q=3.\end{cases}
\tag{10.963}
\]

These inequalities continue to hold at larger ranks, since their left
sides increase. Multiplication of \(cL_r\mathfrak h_r^{m_r}/(r+1)\ge1\)
by the first growth factor, and use of
\(\mathfrak h_{r+2}\ge\mathfrak h_r\), give \(m_{r+2}\ge m_r+d_0\). Thus
\(\mu_{r+2}\le\mu_rq^2/Q_r^{d_0}<\mu_r\), and \(j_r\) is nonincreasing
on each parity sequence. In particular it is not permissible to hold
\(m_r\) fixed while bounding \(q^r\).

Here is a uniform envelope for each parity tail starting at \(b=32\) or
33. In \(K_j\), replace \(1-483/(824a)\) by one, and use the base-rank
values for \(W_r\), \((107/103+3/(8a))a/L_r\), and \(j_r\). In
\(V_\theta\), use the larger common bound

\[
\frac{\theta\mu_b}{c_2}
 +\frac{\xi z}{1+z}\frac{3\lambda l_q}{c_4L_{Q_b}}.
\tag{10.964}
\]

Every replacement increases the arithmetic bound. Indeed \(S_-\)
increases, \(a/L_r\) decreases by the growth inequality,
\(\varepsilon_r,e_r\) decrease from their displayed formulas,
\(\mu_r,j_r\) decrease along the parity sequence, and \(L_Q\) increases
with Q. The initial mean factor was replaced by one because it
increases, rather than decreases, with rank. For the positive mass terms
use the base values of \(\eta\) and \(q^r\); for \(\sigma_r,D_r,1/L_r\)
and negative \(q^{-r}\) terms use their base upper bounds. These
substitutions decrease each comparison. For \(P_7\), the base value of
\(\mathfrak h\) also decreases the comparison. The resulting tail bounds
are again maxima of affine functions of \(1/(1+z)\), so use \(z=a_*\)
and the limit \(z\to\infty\).

The exact calculation substitutes ranks 3 through 33 in the six
odd-prime cases and ranks 4 through 33 in V, and then the two parity
tail envelopes in every case. It checks seven inequalities at each of
the two parameter endpoints: 3,220 rational inequalities in total. The
14 parity steps also check both displayed growth inequalities. The
integers \(m_r,j_r\) are obtained by multiplication until their defining
rational inequalities change sign. These same finite and tail checks
give \(m_r\ge1\), \(j_r\le1\), and the listed strict margins. Together
with the tail argument this proves the claims for every admissible rank.
\(\square\)

\subsubsection{Two integer extensions and the fractional
zeros}\label{two-integer-extensions-and-the-fractional-zeros}

\textbf{Theorem 10.216 (the second-scale fixed-order fractional block).}
Suppose an admissible selected family at depth \(I_1\le I<I_3\) has
every prepared zero on the integer interval \(|s|\le R\) through order
O. It then has every prepared zero at \(s/q\), for integers
\(|s|\le R\), through the same order O. Its coefficients, minimum
valuation, phase and support are unchanged.

\textbf{Proof.} All restrictions retain the combined-frequency
congruence by Theorem 10.176, at every finite depth. The strict disc and
perturbation conditions follow from Theorems 10.180, 10.190 and 10.203.
The original distinguished coefficient \(b_n\) is used in the
perturbation. Since q is prime to p, scaling the additive argument by
\(q^{-I}\) preserves its local integrality and degree. For a full
interval with n known integer nodes, Theorem 10.182 with \(\mu=1\)
gives, at every \(x\in\mathbb Z_p\),

\[
v_p(\Phi_I(x;\boldsymbol t))-\beta
\ge\min\{nN-Q_D,\ \kappa+\chi\widehat U_r-v_p(b_n)\}
\qquad(|\boldsymbol t|\le O).
\tag{10.965}
\]

There is no ordinary-jet loss because \(\mu-1=0\). Equation (10.863),
with hats, separately bounds the original loss:

\[
\frac{c_1v_p(b_n)}{\widehat{\mathcal Z}}
<\frac{\xi}{c_3(1+a_*)\widehat T}
<\frac{\xi}{c_3(1+a_*)L_r}.
\tag{10.966}
\]

In particular \(P_6>0\) gives \(v_p(b_n)<(1-\zeta)\chi\widehat U_r\).
Define two integer radii

\[
R_{\rm s}=\left\lfloor\frac{\zeta\chi\widehat U_r/(qN)-1}{2}\right\rfloor,
\qquad
R_{\rm l}=\left\lfloor\frac{\zeta\chi\widehat U_r/N-1}{2}\right\rfloor.
\tag{10.967}
\]

They are positive. For example
\(\widehat U_r/(qN)=c_1q^{r-1}S\widehat T/a\), while \(c_1\ge7/10\),
\(q^{r-1}\ge4\), \(S>16\), and \(\widehat T/a>(12/29)12^r>700\). Even
the smaller radius therefore exceeds one. With
\(n_{\rm s}=2R_{\rm s}+1\) and \(n_{\rm l}=2R_{\rm l}+1\), the exact
floor inequalities are

\[
\begin{aligned}
\zeta\chi\widehat U_r/q-2N&<n_{\rm s}N
                                      \le\zeta\chi\widehat U_r/q,\\
\zeta\chi\widehat U_r-2N&<n_{\rm l}N
                                      \le\zeta\chi\widehat U_r,\\
qR_{\rm s}&\le R_{\rm l}.
\end{aligned}
\tag{10.968}
\]

The last follows because \(qR_{\rm s}\le(\zeta\chi\widehat U_r/N-q)/2
\le(\zeta\chi\widehat U_r/N-1)/2\) and its left side is integral. No
comparison between R and either boosted radius is required.

First extend to \(R_{\rm s}\). The starting integer interval has
\(2R+1>2q^{j_0+1}S_1-2q\) nodes. Since \(a/T_1>c\eta^{j_0+1}\),
\(q\eta>1\), and \(S_1T_1=S\widehat T\), its first analytic entry,
multiplied by \(c_1/\widehat{\mathcal Z}\), is greater than

\[
2cq^2\eta(q\eta)^{j_0}-2q\sigma_r-C
\ge2cq^2\eta-2q\sigma_r-C.
\tag{10.969}
\]

The second analytic entry is greater than
\(c_1\chi q^{r+1}-\xi/[c_3(1+a_*)L_r]\). We retain this second entry
separately; we do not assume that the original interval's node mass is
smaller than \(\chi\widehat U_r\).

The physical target radius satisfies

\[
\begin{aligned}
q^{-I}R_{\rm s}
&<\frac{\zeta\chi c_1}{2cq\mathfrak h}
                q^r e^{\alpha I_1-I\log q}S\\
&<q^r e^{\alpha I_1-I\log q}S
\le\mu_rS\le q^{j_r}S.
\end{aligned}
\tag{10.970}
\]

Here \(T_1<a/(c\mathfrak h)\), \(S\widehat T=S_1T_1\),
\(I\ge I_1\ge m_r\), and \(P_7>0\) were used. The arithmetic bound is
\(\mathcal H_{\rm s}(A_0)\), by Lemma 10.214. The first analytic entry
exceeds it by \(P_1>0\). The second exceeds it by \(P_5+D_r>0\). Thus
each proposed nonzero integer target is impossible. Orders exceeding the
additive degree already vanish identically. This gives all zeros through
O on \(|s|\le R_{\rm s}\), without changing the family.

Use this smaller boosted interval as the interpolation input for the
extension to \(R_{\rm l}\). Its node mass is at most
\(\zeta\chi\widehat U_r/q\), while the second analytic entry is greater
than \(\zeta\chi\widehat U_r\), by \(P_6>0\). Hence the minimum is at
least \(n_{\rm s}N-Q_D\). Its normalized lower bound is
\(\zeta\chi c_1q^r-2\sigma_r-C\). The physical large radius is less than
q times the real upper bound just used for the small radius. Thus Lemma
10.214 applies with \(\theta=q,j=j_r+1\), giving
\(\mathcal H_{\rm l}(A_0)\). The strict comparison \(P_2>0\) forces
every integer target to zero. All these extensions retain the original
R-block as well.

Finally interpolate from the full \(R_{\rm l}\)-interval. Again the
second entry is greater than its node mass, so the analytic lower bound
is \(n_{\rm l}N-Q_D\). Its normalization is greater than
\(\zeta\chi c_1q^{r+1}-2\sigma_r-C\). For the prescribed fractional
targets,

\[
q^{-I}R/q\le q^r e^{\alpha I_1-I\log q}S\le q^{j_r}S.
\tag{10.971}
\]

They belong locally to \(\mathbb Z_p\). Their global field degree is at
most \(q^rd\); keep this full factor. Lemma 10.214 bounds their
normalized valuation divided by \(q^r\) by \(\mathcal H_{\rm s}(A_0)\).
Dividing the analytic bound by the same \(q^r\), the strict difference
is \(P_3>0\). Thus every such value is zero. All target orders are still
at most O, and every interpolation used multiplicity one. \(\square\)

\subsubsection{Restoring every integer and reaching the
endpoint}\label{restoring-every-integer-and-reaching-the-endpoint}

\textbf{Theorem 10.217 (fixed-order successor and terminal
contradiction).} Under the same hypotheses, a nonzero admissible stage-I
family with its full R-block through O produces such a family at depth
\(I+1\), for every \(I_1\le I<I_3\). Consequently the alternative
\(I_M<I_G\) is impossible.

\textbf{Proof.} Theorem 10.216 provides the entire fractional block.
Apply the exact support and phase extraction of Theorems 10.48 and
10.51, including the extra phase partition when q divides \(G_0\). Keep
the class containing a coefficient of minimum valuation \(\beta\). The
Kummer monomial basis separates the classes for \(q\nmid s\). The
triangular affine change in the prepared indices preserves every total
order through O, with diagonal powers of q. The resulting family has
q-deleted zeros at \(|s|\le R\) through O, widths at most
\(q^{-I-1}\widehat D_i\), additive argument \(q^{-I-1}s\), and integer
values in K. Theorems 10.29, 10.55 and 10.176 preserve the nonzero
polynomial, complete polynomial fibres, original coefficient minimum,
phase and selected combined-frequency congruence. The coefficient height
does not increase.

It remains to fill the multiples of q. Since R is divisible by q, the
deleted interval contains exactly \(n_E=2(q-1)R/q\) nodes. Write
\(B_E=\lfloor\log_p(2R)\rfloor\). Theorem 10.182 with \(\mu=1\) gives
the two analytic entries

\[
n_EN-Q_D,\qquad
\kappa+\chi\widehat U_r-v_p(b_n)-B_E.
\tag{10.972}
\]

There is exactly one deleted-node separation loss in the second entry
and no inverse-jet loss. For an even smaller common lower bound we may
subtract \(B_E\) from the first entry as well. The normalization is
still \(Q_D-\beta\); it has not been discarded in either integer
extension or in this closure.

We bound the separation loss using the fixed starting depth, rather than
the later stage. Since \(j_0\le r\), \(R\le q^{r+1}S_1\), while Theorem
10.203 gives \(\log(2q^rS)<H\). Hence

\[
\log(2R)<H+\log q+\alpha I_1
<\frac{103}{100}H e^{\alpha I_1}.
\tag{10.973}
\]

Indeed \(H>40\), \(\log q<11/10<3H/100\), and \(1+u\le e^u\) for
\(u\ge0\). Applying (10.863) with hats and
\(T_1=\mathfrak h^{I_1}\widehat T\ge a/c\) gives

\[
\frac{c_1B_E}{\widehat{\mathcal Z}}
<\frac{103\xi}{100c_3(1+a_*)T_1}\le D_r.
\tag{10.974}
\]

The exact count satisfies \(n_E>2(q-1)q^{j_0}S_1-2(q-1)\). As in the
preceding proof, \(a/T_1>c\eta^{j_0+1}\) and \(q\eta>1\) show that the
first, conservatively lowered analytic entry has normalized lower bound
\(2cq(q-1)\eta-2(q-1)\sigma_r-C-D_r\). The second has lower bound
\(c_1\chi q^{r+1}-\xi/[c_3(1+a_*)L_r]-D_r\).

For the successor integer target, the physical radius is \(q^{-I-1}R\),
exactly the preceding fractional radius. Its denominator has total depth
\(I+1\), also exactly the one just charged. Its field is K, so there is
no factor \(q^r\) on its arithmetic bound. Its order is unchanged and
its coefficient height has not increased. Lemma 10.214 therefore gives
\(\mathcal H_{\rm s}(A_0)\), using the previous depth I in the triangle.
This remains valid at \(I+1=I_3\); no extension of the triangle beyond
its proved range has been made. The two strict differences are \(P_4>0\)
and \(P_5>0\). Every integer target is consequently zero, through the
same order O.

Lemma 10.211 supplies the actual initial family at \(I_1\). Finite
induction with the successor just proved reaches \(I_3\), retaining all
hypotheses of Proposition 10.212. That proposition forces the torus
degree to be zero and gives more integer roots than the degree of its
nonzero additive polynomial, a contradiction. Thus \(I_M<I_G\) is
impossible. \(\square\)

The argument retains the fixed derivative order, the entire original
integer interval, the original coefficient loss and the complete
root-field factor. Its intermediate integer radii are used only to gain
interpolation mass; the final contradiction concerns the same R-block
that was constructed in Lemma 10.211.

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={The coupled depth triangle and the unchanged-order route through integer and fractional zero sets}]{../figures/yu-second-fixed-order.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.65. Left: the real depth enclosure of Lemma 10.213 for
the illustrative parameters of case III.1, rank three and B = 10. The
two green vertices bound the convex sum of torus-radius and denominator
costs; the upper left vertex has the same exponential term as C and a
smaller linear term. The plotted domain is a parameter enclosure, not a
list of realized number-field families. Right: the zero sets constructed
in Theorems 10.216--10.217, with their exact comparison labels. Every
step retains order O. The fractional comparison uses the complete global
degree bound \(q^rd\); the successor integer comparison uses K itself.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_second_fixed_envelopes.py}{Exact calculation
and figure program}. Human-source context: the auxiliary-function
extrapolation and contraction method in Sections 10--11 of
\href{https://matwbn.icm.edu.pl/ksiazki/aa/aa89/aa8944.pdf}{Yu's freely
accessible 1999 paper}. The bounds and induction used in this figure are
proved in Lemmas 10.213--10.215 and Theorems 10.216--10.217; Solution 86
checks the rational depth example and the node counts.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{73. Closing the two-base case at an unramified prime above
five}\label{closing-the-two-base-case-at-an-unramified-prime-above-five}

The comparison in Theorem 10.189 left one range: two bases, a local
field equal to \(\mathbb Q_5\), global degree greater than one, and a
small coefficient parameter. A two-logarithm determinant handles this
range. Its usual height floor is unnecessarily large here. We first
lower that floor when the interpolation parameter is large, and retain
every rounding and common-coefficient term in the proof.

\subsubsection{A smaller height floor at a large interpolation
parameter}\label{a-smaller-height-floor-at-a-large-interpolation-parameter}

\textbf{Lemma 10.218 (the determinant estimate with a tenth-height
floor).} Use the algebraic, local-unit and independence hypotheses of
Theorem 10.170, with nonzero integer coefficients. Its global degree may
also be the degree of a specified number field containing both bases;
use that field's local residue degree in \(D=d/f\). Choose

\[
\begin{aligned}
L_i&\ge\max\{h(\alpha_i),\log p/(10D)\},\\
B'&=\frac{|b_1|}{DL_2}+\frac{|b_2|}{DL_1},\\
H&\ge\max\{\log B'+\log\log p+2/5,10\log p/D,10\}.
\end{aligned}
\tag{10.979}
\]

If \(B=DH/\log p\ge1200\), then

\[
v_p(\alpha_1^{b_1}-\alpha_2^{b_2})
<24\frac{pgD^4}{(p-1)(\log p)^4}L_1L_2H^2,
\tag{10.980}
\]

where g is the order of the subgroup generated by the residues. The
ordinary floor \(L_i\ge\log p/D\) gives the same weak inequality without
the extra condition \(B\ge1200\), by Theorem 10.170.

\textbf{Proof.} The height calculation in Lemma 10.6 and Theorem 10.169
uses the product formula in a specified field containing the bases. For
a place of local degree ef its weight is \(ef/d\), giving exactly
\(D=d/f\). Neither the polynomial zero lemma nor the local power-law
argument requires the two bases to generate that field. Thus the proof
of Theorem 10.170, including its ordinary-floor conclusion, applies to
these ambient degrees. In particular \(D\ge e\ge1\). This observation
does not identify global and local degrees.

Invert bases and multiply by a local unit to reduce to positive
coefficients, as in that theorem. Put \(a_i=DL_i/\log p\),
\(z=ga_1a_2\), \(k=3,\ell=13/5\), and take the same integers
\(L=\lfloor\ell B\rfloor+2\), \(K=\lfloor3zL\rfloor+1\) and the four
rounded square-root sides of Theorem 10.165. Those side-product and
side-sum inequalities use only \(a_i>0\). Here

\[
a_i\ge1/10,\qquad z\ge1/100,\qquad zB\ge12,
\qquad K>3(1/100)(13/5)1200>85.
\tag{10.981}
\]

The two residue-class cardinalities therefore have the same proof as in
Theorem 10.170, unless linear labels collide. First suppose there is no
such collision. The factorial argument (10.611)--(10.613) uses
\(K\ge85\), not a lower bound on either \(a_i\). Consequently
\(\log\beta<\log(B'\log p)+2/5\le H\).

The compressed height calculation (10.614)--(10.616) is also unchanged.
Its final use of \(a_i\ge1\) must now be replaced: since
\(g(a_1+a_2)=z(1/a_1+1/a_2)\le20z\), it gives

\[
\gamma_1Ra_1+\gamma_2Sa_2
\le\frac23\sqrt3\,zL+\frac43\sqrt{zL}+\frac{40}3z.
\tag{10.982}
\]

Put \(\lambda=14/5\); then \(L\le\lambda B\). The sufficient determinant
inequality (10.618), divided by \(zB^2\), has left side minus the
coefficient term greater than \(3(13/5)^2-3(13/5)-6/B\). The three
height terms are at most

\[
\frac23\sqrt3\lambda^2,
\qquad \frac43\frac{\lambda^{3/2}}{\sqrt{zB}},
\qquad \frac{40\lambda}{3B}.
\tag{10.983}
\]

For the remaining logarithm use \(D/\log p=B/H\le B/10\) and
\(N\le3z\lambda^2B^2+\lambda B\). The function \(\log(az+b)/z\)
decreases when \(az+b>\mathrm e\): its derivative has numerator
\(az/(az+b)-\log(az+b)<0\). At \(z=1/100\), the function
\(\log(3\lambda^2B^2/100+\lambda B)/B\) decreases for \(B\ge1200\),
because its logarithmic derivative times B is less than two and its
logarithm is greater than two. It follows that

\[
\frac{3(D/\log p)\log N}{zB^2}
\le\frac{\log342048}{40}<\frac{13}{40}.
\tag{10.984}
\]

The remaining comparisons are entirely rational. Use \(\sqrt3<26/15\),
\(\lambda^{3/2}/\sqrt{12}<34/25\), and \(B\ge1200\). The strict
sufficient margin is

\[
\frac{3848}{1125}
-\left(\frac1{200}+\frac{136}{75}+\frac7{225}
                            +\frac{13}{40}\right)>0.
\tag{10.985}
\]

Thus Theorem 10.169 applies. Its common coefficient p-part is not
discarded. Since \(\log(b_1/a_2)<H\), \(\log b_1\le\max\{1,a_2\}H\): for
\(a_2\ge1\) use \(\log a_2\le a_2-1\) and \(H\ge1\); otherwise use
\(\log a_2\le0\). Also \(\max\{1,a_2\}\le100z\). Hence
\(u\le100zB/D\le100zB\). With \(C=p/(p-1)\), the final bound \(CN+u\),
divided by \(CzB^2\), is strictly below

\[
3\lambda^2+\frac{\lambda}{12}+\frac{100}{1200}
=\frac{7151}{300}<24.
\tag{10.986}
\]

If linear labels collide, the exact residue-order and power-law argument
(10.625)--(10.627) applies with these same positive heights. Its bound
is \(u+2C\sqrt3\,zL+(D/\log p)\log2\). Dividing by \(CzB^2\), using
\(\sqrt3<7/4\) and \(\log2<7/10\), gives the strict upper bound

\[
\frac1{12}+\frac{49}{6000}+\frac7{1200}
=\frac{73}{750}<1.
\tag{10.987}
\]

Finally \(zB^2=gD^4L_1L_2H^2/(\log p)^4\), as before. This proves the
assertion in both cardinality alternatives. \(\square\)

\subsubsection{Comparing the determinant with the second
constant}\label{comparing-the-determinant-with-the-second-constant}

Set \(t=\log5\), and retain the independent original bases and
coefficient conventions of Theorem 10.152. In this subsection
\(n=2,e=f=1,d>1\), \(A=4+\log(3d)\), \(\Omega=h(a_1)h(a_2)\), and
\(\mathsf h_2\) is (10.643).

\textbf{Lemma 10.219 (three degree ranges).} In this setting,
\(C_1\mathsf h<C_2\mathsf h_2\) for \(d\ge2048\). For \(2\le d\le2047\),
suppose \(\mathsf h_2<G_1(2,d)\) and both coefficients are nonzero. Then

\[
\frac{v_5(\Xi-1)}{C_2\mathsf h_2\Omega}
<\begin{cases}1/2,&2\le d\le64,\\3/5,&65\le d\le2047.
\end{cases}
\tag{10.988}
\]

\textbf{Proof.} Write \(a_0=2+\log7\). The constants independent of d in
\(G_1/3\) and \(G_2/3\) are respectively

\[
g_1=2a_0+471/100+\log(2a_0+471/100+\log2),\qquad
g_2=2a_0+199/100+\log3.
\tag{10.989}
\]

They satisfy \(g_1<76/5\), \(g_2>549/50\). The exact quotient of the
first and second constants is
\(C_2/C_1=(495/1794)(3/8)\mathrm e^2t>123/100\). All three strict
numerical comparisons are proved by the rational logarithm calculation
described below. Since \(\log2>69/100\), at \(d\ge2048=2^{11}\) we have

\[
\frac{123}{100}(549/50+\log d)-(76/5+\log d)
\ge\frac{511}{10000}>0.
\tag{10.990}
\]

Thus \(G_1<(123/100)G_2\). Every other entry in the two coefficient
maxima agrees, so \(\mathsf h\le(123/100)\mathsf h_2\). The strict
quotient proves the large-degree assertion.

For the bounded-degree ranges let \(w=w_K\). Reduction embeds the roots
of unity of K into \(\mathbb F_5^*\): a root reducing to one lies in
\(1+5\mathbb Z_5\), where the deep-unit power laws of Lesson 9 rule out
nontrivial torsion. Therefore \(w\mid4\), and \(w=q^u\). The singleton
and two-element cases of Theorem 10.18 give, with
\(\ell_d=\max\{1,\log d\}\),

\[
h(a_i)\ge\frac{w}{58\mathrm e d^2\ell_d},\qquad
\Omega\ge\frac{w}{29\mathrm e^2d^3\ell_d}.
\tag{10.991}
\]

Use \(\epsilon=1\) for \(d\le64\), and \(\epsilon=1/10\) otherwise. Set
\(L_i=\max\{h(a_i),\epsilon t/d\}\). Separating the cases where neither,
one, or both heights are raised, (10.991) gives

\[
\begin{aligned}
\frac{L_1L_2}{\Omega}
&\le\max\left\{1,\frac{58\mathrm e\epsilon t d\ell_d}{w},
                  \frac{29\mathrm e^2\epsilon^2t^2d\ell_d}{w}\right\}
\\&\le\begin{cases}
29\mathrm e^2t^2d\ell_d/w,&\epsilon=1,\\
(58/10)\mathrm e t d\ell_d/w,&\epsilon=1/10.
\end{cases}
\end{aligned}
\tag{10.992}
\]

For the last comparisons use \(2<\mathrm e t<6\), \(d\ge2\), \(w\le4\)
and \(\ell_d\ge1\). Each selected term exceeds one; the ratio of the
third term to the second is \(\mathrm e\epsilon t/2\).

To apply the two-logarithm estimate to \(\Xi-1\), use
\((\alpha_1,\alpha_2)=(a_1,a_2^{-1})\) and \((b_1,b_2)\) as its
exponents. The resulting difference differs from \(\Xi-1\) by a local
unit. Its residue subgroup has order at most four. Put

\[
c(d)=\log t+3\log\log(3d)+2/5,\qquad H=\mathsf h_2+c(d).
\tag{10.993}
\]

Since the padded reciprocal heights decrease,
\(B'\le R_{\max}/d\le W(d)\exp(\mathsf h_2)\). Thus H satisfies the
logarithmic requirement of Lemma 10.218. We also have
\(\mathsf h_2>33\), \(c(d)>0\), and \(10t/d<10\), since
\(d\ge2,t<13/8\). For \(d\ge65\), \(B=dH/t>65\cdot33/(13/8)=1320>1200\).
The ordinary estimate handles the lower range; Lemma 10.218 handles the
upper one.

The residue maximum in C2 is exactly its second entry:
\(\mathrm e^2t^3/4>5\ge5/\delta(\boldsymbol a)\). The complete formula
(10.642) therefore reduces to

\[
C_2=218295\mathrm e^4d^4\ell_d A/w.
\tag{10.994}
\]

Combining (10.992), the bound with coefficient \(24\), and \(g\le4\),
bounds the desired ratio above by

\[
\begin{cases}
\dfrac{3480}{218295\mathrm e^2t^2},&d\le64,\\[3pt]
\dfrac{696}{218295\mathrm e^3t^3},&65\le d\le2047
\end{cases}
\quad\frac dA\frac{(\mathsf h_2+c(d))^2}{\mathsf h_2}.
\tag{10.995}
\]

Here \(d/A(d)\) increases, since its derivative has the sign of
\(A-1>0\). Write \(\bar G(d)=3(76/5+\log d)\). On \(2\le d\le2048\) we
have \(0<c(d)<15/2<33<\mathsf h_2<G_1<\bar G\). The function
\((x+c)^2/x\) increases in x when \(x>c>0\), and increases in c.~Both
\(\bar G(d)\) and c(d) increase. Consequently it suffices to bound the
two expressions at d=64 and d=2048, respectively. At the first endpoint
use \(A>46/5,\bar G<291/5,c<6\); at the second use
\(A>12,\bar G<687/10,c<15/2\). Together with \(\mathrm e>27/10,t>8/5\),
these give

\[
\frac{3320210000}{7889661549}<\frac12,\qquad
\frac{37419280000}{65596294071}<\frac35.
\tag{10.996}
\]

For completeness, all logarithmic endpoints used here admit one short
exact verification. For rational \(1\le x\le2\), set \(y=(x-1)/(x+1)\).
Integration of the geometric series gives

\[
0\le\log x-2\sum_{j=0}^{N-1}\frac{y^{2j+1}}{2j+1}
\le\frac{2y^{2N+1}}{(2N+1)(1-y^2)}.
\tag{10.997}
\]

Reduce any larger rational argument by a power of two, and use N=64 in
both it and \(\log2\). Monotonicity permits rational interval endpoints
inside the one nested logarithm in (10.989). The positive exponential
series through degree ten supplies the lower bound for \(\mathrm e\).
Substitution checks (10.989)--(10.990), \(8/5<t<13/8\), \(\log t<1/2\),
\(26/5<\log192<53/10\), \(\log(53/10)<17/10\), \(\log64<21/5\),
\(8<\log6144<44/5\), and \(\log(44/5)<11/5\). These are exactly the
endpoint bounds above. The same calculation verifies \(\log342048<13\)
for Lemma 10.218. The editable rational calculation checks these and all
remaining displayed rational margins; the monotonicity argument, not a
sample over degrees, supplies the quantified ranges. \(\square\)

\textbf{Theorem 10.220 (the second estimate for every two-base field
case).} Under the hypotheses of Theorem 10.152 with n=2,

\[
\operatorname{ord}_{\mathfrak p}(\Xi-1)
<C_2\mathsf h_2h(a_1)h(a_2).
\tag{10.998}
\]

\textbf{Proof.} Theorem 10.189 covers every case except
\(p=5,e=f=1,d>1,\mathsf h_2<G_1(2,d)\). When both coefficients are
nonzero, Lemma 10.219 covers that range; for d at least 2048 use Theorem
10.152 after its constant comparison. Since e=1, its p-adic valuation is
the ideal valuation in (10.998).

If exactly one coefficient b is nonzero, put \(a=a_i\) for its base. Its
residue order is at most four. The elementary one-base estimate (10.60),
\(\log|b|\le\mathsf h_2\), and \(\mathsf h_2>33\) give
\(v_5(a^b-1)<(34/33)d\mathsf h_2/t+8d h(a)/t\). Divide by the target,
use the product lower bound in (10.991) for the first term and the
singleton bound for the unused base in the second, and substitute
(10.994). The ratio is strictly less than

\[
\frac{(34/33)29}{218295\mathrm e^2At}
+\frac{464}{218295\mathrm e^3dA\mathsf h_2t}
<\frac1{1000}<1.
\tag{10.999}
\]

The last rational comparison follows already from
\(\mathrm e>2,A>5,t>1,d\ge2,\mathsf h_2>33\). This includes the
zero-coefficient alternative without feeding it into a determinant
requiring two nonzero coefficients. \(\square\)

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={The determinant bounds in two finite degree ranges and the increasing large-degree coefficient gap}]{../figures/yu-second-rank-two-closure.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.66. Left: the two rational upper bounds from (10.996),
valid throughout their indicated integer-degree ranges when both
coefficients are nonzero and \(\mathsf h_2<G_1\). The auxiliary height
floors differ by a factor of ten; Lemma 10.218 proves the second
determinant estimate, and Lemma 10.219 absorbs its actual padding costs.
Right: the lower bound \(511/10000+(23/100)\log(d/2048)\) for the
normalized coefficient-parameter gap is positive for every \(d\ge2048\),
although only the range through 8192 is drawn. Theorem 10.220 includes
the zero-coefficient case by its separate one-base proof. These are
proved bounds, not samples of algebraic inputs.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_second_rank_two_closure.py}{Exact rational
calculation and figure program}. Human-source context: the two-logarithm
determinant in Bugeaud--Laurent's
\href{https://irma.math.unistra.fr/~bugeaud/travaux/logpadicdef.ps}{freely
accessible 1996 paper} and the second constant in
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
freely accessible 2013 paper}. The lower floor and degree comparisons
are proved above.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{74. The second explicit estimate for an arbitrary number of
bases}\label{the-second-explicit-estimate-for-an-arbitrary-number-of-bases}

The second estimate follows by combining the two stopping arguments with
a minimum-rank representation. The coefficients of that representation
are rational. Its coefficient bounds must therefore continue to refer to
the original integers. We first establish the precise weighted form of
the first estimate that is needed at a small intermediate rank.

\subsubsection{The first estimate for integral forms with prescribed
weights}\label{the-first-estimate-for-integral-forms-with-prescribed-weights}

Retain the original field, independent units \(a_1,\ldots,a_n\),
integers \(b_j\), distinguished coefficient \(b_n\), and
\(\mathcal L=\sum_jb_jz_j\) of Theorem 10.152. In particular,
\(b_n\ne0\) has minimum p-adic valuation. Let \(a_0\) generate the
q-primary roots of unity. Suppose \(L_0=z_0,L_1,\ldots,L_r\) are
independent integral forms, \(2\le r\le n\), and

\[
\mathcal L=B_0L_0+\sum_{i=1}^rB_iL_i,\qquad
L_i=\sum_{k=0}^na_{ik}z_k,\qquad B_i\in\mathbb Q.
\tag{10.1002}
\]

Set \(\alpha_i=a_0^{a_{i0}}\prod_{k=1}^na_k^{a_{ik}}\). Choose positive
weights \(\sigma_i\ge\sum_{k=1}^n|a_{ik}|h(a_k)\), and write
\(\Omega_r=\prod_i\sigma_i\). The bases \(\alpha_i\) are independent, by
the argument following (10.277). In constants bearing a rank argument,
use that family's actual saturated residue index.

\textbf{Lemma 10.221 (a weighted first estimate with the original
coefficients).} If

\[
H\ge\max\left\{\log\frac{R_{\max}}{dW(d)},
             \log B_\dagger,G_1(r,d),(r+1)t\right\},
\tag{10.1003}
\]

where \(R_{\max},B_\dagger\) still concern the original \(a_k,b_k\),
then

\[
\operatorname{ord}_{\mathfrak p}(\Xi-1)
<C_1(r,d,p,\boldsymbol\alpha)H\Omega_r.
\tag{10.1004}
\]

No size or integrality hypothesis on the rational \(B_i\) is added.

\textbf{Proof.} We give the minimum-rank argument with the stated
starting forms. This identifies the coefficient and height data at each
step. Choose \(\tau=10^{-26}/(2n)\), and let \(H_*\) be the first-scale
quantity in (10.272), so \(qH_*=\mathfrak a(1+\tau)\). In this proof
only, put \(\Gamma=\mathrm eH_*(r+1)d\). Consider integral forms

\[
F_j=u_{j0}L_0+\sum_{i=1}^ru_{ji}L_i,
\qquad
s_j\ge\sum_{i=1}^r|u_{ji}|\sigma_i,
\tag{10.1005}
\]

independent with \(L_0\), whose rational span with \(L_0\) contains
\(\mathcal L\). For a family of length s require

\[
\prod_{j=1}^ss_j\le
\Gamma^{r-s}
\frac{M_r(\boldsymbol\alpha)}{M_s(\boldsymbol\beta)}\Omega_r,
\qquad
\beta_j=a_0^{u_{j0}}\prod_i\alpha_i^{u_{ji}}.
\tag{10.1006}
\]

Here \(M_j\) is (10.272), so \(t^jM_j=\mathcal R_j\). The initial family
\(F_i=L_i\) with weights \(\sigma_i\) is admissible. There is a least
positive admissible length s.

For every such family, the coefficient triangle inequality gives

\[
\sum_{k=1}^n\left|\sum_i u_{ji}a_{ik}\right|h(a_k)
\le\sum_i|u_{ji}|\sigma_i\le s_j.
\tag{10.1007}
\]

Thus (10.218) holds with the original \(a_k,b_k\). Lemma 10.84 bounds
each integral Euler argument by the original \(R_{\max}\), and Lemma
10.23 uses the same original \(b_n\) to choose the derivations. The
scalar bounds and the common coefficient loss therefore use (10.1003).
Since \(G_1(s,d)\le G_1(r,d)\) and \((s+1)t\le(r+1)t\), all their rank-s
height hypotheses hold. The evaluation of the original form remains
\(P\log\Xi\); no logarithm of a rational coordinate coefficient is
substituted.

Suppose first \(s\ge2\). Form the first-scale parameters
(10.183)--(10.186), using these bases and weights and height H. Under
\(v_p(\Xi-1)\ge U_s\), the kernel and initial zero arguments in the
proof of Theorem 10.152 apply: the principal-unit depth is supplied by
(10.280), the integer coefficient vector by Theorem 10.71, and the
complete subsequent zero blocks by Corollaries 10.140 and 10.150. At the
geometric stop, Theorem 10.102 applies with the fixed reference rank r.
Its proof uses this reference rank only through
\(\Gamma=\mathrm eH_*(r+1)d\) and the inequality \(s\le r\). It produces
a shorter family with product strictly smaller than
\(\Gamma^{s-m}(M_s/M_m)\prod_js_j/700\). Composing its integral rows
with (10.1005) preserves the stronger weight condition there, by the
triangle inequality. Its product is strictly below the right side of
(10.1006) at length m. This contradicts the chosen minimum. At the other
stop, Corollary 10.150 gives the nonzero additive-polynomial
contradiction. The alternatives exhaust the two finite stopping indices.
Hence \(v_p(\Xi-1)<U_s\).

The exact cancellation (10.452)--(10.453) now has r as its fixed
reference rank and \(\Omega_r\) as the initial product. The residue
factor \(M_s\) cancels through (10.1006). The remaining ratio of rank
factors is at most

\[
\left(\frac{\mathrm e}{q}\right)^{r-s}
\frac{s^s/s!}{r^r/r!}
\left(\frac{s+1}{r+1}\right)^{s+1}\le1.
\tag{10.1008}
\]

Consecutive ratios of \(j^j/j!\) are \((1+1/j)^j\ge2\); also
\(\mathrm e<3\) and \(q\ge2\). These prove the displayed inequality. We
have \(A_s\le A_r\), and \((1+\tau)^s<1+10^{-26}\) because \(s\le n\).
Lemma 10.151 therefore gives (10.1004).

If the least length is one, write its nonzero rational coefficient as
\(v/w\) in lowest terms, \(w>0\). The form \(F_1\) is integral in the
original variables. Comparing their coefficients shows \(v\mid b_k\) for
every k, hence \(|v|\le B_\dagger\). As in (10.454), raising to \(wq^u\)
kills the torsion factor and gives \(\Xi^{wq^u}=\beta_1^{vq^u}\ne1\).
The elementary residue-order bound yields

\[
\operatorname{ord}_{\mathfrak p}(\Xi-1)
\le\frac d t\bigl(2H+2eg s_1\bigr),
\tag{10.1009}
\]

where g is the actual residue order of \(\beta_1\). Here
\(\log(2q^uB_\dagger)<2H\) follows from \(q^u<p^f\), \(t>1\),
\(H\ge3t\), and \(H\ge\log B_\dagger\). The weight product of the
initial independent bases satisfies (10.55), since
\(\sigma_i\ge h(\alpha_i)\). Consequently the calculation
(10.455)--(10.456), with r in place of its reference rank, bounds the
two terms in (10.1009) by respectively \(1/400\) and \(1/4000\) of the
target in (10.1004). Its second calculation uses (10.1006), \(M_1>g/t\),
and \(eq^u\le2d\); all are available here. The sum \(11/4000<1\)
completes this case. This proves the lemma. \(\square\)

\subsubsection{Closing the small intermediate
ranks}\label{closing-the-small-intermediate-ranks}

Use the second rank allowance (10.725), and retain the integral-form and
individual weighted conditions (10.1002). Let \(h=\mathsf h_2\) for the
original n-base form, and put

\[
\mathcal T=C_2(n,d,p,\boldsymbol a)h\prod_{k=1}^nh(a_k),
\qquad
\Gamma_2=a^{(2)}\mathrm e^2Pd/q.
\tag{10.1010}
\]

\textbf{Lemma 10.222 (small-rank bounds with the second allowance).} For
\(n\ge3\), a representation of rank two satisfying (10.725) gives

\[
\operatorname{ord}_{\mathfrak p}(\Xi-1)
<\frac{25}{41}\mathcal T.
\tag{10.1011}
\]

For \(3\le r\le n\), the second estimate holds whenever
\(r/t-A_0\le3/5\). In the latter assertion r is the rank of the
representation, rather than the original number of bases.

\textbf{Proof.} Put \(H=\max\{h,G_1(r,d)\}\). The function \(G_2(j,d)\)
increases with j: each positive factor in (10.643) increases. Lemma
10.172 gives \(G_1(r,d)<3G_2(r,d)/2\le3h/2\). Every other entry in
(10.1003) is at most h. Thus Lemma 10.221 applies and \(H\le3h/2\).

After multiplication by the allowance (10.725), the exact quotient
between the rank-r second constant and the rank-n target is at most

\[
\frac{C_2(r,\boldsymbol\alpha)\Omega_r}
     {C_2(n,\boldsymbol a)\prod_kh(a_k)}
\le\left(\frac{\mathrm e}{q}\right)^{n-r}
       \frac{F(r)}{F(n)}\frac{A_r}{A_n}
\le1.
\tag{10.1012}
\]

The actual \(\mathcal R_r\) cancels, even if its residue index differs
from the original one. Lemma 10.187 proves the purely algebraic ratio in
this display for every \(r\ge1\): its proof that
\(F(j+1)/F(j)>\mathrm e\) does not use an optimizer condition. For
\(r<n\), the right side is strictly less than \(q^{r-n}\).

At r=2, the exact quotient (10.648) gives uniformly

\[
\frac{C_2(2,\boldsymbol\alpha)}
     {C_1(2,\boldsymbol\alpha)}>\frac{123}{100}.
\tag{10.1013}
\]

Indeed \(2\le Pt\) gives the stronger \(3/2\) bound in Theorem 10.173.
If \(2>Pt\), the field-case proof of Theorem 10.189 leaves only p=5 or
7, with f=1 and P=1. At seven its two rational comparisons give \(3/2\).
At five and d=1, (10.744) gives \(11/8>123/100\). At five and
d\textgreater1, the exact quotient is
\((495/1794)(3/8)\mathrm e^2\log5>123/100\), proved by the rational
series calculation in Lemma 10.219. These comparisons depend only on the
field data; the common residue maximum cancels. Since \(n\ge3\),
(10.1012) now gives

\[
\frac{\operatorname{ord}_{\mathfrak p}(\Xi-1)}{\mathcal T}
<\frac32\frac{100}{123}q^{2-n}
\le\frac{25}{41}<1.
\tag{10.1014}
\]

For r at least three in the stated range, the constant comparison in
Theorem 10.188 gives \(C_2(r)/C_1(r)>3/2\). Its argument uses only the
rank and field constants, so it applies to this r. Combining that
comparison with Lemma 10.221, \(H\le3h/2\), and (10.1012) proves the
second assertion. This argument has never treated the rational \(B_i\)
as original integer exponents. \(\square\)

\subsubsection{The complete second
estimate}\label{the-complete-second-estimate}

\textbf{Theorem 10.223 (Yu's second explicit estimate, with proof).}
Under every hypothesis of Theorem 10.152, with \(n\ge2\), the second
constant and height parameter (10.642)--(10.643) satisfy

\[
\operatorname{ord}_{\mathfrak p}(\Xi-1)
<C_2\mathsf h_2\prod_{k=1}^nh(a_k).
\tag{10.1015}
\]

\textbf{Proof.} Theorem 10.220 gives n=2, including the exceptional
unramified-five field case. Suppose n is at least three. Among the
integral-form representations in (10.277), replace its product allowance
by (10.725) and choose one of least positive rank r. This set is
nonempty: \(L_i=z_i\) and \(\sigma_i=h(a_i)\) give the original rank-n
representation with equality in the allowance. The original weighted
conditions make its bases independent local units in K, and its nonzero
form gives a nonzero rational coefficient vector.

Rank one is settled by Proposition 10.193. Rank two is settled by Lemma
10.222, which also handles every \(r\ge3\) with \(r/t-A_0\le3/5\). We
may therefore assume \(r\ge3\) and \(r/t>A_0+3/5\). Put
\(h=\mathsf h_2\). Monotonicity of \(G_2\) gives all of (10.806), with
the original \(R_{\max}\), \(B_\dagger\), and minimum-valuation
coefficient \(b_n\) still in place.

Use the hatted construction of Theorem 10.203, with
\(\tau=10^{-26}/(2n)\). Its scale comparison and the cancellation in
(10.1012) give

\[
\chi e\widehat U_r
<C_2(r,\boldsymbol\alpha)h\Omega_r\le\mathcal T.
\tag{10.1016}
\]

Assume for contradiction that
\(\operatorname{ord}_{\mathfrak p}(\Xi-1)\ge\mathcal T\). Then
\(v_p(\Xi-1)\ge\chi\widehat U_r\). Theorem 10.203 constructs an actual
nonzero selected kernel with its initial integer, fractional and
contracted zero blocks. The original integral forms provide (10.218);
Lemma 10.23 chooses the Euler directions using \(b_n\). Their
coefficient vector is nonzero and their hyperplane is the one used in
Theorem 10.25. The phase and support extraction preserves those
directions, independence of the torus bases, a nonzero coefficient of
the original minimum valuation, and complete polynomial fibres. Thus
these are the families and the hyperplane hypotheses in Theorems 10.204
and 10.217.

Let \(I_G,I_M\) be (10.933). They are finite and their defining sets are
nonempty, as proved in Corollary 10.209. If \(I_G\le I_M\), that
corollary produces the complete prepared integer zeros required by
Theorem 10.204. The latter produces \(1\le m<r\), integral vectors
\(u_j\), and weights \(R_j=\sum_i|u_{ji}|\sigma_i\) such that

\[
\prod_{j=1}^mR_j
<10^{-6}\Gamma_2^{r-m}
  \frac{\mathcal R_r(\boldsymbol\alpha)}
       {\mathcal R_m(\boldsymbol\beta)}\Omega_r
\le10^{-6}\Gamma_2^{n-m}
  \frac{\mathcal R_n(\boldsymbol a)}
       {\mathcal R_m(\boldsymbol\beta)}\prod_kh(a_k).
\tag{10.1017}
\]

Compose the original forms with these integral vectors. The rational
span conclusion of Theorem 10.25 preserves the representation of
\(\mathcal L\). Corollary 10.26 gives independence and bases in K. The
coefficient triangle inequality (10.1007) gives each original weighted
condition, and (10.1017) is stronger than its required product
allowance. Hence m is an admissible smaller rank, a contradiction.

If \(I_M<I_G\), Lemma 10.211 supplies the actual starting interval at
its fixed order. Theorems 10.216--10.217 produce every fractional and
integer successor through \(I_3\), retaining that order, the full
root-field factor in fractional comparisons and the original coefficient
loss. Proposition 10.212 then gives a nonzero additive polynomial with
more distinct integer roots than its degree. This is again a
contradiction. Equality of the stopping indices belongs to the first
case, so the two cases exhaust the possibilities. The counterhypothesis
is impossible, proving (10.1015). \(\square\)

Both explicit constants now bound the original form with their own
coefficient parameters. The second estimate retains the field-case
constants, the saturated residue index of the original bases, and the
minimum-valuation choice of the original integer coefficient.

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={The four rank cases and the two stopping contradictions for the second explicit estimate}]{../figures/yu-second-complete-bound.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.67. Theorem 10.223 separates the least admissible rank
into four cases. The one-base ratio is less than \(1/3000\), and the
two-base intermediate ratio is less than \(25/41\) for original length n
at least three. In the small-optimizer case the first constant and its
coefficient parameter combine through Lemma 10.222. The
positive-optimizer case has an actual induction to one of two exhaustive
stops; its geometric branch decreases rank with factor \(10^{-6}\), and
its order branch contradicts the degree of a nonzero polynomial. Every
ratio is relative to the exact original target \(\mathcal T\), except
the explicitly marked rank-product factor. The diagram is a proof map,
not a numerical sample of algebraic inputs.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_second_complete_bound.py}{Exact comparison
and figure program}. Human-source context: the two estimates of Theorem
I in
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
freely accessible 2013 paper} and the extrapolation method in Sections
10--11 of
\href{https://matwbn.icm.edu.pl/ksiazki/aa/aa89/aa8944.pdf}{his freely
accessible 1999 paper}. The weighted extension and the assembly of the
proof are established in Lemmas 10.221--10.222 and Theorem 10.223
above.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{75. A coefficient refinement for independent rational
local-unit
parts}\label{a-coefficient-refinement-for-independent-rational-local-unit-parts}

A separate bound for the last coefficient is useful when the other
coefficients can be much larger. We obtain the numerical bound appearing
in Theorem 2.3 of
\href{https://irma.math.unistra.fr/~bugeaud/travaux/StroblRev1.pdf}{Bugeaud's
freely accessible author paper}, under the explicit independence
hypothesis below. The proof uses the second estimate proved in Theorem
10.223. In particular, the cited statement is not used as a proof.

\subsubsection{Enlarging the height
bounds}\label{enlarging-the-height-bounds}

Write \(\omega(d)=1/(dW(d))\), as in (10.222). The constants \(C_1,C_2\)
concern the actual bases and their saturated residue index; they do not
change when upper bounds for their heights are enlarged.

\textbf{Lemma 10.224 (monotonicity with padded heights).} Retain the
hypotheses of Theorem 10.152. For \(i=1,2\), put

\[
\begin{aligned}
H_i&=\max\{\log B_\dagger,G_i(n,d),(n+1)t\},\\
F_i(\boldsymbol w)&=\left(\prod_{j=1}^nw_j\right)
 \max\left\{H_i,
  \max_{j<n}\log\left[\omega(d)
       \left(\frac{|b_n|}{w_j}+\frac{|b_j|}{w_n}\right)\right]\right\}.
\end{aligned}
\tag{10.1021}
\]

The function \(F_i\) is nondecreasing in each positive coordinate.
Consequently, for arbitrary real \(w_j\ge h(a_j)\),

\[
\operatorname{ord}_{\mathfrak p}(\Xi-1)<C_iF_i(\boldsymbol w).
\tag{10.1022}
\]

\textbf{Proof.} We have \(H_i\ge(n+1)t>1\). Fix all coordinates except
one, denoted by u. Apart from a positive constant factor, the function
to consider is the maximum of finitely many functions

\[
u\max\{H_i,\log(A+D/u)\},\qquad A,D\ge0,
\quad A+D>0.
\tag{10.1023}
\]

The constant logarithmic entries are included by taking D=0. Below the
threshold where the logarithm exceeds \(H_i\), the function is \(uH_i\),
hence increasing. Above that threshold its derivative is
\(\log(A+D/u)-D/(Au+D)\ge H_i-1>0\). At the threshold the two formulas
agree. This proves monotonicity of each function and of their finite
maximum. Apply it successively to the n height coordinates. Theorem
10.152, for i=1, and Theorem 10.223, for i=2, give the strict inequality
at the actual heights. Their constants and bases are unchanged, so the
inequality persists at the larger weights. \(\square\)

\subsubsection{A uniform comparison of the second
constant}\label{a-uniform-comparison-of-the-second-constant}

Let \(a_1,\ldots,a_n\) be independent rational p-adic units, \(n\ge2\).
At an odd prime use \(K=\mathbb Q\). At two use
\(K=\mathbb Q(\zeta_3)\). In this last field, \(d=2,e=1,f=2,q^u=3\).
Indeed the prime-to-two root of unity reduces injectively, by the proof
before (10.183), so its order three forces the residue degree to be
even. Since \(ef\le d=2\), we get f=2 and e=1. The same argument gives
\(q^u\mid2^2-1\), and the field contains \(\zeta_3\), so \(q^u=3\).
Heights of the rational bases do not change, and
\(\operatorname{ord}_{\mathfrak p}=v_2\) on rational numbers.

Define

\[
\mathcal C_n(p)=(16\mathrm e)^{2(n+1)}n^{3/2}
       (\log(2n))^2\frac{p}{(\log p)^2}.
\tag{10.1024}
\]

\textbf{Lemma 10.225 (the rational coefficient constant).} For the
fields just specified, the constant in (10.642) satisfies

\[
2C_2<\mathcal C_n(p)\qquad(n\ge2).
\tag{10.1025}
\]

\textbf{Proof.} All comparisons below hold at every rank. We record the
elementary estimates used in them. The positive exponential series gives
\(8/3<\mathrm e<11/4\): the sum through degree three is 8/3, and the
remaining terms are at most \((1/24)/(1-1/5)=5/96\). Integration of the
geometric series at x=1/3 gives
\(\log2=2\sum_{k\ge0}(1/3)^{2k+1}/(2k+1)\). Its first term and the
geometric upper bound for its tail show \(2/3<\log2<25/36<7/10\). Thus
\(\log3>1\), \(\log5>3/2\), and \(\log(2n)>4/3\) for n at least two. For
the middle assertion, \(\mathrm e^3<(11/4)^3<25\) suffices.

Integrating the increasing function \(\log x\) gives
\((n-1)!\ge((n-1)/\mathrm e)^{n-1}\). Also
\((1+2/(n-1))^{n-1}<\mathrm e^2\), by \(\log(1+x)<x\). Hence

\[
\frac{(n+1)^{n+2}}{(n-1)!}
 <\mathrm e^{n+1}(n+1)^3,\qquad
\frac{(n+1)^{n+1}}{(n-1)!}
 <\mathrm e^{n+1}(n+1)^2.
\tag{10.1026}
\]

For every t\textgreater0 the maximum of \(t^{n+1}\exp(-t)\) is
\((n+1)^{n+1}\exp(-n-1)\). Therefore

\[
\frac{\mathrm e^nt^{n+1}}{n^n}<(n+1)\exp(t).
\tag{10.1027}
\]

First let \(p\ge5\). Here \(P=1,d=f=1,q^u=2\), \(c^{(2)}\le2228/3\), and
\(a^{(2)}=7\). The residue index is at least one. Put
\(A=\max\{4+\log(n+1),t\}\), where t=log p. Since \(\log(n+1)\le n\) and
\(t>3/2\), \(A/t\le n+2\). Substitution into (10.642), followed by
(10.1026) and (10.1027), gives

\[
\frac{2C_2}{\mathcal C_n(p)}
 <\frac{2228}{768\mathrm e}
 \left(\frac7{256}\right)^n
 \frac{(n+1)^3(n+2)}{n^{3/2}(\log(2n))^2}
 <10n^3\left(\frac7{256}\right)^n.
\tag{10.1028}
\]

For the last inequality use \(\mathrm e>2\), \(\log(2n)>1\),
\(n+1\le3n/2\), \(n+2\le2n\), and \(n^{5/2}\le n^3\). The resulting
scalar coefficient is \(20052/2048<10\). The final expression is
245/4096 at n=2, and its successive ratio is at most 189/2048. It is
always below one.

At p=3, \(P=3,d=f=1,q^u=2,c^{(2)}=648\). Now \(A/t\le2(n+2)\). The same
calculation yields

\[
\frac{2C_2}{\mathcal C_n(3)}
 <6n^3\left(\frac{21}{256}\right)^n.
\tag{10.1029}
\]

Here the scalar coefficient before rounding to 6 is 729/128. The right
side starts at 1323/4096 and has successive ratio at most 567/2048.
Again it is below one.

At p=2 use the quadratic field above. We have
\(P=4,t=2\log2,c^{(2)}=406,a^{(2)}=13\) and \(A=4+\log(2(n+1))\). Since
t\textless7/5, the function \((\mathrm e t/n)^n\) decreases for real n
at least two: the derivative of its logarithm is \(\log(t/n)<0\). Thus

\[
\max\left\{4,\frac{\mathrm e^nt^{n+1}}{n^n}\right\}
 <6,
\qquad
\frac{(11/4)^2(7/5)^3}{4}=\frac{41503}{8000}<6.
\tag{10.1030}
\]

Substitute \(d=2,q^u=3\) and \((\log2)^2=t^2/4\) exactly; do not replace
the residue-field cardinality 4 by 2. This gives

\[
\begin{aligned}
\frac{2C_2}{\mathcal C_n(2)}
 &<\frac{203}{256\mathrm e}
       \left(\frac{13}{32}\right)^n
       \frac{(n+1)^2A}{t n^{3/2}(\log(2n))^2}\\
 &<\frac{345303}{262144}n^{3/2}
       \left(\frac{13}{32}\right)^n
 <\frac43n^2\left(\frac{13}{32}\right)^n.
\end{aligned}
\tag{10.1031}
\]

In the second line we used \(\mathrm e>8/3\), t\textgreater1,
\(\log(2n)>4/3\), \(n+1\le3n/2\), and \(A<n+5\le7n/2\). The rational
comparison with 4/3 is strict. The last expression starts at 169/192 and
has successive ratio at most 117/128. This completes all primes and all
n.~\(\square\)

\subsubsection{Inflating just the last
height}\label{inflating-just-the-last-height}

For a nonzero rational \(a=x/y\) in lowest terms, with y\textgreater0,
define its p-unit part by \(a^\circ=a p^{-v_p(a)}\). Removing the
p-power from either x or y proves \(h(a^\circ)\le h(a)\). For integers
\(b_j\), put \(s=\sum_jb_jv_p(a_j)\). If s\textgreater0, then
\(v_p(\prod_ja_j^{b_j}-1)=0\); if s\textless0, that valuation is s.
These follow from the unequal-valuation case of the ultrametric
inequality, proved in Lesson 9. If s=0, the product is exactly
\(\prod_j(a_j^\circ)^{b_j}\).

\textbf{Theorem 10.226 (the coefficient refinement with independent unit
parts).} Let p be prime and n at least two. Let \(a_j=x_j/y_j\) be
nonzero rational numbers, with coprime integers \(x_j,y_j\) and
\(y_j>0\). Assume that \(a_1^\circ,\ldots,a_n^\circ\) are
multiplicatively independent. Choose

\[
\begin{gathered}
A_j\ge\max\{|x_j|,y_j,\mathrm e\},\qquad w_j=\log A_j,\\
b_j\in\mathbb Z\setminus\{0\},\qquad
v_p(b_n)\le v_p(b_j)\quad(1\le j\le n),\\
B\ge\max\{3,|b_1|,\ldots,|b_n|\},\qquad
B\ge B_n\ge |b_n|,\qquad0<\varepsilon\le\tfrac12.
\end{gathered}
\tag{10.1032}
\]

Write \(\Xi=\prod_ja_j^{b_j}\), and set

\[
\begin{aligned}
V&=\prod_{j<n}w_j,&\Omega&=\prod_{j=1}^nw_j,
&r&=B/B_n,\\
D_n&=(n+1)(6n+5),&
T&=2B_n\varepsilon^{-1}\exp(D_n)p^{n+1}V,&L&=\log T.
\end{aligned}
\tag{10.1033}
\]

Then

\[
v_p(\Xi-1)<\mathcal C_n(p)
             \max\{\Omega\log T,\varepsilon B/B_n\}.
\tag{10.1034}
\]

The parameter \(\varepsilon\) is the auxiliary parameter called delta in
the cited rational statement. It is distinct from the saturated residue
index in \(C_2\).

\textbf{Proof.} If s is nonzero, the preceding valuation calculation
proves the conclusion, whose right side is positive. Suppose s=0.
Replace the bases by their p-unit parts. Their heights are bounded by
\(w_j\), every \(w_j\) is at least one, and the product and coefficient
conditions are unchanged. Use the fields of Lemma 10.225. Their
ramification indices are one, so their prime-ideal valuations of
\(\Xi-1\) equal \(v_p(\Xi-1)\).

Enlarge only the final weight, taking

\[
w_n^*=\max\left\{w_n,\frac{\varepsilon r}{VL}\right\}.
\tag{10.1035}
\]

We check the entire maximum in (10.1021) against 2L. In both fields
\(\omega(d)<1\): at d=1 this follows from
\(\omega(1)=\log2\log3/\log6\), and at d=2 from
\(\omega(2)=1/(2\log^3 6)\). For every j\textless n,

\[
\omega(d)\left(\frac{|b_n|}{w_j}
                   +\frac{|b_j|}{w_n^*}\right)
 \le B_n\left(1+\frac{VL}{\varepsilon}\right)
 \le\frac{B_nV}{\varepsilon}(L+\varepsilon)
 <\exp(2L).
\tag{10.1036}
\]

For the last inequality, both \(B_nV/\varepsilon<\exp L\) and
\(L+\varepsilon<\exp L\) hold. The latter follows from \(L>1\),
\(\varepsilon\le1/2\) and the positive exponential series. Also
\(\log B_\dagger\le\log B_n<L\).

The relevant rows of (10.643) satisfy \(a_0^{(2)}<7,a_1^{(2)}<4\) and
\(\log d<1\). For example \(\log52<5\) follows from
\(\mathrm e^5>(8/3)^5>52\); the other rows have smaller logarithmic
arguments. Hence \(G_2(n,d)<(n+1)(8n+5)<2D_n<2L\). Finally f is at most
two, so \((n+1)t\le2(n+1)\log p<2L\). All entries of the coefficient
maximum are therefore at most 2L.

Lemma 10.224 now gives

\[
v_p(\Xi-1)<2C_2VLw_n^*
 =2C_2\max\{\Omega L,\varepsilon r\}.
\tag{10.1037}
\]

Apply Lemma 10.225. The maximum is positive, so strictness is preserved.
\(\square\)

For p=5, the bases (2,3) and coefficients \((5^k,1)\), \(k\ge1\), meet
the hypotheses. Their independence follows by taking the valuations at
two and three. Here \(B=5^k,B_n=1\) and the exact valuation is one:
\(5^k\equiv5\pmod{20}\), \(2^{20}\equiv1\pmod{25}\), and
\(3\cdot2^5-1\equiv20\pmod{25}\). With \(w_1=1,w_2=\log3\), the two
entries in (10.1034) are \((\log3)\log(250\exp(51)/\varepsilon)\) and
\(\varepsilon5^k\). The parameter can be chosen to balance them.

The independence condition in Theorem 10.226 is essential to its proof:
removing p-powers can create multiplicative dependencies or roots of
unity. For instance (2,6,3) at two has unit parts (1,3,3). Theorem
10.232 treats arbitrary rational lists by controlling their relations
while retaining the distinguished coefficient bound.

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={The two coefficient-bound branches and the last-height inflation}]{../figures/yu-coefficient-refinement.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.68. For the exact inputs of Solution 89, the left panel
samples the two factors in the bound; their maximum is multiplied by
\(\mathcal C_2(5)\). The right panel samples the inflated last height
from (10.1035). The switch occurs exactly when
\(\varepsilon r=\Omega L\). These curves are bounds and height choices,
not sampled valuations: the actual valuation is one. Lemma 10.224 proves
the monotonicity, Lemma 10.225 proves the constant comparison at every
rank, and Theorem 10.226 supplies the coefficient estimate.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_coefficient_refinement.py}{Reproducible
figure and exact rational comparisons}. Human-source context: Theorem
2.3 of
\href{https://irma.math.unistra.fr/~bugeaud/travaux/StroblRev1.pdf}{Bugeaud's
free author paper}.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{76. Rational dependencies and the two-base coefficient
refinement}\label{rational-dependencies-and-the-two-base-coefficient-refinement}

Multiplicative dependence can simplify the coefficient refinement when
the bases are rational. In rank one, all their prime-valuation vectors
lie on an integral line. We first construct its primitive generator;
then the elementary power identity from Lesson 9 replaces the
many-logarithm estimate. This proves the coefficient bound for every
rational two-base input, including dependent bases. The target constant
is the one in Theorem 2.3 of
\href{https://irma.math.unistra.fr/~bugeaud/travaux/StroblRev1.pdf}{Bugeaud's
freely accessible author paper}.

\textbf{Lemma 10.227 (a rational generator with exact height weights).}
Let \(c_1,\ldots,c_n\in\mathbb Q^*\) have multiplicative rank one modulo
signs. There are a positive rational \(\beta\ne1\), integers \(m_i\),
and signs \(\sigma_i\in\{1,-1\}\) such that

\[
c_i=\sigma_i\beta^{m_i},\qquad
h(c_i)=|m_i|h(\beta),\qquad h(\beta)\ge\log2.
\tag{10.1038}
\]

If all the \(c_i\) are p-adic units, so is \(\beta\). For integers
\(b_i\), write \(M=\sum_i m_ib_i\). If \(w_i\ge\max\{h(c_i),1\}\),
\(\Omega=\prod_iw_i\), and \(B\ge\max\{1,|b_1|,\ldots,|b_n|\}\), then

\[
\prod_i c_i^{b_i}=\sigma\beta^M,\qquad
\sigma\in\{1,-1\},\qquad |M|<2nB\Omega.
\tag{10.1039}
\]

\textbf{Proof.} Use the finite set of rational primes occurring in the
numerators or denominators of the \(c_i\). The prime-valuation map has
kernel \(\{1,-1\}\). Its vectors span a one-dimensional rational
subspace. Choose a nonzero integral vector on this line and divide its
coordinates by their greatest common divisor, obtaining a primitive
integral vector \(v\).

Every integral vector on the line is an integral multiple of \(v\).
Indeed, there are integers \(u_\ell\) with
\(\sum_\ell u_\ell v_\ell=1\); if the vector is \(av\), this identity
gives \(a=\sum_\ell u_\ell(av_\ell)\in\mathbb Z\). For completeness, the
integer linear combinations of the coordinates form an ideal in
\(\mathbb Z\). Its least positive element divides each coordinate by
division with remainder, so is their greatest common divisor. This
proves the integer-combination assertion just used.

Set \(\beta=\prod_\ell\ell^{v_\ell}>0\). Unique prime factorization
proves the representation in (10.1038). If \(\beta=x/y\) is reduced, its
positive and negative powers have reduced numerator and denominator
\(x^{|m|},y^{|m|}\), in one order or the other. Thus
\(h(\beta^m)=|m|h(\beta)\). Since \(\beta\ne1\), at least one of x,y is
at least two; this proves the height lower bound. The p-coordinate is
zero when all original vectors have p-coordinate zero, proving the unit
assertion.

The sign in (10.1039) is \(\prod_i\sigma_i^{b_i}\). Moreover
\(\log2>1/2\), as proved in Lemma 10.225, and \(w_i\le\Omega\) because
every weight is at least one. Hence

\[
|M|\le B\sum_i|m_i|
\le\frac{B}{h(\beta)}\sum_iw_i
\le\frac{nB\Omega}{h(\beta)}<2nB\Omega.
\tag{10.1040}
\]

At least one \(m_i\) is nonzero, so (10.1038) also gives
\(h(\beta)\le\max_iw_i\le\Omega\). \(\square\)

\textbf{Theorem 10.228 (the coefficient refinement in rational rank at
most one).} Let \(n\ge2\), and retain the rational inputs, height
bounds, coefficient bounds and parameter \(\varepsilon\) of (10.1032),
except that the independence condition is replaced by the condition that
the p-unit parts have rank at most one modulo signs. Assume \(\Xi\ne1\).
Use \(V,\Omega,r,T,L\) of (10.1033), and put

\[
U=\max\{\Omega L,\varepsilon r\}.
\tag{10.1041}
\]

Then

\[
v_p(\Xi-1)<\frac{5p}{(\log p)^2}U
             <\mathcal C_n(p)U.
\tag{10.1042}
\]

In fact this rank-at-most-one argument does not require the minimum
p-valuation condition on the distinguished coefficient.

\textbf{Proof.} The valuation reduction before Theorem 10.226 applies
unchanged. If \(\sum_i b_iv_p(a_i)\ne0\), then \(v_p(\Xi-1)\le0\), and
the conclusion follows. Otherwise replace the bases by their p-unit
parts \(c_i\), without increasing their heights or changing the product.
All weights are at least one.

In rank zero all \(c_i\) are signs. If their product with the given
coefficients is not one, it is minus one, so its distance from one has
valuation at most one. The same conclusion holds in rank one when the
exponent M in Lemma 10.227 is zero. These cases satisfy (10.1042): since
\(U\ge L>3\log p\) and \(\log p<p\), its first upper bound is greater
than 15.

We may therefore assume rank one and \(M\ne0\). Take \(N=2(p-1)\). For
odd p the residue of \(\beta^{p-1}\) is one: multiplication by a nonzero
residue permutes the nonzero residues, and comparison of their products
proves this assertion. At two an odd rational unit has square congruent
to one modulo eight, since each odd integer has that property. Thus the
power identity of Lemma 9.5 applies to \(\beta^N\) at every prime.
Raising a unit to a positive integer power cannot decrease the valuation
of its difference from one, by the factorization of \(X^N-1\). Since N
is even, the sign disappears, and

\[
\begin{aligned}
v_p(\Xi-1)
&\le v_p(\Xi^N-1)
 =v_p(\beta^N-1)+v_p(M)\\
&\le\frac{2(p-1)h(\beta)+\log2+\log|M|}{\log p}.
\end{aligned}
\tag{10.1043}
\]

The last line uses the rational local-height inequality proved within
Theorem 10.21 for \(\beta^N-1\) and the elementary inequality
\(p^{v_p(M)}\le|M|\) for a nonzero integer. Negative M causes no change,
since \(\beta^{-N|M|}-1\) is a p-adic unit times \(\beta^{N|M|}-1\).

It remains to retain the coefficient bound \(B_n\) in this elementary
estimate. Lemma 10.227, \(B=B_nr\), and \(\varepsilon r\le U\) give

\[
\begin{aligned}
\log|M|
&<\log(2n)+\log B_n+\log\Omega+\log r\\
&\le \log(2n)+\log B_n+\log(\varepsilon^{-1})
       +\log\Omega+\log U
 <3U.
\end{aligned}
\tag{10.1044}
\]

Here \(L>\log(2n)+\log B_n+\log(\varepsilon^{-1})\), because
\(D_n>\log n\) and \(V\ge1\). Also \(U\ge\Omega L\ge L>1\),
\(\log\Omega<\Omega\le U\), and \(\log U<U\). These prove every
inequality in (10.1044). We have \(\log2<U\) and
\(h(\beta)\le\Omega<U/((n+1)\log p)\). Substitution in (10.1043) gives

\[
\begin{aligned}
v_p(\Xi-1)
&<\frac{U}{(\log p)^2}
       \left(\frac{2(p-1)}{n+1}+4\log p\right)\\
&<\frac{14p}{3(\log p)^2}U
 <\frac{5p}{(\log p)^2}U.
\end{aligned}
\tag{10.1045}
\]

We used \(n\ge2\) and \(\log p<p-1\), which follows by integrating
\(1/x<1\) from one to p.~Finally the factor
\((16\mathrm e)^{2(n+1)}n^{3/2}\log^2(2n)\) in \(\mathcal C_n(p)\)
exceeds 5: \(n\ge2\), \(\log(2n)>1\), and \(16^6>5\) suffice. This
proves the second strict inequality of (10.1042). \(\square\)

\textbf{Corollary 10.229 (arbitrary two-base rational inputs).} Take
\(n=2\) and all the data and inequalities in (10.1032), with no
independence condition. If \(\Xi\ne1\), then the coefficient bound
(10.1034) holds.

\textbf{Proof.} The two p-unit parts have rank zero, one, or two modulo
signs. Apply Theorem 10.228 in the first two cases. In rank two they are
multiplicatively independent, so Theorem 10.226 applies. A relation with
product minus one would square to a relation with product one, so no
torsion relation is overlooked. \(\square\)

For an exact dependent example, take p=5, the bases (8,4,2), and
coefficients \((5^k,2,1)\), with \(k\ge1\). The primitive generator is
2, the exponent vector is (3,2,1), and

\[
M=3\cdot5^k+5,\qquad \Xi=2^M,
\qquad v_5(\Xi-1)=2.
\tag{10.1046}
\]

Indeed M is divisible by four and has 5-adic valuation one. For k=1 this
is immediate from M=20; for k greater than one write
\(M=5(3\cdot5^{k-1}+1)\). Since \(v_5(2^4-1)=1\), Lemma 9.5 proves
(10.1046). Valid weights are \((\log8,\log4,1)\), and \(B=5^k,B_n=1\).
Thus

\[
V=\Omega=(\log8)(\log4),\qquad
L=92+\log(1250\Omega/\varepsilon),\qquad r=5^k.
\tag{10.1047}
\]

The distinguished coefficient is one. Its p-valuation is minimal, as
required for the common statement, even though the rank-one proof does
not need that condition.

These results complete the rational coefficient refinement for two bases
and for lists of any length whose unit parts have rank at most one.
Theorem 10.232 treats the intermediate dependent ranks for three or more
rational bases. The general-number-field coefficient refinement is a
separate result.

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={An exact rank-one exponent transfer and its coefficient budgets}]{../figures/yu-low-rank-coefficient-refinement.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.69. At p=5, the valuation vectors of (8,4,2) lie on the
integral line generated by the valuation vector of 2. The left panel
shows the exact exponent transfer for \(k=8\). The right panel samples
the two factors defining U in (10.1041), using the exact inputs
(10.1047). Their maximum is multiplied by \(5p/(\log p)^2\) in the
proved rank-one bound; the actual valuation is 2. The plotted curves are
bound factors, not observed valuations. Lemma 10.227 supplies the
generator and Theorem 10.228 supplies the inequality.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_low_rank_coefficient_refinement.py}{Reproducible
figure and exact example checks}. Human-source context: Theorem 2.3 of
\href{https://irma.math.unistra.fr/~bugeaud/travaux/StroblRev1.pdf}{Bugeaud's
free author paper}.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{77. The coefficient refinement for arbitrary rational
inputs}\label{the-coefficient-refinement-for-arbitrary-rational-inputs}

We now remove the independence restriction at every rational rank. The
distinguished base determines the proof. If its valuation vector lies in
the span of the others, we eliminate that base and use the height factor
thereby saved. Otherwise a dependence among the other bases preserves
the minimum-valuation condition on the distinguished coefficient. Both
operations retain the exact constant in (10.1024).

Throughout this section a rank or a span means the rank or rational span
of the prime-valuation vectors, equivalently multiplicative rank modulo
signs. Zero coefficients are allowed; the distinguished coefficient is
nonzero and we set \(v_p(0)=+\infty\).

\textbf{Lemma 10.230 (rational determinants and primitive relations).}
Let \(n\ge2\), let \(c_1,\ldots,c_n\) be nonzero rational numbers, and
let \(w_i\ge\max\{h(c_i),1\}\). Put

\[
K_n=2^n n!,\qquad V=\prod_{i<n}w_i,\qquad
\Omega=Vw_n.
\tag{10.1048}
\]

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\tightlist
\item
  Suppose the first \(n-1\) vectors span all the vectors, with rank
  \(s\ge2\). Choose a basis sublist of s of those first vectors, indexed
  by \(J\), and put \(W=\prod_{j\in J}w_j\). For integers \(b_i\) with
  \(|b_i|\le B\), there are a positive integer \(D\) and integers
  \(M_j\) such that
\end{enumerate}

\[
\left(\prod_i c_i^{b_i}\right)^D
=\prod_{j\in J}|c_j|^{M_j},\qquad
|M_j|\le K_n B\Omega V,\qquad W\le V.
\tag{10.1049}
\]

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{1}
\tightlist
\item
  If the first \(n-1\) vectors are dependent, there is a relation,
  supported on those indices,
\end{enumerate}

\[
\prod_{i<n}c_i^{k_i}=\eta\in\{1,-1\},\qquad
\gcd(k_1,\ldots,k_{n-1})=1,\qquad
|k_i|\le 2^{n-2}(n-2)!V.
\tag{10.1050}
\]

In particular, for every prime p at least one nonzero \(k_i\) is prime
to p.

\textbf{Proof.} Use all rational primes appearing in a numerator or
denominator. Each coordinate of the valuation vector of \(c_i\) has
absolute value at most \(h(c_i)/\log2<2w_i\). Indeed a prime power
dividing the reduced numerator or denominator is at most their maximum.
This proves the coordinate bound also for negative valuations. The
kernel of this vector map is exactly \(\{1,-1\}\).

For (a), choose s rows on which the selected s columns give a
nonsingular integer matrix A. Such rows exist by elementary elimination:
each independent column increases the pivot count, giving s pivot rows.
Write \(d_0=|\det A|\ge1\). Cramer's formula expresses every column in
their rational span with coefficients \(d_{ji}/d_0\), where the signed
integers \(d_{ji}\) are the determinants obtained by replacing column j,
with the sign of \(\det A\) absorbed. The coefficients agree on every
row, since the chosen rows determine a vector in the selected column
span. Cramer's identity itself follows by multiplying by the adjugate:
expansion of a determinant along a row gives
\(A\operatorname{adj}(A)=(\det A)I\).

The determinant expansion has \(s!\) terms. The coordinate bound
therefore gives

\[
|d_{ji}|\le 2^s s!w_i\frac{W}{w_j}.
\tag{10.1051}
\]

Take \(D=2d_0\) and \(M_j=2\sum_i d_{ji}b_i\). Unique prime
factorization gives (10.1049); the even power removes all signs. Since
\(w_i\le\Omega\), \(W/w_j\le V\), and \(s\le n-1\),

\[
|M_j|\le 2^{s+1}s!B\frac{W}{w_j}\sum_iw_i
\le 2^{s+1}s!nB\Omega V\le K_nB\Omega V.
\tag{10.1052}
\]

For (b), choose a nonempty dependent set of smallest size m among the
first \(n-1\) vectors. If m=1, its vector is zero and its single
coefficient can be one. Otherwise this set has rank m-1. Choose m-1
independent rows. Their signed maximal minors form a nonzero kernel
vector, by determinant expansion with a repeated row. Every entry is
nonzero: a zero entry would give a dependence on a proper subset. The
selected and full matrices both have one-dimensional kernels, so this is
a relation among the full vectors. Divide the entries by their greatest
common divisor. The same determinant bound gives \(2^{m-1}(m-1)!\) times
the product of the other column weights, at most the bound in (10.1050),
since \(m\le n-1\) and all weights are at least one. The singleton case
satisfies that bound too. The product has zero valuation at every
rational prime, so is a sign. Primitivity supplies an entry prime to any
specified p. \(\square\)

\textbf{Proposition 10.231 (a redundant distinguished base).} Retain the
rational data and bounds of (10.1032), allowing zero coefficients other
than \(b_n\), and assume \(\Xi\ne1\). If the p-unit part of \(a_n\) lies
in the span of the other p-unit parts, then (10.1034) holds. This case
does not require the minimum-valuation condition on \(b_n\).

\textbf{Proof.} Put \(U=\max\{\Omega L,\varepsilon r\}\), with the
notation of (10.1033). A nonzero total p-valuation gives
\(v_p(\Xi-1)\le0\), so reduce to the p-unit parts \(c_i\). If their rank
is at most one, the proof of Theorem 10.228 applies: it permits zero
coefficients and uses only the nonzero product condition. Hence assume
the rank s is at least two. It is at most n-1. If \(\Xi=-1\), the
valuation is at most one and the same elementary bound from Theorem
10.228 suffices. Thus \(\Xi\notin\{1,-1\}\).

Use Lemma 10.230(a). The independent positive bases \(|c_j|\) are p-adic
units. Their exponent vector \((M_j)\) is nonzero, because a non-torsion
rational number has no positive power equal to one. Put

\[
B^*=K_nB\Omega V,\qquad W=\prod_{j\in J}w_j\le V.
\tag{10.1053}
\]

Choose a nonzero \(M_j\) of minimum p-valuation as the last coefficient.
Theorem 10.223 permits the other coefficients to be zero. Use
\(\mathbb Q\) at odd p and \(\mathbb Q(\zeta_3)\) at two, as in Lemma
10.225. Lemma 10.224 pads the heights to the selected weights. Since
these are at least one and \(\omega(d)<1\), its logarithmic coefficient
entry is at most \(\log(2B^*)\). Also \(\log B_\dagger\le\log B^*\),
\(G_2(s,d)<2D_s\le2D_n<2L\), and \((s+1)f\log p\le2n\log p<2L\). The
parameter bound for \(G_2\) is the one proved in Theorem 10.226.
Consequently, with \(C_2^{[s]}\) denoting the second constant for this
selected list,

\[
v_p(\Xi-1)\le v_p(\Xi^D-1)
<C_2^{[s]}W\max\{\log(2B^*),2L\}.
\tag{10.1054}
\]

The first inequality follows by factoring a positive integral power of a
unit. It needs no assumption on \(v_p(D)\).

We now keep the original distinguished bound inside this estimate. For
every \(n\ge2\),

\[
\log K_n=n\log2+\log(n!)<n+n^2<D_n.
\tag{10.1055}
\]

Here \(n!\le n^n\), \(\log2<1\), and \(\log n<n\). Since
\(r\le U/\varepsilon\), expanding \(B^*=K_nB_nr\Omega V\) gives

\[
\begin{aligned}
\log(2B^*)&\le
 \log(2K_n)+\log B_n+\log(\varepsilon^{-1})
 +3\log V+\log w_n+\log(U/V)\\
&<L+2\log V+\log w_n+\log(U/V).
\end{aligned}
\tag{10.1056}
\]

Multiply by W. We have \(\log V<L\), \(WL\le U\),
\(W\log w_n<Ww_n\le\Omega\le U/L<U\), and \(W\log(U/V)<WU/V\le U\). The
logarithmic inequalities \(\log x<x\) hold for every positive x, and
\(L>1\). Thus

\[
W\log(2B^*)<5U,\qquad 2WL\le2U.
\tag{10.1057}
\]

Lemma 10.225 gives \(2C_2^{[s]}<\mathcal C_s(p)\). Moreover (10.1024)
directly gives

\[
\frac{\mathcal C_{j+1}(p)}{\mathcal C_j(p)}
>(16\mathrm e)^2>256\qquad(j\ge2).
\tag{10.1058}
\]

Since \(2\le s<n\), (10.1054) is less than
\((5/2)\mathcal C_s(p)U<\mathcal C_n(p)U\). This is (10.1034).
\(\square\)

\textbf{Theorem 10.232 (the rational coefficient refinement, all
ranks).} Let p be prime, \(n\ge2\), and take all the rational inputs,
height bounds and coefficient bounds in (10.1032), with no independence
condition. The coefficients may be zero except for \(b_n\ne0\), with
\(v_p(0)=+\infty\) in the minimum-valuation condition. Assume
\(\Xi\ne1\). Then the estimate (10.1034) holds with exactly
\(\mathcal C_n(p),T,\Omega\) and \(\varepsilon\) specified in (10.1024)
and (10.1033).

\textbf{Proof.} As before, reduce to rational p-unit bases \(c_i\),
unless the total valuation already proves the result. Their heights are
at most \(w_i\). Signs cause no height change. We argue by induction on
n.~Independent lists are covered by the proof of Theorem 10.226,
including zero coefficients: its use of Theorem 10.223 and all
inequalities (10.1035)--(10.1037) require only that the distinguished
coefficient be nonzero. Lists of rank at most one are covered by the
proof of Theorem 10.228, again including zero coefficients. These two
cases establish n=2. The torsion product \(\Xi=-1\) is included in the
elementary argument in that proof, even when the list has larger rank.
Hence assume n at least three, intermediate rank, and
\(\Xi\notin\{1,-1\}\).

If the distinguished vector lies in the span of the others, apply
Proposition 10.231. Otherwise the first \(n-1\) vectors are dependent,
since adjoining the distinguished vector increases their rank by one but
the whole list is dependent. Take a primitive relation \((k_i)\) from
Lemma 10.230(b). Choose an index \(x<n\) for which \(p\nmid k_x\), and
reverse the signs of all the \(k_i\) if necessary so that \(k_x>0\). At
every prime put \(\kappa_i=2k_i\). Thus \(\kappa_x\) is positive,
\(v_p(\kappa_x)=v_p(2)\), and every \(\kappa_i\) has valuation at least
\(v_p(2)\). Set \(\kappa_n=0\), and put

\[
K=K_nV,\qquad
b_i'=\kappa_xb_i-\kappa_i b_x\quad(i\ne x),
\qquad b_n'=\kappa_xb_n.
\tag{10.1059}
\]

Lemma 10.230 implies \(|\kappa_i|\le K\), so \(|b_i'|\le2KB\). Every
\(b_i'\) has p-valuation at least \(v_p(b_n)+v_p(2)\), and the
distinguished coefficient has exactly that valuation. It remains
nonzero. This retains the minimum-valuation condition also at two: the
common extra factor two occurs in every transferred coefficient.

The doubled relation has product one, regardless of its original sign.
Eliminating \(c_x\) therefore gives

\[
\Xi'=\prod_{i\ne x}c_i^{b_i'}=\Xi^{\kappa_x}\ne1,
\qquad v_p(b_n')=v_p(b_n)+v_p(2)
=\min_{i\ne x}v_p(b_i').
\tag{10.1060}
\]

The nonvanishing follows because \(\Xi\notin\{1,-1\}\) is rational. Keep
the surviving weights and the distinguished last index, and choose

\[
\begin{aligned}
B'&=2KB,& B_n'&=\kappa_xB_n,&
\varepsilon'&=\frac{\kappa_x}{2K}\varepsilon,\\
V'&=V/w_x,&\Omega'&=\Omega/w_x,&
\varepsilon' B'/B_n'&=\varepsilon B/B_n.
\end{aligned}
\tag{10.1061}
\]

These are valid bounds: \(B'\ge3\), \(B'\ge B_n'\ge|b_n'|\), and
\(0<\varepsilon'\le\varepsilon/2\le1/4\). All hypotheses of the
induction statement hold for this list of length n-1, including possible
zero coefficients.

Let \(L'\) be its parameter (10.1033). Exact cancellation of
\(\kappa_x\) gives

\[
L'=L+\log(2K/w_x)-(D_n-D_{n-1})-\log p<2L.
\tag{10.1062}
\]

Indeed \(D_n-D_{n-1}=12n+5>0\), \(w_x\ge1\), and
\(\log(2K)=\log2+\log K_n+\log V<L\) by (10.1055). Thus
\(U'=\max\{\Omega'L',\varepsilon'B'/B_n'\}\le2U\). Induction, the
factored-power inequality for local units, and (10.1058) now give

\[
v_p(\Xi-1)\le v_p(\Xi'-1)
<\mathcal C_{n-1}(p)U'
\le2\mathcal C_{n-1}(p)U<\mathcal C_n(p)U.
\tag{10.1063}
\]

This completes the induction and all rational ranks. \(\square\)

Theorem 10.232 proves the full rational statement in Theorem 2.3 of
\href{https://irma.math.unistra.fr/~bugeaud/travaux/StroblRev1.pdf}{Bugeaud's
free author paper}, and also permits zero non-distinguished
coefficients. Its proof uses the second independent-base estimate
already proved in Theorem 10.223; no citation substitutes for the
reduction of dependent inputs. The general-number-field refinement has
additional arithmetic data and is not asserted by this rational theorem.

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={Two exhaustive rational rank cases and an exact signed coefficient transfer}]{../figures/yu-rational-coefficient-refinement.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.70. The left panel displays the two cases for the
distinguished valuation vector used in Theorem 10.232. The
redundant-vector case uses Proposition 10.231 and the saved height
product W; the other case uses a primitive relation supported on the
preceding indices. The right panel records the exact p=2 example of
Solution 91, including the common factor in all transferred
coefficients. The displayed valuations are exact; the inequalities
involving U are proved bounds.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_rational_coefficient_refinement.py}{Reproducible
figure and exact reduction checks}. Human-source target: Theorem 2.3 of
\href{https://irma.math.unistra.fr/~bugeaud/travaux/StroblRev1.pdf}{Bugeaud's
free author paper}.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{78. The second application estimate for every multiplicative
rank}\label{the-second-application-estimate-for-every-multiplicative-rank}

The second estimate also admits a bound using the largest coefficient,
even when the bases are dependent. Its height floor differs from the
first estimate's floor, so we prove the coefficient conversion and the
rank reduction with that floor retained.

Keep the field and prime conventions of Sections 52 and 58. Thus the
bases are local units in a number field of degree d; at two the field
contains a primitive cube root of unity. Write \(t=f\log p\),
\(P=p^\kappa\), and \(a=a^{(2)},c=c^{(2)}\), with the constants of
(10.641)--(10.643). Here \(a=7\) at odd primes and \(a=13\) at two. Let
\(\rho\) and \(\ell_d\) have the meaning of (10.55), and let \(w_K\) be
the number of roots of unity in K. The already proved bounds include
\(\rho\le58\), \(q^u\le2d\), and \(w_K\ge q^u\). At two,
\(w_K/q^u\ge2\), since the orders two and \(q^u\) are coprime.

Sort an indexed list \(a_1,\ldots,a_n\) by nondecreasing height and
select its greedy basis \(\boldsymbol\beta\) modulo roots of unity, of
rank \(r\ge1\). At equal heights use any fixed ordering. Let
\(\delta_\beta\) be the saturated residue index of this basis if
\(r\ge2\), and the original residue index if \(r=1\). With
\(\varkappa=25\) at odd primes and \(\varkappa=48\) at two, define

\[
\begin{aligned}
X_n&=\max\{n,t\},\qquad L_n=\frac{X_n}{\varkappa Pdt},\\
H_n(a_i)&=\max\{h(a_i),L_n\},\\
\Omega_n&=\prod_{a_i\in\boldsymbol\beta}h(a_i)
                 \prod_{a_i\notin\boldsymbol\beta}H_n(a_i),\\
M_n&=\max\left\{\frac{p^f}{\delta_\beta},
                 \frac{\mathrm e^nt^{n+1}}{n^n}\right\},\\
A_n&=\max\{4+\log((n+1)d),e,t\},\\
C_n&=\frac{c(a\mathrm eP)^n}{P}
\frac{(n+1)^{n+2}d^{n+2}\ell_d}{(n-1)!q^ut^3}\,M_nA_n,\\
H&=\max\{\log B,t\},\qquad\mathcal T_n=C_n\Omega_nH,\\
\lambda&=\frac{2099}{2100}.
\end{aligned}
\tag{10.1064}
\]

Here e without roman type is the ramification index,
\(B\ge\max\{3,|b_1|,\ldots,|b_n|\}\), and \(\Xi=\prod_i a_i^{b_i}\ne1\).
Repeated bases remain distinct indices. The greedy argument of Lemma
10.155 applies to this floor as well: the chosen basis minimizes the
weighted product, and is unchanged when a discarded base is removed. The
constant \(C_n\) uses the full list length n and the residue index of
this fixed basis.

\subsubsection{The conversion for an independent
list}\label{the-conversion-for-an-independent-list}

We first record elementary envelopes used in both parts of the proof.
Since \(\delta_\beta\le p^f-1\), we have \(M_n>1\). For \(n\ge2\) and
every integer \(1\le j\le n+1\),

\[
\frac{t^j}{M_n}\le
 \left(\frac n{\mathrm e}\right)^{nj/(n+1)}
 <\frac{n^j}{\mathrm e^{j-1}},
\qquad \frac{M_nA_n}{t^2}>\frac1n.
\tag{10.1065}
\]

Indeed, replace \(M_n\) by \(\max\{1,\mathrm e^nt^{n+1}/n^n\}\). The
quotient with numerator \(t^j\) increases up to
\(t_0=(n/\mathrm e)^{n/(n+1)}\) and is nonincreasing after it. Its
maximum is \(t_0^j\). The ratio of this value to \(n^j/\mathrm e^{j-1}\)
is \(\mathrm e^{j/(n+1)-1}n^{-j/(n+1)}<1\). For the last assertion use
\(A_n\ge t\), the first bound at j=1, and \(t_0<n\).

The factorial argument in Lemma 10.153 also gives, for \(n\ge2\),

\[
R_n:=\frac{(n+1)^{n+2}(n-1)^{n-1}}{(n!)^2}
 >\mathrm e^{2n-2}.
\tag{10.1066}
\]

In detail, \(n!\le\mathrm e\sqrt n(n/\mathrm e)^n\), and
\((1+1/n)^{n+2}(1-1/n)^{n-1}>1\), by the integral inequalities proved
there. Substitution proves (10.1066).

\textbf{Lemma 10.233 (second coefficient-height conversion).} Suppose
the list is independent and \(n\ge2\). Put \(\Omega=\prod_i h(a_i)\) and
\(W=\lambda tC_n\Omega/(nd\max_i h(a_i))\). If \(B/\log B>W\), every
entry of the parameter \(\mathsf h_2\) is at most \((n+1)H\), after
moving a nonzero coefficient of minimum p-valuation to the last
position.

\textbf{Proof.} Delete a base of maximal height and apply (10.55) to the
remaining \(n-1\) independent bases. Exact cancellation gives

\[
W\ge\lambda\frac{c\mathrm e}{\rho}\frac{w_K}{q^u}
 \frac{(aP)^n}{P}\,nR_n\frac{M_nA_n}{t^2}\,d.
\tag{10.1067}
\]

Let \(P_0\) be respectively \(3,3,5,1,1,1,4\) in the seven field cases.
Then \(P\ge P_0\): the defining strict inequality for P supplies these
minima, including \(P\ge4\) at two. Let \(c_0\) be the lower bound for c
in the table below, and \(\sigma=2\) in V and one otherwise. Set
\(K_0=4\lambda c_0\sigma/(11\rho P_0)\). Equations (10.1065)--(10.1067),
together with \(\mathrm e<11/4\), imply

\[
W>K_0d\exp(a_0^{(2)}n),\qquad
a_0^{(2)}=2+\log(aP_0).
\tag{10.1068}
\]

The actual degree-one value is \(\rho=17\); the other rows use the
admissible upper bound58. The three numerical comparisons needed below
are finite, one for each row:

\[
\begin{aligned}
K_0a_0^{(2)}&>\exp(a_1^{(2)}),\\
K_0(2a_0^{(2)}+\log d_0+\log K_0)&>3\exp(a_1^{(2)}),\\
4a_0^{(2)}+2\log K_0&>6.
\end{aligned}
\tag{10.1069}
\]

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}
  >{\raggedright\arraybackslash}p{(\linewidth - 12\tabcolsep) * \real{0.1111}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 12\tabcolsep) * \real{0.1481}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 12\tabcolsep) * \real{0.1481}}
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\toprule\noalign{}
\begin{minipage}[b]{\linewidth}\raggedright
Case
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
\(c_0\)
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
\(\rho\)
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
\(P_0\)
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
\(d_0\)
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
\(\sigma\)
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
\(a_1^{(2)}\)
\end{minipage} \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
I.1 & 1438 & 58 & 3 & 2 & 1 & 1.94 \\
I.2 & 648 & 17 & 3 & 1 & 1 & 2.76 \\
II & 690 & 58 & 5 & 2 & 1 & 0.71 \\
III.1 & 495 & 58 & 1 & 2 & 1 & 1.99 \\
III.2 & 557 & 17 & 1 & 1 & 1 & 3.32 \\
IV & 2418 & 58 & 1 & 2 & 1 & 3.58 \\
V & 406 & 58 & 4 & 2 & 2 & 1.48 \\
\end{longtable}
}

Here is a reproducible rational verification, with no rounded decimal
test used as an inference. For each logarithm reduce its positive
argument to \([1,2]\), evaluate the first32 terms of
\(2\sum_{k\ge0}z^{2k+1}/(2k+1)\), and enclose the tail by
\(2z^{65}/(65(1-z^2))\), where \(z=(x-1)/(x+1)\). Bound the exponential
at \(a_1^{(2)}\) from above by the sum through degree64 plus its
geometric tail. These are the enclosures proved in Lemma10.153. The
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_second_application_completion.py}{exact
calculation} records positive rational lower margins for all21
comparisons. It also verifies \(2a_0^{(2)}+\log d_0+\log K_0>1\).

The first two inequalities of (10.1069) give, by a positive linear slope
in n and monotonicity in d,
\(K_0(a_0^{(2)}n+\log d+\log K_0)>(n+1)\exp(a_1^{(2)})\) for every
\(n\ge2,d\ge d_0\). Thus (10.1068) implies
\(\log W+\log\log W>G_2(n,d)/(n+1)\). The large-coefficient assumption
gives \(B>W\log B>W\log W\), because \(\log B>1\), \(W>1\), and the
logarithm increases. It follows that \((n+1)\log B>G_2(n,d)\).

The singleton height bound used in (10.466) gives
\(R_{\max}/(dW(d))\le\rho\mathrm e d^2B\). Also \(\log\rho+1<6\). The
third inequality in (10.1069) implies \(n\log W>2\log d+6\): at n=2 it
is precisely that inequality, and
\(a_0^{(2)}n^2+n\log K_0+(n-2)\log d-6\) increases for \(n\ge2,d\ge1\),
since \(a_0^{(2)}>3\) and \(\log K_0>-1\). Hence the logarithmic ratio
entry is below \((n+1)\log B\). The entries \(\log B_\dagger\) and
\((n+1)t\) are bounded by \((n+1)H\) directly. This proves the lemma.
\(\square\)

\textbf{Theorem 10.234 (independent second application estimate).} For
every independent local-unit list of length \(n\ge2\),
\(\operatorname{ord}_{\mathfrak p}(\Xi-1)<\mathcal T_n\). The singleton
case has the stronger bound \(\mathcal T_1/4000\).

\textbf{Proof.} At length one this is Theorem10.161. For \(n\ge2\), the
full height-product bound (10.55) and (10.1065) give

\[
\frac{t\mathcal T_n}{dH}
 >\frac c\rho\frac{w_K}{q^u}a^nP^{n-1}
       \frac{n^n(n+1)^{n+2}}{(n!)^2}>7900.
\tag{10.1070}
\]

For the last strict inequality the rank factor is at least72, as proved
in (10.464); use \(c\ge200,a\ge7,\rho\le58\). In particular
\(\mathcal T_n>2100(d/t)\log2\). The local height inequality gives
\(\operatorname{ord}_{\mathfrak p}(\Xi-1)
 \le(d/t)(nB\max_i h(a_i)+\log2)\). If \(B/\log B\le W\), its first term
is at most \(\lambda\mathcal T_n\) and its second is strictly less than
\(\mathcal T_n/2100\). Otherwise Lemma10.233 supplies every coefficient
condition, and Theorem10.223 gives the asserted bound, since
\(C_n=(n+1)C_2\). \(\square\)

\subsubsection{Uniform budgets for removing a dependent
base}\label{uniform-budgets-for-removing-a-dependent-base}

\textbf{Lemma 10.235 (second weighted rank budgets).} Let \(2\le r<n\),
and define \(W_n=\lambda tC_n\Omega_n/(ndh(a_n))\). Then

\[
\mathcal T_n>7900\frac d t\log2,\qquad
W_n>200\frac{w_K}{q^u}d\mathrm e^n>\max\{d\mathrm e^n,w_K\}.
\tag{10.1071}
\]

If \(B/\log B>W_n\) and the first base is not torsion, let x be the
first discarded index of the greedy procedure. There is an exact circuit
relation, with positive coefficient at x, whose coefficients have
absolute value at most K, where

\[
K<\begin{cases}B/8,&x<n,\\BdH_n(a_x)/8,&x=n.\end{cases}
\tag{10.1072}
\]

If the list after removal has length \(n-1\) and modified height product
\(\Omega'\), with the same greedy basis, then

\[
\frac{C_n\Omega_n}{C_{n-1}\Omega'}
 >a\mathrm e^2Pt\frac{dH_n(a_x)}{X_n}.
\tag{10.1073}
\]

\textbf{Proof.} The height-product bound (10.55), including the
empty-list case proved in Lemma10.155, is

\[
\prod_{i=1}^j h(\gamma_i)
\ge\frac{w_K}{\rho d^{j+1}\ell_d}
       \frac{j^j}{j!\mathrm e^j}
\tag{10.1074}
\]

for an independent list of length j, with the last ratio one at j=0.
Flooring the other heights and applying (10.1065) with \(j=n-r+1\),
using \(A_n\ge t\), gives

\[
\frac{t\mathcal T_n}{dH}
 >\frac c\rho\frac{w_K}{q^u}
 a^r\left(\frac{a\mathrm e^2}{\varkappa}\right)^{n-r}
 P^{r-1}\frac{(n+1)^{n+2}}{n!}\frac{r^r}{r!}.
\tag{10.1075}
\]

The positive exponential series gives \(\mathrm e>1957/720\), its sum
through degree six. Exact rational squaring shows
\(a\mathrm e^2/\varkappa>2\) in both prime cases. Also
\((n+1)^{n+2}/n!\) increases with n and is at least81/2 for \(n\ge2\),
and \(r^r/r!\ge2\) for \(r\ge2\). The right side of (10.1075) is
therefore greater than \(200\cdot7^2\cdot81/58=396900/29>7900\). Since
\(H>1>\log2\), this proves the first assertion.

For W, remove \(a_n\) from its product. If \(a_n\) is selected, put k=r
and apply (10.1074) to the remaining k-1 selected bases. Otherwise put
k=r+1, use \(H_n(a_n)/h(a_n)\ge1\), and apply it to the same k-1
selected bases. In both cases \(1\le k\le n\), and direct substitution
gives

\[
\begin{aligned}
W_n\ge{}&\lambda\frac c\rho\frac{w_K}{q^u}dA_n
 \frac{a^n\mathrm e^{n-k+1}(n+1)^{n+2}}{n!}
 \frac{(k-1)^{k-1}}{(k-1)!}\\
&\quad\times\frac{P^{k-1}X_n^{n-k}}{\varkappa^{n-k}}
 \frac{M_n}{t^{n-k+2}}.
\end{aligned}
\tag{10.1076}
\]

Use (10.1065), \(X_n\ge n\), and \(n!\le n^{n+1}\mathrm e^{1-n}\), the
latter obtained by bounding \(\sum_{j<n}\log j\) by
\(\int_1^n\log x\,dx\) and then multiplying by n.~Drop the factors
\(P^{k-1}\ge1\) and \((k-1)^{k-1}/(k-1)!\ge1\). The result is

\[
\frac{W_n}{d\mathrm e^n}
 >\lambda\frac{c\mathrm e}{\rho}\frac{w_K}{q^u}
 \frac{A_n}{n}\,a^k
       \left(\frac{a\mathrm e^2}{\varkappa}\right)^{n-k}
 >\frac{67168}{261}\frac{w_K}{q^u}.
\tag{10.1077}
\]

Here \(2^{n-1}\ge n\), \(a\ge7\), \(A_n\ge4\), \(c\ge200\),
\(\rho\le58\), and \(\mathrm e>8/3\). The last rational number
exceeds200. Since \(q^u\le2d\), both remaining assertions of (10.1071)
follow.

For the circuit, choose v earlier selected bases forming a minimal
dependent set with \(a_x\), where \(1\le v\le r\). Theorem10.20 supplies
an exact relation. All earlier heights are at most \(h(a_x)\), and its
coefficient bound is

\[
K\le\rho d^{v+1}\ell_d\frac{v!\mathrm e^v}{v^v}
 \frac{h(a_x)\prod_{a_i\in S}h(a_i)}{\min_{a_i\in S}h(a_i)}.
\tag{10.1078}
\]

Set \(j=r-v+1\) and \(s=n-r-1\ge0\). The selected bases outside S,
together with its smallest-height member, form an independent list of
length j. Apply (10.1074). Then

\[
\frac K{\Omega_n}\le
\frac{\rho^2d^{r+3}\ell_d^2}{w_K}
 \mathrm e^{r+1}\frac{v!j!}{v^vj^j}
 \frac{h(a_x)}{H_n(a_x)}L_n^{-s}.
\tag{10.1079}
\]

We give the full cancellation needed to normalize this bound. At x=n,
divide by \(BdH_n(a_x)/8\), use \(B>W_n\log B\), \(\log B>1\), and
substitute (10.1064). Use \(q^u/w_K\le1\), \(\ell_d/A_n\le1\),
\(h(a_x)/H_n(a_x)\le1\), and \(v!j!/(v^vj^j)\le1\). The normalized ratio
is at most

\[
\frac{8n\rho^2}{\lambda c a^{r+1}}
\frac{(n-1)!}{(n+1)^{n+2}}
\left(\frac{\varkappa}{a\mathrm e}\right)^s
\frac{t^{s+2}}{P^rX_n^sM_n}.
\tag{10.1080}
\]

The degree powers cancel because \(r+3+s=n+2\). If x\textless n and
\(a_n\) is selected, cancel its height before applying (10.1074), using
length j-1 instead of j. The same envelope has an extra factor
\(1/\mathrm e\), including j=1 by its empty-list case. If x\textless n
and \(a_n\) is not selected, cancel its height against \(H_n(a_n)\); now
s is at least one. The same expression has \(\varkappa^s/P^r\) replaced
by \(\varkappa^{s-1}/P^{r+1}\), and \(t^{s+2}/X_n^s\) by
\(t^{s+1}/X_n^{s-1}\). These cancellations cover every position of the
maximal-height base.

In (10.1080), (10.1065) and \(X_n\ge n\) give
\(t^{s+2}/(X_n^sM_n)<n^2\). They give the same bound for the altered
expression, with exponent s+1. Use \((n-1)!\le n^n\mathrm e^{1-n}\), and
\(\varkappa/(a\mathrm e^2)<1/2\). All three cases are bounded by

\[
\frac{8\rho^2n}{\lambda c a^{r+1}\mathrm e^r}\,2^{-s}
\le\frac{24\cdot58^2}
 {\lambda\cdot200\cdot7^3(8/3)^2}
=\frac{68121}{411404}<\frac15<1.
\tag{10.1081}
\]

For the middle inequality the maximum occurs at r=2,s=0,n=3: increasing
s multiplies \(n2^{-s}\) by \((n+1)/(2n)<1\), and increasing r
multiplies \(n/(a^{r+1}\mathrm e^r)\) by \((n+1)/(na\mathrm e)<1\). This
proves (10.1072) uniformly, not just at the sampled ranks.

Finally the two entries of \(M_n\) have ratios one and
\(\mathrm e t(n-1)^{n-1}/n^n>t/n\) compared with \(M_{n-1}\). Their
maxima therefore have ratio at least \(t/X_n\). Also \(A_n\ge A_{n-1}\),
giving

\[
\frac{C_n}{C_{n-1}}
\ge a\mathrm ePd\frac{t}{X_n}
\frac{(n+1)^{n+2}}{(n-1)n^{n+1}}.
\tag{10.1082}
\]

The last factor exceeds \(\mathrm e\), since its logarithm is
\((n+2)\log(1+1/n)-\log(1-1/n)>1+5/(2n)-1/n^2>1\). This uses
\(\log(1+y)\ge y-y^2/2\) and \(-\log(1-y)>y\), proved by integration.
The floor \(L_n\) is nondecreasing in n, so
\(\Omega_n/\Omega'\ge H_n(a_x)\). These prove (10.1073). \(\square\)

\textbf{Theorem 10.236 (the full second application estimate).} For
every indexed local-unit list with positive multiplicative rank,

\[
\operatorname{ord}_{\mathfrak p}(\Xi-1)<C_n\Omega_nH.
\tag{10.1083}
\]

At rank one, the right side may be divided by4000. The constant, floor
and greedy basis are exactly those in (10.1064).

\textbf{Proof.} Rank one, including all torsion patterns, is
Theorem10.163. Full rank is Theorem10.234. Induct on n in the remaining
cases \(2\le r<n\). The small-coefficient alternative
\(B/\log B\le W_n\) follows from the local height inequality and
(10.1071), with its two contributions respectively at most
\(\lambda\mathcal T_n\) and less than \(\mathcal T_n/2100\).

Otherwise \(B>W_n\log B>W_n\), so \(\log B>n+\log d\), \(B>w_K\), and
\(H\ge X_n\). If the first base is not torsion, use the exact circuit of
Lemma10.235. Raising \(\Xi\) to its positive x-th coefficient removes
\(a_x\) and leaves every selected base, hence the same basis and residue
index. Its new integer coefficients are bounded by2BK. If this power is
one, \(\Xi\) is a nontrivial root of unity and Theorem10.21 bounds its
valuation by \((d/t)\log2<\mathcal T_n\).

Otherwise put \(z=dH_n(a_x)\). The bounds (10.1072) give coefficients at
most \(B^2/4\) if x\textless n, and at most \(B^2z/4\) if x=n. As
\(\log u\le u/\mathrm e\) for positive u, a valid new bound is
\(B'=B^2\exp(z/(4\mathrm e))\). Consequently

\[
H'\le H\left(2+\frac{z}{4\mathrm eX_n}\right).
\tag{10.1084}
\]

The ratio (10.1073) is strictly larger than this increase. Indeed
\(z/X_n\ge1/(\varkappa Pt)\), and

\[
a\mathrm e^2-\frac1{4\mathrm ePt}>2\varkappa.
\tag{10.1085}
\]

At an odd prime, \(Pt>1\) and the strict margin is at least
\(7(1957/720)^2-50-3/32=840343/518400>0\). At two, the primitive cube
root forces \(f\ge2\); hence \(Pt\ge4\log4>16/3\). The strict margin
there is at least \(13(1957/720)^2-96-9/512=25049/1036800>0\). Thus
\((a\mathrm e^2Pt-1/(4\mathrm e))z/X_n>2\). Apply the inductive bound to
the shorter list and the elementary factored-power inequality to return
to \(\Xi\). Equations (10.1073)--(10.1085) prove (10.1083).

If the first base is torsion, raise \(\Xi\) to its order \(w\le w_K<B\),
deleting that base. Again a resulting power equal to one is handled by
the torsion bound. Otherwise its coefficients are below \(B^2\), so
\(H'\le2H\). The greedy basis is unchanged. In (10.1073) its deleted
modified height is \(L_n\), so the ratio is greater than
\(a\mathrm e^2/\varkappa>2\). Induction and the same factored-power
inequality finish this case. Every list of positive rank is now covered.
\(\square\)

This proves the second application estimate of Theorem2 in
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
freely accessible 2013 paper}, with its original modified heights and
residue index. The proof retains the stronger rank-one factor already
established in Theorem10.163. Rank zero is separate: a nontrivial
torsion product has valuation at most \((d/t)\log2\), by Theorem10.21.

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={The independent coefficient conversion and the dependent rank reduction}]{../figures/yu-second-application-completion.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.71. The left panel shows the independent-list
coefficient conversion. Its local-height alternative also applies to
dependent lists. The right panel shows the remaining large-coefficient
branch for dependent lists: a circuit or a torsion relation removes one
unselected base while preserving the chosen basis and its residue index.
Lemma10.235 proves that the ratio of successive full budgets exceeds the
complete increase in the coefficient parameter; Theorem10.236 returns
the bound to the original product. The displayed 2-adic margin is an
exact positive rational number, not a numerical sample. Rank one retains
its separate factor1/4000.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_second_application_completion.py}{Reproducible
figure and exact checks}. Human-source context: Theorem2 of
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
freely accessible2013 paper}.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{79. The number-field coefficient refinement for independent
normalized
units}\label{the-number-field-coefficient-refinement-for-independent-normalized-units}

This section uses the absolute logarithmic height and the prime-ideal
valuation. Write \(d=[K:\mathbb Q]\), \(e=e_{\mathfrak p}\),
\(f=f_{\mathfrak p}\), \(N=p^f\), and \(t=\log N\). The symbol
\(\mathrm e\) denotes the exponential constant, not the ramification
index. At two, assume in this section that \(\zeta_3\in K\). The
remaining two-adic fields and dependent normalized lists are not covered
by the theorem below.

For \(n\ge2\), define

\[
\begin{aligned}
Q&=16\mathrm e^2d^2,&D_n&=(n+1)(6n+5),\\
\mathcal A_n&=(16\mathrm e d)^{2(n+1)}n^{3/2}
 \log(2nd)\log(2d)\frac{e^nN}{t^2},\\
c_3(n,d)&=(2d)^{2n+1}\log(2d)\log^3(3d),\\
c_4(n,d)&=2\exp(D_n)d^{3n}\log(2d),&J&=e+d/f.
\end{aligned}
\tag{10.1086}
\]

The target is the number-field estimate stated in Theorem B of the
\href{https://math.leidenuniv.nl/reports/files/2008-16.pdf}{free author
paper by Bérczes, Evertse and Győry}. Here it is proved for the
explicitly specified class of inputs by reducing to Theorem 10.223. The
external statement supplies no proof step.

\textbf{Lemma 10.237 (normalization inside the original number field).}
For \(\alpha_i\in K^*\), put

\[
v_i=\operatorname{ord}_{\mathfrak p}(\alpha_i),\qquad
\beta_i=\alpha_i^e p^{-v_i}.
\tag{10.1087}
\]

Each \(\beta_i\) is a local unit in K and
\(h(\beta_i)\le Jh(\alpha_i)\). For integers \(b_i\), put
\(\Xi=\prod_i\alpha_i^{b_i}\) and \(s=\sum_i b_iv_i\). If \(s\ne0\),
then \(\operatorname{ord}_{\mathfrak p}(\Xi-1)=\min\{s,0\}\). If
\(s=0\), then

\[
\prod_i\beta_i^{b_i}=\Xi^e.
\tag{10.1088}
\]

If this last product is not one, its valuation after subtracting one is
at least \(\operatorname{ord}_{\mathfrak p}(\Xi-1)\).

\textbf{Proof.} The normalization
\(\operatorname{ord}_{\mathfrak p}(p)=e\) gives
\(\operatorname{ord}_{\mathfrak p}(\beta_i)=ev_i-ev_i=0\). The local
contribution to the height of \(\alpha_i\) or its inverse gives
\(|v_i|f\log p\le d h(\alpha_i)\), as in the proof of Theorem 10.21.
Hence the product and power inequalities for height give \[
h(\beta_i)\le e h(\alpha_i)+|v_i|\log p
\le (e+d/f)h(\alpha_i).
\] The unequal-valuation rule gives the assertion for \(s\ne0\).
Multiplication proves (10.1088) when s vanishes. In that case \(\Xi\) is
a unit. The identity \(\Xi^e-1=(\Xi-1)(1+\Xi+\cdots+\Xi^{e-1})\) has an
integral second factor, proving the final assertion. No root extraction,
field extension, or change of valuation normalization is involved.
\(\square\)

\textbf{Lemma 10.238 (absorbing the normalization cost).} For every
independent list of local units in the field cases of Theorem 10.152,
its second constant satisfies

\[
2C_2J^n<\mathcal A_n\qquad(n\ge2).
\tag{10.1089}
\]

\textbf{Proof.} Retain the actual saturated residue index in C2. It is
at least one. Formula (10.1027) gives \[
\max\{N/\delta(\boldsymbol\beta),
       \mathrm e^nt^{n+1}/n^n\}<(n+1)N.
\] The factorial bound (10.1026), and \(J\le2d/f\), therefore give, with
\(a=a^{(2)}\), \(c=c^{(2)}\), and \(P=p^\kappa\),

\[
\frac{2C_2J^n}{\mathcal A_n}
<\frac{c}{128\mathrm e q^u}
 \left(\frac a{128f}\right)^n
 \left(\frac Pe\right)^{n-1}
 \frac{(n+1)^3}{n^{3/2}}
 \frac{\ell_d A_n}{e t\log(2nd)\log(2d)}.
\tag{10.1090}
\]

Suppose first that \(d\ge2\). Then \(\ell_d\le\log(2d)\), and
\(\log(2nd)\ge\log8>2\). Also \(\log((n+1)d)\le\log(2nd)\). The three
entries of \(A_n=\max\{4+\log((n+1)d),e,t\}\) consequently show that

\[
\frac{A_n}{e t\log(2nd)}<
\begin{cases}3,&p\text{ odd},\\9/4,&p=2,\ \zeta_3\in K.
\end{cases}
\tag{10.1091}
\]

For the first entry use \((4+\log(2nd))/\log(2nd)<3\). For the other two
entries use respectively \(1/(t\log(2nd))\) and \(1/(e\log(2nd))\). At
odd primes \(t>1\). At two, injectivity of reduction on the prime-to-two
roots of unity gives \(3\mid2^f-1\), hence \(f\ge2\) and
\(t\ge\log4>4/3\). These verify both rows.

At odd primes \(P/e\le3\), \(q^u\ge2\), \(c\le2418\), and a=7. At two
\(P/e\le4\), \(q^u\ge3\), c=406, and a=13. The bounds on P/e follow
directly from (10.641). Using \(\mathrm e>8/3\), (10.1090) is therefore
bounded by the first two rows below. For \(d=1\), the admissible fields
here have odd p; e=f=1 and J=2. At three we have P=3,c=648; at \(p\ge5\)
we have P=1 and \(c\le2228/3\). Since \(\log(2n)>4/3\), \(\log2>2/3\),
and \(A_n/t\le n+4\) at three and \(A_n/t\le2(n+4)/3\) at primes at
least five, (10.1090) gives the last two rows. Here
\(F_n=(n+1)^3/n^{3/2}\).

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}ll@{}}
\toprule\noalign{}
Field range & Upper bound for \(2C_2J^n/\mathcal A_n\) \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
\(d\ge2\), odd p & \((3627/1024)(21/128)^nF_n\) \\
\(d\ge2\), p=2 & \((1827/8192)(13/64)^nF_n\) \\
d=1, p=3 & \((729/2048)(21/128)^n(n+4)F_n\) \\
d=1, \(p\ge5\) & \((1671/2048)(7/128)^n(n+4)F_n\) \\
\end{longtable}
}

These bounds hold for every integer n at least two. To verify that they
are below one uniformly, use \(F_2=27/(2\sqrt2)<135/14\), since
\(\sqrt2>7/5\). The four resulting rational endpoints are below one.
Their successive ratios are at most, respectively,

\[
\frac7{18},\qquad \frac{13}{27},\qquad
\frac{49}{108},\qquad\frac{49}{324}.
\tag{10.1092}
\]

Indeed \(F_{n+1}/F_n<((n+2)/(n+1))^3\le64/27\), and the extra factor in
the last two rows has ratio \((n+5)/(n+4)\le7/6\). Thus every envelope
decreases. The exact endpoint fractions are recorded by the accompanying
calculation; the argument here covers the infinite rank range.
\(\square\)

\textbf{Theorem 10.239 (the number-field coefficient refinement for
independent normalized units).} Let \(n\ge2\), \(\alpha_i\in K^*\), and
choose real weights and coefficient bounds such that

\[
\begin{gathered}
w_i\ge\max\{h(\alpha_i),Q^{-1}\},\qquad
b_i\in\mathbb Z,\quad b_n\ne0,\quad
v_p(b_n)=\min_i v_p(b_i),\\
B\ge\max_i|b_i|,\qquad B\ge B_n\ge|b_n|,
\qquad0<\varepsilon\le\tfrac12.
\end{gathered}
\tag{10.1093}
\]

Use \(v_p(0)=+\infty\). Assume that the units \(\beta_i\) of (10.1087)
are multiplicatively independent. Put

\[
V=\prod_{i<n}w_i,\qquad\Omega=Vw_n,\qquad r=B/B_n,
\qquad L=\log\left(B_nc_4(n,d)N^{n+1}V/\varepsilon\right).
\tag{10.1094}
\]

Then

\[
\operatorname{ord}_{\mathfrak p}(\Xi-1)
<\mathcal A_n
\max\left\{\Omega L,\frac{\varepsilon r}{c_3(n,d)}\right\}.
\tag{10.1095}
\]

In particular, choosing the smallest allowed weights gives the exact
height floor and constants in the stated number-field target, on this
class of inputs.

\textbf{Proof.} We first verify all scalar comparisons, including the
small height floor. The exponential series gives \(\log16<3\) and
\(\log52<4\). For the second assertion one may use
\(\mathrm e>1957/720>19/7\) and \((19/7)^4>52\). Since
\(\log Q<5+2\log d\) and \(2\log(2d)>1\),

\[
\log(c_4/Q^n)>D_n-5n+n\log d>0.
\tag{10.1096}
\]

For every x\textgreater0, maximizing \((\log x)/x^{1/3}\) gives
\(\log x\le3x^{1/3}/\mathrm e\). At x=3d this implies
\(\log^3(3d)\le81d/\mathrm e^3<5d\). Consequently \[
\frac{c_3}{c_4}=4^n\exp(-D_n)d^{1-n}\log^3(3d)
<5\,4^n\exp(-D_n)<1.
\] The last step uses \(\mathrm e>2\) and \(D_n>2n+3\). Using
\(V\ge Q^{1-n}\), \(B_n\ge1\), and \(\varepsilon^{-1}\ge2\), we also
obtain

\[
L>\log B_n+(n+1)t+D_n-5(n-1)+(n+2)\log d>1.
\tag{10.1097}
\]

Here the discarded constant is \(\log(4\log(2d))>0\). All quantities in
the claimed bound are therefore positive.

If \(\sum b_iv_i\ne0\), Lemma 10.237 proves the conclusion immediately.
Otherwise \(\Xi\) is a unit and \(\prod\beta_i^{b_i}=\Xi^e\ne1\), by
independence and \(b_n\ne0\). Enlarge only the last unnormalized weight:

\[
w_n^*=\max\left\{w_n,
 \frac{\varepsilon r}{c_3 V L}\right\}.
\tag{10.1098}
\]

Use weights \(Jw_j\) for \(j<n\) and \(Jw_n^*\) for the independent
units. Lemma 10.237 verifies these height bounds. Put
\(\omega(d)=1/(dW(d))\). Formula (10.222) implies \(\omega(d)<1\): at
d=1, \(\omega(1)=\log2\log3/\log6<1\); at d at least two,
\(d\log^3(3d)>1\). Since \(J\ge1\), every coefficient entry in Lemma
10.224 is at most the logarithm of

\[
B_n\left(Q+\frac{c_3VL}{\varepsilon}\right)
<\exp(L)(1+L)<\exp(2L).
\tag{10.1099}
\]

For the first strict inequality, (10.1096) and \(Q^{n-1}V\ge1\) give
\(\exp L>B_nQ\); \(c_4>c_3\) gives \(\exp L>B_nc_3V/\varepsilon\). The
second is the positive-series inequality \(1+L<\exp L\).

Also \(\log B_\dagger\le\log B_n<L\), and \((n+1)t<L\). Every row of
(10.643) has \(a_0^{(2)}<6\) and \(a_1^{(2)}<4\), so \[
G_2(n,d)<(n+1)(7n+4+\log d)<2L.
\] To check the final inequality, subtract this upper bound from twice
the non-t term of (10.1097). The difference is
\(5n^2+n+16+(n+3)\log d>0\). Thus the entire padded coefficient maximum
is less than 2L.

Lemma 10.224, Theorem 10.223, and Lemma 10.237 now give

\[
\operatorname{ord}_{\mathfrak p}(\Xi-1)
\le\operatorname{ord}_{\mathfrak p}(\Xi^e-1)
<2C_2J^nVLw_n^*
=2C_2J^n\max\{\Omega L,\varepsilon r/c_3\}.
\tag{10.1100}
\]

Lemma 10.238 proves (10.1095). The original distinguished coefficient,
field degree, residue degree and prime-ideal valuation have all been
retained. \(\square\)

\subsubsection{An exact quadratic
example}\label{an-exact-quadratic-example}

Let \(K=\mathbb Q(\sqrt2)\), p=7, and choose the place with
\(\sqrt2\equiv3\pmod7\). The simple root 3 of \(X^2-2\) modulo seven
lifts to \(\mathbb Q_7\), by the proved lifting lemma, so this place has
e=f=1. Take \[
\alpha_1=7(1+\sqrt2),\qquad
\alpha_2=(6+\sqrt2)/7,\qquad b_1=b_2=1.
\] Their valuations are 1 and -1. The normalized units are
\(\beta_1=1+\sqrt2\), \(\beta_2=6+\sqrt2\), and J=3. Multiplication
gives \(\Xi=8+7\sqrt2\) and \(\Xi-1=7(1+\sqrt2)\), whose valuation is
exactly one. They are independent: conjugation sends their products to
norms -1 and 34. A torsion relation would force the exponent of the
second base to vanish; the real value \(1+\sqrt2>1\) then forces the
first exponent to vanish.

The primitive minimal polynomials of the original inputs are
\(X^2-14X-49\) and \(49X^2-84X+34\). The height formula therefore gives
\[
h(\alpha_1)=\log7,\qquad
h(\alpha_2)=\tfrac12\log(7(6+\sqrt2)),\qquad
h(\beta_1)=\tfrac12\log(1+\sqrt2),\qquad
h(\beta_2)=\tfrac12\log34.
\] For the first polynomial both roots have absolute value greater than
one; for the second only \((6+\sqrt2)/7\) does. These facts verify the
displayed heights directly. With B=Bn=1 and \(\varepsilon=1/2\), the
theorem applies although neither original input is a local unit.

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={The constant comparison and an exact quadratic normalization}]{../figures/yu-number-field-refinement.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.72. The left panel samples the four upper envelopes
proved in Lemma 10.238; the endpoint and successive-ratio argument
covers every rank, beyond the displayed samples. The two-adic row
assumes that the field contains a primitive cube root of unity. The
right panel shows the exact original-field normalization of Lemma 10.237
and the quadratic example: both original inputs are nonunits, their
valuations cancel, and the product minus one has valuation one. Theorem
10.239 retains the distinguished coefficient and absorbs the height cost
J. \href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_number_field_refinement.py}{Reproducible
diagram and exact rational checks}. Human-source target: Theorem B of
\href{https://math.leidenuniv.nl/reports/files/2008-16.pdf}{Bérczes,
Evertse and Győry}.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{80. The number-field coefficient refinement for dependent
inputs}\label{the-number-field-coefficient-refinement-for-dependent-inputs}

We retain \(K,d,e,f,N,t,Q,J,\mathcal A_n,c_3,c_4,D_n\) from Section 79.
Thus p is odd, or p=2 and \(\zeta_3\in K\). We first work with arbitrary
local units \(\beta_i\), with positive weights satisfying
\(h(\beta_i)\le Jw_i\) and \(w_i\ge Q^{-1}\). The units may be dependent
or torsion. Define

\[
u_i=Qw_i\ge1,\qquad
R=\prod_{i<n}u_i=Q^{n-1}V,\qquad
S=Ru_n=Q^n\Omega,\qquad A=58d\ell_d.
\tag{10.1101}
\]

The scalar \(J=e+d/f\) remains fixed when an index is deleted. It is
distinct from the saturation index used below. The bounds
\(B,B_n,\varepsilon\) and the quantities \(V,\Omega,r,L\) have their
meanings in (10.1093)--(10.1094). Put

\[
U=\max\{\Omega L,\varepsilon r/c_3(n,d)\},\qquad
\widehat U=Q^nU.
\tag{10.1102}
\]

In particular \(\widehat U\ge SL>1\). All inequalities for L in Theorem
10.239 depend only on the weights and coefficient bounds, so remain
valid here without independence.

\textbf{Lemma 10.240 (weighted relations and common denominators).} The
following assertions hold for these local units and weights.

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\item
  If the first n-1 bases are dependent, they admit a nonzero exact
  integer relation \(\prod_{i<n}\beta_i^{k_i}=1\), with
  \(|k_i|\le\kappa:=AR\).
\item
  Suppose that all bases lie, modulo roots of unity, in the rational
  span of an independent sublist of the first n-1 bases. Let I be the
  indices of this sublist, of length \(s\ge1\), and put
  \(W_u=\prod_{j\in I}u_j\), \(W=Q^{-s}W_u\). For
  \(Y=\prod_i\beta_i^{b_i}\), there are a positive integer D and
  integers \(M_j\), j in I, such that
\end{enumerate}

\[
Y^D=\prod_{j\in I}\beta_j^{M_j},\qquad
|M_j|\le B^*:=nA^2BSR^2,\qquad W_u\le R.
\tag{10.1103}
\]

\textbf{Proof.} The empty-list height bound in Lemma 10.155 gives
\(w_K\le58d\ell_d=A\). For a minimal circuit of rank a at least one,
Corollary 10.20 bounds each exact relation coefficient by \[
58d^{a+1}\ell_d\frac{a!\mathrm e^a}{a^a}
\left(\frac JQ\right)^a\prod u_j
\le A\left(\frac{\mathrm e dJ}{Q}\right)^a\prod u_j
\le A\prod u_j.
\] The product omits the coefficient's own index. We used \(a!\le a^a\)
and \(\mathrm e dJ/Q\le1/(8\mathrm e f)<1\). Every missing u-factor can
be inserted since it is at least one. A singleton torsion circuit
instead uses its order, at most \(w_K\). Applying these alternatives to
a minimal dependence among the first n-1 indices proves (a).

For (b), write \(\gamma_j=\beta_j\) for j in I. Each base has unique
rational coordinates \(x_{ji}\) in this basis modulo torsion.
Concretely, an integral multiple of its coordinate vector expresses an
integral power of \(\beta_i\) as a product of powers of the
\(\gamma_j\), up to a root of unity. Raising once more by \(w_K\)
removes that root. Thus the coordinate vector belongs to the lattice
\(M_K(\boldsymbol\gamma)\) of Theorem 10.18. Let
\(m=[M_K:\mathbb Z^s]\). The order of each class divides m, by
partitioning the finite group into cosets of its cyclic subgroup. Hence
\(mx_{ji}\in\mathbb Z\). The saturation bound (10.55), with the same
calculation as above, gives

\[
w_Km\le A W_u.
\tag{10.1104}
\]

The quotient \(\beta_i^m/\prod_j\gamma_j^{mx_{ji}}\) lies in K and has a
positive power equal to one. Its order therefore divides \(w_K\). Taking
\(D=w_Km\) and \(M_j=D\sum_i x_{ji}b_i\) gives the exact identity in
(10.1103).

We also need an individual coordinate bound. If \(\beta_i\) is torsion
its coordinates vanish. Otherwise its nonzero coordinates select a
minimal circuit consisting of that base and the corresponding basis
members. The coefficient at \(\beta_i\) is a nonzero integer, so has
absolute value at least one. Dividing the simultaneous bounds of
Corollary 10.20 by this coefficient gives

\[
|x_{ji}|\le A u_iW_u/u_j.
\tag{10.1105}
\]

For a base already in I its coordinates are Kronecker symbols and the
same bound holds. Consequently \[
|M_j|\le A^2B(W_u^2/u_j)\sum_i u_i
\le nA^2BSR^2.
\] Here \(u_i\le S\), \(u_j\ge1\), and \(W_u\le R\). All denominators
and roots of unity have been removed in the original field. \(\square\)

\textbf{Lemma 10.241 (logarithmic budgets and the elementary cases).}
For n at least two the preceding data satisfy

\[
\log R<L,\quad \log(2Q)<L,\quad
\log(2AB_n)<L,\quad
\log\kappa<L-(D_n-5n)-n\log d.
\tag{10.1106}
\]

For the basis in Lemma 10.240(b), put \(F=Q^{n-s}U\). Then

\[
W\log(2QB^*)<8F,\qquad 2WL\le2F.
\tag{10.1107}
\]

Finally, suppose Y is not one. The desired estimate
\(\operatorname{ord}_{\mathfrak p}(Y-1)<\mathcal A_nU\) holds if Y is
torsion; if the list has rank one and its distinguished last base lies
in the span of the others; or if all preceding bases are torsion.

\textbf{Proof.} We have \(\ell_d\le d\), \(\log58<5\), and
\(\log Q<5+2\log d\). Hence \(\log\kappa<5n+2n\log d+\log V\). On the
other hand, the definition gives \(L>D_n+3n\log d+\log V\), since
\(B_n/\varepsilon\ge2\), \(2\log(2d)>1\), and N\textgreater1. This
proves the last inequality of (10.1106), and the first follows since
A\textgreater1. The explicit lower bound (10.1097), together with
\(\log(2Q)<6+2\log d\) and \(\log(2A)<6+2\log d\), proves the other two.

We first bound a scalar coefficient needed for (10.1107):

\[
\frac{nA^2c_3(n,d)}{Q c_4(n,d)}<1.
\tag{10.1108}
\]

Indeed, \(A^2/Q<30\ell_d^2\), and exact cancellation gives
\(c_3/c_4=4^n\exp(-D_n)d^{1-n}\log^3(3d)\). Since \(\ell_d\le\log(3d)\),
maximizing \((\log x)^5/x\) for x at least one yields \[
d^{1-n}\ell_d^2\log^3(3d)
\le\frac{\log^5(3d)}d
\le3(5/\mathrm e)^5
<3(15/8)^5<70.
\] Thus the left side of (10.1108) is less than \(2100n4^n\exp(-D_n)\).
Use \(2100<2^{12}\), \(n\le2^{n-1}\), \(\mathrm e>2\), and \(D_n>3n+11\)
for n at least two to prove (10.1108) for every n. Because
\(Q^nV\ge Q\), it follows that

\[
\log\left(\frac{nA^2c_3B_n}{\varepsilon Q^n}\right)<L.
\tag{10.1109}
\]

The definition of U implies \(r\le c_3U/\varepsilon\). Inserting this
into (10.1103) and writing S=Ru\_n gives

\[
\log B^*\le
\log\left(\frac{nA^2c_3B_n}{\varepsilon Q^n}\right)
+4\log R+\log u_n+\log(\widehat U/R).
\tag{10.1110}
\]

Now \(WL\le F\), since \(W_u\le R\le S\) and \(\widehat U\ge SL\). The
first and the next four logarithmic contributions after multiplication
by W are therefore less than 5F. Also
\(W\log u_n<Wu_n\le S/Q^s\le F/L<F\), and
\(W\log(\widehat U/R)<W\widehat U/R\le F\). These use \(\log x<x\) for
x\textgreater0. We obtain \(W\log B^*<7F\); (10.1106) adds the final F
needed in (10.1107).

For the elementary cases note the uniform lower bound

\[
\frac{\mathcal A_n}{Q^n}
>16^4d^2e^nN/t^2.
\tag{10.1111}
\]

It follows directly from (10.1086): the exact quotient has factor
\(16^{n+2}\mathrm e^2d^2n^{3/2}\log(2nd)\log(2d)\), and
\(\mathrm e^2n^{3/2}\log(2nd)\log(2d)>1\). For example \(n\ge2\),
\(\log(2nd)>4/3\), and \(\log(2d)>2/3\) suffice.

If Y is a nonidentity root of unity, the local height inequality gives
\(\operatorname{ord}_{\mathfrak p}(Y-1)\le(d/t)\log2\). It is below the
target by (10.1111) and \(Q^nU\ge L>1\), using t\textless N.

Suppose next that the list has rank one, that its distinguished base is
redundant, and that Y is not torsion. Lemma 10.240(b) gives
\(Y^D=\gamma^M\ne1\) for one selected preceding base \(\gamma\), with
\(|M|\le B^*\) and \(h(\gamma)\le JW\). Theorem 10.21, its residue-order
bound \(g<N\), and the positive power factorization imply \[
\operatorname{ord}_{\mathfrak p}(Y-1)
\le\frac d t\{\log(2B^*)+2eNJW\}.
\] By (10.1107), \(W\ge Q^{-1}\), and \(W\le Q^{n-1}U/L\), this is
strictly below

\[
\frac d t\left\{8Q^n+
 \frac{2eNJQ^{n-1}}{(n+1)t}\right\}U.
\tag{10.1112}
\]

We retained \(L>(n+1)t\) in the second term; dropping that factor would
lose the required logarithm in the target. Dividing (10.1112) by
\(\mathcal A_nU\), and using (10.1111), bounds its two terms
respectively by \(8/16^4\) and \(4/16^4\). The first uses t\textless N;
the second uses \(J\le2d/f\), Q\textgreater1, e,f at least one. Their
sum is \(3/16384<1\).

Finally suppose that all preceding bases are torsion and Y is not.
Raising to \(w_K\) gives \(Y^{w_K}=\beta_n^{w_Kb_n}\ne1\). Use Theorem
10.21, \(w_K\le A\), \(\log(2AB_n)<L\le Q^nU\), and
\(w_n\le Q^{n-1}U/L\). This gives (10.1112), with its first coefficient
8 replaced by 1, and proves the same conclusion. These alternatives also
cover rank zero. \(\square\)

\textbf{Theorem 10.242 (all normalized local-unit lists).} Let n be at
least two and let \(\beta_i\in K^*\) be arbitrary local units with
\(h(\beta_i)\le Jw_i\), \(w_i\ge Q^{-1}\). Use all the coefficient
hypotheses (10.1093). If \(Y=\prod_i\beta_i^{b_i}\ne1\), then

\[
\operatorname{ord}_{\mathfrak p}(Y-1)<\mathcal A_nU.
\tag{10.1113}
\]

\textbf{Proof.} Independent lists satisfy the result directly by the
padded height proof of Theorem 10.239: use the weights \(Jw_i\) and
\(Jw_n^*\) in Lemma 10.224, apply the scalar inequalities
(10.1096)--(10.1099), and conclude with (10.1089). This argument uses
only the displayed height upper bounds for the local units. Lemma 10.241
handles torsion products and the specified rank-one cases. These cases
establish the statement at n=2.

Induct on n.~Suppose n is at least three, the list is dependent, and Y
is not torsion. If the distinguished base lies in the span of the
preceding ones, choose a basis I among those indices. Its length \(s<n\)
is positive; s=1 was handled in Lemma 10.241, so take \(2\le s<n\).
Lemma 10.240(b) gives \(Y^D=\prod_{j\in I}\beta_j^{M_j}\ne1\). Renumber
the selected bases so that a nonzero \(M_j\) of minimum p-adic valuation
is last. The other coefficients may vanish.

Apply the independent estimate 10.223 to this selected list, with padded
weights \(Jw_j\). Its logarithmic coefficient entry is at most
\(\log(2QB^*)\), because \(w_j\ge Q^{-1}\), J is at least one, and
\(\omega(d)<1\). Its minimum absolute nonzero coefficient is at most B*.
The other two entries satisfy \(G_2(s,d)\le G_2(n,d)<2L\) and
\((s+1)t<2L\). Consequently (10.1107) yields

\[
\operatorname{ord}_{\mathfrak p}(Y-1)
<8C_2^{[s]}J^sQ^{n-s}U
<4\mathcal A_sQ^{n-s}U.
\tag{10.1114}
\]

The first step also uses the factorization of \(Y^D-1\), and the second
is Lemma 10.238 at rank s. Directly from (10.1086),

\[
\mathcal A_j/\mathcal A_{j-1}>16Qe\qquad(j\ge3).
\tag{10.1115}
\]

Since n-s is positive, (10.1114) is less than \(\mathcal A_nU\). This
resolves the redundant distinguished-base case without using induction
or assuming any coordinate relation is prime to p.

Otherwise the distinguished base raises the rank. If every preceding
base is torsion, Lemma 10.241 applies. In the remaining case, the first
n-1 bases are dependent: their rank is one less than the rank of the
whole dependent list. Choose a relation from Lemma 10.240(a), and choose
a nonzero coefficient \(k_x\), x\textless n, with minimum p-adic
valuation. Reverse all its signs if needed to make \(k_x>0\). Set
\(k_n=0\) and eliminate index x by defining

\[
b_i'=k_xb_i-k_i b_x\quad(i\ne x),\qquad b_n'=k_xb_n.
\tag{10.1116}
\]

Then \(\prod_{i\ne x}\beta_i^{b_i'}=Y^{k_x}\ne1\), and
\(|b_i'|\le2\kappa B\), where \(\kappa=AR\). Every new coefficient has
p-adic valuation at least \(v_p(k_x)+v_p(b_n)\), and the distinguished
coefficient has exactly that valuation and is nonzero. This follows from
the minimum-valuation choices of both vectors and the ultrametric
inequality; no coefficient is assumed to be prime to p.

Keep the surviving weights and choose

\[
\begin{aligned}
B'&=2\kappa B,& B_n'&=k_xB_n,\\
\varepsilon'&=\frac{k_x}{2\kappa}\varepsilon
       \frac{c_3(n-1,d)}{c_3(n,d)},&
\frac{\varepsilon'B'}{B_n'c_3(n-1,d)}
 &=\frac{\varepsilon B}{B_nc_3(n,d)}.
\end{aligned}
\tag{10.1117}
\]

These satisfy \(B'\ge B_n'\ge|b_n'|\), all coefficient bounds, and
\(0<\varepsilon'\le1/2\). Exact cancellation gives

\[
L'=L+\log(8\kappa d^2)-(12n+5)-3\log d-t-\log w_x<2L.
\tag{10.1118}
\]

Here \(c_3(n,d)/c_3(n-1,d)=4d^2\), \(D_n-D_{n-1}=12n+5\), and
\(c_4(n-1,d)/c_4(n,d)=\exp(-12n-5)d^{-3}\). For the final inequality use
\(w_x\ge Q^{-1}\), \(\log(8Q)<7+2\log d\), and (10.1106). The excess
above 2L is strictly less than \(-D_n-7n+2-(n-1)\log d-t<0\).

The new height product is \(\Omega' =\Omega/w_x\), so \(U'\le2QU\). The
induction hypothesis, the power factorization for local units, and
(10.1115) give

\[
\operatorname{ord}_{\mathfrak p}(Y-1)
\le\operatorname{ord}_{\mathfrak p}(Y^{k_x}-1)
<\mathcal A_{n-1}U'
\le2Q\mathcal A_{n-1}U<\mathcal A_nU.
\tag{10.1119}
\]

All dependence, torsion and zero-coefficient alternatives have now been
covered. \(\square\)

\textbf{Corollary 10.243 (arbitrary original inputs in the stated
fields).} Let \(\alpha_i\in K^*\) be arbitrary and take every
coefficient and height hypothesis in (10.1093). Assume
\(\Xi=\prod_i\alpha_i^{b_i}\ne1\). Then (10.1095) holds without an
independence or local-unit assumption on the original bases. The field
condition remains: p is odd, or p=2 and \(\zeta_3\in K\).

\textbf{Proof.} Normalize by Lemma 10.237. A nonzero total valuation
makes the left side nonpositive. At total valuation zero, the normalized
product is \(\Xi^e\). If \(\Xi\) is torsion, the elementary torsion
estimate in Lemma 10.241 applies to the original product, using the same
positive target and weights. Otherwise \(\Xi^e\ne1\), and Theorem 10.242
applies to the normalized units with \(h(\beta_i)\le Jw_i\). Factoring
\(\Xi^e-1\) gives the required inequality for \(\Xi-1\). Every field
datum, original weight and original distinguished coefficient is
retained. \(\square\)

\subsubsection{An internal relation in a quadratic
field}\label{an-internal-relation-in-a-quadratic-field}

Use \(K=\mathbb Q(\sqrt2)\) and the place above seven from Section 79.
Write \(a=1+\sqrt2\), \(c=6+\sqrt2\), and take the four local units
\((a,c,ac,3)\), with coefficients \((1,1,1,6)\). Their exact relation
\(a^{-1}c^{-1}(ac)=1\) deletes the third base and gives coefficients
\((2,2,6)\) on \((a,c,3)\). This remaining list is independent: applying
conjugation and multiplication gives norms -1,34,9, so the valuations at
the rational primes seventeen and three force the second and third
relation coefficients to vanish; the real value a\textgreater1 forces
the first to vanish. Thus the distinguished base raises the rank as
required in the internal case.

Take \(B=B_n=6\). Using the actual relation bound \(\kappa=1\), the
parameter transfer (10.1117) gives \(B'=12\), \(B_n'=6\), and
\(\varepsilon'=\varepsilon/32\), since d=2. The second entry of the
maximum is exactly unchanged. Lifting the chosen root of \(X^2-2\) to
108 modulo 343 gives \[
3^6(ac)^2-1\equiv245\pmod{343}.
\] It has valuation exactly two. Both the original and the reduced list
represent that same product. This example verifies a coefficient
transfer and its exact valuation; the proof for all inputs is the
induction above.

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={The two dependence reductions and an exact quadratic example}]{../figures/yu-number-field-dependent.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.73. The upper paths distinguish whether the last base
belongs to the rational span of the preceding bases modulo roots of
unity. The common-denominator path uses Lemmas 10.240--10.241; the
internal-relation path uses (10.1116)--(10.1119). The lower panel gives
an exact instance of the latter path in the quadratic field above seven.
The parameter transfer keeps the second term of the maximum unchanged,
and both expressions have valuation two. Theorem 10.242 proves the
bounds uniformly, including torsion and zero coefficients.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_number_field_dependent.py}{Reproducible
diagram and exact arithmetic checks}. Human-source target: Theorem B of
\href{https://math.leidenuniv.nl/reports/files/2008-16.pdf}{Bérczes,
Evertse and Győry}; the proof here uses the programme results identified
above.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{81. The coefficient refinement at two over every number
field}\label{the-coefficient-refinement-at-two-over-every-number-field}

Adjoining a cube root of unity doubles the global degree. At a prime
above two it can also square the residue-field cardinality. Applying a
bound with these larger parameters would not yet give the estimate with
the original field data. The saturated residue index supplies the needed
compensation. We retain it until the roots-of-unity factor has
cancelled.

Throughout this section, \(K\) does not contain \(\zeta_3\), and \[
L=K(\zeta_3),\qquad d=[K:\mathbb Q],\qquad
e=e_{\mathfrak p},\quad f=f_{\mathfrak p},\quad N=2^f,\quad t=\log N.
\] The exponential constant is \(\mathrm e\). All the constants
\(Q,J,\mathcal A_n,c_3,c_4\) retain the original degree and local data
of (10.1086).

\subsubsection{Cubic saturation and the residue
group}\label{cubic-saturation-and-the-residue-group}

\textbf{Lemma 10.244 (descent of cubic saturation).} Let
\(\gamma_1,\ldots,\gamma_s\) be independent local units in \(K\). Define
the two lattices \(M_3^K\) and \(M_3^L\) by (10.44), using roots in the
indicated field. Then \(M_3^K=M_3^L\). A saturated basis for the
calculation over \(L\) can consequently be represented by exact roots
\(\theta_j\) in \(K\).

At any prime \(\mathfrak P\) of \(L\) above \(\mathfrak p\), the
relative ramification index is one. Write \(f'=\sigma f\) for the
residue degree over \(\mathbb F_2\). Then \(\sigma=1\) for even \(f\),
and \(\sigma=2\) for odd \(f\). If \(f\) is odd and \(m=3^u\) is the
order of the group of 3-primary roots of unity in \(L\), there is a
divisor \(g\) of \(N-1\) such that \[
\delta_L=\frac{N^2-1}{mg},\qquad
\frac{N^2}{m\delta_L}=\frac{N^2g}{N^2-1}\le\frac{N^2}{N+1}<N.
\]

\textbf{Proof.} The global extension has degree two, since \(X^2+X+1\)
is irreducible over \(K\). Suppose \(\boldsymbol v\in M_3^L\). Choose
\(T=3^a\) such that \(T\boldsymbol v\) is integral and
\(\beta^T=A=\prod_i\gamma_i^{Tv_i}\) for some \(\beta\in L\). Let
\(z=\beta\overline\beta\in K\), where the bar is the nonidentity
\(K\)-automorphism of \(L\). Then \(z^T=A^2\). Because \(T\) is odd, the
element \[
\theta=z^{(T+1)/2}A^{-1}\in K,\qquad
\theta^T=A^{T+1}A^{-T}=A
\tag{10.1120}
\] proves that \(\boldsymbol v\in M_3^K\). The reverse inclusion is
immediate. Lemma 10.12 supplies a finite lattice basis, and this
argument supplies its roots in \(K\). By Lemma 10.15, root choices and
integral changes of basis do not change the residue group over \(L\):
their extra factors are powers of its 3-primary root of unity.

Reduction is injective on roots of odd order, as proved in Section 1. If
\(f\) is even, the cyclic group \(\mathbb F_N^*\) contains an element of
order three. Its unique prime-to-two root-of-unity lift from Section 1
has order three: its cube has residue one, so injectivity makes that
cube one. Thus the two roots of \(X^2+X+1\) already lie in
\(K_{\mathfrak p}\), and every completion \(L_{\mathfrak P}\) equals
\(K_{\mathfrak p}\).

If \(f\) is odd, \(3\nmid N-1\), so the reduction of this polynomial has
no root in \(\mathbb F_N\). Its root in the residue field of
\(L_{\mathfrak P}\) gives a residue extension of degree at least two.
The local extension degree is at most two. The elementary inequality
between local degree and ramification times residue degree therefore
forces residue degree two and ramification one. To recall the
inequality, take lifts of a residue-field basis and multiply them by
powers of a uniformizer in distinct valuation classes modulo the base
value group. A linear relation over the base field, after scaling its
coefficients to their least valuation, has a unique least valuation
class; reducing within that class contradicts independence of the
residue basis. These products are linearly independent, proving the
inequality. We have established \[
\begin{gathered}
{}[L:\mathbb Q]=2d,\qquad e_{\mathfrak P}=e,\qquad
f'=\begin{cases}f,&f\text{ even},\\2f,&f\text{ odd},\end{cases}\\
\operatorname{ord}_{\mathfrak P}|_K=\operatorname{ord}_{\mathfrak p}.
\end{gathered}
\tag{10.1121}
\]

Now assume \(f\) is odd. The reductions of the chosen \(\theta_j\)
belong to \(\mathbb F_N^*\); let their generated group have order
\(g\mid N-1\). Reduction embeds the group of \(m\)-th roots of unity
into \(\mathbb F_{N^2}^*\). Thus \(m\mid N^2-1\), and \(3\nmid N-1\)
gives \(m\mid N+1\). The two residue groups have coprime orders, so
their generated group has exactly \(mg\) elements. The definition
(10.49) now gives \[
m\delta_L=\frac{N^2-1}{g}\ge N+1,\qquad
\frac{N^2}{m\delta_L}\le\frac{N^2}{N+1}<N.
\tag{10.1122}
\] This proves the assertions. Only 3-primary saturation has been used;
no claim about full saturation in a quadratic extension is needed.
\(\square\)

\subsubsection{The constant with the original degree and residue
size}\label{the-constant-with-the-original-degree-and-residue-size}

\textbf{Lemma 10.245 (the quadratic transfer constant).} For an
independent list of \(n\ge2\) local units belonging to \(K\), let
\(C_2(L)\) be (10.642) calculated over \(L\), with the actual saturated
residue index of that list. Then \[
2C_2(L)J^n<\mathcal A_n(K).
\tag{10.1123}
\]

\textbf{Proof.} Put \(\ell'=\max\{1,\log(2d)\}\),
\(A'=\max\{4+\log(2(n+1)d),e,\sigma t\}\), and \[
\begin{gathered}
H=\max\left\{\frac{N^\sigma}{m\delta_L},
             \frac{\mathrm e^n(\sigma t)^{n+1}}{mn^n}\right\},\\
T_n=\frac{(n+1)^{n+1}}{(n-1)!\,n^{3/2}},\qquad
Z=\frac{\ell'A'}{et\log(2nd)\log(2d)}.
\end{gathered}
\tag{10.1124}
\] Here \(m\ge3\). Formula (10.641) at two gives \(P/e\le4\). Also
\(ef\le d\), so \(J=e+d/f\le2d/f\). Substituting \(2d,\sigma t\) in
(10.642), and cancelling the powers of \(d\), \(t\), and \(\mathrm e\),
gives \[
\frac{2C_2(L)J^n}{\mathcal A_n(K)}\le
\frac{406}{128\sigma^3\mathrm e^{n+2}}
\left(\frac{13}{16f}\right)^nT_nZ\,\frac HN.
\tag{10.1125}
\] In particular, the residue contribution here is
\(N^\sigma/(m\delta_L)\), not \(N^\sigma/\delta_L\). To check the
cancellation directly, before using \(P/e\le4\) the power of two in the
denominator is \(2^{6n+5}\), and the remaining ramification factor is
\((P/e)^{n-1}/e\). Multiplying by \(4^{n-1}\) changes that denominator
to \(2^{4n+7}\), as in (10.1125); the \(1/e\) is included in \(Z\).

We use the elementary inequalities from Lemma 10.225:
\(8/3<\mathrm e<11/4\), \(2/3<\log2<7/10\), and \(\log(3/2)<1/2\). If
\(d\ge2\), then \(\ell'=\log(2d)\) and \(x=\log(2nd)>2\). Consequently
\(4+\log(2(n+1)d)<x+9/2\). Bounding the three entries of \(A'\)
separately gives \[
Z<\begin{cases}
39/16,&d\ge2,\ f\text{ even},\\
13/8,&d\ge2,\ f\ge3\text{ odd},\\
39/8,&d\ge2,\ f=1,\\
81/8,&d=1.
\end{cases}
\tag{10.1126}
\] For the first three rows the entries are bounded by \(39/(8ef)\),
\(1/(tx)\), and \(\sigma/(ex)\), respectively. Use \(e\ge1\),
\(t>2f/3\), \(x>2\), and \(\sigma\le2\). In the last row \(e=f=1\),
\(\sigma=2\), and \(\ell'=1\). The first entry of \(A'\) is the largest.
Now \(x=\log(2n)>4/3\) and \(A'/x<1+(9/2)/(4/3)=35/8<9/2\). Since
\((\log2)^{-2}<9/4\), the last bound follows.

For even \(f\), (10.1027) and \(\delta_L\ge1\) give \(H/N<(n+1)/3\). For
odd \(f\ge3\), combine Lemma 10.244 with (10.1027) to get
\(H/N<2^{n+1}(n+1)/3\). If \(f=1\), the residue entry is less than two.
The other entry is also less than two for every \(n\ge2\): the
derivative of \(n\log(\mathrm e(2t)/n)\) is \(\log(2t/n)<0\), and at
\(n=2\) an upper bound is \((11/4)^2(7/5)^3/12=41503/24000<2\). Thus \[
\frac HN<W_n,\qquad
W_n=\begin{cases}
(n+1)/3,&f\text{ even},\\
2^{n+1}(n+1)/3,&f\ge3\text{ odd},\\
1,&f=1.
\end{cases}
\tag{10.1127}
\]

Retaining the factorial makes one endpoint calculation sufficient in
each row. Indeed, \[
\frac{T_{n+1}}{T_n}
<\mathrm e\,\frac{(n+2)\sqrt n}{(n+1)^{3/2}}
\le\mathrm e\sqrt{\frac{32}{27}}<\frac{11\mathrm e}{10}.
\tag{10.1128}
\] The first inequality is \((1+1/(n+1))^{n+1}<\mathrm e\). The
logarithmic derivative of the remaining factor is
\((2-n)/(2n(n+1)(n+2))\), which is nonpositive for \(n\ge2\); its
maximum is therefore the stated value at two. Finally \(32/27<121/100\).

Let \(E_n\) be the right side of (10.1125) with \(Z\) and \(H/N\)
replaced by their row bounds. Its successive ratio is at most
\(143/240\) for even \(f\ge2\), at most \(143/180\) for odd \(f\ge3\),
and at most \(143/160\) for \(f=1\). These follow by multiplying
\(143/(160f)\) by the successive ratio of \(W_n\), which is at most
\(4/3\), \(8/3\), or one. All are less than one. At \(n=2\), use
\(T_2=27/(2\sqrt2)<135/14\) and \(\mathrm e^{-4}<(3/8)^4\). The
resulting exact bounds are \[
\begin{array}{c|c|c}
\text{field range}&\text{endpoint bound}&\text{ratio bound}\\ \hline
d\ge2,\ f\text{ even}&2090104965/8589934592&143/240\\
d\ge2,\ f\ge3\text{ odd}&77411295/1073741824&143/180\\
d\ge2,\ f=1&2090104965/8589934592&143/160\\
d=1&4340987235/8589934592&143/160
\end{array}
\tag{10.1129}
\] Every endpoint is below one. The decreasing envelopes prove the
comparison at all ranks, including the rational field. \(\square\)

\subsubsection{The estimate for arbitrary
inputs}\label{the-estimate-for-arbitrary-inputs}

\textbf{Theorem 10.246 (the coefficient refinement over every number
field).} Let \(K\) be any number field, \(\mathfrak p\) a prime above
any rational prime \(p\), and \(n\ge2\). Let \(\alpha_i\in K^*\),
\(\Xi=\prod_i\alpha_i^{b_i}\ne1\), and use the original field data and
constants (10.1086). Assume all the weight and coefficient conditions
(10.1093), including \(w_i\ge\max\{h(\alpha_i),Q^{-1}\}\), \(b_n\ne0\),
\(v_p(b_n)=\min_i v_p(b_i)\), \(B\ge B_n\ge|b_n|\), \(B\ge\max_i|b_i|\),
and \(0<\varepsilon\le1/2\). With \(V,\Omega,r,L\) defined in (10.1094),
one has \[
\operatorname{ord}_{\mathfrak p}(\Xi-1)
<\mathcal A_n
\max\left\{\Omega L,\frac{\varepsilon r}{c_3(n,d)}\right\}.
\tag{10.1130}
\] There is no independence, local-unit, or root-of-unity hypothesis on
the original list. Zero nondistinguished coefficients are allowed.
Taking the smallest allowed \(w_i\) gives the original height floor and
constants of Theorem B in the
\href{https://math.leidenuniv.nl/reports/files/2008-16.pdf}{free paper
of Bérczes, Evertse and Győry}. The proof here uses the preceding
programme proofs.

\textbf{Proof.} Corollary 10.243 covers odd \(p\) and the fields
containing \(\zeta_3\) at two. It remains to prove the result when
\(p=2\) and \(\zeta_3\notin K\). In the rest of this proof the extension
field is denoted by \(K'=K(\zeta_3)\); \(L\) continues to denote the
real logarithmic expression (10.1094).

First consider independent local units \(\beta_i\in K\) with
\(h(\beta_i)\le Jw_i\), \(w_i\ge Q^{-1}\). Enlarge only \(w_n\) to
\(w_n^*\) as in (10.1098), and apply Lemma 10.224 and Theorem 10.223
over \(K'\) with padded weights \(Jw_i\), \(Jw_n^*\). Independence is
unchanged by a field extension. Since \(\omega(2d)<1\), the coefficient
logarithms are still less than \(2L\) by (10.1096)--(10.1099), using the
original \(Q,c_3,c_4\). The two remaining comparisons are \[
\begin{gathered}
(n+1)f'\log2\le2(n+1)t<2L,\\
G_2(n,2d)<(n+1)(7n+4+\log d+\log2)<2L.
\end{gathered}
\tag{10.1131}
\] For the last step the lower bound (10.1097), after subtraction,
leaves \(5n^2+n+16+(n+3)\log d-(n+1)\log2>0\). This is positive since
\(\log2<1\), \(n\ge2\), and \(d\ge1\). Also
\(\log B_\dagger\le\log B_n<L\). The valuation is unchanged by Lemma
10.244. Lemma 10.245 therefore gives
\(\operatorname{ord}_{\mathfrak p}(\prod_i\beta_i^{b_i}-1)
<2C_2(K')J^nU<\mathcal A_nU\), where
\(U=\max\{\Omega L,\varepsilon r/c_3\}\).

We spell out why the dependence argument uses the same original
parameters. Lemmas 10.240--10.241 use height counts, exact
multiplicative relations in \(K\), and the elementary one-logarithm
bound; none uses \(\zeta_3\in K\). Torsion products, rank-one products,
and the two-base cases are handled by their same estimates, in \(K\).
For a nontorsion product of larger rank, proceed by induction on the
list length, as in Theorem 10.242.

If the distinguished last base is redundant, choose an independent basis
among the preceding bases. Its members still belong to \(K\). Lemma
10.240 supplies the exact common power \(Y^D\), integer coefficients
\(M_j\), and \(B^*\). Renumber this selected list so that a nonzero
\(M_j\) of least two-adic valuation is last. If its rank is \(s\ge2\),
apply the second estimate over \(K'\). Every coefficient logarithm is at
most \(\log(2QB^*)\), since the original weights are at least \(Q^{-1}\)
and \(\omega(2d)<1\). Also \(\log B_\dagger\le\log B^*\). The
noncoefficient terms are less than \(2L\), by (10.1131) and monotonicity
in \(s\le n\). Writing \(W=\prod_{j\in I}w_j\), Lemma 10.241 gives
\(W\log(2QB^*)<8Q^{n-s}U\) and \(WL\le Q^{n-s}U\). Hence the padded
coefficient maximum times \(W\) is less than \(8Q^{n-s}U\). The power
identity and Lemma 10.245 give \[
\operatorname{ord}_{\mathfrak p}(Y-1)\le
\operatorname{ord}_{\mathfrak p}(Y^D-1)
<8C_2(K')J^sQ^{n-s}U
<4\mathcal A_sQ^{n-s}U<\mathcal A_nU.
\tag{10.1132}
\] The last inequality is the rank absorption proved in Theorem 10.242.

If the last base raises the rank, use the relation supported on the
preceding bases from Lemma 10.240. The exact transfer
\(b_i'=k_xb_i-k_ib_x\), \(b_n'=k_xb_n\), with \(k_x>0\) of minimum
two-adic valuation among its nonzero entries, preserves the
distinguished minimum valuation. Formula (10.1117) preserves the second
target entry exactly, and (10.1119) gives \(U'\le2QU\). The shortened
bases still belong to \(K\). Induction and
\(2Q\mathcal A_{n-1}<\mathcal A_n\) prove the bound. These two branches
exhaust every dependent list, and use no relation chosen outside the
original field.

Finally apply Lemma 10.237 to the original \(\alpha_i\). If their total
valuation is nonzero, the required valuation is nonpositive. If it is
zero, their normalized product is \(\Xi^e\). For nontorsion \(\Xi\),
apply the unit-list estimate just proved and the integral power factor.
For torsion \(\Xi\ne1\), use the torsion estimate of Lemma 10.241
directly in \(K\), including when \(\Xi^e=1\). This completes all cases.
\(\square\)

\subsubsection{A residue field of eight
elements}\label{a-residue-field-of-eight-elements}

Let \(a\) be the real root of \(X^3-X-1\) between one and two, and put
\(K=\mathbb Q(a)\). The polynomial has no rational root, so \(d=3\) and
\(K\) cannot contain a quadratic subfield. Its reduction \(X^3+X+1\) has
no root over \(\mathbb F_2\), hence is irreducible. At a prime above two
the residue degree is three and the ramification index is one, by the
degree inequality used in Lemma 10.244. Thus \(N=8\), whereas
\(K(\zeta_3)\) has residue cardinality \(64\).

The units \(a,3\) are independent. Their norms are one and \(27\), so a
relation forces the exponent of three to vanish; the real inequality
\(a>1\) then forces the other exponent to vanish. The reduction of \(a\)
is a nonidentity element of \(\mathbb F_8^*\), so has order seven. The
descended saturated residue group therefore has \(g=7\). Indeed the
integral vector of \(a\) is an integer combination of the saturated
lattice basis. The corresponding product of the \(\theta_j\) equals
\(a\) up to a 3-primary root of unity in \(K\), which is necessarily
one. Thus this residue group contains the original residue of \(a\), and
is contained in a group of order seven. Consequently \[
\begin{gathered}
m\delta_L=9,\qquad \frac{64}{m\delta_L}=\frac{64}{9}<8,\\
3a^7-1=2(3a^2+3a+1),\qquad
\operatorname{ord}_{\mathfrak p}(3a^7-1)=1.
\end{gathered}
\tag{10.1133}
\] For the valuation, use \(a^3=a+1\), hence \(a^7=2a^2+2a+1\). The
second factor has nonzero reduction, since a polynomial of degree two
cannot vanish at the residue of \(a\), whose minimal polynomial has
degree three. This gives a concrete application with \(n=2\),
coefficients \((7,1)\), \(B=7\), \(B_n=1\), and \(\varepsilon=1/2\).

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={Quadratic extension, descended cubic saturation and the exact residue-factor cancellation in the cubic example}]{../figures/yu-two-adic-coefficient.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.74. Cubic saturation is represented by roots in the
original field. In the example, the residue field grows from eight to
sixty-four elements, while the product \(m\delta_L=9\) reduces its
contribution to \(64/9<8\). Lemmas 10.244--10.245 prove the general
transfer; Theorem 10.246 retains the original field constants.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_two_adic_coefficient.py}{Reproducible figure
and exact arithmetic}.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{A bound using only the largest
coefficient}\label{a-bound-using-only-the-largest-coefficient}

The coefficient parameter in the second explicit estimate contains fixed
field terms as well as logarithms of the integer coefficients. A local
height estimate treats the small-coefficient range. In the complementary
range, the coefficient itself is large enough to absorb every fixed
term. An exact relation then removes a dependent base. Choosing a base
of largest weight in that relation makes its coefficient cost fit inside
the decrease in the height-product constant. The resulting bound uses
the original field degree, for arbitrary nonzero algebraic bases.

Write \(d=[K:\mathbb Q]\), \(e=e_{\mathfrak p}\), \(f=f_{\mathfrak p}\),
\(t=f\log p\), and \(Q=16\mathrm e^2d^2\). The roman \(\mathrm e\) is
the exponential constant. Set \(J=e+d/f\), \(A=58d\max\{1,\log d\}\).
For \(n\ge1\), put \[
\begin{aligned}
x&=\log(2nd),\qquad y=\log(2d),\\
\mathcal G_n&=(16\mathrm e d)^{2(n+1)}n^{5/2}x^2d^n
                  \frac{p^d-1}{(\log p)^2},\\
Z_n&=\frac{et\mathcal G_n}{2dJn}Q^{1-n}
 =\frac{16^{n+3}\mathrm e^4ef}{2J}
      n^{3/2}x^2d^{n+3}\frac{p^d-1}{\log p}.
\end{aligned}
\tag{10.1134}
\] For \(n\ge2\), also put \[
R_n=\frac{e\mathcal G_n}{\mathcal A_n}
 =\frac{nd^nf^2}{e^{n-1}}
   \frac{p^d-1}{p^f}\frac{x}{y},
\] where \(\mathcal A_n\) is the original-field constant (10.1086).

\textbf{Lemma 10.247 (the coefficient threshold).} For every such field,
prime and integer \(n\ge2\), \[
\begin{gathered}
Z_n>2Q,\qquad R_n\ge nd,\\
\log Z_n>\max\{n,\log(2A)\},\\
(n+1)(7n+5+\log d)<2R_n\log Z_n,\\
(n+1)t<R_n\log Z_n.
\end{gathered}
\tag{10.1135}
\]

\textbf{Proof.} The positive logarithm series
\(\log2=2\sum_{j\ge0}1/((2j+1)3^{2j+1})\) gives \(11/16<\log2<3/4\). For
the lower bound its first three terms sum to \(842/1215>11/16\). For the
upper bound, replace every denominator \(2j+1\), \(j\ge1\), by three;
the resulting sum is \(25/36<3/4\). In particular \(11/4<\log16<3\), and
\(\log2>2/3\).

Since \(p^d-1\ge d\log p\), the formula for \(Z_n\), \(ef\ge1\),
\(J\le2d\), \(x\ge\log4>1\), and \((3/2)\log n\ge(3/2)\log2>1\) give \[
\log Z_n>\frac{11n}{4}+\frac{47}{4}+(n+3)\log d.
\tag{10.1136}
\] This exceeds \(\log(2Q)=\log32+2+2\log d\), because \(\log32<15/4\).
Thus \(Z_n>2Q\). It also exceeds n and \(\log(2A)\). For the latter, use
\(\max\{1,\log d\}\le d\) and \(\log116<5\), so \(\log(2A)<5+2\log d\).
The exponential series gives \(\mathrm e>8/3\), and \((8/3)^5>116\),
proving the stated logarithmic comparison.

Use \(ef\le d\) to bound the exact ratio in (10.1134): \[
R_n\ge nd f^{n+1}\frac{p^d-1}{p^f}.
\tag{10.1137}
\] If \(d\ge2\) and \(f=1\), the last quotient is
\(p^{d-1}-p^{-1}\ge3/2\), so \(R_n\ge3n\). If \(f\ge2\), that quotient
is at least \(1-p^{-d}\ge3/4\), and again \(R_n\ge3n\). Both estimates
also imply \(R_n\ge nd\). If \(d=1\), then \(e=f=1\), and the exact
expression is \[
R_n=n(1-p^{-1})\frac{\log(2n)}{\log2}\ge n.
\tag{10.1138}
\] It is at least \(2n\) for \(n\ge8\).

Whenever \(d\ge2\) or \(n\ge8\), we therefore have \[
\frac{(n+1)(7n+5+\log d)}{2R_n}
\le\frac38(7n+5+\log d)<\log Z_n
\] by (10.1136). In the remaining cases \(d=1\), \(2\le n\le7\), use
\(R_n\ge n\). The required comparison with (10.1136) reduces to \[
-3n^2+23n-10>0.
\] This concave polynomial is at least the smaller of its endpoint
values 24 and 4 on \([2,7]\). The coefficient-parameter assertion
follows.

For the last assertion, if \(p=2\) or \(3\), then \(t<2d\) and
\(R_n\ge nd\), so \((n+1)t/R_n<3<\log Z_n\). If \(p\ge5\), formulas
(10.1137)--(10.1138) give \(R_n\ge8n/5\). Consequently \[
\frac{(n+1)t}{R_n}\le\frac{15}{16}d\log p.
\] For every \(l>0\), maximization of \(\log l-l/16\) gives
\(\log l\le l/16+\log16-1<l/16+2\). The factor preceding
\((p^d-1)/\log p\) in the formula for \(Z_n\) is greater than
\(\mathrm e^4\). Also \(p^d-1\ge p^d/2\). Taking \(l=\log p\) yields \[
\log Z_n>4+dl-\log2-\log l
 >\frac{15}{16}dl.
\] This proves the last assertion. \(\square\)

\textbf{Lemma 10.248 (one base and the height floor).} For every
\(n\ge1\), \[
\mathcal G_nQ^{-n}>\frac{1000d}{\log p}.
\tag{10.1139}
\] If \(\beta\) is a local unit, \(h(\beta)\le Jw\), \(w\ge Q^{-1}\),
\(B\ge\max\{3,|b|\}\), and \(\beta^b\ne1\), then \[
v_p(\beta^b-1)<\mathcal G_1w\log B.
\tag{10.1140}
\]

\textbf{Proof.} Exact cancellation gives \[
\frac{\mathcal G_n}{Q\mathcal G_{n-1}}
 =16d\left(\frac n{n-1}\right)^{5/2}
       \left(\frac{\log(2nd)}{\log(2(n-1)d)}\right)^2
 >16d\qquad(n\ge2).
\tag{10.1141}
\] At n=1, writing \(l=\log p\) and \(y=\log(2d)>2/3\), \[
\frac{l\mathcal G_1}{Q}
 =4096\mathrm e^2d^3y^2\frac{p^d-1}{l}
 \ge4096\mathrm e^2d^4y^2>1000d.
\] This and (10.1141) prove (10.1139).

For n=1, the prefactor of \((p^d-1)/l\) in \(Z_1\) is at least
\(16^4\mathrm e^4d^3y^2/4>2\). Consequently \(Z_1>2Q\): divide its lower
bound, using \((p^d-1)/l\ge d\), by \(2Q\) to obtain
\(512\mathrm e^2d^2y^2>1\). Also \(p^d-1\ge p/2\), so
\(\log Z_1>l-\log l>l/2\). The last inequality follows by maximizing
\(\log l-l/2\), whose maximum is \(\log2-1<0\).

Put \(T=\mathcal G_1w\log B\). If \(B\le Z_1\log B\), the local height
bound gives \[
v_p(\beta^b-1)\le\frac d{et}(\log2+JBw)<T.
\] Indeed the variable term is at most \(T/2\) by the definition of
\(Z_1\), while \(T>1000d\log B/l\) makes the constant term less than
\(T/2\).

Otherwise \(\log B>\log Z_1>l/2\). Let g be the order of the residue of
\(\beta\), so \(g\le p^f-1\). Theorem 10.21 yields \[
v_p(\beta^b-1)
 \le\frac{\log B}{l}
    +\frac{dJg p}{f l(p-1)}w+\frac{d\log2}{et}.
\] The first and third terms are each less than \(T/3\), by (10.1139).
The middle term is at most \(4d^2(p^d-1)w/l\), since \(J\le2d/f\) and
\(p/(p-1)\le2\). It too is less than \(T/3\), because \(\log B>l/2\) and
\(16^4\mathrm e^4d^3y^2>24\). This proves (10.1140), including torsion
bases allowed in Theorem 10.21. \(\square\)

\textbf{Theorem 10.249 (the maximum-coefficient bound for local units).}
Let \(\beta_1,\ldots,\beta_n\in K^*\) be local units, \(n\ge1\), with no
independence assumption. Suppose \[
h(\beta_i)\le Jw_i,\quad w_i\ge Q^{-1},\quad
B\ge\max\{3,|b_1|,\ldots,|b_n|\},\quad
Y=\prod_i\beta_i^{b_i}\ne1,\quad \Omega=\prod_iw_i.
\] Then, with \(v_p(p)=1\), \[
v_p(Y-1)<\mathcal G_n\Omega\log B.
\tag{10.1142}
\]

\textbf{Proof.} We induct on n.~Lemma 10.248 proves n=1. For every n, a
torsion product has height zero, so \(v_p(Y-1)\le d\log2/(et)\);
(10.1139) places this strictly below the target. Hence assume n at least
two and Y nontorsion. Put \(w=\max_iw_i\). The local height inequality
and the sum and power inequalities for height give \[
v_p(Y-1)\le\frac d{et}
  \left(\log2+B\sum_i h(\beta_i)\right)
\le\frac d{et}(\log2+JnBw).
\tag{10.1143}
\] The inequality is immediate if the valuation is nonpositive;
otherwise it follows from the local contribution to the height of
\((Y-1)^{-1}\), just as in Theorem 10.21.

Suppose first that \(B\le Z_n\log B\). Since \(\Omega\ge wQ^{1-n}\), the
definition of \(Z_n\) gives \[
\frac d{et}JnBw\le\frac12\mathcal G_n\Omega\log B.
\] Moreover \(\Omega\ge Q^{-n}\) and \(Z_n>2Q\), so \[
\frac{et}{2d}\mathcal G_n\Omega\log B
\ge\frac{JnZ_n}{Q}\log B>\log2.
\] The constant term in (10.1143) is strictly below the other half of
the target. This proves (10.1142) in the first case.

Now let \(B>Z_n\log B\). As \(B\ge3\), we have \(\log B>\log Z_n\).

First suppose the bases are independent. Apply the padded-weight form of
Theorem 10.223 supplied by Lemma 10.224, with weights \(Jw_i\). Renumber
so a nonzero coefficient of minimum p-adic valuation is last. No
coefficient or height bound changes. Use the same field unless \(p=2\)
and \(\zeta_3\notin K\); in that case use \(K'=K(\zeta_3)\). Its degree
is \(2d\), its local ramification index remains \(e\), its residue
degree is at most \(2f\), and its prime-ideal valuation restricts to the
original one, by Lemma 10.244.

In either case Lemma 10.238 or 10.245 gives \(2C_2J^n<\mathcal A_n\).
The coefficient parameter satisfies \[
\mathsf h_2\le\max\left\{
 \log(2BQ),\ (n+1)(7n+5+\log d),\ 2(n+1)t
\right\}.
\tag{10.1144}
\] Indeed \(1/(d'W(d'))<1\), \(J\ge1\), and \(w_i\ge Q^{-1}\) give the
first entry. Also \(\log B_\dagger\le\log B<\log(2BQ)\). In every row of
(10.643), \(a_0^{(2)}<6\) and \(a_1^{(2)}<4\). Using \(d'\le2d\),
\(\log(n+1)\le n\), and \(\log2<1\) gives the second entry. The
residue-degree bound gives the third entry.

Lemma 10.247 bounds the last two entries strictly by \(2R_n\log B\). For
the first, \(B>Z_n>2Q\) implies \(\log(2BQ)<2\log B\le2R_n\log B\).
Therefore \[
v_p(Y-1)
 <\frac{\mathcal A_n}{2e}\Omega\mathsf h_2
 <\frac{\mathcal A_n}{e}R_n\Omega\log B
 =\mathcal G_n\Omega\log B.
\tag{10.1145}
\] This proves the independent case.

If the bases are dependent, choose a minimally dependent sublist C. Put
\(u_i=Qw_i\ge1\), and choose \(x\in C\) maximizing \(u_x\) on C. There
is an exact integer relation, supported on C, \[
\prod_i\beta_i^{k_i}=1,\qquad k_x>0,\qquad
|k_i|\le A u_x^{n-1}.
\tag{10.1146}
\] For a singleton torsion sublist, use its order, at most A by Lemma
10.155. Otherwise put \(a=|C|-1\). The simultaneous bounds of Corollary
10.20, as calculated in the proof of Lemma 10.240, give \[
|k_i|\le A\left(\frac{\mathrm e dJ}{Q}\right)^a
          \prod_{j\in C\setminus\{i\}}u_j
 \le A u_x^a\le A u_x^{n-1}.
\] Every coefficient on a minimal circuit is nonzero; reversing their
signs makes \(k_x>0\). The relation is exact, since Corollary 10.20 has
already removed the root of unity.

Delete the x-th base by raising the original product to \(k_x\): \[
Y^{k_x}=\prod_{i\ne x}\beta_i^{b_i'},\qquad
b_i'=k_xb_i-k_ib_x,\qquad
|b_i'|\le B':=2ABu_x^{n-1}.
\tag{10.1147}
\] Here \(B'\ge3\), and \(Y^{k_x}\ne1\) because Y is nontorsion. The
field, e, f, Q and J are unchanged. Since
\(\log B>\log Z_n>\max\{n,\log(2A)\}\), we have \[
\frac{\log B'}{\log B}
 =1+\frac{\log(2A)+(n-1)\log u_x}{\log B}
 <2+\log u_x\le2u_x.
\tag{10.1148}
\] The last inequality follows from \(\log u\le u-1\) for \(u\ge1\).
Induction and the integral geometric-sum factor therefore give \[
\begin{aligned}
v_p(Y-1)&\le v_p(Y^{k_x}-1)\\
 &<\mathcal G_{n-1}\frac{\Omega}{w_x}\log B'
 <2Q\mathcal G_{n-1}\Omega\log B
 <\mathcal G_n\Omega\log B,
\end{aligned}
\] where (10.1141) supplies the last comparison. Each deletion reduces
n, so the induction includes every dependence and torsion pattern.
\(\square\)

\textbf{Corollary 10.250 (arbitrary algebraic bases and the original
degree).} Let \(K\) be any number field of degree d, \(\mathfrak p\) a
prime above any rational prime p, and
\(\alpha_1,\ldots,\alpha_n\in K^*\), \(n\ge1\). For integers \(b_i\),
suppose \(\Xi=\prod_i\alpha_i^{b_i}\ne1\). If \[
w_i\ge\max\{h(\alpha_i),(16\mathrm e^2d^2)^{-1}\},\qquad
B\ge\max\{3,|b_1|,\ldots,|b_n|\},
\] then \[
v_p(\Xi-1)
 <(16\mathrm e d)^{2(n+1)}n^{5/2}(\log(2nd))^2d^n
       \frac{p^d-1}{(\log p)^2}
       \left(\prod_iw_i\right)\log B.
\tag{10.1149}
\] The original bases may be dependent, torsion, or nonunits at
\(\mathfrak p\); some coefficients may vanish.

\textbf{Proof.} A nontrivial torsion product is handled by the
height-zero argument in Theorem 10.249. Otherwise apply Lemma 10.237,
putting \(v_i=\operatorname{ord}_{\mathfrak p}(\alpha_i)\) and
\(\beta_i=\alpha_i^ep^{-v_i}\). If \(\sum_i b_iv_i\ne0\), the required
valuation is nonpositive. If the sum is zero, the \(\beta_i\) are local
units satisfying \(h(\beta_i)\le Jw_i\), and their product is
\(\Xi^e\ne1\). Lemma 10.237 and Theorem 10.249 give
\(v_p(\Xi-1)\le v_p(\Xi^e-1)<\mathcal G_n\Omega\log B\). All constants
use the original field. \(\square\)

The free
\href{https://www.birs.ca/workshops/2012/12ss131/files/bugeaud_LFL.pdf}{BIRS
notes of Bugeaud, Theorem G} give this maximum-coefficient constant with
\(p^d\) in place of \(p^d-1\). Here the bound \(B\ge3\) is explicit: a
positive logarithmic coefficient factor is necessary even when every
nonzero coefficient has absolute value one. Corollary 10.250 has the
same height floor and requires no independence. Its proof uses the
programme results above, with the second explicit constant of
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
freely accessible 2013 paper}.

\textbf{Example.} In \(K=\mathbb Q\) at \(p=5\), take bases 2 and 3,
weights \(\log2,\log3\), and coefficients \(a,-b\). Independence follows
by comparing prime exponents. With \(B=\max\{3,|a|,|b|\}\) and
\(2^a3^{-b}\ne1\), the theorem gives \[
v_5(2^a3^{-b}-1)
 <(16\mathrm e)^6\,2^{5/2}(\log4)^2
       \frac4{(\log5)^2}\log2\log3\log B.
\tag{10.1150}
\] For \((a,b)=(3,1)\), the argument of the valuation is \(5/3\), so the
exact valuation is one.

\textbf{Exercise.} Apply Corollary 10.250 over \(\mathbb Q\) at two to
\((6,10)\) with coefficients \((1,-1)\), and then to \((6,10,15)\) with
coefficients \((1,-1,1)\). Compute the normalized units and exact
valuations. For the second list, exhibit the relation and the two-base
product after deleting the normalized part of 15.

\textbf{Solution.} Here \(d=e=f=1\). The normalized units in the first
list are 3 and 5; the product is \(3/5\), giving \(v_2(3/5-1)=1\). In
the second list the valuations are \((1,1,0)\) and the normalized units
are \((3,5,15)\). Their exact relation is \(3^{-1}5^{-1}15=1\). Choosing
weights \((\log6,\log10,\log15)\) makes the third weight largest.
Formula (10.1147), with \(k=(-1,-1,1)\), replaces the coefficient vector
by \((2,0)\). Both products equal 9, and \(v_2(9-1)=3\). In both cases
take B=3; the original valuation sums vanish. The original bases include
nonunits, the second normalized list is dependent, and the original
field has no primitive cube root of unity. Lemmas 10.244--10.245 and
Corollary 10.250 nevertheless retain \(d=1\) in the bound.

\begin{minipage}[t]{\linewidth}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={The coefficient threshold, independent estimates and exact dependent-base elimination}]{../figures/yu-maximum-coefficient.png}}
\par\smallskip
\begin{flushleft}\small
\emph{Figure 10.75. The left panel samples the three quantities in Lemma
10.247 for the rational field at two. Their inequalities for every rank,
degree and prime are proved in the lemma. The right panel records the
alternatives in Lemma 10.248 and Theorem 10.249. A dependent list loses
one base while the increase in its logarithmic coefficient parameter is
less than the decrease in the target constant. The displayed three-base
example has exact valuation three. Corollary 10.250 returns to arbitrary
original bases with the original field degree.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_maximum_coefficient.py}{Reproducible figure
and exact comparisons}. Human-source context: Theorem G of the
\href{https://www.birs.ca/workshops/2012/12ss131/files/bugeaud_LFL.pdf}{free
BIRS notes} and Theorem I of
\href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Yu's
free 2013 paper}.}
\end{flushleft}
\end{minipage}\par\medskip


\subsection{82. Exercises with
solutions}\label{exercises-with-solutions}

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
  \textbf{Easy.} Explain exactly how nonunit bases can be reduced to
  unit bases. Give an example showing that the conclusion of Theorem
  10.8 cannot be extended unchanged to nonunits.
\item
  \textbf{Medium.} Derive (10.34) from (10.22), including the maximum in
  \(H\).
\item
  \textbf{Medium.} Prove (10.36) using the residue order and the power
  laws. Explain where the assumption that \(p\) is odd enters.
\item
  \textbf{Medium.} For \(K=3,L=4,p=3,e=1,t=0,u=0\), find the lower bound
  supplied by (10.10). Find the valuation precision required in its
  hypothesis, and the exact excess in the row expansion when \(n\)
  ordinary rows are used.
\item
  \textbf{Easy.} Show that (10.6) is attained for \(K=2,L=4,R=3\).
  Identify the maximizing list of first coordinates.
\item
  \textbf{Medium.} A proposed assertion for every prime \(p\) and pair
  of coprime positive integers \(x,y\) bounds \(v_p(x^{p-1}-1)\) by
  \(283(p-1)\ln y\ln(xy)/(\ln p)^2+4\). Show why it is false as stated
  when \(y=1\), without using a logarithm estimate.
\item
  \textbf{Hard.} Adapt the determinant argument to the hypotheses of
  Theorem 10.11. Explain why its first-base precision, unit coefficient
  after removing the common p-part, and second-base condition at two are
  all recorded. Derive the power \(E^{-3}\) in its final bound.
\item
  \textbf{Medium.} Compute \(M_2(4,9)\) over \(\mathbb Q\), prove
  (10.48) in this example, and compute the residue index (10.49) at
  five.
\item
  \textbf{Hard.} Derive the constant 58 in (10.55) directly from
  (10.53), including the roles of the chosen height radius and the
  selected archimedean weight. For the rational bases 4,9, compute the
  full saturation index, its root-of-unity factor, and the diamond
  average in Figure 10.2.
\item
  \textbf{Medium.} Apply Theorem 10.21 to the rational unit 2 at three.
  Determine its residue order and exact valuation for every nonzero
  exponent with a nonzero power difference. Compare the bound with the
  normalization in a field of degree two having local degrees e=1, f=2.
\item
  \textbf{Medium.} At p=5 take \(L_1=5z_1+z_3\), \(L_2=5z_2+z_3\),
  \(L_3=-5z_1-5z_2-z_3+z_4\), and \(L=L_1+L_2+L_3=z_3+z_4\). Compute the
  matrix in Lemma 10.23 and all its two-column upper minors. Choose a
  minimum-valuation derivative basis. Explain the denominator introduced
  by the other natural basis.
\item
  \textbf{Hard.} Set \(r=2,S=T=90\), \(D_0=32,D_1=2,D_2=3\), and
  \(S_i=T_i=30\). Show that Theorem 10.25 implies that a polynomial of
  these bounds with all the stated jets zero at \((s,2^s,3^s)\),
  \(0\le s\le90\), is zero, for every nonzero rational vector
  \(\boldsymbol B\). Check the integer degree inequalities and both
  possible one-dimensional relation bounds.
\item
  \textbf{Medium.} In Theorem 10.27 take \(K=\mathbb Q,p=5,n=2\), bases
  \(2,3\), coefficients \(1,1\), and \(L_1=z_1+z_2\). Use
  \(\sigma=\ln6,H=10\ln5\). Compute the original saturated residue
  index, \(w_q,R_2,M_\gamma\), and verify the one-base height-product
  hypothesis without a numerical approximation. What valuation is being
  bounded?
\item
  \textbf{Hard.} For \(k=L=2\), compute the four shifted-power
  polynomials in Figure 10.4 and the determinant of their coefficients
  of \(X,X^2,X^3,X^4\). For \(k=\ell=2\), evaluate
  \(v(k)^m\Theta(1/3;k,\ell,m)\), \(0\le m\le4\), and check the exact
  q-power clearing bound with \(q=3,I=1\).
\item
  \textbf{Medium.} Use Theorem 10.32 over \(K=\mathbb Q\) with
  \(r=2,k=L=1,S=T=0\), \(\vartheta=(2,3)\), \(A_1=A_2=1\),
  \(D_1=Y_1\partial_{Y_1}-Y_2\partial_{Y_2}\), and exponents
  \((0,0),(1,0),(0,1)\). Compute \(N,M,\mathcal B\), exhibit a nonzero
  integral coefficient vector, and compare its exact projective height
  with (10.96). Form the ordinary polynomial \(P\) and identify which
  jets are imposed.
\item
  \textbf{Hard.} For the full node set with \(p=3,R=3,\mu=2,\theta=1\),
  what precision in (10.100) suffices for conclusion
  \(v_3(F(\rho x))\ge14\)? For \(p=2,q=3,R=3\), compute \(L_{-2}(0)\) in
  the q-deleted set and its valuation. Finally, why is the restriction
  \(x\in\mathbb Z_p\subset\mathbb Q_p\) essential to the full-set
  cardinal bound? Use \(p=2,R=1\) and an algebraic \(x\) with \(x^2=2\).
\item
  \textbf{Medium.} In Proposition 10.37 take
  \(p=3,q=2,I=0,k=L=2,\rho=3\), and a single nonzero coefficient \(c=1\)
  at \(\boldsymbol m=0,a=0,\ell=2\). At prepared index zero, compute
  \(\gamma,\Gamma\), and the polynomial \(F(Z)\). Check (10.107) for
  \(s=0,j=1\). If a torus factor with slope \(w\) is added, compare the
  sufficient slope hypothesis for \(w=9\) and \(w=3\).
\item
  \textbf{Hard.} At two, take the additive prepared polynomial
  \(f(y)=\Delta(y;2)=(y+1)(y+2)/2\), with \(I=0,k=2,L=1\) and no torus
  dependence. Show that its prepared values through order two at zero
  are integral, whereas its first ordinary divided jet has valuation
  \(-1\). Explain the exact role of \(C_0\). Next take full
  interpolation nodes with \(p=3,R=3,\mu=2,\theta=1\), and weighted jet
  precision \(\Lambda-jC\) with \(\Lambda=16,C=2\). Compare Theorems
  10.34 and 10.39.
\item
  \textbf{Medium.} At \(p=3\), take \(q=2,k=L=1,I=0,J=1,c=1\), no torus
  dependence, and \(x=1/2\). Compute the prepared value \(\Delta(x;1)\),
  the clearing exponent in (10.115), and the bound (10.116) with
  \(X=1/2,T=0\). Distinguish the denominator cost at two from the
  valuation at three.
\item
  \textbf{Hard.} Let \(K=\mathbb Q,p=3,q=2,J=1,\vartheta=10\). Use the
  exponential and logarithm to choose a square root
  \(\eta\in\mathbb Q_3\) congruent to one. Determine the coefficient
  (10.117) for \(F=\mathbb Q(\eta)\). At \(s=1,x=1/2\), with
  \(k=L=1,I=0\), take exponents \(1,3\) and coefficients \(1,-1\).
  Factor the prepared value into its common local unit and a rational
  value, and compute its valuation. Compare the bounds from the root
  field and the common-factor rule, with \(X=1/2,T=0\).
\item
  \textbf{Medium.} Use \(I=0,k=2,L=1,X=2,r=2,\Omega_1=1,T=2\). Compute
  \(\mathcal P_{0,2}(Z)\), \(\mathcal B_0^*(2,2)\), and the earlier
  \(\mathcal B_0(2,2)\). At \(x=2,a=1,\ell=1,\omega_1=1\), show that the
  three additive bounds are attained.
\item
  \textbf{Hard.} In Corollary 10.45, take
  \(K=\mathbb Q,r=1,k=2,L=1,S=2,T=0\), \(\vartheta=2\), and
  \(\mathcal M=\{0,1,2\}\), so \(A=2\). Compute the separate-row bound
  (10.127) and the uniform averaged bound (10.128). Order the six
  columns by \((m,a)=(0,0),(0,1),(1,0),(1,1),(2,0),(2,1)\). Verify that
  \((2,-18,-57,36,1,0)\) is an integral kernel vector at all five nodes,
  and compute its two projective heights. Explain why the zero row at
  \(s=-2\) does not invalidate either bound.
\item
  \textbf{Medium.} In Theorem 10.47, take
  \(q=2,r=1,F=\mathbb Q,\vartheta=2,\eta=\sqrt2\). Show that at \(s=2\),
  the coefficients \(A_0=-2,A_1=1\) give a zero total sum whose separate
  coset coefficients are nonzero. Give a similar example with odd \(s\)
  in which the full-degree hypothesis fails. Finally, at \(p=7\), use
  the local square root \(\eta\equiv3\pmod7\) of two to show that finite
  valuation precision of a total sum need not transfer to its cosets.
\item
  \textbf{Hard.} Let \(q=2,p=3,r=2\), use one Euler argument
  \(\omega(\boldsymbol m)=m_1\), and retain \(m\equiv1\pmod2\), with
  \(m_*=1\). Compute the affine binomial change \(\Delta(2Z+1;t)\)
  through order two and its inverse. Verify the minimum-valuation
  identity on the new jet vector \((1,3,9)\). Explain why it preserves
  total order when combined with an unchanged additive index. Then apply
  Corollary 10.49 over \(K=\mathbb Q\) with
  \(\alpha_0=-1,\theta_1=2,\theta_2=3,P=3,d_1=1,d_2=0\): identify \(F\),
  the two phased roots and bases, their heights, and the root degree
  over \(F\).
\item
  \textbf{Hard.} Use the divided support in Figure 10.8. Compute
  \(a_0,H\), both subclasses and their phase factors after
  multiplication by \(\zeta^s\). If a coefficient of minimum valuation
  belongs to \(\lambda_0=(-1,0)\), state the retained phase congruence
  and normalized root-of-unity factor. Explain why the subclass is
  chosen independently of \(s\) and the derivative index. Compare the
  coprime case \(q=2,G_0=5,a_*=2\).
\item
  \textbf{Hard.} Over \(\mathbb Q_5\), apply the Newton construction to
  \(f(X)=X^2+1\), starting at \(x_0=2\). Compute \(x_1,x_2\), the three
  error valuations and the primitive fourth root \(\zeta\) modulo
  \(625\). Take \(K=\mathbb Q,\alpha_0=-1,q=2,\theta_1=2\), and
  \(g=2/\zeta\). Prove \(v_5(g-1)=1\). For the target \(\vartheta=3\),
  verify the choices \(t=0,j=3,P=125,d_1=-125\) in Lemma 10.53 and
  compute the exact valuation of \(\gamma_1-1\).
\item
  \textbf{Medium.} In Corollary 10.54, assume the other hypotheses and
  \(\mathcal A_{\mathrm{ph}}(4,4)=11\). Take
  \(p=3,q=2,R=4,T=5,\mu=2,k=L=2,\theta=3,C_0=0\). Compute the exact
  input node set, \(n,B,M_0,T'\), and the least \(U\) meeting the exact
  first budget. State the resulting zeros. Compare both lines of the
  alternative larger-scale budget; explain why its first line alone does
  not give the same zero conclusion.
\item
  \textbf{Hard.} In Figure 10.9, verify that the individual \(\gamma_i\)
  lie outside \(\mathbb Q\), whereas every normalized row entry is
  rational. Determine \(N,M_*,H_{\mathrm{ph}}\), and check the vector
  \((-b,C,-C,b)\) at all three nodes. Factor its torus polynomial,
  compute its two projective heights and its minimum 5-adic coefficient
  valuation, and compare its height with (10.152). Explain why the zero
  first row does not invalidate the bound.
\item
  \textbf{Hard.} Use the data of Figure 10.10. Compute the cancelled
  congruence and \(h_\Lambda\), prove the degrees and chosen completions
  of \(E\) and \(F\), and evaluate both 5-adic valuations of \(V\). For
  \(I=0,k=L=1,\lambda_0=0,c=(1,-1),s=1,x=1/2,\boldsymbol t=0\), compute
  every term in (10.156). Compare the support-field budget with the
  budget from the full root field, and explain why equal local degrees
  do not justify using the original global degree.
\item
  \textbf{Medium.} Verify the auxiliary-prime choices in Corollary 10.9
  for \((a,p)=(8,3),(3,2),(25,2)\). Compute the exact original and
  transformed valuations, the rational heights and the cutoff \(H\).
  Derive the all-power bound when \(b=1\), retaining the same
  \(250000\ln(5a)\ln5\) coefficient. Explain why simply setting the
  second base to one would not prove this case.
\item
  \textbf{Hard.} Verify the two support heights in Figure 10.11 by
  clearing denominators in the monomial vectors and finding their
  primitive integer coordinates. Compute both signed heights of
  \(\{(0,0),(1,0),(0,1)\}\). Extract the even-even coset from the
  diamond at \(q=2\), use reference \((-2,0)\), and verify (10.162). At
  \(P=1,S=1,k=L=1,T=0\), compare the separate-row coefficient budgets
  for the seven-point diamond support using its exact height and its
  coordinate box. Explain how a shifted original-coordinate box is
  handled when its centre is outside the image of \(B\).
\item
  \textbf{Hard.} Verify the two initial kernels in Figure 10.12 and
  determine their exact row fields. Prove the discriminants \(5\) and
  \(24\) using the full integral bases of the two quadratic fields.
  Compute both projective coefficient heights, explain why their
  finite-place contributions vanish, and evaluate (10.171) with the
  actual row-field discriminant for each example. For \(t=\sqrt6\),
  compare the one-generator discriminant bound with the bound using only
  the maximum support height. Explain what happens to the row field when
  \(S=0\), and why \(N>M_*\) and the later analytic inequalities still
  need checking.
\item
  \textbf{Medium.} Verify every lattice and phase coordinate in Figure
  10.13. Compute \(J,V/\Delta,g,H\), the guaranteed class size, and both
  sides of (10.180) at \(k=2,L=1\). At \(S=1,T=0\), compare the
  coefficient surplus certified by the exact class count with the
  coarser volume criterion. Then take \(T=1\) and one nonzero Euler
  direction: compute \(M_*\), and state precisely what the counting
  criterion does and does not prove. Explain why replacing the closed
  box by an open one would invalidate the integer-volume argument.
\item
  \textbf{Hard.} Use \(K=\mathbb Q,p=3,r=2\),
  \((\alpha_1,\alpha_2)=(2,5)\), and
  \(c_0=1.9,c_1=1.4494,c_3=1.3852,c_4=20.8\). Put \(a_0=2+\ln14\),
  \(a_1=4.79\), \(g_0=3(2a_0+a_1+\ln(2a_0+a_1))\), \(h=g_0\), and
  \(\tau=1/(4\cdot10^{26})\). Use \(\widehat\vartheta=3/2\) and
  \(\sigma_1=\ln2,\sigma_2=\ln5\). Prove that \(J=1,u=1,H=1\). Verify
  the floors in Figure 10.14 with rigorous bounds, construct a support
  of exactly \(\lfloor D_1D_2\rfloor+1\) points without listing it, and
  compute \(N,M_{\rm full},M_*\). Use the integral identity
  \(20N-57M_{\rm full}>0\) to verify the strict surplus. Identify the
  actual coefficient field and its discriminant cost.
\item
  \textbf{Hard.} First take \(K=\mathbb Q(\zeta_3,\sqrt2)\) and \(p=2\).
  Prove that \(d=4,e=2,f=2,u=1\), and verify equality in (10.192). Then
  return to all the rational parameters of Solution 34. Compute
  \(\Omega_1\), the actual clearing integer, both maximizing additive
  orders in (10.199) and (10.201), and the exact mean orders in
  (10.204). Reconstruct each signed component and net height bound in
  Figure 10.15, using (10.174) with the actual ratio \(M_* /(N-M_*)\).
  Include the row-scaling bounds of Corollary 10.80. Explain why the
  coefficient field and all zeros are preserved by the alternative
  global clearing.
\item
  \textbf{Medium.} Reconstruct every exact lower-bound path and final
  depth in Figure 10.16. Distinguish these lower bounds from the actual
  valuation path for \(\beta=1+\sqrt2\) in \(\mathbb Q_2(\sqrt2)\). Next
  let \(\pi^3=3\), \(L=\mathbb Q_3(\pi)\), and \(\beta=1+\pi\). Show
  that the bound (10.210) is attained at \(P=9\), and use the
  exponential coefficients to prove that the closed boundary radius
  \(3^{\vartheta}\) cannot replace \(3^\theta\) in Corollary 10.82.
  Describe the coherent principal square root of the phased coordinate
  for \(\alpha=-\beta,\zeta=-1\).
\item
  \textbf{Hard.} Reconstruct Figure 10.17 over \(K=\mathbb Q\) at
  \(p=5\). Prove that the q-primary saturation lattice of \(4,9\) has
  basis \(B=\tfrac12I\) and index \(J=4\), and identify the even phase
  class of its generators \(2,3\). Compute the exact prepared Euler
  range at reference \(0\), and again at reference \((8,0)\). Explain
  why the full box widths give a bound without a factor two, and why
  omitting \(J\) can even destroy integrality. Then return to Solution
  34 and use (10.222)--(10.228) to certify the uniform initial height
  bound, retaining every field and scalar cost.
\item
  \textbf{Hard.} Reconstruct Figure 10.18 for \(K=\mathbb Q(\sqrt5)\),
  with original bases \(5,7\) and coefficients \(b_1=b_2=1\). Determine
  the full saturation lattice and its q-primary index at \(q=2\), verify
  \(B'=BU\), and compute both lists of generator heights. Identify the
  field and its discriminant. At \(p=3\), determine \(e,f,u,\nu,P\) and
  the phase classes. Use the constants of Solution 34, put
  \(g_0=3(2a_0+a_1+\ln(2a_0+a_1)+\ln2)\), \(h=g_0\),
  \(\sigma_1=\ln5,\sigma_2=\ln7\), and retain its \(\tau\). Certify the
  original \(g_{11},g_{12}\), a strict initial column surplus, and the
  full bound (10.235). Compare the generator field with the smallest
  field of the actual rows.
\item
  \textbf{Hard.} Reconstruct Figure 10.19 for the entire coefficient
  family (10.245). Obtain a rigorous common Euler-argument bound and a
  bound for \(v_3(b_2)\), then determine every
  \(R_j,T_j,\mu_j,n_j,B_j\). Certify both comparisons at all three
  integer steps, retaining the complete pointwise factorial and lcm
  costs. Explain how a ratio search bounds the enormous Euler binomial
  without constructing it, and why these integer comparisons do not
  establish the fractional descent.
\item
  \textbf{Hard.} Derive the five Hermite basis polynomials for
  \(R_0=0,R_1=1,\mu_0=3,\mu_1=1\), checking every divided jet.
  Reconstruct the nested weights, total multiplicity and both strict
  comparisons in Theorem 10.95. Explain why equality of the chosen
  completions does not remove the quartic arithmetic factor, and why
  evaluating the denominator cost only at the scalar-maximizing order
  need not bound the joint maximum.
\item
  \textbf{Hard.} Expand \(\binom{2n+a}{2}\) in the basis
  \(1,n,\binom n2\), and give the inverse expansion of \(\binom n2\) in
  the old argument. Explain the relevant integrality at three and two.
  Reconstruct all seventeen rows of the table in Theorem 10.98,
  retaining the odd-node loss, the true contracted Euler bound and both
  global field factors. Give the stage-one and stage-seventeen radii,
  orders and mixed multiplicities. Complete the polynomial
  contradiction.
\item
  \textbf{Hard.} Verify the smallest-rank rational constants in
  Theorem10.100 and prove that every displayed bound increases with its
  rank or subset-rank index. Explain why using \(a_*=7\) in every case
  would omit an original parameter case. Check the complete cancellation
  in the intermediate and final allocation ratios, retaining both
  factorials, all \(q^\nu\) and \(e\theta\) factors, and the
  real-to-integer floors. Explain how Corollary10.101 returns the new
  bases to the original field.
\item
  \textbf{Hard.} Derive (10.274) directly from (10.270) and the complete
  original parameter formulas. Identify where the old and new residue
  indices occur and why neither may be replaced by a fixed value. Prove
  the all-rank bound on \(t^jM_j\), including the field condition at
  two, and reconstruct the bound \(1/700\). Show that every independence
  and weighted-height condition in (10.277) survives composition with
  the integral vectors \(u_i\). Verify the rank-two example and explain
  why its actual minimum rank is two.
\item
  \textbf{Hard.} Check the four minima in (10.281) by exact rational
  arithmetic. Prove the bound (10.285) by adding its three separate
  contributions. Enumerate the full rank-two derivative simplex through
  order six and show that its 28 rows have mean additive and Euler order
  two even when the additive degree is two. Derive (10.290), and explain
  why zero rows may be retained in the integral kernel argument. Finally
  take \(k=4,L=1,p=3,u=2,s=0,I=J=0\) and the scalar column
  \(F_{0,1}(X)\). Compute its prepared value, the value \(W\) of
  (10.298), and their exact valuation difference. Explain the
  cancellation in (10.297) and why it does not remove the global degree
  \(d_F\).
\item
  \textbf{Hard.} Reproduce the certificate of Lemma 10.111 using only
  integer and rational arithmetic. Explain why checking
  \(\psi(n)>25n/26\) would be insufficient for the real-argument lower
  induction. Show that \(F\) has period 30030 and that its reflection
  sum is three. Derive \(C_r^{\sharp}\), and quantify the improvement
  over \(C_r\) at rank two.
\item
  \textbf{Hard.} Derive (10.309) directly from the positive geometric
  series, retaining a prescribed additive order. Reconstruct every
  scalar and bound in Figure 10.26 with exact rational arithmetic before
  taking logarithms. Show that its peak is at \(u=1\), and explain why
  the remainder in (10.310) has a separate additive term. Verify the
  seven rational products in its proof and the two sharp-regime gaps in
  Lemma 10.114.
\end{enumerate}

\textbf{47 (additive divisibility and exact jet envelopes).} At
\(p=3,k=4,L=2\), compute \(D_{\mathrm a},\ell_p,\chi_p\). With
\(\beta=0,\vartheta=1,B=0,M=7\), evaluate the full integer minimum
\(K_u(M,B)\) for \(0\le u\le8\), and compare it with the linear-loss
bound from \(C_0\). Explain why no further lcm loss occurs after the
breakpoint in the left panel of Figure 10.27. Next set
\(\beta=3,B=1,M=5\), keeping the other data, and evaluate all nine
envelopes again. Compare with the linear consequence of (10.318).
Finally explain why a local q-unit in the jet proof does not allow
deletion of \(\Xi_{I,J,u}\ln q\) or of the global field degree in the
zero criterion.

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\setcounter{enumi}{47}
\item
  \textbf{Hard.} Reconstruct the strict original integer-extension
  comparison (10.332) in every field case. Explain why rank eight
  supplies a uniform estimate for all larger ranks, whereas checking
  finitely many ranks alone would not. Calculate the exact uniform I.2
  margin and its difference from \(1/200\). Prove why the first
  q-denominator indicator does not spoil the increase from step zero to
  step one. Finally explain why the continuous depth curve in Figure
  10.28 does not assert existence of an auxiliary stage at every plotted
  integer depth.
\item
  \textbf{Hard.} Use the exact original floor data of Solution 39 to
  calculate all three nested multiplicities in (10.336) at output order
  \(O=993090\). Calculate \(N_*\) both by annulus counts and by the
  telescoping formula. Compare it with the last zero block alone, and
  explain why this comparison concerns interpolation precision rather
  than actual nonzero function valuations. Prove the uniform four-term
  lower bound in (10.342), including its radius-floor correction.
\item
  \textbf{Hard.} Follow the complete first contraction through
  (10.344)--(10.348). In the original rational example, verify
  \(R=1374\), \(M=178095\), \(B=7\), and the q-deleted cardinality. At
  fixed additive order zero and \(\beta=0\), compute the two separate
  Hermite input losses. Explain why the fractional step retains
  \(q^{h_\Lambda}\), whereas the contracted integer step uses the field
  \(K\), even though both selected local completions may already
  coincide.
\item
  \textbf{Hard.} In the small profile (10.355), list the nodes and their
  multiplicities, compute the four deleted-node residue counts modulo
  four, and derive the three cardinal-factor bounds from (10.354).
  Verify the total degree by both the annular and telescoping formulas.
  Explain why replacing the discrepancy two by one changes the proof.
\item
  \textbf{Continuous-depth comparisons and terminal closure (advanced).}
  Prove that a concave function on \([a,b]\) is bounded below by the
  lesser of its endpoint values. Explain why the positive coefficients
  of \(\tau_j\) in (10.367) are needed when replacing them by
  conservative endpoint bounds. Derive both geometric sums used in
  (10.373). Finally, explain why the phase induction needs a closure
  estimate on \([1,L_D+1]\), and compute the added clearing cost at its
  last endpoint.
\item
  \textbf{{[}Challenging: a complete original mixed profile.{]}} Take
  the illustrative rank-two III.2 parameters
  \(q=2,c=0.5267,S_I=100,T_I=1000\). Compute H, the full integer radii,
  all original derivative cutoffs, the fractional output cutoff O, the
  three multiplicities \(\mu_0,\mu_1,\mu_2\), and the deleted increment
  E. Count the interpolation degree by disjoint node types and verify
  (10.356). Explain why the terminal closure at \(J=L_D+1\) uses a
  clearing cost \(\ell(L_D+2)\), and why a local completion of degree
  one does not remove \(q^r\) from the fractional comparison.
\item
  \textbf{The geometric endpoint and the preserved rank allowance
  (intermediate).} (a) For arbitrary \(Q>1,D>0\), show that
  \(I^*=\lfloor3D/\ln Q\rfloor+1\) is the least integer for which
  \(Q^{I^*}>\exp(3D)\). Evaluate this integer when
  \(Q=9/8,D=64\ln(9/8)\), and explain why equality \(I^*=I_1\) is
  included in Theorem 10.141. These numbers test the stopping
  definition, rather than claim a complete choice of contradiction
  parameters. (b) Explain why the Laurent monomial shift in Proposition
  10.31 preserves all ordinary hyperplane jets through the same total
  order. (c) If \(L_i'=\sum_ju_{ij}L_j\) and
  \(R_i=\sum_j|u_{ij}|\sigma_j\), prove the weighted-height condition in
  (10.277) for \(L_i'\). Then verify that the factor
  \(\Gamma_n^{r-m}M_r/M_m\) cancels the original rank allowance to give
  precisely \(\Psi_m\), including \(m=1\).
\item
  \textbf{The multiplicity endpoint and final support (intermediate).}
  (a) For \(a=3,c=269/500,\eta=1-c/a\), set \(T_1=a/(c\eta)\). Find the
  unique \(r_1\) of Lemma 10.142 and compute \(cT_\flat/a\). These
  scalar values illustrate the stopping definition, rather than a full
  choice of original contradiction parameters. (b) Show why replacing
  floor plus one by a ceiling in the definition of \(I_2\) fails at an
  integer quotient. (c) Verify the rational equality
  \(5220/(7\cdot9\cdot2^{13})=145/14336\), and explain why the strict
  coordinate bound in Proposition 10.143 leaves only the origin in the
  integral torus support. Explain why the resulting nonzero additive
  polynomial cannot have the stated number of integer roots.
\item
  \textbf{The retained terminal jets (advanced).} (a) Compute
  \(R_0,N_*\) for \(R_b=25\) and \(R_b=26\), and prove the two mass
  inequalities for every positive integer \(R_b\). (b) Explain why the
  outer cardinal factors lose at most \(B\), rather than zero, and why
  this is already charged in the \(M=2,E=0\) input bound. (c) Starting
  from \(1\le c\eta^rT_1/a\) and \(T_1<a/(cH)\), derive the upper input
  bound \(P_r\) and strict mass lower bound \(G_r\) in Theorem 10.144.
  Identify the term in \(C_r\) that retains the full torus and
  q-denominator cost. (d) Explain why the eight rank-eight endpoint rows
  cover every larger rank, and why a finite list of successful sampled
  ranks would not give that conclusion.
\item
  \textbf{Values without additional ordinary jets (advanced).} (a)
  Reconcile the zero-extra-jet convention with multiplicity \(M=1\) in
  Theorem 10.119. Explain why the cardinal and inverse-polynomial losses
  vanish, while the coefficient minimum \(\beta\) remains. (b) Derive
  \(c_1W_*\beta\le1/(c_3Y_-g_4)\) directly from the original formulas
  for \(S,T,W_*\) and \(ef\le d\). (c) For \(p=5,q=2,R=2,x=1/2\),
  compute all five \(L_s(x)\) and their valuations, and verify the exact
  root-polynomial calculation in Figure 10.36. These values illustrate
  full-node normal interpolation; they are not a complete choice of
  logarithmic contradiction parameters. (d) Explain why the local
  analytic conclusion of Proposition 10.145 alone does not prove the
  fractional zero, even if its value belongs to the selected completion.
\item
  \textbf{The successor integer block (advanced).} (a) Prove that the
  multiplicity-stop index is at least eight for odd p and at least six
  in V, using the explicit band bounds on H in Theorem10.146. Show why
  this implies \(Q^{I_1}>q\). (b) Take \(p=2,q=3,R=3\). Compute the four
  q-deleted cardinals at zero and their valuations; compare their actual
  loss with \(B=\lfloor\log_2(2R+1)\rfloor\). (c) Derive the bound on
  \(I_3\ln q\) in the theorem, retaining its floor-plus-one endpoint.
  Identify the denominator term that would be lost by using the old
  geometric range for J. (d) Explain why the theorem includes \(J=I_3\),
  and what must still be proved to construct that family.
\end{enumerate}

\textbf{Exercise 59 (medium: the field of a prepared value).} Use the
exact phase data and principal roots of Theorem10.147. (a) Compute the
surviving component spans and exact degrees of \(U_1U_2\), \(1+U_1\) and
\(V=2-U_1-U_2\). (b) Prove the displayed minimal polynomial of V and its
norm. (c) Compute its four valuations above73 without decimal root
approximations. Explain why a selected completion equal to
\(\mathbb Q_{73}\) does not identify those valuations.

\textbf{Exercise 60 (advanced: extra integer zeros at a fixed order).}
(a) For the illustrative floor data \(q=3,\theta=1,U=1001\),
\(\xi=24/25\), compute \(R_{\rm s},R_{\rm l}\) and their node counts in
Theorem10.149. Check \(qR_{\rm s}\le R_{\rm l}\) and the strict
normal-mass lower bounds. (b) Explain why all three stages use the same
selected order, while the two integer comparisons use the original field
and the fractional comparison retains \(q^r\). Locate the term which is
added outside that field factor. (c) Prove that the two-step depth
increments in Lemma10.148 give an enclosure of every rank at least32.
Why would a fixed value of \(m_{32}\), or finite positive rank
comparisons alone, not suffice?

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\setcounter{enumi}{60}
\item
  \textbf{Hard: the explicit bound after a change of rank.} (a) Derive
  the exact expression for \(eU_r\) in Theorem 10.152 from (10.186),
  retaining \(\gamma\), \(q^\nu\), \(q^u\) and both maxima. Substitute
  the profile (10.277) and identify every factor that cancels. Prove the
  bound on the remaining rank ratio for every \(1\le r\le n\). (b)
  Explain why the actual rank-one residue order gives \(M_1>g/t\), and
  verify the two strict fractions \(1/400\) and \(1/4000\) in that
  proof. (c) In the III.2 row, compute \(L_2,G_2\), and show by rational
  arithmetic that its gap exceeds \(4059/10^6\). Explain why this
  rank-two calculation controls every rank, and why retaining the lower
  bound \(g_1>509/100\) matters.
\item
  \textbf{Hard: replacing the coefficient-height parameter.} (a) Prove
  \(n!\le\mathrm e\sqrt n(n/\mathrm e)^n\) from the concavity of the
  logarithm, without a Stirling formula. Use it to prove the
  infinite-rank part of Lemma 10.153. Explain what the exact finite
  certificate covers and what it cannot prove alone. (b) Verify both
  alternatives \(B/\ln B\le W_B\) and \(B/\ln B>W_B\) in Theorem 10.154,
  including the small remainder \((d/t)\ln2\), the coefficient
  renumbering and the singleton-height bound. (c) Over \(\mathbb Q\),
  take \(p=3,a_1=4,b_1=3^k\) with \(k\ge1\). Compute
  \(g,\delta,q^u,M_1,C_1^*,\Omega,H\), and the exact valuation. Derive
  the displayed bound with factor \(1/2100\). Verify the two rational
  fractions used in the general rank-one proof.
\item
  \textbf{Hard: eliminating a dependent base with its weights.} (a) For
  \((2,3,4)\subset\mathbb Q^*\) at five, compute the greedy basis,
  \(L_3,L_2\), the two modified height products and the residue index.
  Transfer coefficients \((1,1,2)\) through the exact circuit, and
  compute the valuation. Identify its coefficient-height branch. (b) In
  the general proof of Lemma 10.155 derive the normalized circuit
  envelope, first for terminal removal and then for both internal cases.
  Include the empty-list height bound and every degree power. (c) Derive
  the polynomial lower bound for \(\ln(J_n/(n+5))\), and explain how the
  fourteen rational endpoints combine with it. Verify the final
  absorption inequality of Theorem 10.156. (d) Explain why removing
  either the first discarded base or an initial root of unity preserves
  both the greedy basis and its residue index; state the direct bound
  when the original list has rank zero.
\item
  \textbf{Hard.} For the indexed list \((4,16)\) at five, with
  coefficients \((0,5^k)\), compute the greedy rank-one basis, its
  original residue index, the length-two floor and weighted product.
  Give an exact one-base reduction and valuation. Then prove the
  cancellations in Lemma 10.157, explain why the maximum in the
  definition of s in the length-two scalar comparison is necessary, and
  verify the complete infinite-rank estimate in Theorem 10.158. State
  where a torsion complement changes the exponent budget.
\item
  \textbf{Medium.} For \(S=\{3,5,7\},p=3\), compute the common factor,
  coprime exponents and exact valuation of \(3^N25-3^N7\). Verify the
  residue index and constant displayed in Theorem 10.159, and show why a
  bound without the common-factor term would be false. For
  \(x=25,y=-7,p=2\), verify the field data, sign coefficient,
  length-three floor and weighted product. Finally prove the exact
  \(n=0\) clause and derive Corollary 10.160.
\item
  In Theorem 10.161, compute the exact III.1 budget by using
  \(d_{\min}=2\), \(\rho=58\), \(c_{\min}^{(2)}=495\), \(a^{(2)}=7\),
  and \(w_K/q^u\ge1\). Prove its strict comparison with \(1/4000\). Then
  determine the parameters for \(K=\mathbb Q,p=3,\beta=4,b=3^k\),
  \(k\ge1\), and verify both the exact valuation and the displayed
  second-constant estimate.
\item
  For the rank-one list \((4,16)\) at five in \(\mathbb Q\), find the
  second-constant height floor, the minimizing selected base, and its
  original residue index. For coefficients \((0,5^k)\), \(k\ge0\), prove
  the exact valuation. Then prove the all-length inequality
  \(\alpha_n>3n/2\) from the two scalar facts \(\alpha_2>3\) and
  \(F_{n+1}>3F_n\), and explain how the strict exponent and
  residue-height budgets add to \(1/4000\).
\item
  \textbf{(Medium: ordering, residue order and the two cutoffs.)} For
  \(p=5\), \(\alpha_1=1+5^m\), \(\alpha_2=1+2\cdot5^m\), and
  \(b_1=b_2=1\), verify all hypotheses of Theorem 10.165, determine
  \(g,E,u,H_6,H_4\), and calculate the exact difference valuation. Which
  displayed bound is smaller? Explain why the coefficient-ordering
  condition is automatic when both bases have the same deep precision,
  and why it cannot be dropped solely by interchanging a pair whose
  first base alone has that precision.
\item
  \textbf{(Medium: the two 2-adic unit classes.)} Apply the sign
  normalization of Corollary 10.166 to \((\alpha_1,\alpha_2)=(3,5)\).
  Compute the valuations for \((b_1,b_2)=(1,2)\) and \((2,2)\). Prove
  that every odd \(b_1\) gives valuation one, for every nonzero integer
  \(b_2\). Explain how the limiting argument permits a prescribed
  logarithmic height equal to \(\ln2\).
\item
  \textbf{(Medium: fixed rectangles and the one-base case.)} Verify the
  exact margin \(22/7-96/35=2/5\) and the coefficient
  \(539/2+7/3=1631/6<283\) in Theorem 10.167. For \(a=8,b=1,p=3\),
  choose an allowed auxiliary prime \(c\), check the
  valuation-preserving identity and calculate the exact
  Fermat-difference valuation. Determine \(v_3(8^n-1)\) for all positive
  integers \(n\).
\item
  \textbf{(Medium: congruence capacity and field data.)} In the
  \(8\)-by-\(8\) rectangle with \(r+s\equiv0\pmod2\), select all 32
  admissible pairs and put \(K=4,L=8\). Compute the exact maximum
  centered first-coordinate sum, the compressed occupancy bound and the
  relaxed bound \(G_1\) of Lemma 10.168. Then take the independent
  rational bases \(2,3\) at \(p=5\): determine \(d,e,f,D,g\) in Theorem
  10.170 and explain why no unramifiedness assumption or
  coefficient-ordering assumption is needed in its general statement.
\item
  \textbf{Hard: constants and coefficient imbalance.} (a) Derive the
  exact quotient \(C_2/C_1\), and verify its seven rational lower bounds
  under \(n\le Pt\). Calculate the smallest gap above \(3/2\). Why does
  the comparison also require Lemma 10.172? Determine P over
  \(\mathbb Q\) at three and at an unramified prime at least five. (b)
  For \((2,5)\) at three with exponents \((3^k,1)\), compute the
  coefficient ordering, residue index, height factor X and exact
  valuation. State which second estimate applies. (c) Prove the scalar
  logarithm bound used in Lemma 10.174, verify the exact positive
  exponential-series margin, and explain why the scalar inequality can
  be weak although Theorem 10.175 is strict. In the model of Figure
  10.51, express both branches in terms of \(z=\varepsilon r\), and show
  that \(z\ge0\) alone does not specify the allowable parameter
  interval.
\item
  \textbf{Hard: the congruence cost and the analytic gain.} (a) For
  \(u_1=10,u_2=19\) in \(\mathbb Q_3\), with j=2,m=1, compute the
  unit-product homomorphism on \(\{0,\ldots,5\}^2\). Give its three
  fibres and the even-coordinate contraction of the zero fibre. Explain
  why the entire set of auxiliary polynomial indices E must be retained
  independently of that homomorphism. (b) Derive the rounding threshold
  of Lemma 10.177 and the bound on \(p^{fm}(A+m)^{-r}\). In the
  \(A_0=1/2\) row, verify both the rational gap between \(1.3701\) and
  its H endpoint and the gap between its F endpoint and \(1/1.0568\).
  Explain why endpoint calculations alone are insufficient. (c) In
  Corollary 10.179 take \(p=3,q=2,I=1,a=1,k=\ell=2,t_0=1\), with an
  empty product of the other prepared factors. Compute P exactly,
  including its coefficient valuations. Use N=2 and \(c_\lambda=1\) on
  the zero congruence fibre, with zero perturbations, to compare the
  uniform normalizing valuation with the sharper termwise one. Compute
  f(0), and explain the original field degree used in a later global
  height argument.
\item
  \textbf{Advanced: selection, projection and a complete input budget.}
  At three take \(a_1=10\), \(a_2=3281/500\), \((b_1,b_2)=(3,1)\),
  \(P=1\), \(q=2\), \(\nu=0\), and \(L_1(Z)=Z_1,L_2(Z)=Z_2\).
\end{enumerate}

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\tightlist
\item
  Determine the exact valuations of \(a_1-1,a_2-1,\Xi-1\), and all
  quotient classes modulo 27 on \(\{0,\ldots,5\}^2\). For the class
  containing zero, find the projected frequency and explain why the
  closed disc with \(N=2\) is allowed even though the first individual
  logarithm has valuation two.
\item
  Take \(k=L=2\), the full nodes \(-1,0,1\), and every prepared index of
  total degree at most two. Count the columns and the whole row list,
  prove that a nonzero integral kernel exists, and identify its row
  field. Explain how the actual count improves the height row factor to
  \(3/5\). Verify one primitive integral kernel using the reproducible
  exact calculation supplied with Figure 10.53.
\item
  Take \(\mu=3\) and base prepared index zero. Determine
  \(B,C_0,\Gamma\) and the output precision from Theorem 10.182. Explain
  which equalities in the numerical budget are harmless and which disc
  inequalities must be strict. Finally derive the growing Taylor
  precision in Lemma 10.181 directly from its exponential difference.
\end{enumerate}

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\setcounter{enumi}{74}
\item
  \textbf{Second-scale support and its exact integer floors (hard).}
  Over \(K=\mathbb Q\), take \(p=3\), the four independent bases
  \(2,5,7,11\), weights equal to their logarithmic heights, and
  \(h=150\). Use the parameters of Theorem 10.184. Determine the
  optimizing depth, all four integer side lengths, the selected
  congruence class modulo 81, and the integers
  \(k,L,D_{\mathrm a},\lfloor S\rfloor,\lfloor T\rfloor\). Count the
  columns and the full prepared rows exactly and prove their required
  surplus. Check also the first full-node local input budget of Theorem
  10.182 when the original distinguished coefficient is a 3-adic unit.
  Give a reproducible rational calculation and state precisely which
  assertion about the full valuation estimate remains outside this
  finite check.
\item
  \textbf{Closing the small optimizer and retaining the first input
  (hard).} \textbf{(a)} At eleven take the independent rational bases
  \(2,3,5\). Prove, using elementary exponential bounds, that Theorem
  10.188 applies while the numerical range \(n\le Pt\) in Theorem 10.173
  does not. \textbf{(b)} At five take two independent rational units.
  Derive the strict \(11/8\) comparison of their coefficient parameters
  and the exact remaining constant gap. Explain why the same calculation
  does not settle all higher-degree fields unramified at five.
  \textbf{(c)} For the four-base support of Solution 75, with original
  \(b_n=B_\dagger=1\), check the additional height condition of Theorem
  10.190 and bound its first local input cost as a fraction of \(U_r\).
  Keep the actual \(\mu,B,C_0\). Distinguish this analytic calculation
  from a proof of an algebraic zero.
\end{enumerate}

\textbf{Exercise 77 (degree allocations and the second rank allowance;
intermediate).} Use the parameters of Theorems 10.184 and 10.191.
\textbf{(a)} Derive the exact unrounded lower expressions for the
rank-zero and intermediate allocation ratios. Explain why the polynomial
degree is \(kL\), rather than \(\mathcal D\), and check the factor
\(((m+1)!)^2\). \textbf{(b)} Verify the final universal bounds
\(2^{20}/29>8\), \(1050/2^{36}<10^{-6}\), and explain which conclusion
still requires prepared zeros at the stopping depth. \textbf{(c)} For
the rank-one argument, prove the inequality
\(\mathcal R_nA_n/t^2\ge1/n\) in both ranges of t. Recover the number
9261 and compare the two elementary contributions with
\(C_2h\Omega_n/3000\).

\textbf{Exercise 78 (complete averages and exact zeros; advanced).} For
the second-scale parameters of Section 64: (a) derive the coefficient
\(C_r\) by averaging the additive, Euler and slack coordinates, and
prove its stated deficit from one; (b) explain why using the Euler
parameter \(a_E\) in the remainder allows the lcm coefficient
\(107/103\) to be kept separately; (c) in case I.2, at
\(y=1/4,A_0=3/2\), evaluate the analytic and arithmetic bounds in
Theorem 10.197 and prove their difference exceeds \(41/1000\); (d) prove
the floor inequality that supplies all ordinary input jets, and identify
the two additional costs that prevent this integer argument from proving
a fractional extension by itself.

\textbf{Exercise 79 (uniform integer stages; advanced).} (a) Derive the
Bernstein coefficient identity in Theorem 10.198 and use it to prove
\(1-2y+3y^2\ge11/16\) on \(0\le y\le1/4\). (b) For r = 6 and q = 2,
compute the final radius and derivative cutoff as exact multiples of S
and T. Explain how the floors enter. (c) Verify both strict local-input
margins below \(15U_r/16\) and count the polynomials and coefficients in
the finite comparison. (d) Prove that the stated positive tail increment
at five controls every subsequent integer j. Explain why the result does
not by itself supply the first fractional block.

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\setcounter{enumi}{79}
\tightlist
\item
  \textbf{The full first fractional budget (advanced).} (a) In case I.2,
  show directly that the actual selected-column ratio is greater than
  \(27/2\), so the row-to-surplus ratio is less than \(2/25\).
\end{enumerate}

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{1}
\tightlist
\item
  Verify the reserve in Lemma 10.200 for that case, including its strict
  remaining rational margin. (c) For the illustrative nested counts
  \(S=21/2,T=1200,r=3,q=2,c_5=27/50\), compute all four radii,
  derivative cutoffs and multiplicities of Theorem 10.201, and their
  total interpolation mass. These are a counting example, not an
  assertion that this T meets the second-scale construction. Compare the
  mass with that of the final interval alone.
\item
  In \(\mathbb Q_7\), let \(\alpha^2=2\) with \(\alpha\equiv3\pmod7\).
  Prove that \(v_7(\alpha-3)=1\) and \(v_7(-\alpha-3)=0\). Explain what
  this says about replacing the global root-field degree by its chosen
  local degree. (e) Explain which inequalities in the proof make the
  seven rank-eight comparisons cover all larger ranks.
\end{enumerate}

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\setcounter{enumi}{80}
\item
  \textbf{The first contracted block.} Use the exact counting data of
  Solution 80: \(S=21/2\), T = 1200, r = 3, q = 2, \(c=27/50\) and
  \(\eta=173/200\). Compute R, the input and output order floors, the
  interpolation multiplicity M, the number of q-deleted nodes and the
  cardinal degree. Check the available ordinary-jet order. Explain why
  the actual value field at an integer of the contracted family is K and
  why this does not remove its rational q-denominator.
\item
  \textbf{Saturation and the depth cost (advanced).} (a) For
  \(h>40,A>5,x\ge0\), prove \((h+x)(A+x)\le hA\exp x\) by comparing the
  coefficients of a quadratic. (b) At \(q=2,r=3\), use
  \(c_2=8750/5101\), \(c_4=14\), \(\lambda=59/58\) and \(\xi=101/100\)
  to verify exactly that the linear endpoint ratio in Lemma 10.205 is
  below \(3/5\). (c) Explain why the geometric assertions in Theorem
  10.204 alone do not supply the zeros assumed in its rank conclusion.
\item
  \textbf{Hard.} In the counting illustration of Figure 10.62 take
  \(q=2,r=3,I=2,S=21/2,\widehat T=1200,c=27/50\). Compute the three
  integer radii and derivative cutoffs in Theorem 10.206, both
  interpolation multiplicities, and both cardinal degrees. Verify the
  exact floor inequalities. Explain why the term \((\mu_j-1)M_*\) has a
  bound independent of I after expressing it in terms of \(\widehat T\),
  while the original coefficient loss contributes \(1/\widehat T\). Do
  these counting data alone establish the existence of a depth-two
  auxiliary family?
\item
  \textbf{Mixed profiles (advanced).} In Figure 10.63 use
  \(q=2,r=3,I=1,S=21/2,\widehat T=1200,c=27/50\). Compute the three
  radii, four multiplicities, deleted increment, and total interpolation
  degree, both by disjoint node types and by nested layers. Verify every
  required jet-order floor. Explain why the finite-rank arithmetic
  comparison in Theorem 10.207 uses \(A_0\le r\), and why Theorem 10.208
  can supply the first integer block beyond \(I\log Q\le3B\) without
  discarding a denominator. Does Corollary 10.209 handle a family that
  reaches the derivative-order threshold first?
\item
  \textbf{The last integer block (advanced).} Use the counting data
  \(q=2,r=3,c=27/50,\eta=173/200,S_1=40,T_1=12\). Find \(j_0,R,O\) from
  Lemma 10.211 and check the final jet-order floor in Theorem 10.210.
  Explain why \(\mu=1\) would also be allowed. Derive the constants
  \(609/163840\) and \(1305/236\) in Proposition 10.212. Which
  additional assertion is needed to turn that proposition into a
  contradiction for the starting family?
\item
  \textbf{A fixed order and two different integer intervals (advanced).}
\end{enumerate}

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\tightlist
\item
  In case III.1 at r = 3, calculate \(L_r,m_r,j_r\) from Lemma 10.213
  using exact rational comparisons. Identify the two vertices sufficient
  for the convex bound in Lemma 10.214 when B = 10.
\item
  For the counting parameter \(u=\zeta\chi\widehat U_r/N=201/2\) and q =
  2, calculate \(R_{\rm s},R_{\rm l},n_{\rm s},n_{\rm l}\). Check all
  three floor inequalities in Theorem 10.216.
\item
  For a q-deleted interval of radius R = 40 with q = 2 and p = 5, count
  its nodes and find its separation bound. Write both analytic entries
  at multiplicity one. Explain why the derivative order is unchanged and
  why the original coefficient loss remains. Parts (b) and (c) use
  node-count examples, not complete auxiliary kernels.
\end{enumerate}

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\setcounter{enumi}{86}
\item
  \textbf{A smaller height floor (advanced).} (a) For the boundary
  parameters \(a_1=a_2=1/10,g=1,B=1200\) in Lemma 10.218, compute
  z,L,K,N. Check the lower bound on K and the strict determinant and
  coefficient margins. These are interpolation parameters, not an
  asserted example of algebraic numbers attaining all boundaries. (b)
  Separate the three possible padding cases in (10.992) and derive their
  bounds from the singleton and two-base height inequalities. Explain
  why squaring the singleton bound would be weaker. Verify the two exact
  fractions in (10.996) and the degree 2048 margin. (c) Why must a zero
  coefficient be treated separately? Derive both terms of (10.999),
  identifying which height bound absorbs each term.
\item
  \textbf{Original coefficients through a change of rank (advanced).}

  \begin{enumerate}
  \def\labelenumii{(\alph{enumii})}
  \tightlist
  \item
    For a rank-r integral-form representation, prove (10.1007) after a
    second integral change of forms. Explain why clearing denominators
    of its rational coefficient vector does not, by itself, give the
    coefficient-height hypotheses needed for Theorem 10.152.
  \item
    Derive the rank quotient (10.1012), including the factorial and
    residue factors, and recover the uniform fraction \(25/41\) when r=2
    and n is at least three.
  \item
    At five take the independent rational units \((2,3,11)\), original
    coefficients \((1,1,1)\), and forms \(L_1=z_1+z_2,L_2=z_3\). Compute
    the two saturation and residue indices, prove that weights
    \((\log6,\log11)\) satisfy (10.725), and find \(v_5(\Xi-1)\). Does
    this example prove that two is the least admissible rank?
  \end{enumerate}
\item
  \textbf{Hard: padded heights and a separate coefficient bound.}

  \begin{enumerate}
  \def\labelenumii{(\alph{enumii})}
  \tightlist
  \item
    Prove the monotonicity in Lemma 10.224 also when one of the
    undistinguished coefficients is zero. Explain why increasing only
    the height product without updating the logarithmic coefficient
    parameter would not establish that lemma.
  \item
    Derive the three rational envelopes in Lemma 10.225 directly from
    (10.642), including the unramified quadratic field at two. Prove
    their decrease for every integer n at least two.
  \item
    Take \(p=5,(a_1,a_2)=(2,3),(b_1,b_2)=(5^8,1)\),
    \((w_1,w_2)=(1,\log3)\), and \(B_n=1\). Write the two branches and
    the inflated last weight for \(\varepsilon=1/2\) and
    \(\varepsilon=5^{-8}\). Prove the exact valuation. Explain precisely
    why the same proof does not cover arbitrary dependent unit parts.
  \end{enumerate}
\item
  \textbf{Medium to hard: rational dependencies with a distinguished
  coefficient.}

  \begin{enumerate}
  \def\labelenumii{(\alph{enumii})}
  \tightlist
  \item
    Find a primitive rational generator for \((4/9,8/27)\). Compute the
    combined exponents for coefficients \((3,-2)\) and \((2,-1)\), and
    explain which product is excluded by the nonvanishing condition. At
    p=2 compute the valuation for bases \((-3,9)\) and coefficients
    \((1,1)\); explain why the even power in Theorem 10.228 handles its
    sign.
  \item
    Derive \(\log|M|<3U\) retaining \(B_n\), and prove the factor
    \(14/3\) in (10.1045) for every prime and every \(n\ge2\). Identify
    where the proof uses rationality.
  \item
    Prove (10.1046) for every \(k\ge1\), and compute U from (10.1047) at
    \(\varepsilon=1/2\) and \(\varepsilon=5^{-k}\). Explain why the rank
    cases exhaust two rational bases but leave a case for \((2,3,6)\) at
    five.
  \end{enumerate}
\item
  \textbf{Hard: all rational ranks and the sign at two.} (a) For p=5,
  the bases (2,3,6), and coefficients (2,1,1), compute the determinant
  reduction in Lemma 10.230(a). Verify the exact product and its
  distance from one. (b) At p=2, use the bases (-3,27,5) and
  coefficients (1,1,1). Find a primitive relation among the first two
  vectors, choose an odd primitive coefficient for elimination, and
  double the whole relation as in Theorem 10.232. Check both valuations
  in the factored-power inequality. Repeat for coefficients (2,2,2),
  tracking the common additional factor two in the transferred
  coefficients. (c) Derive (10.1062), including the cancellation of the
  chosen relation coefficient. Explain why doubling the whole relation
  at p=2 preserves the minimum-valuation condition even though its
  minimum increases.
\end{enumerate}

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{3}
\tightlist
\item
  Prove that the two distinguished-vector cases exhaust every
  intermediate rational rank, and explain why the determinant and
  circuit arguments do not by themselves prove the number-field
  refinement.
\end{enumerate}

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\setcounter{enumi}{91}
\item
  \textbf{Hard: the second estimate and an unchanged basis.} (a) For
  \(K=\mathbb Q,p=5\), use the sorted list \((2,3,4)\), coefficients
  \((1,1,2)\), and B=3. Find the greedy basis and its saturated residue
  index. Compute \(\Omega_3\), determine the two maxima \(M_3,A_3\) in
  (10.1064), and evaluate the product's distance from one. Eliminate4 by
  an exact circuit and give the new coefficients. (b) Repeat the
  torsion-removal step for \((-1,2,3,4)\) with all coefficients one.
  Track the two valuations and explain why the basis is unchanged. (c)
  Prove (10.1065) by maximizing its quotient in t. Use the exact sum
  \(1957/720\) to verify the positive margin in (10.1085) at two,
  explaining where the hypotheses on P and the residue degree enter. (d)
  In the circuit calculation, explain separately the cases in which the
  largest-height base is deleted, is a surviving basis member, or is a
  surviving nonbasis member. Why is the last case impossible when s=0?
\item
  \textbf{Hard: normalization without changing the field.} For the
  quadratic example in Section 79, verify the two original heights from
  their primitive minimal polynomials and verify the valuation of the
  product minus one by lifting the root of \(X^2-2\) modulo 49. Derive
  the rational upper endpoint for the first row of Lemma 10.238 and
  explain why it controls every rank. Finally, at the same place, take
  the bases \((7,1+\sqrt2)\) and coefficients \((0,1)\). Check the
  coefficient condition and explain exactly why Theorem 10.239 does not
  apply to this last list.
\item
  \textbf{Dependence without a prime-to-p relation.} In the quadratic
  example of Section 80, use \(\varepsilon=1/2\), \(B=B_n=6\), and any
  allowed height weights. (a) Compute the norm of each of \(a,c,3\),
  prove their independence, and eliminate the third base from
  \((a,c,ac,3)\) using \((-1,-1,1,0)\). Give \(B',B_n',\varepsilon'\)
  and verify the equality of the second entries in (10.1117). (b) Repeat
  with the exact relation \((-7,-7,7,0)\), using the relation bound
  \(\kappa=7\). Check the minimum seven-adic valuation of the
  transformed coefficients. Explain why the proof does not need a
  relation coefficient prime to seven. (c) Lift the root 3 of \(X^2-2\)
  modulo seven first to 49 and then to 343, and compute the valuation of
  \(3^6(ac)^2-1\). With coefficients \((1,-1,1,6)\), show instead that
  elimination produces a zero coefficient while preserving the
  distinguished minimum valuation.
\item
  \textbf{Quadratic descent at two (hard).} Let \(K\) be a number field
  not containing \(\zeta_3\), and let \(L=K(\zeta_3)\). If \(T\) is odd,
  \(A\in K^*\), and \(\beta^T=A\) with \(\beta\in L\), construct a
  \(T\)-th root of \(A\) in \(K\). For odd residue degree \(f\), explain
  why the saturated residue factor satisfies \(N^2/(m\delta_L)<N\).
  Verify the all-rank ratio in (10.1128) and the largest endpoint in
  (10.1129). Finally apply Theorem 10.246 to the rational list
  \((6,11/2,33)\), with all three coefficients one, \(B=B_n=1\), and
  \(\varepsilon=1/2\): determine the normalized units, their rank, and
  the exact valuation of the product minus one.
\end{enumerate}

\textbf{Solution 1.} Put \(q_i=v(\alpha_i)\). If \(b_1q_1\ne b_2q_2\),
the ultrametric inequality gives the exact valuation
\(\min(b_1q_1,b_2q_2)\), so no cancellation estimate is needed. If they
are equal, choose \(\pi=p^{1/e}\) in an algebraic extension and set
\(\epsilon_i=\alpha_i\pi^{-e q_i}\). The exponents \(e q_i\) are
integers, the \(\epsilon_i\) are units, and

\[
\alpha_1^{b_1}-\alpha_2^{b_2}
=\pi^{e b_1q_1}(\epsilon_1^{b_1}-\epsilon_2^{b_2}).
\]

The valuation is \(b_1q_1\) plus the unit-difference valuation. The
field degree, heights, and any dependence of the new bases must be
checked anew. The linear offset cannot be discarded. For instance, \(3\)
and \(6\) are multiplicatively independent but nonunits at three, and
\(v_3(3^m-6^m)=m+v_3(1-2^m)\ge m\). A bound by a fixed multiple of
\((1+\ln m)^2\) cannot hold. Thus unit normalization reduces the
cancellation problem; it does not preserve the unadjusted bound.

\textbf{Solution 2.} Use rational heights \(h(a)=\ln a\),
\(h(b)=\ln b\), and the choices \(\ln A_1=\ln(5a)\),
\(\ln A_2=\ln(5b)\). Coprimality makes both exponents vanish in a
multiplicative relation. Also

\[
\ln B'+\ln\ln p
\le\ln(p-1)+\ln(2/\ln5)+\ln\ln p
\le3\ln p.
\]

Hence the maximum is \(10\ln p\), not the smaller first term. Squaring
it cancels two of the four denominator powers of \(\ln p\);
\(2500\cdot100=250000\). This proves (10.34).

\textbf{Solution 3.} The residue field group has order \(p-1\), so the
order \(q\) of \(a/b\) divides it. A nonmultiple of \(q\) gives
valuation zero. For \(n=qm\), prime-to-p powers preserve the valuation
of \((a/b)^q-1\), and each p-th power adds one because its initial
valuation is an integer at least one. For odd \(p\), one is strictly
greater than \(1/(p-1)\), so the boundary case does not occur. The same
calculation for \(p-1=q((p-1)/q)\) has no p-part. Combine the two
identities and Corollary 10.9. At \(p=2\), valuation one lies on the
boundary; for example \(v_2(3^2-1)=3\), whereas \(v_2(3-1)+v_2(2)=2\).

\textbf{Solution 4.} Here \(N=12\), \(c_0=1\), and \(\sum_i k_i=12\).
Formula (10.10) gives

\[
v_3(\Delta)\ge54-\frac{12\ln12}{\ln3}.
\]

Its hypothesis is \(v_3(\Lambda)\ge23/2\), equivalently at least twelve
when the difference lies in \(\mathbb Q_3\). Before the common
logarithmic loss, the contribution from \(n\) ordinary rows is at least

\[
(12-n)\frac{23}{2}+\frac{n(n-1)}2-12
=54+\frac{(12-n)^2}2.
\]

Thus its excess over the common minimum is exactly \((12-n)^2/2\). This
is a lower bound for a summand's valuation, rather than an equality for
that valuation.

\textbf{Solution 5.} Order the row labels as \(0,0,1,1,2,2,3,3\), and
take \(r_i=0,0,0,0,2,2,2,2\). Then \(\sum_i r_i=8\),
\(\sum_i l_ir_i=20\), and the center is \(12\). The error is eight,
equal to \(NL(R-1)/8=8\). Reversing the chosen zero and two coordinates
attains the negative error.

\textbf{Solution 6.} Take \(p=3\), \(x=3^{10}+1\), \(y=1\). The pair is
coprime and \(p\nmid x\). Factor \(x^2-1=(x-1)(x+1)\). Its valuation is
ten, because \(x-1=3^{10}\) and \(x+1\equiv2\pmod3\). The proposed right
side is four, since \(\ln y=0\). This disproves the unrestricted
statement. All the estimates used above have their hypotheses proved and
stated separately.

\textbf{Solution 7.} Remove residue roots in the selected congruence
class, and remove the common p-part of the coefficients by the deep-unit
power law. The chosen base has \(v(\log\epsilon_1)\ge E\). The
assumption \(p\nmid b_2'\) gives integral nodes \(r+(b_1'/b_2')s\); it
cannot be dropped from this construction when the first base alone
supplies \(E\). The second principal unit must already lie in the
exponential domain because the exponent multiplier is now one. At odd
primes its positive integer valuation does this automatically; at two
the additional depth condition supplies it.

The pure minor of order \(n\) has the bound in the proof of (10.40),
while complementary error rows cost at most \(E\sum k_i\). A difference
precision \(E(N-1/2)\) and the identity (10.14) give
\(ENK(L-1)/2-N\ln N/\ln p\). Compare this with the same centered
polynomial height bound, now without any exponent multiplier. Use
\(a_i=\ln A_i/(E\ln p)\), \(B=H_E/(E\ln p)\), and \(G=p-1\). The
cardinalities and inequalities are exactly those checked in
(10.40)--(10.42); the collision branch uses the residue order before
applying \(X-1\) Liouville. The final budget is

\[
EzB^2
=E(p-1)\frac{\ln A_1\ln A_2}{E^2(\ln p)^2}
\frac{H_E^2}{E^2(\ln p)^2}
=\frac{p-1}{E^3(\ln p)^4}\ln A_1\ln A_2H_E^2.
\]

This accounts for every power of \(E\); replacing the established
coefficient \(1300\) by \(88000\) gives the asserted weaker version.

\textbf{Solution 8.} If a rational \(\beta\) realizes \(\boldsymbol v\)
in (10.44), valuation at two gives \(v_2(\beta)=2v_1\), and valuation at
three gives \(v_3(\beta)=2v_2\). Hence both coordinates lie in
\(\tfrac12\mathbb Z\). Conversely every such vector is realized by
\(\beta=2^{2v_1}3^{2v_2}\), with a common denominator \(T\) in (10.44).
Thus the stated saturated lattice and the bases \(2,3\) are exact. If
\((-1)^{a_0}2^{a_1}3^{a_2}\) is a rational square, prime valuations make
\(a_1,a_2\) even, and positivity makes \(a_0\) even. The three square
classes are independent, so the proved Kummer degree lemma gives degree
eight. At five, the residue of two has order four, since
\(2^2\equiv-1\pmod5\). The residue group generated by \(-1,2,3\) is
therefore the whole group of order four. Its index is one.

\textbf{Solution 9.} The choice \(H=\exp(n/(2d))\) cancels the \(2^n\)
in the diamond volume and the \(H^{2d}=\mathrm e^n\) in the count. The
mixed complex term contributes the additional factor
\(\mathrm e^{-n/2}\le1/\sqrt{\mathrm e}<2/3\). Its two coefficients in
(10.53) are less than 31 and 32, so their total is less than
\(31(2/3)+32=158/3<58\). The real coefficient is less than 33. The
selected archimedean weight is \(f\ell/d\); only \(\ell(d-f)/d\) remains
for the other archimedean Hadamard factors. This exact division gives
(10.50), before any numerical constant is chosen.

For rational 4,9, the valuations at two and three in (10.54) force
\(2v_1,2v_2\) to be integers, for any realizing integer \(T\).
Conversely \(\beta=2^{2v_1}3^{2v_2}\) realizes every half-integral
vector. Hence \(M_{\mathbb Q}=\tfrac12\mathbb Z^2\), its index is four,
and \(w_{\mathbb Q}=2\), since the only rational roots of unity are
\(1,-1\). Thus the left side of (10.55) is eight. For \(H=\mathrm e\),
(10.56) gives \(V=2/(\ln4\ln9)\) and average \(4V=8/(\ln4\ln9)\). The
three points \((-1/2,0),(0,0),(1/2,0)\) lie in the open diamond, while
the other points of its zero lattice coset lie outside. Their values are
\(1/2,1,2\), and the six signed values are distinct and have
multiplicative height at most two, which is less than \(\mathrm e\).

\textbf{Solution 10.} The residue of 2 at three is -1, of order two. An
odd exponent gives valuation zero. For even b, factor through 2\^{}2=4;
since v\_3(4-1)=1, the proved power law gives \(v_3(2^b-1)=1+v_3(b)\),
also for negative b by unit inversion. The height is \(\ln2\), so
(10.59) with \(d=e=f=1\) gives \(v_3(2^b-1)\le v_3(b)+4\ln2/\ln3\). In a
field with \(d=2,e=1,f=2\), the ratio \(d/f\) is still one, the residue
of the rational unit is still -1, and
\(\operatorname{ord}_{\mathfrak p}=v_3\). The same expression therefore
follows. A degree-only bound that discarded f would be larger, and a
normalization omitting e in a ramified field would instead be incorrect.

\textbf{Solution 11.} Here \(b=(0,0,1,1)\), \(B=(1,1,1)\), and \[
C=\begin{pmatrix}5&0&1\\0&5&1\\-5&-5&-2\end{pmatrix}.
\] The upper minors for column pairs \((1,2),(1,3),(2,3)\) are
\(25,5,-5\), with valuations \(2,1,1\). Choose columns \(1,3\). The
derivations are \(\delta_1=5(E_1-E_3)\), \(\delta_2=5(E_2-E_3)\),
\(\delta_3=E_1+E_2-2E_3\), and \(\delta_2=-\delta_1+5\delta_3\). Their
coefficients are 5-adically integral. Choosing columns \(1,2\) would
instead give \(\delta_3=(\delta_1+\delta_2)/5\). For example
\(\delta_2^2=\delta_1^2-10\delta_1\delta_3+25\delta_3^2\), so the
minimum-minor change preserves integrality for the second-order jets
too. The rank and Cramer arguments prove the assertion at every order.

\textbf{Solution 12.} Both sums of allocations are \(90\), and
\(D_0=T_2+2=32\) is largest. The \(m=0\) inequality is \(31^2=961>32\).
The \(m=1\) inequality is \(31\binom{32}{2}=15376>2\cdot32\cdot3=192\).
The last inequality is \(15376>6\cdot32\cdot2\cdot3=1152\). The bases
\(2,3\) are independent by their prime valuations. If the polynomial
were nonzero, the theorem would supply \(\nu=1\). Choose \(A_1=A_2=1\);
then \(C_{2,1}=2\), and any nonzero integral relation vector has
\(R_1\ge1\). Its first possible bound is \(31R_1\le6\), and its second
is \(961R_1\le384\). Both contradict \(R_1\ge1\). Thus no nonzero
polynomial can have the proposed jets.

\textbf{Solution 13.} Prime valuations show that the q-primary saturated
lattice of \(2,3\) over \(\mathbb Q\) is \(\mathbb Z^2\): any realizing
power forces both rational exponents to be integers. At five their
residues, together with \(-1\), generate all four nonzero classes. Thus
\(\delta(\boldsymbol\alpha)=1\), while \(w_q=2\), \(q=2\),
\(e_{\mathfrak p}=f=d=1\). We have \(a=28/3\) and
\(R_2=14\mathrm e(1+\varepsilon_2)\).

Here \(\gamma=6\) has residue order one, so \(\delta_\gamma=4\) and
\(M_\gamma=\max\{5/(4(\ln5)^2),\mathrm e\}=\mathrm e\). Indeed
\(\ln5>1\) since \(\mathrm e<3\), and \(\mathrm e>2\). Also
\(M_2\ge\mathrm e^2/4\). Consequently \[
R_2\frac{M_2}{M_\gamma}\ln2\ln3
\ge\frac72\mathrm e^2(1+\varepsilon_2)\ln2\ln3>7.
\] The last inequality uses \(\mathrm e>2\), \(\ln2>1/2\) from its
integral, and \(\ln3>1\). On the other hand \(\sigma=\ln6<2\), because
the first four exponential terms give \(\mathrm e>8/3\), hence
\(\mathrm e^2>6\). This proves (10.77). The height of the rational
number 6 is exactly \(\ln6\). The chosen \(H\) exceeds all three
required lower bounds. The valuation is \(v_5(6-1)=1\). Taking
\(c=1790\) matches the rational unramified case of the free preparation;
Theorem 10.27 supplies its fully proved numerical bound.

\textbf{Solution 14.} Put \(F_0=(X^2+3X+2)/2\), \(F_1=(X^2+5X+6)/2\).
The four polynomials are \(F_0,F_1,F_0^2,F_1^2\), and their coefficient
matrix is \[
\begin{pmatrix}3/2&5/2&3&15\\1/2&1/2&13/4&37/4\\0&0&3/2&5/2\\0&0&1/4&1/4\end{pmatrix}.
\] Its lower left block is zero; the diagonal block determinants are
\(-1/2,-1/4\), giving \(1/8\). Thus the polynomials are independent,
despite their common root \(-2\).

For the Taylor calculation, \(v(2)=2\). At \(z=1/3\),
\(\Delta(z;2)=14/9\), \(\Delta'(z;2)=11/6\), and \(\Delta''(z;2)=1\).
Expanding its square at that point gives the prepared coefficients \[
\left(\frac{196}{81},\frac{308}{27},\frac{59}{3},\frac{44}{3},4\right).
\] The exponent in (10.87) is \(\ell(kI+v_3(k!))=2(2+0)=4\).
Multiplication by \(3^4\) gives the integers \(196,924,1593,1188,324\),
as required. There is no denominator at five, so this clearing factor
has no cost in the 5-adic valuation.

\textbf{Solution 15.} Here \(v(1)=1\), \(N=3\), \(M=1\), \(\Omega=1\),
and \(\mathcal B=2\). The only equation is \(c_0+c_1+c_2=0\) at
\((X,Y_1,Y_2)=(0,1,1)\). Take \(c=(-1,1,0)\). Then \(h_\infty(c)=0\) and
\(h_2(c)=\frac12\log2\); the finite maximum norms are one. Formula
(10.96) is \(h_2(c)\le\frac34\log3+\frac12\log2\), which holds strictly.
Since \(\boldsymbol d=(1,1)\), the ordinary polynomial is
\(P=Y_1Y_2(X+1)(Y_1-1)\). It is nonzero and vanishes at \((0,1,1)\).
Only the order-zero value is imposed: the \(D_1\)-derivative there
equals one, so no first-order vanishing is asserted. Every nonzero
coefficient is a local unit at every prime, so the optional
normalization changes nothing.

\textbf{Solution 16.} In the first case \(n=7,B=1,\kappa=0\). Theorem
10.34 gives \(v_3(F(\rho x))\ge\min\{14,\Lambda-1\}\), so
\(\Lambda\ge15\) suffices. The larger logarithmic cost in Corollary
10.35 is sufficient but is not needed in this example.

The q-deleted set is \(\{-2,-1,1,2\}\), and \[
L_{-2}(0)=\frac{(0+1)(0-1)(0-2)}{(-2+1)(-2-1)(-2-2)}=-\frac16.
\] Its 2-adic valuation is \(-1\). Thus the deleted-node cardinal
factors need not be integral, whereas the full-node factors are integral
at every \(x\in\mathbb Z_p\). The bound \(-B=-2\) is valid here.

For the last case the full nodes are \(-1,0,1\), and
\(L_1(X)=X(X+1)/2\). The equation \(x^2=2\) gives \(v_2(x)=1/2\); since
\(v_2(x)>0\), the ultrametric inequality gives \(v_2(x+1)=0\). Therefore
\(v_2(L_1(x))=-1/2\), although \(x\) is integral in its extension. It
lies outside \(\mathbb Q_2\), whose nonzero valuations are integers. The
residue-count proof of Theorem 10.34 applies to \(\mathbb Z_p\), and
does not assert this cardinal integrality on the entire integral ball of
\(\mathbb C_p\).

\textbf{Solution 17.} Here \(D=4,\delta=0\), \(\gamma=3^4(2!)^2=324\),
and \(\Gamma=4\). The unscaled polynomial is \(f(y)=((y+1)(y+2)/2)^2\),
so \[
F(Z)=(Z+3)^2(Z+6)^2=Z^4+18Z^3+117Z^2+324Z+324.
\] Its coefficients are 3-adically integral and its finite tail is zero,
so it is normal. We have \(f_1(0)=3\) and \(F_1(0)=324\). Thus
\(v_3(F_1(0))+1=5=\Gamma+v_3(f_1(0))\). The sufficient slope threshold
is \(\theta+1/(p-1)=3/2\). It holds for \(w=9\), whose valuation is two,
and fails for \(w=3\), whose valuation is one. Under scaling, the
corresponding exponential is \(\exp(wZ/3)\); only the former meets the
stated normal-series hypothesis.

\textbf{Solution 18.} Here \(v(2)=2\), and the prepared values at zero
are \[
f(0)=1,\qquad v(2)f_1(0)=2\cdot\frac32=3,
\qquad v(2)^2f_2(0)=4\cdot\frac12=2.
\] Their 2-adic valuations are \(0,0,1\). Take \(P_0=0\) and
\(\beta=0\); the additive part of Lemma 10.38 has \(C_0=v_2(v(2))=1\).
It gives \(v_2(f_1(0))\ge-1\), with equality. Prepared integrality does
not imply ordinary divided-jet integrality. For order two the bound
\(-2\) is sufficient, while the actual valuation is \(-1\). This also
illustrates why the lcm cost must be retained.

For the interpolation example \(n=7,B=1,\kappa=0\). Theorem 10.39 gives
\(\min\{14,16-\max\{1,2\}\}=14\). If one first discards the difference
between the two jet precisions, their common bound is only
\(\Lambda-C=14\). Theorem 10.34 then gives \(\min\{14,14-1\}=13\).
Retaining each jet's precision saves the extra unit.

\textbf{Solution 19.} The value is \(3/2\), with \(v_3(3/2)=1\). Here
\(D=N=1\), \(v_2(1!)=v_2(v(1))=0\), and \(t_0=0\), so \(\Xi_{0,1,0}=1\).
The clearing factor is two. The coefficient heights and torus term are
zero, while \(\mathcal B_0(1/2,0)=5/2\). Thus (10.116) gives \[
v_3(3/2)\le\frac{\ln(5/2)+\ln2}{\ln3}=\frac{\ln5}{\ln3}.
\] This exceeds one, since \(5>3\). The denominator at two contributes
to the global height budget, although it causes no loss in the local
valuation at three.

\textbf{Solution 20.} Since \(v_3(10-1)=2>1/2\), the proved logarithm
and exponential laws give \(\eta=\exp(\tfrac12\log10)\in\mathbb Q_3\),
\(\eta^2=10\), and \(v_3(\eta-1)=2\). The defining series have rational
coefficients and converge in \(\mathbb Q_3\). The number ten is not a
rational square, by its odd prime valuations, so \([F:\mathbb Q]=2\).
The chosen completion is \(\mathbb Q_3\): the embedded field contains
the dense subfield \(\mathbb Q\) and lies in \(\mathbb Q_3\). Hence
\(e_F=f_F=1\), and (10.117) is two, whereas \(d/(ef)=1\) for \(K\).

Both exponents are congruent to \(m_*=1\) modulo two. Their differences
give exponents \(0,1\) after division by two. The common-factor identity
is \[
V=\frac32(\eta-\eta^3)
=\eta\frac32(1-10)=-\eta\frac{27}{2},
\qquad V_0=-\frac{27}{2}\in\mathbb Q.
\] The root is a local unit, so both values have 3-adic valuation three.

Here \(N=2\), \(h_\infty(c)=0\), \(h_2(c)=\tfrac12\ln2\), so the
coefficient norm term is \(\ln2\). Also \(\mathcal B_0(1/2,0)=5/2\),
\(\Xi=1\), \(h(10)=\ln10\), and the original exponent bound is \(A=3\).
The root-field bound is therefore
\(2(\ln2+\ln5+3\ln10)/\ln3=8\ln10/\ln3\). For the common-factor rule
\(\widehat A=2\); it keeps the original field and gives
\((\ln2+\ln5+2\ln10)/\ln3=3\ln10/\ln3\). Both bounds exceed the actual
valuation three. The different local-degree coefficients are accounted
for explicitly; membership in the same local field alone does not remove
the global field-degree factor.

\textbf{Solution 21.} Here \(v(2)=2\), and \[
\mathcal P_{0,2}(Z)=\frac{(4+2Z)(5+2Z)}2=10+9Z+2Z^2.
\] Here \(E_h(1)=h+1\). The maximum over total order at most two is
\(\mathcal B_0^*(2,2)=30\), attained at additive order zero and Euler
order two. The other allocations give at most \(9\cdot2=18\) and \(2\).
The old bound is \(\mathcal B_0(2,2)=(7^2/2)3^2=441/2\). At the
specified row, the polynomial whose coefficients are the prepared
additive derivatives is exactly the displayed polynomial. Thus the
values for \(t_0=0,1,2\), with zero Euler order, are \(10,9,2\). The
order-dependent bounds are attained.

\textbf{Solution 22.} Here \(D=2,N=6,J_0=1,M_*=5,H=2\ln2\), and the
discriminant is one. The scalar majorants for \(|s|=0,1,2\) are
\(3,6,10\), respectively. Since \(\sum_{s=-2}^2|s|=6\), (10.127) gives
\[
h_2(c)\le5\ln6+\ln3+2\ln10+24\ln2.
\] The simpler (10.128), using \(\mathcal B_0^*(2,0)=10\), is
\(h_2(c)\le3\ln6+5\ln10+24\ln2\). The first right side is smaller by
\(\ln(1000/108)>0\). Using the maximum node size instead of its average
would replace \(24\ln2\) by \(40\ln2\).

At the five nodes the row matrix is \[
\begin{pmatrix}
0&0&0&0&0&0\\
0&1&0&1/2&0&1/4\\
1&3&1&3&1&3\\
3&6&6&12&12&24\\
6&10&24&40&96&160
\end{pmatrix}.
\] Multiplication by the given vector gives zero in every row. Its
coordinates have greatest common divisor one, so all finite maximum
norms are one. Consequently \(h_\infty(c)=\ln57\) and
\(h_2(c)=\tfrac12\ln4874\), which satisfy both bounds. The first row is
zero because the two shifted additive polynomials share the root \(-2\).
The weighted Siegel lemma allows dependent and zero rows; \(M_*=5\)
remains a valid upper bound for the number of equations.

\textbf{Solution 23.} At \(s=2\), the total sum is \(-2+\eta^2=0\). The
coefficients in (10.131) are \(-2\) in the even class and \(1\) in the
odd class; both are nonzero. Multiplication by the even node does not
permute the residue classes. For failure of full degree take
\(\vartheta=4,\eta=2,s=1\), with the same coefficients. Again the total
sum is zero and the two coset coefficients are nonzero, while
\([F(\eta):F]=1\).

At seven, \(\eta=3\exp(\tfrac12\log(2/9))\in\mathbb Q_7\): the argument
\(2/9\) lies in \(1+7\mathbb Z_7\), so Proposition 9.4 proves
convergence and \(\eta^2=2,\eta\equiv3\pmod7\). The sum \(\eta-3\) has
valuation one. Indeed \((\eta-3)(\eta+3)=-7\), and
\(\eta+3\equiv6\pmod7\) is a unit. Its even and odd coset terms
\(-3,\eta\) both have valuation zero. The degree over \(\mathbb Q\) is
two, but a finite precision statement for their sum does not imply that
precision for each term.

\textbf{Solution 24.} The three changed binomials are \[
1,\qquad 2\Delta(Z;1),\qquad
4\Delta(Z;2)-\Delta(Z;1).
\] Thus, with rows and columns ordered by \(0,1,2\), the change and
inverse are \[
D=\begin{pmatrix}1&0&0\\0&2&0\\0&-1&4\end{pmatrix},
\qquad
D^{-1}=\begin{pmatrix}1&0&0\\0&1/2&0\\0&1/8&1/4\end{pmatrix}.
\] Both are 3-adically integral. The vector \((1,3,9)\) becomes
\((1,6,33)\); both have minimum valuation zero. In general p-integrality
in both directions proves the equality for every vector, not just this
example. Each row uses only indices no larger than its own, so keeping
the additive index fixed preserves the total-order cutoff.

Here \(F=\mathbb Q(i)\). With \(\rho_1=\sqrt2,\rho_2=\sqrt3,\xi=i\), the
phased roots are \(\eta_1=i\rho_1^3\) and \(\eta_2=\rho_2^3\), and their
squares are \(\vartheta_1=-8,\vartheta_2=27\). Their heights are
\(3\ln2,3\ln3\). The saturated Kummer theorem gives degree eight over
\(\mathbb Q\), and Corollary 10.49 gives degree four over \(F\).
Concretely \(\rho_1=(\eta_1/i)/2\) and \(\rho_2=\eta_2/3\), so the
phased and unphased root fields over \(F\) coincide. The original two
bases are local units at five or seven; at three the second one is not,
so the local-unit version of Theorem 10.48 would require a different
prime in this last example.

\textbf{Solution 25.} Here \(a_0=-1,H=3\), and the possible coordinate
sums are \(-4,-1,2\). The two classes are \[
\begin{aligned}
b=0:\ &(-2,1),(-1,0),(0,-1),(1,-2),\\
b=1:\ &(-2,-2),(0,2),(1,1),(2,0).
\end{aligned}
\] For a point of sum \(u=-1+3b+6v\), multiplication of its original
phase \(\zeta^{su}\) by \(\zeta^s\) gives \(i^{sb}(-1)^{sv}\). The red
class has \(v=0\). In the blue class \(v=-1\) at \((-2,-2)\) and \(v=0\)
at the other points. Thus the two grouped contributions are \(Z_0\) and
\(i^sZ_1\), with the sign at the first blue point included in \(Z_1\).

The chosen minimum coefficient belongs to the red class, so
\(\varepsilon'=-1\pmod6\). Centering at \(\lambda_0=(-1,0)\) makes every
red exponent difference have coordinate sum zero; its normalized phase
is exactly one. The minimum coefficient and its congruence class are
fixed data of the coefficient vector. Extraction proves its zeros for
every specified \(s,\boldsymbol t\), so no new choice is made at each
node or order. In the coprime example, \(2^{-1}=3\pmod5\), and
\(u\equiv-3\cdot2\equiv4\pmod5\). There is just one divided phase class
and no second extraction.

\textbf{Solution 26.} Newton's formula gives \[
x_1=2-\frac5{4}=\frac34,\qquad
x_2=\frac34-\frac{25/16}{3/2}=-\frac7{24}.
\] The errors \(f(x_0),f(x_1),f(x_2)\) are \(5,25/16,625/576\), with
valuations \(1,2,4\). The unique root \(\zeta\equiv2\pmod5\) therefore
satisfies \(\zeta\equiv-7/24\equiv182\pmod{625}\). It has order four,
since its square is \(-1\) and its reduction is neither \(1\) nor
\(-1\). Here \(G=4,G_0=2,\zeta^2=\alpha_0=-1\).

The factorization \((2-\zeta)(2+\zeta)=5\) and the unit
\(2+\zeta\equiv4\pmod5\) give \(v_5(2-\zeta)=1\). As \(\zeta\) is a
unit, \(v_5(g-1)=1\). The least \(t\) with \(5^t>1/4\) is zero; with
\(j=3\), the lower depth \(1+3=4\) is strictly greater than
\(\vartheta+1/4=13/4\). Therefore \(P=5^3=125\), \(d_1=-125\), and
\(\gamma_1=2^{125}\zeta^{-125}=g^{125}\). The exact p-power law,
starting strictly above \(1/4\), gives \(v_5(\gamma_1-1)=1+3=4\), not
merely its lower bound. The individual phase is algebraic outside
\(\mathbb Q\); a support congruence modulo \(G_0=2\) makes its
normalized powers rational, as Theorem 10.51 proves.

\textbf{Solution 27.} The input nodes are \(-3,-1,1,3\), so
\(n=4=2(1-1/2)R\). Here \(B=\lfloor\log_3 8\rfloor=1\), \(M_0=1\),
\(T'=4\), \(D=4\), and \(Lv_3(k!)=0\). The exact first budget is
\(U+12\ge24+4=28\), hence \(U\ge16\). The exact output lower bound is
\((8-4)3=12>11\). Therefore every integer \(-4\le x\le4\) has all
prepared zeros through total order four, including the previously
omitted \(-4,-2,0,2,4\).

For the larger scale, the first line is \(U+14\ge28\), requiring only
\(U\ge14\). Its corresponding lower bound is \(12-4/2=10\), which does
not exceed eleven. Thus the second line fails; this alternative pair
does not certify the same zeros. These computations evaluate the stated
conditional budgets, rather than construct their auxiliary coefficients.

\textbf{Solution 28.} Since \(P=125\) is odd, the two powers of
\(\zeta\) are \(\pm\zeta\); neither nonzero rational multiple belongs to
\(\mathbb Q\), because \(\zeta^2=-1\) has no rational root. Their
product phase is \(\zeta^{-4P}=1\), giving \(\gamma_1\gamma_2=b=6^P\).
On the support \((j,j)\), \(d\cdot(j,j)=-4Pj\) is divisible by
\(G_0=2\), so (10.151) gives \(b^{sj}\) at every positive or negative
integer node.

Here \(N=4,D=1,J_0=1,M_*=3\), and \(A_1=A_2=3\), so
\(H_{\mathrm{ph}}=3P(\ln2+\ln3)=3\ln b\). The torus polynomial is \[
A(Y)=-b+CY-CY^2+bY^3
=b(Y-b^{-1})(Y-1)(Y-b).
\] It vanishes at \(b^{-1},1,b\). Consequently
\(\Psi_0(s;0)=(s+1)A(b^s)\) is zero for \(s=-1,0,1\). At \(-1\) the
additive factor alone already gives zero, and the vector supplies the
additional torus zero. Directly the other two rows give \(-b+C-C+b=0\)
and \(-b+Cb-Cb^2+b^4=0\).

Since \(C=b^2+b+1\equiv1\pmod b\), the integer coordinates have greatest
common divisor one. Their heights are therefore \[
h_\infty(c)=\ln C,\qquad
h_2(c)=\tfrac12\ln\bigl(2b^2+2C^2\bigr).
\] Also \(b\equiv1\pmod5\), \(C\equiv3\pmod5\), so every coordinate is a
5-adic unit and the minimum valuation is zero. With \(k=L=1,T=0\), the
scalar majorant is \(\mathcal H_0(|s|;0)=|s|+1\). The separate-row bound
(10.152) is exactly \[
\ln2+\sum_{s=-1}^1\bigl(\ln2+\ln(|s|+1)+6|s|\ln b\bigr)
=6\ln2+12\ln b.
\] It bounds the displayed height: \(C\le3b^2\) and \(b\ge1\) give
\(h_2(c)\le2\ln b+\tfrac12\ln20\le12\ln b+6\ln2\), since \(20<2^{12}\).
The weighted Siegel proof permits zero and dependent rows. Counting the
first row in \(M_*=3\) merely keeps a sufficient budget; it does not
assert that the matrix has rank three.

\textbf{Solution 29.} Here \(g=\gcd(2,0)=2\), \(G_*=1,d_{*1}=0\), so the
cancelled relation is \(w=0\). The two support vectors are
\((0,0),(0,1)\) modulo two, and \(h_\Lambda=1\). Theorem 10.56 gives
\(E=\mathbb Q(\eta)\), with \(\eta^2=6\), and degree two. A rational
square has even prime valuations, whereas \(v_2(6)=1\), so \(\sqrt6\) is
not rational. The square classes of \(-1,6\) are independent: a relation
\((-1)^a6^b\) that is a rational square first forces \(b\) even by
valuation at two, then \(a\) even by positivity. Thus the proved Kummer
theorem gives \([F:\mathbb Q]=4\).

The simple residue root of \(X^2-6\) at \(1\) modulo five has unit
derivative two, and the complete Newton proof in Lemma 10.53 supplies
\(\eta\in\mathbb Q_5\), \(\eta\equiv1\pmod5\). That same proof gives
\(i\in\mathbb Q_5\), \(i\equiv2\pmod5\). Hence the chosen closures of
\(E,F\) are both \(\mathbb Q_5\); their absolute indices are \(e=f=1\).
Their global degrees are still two and four.

The normalized torus terms are \(1,\eta\), and the prepared additive
factor is \(\Delta(1/2;1)=3/2\). Hence \(V=\tfrac32(1-\eta)\). The
identity \((1-\eta)(1+\eta)=1-6=-5\) and the unit
\(1+\eta\equiv2\pmod5\) give \(v_5(V)=1\). Under the other embedding
\(\eta\mapsto-\eta\), the value is \(\tfrac32(1+\eta)\), a unit, with
valuation zero. In particular a lower bound at one embedding cannot be
counted twice.

Here \(\delta=0,N=2,A_1=1\), \(h_\infty(c)=0,h_2(c)=\tfrac12\ln2\),
\(\mathcal H_0(1/2;0)=3/2\), and \(\Xi_{0,1,0}=1\). The torus term is
\(2PXh(6)=\ln6\). Therefore (10.156) is exactly \[
v_5(V)\le \frac2{\ln5}
\left(\ln2+\ln(3/2)+\ln2+\ln6\right)
=\frac{2\ln36}{\ln5}.
\] It bounds the exact valuation one, since \(36^2>5\). Applying the
same row estimate over \(F\) would instead give \(4\ln36/\ln5\). The
support field halves this sufficient budget. Its unchanged completion
removes no further factor: \(V\notin\mathbb Q\), and the product formula
still sums over a global field of degree two.

\textbf{Solution 30.} Choose the smallest admissible prime. For
\(a=8,p=3\), two is dependent with eight and three is not a unit, so
\(c=5\). The transformed difference is \(63/25\), and
\(v_3(8^2-1)=v_3(63/25)=2\). The heights are \(\ln8,\ln5\), bounded by
\(\ln40,\ln5\). For \(a=3,p=2\), two is excluded by \(p\) and three by
dependence, so again \(c=5\). The difference is \(3/5-1/5=2/5\), with
valuation one, equal to \(v_2(3-1)\). Here the heights are both
\(\ln5\), bounded by \(\ln15,\ln5\). For \(a=25,p=2\), the smallest
admissible prime is three. The difference is \(25/3-1/3=8\), with
valuation three, equal to \(v_2(25-1)\). The heights are \(\ln25,\ln3\),
bounded by \(\ln125,\ln5\).

For all these pairs, the proof gives \(\ln(B'\ln p)\le3\ln p\), so
\(H=10\ln p\). For the all-power conclusion let \(q\mid p-1\) be the
residue order of \(a\). If \(q\nmid n\), the valuation of \(a^n-1\) is
zero. Otherwise the proved prime-to-p and deep p-power identities at odd
\(p\) give \[
v_p(a^n-1)
=v_p(a^{p-1}-1)+v_p(n)
\le250000\ln(5a)\ln5\,\frac p{(\ln p)^2}+v_p(n).
\] The prime \(p\) is odd in this last assertion, exactly as in
Corollary 10.10. A base equal to one would be multiplicatively dependent
with every base; Theorem 10.8 requires independence. The two rational
bases constructed here satisfy that requirement while preserving the
original valuation.

\textbf{Solution 31.} Order the diamond monomials as
\(1/4,1/2,1,2,4,1/3,3\). Multiplication by \(12\) gives
\((3,6,12,24,48,4,36)\). Its coordinates have gcd one and maximum
\(48\), so the projective height is \(\ln48\). For the box the smallest
and largest monomials are \(1/12\) and \(12\). Multiplication by \(12\)
gives integer coordinates of gcd one, since one coordinate is one, and
maximum \(144\). Its height is \(\ln144\). Direct local maxima give the
same products \(4\cdot4\cdot3=48\) and \(12\cdot4\cdot3=144\).

For the asymmetric three-point support the vectors \((1,2,3)\) and
\((6,3,2)\) are primitive; their heights are \(\ln3\) and \(\ln6\). They
show why a negative node needs the negative support.

The even-even diamond points are \((-2,0),(0,0),(2,0)\). Subtracting
\((-2,0)\) and dividing by two gives \(M_1=\{(0,0),(1,0),(2,0)\}\). Its
two heights are both \(\ln4\). It is contained in \(C/2+(1,0)\), and
\(\ln4\le\tfrac12\ln48\) follows exactly from \(16\le48\). Translation
of the envelope is harmless; it does not change its height.

Here \(N=7,J_T=1,M_*=3,\Delta_{\mathbb Q}=1\). The additive scalar
majorants at \(s=-1,0,1\) are \(2,1,2\), while the exact signed torus
costs are \(\ln48,0,\ln48\). Thus (10.164) gives \[
h_2(c)\le\frac12\ln7+
\frac14\left(\frac32\ln7+2\ln2+2\ln48\right)
=\frac78\ln7+\frac12\ln96.
\] The coordinate-box calculation replaces \(48\) by \(144\); it gives
\(\tfrac78\ln7+\tfrac12\ln288\). The exact-support bound improves this
coefficient budget by \(\tfrac12\ln3\). The mean-node bound (10.165)
also uses \(\ln48\), but its uniform scalar majorant \(2\) loses a
further \(\tfrac14\ln2\) relative to the separate-row bound. The actual
row at \(s=-1\) is zero; allowing it the displayed factor is legitimate
and does not assert that the row is nonzero.

Finally, for \(|(Bu)_j-z_j|\le D_j\), the local bound includes the
linear term \(\sum_j z_j\ln|\alpha_j|_v\). It sums to zero by the
product formula. No vector \(a\) with \(Ba=z\) is required.

\textbf{Solution 32.} Put \(C=t^2+t+1\). The vector \((-t,C,-C,t)\) has
coordinate sum zero, and \[
-t+Ct-Ct^2+t^4=0.
\] Therefore it annihilates both nonzero rows in the figure, and also
the zero row. Its torus polynomial is \[
-t+CY-CY^2+tY^3=t(Y-t^{-1})(Y-1)(Y-t).
\] For \(t=6\) all row entries are rational, so \(E_0=\mathbb Q\), even
with ambient field \(\mathbb Q(\sqrt5)\). For \(t=\sqrt6\), the entry
\(2t\) in the \(s=1\) row recovers \(t\); hence
\(E_0=\mathbb Q(\sqrt6)\), of degree two because six is not a rational
square.

Here are the integral-basis checks. For square-free \(m\), write an
algebraic integer in \(\mathbb Q(\sqrt m)\) as \(\beta=x+y\sqrt m\). Its
trace \(a=2x\) and norm are rational integers, by the earlier
norm-and-trace proof. Thus \(m(2y)^2=a^2-4N(\beta)\) is an integer. If
\(2y=c/b\) in lowest terms, this implies \(b^2\mid m\), so \(b=1\).
Hence \(\beta=(a+c\sqrt m)/2\) with integers \(a,c\) and
\(a^2-mc^2\equiv0\pmod4\). For \(m=5\) this is exactly
\(a\equiv c\pmod2\); for \(m=6\) it forces both \(a,c\) even. Conversely
\(u=(1+\sqrt5)/2\) and \(t=\sqrt6\) are integral, satisfying
\(u^2-u-1=0\) and \(t^2-6=0\). Their integer spans are integral and
include every possible \(\beta\). The full integral bases are
consequently \((1,u)\) and \((1,t)\). Their trace matrices are \[
\begin{pmatrix}2&1\\1&3\end{pmatrix},
\qquad
\begin{pmatrix}2&0\\0&12\end{pmatrix},
\] of determinants \(5\) and \(24\).

In the rational kernel, \(43-7\cdot6=1\), so its integer coordinates
have gcd one. Its maximum is \(43\) and its Euclidean norm is
\(\sqrt{3770}\). Thus \(h_\infty(c)=\ln43\) and
\(h_2(c)=\tfrac12\ln3770\).

In the second kernel \(C=7+t\), and the identity \(C-(1+t)t=1\) shows
that the integral coordinates generate the unit ideal. At every finite
place their maximum norm is therefore one. At the two real embeddings
the maximum coordinates are \(7+t\) and \(7-t\): the latter exceeds
\(t>0\), since \(49>24\). Their product is \(43\), giving
\(h_\infty(c)=\tfrac12\ln43\). The squared Euclidean norms are
\(122+28t\) and \(122-28t\); their product is \(10180\). With both the
square root and field-degree normalization, \(h_2(c)=\tfrac14\ln10180\).

Both examples have \(N=4,J_T=1,M_*=3\). Their scalar majorants at
\(-1,0,1\) are \(2,1,2\). For \(t=6\), the signed support heights are
\(3\ln6\). Thus the bound retaining the actual discriminant in (10.171)
is \(6\ln2+6\ln6=6\ln12\). An ambient-field application over
\(\mathbb Q(\sqrt5)\) would add \(\tfrac14\ln5\).

For \(t=\sqrt6\), the monomial vector is \((1,t,6,6t)\); its signed
support heights are both \(\tfrac32\ln6\). Equality of the two signs
also follows from translating the inverse support by three. The
corresponding actual-discriminant bound is \[
h_2(c)\le6\ln2+3\ln6+\tfrac14\ln24.
\] These inequalities exceed the exact heights just calculated: for the
first, \(3770<12^{12}\); for the second, \(10180<2^{24}6^{12}\cdot24\).

The tower with the single generator \(z_1=t\) has \(e_1=2\) and
\(h(t)=\tfrac12\ln6\), so (10.170) bounds its discriminant cost by
\(\tfrac12\ln2+\tfrac12\ln6=\tfrac12\ln12\). It bounds the exact cost
because \(24<12^2\). Using only the support maximum instead would give
\(\tfrac12\ln2+\tfrac32\ln6\), a loss of \(\ln6\).

If \(S=0\), every torus factor is \(W_\lambda^0=1\), so every row entry
is rational even in the second example; the exact row field is
\(\mathbb Q\). In neither case does choosing a smaller row field change
the number of required equations. The surplus \(N>M_*\) still supplies
the kernel, and extending its zeros still requires all the stated
analytic and strict arithmetic comparisons.

\textbf{Solution 33.} The basis matrix and its inverse are \[
B=\begin{pmatrix}1&1/2\\0&1/2\end{pmatrix},\qquad
B^{-1}=\begin{pmatrix}1&-1\\0&2\end{pmatrix}.
\] Thus \(B(\lambda_1,\lambda_2)=(\lambda_1+\lambda_2/2,\lambda_2/2)\),
its covolume is \(1/2\), and the index over \(\mathbb Z^2\) is \(J=2\).
Equivalently its nontrivial coset is represented by \((1/2,1/2)\), whose
double is integral. The box has volume two and normalized volume four,
so Theorem 10.66 supplies five points. The five displayed points are
\((0,0),(1/2,1/2),(1,0),(3/2,1/2),(2,0)\), with exactly the stated
integer basis coordinates.

Here \(g=\gcd(4,2,2)=2\) and \(H=2\). Their phases
\(2(\lambda_1+\lambda_2)\pmod4\) are \(0,2,2,0,0\). The zero class
retains \((0,0),(1,1),(2,0)\), of size three, equal to the guarantee
\(\lceil5/2\rceil\). At \(kL=2\), (10.180) gives \(6\le N\le10\), and
the actual chosen support has \(N=6\). Its strict volume bound is
\(N>4\).

For \(S=1,T=0\), \(J_T=\overline J_T=1\) and \(M_*=3\). The actual class
count therefore certifies \(N\ge2M_*\), so \(c_*=2\) can be used in
(10.176). The sufficient volume criterion certifies \(c_*=4/3\), since
\(4=(4/3)\overline M\); it does not certify \(c_*=2\). This is a
limitation of that coarser bound, not a failure of the exact class
count.

At \(T=1\), with one nonzero Euler direction, the representative indices
are \((0,0),(1,0),(0,1)\), so \(J_T=3\) and \(M_*=9\). The six columns
no longer satisfy \(N>M_*\). Thus this counting criterion does not
supply the kernel for those requested jets. It makes no claim that an
actual matrix of these rows has rank nine or that every kernel is
impossible.

Finally an open interval of length one cannot contain two integers:
their distance would be at least one, whereas any two points of that
interval are at distance strictly less than one. Its volume nevertheless
equals the covolume of \(\mathbb Z\). The closed-body limiting argument
is therefore essential at integer normalized volume.

\textbf{Solution 34.} Prime valuations at two and five prove both
independence and saturation: if a rational \(\beta\) has
\(\beta^T=2^{Ta}5^{Tb}\) with \(Ta,Tb\) integers, then \(a=v_2(\beta)\)
and \(b=v_5(\beta)\) are integers. Thus the full and q-primary lattices
are \(\mathbb Z^2\), and \(J=1,\nu=0\). The only rational roots of unity
are \(1,-1\), since a rational algebraic integer whose inverse is also
integral is \(1\) or \(-1\); hence \(q^u=2\). At three,
\(G=2,G_0=1,\delta=1,H=1\), regardless of the phase coefficients. Here
\(e=f=d=1,P=3,c_2=7/4,\theta=(3/2)/(1+\tau)\), \(\rho=17\),
\(A=g_1=4+\ln3\), and \(\gamma=1\). The specified \(g_0>39\) and
\(h\ge3\ln3\), as the bounds below also verify.

For a positive \(z>1\), put \(v=(z-1)/(z+1)\). Integrating the geometric
series for \(2/(1-v^2)\) gives \[
\ln z=2\sum_{j=0}^{n-1}\frac{v^{2j+1}}{2j+1}+R_n,\qquad
0<R_n\le\frac{2v^{2n+1}}{(2n+1)(1-v^2)}.
\] The remainder estimate follows by replacing every later denominator
by \(2n+1\) and summing the remaining geometric powers. For an interval
of positive \(z\)'s, the monotonicity of \(\ln z\) bounds it by applying
these rational bounds at the two endpoints. Similarly the positive
exponential series at two has \[
0<\mathrm e^2-\sum_{j=0}^{64}\frac{2^j}{j!}
\le\frac{2^{65}}{65!}\frac1{1-2/66},
\] because the ratio of each successive omitted term is at most
\(2/66\). Using \(n=512\) in the logarithm bounds, exact rational
interval operations in (10.185)--(10.186) give the following enclosures.
Rounding each lower endpoint down and each upper endpoint up to a
multiple of \(10^{-60}\) preserves the inequalities at every operation.
The reproducible source of Figure 10.14 implements these rational
operations and tails.

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}lrr@{}}
\toprule\noalign{}
Quantity & Lower bound & Upper bound \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
\(g_0=h\) & 50.136076 & 50.136077 \\
\(S\) & 379.288456 & 379.288457 \\
\(\mathcal D\) & 715255.319062 & 715255.319063 \\
\(T\) & 1796752.997002 & 1796752.997003 \\
\(\widetilde D_0\) & 35298.572063 & 35298.572064 \\
\(D_1\) & 67804.431507 & 67804.431508 \\
\(D_2\) & 29201.779183 & 29201.779184 \\
\(D_1D_2\) & 1980010036.552128 & 1980010036.552129 \\
\end{longtable}
}

Thus
\(s_0=379,t_0=1796752,k=50,L=35299,m=\lfloor D_1D_2\rfloor=1980010036\).
A concrete support of size \(m+1\) is \[
\Lambda=\left\{\left(\left\lfloor\frac j{29202}\right\rfloor,\,
j-29202\left\lfloor\frac j{29202}\right\rfloor\right):0\le j\le m\right\}.
\] The enclosing integer grid has \(67805\cdot29202\) points, at least
\(m+1\), because each side count exceeds its corresponding real width.
Integer division gives distinct points within \([0,D_1]\times[0,D_2]\).
Every point has the same phase, since \(H=1\), and \((0,0)\) is a
permissible reference.

Take the nonzero Euler direction
\(Y_1\partial_{Y_1}-Y_2\partial_{Y_2}\); its exponent form
\(\lambda_1-\lambda_2\) is nonzero on this support. The additive degree
cap is \(kL=1764950<t_0\). Therefore the representative count from
(10.126) is \[
J_T=\binom{t_0+2}{2}-\binom{t_0-kL+1}{2}
=1613655870378.
\] This subtraction counts the excluded indices with additive order at
least \(kL+1\), after shifting that order down by \(kL+1\). The exact
column and row counts are \[
\begin{aligned}
N&=50\cdot35299\cdot1980010037=3494618714803150,\\
M_{\rm full}&=759\binom{1796754}{2}=1225148631539679,\\
M_*&=759J_T=1224764805616902,\\
20N-57M_{\rm full}&=58902298301297>0.
\end{aligned}
\] Since \(c_{01}=1.9\cdot3/2=57/20\), the last integer identity
verifies the stronger full-row surplus in (10.190). The representative
rows have a still larger surplus. This is a count of equations, not a
claim that these rows are linearly independent.

The saturation basis is the identity, so the normalized integer torus
values are rational powers of the rational bases, with a possible
rational sign. Lemma 10.63 therefore gives \(E_0=\mathbb Q,d_0=1\); its
discriminant is one. Both the discriminant and the generator
contribution vanish. The scalar row-height contribution in (10.191)
remains and must be retained.

\textbf{Solution 35.} At two, \(F=\mathbb Q(\zeta_3)\) has unramified
completion of degree two, since \(2\) has order two modulo three; use
the prime-to-\(p\) proof cited in Lemma 10.73. Over this completion
\(X^2-2\) is Eisenstein: two is still a uniformizer, its constant
coefficient has valuation one, and its middle coefficient is zero.
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/NT-LOC/NT-LOC-04.html\#5-root-valuations-in-the-coefficients}{Extensions
of complete valued fields}, Corollary 5.2, gives a totally ramified
quadratic extension. Thus the resulting completion has degree four,
ramification index two and residue degree two. The global field is
generated by two quadratics, so its degree is at most four; the
local/global dimension argument of Lemma 10.73 forces it to be at least
four. Hence \(d=4\). The field contains \(\zeta_3\), but cannot contain
\(\zeta_9\), whose degree six would have to divide four. Therefore
\(q^u=3,u=1\), and \(d/e=2=3^{u-1}(3-1)\).

For the rational example the support contains \((67804,0)\) and
\((0,29201)\). Every coordinate is between the corresponding endpoints,
so the Euler argument \(\lambda_1-\lambda_2\) has exact maximum absolute
value \(\Omega_1=67804\). Thus \(m=67804\), \(R=379+2\cdot50=479\),
\(D_{\mathrm a}=1764950\), and \(t_0=1796752\). Prime powers give \[
\mathfrak v=\operatorname{lcm}(1,\ldots,50)
=3099044504245996706400,\qquad
3^{50}=717897987691852588770249.
\] Indeed the prime-power factors are
\(2^5,3^3,5^2,7^2,11,13,17,19,23,29,31,37,41,43,47\). Their product is
the displayed integer. The prime list is checked by trial division by
two, three, five and seven: every composite at most fifty has a prime
divisor at most its square root, hence at most seven.

The separated and sharper maxima both occur at \(u=D_{\mathrm a}\). For
the sharper one it suffices to check the last step, since its ratios
decrease:

\[
\mathfrak v\cdot31803
>479\cdot1764950\cdot99607.
\]

This holds for both \(\mathfrak v\) and the larger \(3^{50}\). The
remaining Euler order is \(31802\), and the sharper logarithm is \[
\ln\mathcal H_{\rm fine}
=1764950\ln\mathfrak v-35299\ln(50!)
+\ln\binom{99606}{31802}.
\] The factorial contribution is negative; omitting its sign would
overstate this upper bound. The mean orders are obtained by summing
\(J_u=t_0-u+1\): \[
\overline u=\frac{820859713025}{1371417},
\qquad
\overline h=\frac{1643236524559}{2742834}.
\] Use \(\sum_{u=0}^Uu=U(U+1)/2\) and \(\sum_{u=0}^Uu^2=U(U+1)(2U+1)/6\)
in \(\sum u(t_0-u+1)\). The two polynomial sum identities follow by
telescoping the consecutive squares and cubes, respectively.

Here \(E_0=\mathbb Q,d_0=1\), so the discriminant and field-generator
terms vanish. The exact ratio is \(M_*/(N-M_*)\), with both integers
from Solution 34, and \(Z=S\mathcal D\). Since \(\sigma_i=h(\alpha_i)\),
\(PL_\alpha=\mathcal D/(c_1c_2)\). The mean torus contribution divided
by \(Z\) is therefore exactly \[
\frac{s_0(s_0+1)}{(2s_0+1)c_1c_2S}.
\] Substitute the sharper logarithm above, or (10.205), into the
weighted coefficient bound. The rational series and outward interval
operations from Solution 34 give these net upper bounds:

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}lr@{}}
\toprule\noalign{}
Scalar estimate & \(h_2(\boldsymbol c)/Z\) is at most \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
Sharper peak, actual lcm & 0.269837353526 \\
Sharper peak, \(3^{50}\) & 0.288952296008 \\
Row mean, actual lcm & 0.172796821752 \\
Row mean, \(3^{50}\) & 0.179279284082 \\
Alternative global rows, peak & 0.128675661391 \\
Alternative global rows, mean & 0.124309602511 \\
\end{longtable}
}

For the alternative peak, the exact integer rule (10.201) with clearing
factor one gives \(\widehat u=3543\). Check its last increasing step and
its next decreasing step with integer products. The alternative mean
replaces \(A_{\rm mean}\) by \(A_{\rm mean}+35299\ln(50!)\), and removes
\(\overline u\ln\mathfrak v\), exactly as (10.207) prescribes.

Large binomial logarithms need no approximation of their integer
coefficients: for a positive integer \(n\), write \(n=2^b y\) with
\(1\le y<2\), and use \(\ln n=b\ln2+\ln y\). The logarithm series in
Solution 34 then has ratio at most \(1/3\), even for these enormous
integers. The figure program uses this reduction and exact rational
outward intervals, retaining each component. The separate logarithmic
contribution is less than \(0.000000101556\).

Finally, multiplication of a row by \((50!)^{35299}/\mathfrak v^u\) is
multiplication by a nonzero rational number. It changes neither its zero
equations nor its rational coefficient field. Its normalized projective
height is invariant by the product formula. These global row estimates
therefore preserve all required zeros; later local prepared derivatives
retain their own original clearing convention.

\textbf{Solution 36.} Iterating \(c\mapsto\min\{pc,c+1\}\) from \(1/e\)
gives the following exact lower bounds. The inequalities (10.209)
determine the power; the separate unramified clause determines
\(\vartheta\) in the last row.

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}
  >{\raggedright\arraybackslash}p{(\linewidth - 8\tabcolsep) * \real{0.1667}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 8\tabcolsep) * \real{0.2222}}
  >{\raggedright\arraybackslash}p{(\linewidth - 8\tabcolsep) * \real{0.1667}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 8\tabcolsep) * \real{0.2222}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 8\tabcolsep) * \real{0.2222}}@{}}
\toprule\noalign{}
\begin{minipage}[b]{\linewidth}\raggedright
\(p,e\)
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
\(P\)
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
Guaranteed valuation path
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
\(\vartheta\)
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
Final required depth
\end{minipage} \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
\(2,2\) & 8 & \(1/2\), \(1\), \(2\), \(3\) & 2 & 3 \\
\(2,3\) & 8 & \(1/3\), \(2/3\), \(4/3\), \(7/3\) & \(4/3\) & \(7/3\) \\
\(3,2\) & 3 & \(1/2\), \(3/2\) & \(3/4\) & \(5/4\) \\
\(3,3\) & 9 & \(1/3\), \(1\), \(2\) & \(3/2\) & 2 \\
\(7,2\) & 1 & \(1/2\) & \(1/4\) & \(5/12\) \\
\(5,1\) & 1 & 1 & \(3/4\) & 1 \\
\end{longtable}
}

For the first actual example, \(X^2-2\) is Eisenstein, so the complete
local-field proof linked in Solution 35 gives ramification index two.
With \(\beta=1+\sqrt2\), the first difference is
\(\beta^2-1=2(1+\sqrt2)\), of valuation one. The next factor is
\(\beta^2+1=4+2\sqrt2=2\sqrt2(1+\sqrt2)\), of valuation \(3/2\).
Consequently \(v_2(\beta^4-1)=5/2\). The last step is above the
threshold one, so Lemma 9.5 gives \(v_2(\beta^8-1)=7/2\). The actual
path is therefore \(1/2,1,5/2,7/2\). Its extra gain at the boundary is
permitted by the lower-bound path, but is not asserted for all units.

For the second example \(X^3-3\) is Eisenstein, giving \(e=3\) and
\(v_3(\pi)=1/3\). Direct expansion gives \[
\beta^3-1=3(1+\pi+\pi^2),\qquad v_3(\beta^3-1)=1.
\] The parenthesis is a unit since its residue is one. The next step is
strictly above \(1/(3-1)=1/2\), hence \(v_3(\beta^9-1)=2\). Formula
(10.209) gives \(\kappa=2,P=9,\vartheta=3/2\); thus the final depth
\(\vartheta+1/2=2\) is attained.

Put \(\ell=\log(\beta^9)\); its valuation is two. At the indices
\(n=3^j\), the exponential coefficient \(A_n=\ell^n/n!\) has \[
v_3(A_n)=2n-\frac{n-1}{2}=\frac32n+\frac12.
\] Thus \(v_3(A_n)-n\vartheta=1/2\) along infinitely many indices. These
weighted coefficients do not tend to zero, so the series is not in
\(\mathcal A_{3^{3/2}}\). At any \(z\in\mathbb C_3\) of valuation
\(-3/2\), the terms of those indices still have valuation \(1/2\), and
the series does not converge. Such a point exists, for example a square
root of \(3^{-3}\). Every smaller radius with \(\theta<3/2\) instead has
the positive linear decay (10.213).

Finally the residue field here is \(\mathbb F_3\). For \(\alpha=-\beta\)
and \(\zeta=-1\), choose \(a=1\), so the phased coordinate is
\(V=\alpha^9\zeta^{-9}=\beta^9\). Its principal square root is
\(\eta=\exp(\ell/2)\in L\), of depth two. Fix \(\xi^2=-1\), and begin
with any \(\rho_0^2=\alpha\). The ratio \(\eta/(\rho_0^9\xi^{-9})\) is
\(1\) or \(-1\). Replacing \(\rho_0\) by its negative when the ratio is
\(-1\) changes its ninth power by that sign and leaves its square
unchanged. It therefore gives the exact choice \(\rho^9\xi^{-9}=\eta\)
required in (10.217).

\textbf{Solution 37.} If a rational power combination of \(4,9\) has a
rational root, its valuations at two and three show that both
coordinates of the corresponding exponent vector are half-integers.
Conversely every half-integer pair produces a rational number
\(2^{2x}3^{2y}\). Thus the full saturation lattice, and hence its
q-primary part at \(q=2\), is \((\tfrac12\mathbb Z)^2\). Its four cosets
modulo \(\mathbb Z^2\) prove \(J=4\), with basis \(B=\tfrac12I\); the
corresponding generators are \(2,3\).

The rational roots of unity are \(1,-1\), so \(q^u=2\). At five the
residue group has order four. Choose the primitive residue lift
\(\zeta\) with residue two, whose existence and primitivity are proved
in Lemma 10.53. Then \(G_0=4/2=2\), and \(2,3\) have residues
\(\zeta,\zeta^3\), respectively. With \(P=1\), their phase coefficients
are \(-1,-3\); the zero phase class is therefore
\(\lambda_1+\lambda_2\equiv0\pmod2\).

There are \(5\cdot4+4\cdot3=32\) such points in
\(\{0,\ldots,8\}\times\{0,\ldots,6\}\). Their original coordinates are
\((\lambda_1/2,\lambda_2/2)\), with full widths \(4,3\). For
\((b_1,b_2)=(2,3)\) and the identity matrix \(A_0\), formula (10.219)
gives \[
\omega=4(3\mu_1-2\mu_2)=6\lambda_1-4\lambda_2.
\] At reference zero its range is \([-24,48]\), and both extremes occur
at allowed points, \((0,6)\) and \((8,0)\). Hence \(\Omega=48\). At
reference \((8,0)\), subtract its value \(48\); the range becomes
\([-72,0]\) and the exact maximum absolute value becomes \(72\). The
direct full-width estimate is \(4(3\cdot4+2\cdot3)=72\). It covers every
reference in the same box because each difference of two original
coordinates is bounded by its full width. The weighted estimate in Lemma
10.84 also holds, using \(R_{\max}=3/\ln4+2/\ln9\) and
\(\sum_iD_i\sigma_i=4\ln4+3\ln9\).

Omitting \(J\) would give \(3\mu_1-2\mu_2\). At the allowed point
\((\lambda_1,\lambda_2)=(1,1)\), this is \(1/2\); its first rising
binomial is \(3/2\), which is not an integer. Thus the original
denominator clearing is part of the integral prepared system.

For the mean-order panel, put \(a=\overline u/t_0\). At \(r=2\),
(10.204) gives \(\overline h/t_0=(1-a)/2\) and \(0\le a\le1/3\). The
joint multiplier is \((1-a)/2+2a=(1+3a)/2\), between \(1/2\) and one.
These are exactly the plotted lines; no numerical approximation of the
means is used.

For the original rational parameters of Solution 34, \(d=e=f=1,\nu=0\),
\(R_{\max}=1/\ln2+1/\ln5\), and \[
W(1)=\frac{\ln6}{\ln2\,\ln3},\qquad
g_{91}=1+\frac{1+\ln W(1)}{g_0}.
\] The condition \(h\ge\ln(R_{\max}/W(1))\) holds at the stated
\(h=g_0\). Its exact outward logarithm intervals are given by the series
proved in Solution 34. The coefficient field is \(\mathbb Q\), so the
field-degree logarithm and generator/discriminant costs are zero. Keep
\(c_{01}=57/20\), all \(g_2,g_5,g_8\), and the strict \(\theta\) from
that solution. Substitution into (10.228) gives the following certified
upper bounds.

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}
  >{\raggedright\arraybackslash}p{(\linewidth - 2\tabcolsep) * \real{0.4286}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 2\tabcolsep) * \real{0.5714}}@{}}
\toprule\noalign{}
\begin{minipage}[b]{\linewidth}\raggedright
Quantity
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
Upper bound
\end{minipage} \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
\(g_{91}\) & 1.037012532704 \\
\((E_{\rm mean}+\overline u\ln\mathfrak v)/(t_0h)\), actual lcm &
0.336168252156 \\
The same combined cost using \(3^{50}\) & 0.372348992541 \\
The coefficient height \(h_2(\boldsymbol c)/Z\) in (10.228) &
0.310817025826 \\
\end{longtable}
}

For the two combined costs use the exact \(\overline u,\overline h\)
from Solution 35 and
\(E_{\rm mean}=\overline h\ln[\mathrm e(1+67804/\overline h)]\). For the
last line, the exact expression enclosed is \[
\frac{g_8}{2}+\frac{20}{37}
\left[\frac{g_8}{2}+\frac{1+1/g_5}{c_1c_4}
+\frac{g_{91}}{c_1c_3\theta}
+\frac{1+1/(2g_2)}{2c_1c_2}\right].
\] The first logarithmic term is outside the row ratio; all other scalar
and torus terms are inside it. The same rational outward interval rules
apply to each positive logarithm and arithmetic operation, proving the
displayed bounds.

\textbf{Solution 38.} Suppose a rational exponent vector belongs to the
full saturation lattice. After taking a common positive integer power,
its realizing element \(\beta\in K^\times\) has a rational power
\(\beta^T=5^a7^b\). Conjugation gives \((\overline\beta/\beta)^T=1\).
This quotient is a root of unity in the real quadratic field, hence is
\(1\) or \(-1\). Writing \(\beta=x+y\sqrt5\) shows that \(\beta\) is
either rational or a rational multiple of \(\sqrt5\).

In the rational case, the ordinary prime valuations at five and seven
make both exponent coordinates integers. In the second case, dividing by
\(\sqrt5\) makes the first coordinate an integer plus \(1/2\), and the
second an integer. Conversely every such vector is realized by
\((\sqrt5)^{2v_1}7^{v_2}\). Thus \[
M=(\tfrac12\mathbb Z)\times\mathbb Z,\qquad
J=2,\qquad
B'=\begin{pmatrix}1/2&0\\0&1\end{pmatrix}.
\] This is already q-primary. Direct multiplication gives \[
\begin{pmatrix}1/2&5\\0&1\end{pmatrix}
\begin{pmatrix}1&-10\\0&1\end{pmatrix}
=B',\qquad
U^{-1}=\begin{pmatrix}1&10\\0&1\end{pmatrix}.
\] The new generators are \(\theta'_1=\sqrt5\) and
\(\theta'_2=(\sqrt5)^{-10}(5^5\cdot7)=7\). Their heights are
\(\frac12\ln5,\ln7\); the old heights are \(\frac12\ln5,5\ln5+\ln7\).
For \(\lambda'= (a,b)\), the old label is \((a-10b,b)\), and both
monomials are \((\sqrt5)^a7^b\). Both generator lists give \(E=K\). The
integral-basis proof in Solution 32 gives \(\Delta_E=5\).

At three, \(X^2-5\) has irreducible separable reduction \(X^2+1\). The
complete quadratic proof in
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/NT-LOC/NT-LOC-12.html\#3-quadratic-extensions-from-square-classes}{Cyclotomic,
quadratic and Kummer extensions of local fields}, Section 3, makes its
completion unramified quadratic. Hence \(e=1,f=2,d=2\); both original
bases are units. The only global roots of unity are \(1,-1\), so
\(u=1,\nu=1\). The first admissible power is \(P=3\), since
\(p-1=2\le2e\) and \(3(p-1)=6>2e\). Thus \(\vartheta=3/2\), with the
strict \(\theta\) of Solution 34.

The residue field has order nine. Its element \(\overline{\sqrt5}\) has
square \(-1\), so order four, whereas \(7\) has residue one. Choose the
primitive residue lift \(\zeta\) of order eight from Lemma 10.53 with
\(\overline{\sqrt5}=\overline\zeta^{\,2}\). Then \(G_0=8/2=4\); the
centered generator phases can be taken to be \(d'=(-6,0)\). Consequently
\(\delta=2,H=2\), and the two phase classes distinguish even and odd
first coordinates. In the old basis the second phase is \(-60\);
\(U^{\mathsf T}(-6,-60)=(-6,0)\). Thus both bases have exactly the same
phase at corresponding labels. The blue and brown points in the figure
display the two classes.

Apply the outward rational interval method proved in Solution 34 to the
parameters specified in the exercise, with \(t=2\ln3,\rho=58,x=\ln2\),
and \(\delta=2\). The basic linear-form matrix is the identity and
\(R_{\max}=1/\ln5+1/\ln7<2\). Since \(dW(d)=2(\ln6)^3>2\), the height
condition of Theorem 10.86 holds at positive \(h\). In the maximum
defining \(\mathcal D\), the second member exceeds the first. The
resulting integer data are

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}lr@{}}
\toprule\noalign{}
Quantity & Exact value \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
\(\lfloor S\rfloor,\lfloor T\rfloor\) & \(400,\ 36977365\) \\
\(k,L\) & \(52,\ 579557\) \\
\(\lfloor2D_1D_2\rfloor+1\) & \(597441303726\) \\
Selected points in either phase & \(298720651863\) \\
\(N\) & \(9002533531251763932\) \\
\(M_{\rm full}\) & \(547613916126766461\) \\
\(80N-171M_{\rm full}\) & \(626560702842464049729\) \\
\end{longtable}
}

Here is an explicit selection without listing the points. The
integer-coordinate grid has \(1201952\) first-coordinate choices and
\(497060\) second-coordinate choices. Take its first \(597441303726\)
points, ordering the first coordinate fastest. For each full row,
exactly half the first coordinates are even; the remaining initial
segment also has even length. Each phase therefore contains the stated
number of points. Their original coordinates lie in the box of widths
\(D_1,D_2\). Since \(c_{01}=1.9\cdot9/8=171/80\), the last line proves
\(N>c_{01}M_{\rm full}\), with all floors retained.

The same outward interval arithmetic gives

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}lr@{}}
\toprule\noalign{}
Quantity & Certified upper bound \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
\(g_{11}\) & \(2975665.522410184771\) \\
\(g_{12}=g_1/(2g_7)+1/g_{11}\) & \(0.000001658758\) \\
\(g_{91}\) & \(1.052658564765\) \\
Generator-field cost \(\ln5/(4Z)\) & \(0.000000000069\) \\
Explicit tower bound \(\ln10/(2Z)\) & \(0.000000000196\) \\
Centered bound \((\frac12\ln2+\ln7)/Z\) & \(0.000000000390\) \\
Full \(h_2(c)/Z\) bound (10.235) & \(0.510159606689\) \\
\end{longtable}
}

For a lower bound on \(g_{11}\), the same calculation gives
\(2975665.522410184770\). Also \[
5891891927.130371871238\le Z\le
5891891927.130371871239.
\] The explicit tower first adjoins \(\sqrt5\), of degree two and height
\(\frac12\ln5\); the generator \(7\) has degree increment one. Equation
(8.33) therefore gives \(\frac12\ln2+\frac12\ln5=\frac12\ln10\) as its
discriminant bound. The centered maximum bound in (10.236) is
\(\frac12\ln2+\ln7\). Both are larger than the exact cost
\(\frac14\ln5\). The interval programme enclosed in the figure source
uses only the rational positive series and geometric tail bounds of
Solution 34, and verifies both instances of (10.232), the strict integer
surplus, and every term in the last line.

Finally, within one phase, \(\lambda_1-\lambda_{01}\) is even. Since
\(P=3\), each normalized torus monomial has a rational power of \(5\), a
rational power of \(7\), and a rational sign. Every initial row is
therefore rational, and its smallest field is \(E_0=\mathbb Q\), of
discriminant one. The construction over the generator field \(E\) is
valid and satisfies the stated uniform bound; applying Theorem 10.63
over \(E_0\) improves its field cost to zero. The generator field and
the actual row field are distinct objects.

\textbf{Solution 39.} Start with the real parameters and exact integer
support of Solution 34, without replacing \(S,T\) by their floors inside
the parameter formulas. Here \(\eta=1231/1500\), and the certified floor
data are

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}lrr@{}}
\toprule\noalign{}
\(j\) & \(R_j=\lfloor2^jS\rfloor\) & \(T_j=\lfloor\eta^jT\rfloor\) \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
0 & 379 & 1796752 \\
1 & 758 & 1474535 \\
2 & 1517 & 1210101 \\
3 & 3034 & 993090 \\
\end{longtable}
}

For the three extensions, \(\mu_j=T_j-T_{j+1}+1\) is
\(322218,264435,217012\), and \(n_j=2R_j+1\) is \(759,1517,3035\). These
choices use all available input derivative orders. Integer prime powers
give \(B_j=6,6,7\). The actual \(v_3(\mathfrak v)=3\), and \[
v_3(50!)=\lfloor50/3\rfloor+\lfloor50/9\rfloor+\lfloor50/27\rfloor=22.
\]

For every allowed pair \(b_1,b_2\), (10.245) bounds \(|b_2|\) by
\((\ln2)W\exp(h)\). Taking logarithms, dividing by \(\ln3\), and
applying the outward rational intervals of Solution 34 bounds the
integer \(v_3(b_2)\) by46. Thus \(C_0\le46\) and \(M_0=46\) is valid in
every step.

The full-width Euler bound is the same one for the whole permitted
family: \[
\Omega\le
\left\lceil\frac{\mathcal D}{c_1c_2P}W\exp(h)\right\rceil
\le1313846807806555244846459051.
\] The figure programme encloses \(\exp(h)\) without a floating-point
exponential. Beginning with term one, successively multiply by the
outward interval for \(h\) and divide by \(j\), through \(j=512\). If
the next term is \(a_{513}\), the remaining positive terms are at most
\(a_{513}/(1-h/514)\), since every following ratio is at most
\(h/514<1\). Combine this with the rational logarithm intervals from
Solution 34, and round the upper endpoint of the Euler bound up to an
integer.

Use the proved \(\mathfrak v<3^{50}\) for each individual scalar, with
no averaging of its additive order. At step \(j+1\), put \[
R=R_{j+1}+99,\quad m=\Omega,\quad
D_{\mathrm a}=1764950,\quad T'=T_{j+1}.
\] The exact binary search (10.242), using the larger multiplier
\(3^{50}\), gives maximizing additive orders \(621676,384993,211681\).
The respective remaining Euler orders are \(852859,825108,781409\). All
steps before these indices are increasing; the next step is decreasing.
These tests involve only integer products, not factorials of the
enormous Euler argument.

To bound the logarithm at that maximizing index, use \[
\begin{aligned}
\ln\binom{D_{\mathrm a}}u
&\le D_{\mathrm a}\ln D_{\mathrm a}
-u\ln u-(D_{\mathrm a}-u)\ln(D_{\mathrm a}-u),\\
\ln\binom{m+H}H
&\le H\left[1+\ln(1+m/H)\right]\qquad(H>0).
\end{aligned}
\] The first follows from the \(u\)-th term of the binomial expansion of
\((t+(1-t))^{D_{\mathrm a}}=1\), taking \(t=u/D_{\mathrm a}\); endpoint
binomials equal one. For the second, the numerator product is at most
\((m+H)^H\), and the integral proof in Lemma 10.76 gives
\(H!\ge(H/\mathrm e)^H\). At \(H=0\), its logarithm is zero. The exact
maximum of the factorial expression has been located first; bounding its
logarithm at that index therefore bounds the whole maximum.

Retain the negative term \(-L\ln(50!)\). The upper bounds on
\(\ln\mathcal H_0^{\rm fine}/Z\), in step order, are \[
0.295755251977,\qquad0.251149033062,\qquad0.215672455561.
\] For all three steps, the coefficient/logarithmic contribution is \[
\frac{h_2(c)/Z+g_8/2}{\ln3}
\le0.282940661808.
\] The normalized arithmetic bound at step \(j+1\) is \[
\frac1{\ln3}\left[
\frac{h_2(c)}Z+\frac{g_8}{2}
+\frac{\ln\mathcal H_0^{\rm fine}(R_{j+1},T_{j+1})}Z
+\frac{R_{j+1}}{c_1c_2S}\right].
\] There is no q-denominator at these integer arguments. The full-width
torus terms, after division by \(\ln3\), have upper bounds
\(0.717181076407,1.435308301992,2.870616603983\). Their sum with the
coefficient and scalar terms gives exactly the arithmetic bounds in
Theorem 10.93.

For the input precision, the common available lower bound is \[
\frac{U+G_0}{Z}\ge7.294535093395.
\] The three required upper bounds are
\(1.406870557237,2.262855922296,3.678486939799\), respectively. Thus the
sufficient inequality (10.113) holds at every step, including the full
derivative loss \(M_0=46\). The analytic gains are computed from \[
\frac{(n_j\mu_j-D_{\mathrm a})\theta-22L}{Z}.
\] The gaps over the arithmetic upper bounds exceed, in step order, \[
0.070283757014,\qquad0.258541995694,\qquad0.279197996096.
\] These certified positive gaps prove the three zero-forcing
comparisons; no numerical estimate of an actual valuation was
substituted.

Finally, at a first fractional argument, the scalar has the clearing
cost (10.243), and its normalized torus value may belong to a larger
global field. Both enter (10.244). Equality of the chosen completions
does not remove that global degree, as shown in Theorems 10.42 and
10.56. The present comparisons use \(I=J=0,F=\mathbb Q\), so they prove
the specified integer block. They do not verify these additional
fractional costs or the subsequent support extraction and multiplicity
contradiction.

\textbf{Solution 40.} The three nodes are \(-1,0,1\) with respective
multiplicities \(1,3,1\), giving total degree five. The middle cardinal
polynomial is \(L_0(X)=1-X^2\); its inverse through degree two at zero
is \(1+X^2\). Thus its three basis polynomials are \[
H_{0,0}=1-X^4,\qquad H_{0,1}=X-X^3,\qquad H_{0,2}=X^2-X^4.
\] Each has the required one middle divided jet, with the other two
equal to zero, and all vanish at the two end nodes. At the end nodes no
inverse coefficient beyond degree zero is needed. Their cardinal
polynomials are \[
H_{-1,0}=(X^4-X^3)/2,\qquad H_{1,0}=(X^4+X^3)/2.
\] They vanish to order three at zero, take value one at their own end
node and zero at the other. All five degrees are at most four, so these
jet checks and uniqueness prove the full interpolation formula.

For the original parameter family, subtract \(T'=993090\) from each
available order and add one. This gives \(803663,481446,217012,1\). Cap
the first multiplicity at \(780000\), leaving the other three unchanged.
The integer node counts in the successive rings are
\(759,758,1518,3034\). Therefore \[
N_*=759\cdot780000+758\cdot481446+
1518\cdot217012+3034=1286383318.
\] The same interval program of Solutions 34 and 39 gives the following
outward bounds.

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}lr@{}}
\toprule\noalign{}
Quantity divided by \(Z\) & Relevant certified bound \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
Available input \(U+G_0\) & at least 7.294535093395 \\
Required input \(N_*\theta+780000\cdot46\) & at most 7.244899709935 \\
Analytic excess \(N_*\theta-G_0\) & at least 7.100020502185 \\
Arithmetic excess over the quartic field & at most 4.522335429896 \\
\end{longtable}
}

The first comparison has margin greater than \(0.049635383461Z\), and
the second has margin greater than \(2.577685072290Z\). No earlier zeros
or their extra inner derivatives have been discarded.

The bounds on \(\eta^{-3}S\) lie strictly between \(686\) and \(687\),
so \(\lfloor\eta^{-3}S\rfloor=686\) and every target argument has
absolute value at most \(687\). With \(R=786\) and the common \(\Omega\)
of Theorem 10.93, the exact consecutive-ratio search gives additive
order \(470396\) and Euler order \(522694\). Use the entropy bounds of
Solution 39 at this maximizing index and retain \(-35299\ln(50!)\). This
gives the scalar-logarithm upper bound \(0.208379911543Z\).

For the denominator, (10.243) here is \[
\Xi_{0,1,u}=\max\{0,3424003-6u\},
\] because \(v_2(50!)=47\) and \(v_2(\mathfrak v)=5\). Its supremum is
\(3424003\), contributing at most \(0.008748404912Z\) to the logarithmic
bracket. A scalar-maximizing index does not necessarily maximize the sum
of its scalar logarithm and this decreasing, piecewise linear
denominator term. The common supremum bounds every order and avoids that
unsupported inference.

The remaining bracket terms have these upper bounds.

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}lr@{}}
\toprule\noalign{}
Term & Upper bound \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
Coefficient height and \(\tfrac12\ln N\) & \(0.310842088027Z\) \\
Full-width torus height & \(0.714102914710Z\) \\
\end{longtable}
}

Multiply their sum with the scalar and denominator terms by \(4/\ln3\),
using the unrounded rational enclosures. This gives the stated
arithmetic upper bound. The global quartic field is proved in Theorem
10.95 even though its selected completion is \(\mathbb Q_3\). The
product formula sums over the global field; its degree cannot be
replaced by the degree of that one completion. The result proves the
exact first fractional range (10.250). Proposition 10.97 and Theorem
10.98 supply the subsequent contractions and comparisons.

\textbf{Solution 41.} Direct expansion gives \[
\binom{2n+a}{2}=4\binom n2+(2a+1)n+\binom a2.
\] Writing \(z=2n+a\), the inverse is \[
n=\frac{z-a}{2},\qquad
\binom n2=\frac14\binom z2-\frac{2a+1}{8}z+\frac{a(a+2)}8.
\] All denominators are powers of two, hence units at three. The inverse
is not in general integral at two; at \(a=0\) its linear coefficient is
\(-1/8\). In higher degree the finite-difference proof of Lemma 10.30
supplies the same triangular filtration. Polynomial coefficients are
retained as a subvector, so this inverse change does not enlarge their
projective height.

For the numerical table, use the exact rational interval algebra of
Solution 34 with \(B_{\rm arith}=10^{60}\). This precision parameter is
distinct from the separation integer \(B\) in the interpolation lemmas.
Multiplication and division round outwards. For logarithms reduce an
argument by exact powers of two to the interval \([1,2]\), put
\(y=(x-1)/(x+1)\), and use the positive series \[
\ln x=2\sum_{j=0}^{511}\frac{y^{2j+1}}{2j+1}
+\mathcal R,\qquad
0\le\mathcal R\le\frac{2y^{1025}}{1025(1-y^2)}.
\] The same formula bounds \(\ln2\); integer powers restore the original
logarithm. These remainder bounds and the earlier exact bounds for
\(\exp(h)\) make all parameter floors decisive. Use \(k=50,L=35299\),
\(v_3(50!)=22\), \(v_2(50!)=47\), and the proved bound
\(\mathfrak v<3^{50}\).

For each of \(I=1,\ldots,17\), calculate (10.255) from the unfloored
original \(S,T\), then form the three input and analytic bounds
(10.256)--(10.258). The mixed multiplicities satisfy all stated order
inequalities at each stage. For each respective arithmetic output set
\(r=X/2^I+99\), \(D=1764950\), \(\Omega=\lfloor\Omega_0/2^I\rfloor\),
and \(V=3^{50}\).

If \(f(u)\) is the summand in the maximum (10.259), its exact ratio is
\[
\frac{f(u+1)}{f(u)}
=\frac{(D-u)V(O-u)}{r(u+1)(\Omega+O-u)}.
\] This decreases in \(u\): both \((D-u)/(u+1)\) and
\((O-u)/(\Omega+O-u)\) decrease, with the second constant when
\(\Omega=0\). Comparing the numerator and denominator as rational
integers locates the maximum by binary search, including endpoints. At
its order \(u\), put \(H=O-u\), and use the proved binomial entropy
bounds of Solution 39: \[
\begin{aligned}
\ln f(u)\le{}&(D-u)\ln r-L\ln(50!)
+D\ln D-u\ln u-(D-u)\ln(D-u)\\
&+50u\ln3+H\bigl(1+\ln(1+\Omega/H)\bigr).
\end{aligned}
\] Take \(0\ln0=0\) and the last term zero for \(H=0\). Retain the
negative factorial term. Insert this upper bound, the coefficient
bracket bound \(0.310842088027Z\), the complete common denominator
\(((I+J)D+47L)\ln2\), and the full torus term into (10.260). Use factor
one for both integer outputs and four for the fractional output.

The following endpoint data make the calculation directly checkable.

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}lrr@{}}
\toprule\noalign{}
Quantity & Stage 1 & Stage 17 \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
\(A_I\) & 1374 & 18092274 \\
\(R_I\) & 2744 & 36184546 \\
\(X_I\) & 1242 & 16366762 \\
\(O_0,O_1\) & 993090, 814996 & 75, 61 \\
\(O_2,O_3\) & 668840, 548894 & 50, 41 \\
\(\mu_0,\mu_1,\mu_2\) & 431116, 266102, 119946 & 32, 20, 9 \\
Mixed \(N_*\) & 1286895674 & 1266459164 \\
Mixed separation \(B\) & 7 & 16 \\
\end{longtable}
}

At stage one the respective scalar maximizing orders are
\(517016,343527,372590\); at stage seventeen they are \(61,50,41\). The
stage-one common denominator exponents are \(3424003,3424003,5188953\);
the stage-seventeen ones are \(31663203,31663203,33428153\).

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}
  >{\raggedright\arraybackslash}p{(\linewidth - 6\tabcolsep) * \real{0.2000}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 6\tabcolsep) * \real{0.2667}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 6\tabcolsep) * \real{0.2667}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 6\tabcolsep) * \real{0.2667}}@{}}
\toprule\noalign{}
\begin{minipage}[b]{\linewidth}\raggedright
Comparison at stage 1
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
Required input / \(Z\), upper
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
Analytic gain / \(Z\), lower
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
Arithmetic cost / \(Z\), upper
\end{minipage} \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
Odd-to-full closure & 1.387797139482 & 1.340382458724 &
1.100243509064 \\
Full integer extension & 2.246326479139 & 2.208922606590 &
1.727090951833 \\
Mixed fractional step & 7.192833169904 & 7.102853409486 &
3.989657490226 \\
\end{longtable}
}

Use the unrounded interval endpoints when subtracting, then round the
resulting gap down. At each stage, taking the maximum of the three
required-input upper bounds and the minimum of the three gap lower
bounds produces exactly the seventeen-row table in Theorem 10.98. Its
largest required input is \(7.192833169904\), below the available lower
bound \(7.294535093395\); the smallest zero-forcing gap is at least
\(0.240138949661\). Thus every comparison passes with a strict margin.
The \href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/first_contraction_budget.py}{reproducible
figure program} implements these rational interval operations and prints
all fifty-one comparisons.

Finally the original widths are below \(67805,29202\). Dividing by
\(2^{17}\) puts both below one; since they are integer widths, both are
zero. The retained coefficient vector is nonzero and gives the nonzero
degree-at-most-\(1764950\) polynomial (10.262). The stage-seventeen full
closure gives \(36184549\) distinct integer roots. The elementary
polynomial root bound contradicts this. No restriction to a single
coefficient pair was made: all coefficient-height, Euler and precision
estimates used the whole family (10.245).

\textbf{Solution 42.} The unramified case \(p\ge5,e=1\) has \(a_*=7/2\)
in (10.184). It must be retained in a uniform argument. With this weaker
value, the common lower bound for \(g_5\) at rank two is \[
\frac{893}{7250}\,3\left(\frac72\right)^2
=\frac{131271}{29000}>4.
\] Dividing that bound by \(r\), its consecutive ratio is
\((7/2)r(r+2)/(r+1)^2\ge28/9>1\). Thus \(g_5>2r\) at every rank
\(r\ge2\), and \(1+g_5^{-1}<5/4\). For \(g_4/[r(r+1)]\), the rank-two
lower bound is \[
\frac{57}{116}\frac{(7/2)^2}{2}
=\frac{2793}{928}>1.
\] The consecutive ratio of \((7/2)^r/r\) is \((7/2)r/(r+1)\ge7/3>1\).
This gives \(1+\epsilon<\exp(1/2)<2\).

The coarse degree-dominance comparison is \[
\frac{135143424}{6699000}>20,
\] and the additive degree/output-order comparison follows from \[
\frac{47\cdot49\cdot39\cdot32768}
{10000\cdot21\cdot44}>4.
\] The complete singleton height bound from Theorem10.18 gives
\(d\sigma_i>1/(87d^2)\). The stopping exponential contributes
\(\mathrm e^8(r+1)^2d^2\); its \(d^2\) cancels that denominator before
any constant is discarded. The remaining \(\exp A/[A(A+39)]\) increases
for \(A\ge5\).

For the intermediate allocation, let \(\lambda=1+g_5^{-1}\),
\(E=(q\eta^{r+1})^I\), and use the upper additive degree \[
D_+=\lambda\frac{S\mathcal D}{c_1c_4d(A+x)}.
\] The unrounded lower numerator is \[
\frac{2\overline S}{3r}
\frac{(\overline T/(r+1))^{m+1}}{(m+1)!}.
\] Divide by \((m+1)!D_+\prod_{j\in J}\mathsf D_j\). Inserting
\(\overline S,\overline T,\mathsf D_j\) and cancelling produces \[
\frac{2B_0}{r}\,
\frac{\prod_{j\in J}\sigma_j}{((m+1)!)^2}
\left(\frac{\eta c_2qPrd}{2e\theta t}\frac E{q^\nu}\right)^m.
\] One factorial comes from the binomial lower bound, and the other from
the multiplicity theorem; neither may be removed. Every \(m\)-element
subfamily satisfies its own height-product bound. After that bound is
inserted, the stopping inequality and the proved elementary exponential
bounds give \[
\frac{72}{725}\frac{(2048/3)^m}{(m+1)^3}.
\] At \(m=1\) this equals \(147456/17400>8\). Its consecutive ratio is
larger than \((2048/3)(2/3)^3>1\), so the comparison holds for every
\(m\ge1\).

For the final allocation use the lower numerator \[
\frac{2\overline S}{3r}
\frac{(\overline T/(r+1))^r}{r!}.
\] Divide by \((r+1)!D_+\prod_{j=1}^r\mathsf D_j\), and substitute the
full expression for \(\mathcal D\). The exact cancellation, before lower
bounds on the stopping exponential, is \[
\frac{2q\eta^r q^{\nu+u}}{\gamma}\,
\frac{q^{rI-r\nu}\eta^{(r-1)(r+1)I}}
{3r(r+1)!\lambda(1+\epsilon)(2+g_2^{-1})c_0
[2(r+1)]^r t^rM\ell_d}.
\] The factor \(r!\) cancels against the one in \(\mathcal D\). The
factor \((r+1)!\) remains. Use \(q^{rI}\eta^{(r-1)(r+1)I}>E^r\),
\(E>\exp(3(A+x))\), and \(q^{\nu+u}/\gamma\ge1\) before dropping any
saturation factor. The resulting rank-dependent lower bound is \[
\frac{8(192/125)^r16^{2r-1}}{135r(r+1)^2}.
\] At rank two this is \[
\frac{8\cdot192^2\cdot4096}{125^2\cdot135\cdot2\cdot9}>31.
\] Its consecutive ratio is at least \((192/125)256(3/8)>1\). These are
analytical bounds for all ranks, rather than an inference from a finite
rank check.

The floor allocations are formed from the real
\(\overline S,\overline T\). Their sums are integers at most those real
quantities and thus at most their floors. Also \(\lfloor y\rfloor+1>y\),
including when \(y\) is an integer. This explains the strict
inequalities used in every numerator.

Finally the multiplicity theorem supplies integral \(u_i\), not an
asserted integral basis of the whole relation lattice. They are
rationally independent, span \(\boldsymbol B\), and give
\(\beta_i=\prod_j\alpha_j^{u_{ij}}\in K\). Each is a unit at the chosen
prime, and the height power and product laws give \(h(\beta_i)\le R_i\).
A relation among the \(\beta_i\) would be a relation among the original
independent \(\alpha_j\); therefore the new bases are independent.
Clearing denominators in the rational-span relation and killing the
original finite root-of-unity phase proves (10.271). The arithmetic
field of a fractional zero used earlier in the descent remains a
separate quantity; it is not changed by this return of the new bases to
\(K\).

\textbf{Solution 43.} At two the reduction of \(\zeta_3\) has order
three, so \(f\) is even and at least two. Hence \(t>1\) in every case.
For any \(j\ge1\), the first member of \(t^jM_j\) is
\(p^f/\delta\le p^ft\). The second is bounded by \(p^ft\) because
\(\ln(t/j)\le t/j-1\). This applies to the actual order at rank one as
well as to the saturated residue index at larger ranks.

Write \(D_+=\lambda S\mathcal D/(c_1c_4d(A+x))\). The lower denominator
in the bound (10.270) is

\[
\frac{2\overline S}{3r}\frac{(\overline T/(r+1))^m}{m!}.
\]

Before substituting \(\mathcal D\), the resulting upper bound for the
new weight product is

\[
\frac{3r}{2q}(m+1)!m!\lambda
\frac{\mathcal D}{c_1c_4d(A+x)}
\left(\frac{2e\theta t}{c_2qPd}\right)^m
q^{m\nu}\frac{\eta^{(r+1)I-m}}{E^m}.
\]

Substitute \(\mathcal D/(c_1c_4d(A+x))
=(\gamma/q^{\nu+u})(1+\epsilon)(2+g_2^{-1})c_0
[H_*r(r+1)d]^rM_r\ell_d\prod_j\sigma_j/r!\). Division by
\(\Gamma_n^{r-m}(M_r/M_m)\prod_j\sigma_j\) cancels all \(H_*\), \(d\),
and old \(M_r\) factors, giving (10.274). The new \(M_m\) remains, so
its actual residue index is still present before the uniform bound is
applied. The two factorials arise separately from the multiplicity
output and the binomial lower bound.

The factorial integral gives \(r^r/r!<\mathrm e^r\), while \(n\ge r\),
\(\gamma\le q^\nu\), and \(q^u\ge1\). The geometric stop contributes
\(\exp(-3(m+1)A-(2m+3)x)\). Combining these with \(t^mM_m\le\exp(A)A\),
\(C<45/2\), and \(2\mathrm e/\eta<25/3\) gives (10.275). At
\(A\ge4+\ln((r+1)d)\), its exponential factor is decreasing. Use
\(4+\ln((r+1)d)\le5(r+1)d\), \(\ell_d\le d\), and

\[
\frac{r(m+1)!m!}{(r+1)^{2m+1}}
<m+1\le2^m.
\]

For the strict inequality, first maximize \(r/(r+1)^{2m+1}\) over
\(r\ge m+1\), then use \((m+1)!m!\le(m+1)^{2m+1}\). Finally
\(\mathrm e>2\) gives the exact envelope \((675/2048)(25/6144)^m\). Its
largest value is \(5625/4194304\), at \(m=1\), and
\(5625\cdot700=3937500<4194304\).

For the composed forms, independence modulo \(z_0\) follows from
independence of both the original projected rows and the integral
vectors \(u_i\). Rational span gives the original form representation.
Coefficientwise triangle inequalities bound every new weighted row by
\(R_i\); integer products preserve the original field and chosen-prime
units. The old product bound and the new one multiply with exact
cancellation of \(M_r\), leaving \(\Psi_m\) with the new residue index.
Since \(m<r\), this contradicts the original minimum rank.

In the example \(\Gamma_2<3\cdot7\cdot(101/100)\cdot3<64\). The two
parts of \(M_2\) are less than \(3\) and \(9(11/10)/4<3\), respectively.
Every \(M_1\) is greater than \(2\), even when a torsion factor changes
its actual residue order. Also \(h(2)h(5)<(7/10)2=7/5\). Thus the
rank-one profile allows a weight less than \(64(3/2)(7/5)=672/5\). The
Euclidean calculation \(103-101=2\), with 101 odd, gives
\(\gcd(101,103)=1\). Any proportional integral row is consequently a
nonzero integer multiple of \((101,103)\). Its required weight exceeds
\(511/3\), and \(511\cdot5>672\cdot3\). No rank-one representation is
admissible; the standard two rows have exactly the permitted product, so
the minimum rank is two.

\textbf{Solution 44.} The minima, in the order in (10.281), are
\(20838501411/25000000\) in III.2, \(1273657/6250\) in III.1,
\(42419/40\) in V, and \(14271/10000\) in V. They exceed respectively
\(800,202,1000,7/5\). These are obtained by multiplying the exact
terminating decimals in each of the seven rows and comparing their
rational numerators after clearing denominators.

The three contributions in Lemma 10.105 are, with common denominator
\(7(r+1)^2b_r\), less than \(10\), \((3r+2)(r+1)\), and \(5(3r+1)\).
Their sum is \(3r^2+20r+17=(r+1)^2[3+14/(r+1)]\). At rank two the result
is \(667/151200<1/225\); the residual factor \(3+14/(r+1)\) decreases,
giving the complete all-rank bound with \(14^{2-r}\).

For the simplex, at additive order \(u=0,\ldots,6\), there are \(7-u\)
choices of Euler order \(h_{\mathrm E}=0,\ldots,6-u\). Their total is
\(7+6+\cdots+1=28\). The sum of the additive orders is
\(0\cdot7+1\cdot6+2\cdot5+3\cdot4+4\cdot3+5\cdot2+6=56\), so their mean
is two. Swapping the additive and Euler coordinates gives the same mean
for the Euler order. The ten rows with \(u>2\) are zero when the
additive degree is two, but retaining their equations imposes no new
restriction. Assigning positive upper bounds to their zero entries still
supplies legitimate row majorants, and Theorem 10.71 already supplies a
strict column surplus for the full 28-row pattern.

Subtracting (10.289) from one over the common denominator \(120(r+1)^2\)
gives \(108(r+1)-85(r-1)=23r+193\). Thus (10.290) holds for every rank,
not just the plotted ranks.

For the scalar calculation,

\[
F_{0,1}(X)=\frac{X^4+10X^3+35X^2+50X+24}{24},
\qquad \mathfrak v=12,
\qquad \mathfrak v^2\frac{F_{0,1}''(0)}{2!}=210.
\]

The factor \(24/12^2\) in (10.298) gives \(W=35\). Hence \(v_3(V)=1\),
\(v_3(W)=0\), and \(Lv_3(4!)-uv_3(12)=1-2=-1\), exactly the stated
valuation difference. For any permitted \(\theta\), \(G_u=2\theta-1\),
so \(v_3(V)+G_u=2\theta=v_3(W)+2\theta\). The cancellation arises from
multiplying the whole value by one rational scalar. This leaves its
number field unchanged. The product formula still sums all embeddings
and places of that global field, so the factor \(d_F/(e_Ff_F\ln p)\)
remains.

\textbf{Solution 45.} The linked certificate starts from the fifteen
displayed weights and the interval lists. Its exact rational check gives
\(\sum v_r/r=0\). Each denominator divides 30030, so increasing an
integer argument by 30030 changes \(F\) by \(30030\sum v_r/r=0\). At the
reflected argument \(30030-1-n\), each pair of floors sums to
\(30030/r-1\). Summing gives \(-\sum v_r=3\), not one. The coefficient
sum is \(-3\); the mean value of \(F\) on a period is therefore \(3/2\).

For logarithms, the sixteen down-rounded terms of (10.304) give a lower
endpoint; add seventeen units of \(1/10^{12}\) for an upper endpoint.
Normalize each prime by its largest power of two and add the
corresponding enclosure of \(\ln2\). The sieve and the assignments at
every prime power produce cumulative \(L_n,U_n\). Direct integer checks
give the four strictly positive minima in Lemma 10.111. The lower check
uses \(26L_n>25Q(n+1)\): for \(x\) just below \(n+1\),
\(\psi(x)=\psi(n)\), and a comparison only with \(25n/26\) would not
imply \(\psi(x)>25x/26\). The two rational recurrence margins are
respectively \(6731/26000000\) and \(11523/26000000\). They prove both
induction inequalities for every later real interval.

The new mean coefficient is \((r-1+107/103)/(r+1)+17(r-1)/[24(r+1)^2]\),
which is \(C_r^{\sharp}\). At rank two it is \(16871/22248\), while
\(C_2=841/1080\). Their difference is \(21/1030\); it is an absolute
improvement in the coefficient of \(TH_1\), with the same full row set
and coefficient field.

\textbf{Solution 46.} Multiplying \(m+1\) positive geometric series
gives coefficient \(\binom{m+h}{h}\) at \(z^h\). Its single contribution
is no larger than the full sum. If \(v(k)z\le1\), multiplication by
\(v(k)^uz^{u+h}\) cannot increase this inequality. This proves (10.309)
after setting \(z=\exp(-a)\), without using an average derivative order.

In the figure, \(v(4)=12\), \(z=1/13\), and the exact scalar at order
\(u\) is \(12^u\binom8u15^{8-u}\binom{27-u}{10-u}/24^2\). The
consecutive ratio is \(12(8-u)(10-u)/[15(u+1)(27-u)]\). At zero its
numerator is 960 and denominator 405, so the scalar increases. At one
its numerator is 756 and denominator 780, so it decreases; both
component ratios decrease thereafter. The maximum is therefore at one
and equals \(88976443359375\).

The fixed-allocation upper bound before taking logarithms is
\(\binom8u15^{8-u}13^{10}(13/12)^{18}/24^2\). The uniform bound replaces
its first two factors by \(16^8\), and is
\(15502932802662396215269535105521/3570467226624\). These formulas
determine the entire left panel. The right-panel coefficient is
\((1000/999)(3/8)(r-1)/[15(r+1)^2 2^I]\). Its maximum over real
\(r\ge2\) occurs at three, because the difference from \(1/8\) is
\((r-3)^2/[8(r+1)^2]\). The maximum at depth zero is \(25/7992<1/300\),
and each further depth halves it in this illustration.

The additive term comes from the \(r-1\) geometric factors in
\(m_I+1=\sum\Omega_j+r-1\). It remains even when their integer arguments
are zero; an estimate only for \(\sum\Omega_j\) would omit it. The seven
products follow by multiplying the displayed \(c_1,c_3,q\), \(c_2=7/4\)
or \(13/9\), and the justified minimum \(P\). Their smallest is
\(18772369/1250000>15\). Finally the exact differences are
\(4/103-125/3228=37/332484\) and \(4/103-1/27=5/2781\), which prove the
two caps used in Lemma 10.114.

\textbf{Solution 47.} We have \(D_{\mathrm a}=8\), \(\mathfrak v=12\),
\(\ell_p=1\), and \(\chi_p=2v_3(24)=2\). In the first instance
\(\vartheta-\beta+B=1\), so the Euler-order minimum in (10.319) is at
zero. The remaining expression is \(\max\{-a,u-2\}\), for
\(0\le a\le\min\{6,8-u\}\). For \(u=0,1\), the allowed range reaches the
breakpoint \(2-u\). For \(u\ge2\), the constant \(u-2\) dominates
already at zero. Thus the nine envelopes are

\[
(K_0,\ldots,K_8)=(-2,-1,0,1,2,3,4,5,6).
\]

Here \(C_0=1\); with \(B=0,M=7\), the older contribution is
\(-(M-1)C_0=-6\) at every order. The new finite bound improves it by
\(u+4\). The left breakpoint occurs because \(\gamma_{u+a}\) acquires
one unit of valuation for each further additive derivative. This exactly
compensates that derivative's lcm factor. A higher Euler derivative
still has its separate slope cost.

For the second instance, \(\vartheta-\beta+B=-1\). Formula (10.319),
with \(H=\min\{4,8-u\}\), becomes

\[
K_u(5,1)=\min_{0\le a\le H}
\{\max\{-a,u-2\}+2a-4\}.
\]

Both affine pieces strictly increase with \(a\), so zero gives its
minimum. Therefore

\[
(K_0,\ldots,K_8)=(-4,-4,-4,-3,-2,-1,0,1,2).
\]

The interpolation contribution excluding \(U-\beta+G_u\) and the
separate deleted-node cost is \(-(M-1)B+K_u=-4+K_u\). The original
common loss gives \(-4\max\{1,3\}=-12\). The sharpened linear loss gives
\(C_{\mathrm{sharp}}=2\), hence \(-8\). The first three new
contributions equal \(-8\); the later ones are strictly larger. No
minimum here is being asserted to equal an actual function's valuation.

The q-powers used in differentiation have p-adic valuation zero because
\(p\ne q\). Their rational denominators contribute at primes above q in
the global product formula; their clearing factor therefore stays in
(10.312). Likewise lying in an old local completion does not change the
degree of the actual number field containing the algebraic value. Both
costs remain in \(\mathcal A_u(x)\), and both strict numerical
comparisons in Corollary 10.120 remain necessary.

\textbf{Solution 48.} All the input data are rational after Lemmas
10.121--10.123: the lower depths include the factor \(1+10^{-6}\), not a
rounded decimal. For each of the eight field rows and each
\(r=2,\ldots,7\), substitute the original case constants, its actual
\(c_5\) band, and \(C_r^{\sharp}\) into (10.331)--(10.332) at \(j=0\).
These 48 differences give the first column of Lemma 10.125. Each larger
rank uses the last \(c_5\) band; the inequalities \(C_r^{\sharp}<1\),
\(\eta\le1\), decreasing \(\epsilon_r,\bar g_{12}\), increasing
\(b_r,S_-\), and \(\bar g_9=107/103\) give the second column from eight
rational substitutions at rank eight. These monotonic inequalities
supply the missing universal quantifier.

For I.2 the uniform data are

\[
\begin{aligned}
c_1&=7247/5000,&c_3&=3463/2500,&c_4&=104/5,\\
C&=57/20,&Y_-&=1500000/1000001,&c_2&=7/4,&c_5&=14/25.
\end{aligned}
\]

At rank eight, \(b_8=54212659200/29\),
\(\epsilon_8=1189/3415397529600\), \(S_-=53856576/6875\), and
\(\bar g_{12}=0\). Set \(C_r^{\sharp}=\eta=1\) only in the arithmetic
upper bound, keeping \(c_5=14/25\) in the gain. Direct substitution
gives

\[
\begin{aligned}
\mathcal L_{\mathrm{I.2},\ge8}(0)
&=\frac{46304258427679214506849307}
{8985131362373607395136000000},\\
\mathcal L_{\mathrm{I.2},\ge8}(0)-\frac1{200}
&=\frac{1378601615811177531169307}
{8985131362373607395136000000}>0.
\end{aligned}
\]

Thus the smallest uniform margin exceeds \(1/200\) strictly, although
the lower-bound marker in the figure is exactly \(5/1000\). The
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/integer_comparison_bounds.py}{exact rational
program} lists all 56 enclosing differences without floating-point
decisions.

Write \(a=\bar\lambda\ell_q/(5c_4)\). The increasing continuous
difference in Lemma 10.124 excludes the indicator cost. On \([0,1]\),
its increase exceeds \(2(7/500)-1/125>1/125\) at odd primes, and
\(3/20-3/50>3/100\) in V. These bounds exceed \(a/(q-1)\), so
subtraction of that full indicator still leaves a strict increase. On
every later integer step the indicator is constant. This explains why it
cannot simply be omitted at the first step.

Finally Figure 10.28 displays the geometric restriction \(I\ln Q\le3D\),
under which the convex function \(a_0Q^{-I}+b_0I\) is bounded by its
endpoint chord. Actual stages must additionally satisfy
\(c_5T_{I+1}/(r+1)\ge1\), and a contracted stage must obtain its input
zeros through fractional interpolation and coset extraction. A valid
inequality at a candidate depth is different from existence of that
stage. Corollary 10.127 provides the complete initial integer block
without needing such later stages.

\textbf{Solution 49.} With the original floors of Solution 39, the
output order is \(O=993090\). Thus

\[
(\mu_0,\mu_1,\mu_2)=(803663,481446,217012).
\]

The symmetric node counts in the three annuli are \(759,758,1518\).
Therefore

\[
\begin{aligned}
N_*&=759(803663)+758(481446)+1518(217012)\\
&=1304340501\\
&=3035(217012)+759(322217)+1517(264434).
\end{aligned}
\]

The last block alone contributes \(3035(217012)=658631420\). The added
inner derivatives nearly double that degree. These are exact cardinal
degrees in the normal quotient, which then supplies \(N_*\theta\); no
actual nonzero auxiliary valuation is inferred from the diagram.

For \(r\ge8\), reverse the geometric sum:
\(q^{-r}\sum_{j=0}^r(q\eta)^j=\eta^r\sum_{k=0}^r(q\eta)^{-k}\). Its
first four reversed terms are at least \(\eta^r\sum_{k=0}^3q^{-k}\).
Integration gives \(-\ln(1-c/(r+1))\le c/(r+1-c)\), hence
\(\eta^r>\exp(-c)>1/E_+(c)\). For the radius correction, \(1-H<1\) and
every \(\eta^j-H<1\); consequently \(F_r\le q(r+1)(2r+1)/(q^rS_-(r))\).
Divide \(S_-\) by \((r+1)^2\). Its remaining factor is nondecreasing in
each original case, while \((2r+1)/((r+1)q^r)\) decreases. Its maximum
at rank eight is exactly \(153q/(q^8S_-(8))\). Combining the two
estimates proves (10.342) for every larger rank.

\textbf{Solution 50.} The original certified first-stage radius has
\(\lfloor S_1\rfloor=686\), so \(R=2(686+1)=1374\). The floor
\(\lfloor T_1\rfloor=993090\) alone gives \(178094<cT_1/3<178095\),
since \(c=269/500\). Thus \(M=178095\). The earlier exact intervals also
give \(\lfloor\eta T_1\rfloor=814996\), and \(814996+M-1=993090\),
verifying the required ordinary jets.

Because \(R\) is even, its q-deleted nodes are precisely its 1374 odd
integers. Their total cardinal degree is \(N_*=1374(178095)=244702530\).
At three, \(3^7=2187<2R=2748<6561=3^8\), so \(B=7\). At additive order
zero with \(\beta=0\), the jet minimum \(K_0(M,B)\) is zero: every
allowed positive additive increment contributes at least
\(a(B-\ell_p)\ge0\), where \(\ell_p=3\), and the remaining Euler
contribution is nonnegative. The zero increment attains zero. The normal
interpolation entry consequently has the two losses

\[
MB=1246665,\qquad (M-1)B=1246658,
\]

whose sum is \(2493323\). The first is the deleted-node cardinal cost;
the second comes from the inverse coefficients paired with the jets.
They have separate origins and both occur in (10.320).

Before extraction, Theorem 10.56 gives the actual normalized support
field of degree \(q^{h_\Lambda}\) over \(K\). Its unchanged selected
completion does not alter that global factor, which belongs in (10.340).
After extraction, (10.141) and the phase congruence write each new
integer torus term as a product of integral powers of the original
\(\theta_i^P\) and \(\alpha_0\); hence the value belongs to \(K\). This
algebraic identity justifies the return to degree \(d\) in (10.346). A
statement about a selected local completion would not justify it. The
separate q-denominator and normal-term bounds continue to apply in that
original field.

\textbf{Solution 51.} The deleted inner nodes are
\(-8,-7,-5,-4,-2,-1,1,2,4,5,7,8\), each of multiplicity five. The other
inner nodes are \(-9,-6,-3,0,3,6,9\), each of multiplicity three. The
outer nodes are the nine positive integers 10--18 and their negatives,
each of multiplicity one. The annular degree is \(12(5)+7(3)+18=99\).
The telescoping degree is
\((2\cdot18+1)(1)+(2\cdot9+1)(3-1)+12(5-3)=99\).

Modulo four the deleted counts are \(4,3,2,3\), in classes \(0,1,2,3\).
For example, the class zero has \(-8,-4,4,8\), while the class two has
only \(-2,2\). Also \(2^5=32\le36<64=2^6\), so \(B=5\), and \(M=5,E=2\).
Formula (10.354) gives \(-5(5-5+2)=-10\), \(-5(5-3+2)=-20\), and
\(-5(5-1+2)=-30\). Each Taylor-inverse coefficient of degree a has the
separate bound \(-5a\). The combined Hermite bound at \(C=0\) is
consequently \(\Lambda-EB-(M-1)B=\Lambda-30\).

If a deleted increment contains the cardinal node, residue-count
discrepancy two permits a loss of one after omitting that node; if it
does not contain the node, the loss can be two. A claim of discrepancy
one would replace these bounds by zero and one, respectively, and would
discard the \(EB\) cost. The displayed counts disprove that claim. No
change to the degree count or the Taylor-inverse loss justifies
discarding the deleted cost.

\textbf{Solution 52.} Write \(x=(1-t)a+tb\), \(0\le t\le1\). Concavity
gives \(f(x)\ge(1-t)f(a)+tf(b)\ge\min\{f(a),f(b)\}\). For a twice
differentiable function, \(f''\le0\) gives this concavity by subtracting
the affine chord and observing that an interior negative minimum would
contradict the decreasing derivative; alternatively integrate its
decreasing derivative on the two sides. Each \(\tau_j\) is a constant
minus a positive sum of convex exponential-plus-affine functions. The
mass is a positive affine combination of these cutoffs. Thus subtracting
the convex arithmetic envelope gives the required concave gap. At higher
ranks the last-block mass has the same property, because its coefficient
of \(\tau_r\) is positive.

Replacing a cutoff by a lower bound decreases its mass contribution only
when that coefficient is nonnegative. In (10.367), the coefficients for
j=2 and j at least three are positive because \(2(q^2-q)-2(q+1)/S_->0\)
and \(2(q^j-q^{j-1})-2/S_->0\). The first coefficient has no subtracted
floor term. This justifies the direction of every finite endpoint
substitution.

For \(z=1/q\), the finite geometric sum has remainder tending to zero,
so \(\sum_{j\ge1}q^{-j-1}=z^2/(1-z)=1/[q(q-1)]\). Write
\(j=\sum_{k=1}^j1\); all terms are nonnegative, and grouping them gives
\(\sum_{j\ge1}jz^{j+1}=z^2/(1-z)^2=1/(q-1)^2\). At q=3 these are
\(1/6,1/4\). Summing the defining \(\Delta_j\) with these upper bounds
gives (10.373), including its separate \((r-1)Q^{-I}\) torus cost.

Fractional zeros are produced only at \(I<I_0\), so
\(I\le I^*-1\le L_D\). Their phase passage constructs the successor
J=I+1, which can equal \(I^*=\lfloor L_D\rfloor+1>L_D\). Its deleted
zeros still need full closure at order \(\lfloor\eta T_J\rfloor\). The
late endpoint arithmetic cost is bounded by the former late \(B_0\)
bound plus one extra \(\ell_-\), or \(\ell_+\) in the uniform rank
range, because \(\ell(L_D+1)=\ell L_D+\ell\). Its torus term remains
less than \(1/1000\), since \(Q^{-(L_D+1)}<\exp(-3D)\). Its input loss
uses only \(cT_J/a\ge1\), giving \(3c/(ac_3Y_-)\). Formula (10.369)
therefore supplies the actual final full block, without extending the
fractional comparison beyond its proved interval.

\textbf{Solution 53.} We have \(\eta=24733/30000\) and
\(H=15129702640837/27000000000000\). The full radii are
\(R_1=2(\lfloor100\rfloor+1)=202\) and
\(R_2=\lfloor4\cdot100\rfloor=400\). The full derivative cutoffs are 824
and 679, the deleted cutoff is 1000, and \(O=\lfloor1000H\rfloor=560\).
Hence \(\mu_0=441,\mu_1=265,\mu_2=120,E=176\).

There are 202 deleted inner nodes, 203 inner multiples of 2, and 396
nodes in the outer annulus. Their respective multiplicities are 441,
265, 120. Thus \[
N=202\cdot441+203\cdot265+396\cdot120=190397.
\] Equivalently, (10.356) gives
\(801\cdot120+405(265-120)+202\cdot176=190397\). Every output of total
order at most 560 has all these additional ordinary jets, including at
each input cutoff: its largest resulting orders are 1000, 824, 679,
respectively. The fractional numerator radius is
\(2(\lfloor100/H\rfloor+1)=358\).

Integer clearing at depth J costs at most \(\ell(J+1)\). Substitution of
\(J=L_D+1\) therefore gives \(\ell L_D+2\ell\); keeping only one
residual \(\ell\) would omit the extra terminal step. Finally Theorem
10.56 bounds the global degree by \(dq^{h_\Lambda}\), while local
degrees describe only the selected completion. Local degree one does not
bound \(h_\Lambda\) by zero. The proved comparison uses
\(h_\Lambda\le r\) and retains the normalization \(c_1W_*/q^r\)
throughout.

\textbf{Solution 54.} (a) Taking logarithms, the required strict
inequality is \(I>3D/\ln Q\). The least integer with this property is
its floor plus one. For the displayed values the quotient equals 192, so
\(I^*=193\): integer 192 gives equality and does not suffice. If
\(I^*=I_1\), the minimum defining \(I_0\) is still \(I^*\). Corollary
10.140 includes its terminal stage, so the same zeros and the same
strict geometric inequality apply.

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{1}
\item
  Write \(P=Y^{\boldsymbol d}Q\). The torus Euler operator \(D_j\) acts
  on the factor \(Y^{\boldsymbol d}\) by the scalar
  \(c_j=\sum_iC_{ij}d_i\). Thus (10.93) replaces each \(D_j\) acting on
  \(Q\) by \(D_j+c_j\). Expanding a product produces jets of no greater
  total order, with diagonal coefficient one. Replacing \(c_j\) by
  \(-c_j\) gives the inverse change. The additive derivative commutes
  with this multiplication. At the evaluation point all torus
  coordinates are nonzero, so both multiplication by
  \(Y^{\boldsymbol d}\) and its inverse are defined. Consequently both
  sets of jets vanish together through the identical cutoff.
\item
  Let \(a_{jk}\) be the coefficient of \(z_k\) in \(L_j\). By the
  triangle inequality and the old weighted-height condition,
\end{enumerate}

\[
\sum_{k=1}^n\left|\sum_{j=1}^ru_{ij}a_{jk}\right|h(a_k)
\le\sum_{j=1}^r|u_{ij}|\sum_{k=1}^n|a_{jk}|h(a_k)
\le\sum_{j=1}^r|u_{ij}|\sigma_j=R_i.
\]

Theorem 10.102 and (10.277) now bound the new weight product strictly by

\[
\frac1{700}\,
\Gamma_n^{r-m}\frac{M_r}{M_m}
\Gamma_n^{n-r}\frac{M_n}{M_r}\prod_kh(a_k)
=\frac1{700}\Psi_m\prod_kh(a_k).
\]

Every \(M_j\) is positive, so the cancellation is legitimate. At rank
one \(M_1\) uses the actual rank-one residue order specified in
(10.272); keeping that factor in both expressions gives the same
cancellation. Independence of the new forms and representation of the
same nonzero linear form follow from Corollary 10.103. Thus this is a
smaller admissible representation, which contradicts the least-rank
choice.

\textbf{Solution 55.} (a) Here \(\eta=1231/1500\). The value of
\(c\eta^jT_1/a\) is \(\eta^{j-1}\): it is greater than one at \(j=0\),
equals one at \(j=1\), and is less than one from \(j=2\) onward. Thus
\(r_1=1\) and \(cT_\flat/a=\eta=1231/1500\). This includes the lower
endpoint in Lemma 10.142.

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{1}
\item
  If \((3D-I_1\ln Q)/\ln q=m\) is an integer, its ceiling is \(m\). That
  choice gives \(Q^{I_1}q^m=\exp(3D)\), whereas floor plus one gives
  \(m+1\) and the required strict inequality. The latter also covers
  nonintegral quotients.
\item
  Since \(5220=36\cdot145\), \(9\cdot2^{13}/36=2048\), and
  \(7\cdot2048=14336\), the fractions agree. The sharper coordinate
  bound is \(|m_j|<145/28672<1\) for integers \(m_j\). Every coordinate
  is therefore zero. Proposition 10.31 preserves a nonzero polynomial,
  and the shifted-power independence of Lemma 10.29 preserves
  nonzeroness after this support collapse. Its degree is at most
  \(D_{\rm a}=kL\), while it has \(2R+1>(171/58)D_{\rm a}>D_{\rm a}\)
  distinct roots. Repeated division by \(X-s\) at distinct roots makes
  their product divide the polynomial; its degree would be at least
  their number. This is impossible.
\end{enumerate}

\textbf{Solution 56.} (a) At \(R_b=25\), \(R_0=24\) and \(N_*=100\). At
\(R_b=26\), \(R_0=24\) and \(N_*=102\). Generally
\(24R_b/25-1<R_0\le24R_b/25\). Substitution into \(N_*=2R_b+2R_0+2\)
gives \((98/25)R_b<N_*\le(98/25)R_b+2\). The strict lower bound
therefore includes both integral and nonintegral floor arguments.

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{1}
\item
  For an outer node, its multiplicity is one while the maximum
  multiplicity is two. The cardinal estimate proved in Lemma 10.94 is
  \(-(M-\mu(s))B=-B\). For an inner node that part is zero, but the
  first-jet Taylor inverse can still cost \(B\). Theorem 10.119 charges
  the combined loss with \((M-1)B=B\). Both intervals are full, so its
  separate q-deleted term is \(EB=0\).
\item
  Put \(z=c\eta^rT_1/a\ge1\). Multiply the mass upper bound by \(z\),
  use \(R_b\le q^rS_1\), and substitute \(c_1W_*S_1T_1\theta=qa\).
  Dividing by \(q^a\) gives the first two terms of \(P_r\). The
  coefficient bound contributes \(c_1/(7ab_rq^a)\). Since
  \(M-1=1\le z\), the separation and retained-jet bound contributes at
  most \(c\eta^r/(c_3Y_-aq^a)\), the last term. For the lower bound use
  \(R_b>q^rS_1-1\) and then \(a/T_1>cH\); this gives exactly \(G_r\).
  The term \(1/c_2\) in \(C_r\) is the full normalized \(q^a/c_2\) cost.
  It must be retained along with the separate linear depth-indicator
  term.
\item
  In each rank band at least eight, \(c\) is fixed. The proof gives
  increasing \(H,S_-,b_r\), decreasing
  \(\eta^r,\epsilon_r,\bar g_{12}\), the uniform \(C_r^\sharp<1\) and
  \(\bar g_9=107/103\), and decreasing \(r/q^{r+1}\). These enclose
  every rank by the displayed endpoint formulas at eight. Exact rational
  substitution then checks those enclosing formulas. Successful
  evaluations at finitely many ranks alone would leave all other ranks
  unproved; it is the analytic enclosure that covers them.
\end{enumerate}

\textbf{Solution 57.} (a) The only required divided jet at each node is
order zero, which is the prepared value of the target selected order.
Thus each node has multiplicity one. Since \(M-\mu(s)=0\), the
full-interval cardinal is integral, and the inverse Taylor polynomial
has degree \(M-1=0\). Both of those losses are zero, and \(E=0\) because
no node is deleted. The normal baseline still contains \(U-\beta\);
removing the jet losses does not remove that coefficient cost.

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{1}
\item
  Use \(\beta\ln p\le h\le H_1\) and \(W_*=et/(S\mathcal D)\).
  Substituting \(S=c_3qa dH_1/t\) and \(\mathcal D=c_1\theta etT/(qa)\)
  gives \(c_1W_*\beta\le f/(c_3d\theta T)\). The inequality \(ef\le d\)
  bounds this by \(1/(c_3e\theta T)=1/(c_3YT)\). Finally \(Y\ge Y_-\)
  and \(T\ge g_4\) give the claimed bound. All these are the original
  simultaneous field parameters.
\item
  Direct substitution into \(L_s(x)=\prod_{t\ne s}(x-t)/(s-t)\), in node
  order \(-2,-1,0,1,2\), gives \(3/128,-5/32,45/64,15/32,-5/128\). Their
  sum is one, and their 5-adic valuations are \(0,1,1,1,1\). For
  \(W(Z)=\prod_{s=-2}^2(Z-5s)\), the value is
  \(W(5/2)=5^5(45/32)=140625/32\). Since \(140625=9\cdot5^6\) and the
  denominator is prime to five, its valuation is six, at least the
  normal root-polynomial bound \(N\theta=5\).
\item
  The inequality concerns one selected analytic valuation. A zero
  criterion must compare it with an arithmetic upper bound for the
  actual algebraic value, including the full global degree, local
  degrees, scalar denominators and coefficient data. Membership in a
  selected completion neither supplies that bound nor implies equal
  valuations at other primes. Those are separate mathematical assertions
  which require their own proofs.
\end{enumerate}

\textbf{Solution 58.} (a) At odd p, \(cTH^8/a>(4500/29)(11/20)^8>1\), so
the last index with \(cTH^I/a\ge1\) is at least eight. Also \(Q>11/10\),
and \((11/10)^8>2\). In V, \(cTH^6/a>(6750/29)(5/12)^6>1\), so
\(I_1\ge6\); here \(Q>5/4\), and \((5/4)^6>3\). These are exact rational
comparisons, including the original three c bands.

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{1}
\item
  The nodes in increasing order are \(-2,-1,1,2\). Substitution in
  \(L_s(0)=\prod_{t\ne s}(-t)/(s-t)\) gives \(-1/6,2/3,2/3,-1/6\), whose
  sum is one. Their 2-adic valuations are \(-1,1,1,-1\). Thus a deleted
  cardinal can have a negative valuation. The uniform B is two, whereas
  the actual largest loss here is one. With \(W(Z)=\prod_s(Z-2s)\),
  \(W(0)=64\), whose valuation six exceeds the normal bound
  \(4\theta=4\) at \(\theta=1\). This is the exact illustration in
  Figure10.37, rather than a full logarithmic-contradiction parameter
  choice. There is one value condition per node, and no further ordinary
  jet.
\item
  The floor plus one gives \(I_2\ln q\le3D-I_1\ln Q+\ln q\). Add
  \(I_1\ln q\), and use \(I_1\le3D/\ln Q\), to get
  \(I_3\ln q\le3D\ln q/\ln Q+\ln q\). In particular the extra \(\ln q\)
  endpoint remains. The complete denominator is
  \(D_{\rm a}[J+1/(q-1)]\); J can exceed the earlier geometric range.
  Keeping only that old range would discard part of the depth cost. Its
  correct normal upper bound contributes
  \(\bar\lambda[3\ell_q/\mathcal L(Q)+(1+1/(q-1))\ell_q/5]/c_4\).
\item
  The scalar, denominator and torus estimates use the proved larger
  interval \(I_1+1\le J\le I_3\), and both strict comparisons include
  its endpoint. Thus every missing integer zero follows for an
  admissible stage-\(I_3\) family with the prescribed q-deleted zeros.
  Existence of such a family requires the original fractional zero at
  each preceding depth and its phase extraction. A local analytic
  valuation by itself does not prove that fractional zero or give its
  valuations at other primes. Theorem10.149 supplies the fractional
  zeros, and Corollary10.150 combines the actual phase construction with
  the integer closure.
\end{enumerate}

\textbf{Solution 59.} (a) The supports are respectively \(\{(1,1)\}\),
\(\{(0,0),(1,0)\}\) and \(\{(0,0),(1,0),(0,1)\}\) in \(\mathbb F_2^2\).
Their span dimensions are one, one and two. Theorem10.147 gives degrees
two, two and four. The constant term does not change a span.

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{1}
\item
  Put \(W=U_1+U_2\). Then \(W^2=221+2U_1U_2\) and \((U_1U_2)^2=10878\),
  so \((W^2-221)^2=43512\). Since \(W=2-V\), expansion gives the
  polynomial in Theorem10.147. It has degree four, already proved to be
  the degree of V, so it is its minimal polynomial. Its constant term is
  the product of the four sign conjugates and is \(3577=49\cdot73\).
\item
  Both bases are congruent to one modulo73. Lesson9's exponential root
  law gives their roots congruent to one. Modulo \(73^2\), their values
  are \(U_1\equiv2702\) and \(U_2\equiv74\): squaring gives 74 and147
  respectively, and each representative is congruent to one modulo73.
  Uniqueness at this precision follows by subtracting squares; the sum
  of two roots congruent to one is a unit at73. Thus
  \(V\equiv2-2702-74=-2774=-38\cdot73\pmod{73^2}\), of exact valuation
  one. In the other three sign embeddings its residues are
  respectively2,2 and4, hence its valuations are zero. All four
  completions are \(\mathbb Q_{73}\), since both signs of both roots lie
  in that field. Their embeddings of the global value are nevertheless
  different. The sum of the four valuations is one, as also follows from
  the norm; replacing it by four times the selected valuation would be
  incorrect. This example asserts no prescribed interval of auxiliary
  integer zeros.
\end{enumerate}

\textbf{Solution 60.} (a) Here \(\xi U=24024/25\). The radii are
\(R_{\rm s}=\lfloor7983/50\rfloor=159\),
\(R_{\rm l}=\lfloor23999/50\rfloor=479\), with node counts319 and959. We
have \(3R_{\rm s}=477\le479\). Their normal masses are respectively319
and959; they exceed \(\xi U/3-2=7958/25\) and \(\xi U-2=23974/25\). They
are also at most \(\xi U/3\) and \(\xi U\). These data check the floor
mechanism, not a complete original logarithmic-contradiction parameter
choice.

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{1}
\item
  Full-node interpolation with multiplicity one requires only the
  prepared zero at the target order itself. Its inverse Taylor
  polynomial has degree zero, so no additional ordinary jet is consumed.
  Both integer values lie in K by the retained phase. At a fractional
  node, the value lies in the support field of Theorem10.56, whose
  selected completion has the same e,f but whose global degree can be
  \(q^r\). Theorem10.147 can improve the factor for a particular value,
  but Theorem10.149 keeps its full valid upper factor. The normal
  addition \(G_u\) is bounded by \(\Gamma\) and is added after that
  product-formula arithmetic bracket. Multiplying \(\Gamma\) again by
  \(q^r\) would charge a different quantity.
\item
  From the definition of \(m_r\), multiply \(cL_rH_r^{m_r}/(r+1)\ge1\)
  by \(a_*^2(13/2)^2H_{32}^{d_0}>1\). Increasing H and the exact
  two-rank ratio give \(cL_{r+2}H_{r+2}^{m_r+d_0}/(r+3)>1\), hence
  \(m_{r+2}\ge m_r+d_0\). Increasing Q then gives
  \(\mu_{r+2}\le\mu_rq^2/Q_r^{d_0}<\mu_r\). Thus even ranks are bounded
  by32 and odd ranks by33; the remaining cost terms decrease on those
  same sequences. Keeping \(m_{32}\) fixed would instead leave a growing
  numerator \(q^r\), so would not be the proved enclosure. A finite list
  without the recurrence would leave all later ranks unchecked. The
  coefficient, denominator and phase hypotheses are still needed to
  construct each actual successor; Corollary10.150 supplies that
  induction from144,149,51 and146.
\end{enumerate}

\textbf{Solution 61.} (a) Multiplication of
\(U_r=q^{r+1}S\mathcal D/(et)\) by \(e\), followed by substitution for
\(S\), gives the prefactor \(c_3q^{r+2}(r+1)d(\mathsf h+x)/t^2\). The
formula for \(\mathcal D\) then gives
\(F_0(qH_*)^r r^r(r+1)^{r+1}d^{r+2}\ell_d M_r A_r\mathsf h/(r!q^ut^2)\)
times the weight product, because
\(\gamma(\mathsf h+x)(A_r+x)/q^\nu=\mathsf hA_r\). The profile
multiplies this by \((\mathrm eH_*(n+1)d)^{n-r}M_n/M_r\) and by the
original height product. Thus \(M_r\) disappears, the powers of \(H_*\)
become \(H_*^n\), and the powers of \(d\) become \(d^{n+2}\). Dividing
by the rank-n expression leaves exactly
\((\mathrm e/q)^{n-r}(r^r/r!)/(n^n/n!)\) times \(((r+1)/(n+1))^{r+1}\).
Consecutive ratios of \(k^k/k!\) are at least two, and \(\mathrm e<2q\);
all three factors together are at most one. This algebraic rank-ratio
estimate includes rank one, although its actual valuation estimate is
established separately.

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{1}
\item
  The rank-one definition is \(\delta=(p^f-1)/g\), so the first entry of
  its maximum is \(p^fg/((p^f-1)t)>g/t\). The rank-one power identity
  kills precisely the q-primary torsion; its nonzero exponent numerator
  divides every original coefficient. The height lower bound makes the
  stated target greater than \(800d\mathsf h/t\), so its summand
  \(2d\mathsf h/t\) is less than one four-hundredth of the target. After
  inserting the profile in its other summand and using \(eq^u\le2d\),
  the ratio is less than \(8/(c\mathfrak a(n+1)^3d\ell_d)\). At the
  universal minima this is at most \(8/37800<1/4000\), since
  \(32000<37800\). The two terms sum to less than \(11/4000\) of the
  target, hence prove the strict bound with ample room.
\item
  The exact III.2 values are \(L_2=1599907960287/85000000\) and
  \(G_2=1467/25\). Put \(\Delta_2=3/(2L_2)\) and multiply
  \((1+10^{-26})(1+\Delta_2)^2(2+1/G_2)\) by
  \(2.6\cdot1.432\cdot3.26\cdot18.2757\cdot4\). Subtracting from 1790
  gives a positive rational greater than \(4059/10^6\), as
  multiplication of its positive denominators verifies. The exact
  calculation file preserves that fraction. The ratio of successive
  \(F_r\) is greater than four, so \(\Delta_{r+1}<\Delta_r/14\); \(G_r\)
  increases and \(\Delta_2<1\). Hence the enclosing coefficient
  decreases for every later rank, as proved in Lemma 10.151. The
  stronger lower bound for \(g_1\) belongs to the actual III.2 parameter
  formula; discarding it needlessly enlarges the endpoint in the case
  with the smallest gap.
\end{enumerate}

\textbf{Solution 62.} (a) On \([j,j+1]\), the chord of the concave
logarithm lies below its graph. Sum the chord integrals from 1 to n:
their value is \(\ln(n!)-\tfrac12\ln n\), whereas the integral of the
logarithm is \(n\ln n-n+1\). Exponentiation proves the asserted
factorial bound. Also \((n+2)\ln(1+1/n)>1\) and \((n-1)\ln(1-1/n)>-1\),
so the product of their two exponentials exceeds one. In the lower bound
for \(W_B\), these facts leave
\(W_B/(d\exp(a_0n))>(7899c/(7900\rho\mathrm e))A_n\). For \(n\ge512\),
\(\exp6<513\) makes \(A_n>10\). The seven exact exponential upper
intervals now establish its strict comparison with \(\exp a_1\). The
3570 finite checks cover all seven rows and all integer ranks 2--511.
The analytic argument, rather than a finite extrapolation, supplies
every remaining rank.

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{1}
\item
  Put \(\mathcal T=C_1^*\Omega H\). In the first branch, multiplication
  of the definition of \(W_B\) by \(nd\ln B/t\) bounds
  \((d/t)nB\max h(a_j)\) by \((7899/7900)\mathcal T\), using
  \(\ln B\le H\). The strict lower target bound makes the remaining
  \((d/t)\ln2\) less than \(\mathcal T/7900\). Their sum is strictly
  below the target. In the other branch, \(B>W_B\ln W_B\) gives
  \(\ln B>\ln W_B+\ln\ln W_B\); Lemma 10.153 therefore bounds
  \(G_1(n,d)\) by \((n+1)\ln B\). Choose the last coefficient to have
  minimum p-adic valuation, as required by Theorem 10.152. The singleton
  height bound yields \(R_{\max}/(dW(d))\le\rho\mathrm e d^2B\). Since
  \(\ln\rho+1<6<n\ln B-2\ln d\), this logarithmic entry is also less
  than \((n+1)\ln B\). Finally \(\ln B_\dagger\le\ln B\) and
  \((n+1)t\le(n+1)H\). The complete maximum \(\mathsf h\), with the new
  coefficient ordering, is thus at most \((n+1)H\).
\item
  The raw residue of four at three is one, so \(g=1\), \(\delta=2\). The
  rational q-primary roots are \(\{1,-1\}\), giving \(q^u=2\). In
  \(M_1\) the entries are \(3/(2\ln3)\) and \(\mathrm e\ln3\); the
  latter is larger. We have \(\Omega=\ln4\), \(H=k\ln3\), and
\end{enumerate}

\[
C_1^*=\frac{4\cdot636\cdot14\,\mathrm e(4+\ln2)}{\ln3}.
\]

Thus \(C_1^*\Omega H/2100\) is exactly the right side of the worked
example. The deep-unit power law gives \(v_3(4^{3^k}-1)=1+k\). Its bound
exceeds \(128k>1+k\). In the general rank-one proof the logarithmic term
is less than \(\mathcal T/2688\), and the residue term is less than
\(\mathcal T/10000\). Their exact sum is \(793/1680000\), and
\(793\cdot2100=1665300<1680000\); hence the sum is less than \(1/2100\).
The raw residue index two in this example belongs to the rank-one
theorem; replacing it by the saturated-basis residue index would change
the parameter being proved.

\textbf{Solution 63.} (a) The prime-valuation vectors are
\((1,0),(0,1),(2,0)\). Thus the selected basis is \((2,3)\), and at five
its saturated residue group, including \(-1\), has order four, so
\(\delta_\beta=1\). The III.2 value is \(\varkappa=12\). Since
\(\ln5<2\), \(L_3=3/(12\cdot8)=1/32\) and \(L_2=2/(12\cdot7)=1/42\).
Every actual height exceeds both floors. Consequently
\(\Omega_3=(\ln2)(\ln3)(\ln4)\) and \(\Omega'=(\ln2)(\ln3)\). The
relation \(4\cdot2^{-2}=1\) has coefficient at four equal to one. Its
transfer gives \(b'_1=1-2(-2)=5\), \(b'_2=1\), and \(\Xi=2^5\cdot3=96\).
Since \(95=5\cdot19\), its valuation is one. Here B=3 and
\(B/\ln B<3<8<\mathcal W_3\), so the local-height branch applies.

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{1}
\item
  Write j=r-v+1, s=n-r-1. The circuit bound and height-product bound
  give the exact envelope for \(K/\Omega_n\) displayed in the proof. At
  x=n, normalize by \(B dH_n(a_x)/8\). The two appearances of
  \(h(a_x)/H_n(a_x)\) are at most one. Insert \(L_n^{-s}\), the second
  entry of the maximum in \(C_n\), and \(B>\mathcal W_n\ln B\). The
  degree identity \(r+3+s=n+2\) cancels all powers of d.~The remaining
  rank expression is the terminal normalized envelope. If the
  largest-height base is selected and x\textless n, cancel its height
  and use the bound at length j-1; the envelope gains \(1/\mathrm e\).
  At j=1 that height product is empty, and the root count of Theorem
  10.17 supplies its bound. If the largest-height base is unselected,
  cancel its modified height, use s-1 floors, and gain
  \(1/(\mathfrak a\mathrm e)\). In all cases the exponential base is
  less than \((1+5/n)/2\), its power is at most n-2, and the result is
  strictly below \(107648/132237<1\).
\item
  At n at least sixteen put y=1/n.~The lower quadratic bounds for
  \(\ln(1+y)\), \(-\ln(1-y)\), \(\ln(1+4y)\), and the upper cubic bound
  for \(\ln(1+5y)\), after multiplication by their positive
  coefficients, sum to \(1+(5/2)y-(119/3)y^2+(125/3)y^3\). Since
  \(5/2-119/(3\cdot16)=1/48\), this exceeds \(1+y/48\). The fourteen
  exact checks at 2--15 give \(J_n/(n+5)>11/4>\mathrm e\), completing
  every integer n.~For the final absorption,
  \(\mathfrak a/\varkappa\ge13/17\), \(\mathrm e>8/3\),
  \(\varkappa\ge9\) and n+5 at least seven give \(104/51-1/672>2\). This
  is the inequality after subtracting the full additive
  coefficient-growth term \(z/(4\mathrm e X_n)\); it is the required
  parameter comparison, not a test at one rank.
\item
  The first discarded base is in the span of earlier retained bases, so
  removing it changes no later span or greedy rank choice. An initial
  root has zero rank and is discarded for the same reason. The ordered
  retained bases therefore remain identical, even as the complementary
  height floor changes; their actual residue index is unchanged. If a
  power of \(\Xi\ne1\) becomes one, the torsion part of Theorem 10.21
  bounds its valuation by \((d/t)\ln2\). The same direct bound applies
  when the full original list has rank zero.
\end{enumerate}

\textbf{Solution 64.} Prime valuations show the list has rank one and
first non-torsion base four. Its residue has order two at five, so the
raw index is \((5-1)/2=2\). Here \(\varkappa=12\), \(X_2=2\),
\(L_2=1/42\), and both heights exceed the floor. Thus
\(\Omega_2=(\ln4)(\ln16)\). The relation \(16\cdot4^{-2}=1\) gives P=1
and \(M=2\cdot5^k\). Since \(v_5(16-1)=1\), Lemma 9.5 gives
\(v_5(16^{5^k}-1)=1+k\).

For \(U\), the powers of d cancel as
\(d^{n+2}d^{-2}d^{-(n-1)}d^{-1}=1\). The first factor is in \(C_n\), the
second is the singleton height bound, the third is the complementary
floors, and the fourth comes from \(U=t\mathcal
T_n/d\). The factor \(\ell_d\) cancels between the constant and height
bound. The second maximum entry and the height bound leave
\(\mathrm e^{n-1}\), while the root factor becomes \(w_K/q^u\). This
gives \(F_nA_nHR\). For the residue-height contribution, the first
maximum entry contributes a strict factor g, and \(t^2\le A_nH\),
\(eq^u\le2d\) give exactly \(E_n/(d\ell_dR)\).

At two, \(E_2=14\varkappa/(81c\mathfrak a^2)\). The seven last-column
bounds use the actual degree restrictions; for example I.2 gives
\(1/40068<1/40000\), and II gives \(1/63630<1/40000\). All factors of
the original residue order were retained. At every \(n\ge6\), the other
power exceeds one, the logarithmic power exceeds \(\mathrm e\), and
\((n+1)/(n+6)\ge7/12\), giving \(1456/459>3\). This proves the claim for
infinitely many ranks, rather than extrapolating the four finite checks.

The maximum in s is needed because \(D_0\ge1\). In I.2 the unmodified
coefficient \(2\rho\mathrm e/(7\varkappa)=17\mathrm
e/63\) is less than one, so it cannot by itself give the asserted bound
on \(D_0\). Taking its maximum with one keeps the logarithmic comparison
valid in every case.

For the logarithmic comparisons, the positive exponential series through
degree five is \(163/60>27/10\). To upper-bound a positive exponential
at x, sum through degree 32 and bound the tail by its first term divided
by \(1-x/34\). Lower bounds need only the positive partial sum. The
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_rank_one_refinement_enclosures.py}{exact
calculation} checks \(\exp(8/5)<5\), \(\exp(7/10)>2\), \(\exp3>18\),
\(\exp4>45\), \(\exp(21/10)>8\), \(\exp(14/5)>16\),
\(\exp(5/3)>319/63\), and \(\exp5>58\), with strictly positive rational
differences. This supplies all numerical logarithmic comparisons used in
the proof.

In the infinite range \(n\ge3\), the coefficient of \(\ln d\) in
\(\alpha_nA_nH\) is larger than \(9n/8\). The remaining comparison at
H=1 has difference at least \(181n/90-5\ge31/30>0\); its derivative with
respect to H is positive. Hence it holds for every H at least one.
Finally, \(\ln R\le R-1\) and \(\alpha_nA_nH>2\) absorb the full product
of modified heights. With no torsion bases, P is just the product of the
exact relation coefficients. A torsion complement adds \(w_K\), but
removes at least one non-torsion coefficient factor. Its orders all
divide \(w_K\), so the resulting power eliminates every torsion base
simultaneously. The torsion case \(\Xi^P=1\) is handled by Theorem 10.21
and Lemma 10.155.

\textbf{Solution 65.} Coprimality of 25 and seven gives \(g_0=3^N\); the
coprime exponents at five and seven are \((2,-1)\). Their difference is
18, so the valuations are two for the coprime parts and N+2 for the
original integers. At three the residue of five is minus one and seven
is one; the group including minus one has order two, so its index is
one. For m=n=2 the rank factor in \(C_2\) is
\(2^2\cdot3^4/(2!\cdot2)=81\), and the remaining field factors are
exactly those in the displayed constant. The common factor cannot be
omitted: N+2 grows linearly in N, whereas the remaining height parameter
grows only like \(\ln N\). For example \(\ln N/N\to0\) follows by
integrating \(1/x\le1/\sqrt x\) for \(x\ge1\).

At two, the residue polynomial \(X^2+X+1\) is irreducible over
\(\mathbb F_2\), so degree two forces \(e=1,f=2\). The primitive cube
root's residue generates all three units in the residue field. Thus the
rank-two saturated index is one, even though both rational odd prime
bases reduce to one. The sign base has coefficient one, the others have
coefficients two and minus one, and the list length is three. Since
\(2\ln2<3\), \(L_3=3/[34(3+5)2]=3/544\). The sign has height zero, so
its modified height is this floor. Multiplication by \(\ln5\ln7\) gives
the weighted product. The actual difference is 32, of valuation five.

If n=0, the coprime magnitudes are powers of p.~If one has positive
p-valuation, the other has zero valuation, so their signed difference is
a unit. If both have zero valuation, both magnitudes are one;
distinctness makes their signs opposite and their difference is
\(\pm2\). This proves the stated exact alternatives, including the
two-adic torsion case. Finally apply the decreasing function
\(z\mapsto p^{-z}\) to the strict excess-valuation bound. Its value at
\(K\max\{\ln B_M,t\}\) is \(\min\{B_M^{-K\ln p},p^{-Kt}\}\), proving the
corollary.

\textbf{Solution 66.} In III.1 the strict lower bound for \(U_2/H\) is
\(8\cdot495\cdot7\cdot(27/10)\cdot(16/3)/58
=199584/29\). With \(\mu=23/16\), the logarithm share is
\(667/3193344\), and the original residue-order height share is less
than \(1/37422\). Their sum is \(2257/9580032\). Exact subtraction gives
\(1/4000-2257/9580032=17251/1197504000>0\). The factor \((p-1)/(p-2)>1\)
increases the actual constant, so this covers all primes in III.1.

For the rational example, \(d=e=f=1\), \(t=\ln3\), \(q^u=2\), \(g=1\),
\(\delta_\beta=2\), \(c^{(2)}=648\), \(a^{(2)}=7\), and \(h(4)=\ln4\).
Since \(\ln3<11/10\), \(A_1=4+\ln2\). Take \(B=3^k\), so \(H=k\ln3\).
The maximum selects \(\mathrm e(\ln3)^2\), because \(\ln3>1\) and
\(\mathrm e>8/3\). Substitution cancels all three factors of \(\ln3\),
giving the displayed coefficient of \(k\ln4\). Lemma 9.5 gives
\(v_3(4^{3^k}-1)=k+1\). The bound exceeds \(126k\), and
\(k+1\le2k<126k\) for every \(k\ge1\), verifying its strict inequality.

\textbf{Solution 67.} Here \(d=e=f=1\), \(t=\ln5\), \(P=1\),
\(\varkappa_2=25\), and \(X_2=2\), since \(1<\ln5<2\). Thus
\(L_2^{(2)}=2/(25\ln5)\). Both \(\ln4\) and \(\ln16\) exceed this floor.
The selected base is four, with residue order two and index
\((5-1)/2=2\). The exact relation is \(16=4^2\), giving
\(\Xi=4^{2\cdot5^k}\); Lemma 9.5 proves its valuation \(1+k\). The
minimizing weighted product is \((\ln4)(\ln16)\).

The constant factor defining \(\alpha_n\) is independent of n, so
\(\alpha_n>3\cdot3^{n-2}\). Induction proves \(2\cdot3^{n-2}\ge n\):
equality holds at two, and \(3n\ge n+1\) propagates the assertion. Hence
\(\alpha_n>3n/2\) for every \(n\ge2\). Theorem 10.163 bounds
\(\ln(2|M|)/U\) strictly by \(1/4000-1/100000=3/12500\), and Lemma
10.162 bounds \(2eg h(\beta)/U\) strictly by \(1/100000\). Their sum is
strictly below \(1/4000\). Multiplication by \(dU/t=\mathcal T_n^{(2)}\)
gives the asserted valuation bound. The original residue order and the
second height floor have both remained in the proof.

\textbf{Solution 68.} Both bases are positive 5-adic units and are
congruent to one modulo five, so \(g=1\). Their difference from one has
valuation \(m\), so \(E=m>1/4\). They are coprime positive integers
larger than one: subtracting twice the first from the second gives
\(-1\). Any multiplicative relation is trivial by comparing a prime
dividing each base. Their coefficients give \(u=0\), with the second
coefficient a unit. Rational heights are their logarithms, both larger
than \(m\ln5\). Therefore \[
B' m\ln5
=\frac{m\ln5}{\ln(1+2\cdot5^m)}
+\frac{m\ln5}{\ln(1+5^m)}<2.
\] Thus the logarithmic cutoff is less than \(\ln2+2/5<11/10\), while
\(4m\ln5>5\). It follows that \(H_6=6m\ln5\) and \(H_4=4m\ln5\). The
difference is \(-5^m\), so its valuation is \(m\). The two coefficient
factors are \(36.1\cdot6^2=1299.6\) and \(53.8\cdot4^2=860.8\),
respectively. The latter bound is smaller. Both estimates still exceed
\(m\), since the product of the two logarithms exceeds \(m^2(\ln5)^2\).

After division of both nonzero coefficients by their common p-part, at
least one is a unit. If both bases have precision \(E\), swapping the
pair if necessary preserves the precision and makes the second
coefficient a unit. If only the first base has that precision, the same
swap may assign the analytic parameter \(E\) to a logarithm with smaller
valuation. The determinant argument then loses its stated Taylor-order
gain. The precision condition must be checked for the newly ordered
first base; the swap itself does not establish it.

\textbf{Solution 69.} Choose \(\eta_1=-1,\eta_2=1\), so
\(\beta_1=-3,\beta_2=5\) are both one modulo four and have precision
two. They are independent by the prime exponents of three and five. The
remaining sign is \((-1)^{b_1}\). If \(b_1\) is odd,
\(3^{b_1}\equiv3\pmod4\) and \(5^{b_2}\equiv1\pmod4\), including
negative powers because inverses preserve these unit classes. The
difference is two modulo four and its valuation is exactly one. In
particular \(3-25=-22\) has valuation one. For even \(b_1\), the sign is
one and the deep-unit estimate applies. The example \(3^2-5^2=-16\) has
valuation four.

If a logarithmic height is prescribed to be \(\ln2\), assigning \(E=1\)
directly would violate the strict exponential-domain condition \(E>1\)
in Theorem 10.165. Add \(\varepsilon>0\) to both prescribed heights and
use
\(E_\varepsilon=\min\{2,(\ln A_1+\varepsilon)/\ln2,(\ln A_2+\varepsilon)/\ln2\}>1\).
Both sign-normalized bases have precision at least two. The uniform
coefficient in the proof is independent of \(\varepsilon\), and the
height product and coefficient cutoff are continuous as \(\varepsilon\)
decreases to zero. Passing to the limit therefore proves the stated
closed-boundary estimate.

\textbf{Solution 70.} The difference of the two normalized determinant
budgets is \(110/35-96/35=14/35=2/5\). The determinant coefficient is
\((1617+14)/6=1631/6\), whose gap below \(283\) is \(67/6>0\). For
\(a=8,p=3\), the candidate two is dependent on \(a\), and three is
excluded by the unit condition, so take \(c=5\). The rational bases
\(8/5,1/5\) are independent by comparing the prime exponents of two and
five. The identity \[
(8/5)^2-(1/5)^2=63/25
\] has valuation \(v_3(63)=2\); its denominator is a unit. The original
difference \(8^2-1=63\) also has valuation two. The order of eight
modulo three is two. If \(n\) is odd, \(8^n-1\equiv1\pmod3\), so its
valuation is zero. If \(n\) is even, the deep power law gives
\(v_3(8^n-1)=v_3(8^2-1)+v_3(n/2)=2+v_3(n)\).

\textbf{Solution 71.} Every first coordinate \(r=0,\ldots,7\) appears
four times. Pairing the increasing row labels \(0,\ldots,7\), also each
repeated four times, with the increasing coordinates maximizes the
weighted sum. Indeed swapping an inverted pair changes it by the
nonnegative product of the two coordinate differences. Thus the exact
centered maximum is \[
4\sum_{r=0}^{7}(r-7/2)r
=4(140-98)=168.
\] The congruence gives \(g_0=1,g_1=2\), so \(R'=8,S'=4\). Lemma 6.4
yields \[
\frac{32\cdot8\cdot8}{8}
-\frac{32^2\cdot8}{24\cdot4}
=\frac{512}{3}.
\] The relaxed residue-class bound is \[
G_1=\frac{32\cdot8\cdot9}{8}
-\frac{2\cdot32^2\cdot8}{24\cdot9}
=\frac{5728}{27}.
\] It is less than the no-capacity bound \(32\cdot8\cdot7/8=224\),
though relaxing the compressed parameters loses some of the stronger
bound \(512/3\).

For the bases \(2,3\), the global field and the local field are
\(\mathbb Q\) and \(\mathbb Q_5\), so \(d=e=f=D=1\). The residue of two
has order four and already generates the whole residue group, so
\(g=4\). Independence follows by comparing exponents at the primes two
and three. The general proof keeps \(e\) throughout the analytic and
one-place height estimates and uses only \(ef\le d\). It handles
whichever coefficient becomes a unit after removing their common p-part
by interchanging the bases inside the analytic argument; both principal
factors have the same universally guaranteed precision \(p^t/e\). Hence
neither unramifiedness nor an asymmetric coefficient condition is
imposed by Theorem 10.170.

\textbf{Solution 72.} (a) The first maximum becomes the second on
multiplication by \(t^{n+1}\). The factorial quotient is n, so the
constant quotient is exactly
\((c^{(2)}/c)(a^{(2)}\mathrm e/\mathfrak a)^n(Pt/n)^{n-1}\). With
\(Pt/n\ge1\), the least power outside III is the square of
\(\mathrm e/2\); use \(\mathrm e>27/10\). In III write
\(x=(p-1)/(p-2)\), with \(1<x\le4/3\). The expression
\((c_*/c)\mathrm e^n x^{1-n}\) increases with n for each x, and its n=2
value decreases with x. Using x=4/3 and \(\mathrm e>27/10\) gives the
two III fractions. Subtracting \(3/2\) from the seven fractions gives,
in order,

\[
\frac{80817}{62600},\quad\frac{3783}{10600},\quad
\frac{20001}{20200},\quad\frac{411}{47840},\quad
\frac{144159}{716000},\quad\frac{77361}{536000},\quad
\frac{86187}{41200}.
\tag{10.657}
\]

Cross-multiplication shows that the smallest is \(411/47840\). The
coefficient parameters are different; comparing \(C_1,C_2\) alone would
not compare the full bounds. Lemma 10.172 supplies
\(\mathsf h\le3\mathsf h_2/2\), which is absorbed by the strict constant
quotient. At three with e=1, the inequalities are \(2\le2<6\), giving
\(\kappa=1,P=3\). At an unramified prime \(p\ge5\), they are
\((p-1)/p\le2<p-1\), giving \(\kappa=0,P=1\).

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{1}
\tightlist
\item
  The coefficients have valuations k and zero, so the last coefficient
  is correctly ordered. The prime valuations of two and five prove
  independence and that a rational element of their 2-primary saturation
  has integral prime exponents. Including minus one, the residues
  generate \(\mathbb F_3^*\); thus \(\delta(2,5)=1,q^u=2,d=e=f=1\). The
  heights are \(\ln2,\ln5\). Substitution of W(1) shows
\end{enumerate}

\[
\frac{1/h_{\min}+1/h(5)}{W(1)}
=\frac{(\ln3)(\ln10)}{(\ln5)(\ln6)}<1,
\tag{10.658}
\]

because cancelling the common \((\ln3)(\ln5)\) term reduces the
comparison to \(\ln3<\ln5\). Hence X=1. For k at least one,
\(3^k\equiv3\pmod6\), and the powers of two modulo nine have period six.
Therefore \(5\cdot2^{3^k}-1\equiv3\pmod9\), of exact valuation one.
Since n=2 is less than \(3\ln3\), Theorem 10.173 applies with the I.2
row, \(c^{(2)}=648,a^{(2)}=7\), the original residue index one and the
same field. Theorem 10.175 applies as well, with \(B_n=1,B=3^k\).

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{2}
\tightlist
\item
  For s positive, the derivative of \(u\exp(-su)\) is
  \(\exp(-su)(1-su)\). Its maximum for u nonnegative is
  \(1/(\mathrm e s)\), at u=1/s. Taking u=ln y gives the claimed bound.
  The partial exponential sum through degree six is \(1957/720\), whose
  difference from \(125/46\) is \(11/16560>0\); therefore
  \(8\eta\mathrm e>1\). The scalar equality in the constant branch can
  occur when \(A=\exp(H_0/(1+\eta))\), so a universal strict scalar
  statement is unnecessary. The strict inequality supplied by Theorem
  10.152 persists after composing it with this weak scalar estimate.
\end{enumerate}

In the plotted model \(\Phi=523/500\) and
\(Z=\max\{8\Phi,\exp(1000/523)\}\). The branches are
\(\Phi\ln(Z\exp20/z)\) and z, and the scalar bound is 20. The parameter
condition is exactly \(0<z\le r=\exp20\), since \(0<\varepsilon\le1\).
Only a bounded subinterval of that positive interval is drawn; the
proof, rather than the plot, supplies every allowable value.

\textbf{Solution 73.} (a) Modulo 27 the monomial is
\(1+9(\lambda_1+2\lambda_2)\). For \(\lambda_1\) congruent to zero, one
or two modulo three, the allowed \(\lambda_2\) in the zero fibre are
respectively congruent to zero, one or two. Thus that fibre consists of
the four pairs in each of \(\{0,3\}^2,\{1,4\}^2,\{2,5\}^2\). More
generally, the fibre with image b is exactly
\(S_b=\{(x,y):0\le x,y\le5,\ y\equiv x-b\pmod3\}\) for b=0,1,2. The
other two fibres are obtained by assigning the indicated residue to
\(\lambda_1+2\lambda_2\); for each fixed first coordinate there are
exactly two possible second coordinates in each fibre. Every fibre
therefore has twelve points. Its even-even zero part is
\(\{(0,0),(2,2),(4,4)\}\), which contracts to \(\{(0,0),(1,1),(2,2)\}\).
The new monomials are \(190^a\), with \(a=0,1,2\), and are still one
modulo 27.

The finite quotient has three possible monomial images, independent of
E. Selecting one image therefore retains E in full. Fixing an auxiliary
index as part of the equivalence relation would multiply the possible
number of classes by \(|E|\) and would prevent a selected class from
containing \(E\times\{\lambda^0\}\) when \(|E|>1\). That is a different
selection and has a different counting cost.

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{1}
\tightlist
\item
  For \(x_0=k+v>1\), comparison of
  \(\chi(k+1)/\chi(k)=\exp(-\rho)(A+k+1)/(A+k)\) with one gives
  \(v\ge1/\rho-1/(\exp\rho-1)\) for choosing k+1; at equality either
  choice has the same value. The continuous maximum is
  \(\rho^{-1}\exp(-1+\rho A)\), so \(\chi(m)>\chi(x_0)/\varsigma_2\)
  gives
\end{enumerate}

\[
\frac{p^{fm}}{(A+m)^r}
<\varsigma_2^r\rho^r\exp(r-\rho Ar)
=\varsigma_2^r(\mathrm e t/r)^r p^{-fA}.
\tag{10.678}
\]

The two requested positive gaps are exactly

\[
\frac{13701}{10000}-\frac{1370098893207}{10^{12}}
=\frac{1106793}{10^{12}},\qquad
\frac{47314037827}{50000000000}-\frac{1250}{1321}
=\frac{1843969467}{66050000000000}.
\tag{10.679}
\]

Finite endpoint calculations certify these numbers only. The derivative
comparisons for H, z and F, the decreasing forced-ratio function, and
the increase/decrease of the two rounding alternatives prove that these
endpoints cover every allowed A and \(\rho\).

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{2}
\tightlist
\item
  We have \(\Delta(z/2+1;2)=(z^2+10z+24)/8\) and \(V(2)=2\). After
  squaring, differentiating and multiplying by two,
\end{enumerate}

\[
P(z)=\frac{(z^2+10z+24)(z+5)}8
=\frac{z^3+15z^2+74z+120}{8}.
\tag{10.680}
\]

The coefficient valuations, in increasing powers of z, are
\((1,0,1,0)\), so all coefficients are integral. The uniform choice uses
D=4 and \(C=D/(p-1)=2\). At N=2 and \(\beta=0\) it gives a normalizing
scalar of valuation \(4(2+1/2)=10\). Using the actual degree three and
coefficient bound C=0 instead, Theorem 10.178 gives the sharper termwise
sufficient valuation six. For every exponent in the zero fibre, the
logarithmic frequency has valuation at least three, so both applications
have \(2<3-1/(3-1)\) as required. Since every exponential equals one at
zero and the fibre has twelve terms, \(f(0)=12P(0)=180\), with valuation
two. The normalization concerns a single p-adic norm. A global height or
product-formula estimate continues to use the actual original
coefficient field and its degree; the choice of the local scalar does
not replace either of them.

\textbf{Solution 74.} (a) The three differences are
\(9,2781/500=27\cdot103/500\), and \(\Xi-1=3^8\), because
\(10^3(3281/500)=6562=1+3^8\). Their valuations are therefore two, three
and eight. Modulo 27 the monomial is \(1+9\lambda_1\); the three classes
are distinguished by \(\lambda_1\pmod3\), with twelve points each. The
zero class is \(S_0=\{0,3\}\times\{0,\ldots,5\}\). Take the reference
\((0,0)\). With \(z_i=\log a_i\), the original and projected frequencies
are

\[
\gamma_\lambda=\lambda_1z_1+\lambda_2z_2,\qquad
\epsilon_\lambda=-\lambda_2\log\Xi,\qquad
\gamma_\lambda+\epsilon_\lambda
 =(\lambda_1-3\lambda_2)z_1.
\tag{10.697}
\]

Every quotient frequency has valuation at least three. This lower bound
is attained at \((3,0)\), since \(v_3(3\log10)=3\). The perturbations
have valuation at least eight. Here \(j=2,m=1,e=1\), so \(2<3-1/2\) and
\(8>2+1/2\) are both strict. These are the combined-frequency
conditions. An assumption that each individual slope had valuation
greater than \(5/2\) would fail for \(z_1\); Theorems 10.180 and 10.182
do not make that assumption. The circled pair in the figure has
difference \((3,1)\): its original quotient is \(\Xi\), and its
projected frequency is zero.

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{1}
\tightlist
\item
  There are four polynomial indices at each of twelve exponents, hence
  48 columns. The six indices
  \((t_0,t_1)=(0,0),(0,1),(0,2),(1,0),(1,1),(2,0)\) at each of three
  nodes give 18 rows. Every row entry is the rational number
\end{enumerate}

\[
\frac{2^{t_0}}{t_0!}
 \left[\frac{d^{t_0}}{dy^{t_0}}
  \left(\frac{(y+a+1)(y+a+2)}2\right)^\ell\right]_{y=s}
 \Delta(\lambda_1-3\lambda_2;t_1)
 \left(10^{\lambda_1}(3281/500)^{\lambda_2}\right)^s.
\tag{10.698}
\]

Here \(a=0,1\), \(\ell=1,2\), \(s=-1,0,1\), and \(\lambda\in S_0\). Thus
the row field is \(\mathbb Q\), and its discriminant cost is zero.
Elimination of an 18-by-48 rational matrix gives a nonzero rational
kernel because there can be at most 18 pivot columns. Clearing
denominators and then dividing the common integer gcd gives a primitive
integral vector. Its coefficient minimum at three is zero. The exact
program builds this full matrix, produces one such vector, and
substitutes it into all eighteen original rational equations; every sum
is exactly zero. The row factor in (10.174) is \(18/(48-18)=3/5\). This
uses the actual finite counts. The volume-only bound in Proposition
10.183 is smaller: its column lower bound is \(4\cdot25/3=100/3\), which
still exceeds \((11/6)18=33\).

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{2}
\tightlist
\item
  Distinct nodes have nonzero differences one or two, so B is zero. Also
  \(V(2)=2\) and \(b_n=1\), so \(C_0=M_0=0\). The uniform degree bound
  is \(D=4\), and the primitive coefficient vector has \(\beta=0\).
  Therefore
\end{enumerate}

\[
\Gamma=4(2+1/2)=10,\qquad
H+4(2+1/2)=8+10=18=3\cdot3\cdot2,
\qquad v_3(f(x;0)),v_3(\Phi(x;0))\ge8.
\tag{10.699}
\]

The equality of the two sides of the sufficient interpolation budget is
permitted. The normality hypotheses remain the two strict inequalities
verified in (a). No global height comparison or final logarithmic bound
is inferred from this local example.

For the last assertion, let l be the exponential Taylor order. At order
zero its error has valuation at least H. At positive order l, the
binomial expansion in Lemma 10.181 has at least one perturbation in
every term outside the first difference. It gives
\(H+(l-1)L-v_p(l!)\ge H+(l-1)N\), because \(L>N+1/(p-1)\) and
\(v_p(l!)\le(l-1)/(p-1)\). Multiplication by a polynomial of degree D
leaves exponential order at least \(t-D\) in an ordinary divided
derivative of order t. Thus its lower bound is
\(\beta-C+H+\max\{0,(t-D-1)N\}\), as asserted.

\textbf{Solution 75.} Here \(n=r=4,d=e=f=1,P=3,q=2,u=1\),
\(\vartheta=3/2\), and \(J=\delta_r=1\). Unique factorization proves
independence of the four bases and shows that their exponent lattice is
already saturated: a rational power lying in the subgroup has integer
valuations at the four indicated primes. Their heights are respectively
\(\log2,\log5,\log7,\log11\).

The real optimizer of Lemma 10.177 is \(x_0=4/\log3-3/2\), between two
and three. To compare the two adjacent integers, the quotient of the
objective at three by its value at two is \((9/7)^4/3<1\). Thus the
maximizing depth is \(m=2\), and

\[
N=\frac{7/2}{1+10^{-26}/8},\qquad
A_r=4+\log5,\qquad \mathcal R_r=3.
\tag{10.729}
\]

The last identity follows by bounding \((\mathrm e/4)^4(\log3)^5<3\).
All these comparisons can be made with rational intervals, as described
below. The deep units are \(-8,-125,343,-1331\). They belong to
\(1+9\mathbb Z_3\) and, modulo 81, have the form \(1+9c_i\) with

\[
(c_1,c_2,c_3,c_4)\equiv(8,4,2,5)\pmod9.
\tag{10.730}
\]

Consequently the actual unit-quotient image has order nine: the first
coordinate already generates \(\mathbb Z/9\mathbb Z\). The phase map is
trivial. Substitution in Theorem 10.184 gives the following integer
quantities, with every floor fixed by the rational intervals in the
editable calculation:

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}lr@{}}
\toprule\noalign{}
Quantity & Exact value \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
\(\lfloor D_1\rfloor+1\) & 9530854280 \\
\(\lfloor D_2\rfloor+1\) & 4104715517 \\
\(\lfloor D_3\rfloor+1\) & 3394958794 \\
\(\lfloor D_4\rfloor+1\) & 2755034737 \\
k & 150 \\
L & 8105949489 \\
\(D_{\mathrm a}=kL\) & 1215892423350 \\
\(\lfloor S\rfloor\) & 1638 \\
\(\lfloor T\rfloor\) & 353654158504 \\
\end{longtable}
}

Here is an exact way to count the selected columns without traversing
the box. A side containing the integers \(0,\ldots,a_i-1\) contains
\(\lfloor a_i/9\rfloor+1\) integers in residue j when \(j<a_i\bmod9\),
and \(\lfloor a_i/9\rfloor\) otherwise. Start with the residue count
vector \((1,0,\ldots,0)\) and convolve successively with these four
nine-entry vectors, placing residue j of side i at \(c_i j\bmod9\). This
is simply partitioning all sums according to their residue, so it counts
each box point once. The classes \(1,3,6,8\) have the largest size,

\[
|S_0|=40656872678227694342770670245800232365.
\tag{10.731}
\]

All other classes have one fewer point. Select class one and reference
\((8,0,0,0)\), since \(8\cdot8\equiv1\pmod9\). Every one of the \(kL\)
polynomial pairs is retained. The exact column count and the full row
count are therefore

\[
\begin{aligned}
N_c&=49434383446562676057514515798469784690272951722750,\\
M&=2135896855592220094737330490491655575834866435515.
\end{aligned}
\tag{10.732}
\]

Direct integer multiplication gives \(10N_c>27M\), the claimed
\(c_0=27/10\) surplus. Theorem 10.184 then supplies a nonzero
global-integral kernel for every one of these original equations; this
count does not discard any row to improve a height average.

For the first full-node input take
\(\mu=\lfloor T/10\rfloor+1=35365415851\), and all
\(2\lfloor S\rfloor+1=3277\) integer nodes. Their separation bound is
\(B=\lfloor\log_3(3276)\rfloor=7\), since \(3^7\le3276<3^8\). The
polynomial jet loss is \(C_0=v_3V(k)=\lfloor\log_3 150\rfloor=4\), by
the proved least-common-multiple formula for \(V(k)\). For a 3-adic unit
distinguished coefficient this gives \(M_0=\max\{B,C_0\}=7\). With
\(U=32S\mathcal D/t\), the sufficient inequality in Theorem 10.182 is

\[
1+U+D_{\mathrm a}(N+1/2)
\ge3277\mu N+7\mu.
\tag{10.733}
\]

Under the counterhypothesis \(v_3(\Xi-1)\ge U\), the strict normality
hypothesis is also valid. The exact interval calculation bounds left
minus right strictly above \(10^{15}\). It also gives \(1+U>N+1/2\), as
required for the strict normality inequality. Here the node set contains
all its nodes, so \(\eta_E=0\). This verifies this first local input
budget, including the degree, node and jet costs. It does not check
every later stage of a global zero induction or the preservation of the
weight allowance in Lemma 10.187. In particular it does not by itself
prove the full second valuation estimate.

For completeness, the rational enclosures used in this solution have an
elementary derivation. For \(1\le x\le2\) put \(z=(x-1)/(x+1)\), so
\(0\le z\le1/3\). Integrating the geometric series for \(2/(1-z^2)\)
gives

\[
0\le\log x-2\sum_{j=0}^{63}\frac{z^{2j+1}}{2j+1}
\le\frac{2z^{129}}{129(1-z^2)}.
\tag{10.734}
\]

Indeed each remaining denominator is at least 129 and the remaining
powers form a geometric series. Reduce an integer a to \(a=2^k x\), with
\(1\le x<2\), and add k copies of the enclosure for \(\log2\). Similarly
the exponential series and its consecutive-term ratios give

\[
0<\mathrm e-\sum_{j=0}^{64}\frac1{j!}
<\frac{1}{65!}\frac1{1-1/66}.
\tag{10.735}
\]

Every arithmetic operation is rounded outwards to multiples of
\(10^{-36}\); multiplication takes the four endpoint products and
division inverts a positive interval before multiplying. Thus interval
inclusion is preserved by induction through every operation. The
\href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_second_scale.py}{editable calculation} checks
that the two endpoints of each interval have the same floor, performs
the nine-residue convolution with integers, and verifies the exact
inequalities stated above. No sampled floating-point values are used for
these certificates. \(\square\)

\textbf{Solution 76.} (a) Over \(\mathbb Q\) at eleven, \(n=3,P=e=f=1\),
\(t=\log11\), and \(\vartheta=9/10\). The three distinct prime
valuations prove independence of the bases. The exponential bounds
already proved give

\[
\exp2<9<11<\left(\frac{27}{10}\right)^3<\exp3.
\tag{10.762}
\]

Thus \(2<t<3\), so \(n>Pt\). But \(3/t<3/2=\vartheta+3/5\), as required
by Theorem 10.188. Equivalently the real optimizer \(3/t-9/10\) is less
than \(3/5\). The theorem proves the second valuation estimate for every
original integer coefficient vector with its stipulated
minimum-valuation ordering, not merely for a sample choice.

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{1}
\tightlist
\item
  In III.2 at five, \(\vartheta=3/4\), and the height parameters use
  \(a_0=2+\log7\), \(a_1=a_2=146/25\), and \(a_1^{(2)}=83/25\). The
  elementary bounds of Theorem 10.189 give \(G_1/3<827/50\),
  \(G_2/3>303/25\), and therefore
\end{enumerate}

\[
\frac{11}{8}G_2-G_1>\frac38>0,
\qquad
\frac{C_2}{C_1}-\frac{11}{8}>
\frac{899}{1754200}>0.
\tag{10.763}
\]

The identical other entries of the two maxima preserve the \(11/8\)
comparison for \(\mathsf h,\mathsf h_2\). This proves the full second
estimate over \(\mathbb Q\) at five. The III.1 constants are different:
\(c=1794,c_*=495,a_1=4.71\), and \(a_1^{(2)}=1.99\). Substituting the
III.2 constants in that row would change the theorem being proved. No
higher-degree conclusion follows from the rational-field comparison
alone.

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{2}
\tightlist
\item
  For \(n=r=4,d=1,p=3\), the relevant row is I.2. Since \(\log21<4\) and
  \(\log5<2\),
\end{enumerate}

\[
G_2(4,1)=5\bigl(4(2+\log21)+2.76+\log5\bigr)<145<150.
\tag{10.764}
\]

Also \(5\log3<150\), since \(\log3<2\), and \(\log B_\dagger=0\). Thus
the stated h meets every additional hypothesis. Retain the exact
integers from Solution 75: \(s_0=1638\), \(\mu=35365415851\),
\(B=7,C_0=4,M_0=7\), and \(N=(7/2)/(1+10^{-26}/8)\). The cost fraction
is exactly

\[
\frac{35365415851\,(3277N+7)}{32S\mathcal D/\log3}.
\tag{10.765}
\]

The rational logarithm intervals proved in Solution 75, with outward
rounding through this expression, put it strictly between \(8/100\) and
\(9/100\). The \href{https://kokunoyumeto.github.io/open-math-courses-public/courses/TR-BAKER/figure_sources/yu_second_closure.py}{editable
calculation} records its rational endpoints. The all-rank theorem also
bounds it by \(437/2016<1/4\). The original kernel supplies the initial
prepared zeros, so Theorem 10.190 yields the stated local valuation
precision at every \(x\in\mathbb Z_3\) under its counterhypothesis. To
force an algebraic value to vanish one must still compare this precision
strictly with its global height and the actual degree of its value
field. A large local lower bound alone is not that comparison.
\(\square\)

\textbf{Solution 77.} \textbf{(a)} Write
\(D_* = S\mathcal D/(c_1c_4(N/\vartheta)dA)\). Corollary 10.185 gives
\(D_*<kL<\lambda D_*\). The powers of the prepared polynomial have
degree at most \(kL\); replacing this integer by \(\mathcal D\) would
not be justified by its definition. The strict floor inequalities give

\[
\frac{(\bar S/3)(\bar T/(r+1))}{\lambda D_*}
=\frac{q^2\eta c_4dA}{3\lambda e\vartheta t}=B_0.
\tag{10.802}
\]

For an m-element subset J', divide
\((2\bar S/(3r))(\bar T/(r+1))^{m+1}/(m+1)!\) by
\((m+1)!\lambda D_*\prod_{j\in J'}(2Jq^{-I}D_j)\). The result is

\[
\frac{2B_0}{r}\frac{\prod_{j\in J'}\sigma_j}{((m+1)!)^2}
\left(\frac{\eta c_2qrPd}{2eNt}\frac EJ\right)^m.
\tag{10.803}
\]

One factorial comes from the binomial product; the other comes from the
coordinate-degree side of the multiplicity inequality. Neither can be
dropped during the cancellation.

\textbf{(b)} The first comparison is \(1048576>232\). For the second,
\(1050\cdot10^6=1050000000<68719476736=2^{36}\). The consecutive-ratio
proof in Theorem 10.191 applies to all intermediate ranks, and the
factor \(35/2^{20}<1\) in Corollary 10.192 applies to all output ranks.
Thus these endpoint computations are supported by uniform monotonicity
proofs. The degree inequalities hold from the parameters alone. The
production of lower-rank bases with the asserted height product also
requires the nonzero polynomial and prepared zeros stipulated in
Corollary 10.192.

\textbf{(c)} If \(t\le n\), use \(\mathcal R_n\ge1\) and \(A_n\ge t\),
giving \(\mathcal R_nA_n/t^2\ge1/t\ge1/n\). If \(t>n\), use the second
entry of the maximum:

\[
\frac{\mathcal R_nA_n}{t^2}
\ge\left(\frac{\mathrm et}{n}\right)^n>1>1/n.
\tag{10.804}
\]

The height-product lower bound (10.55) and the exact formula for \(C_2\)
then give \(\mathcal T=C_2h\Omega_n\ge9261dh/t\), since
\(406\cdot49\cdot27/58=9261\). The logarithmic contribution \(2dh/t\) is
less than \(\mathcal T/4000\), as \(8000<9261\). For the height
contribution use the actual residue order \(g<\mathcal R_1\), the second
rank allowance and \(eq^u\le2d\). Its ratio to \(\mathcal T\) is less
than

\[
\frac4{406\cdot7\cdot2\cdot27\cdot3}
=\frac4{460404}<\frac1{100000}.
\tag{10.805}
\]

The sum is therefore less than \(1/4000+1/100000=13/50000<1/3000\); the
final comparison is \(39000<50000\). All coefficient-height inputs
concern the original integer coefficients, as required in Proposition
10.193. \(\square\)

\textbf{Solution 78.} (a) Put \(t_0=\lfloor T\rfloor\). The additive
coordinate, the \(r-1\) Euler coordinates, and the slack are \(r+1\)
symmetric nonnegative coordinates summing to \(t_0\). Their means are
each \(t_0/(r+1)\). This remains true for the full equation list because
its zero rows have been retained. The mean lcm contribution is at most
\((107/103)Ht_0/(r+1)\); the mean Euler contribution is at most
\((1+3/[8(r+1)])H(r-1)t_0/(r+1)+E_0\). Adding, and using \(t_0\le T\),
gives

\[
C_r=\frac{r+4/103}{r+1}+\frac{3(r-1)}{8(r+1)^2},
\qquad
1-C_r=\frac{483r+1101}{824(r+1)^2}.
\tag{10.833}
\]

Subtracting \(483/[824(r+1)]\) from the deficit leaves
\(618/[824(r+1)^2]>0\). Deleting zero rows before this average would
destroy the permutation argument.

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{1}
\item
  With \(z=\exp(-a_E)\), positivity of the binomial series bounds the
  Euler logarithm by \(a_Ev+E_0\), independently of the additive order.
  The additive lcm logarithm is bounded by \((107H/103)u\). A single
  larger coefficient a suffices for individual values; the separate
  coefficients give the smaller mean in part (a). Replacing the Euler
  remainder by the smaller remainder at \(\exp(-a)\) while keeping
  coefficient \(a_E\) would not be justified.
\item
  Here \(q=2,c_1=7/10,c_3=6/5,c_4=14\), \(c_2=8750/5101\),
  \(c_5=27/50\), and \(\gamma=12/5\). Use the exact \(e_0\),
  \(\varepsilon=13/18000\), and
  \(\mathcal N=\lambda(3/2+101/200)/(c_1c_4)\) from the theorem.
  Rational substitution gives
\end{enumerate}

\[
\mathcal L=\frac{3411}{1120},\qquad
\mathcal A=\frac{47193470266316287}{15709515479280000},
\qquad
\mathcal L-\mathcal A
=\frac{650420179955213}{15709515479280000}.
\tag{10.834}
\]

To compare the last fraction with \(41/1000\), multiply its positive
denominator by 41 and its numerator by 1000. Their difference is
\(6330045304733000>0\). The other three endpoint differences, and the
concavity reduction, are necessary for the uniform case-I.2 conclusion;
a single rank-three computation alone would not prove it.

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{3}
\tightlist
\item
  For real \(x,z\), writing each as its integer part plus a number in
  \([0,1)\) gives
  \(\lfloor x\rfloor+\lfloor z\rfloor\le\lfloor x+z\rfloor\). Apply this
  with \(x=\eta T\), \(z=(1-\eta)T\). Since
  \(\mu-1=\lfloor(1-\eta)T\rfloor\), every target of order at most
  \(\lfloor\eta T\rfloor\) has all ordinary jets through \(\mu-1\) among
  the original prepared equations. At a fractional node, the scalar
  q-denominator in (10.243) may be nonzero, and the consistent torus
  roots may generate a larger global value field. Both costs belong in
  (10.244). Neither has been bounded by this depth-zero integer
  calculation.
\end{enumerate}

\textbf{Solution 79.} (a) For \(u=y/b\), expand the proposed Bernstein
sum and interchange its two finite sums. The contribution of \(a_kb^k\)
is

\[
\sum_{i=k}^n\frac{\binom ik}{\binom nk}\binom ni
 u^i(1-u)^{n-i}
=u^k\sum_{i=k}^n\binom{n-k}{i-k}
 u^{i-k}(1-u)^{n-i}=u^k.
\tag{10.850}
\]

This proves the identity. The weights are nonnegative on the unit
interval and their sum is one. For the proposed quadratic, b = 1/4 and n
= 2 give coefficients \(1,3/4,11/16\). The polynomial is therefore at
least \(11/16\), attained at y = 1/4.

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{1}
\item
  Here \(\eta=1-(27/50)/7=323/350\). The final real radius is \(64S\)
  and the derivative cutoff is \((323/350)^6T\). The actual nodes are
  the integers of absolute value at most \(\lfloor64S\rfloor\), and the
  prepared indices have total order at most
  \(\lfloor(323/350)^6T\rfloor\). At each prior step the difference of
  the real cutoffs is \(c_5\eta^jT/(r+1)\). Taking its floor and adding
  one gives \(\mu_j\), and the floor-sum inequality supplies every
  required jet through \(\mu_j-1\).
\item
  Direct subtraction gives
\end{enumerate}

\[
\frac{15}{16}-\frac{20907}{22400}=\frac{93}{22400}>0,
\qquad
\frac{15}{16}-\frac{2323}{2520}=\frac{79}{5040}>0.
\tag{10.851}
\]

There are seven field cases, four step numbers, and two endpoints of the
fractional-linear function: \(7\cdot4\cdot2=56\) polynomials. Their
degrees are at most 3,4,5,6, hence their Bernstein lists have 4,5,6,7
entries. The total is \(7\cdot2(4+5+6+7)=308\). These are coefficient
checks; the convex-combination identity, rather than a list of sampled
ranks, proves positivity throughout each required interval.

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{3}
\tightlist
\item
  The tail has the form
  \(F_j=\mathfrak a q^{j+1}-\mathfrak b-\mathfrak c(j+1)\) with
  \(\mathfrak a>0\) and q greater than one. Thus
\end{enumerate}

\[
F_{j+1}-F_j=\mathfrak a(q-1)q^{j+1}-\mathfrak c
\ge\mathfrak a(q-1)q^6-\mathfrak c>0
\quad(j\ge5).
\tag{10.852}
\]

The lower bound for \(F_5\) controls the entire tail. In the local-input
proof the step number satisfies \(j\le r-1\), which supplied
\(q^{j-r}\le1/q\). A further step has not been included in that
estimate. Moreover fractional prepared scalars may have a q-denominator
and their torus roots may enlarge the global field of the value. The
arithmetic comparison here used neither cost because every target was an
integer at depth zero. They must be included before a fractional
extension can be concluded.

\textbf{Solution 80.} (a) Here \(p=3,A_0\ge3/2\). Since \(27>25\), one
has \(3\sqrt3>5\). Lemma 10.199 therefore gives

\[
\frac{N_c}{M}>\frac{27}{10}3\sqrt3>\frac{27}{2},
\qquad \frac{M}{N_c-M}<\frac1{27/2-1}=\frac2{25}.
\tag{10.869}
\]

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{1}
\tightlist
\item
  The upper prefactor multiplied by the proposed reserve is
\end{enumerate}

\[
631\frac{51}{50}=\frac{32181}{50},\qquad
648-\frac{32181}{50}=\frac{219}{50}>0.
\tag{10.870}
\]

Thus the multiplier still fits strictly below the stated second
constant, in every rank covered by the original prefactor bound.

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{2}
\tightlist
\item
  Here \(\eta=173/200\). Exact floors give
\end{enumerate}

\[
\begin{aligned}
(R_0,R_1,R_2,R_3)&=(10,21,42,84),\\
(\lfloor T\rfloor,\lfloor\eta T\rfloor,
 \lfloor\eta^2T\rfloor,\lfloor\eta^3T\rfloor)
 &=(1200,1038,897,776),\\
\lfloor\eta^4T\rfloor&=671,\\
(\mu_0,\mu_1,\mu_2,\mu_3)&=(530,368,227,106),\\
N_*&=21(530)+22(368)+42(227)+84(106)=37664.
\end{aligned}
\tag{10.871}
\]

Using only the final interval would give \(169(106)=17914\), omitting
19750 valid jet conditions. The inequalities
\(\lfloor\eta^4T\rfloor+\mu_j-1
=\lfloor\eta^jT\rfloor\) check every required order exactly.

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{3}
\item
  The root exists by successive lifting: if \(a_m^2\equiv2\pmod{7^m}\)
  and \(a_m\equiv3\pmod7\), there is a unique digit
  \(t\in\{0,\ldots,6\}\) such that \(a_{m+1}=a_m+t7^m\) solves the
  congruence modulo \(7^{m+1}\), because \(2a_m\equiv6\pmod7\) is
  invertible. The compatible sequence converges in \(\mathbb Q_7\) to
  the indicated root. Now \((\alpha-3)(\alpha+3)=-7\) and
  \(\alpha+3\equiv6\pmod7\). The second factor is a unit, so the first
  has valuation one. Also \(-\alpha-3\equiv1\pmod7\) is a unit. The
  global field \(\mathbb Q(\sqrt2)\) has degree two, while the chosen
  completion has degree one; its two embeddings give different
  valuations for the same algebraic expression. Chosen-completion
  membership alone therefore supplies no equality of conjugate
  valuations that would cancel the global factor.
\item
  The function \((1-c/v)^v\) increases, and its limit is bounded using
  the explicit positive exponential series. These facts bound both
  \(\mathfrak h\) and its inverse in the required directions. The
  radius-floor loss decreases with rank after division by \(q^rS_-\),
  while \(S_-,T_-,b_r,q^r\) increase and \(\varepsilon_r,e_r\) decrease.
  The proof keeps the last four terms of a positive reversed geometric
  sum. Finally the fractional-linear dependence on \(A_0\) requires only
  its two endpoints. These are uniform inequalities, so the tail rows
  prove all larger ranks rather than sampling them.
\end{enumerate}

\textbf{Solution 81.} The fraction is
\(\mathfrak h=(173/200)^4=895745041/1600000000\). Consequently

\[
\begin{aligned}
S_1&=\frac{16800000000}{895745041},&
T_1&=\frac{2687235123}{4000000},\\
R&=2(18+1)=38,&
\lfloor T_1\rfloor&=671,\\
M&=\left\lfloor\frac{27T_1}{200}\right\rfloor+1=91,&
\lfloor\eta T_1\rfloor&=581.
\end{aligned}
\tag{10.883}
\]

There are 19 positive odd nodes and 19 negative odd nodes in the
interval, so \(n_E=38\) and \(n_EM=3458\). The largest required total
jet order is \(581+91-1=671\), exactly the available input order. Among
the 77 integers in the interval, the other 39 are even, including zero.
Theorem 10.202 fills these at order 581 while retaining the odd-node
zeros through order 671. These counts illustrate the interpolation
formula; the small S and T are not asserted to satisfy the full
parameter construction.

For the field assertion, Theorem 10.51 first separates the actual
root-monomial classes, including its second phase partition when needed.
After restriction and division of the exponents, the remaining phases
are powers of the root of unity in K, and the torus bases belong to K.
Theorem 10.55 therefore puts the integer prepared value in K. The
additive polynomial is nevertheless evaluated at s/q, so its rational
denominator must still be cleared. The bound
\(\Xi_{1,0,u}\le D_{\mathrm a}(1+1/(q-1))\) is included in the
arithmetic comparison. Membership of a rational number in K imposes no
integrality condition on that number.

\textbf{Solution 82.} Dividing the first inequality by hA gives
\(1+(1/h+1/A)x+x^2/(hA)\). Its linear coefficient is smaller than
\(9/40<1\), and its quadratic coefficient is smaller than \(1/200<1/2\).
The positive exponential series gives the required inequality, strictly
for \(x>0\) and with equality at zero.

For (b), \(Q\ge2(173/200)^4=895745041/800000000\). Consequently
\(\log Q>191490082/1695745041\). An exact upper bound for the linear
endpoint ratio is \((3\cdot59\cdot101\cdot8750\cdot7\cdot1695745041)/
(58\cdot100\cdot5101\cdot10\cdot14\cdot4\cdot191490082)\).
Multiplication by the positive denominator verifies that this is less
than \(3/5\). Adding the bound \(1/1000\) for the exponential endpoint
term still gives less than one. Convexity then covers every intermediate
depth; the separate \(1/(q-1)\) denominator term is retained in Lemma
10.205.

For (c), degree and allocation inequalities control the possible
multiplicity output. They do not assert that the auxiliary polynomial
vanishes at any particular point. The rank conclusion explicitly uses
the full prepared integer zeros through \(\bar S\) and order \(\bar T\),
and those zeros require their analytic and arithmetic comparisons. The
parameter construction supplies only the specific blocks stated in
Theorem 10.203.

\textbf{Solution 83.} Here \(\eta=173/200\),
\(\mathfrak h=(173/200)^4\), and the exact contracted parameters are

\[
S_2=\frac{26880000000000000000}{802359178476091681},\qquad
T_2=\frac{2407077535428275043}{6400000000000000}.
\tag{10.915}
\]

Integer division of the positive numerators by their denominators gives
the following values. All radii count both endpoints.

{\def\LTcaptype{none} % do not increment counter
\begin{longtable}[]{@{}
  >{\raggedright\arraybackslash}p{(\linewidth - 8\tabcolsep) * \real{0.1579}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 8\tabcolsep) * \real{0.2105}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 8\tabcolsep) * \real{0.2105}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 8\tabcolsep) * \real{0.2105}}
  >{\raggedleft\arraybackslash}p{(\linewidth - 8\tabcolsep) * \real{0.2105}}@{}}
\toprule\noalign{}
\begin{minipage}[b]{\linewidth}\raggedright
j
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
\(\lfloor2^jS_2\rfloor\)
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
\(\lfloor\eta^jT_2\rfloor\)
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
\(\mu_j\) for the next step
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft
Cardinal degree
\end{minipage} \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
1 & 67 & 325 & 44 & 5940 \\
2 & 134 & 281 & 38 & 10222 \\
3 & 268 & 243 & --- & --- \\
\end{longtable}
}

The two degrees are \(135\cdot44\) and \(269\cdot38\). The jet
requirements are \(281+43=324\le325\) and \(243+37=280\le281\). Thus no
unprovided derivative is used.

At general depth, the two opposite powers cancel:

\[
(\mu_j-1)M_*
\le\frac{c\eta^j\mathfrak h^I\widehat T}{r+1}
       \frac{\mathfrak h^{-I}H}{\log p}
=\frac{c\eta^j\widehat TH}{(r+1)\log p}.
\tag{10.916}
\]

The separate original bound is \(v_p(b_n)\le H/\log p\). Factoring
\(\widehat TH/\log p\) out of the sum therefore leaves
\(c\eta^j/(r+1)+1/\widehat T\), exactly the bracket in the input
estimate of Theorem 10.206. Replacing \(\mu_j-1\) by an unnecessarily
rounded upper bound would lose this exact cancellation.

These data demonstrate integer counting and jet availability. They do
not construct the required number field, support or kernel at depth two.
The theorem's application there also needs its starting integer block
and the stipulated phase-family hypotheses. The depth-one starting block
is supplied by Theorems 10.202--10.203; the stronger q-deleted data
survive the integer extensions.

\textbf{Solution 84.} The exact parameters are

\[
\eta=\frac{173}{200},\qquad
S_1=\frac{16800000000}{895745041},\qquad
T_1=\frac{2687235123}{4000000},\qquad
O=\lfloor T_2\rfloor=376.
\tag{10.934}
\]

The radii are 38, 75 and 150. The input floors are 671, 581, 502 and
434, so the four multiplicities are 296, 206, 127 and 59. Their
jet-order checks are the exact equalities \(376+295=671\),
\(376+205=581\), \(376+126=502\), and \(376+58=434\). The deleted
increment is \(296-206=90\).

There are 38 odd inner nodes, 39 even inner nodes including zero, 74
nodes in the first annulus and 150 in the second. Thus

\[
\begin{aligned}
N_I&=38\cdot296+39\cdot206+74\cdot127+150\cdot59\\
   &=301\cdot59+151(127-59)+77(206-127)+38\cdot90\\
   &=37530.
\end{aligned}
\tag{10.935}
\]

The first line counts disjoint node types. The second starts with the
full outer layer, then adds two full inner increments and the deleted
increment. In particular the inner full layer contains \(2\cdot38+1=77\)
nodes, not just the 39 multiples of q. The multiplicity hypothesis holds
because \(T_2>376\) and \((27/50)376/4>1\). These are counting data;
they do not specify a complete number field and auxiliary kernel.

The optimizer gives \(r/t>A_0+3/5\), with \(t>1\), hence \(A_0<r\). A
fractional-linear arithmetic bound can therefore be checked at the two
endpoints \(a_*\) and r for each fixed small rank. An unrestricted
endpoint at infinity would ignore this hypothesis and may give an
insufficient bound.

The depth-I fractional value and the following depth-\(I+1\) integer
value have the same scaled radius and the same total denominator depth
\(I+1\) in (10.243). The latter has a smaller derivative order and lies
in K. Thus the already proved bound at depth I supplies its arithmetic
cost, even if the new depth is the first one beyond the logarithmic
threshold. No denominator is removed in this passage.

Corollary 10.209 constructs the family only through \(\min(I_G,I_M)\).
If \(I_M<I_G\), it has not justified a further application of the mixed
step: the required derivative order can be too small. The distinct
fixed-order argument of Theorems 10.216--10.217 supplies the
continuation to its endpoint.

\textbf{Solution 85.} The five successive threshold quantities are

\[
\frac{c\eta^jT_1}{4}=
\frac{81}{50},\quad \frac{14013}{10000},\quad
\frac{2424249}{2000000},\quad
\frac{419395077}{400000000},\quad
\frac{72555348321}{80000000000}
\quad(j=0,1,2,3,4).
\tag{10.950}
\]

The last one at least 1 occurs at \(j_0=3\). Hence
\(R=2\lfloor8\cdot40\rfloor=640\) and
\(O=\lfloor(173/200)^4\cdot12\rfloor=6\). The input interval has radius
\(\lfloor8\cdot40\rfloor=320\) and order
\(\lfloor(173/200)^3\cdot12\rfloor=7\). Its interpolation multiplicity
is \(\mu=\lfloor419395077/400000000\rfloor+1=2\), so the jet-order test
is exactly \(6+2-1=7\). These are counting data, not a complete
number-field auxiliary-kernel instance.

If \(\mu=1\), only the ordinary jet of order zero is used at each node.
The floor inequality still holds, and the loss \((\mu-1)M_*\) is zero.
The separate original \(v_p(b_n)\) loss remains; it must not be
discarded with the jet loss.

The torus-degree estimate is

\[
\frac{2436}{5}d^2\exp(-3A)
 <\frac{2436}{5\cdot16\cdot2^{13}}
 =\frac{609}{163840}<1.
\tag{10.951}
\]

Here the factors \(16,2^8,2^5\) come respectively from \(a^2\ge16\),
\(\exp(8)>2^8\), and \(\exp(A)>2^5\). For the root count,
\(\lambda=59/58\) gives
\(45/(8\lambda)=45\cdot58/(8\cdot59)=1305/236>1\). Both estimates are
strict. The first leaves a nonzero polynomial in the additive variable;
the second gives more distinct roots than its degree.

One must construct every intervening selected family through \(I_3\),
retaining a nonzero coefficient subvector, complete fibres and the full
integer interval through order O. Lemma 10.211 gives the starting family
at \(I_1\). Theorems 10.216--10.217 supply the intervening fractional
zeros and full integer successors; the endpoint contradiction alone does
not prove those assertions.

\textbf{Solution 86.} (a) Here \(a=4,c=27/50,c_0=27/10,c_4=14\),
\(c_2=3500/2083\), \(P_0=1\), and \(\max(1,5(a_*+3/5)/r)=9/4\).
Substitution gives

\[
L_3=\frac{11202105600000000}{262099267823},\qquad
\mathfrak h=(173/200)^4=\frac{895745041}{1600000000}.
\tag{10.975}
\]

Multiplying rational numbers and comparing integer numerators gives

\[
\frac{17}{10}<\frac{cL_3\mathfrak h^{14}}4<\frac{43}{25},
\qquad
\frac{95}{100}<\frac{cL_3\mathfrak h^{15}}4<\frac{97}{100}.
\tag{10.976}
\]

The powers decrease strictly, so \(m_3=14\). Also
\(164/100<8/(2\mathfrak h)^{14}<165/100\), which places \(\mu_3\)
strictly between 1 and 2. Thus \(j_3=1\). For B = 10 the two required
vertices are \((14,14)\) and \((30/\log Q,30/\log Q)\). The third vertex
is \((14,(30+14\alpha)/\log2)\), with \(\alpha=-\log\mathfrak h\). Its
exponential term is the same as that at the second required vertex, and
its linear term is smaller. Convexity on their triangle gives the entire
enclosure.

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{1}
\tightlist
\item
  The two radii are \(R_{\rm s}=\lfloor197/8\rfloor=24\) and
  \(R_{\rm l}=\lfloor199/4\rfloor=49\). Their node counts are 49 and 99.
  Precisely,
\end{enumerate}

\[
\frac{193}{4}<49\le\frac{201}{4},\qquad
\frac{197}{2}<99\le\frac{201}{2},\qquad 2\cdot24\le49.
\tag{10.977}
\]

Multiplication of the first two comparisons by N gives the exact mass
inequalities. These formulas do not order either new radius relative to
the original R. Each known interval can be used as its own interpolation
input, so such an ordering is unnecessary.

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{2}
\tightlist
\item
  There are forty odd integers in \([-40,40]\), twenty of each sign.
  Since \(5^2\le80<5^3\), the separation bound is two. The raw analytic
  entries of Theorem 10.182, after subtracting \(\beta\), are
\end{enumerate}

\[
40N-Q_D,\qquad
\kappa+\chi\widehat U_r-v_p(b_n)-2.
\tag{10.978}
\]

Only ordinary jets of order zero are used: \(\mu-1=0\). Thus every
output prepared order up to O was already present in the input. The
q-deleted separation loss occurs once. The projection error still
contains the original \(v_p(b_n)\); setting the ordinary-jet loss to
zero does not remove that distinct term.

\textbf{Solution 87.} (a) We have \(z=1/100,L=3122,K=94,N=293468\). Thus
\(K>85\), and the factorial comparison used to bound \(\log\beta\)
applies. The determinant margin is
\(3848/1125-(1/200+136/75+7/225+13/40)>0\). The bound including the
common p-part is \(588/25+7/30+1/12=7151/300<24\); the collision
coefficient is \(1/12+49/6000+7/1200=73/750<1\).

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{1}
\tightlist
\item
  With no padding the ratio is one. With one padded height, say the
  first, it is \(\epsilon t/(dh(a_1))\), bounded by
  \(58\mathrm e\epsilon t d\ell_d/w\). If both heights are padded, the
  ratio is \(\epsilon^2t^2/(d^2\Omega)\), bounded by
  \(29\mathrm e^2\epsilon^2t^2d\ell_d/w\). Squaring the singleton bound
  would introduce \(d^2\ell_d^2/w^2\) instead of \(d\ell_d/w\), and
  would miss the independent two-base height-product information.
\end{enumerate}

At d=64 the sufficient upper bound is \[
\frac{3480}{218295(27/10)^2(8/5)^2}
\frac{64}{46/5}\frac{(291/5+6)^2}{291/5}
=\frac{3320210000}{7889661549}<\frac12.
\tag{10.1000}
\] At the upper endpoint of the middle range it is \[
\frac{696}{218295(27/10)^3(8/5)^3}
\frac{2048}{12}\frac{(687/10+15/2)^2}{687/10}
=\frac{37419280000}{65596294071}<\frac35.
\tag{10.1001}
\] Multiplying positive denominators checks both inequalities. For large
degrees, substitute \(\log d\ge11(69/100)\) in
\((123/100)(549/50+\log d)-(76/5+\log d)\); the result is \(511/10000\).
Its slope as a function of \(\log d\) is \(23/100>0\), so it remains
positive for every larger degree.

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{2}
\tightlist
\item
  The two-logarithm determinant was stated for two nonzero coefficients,
  and its collision argument divides by their gcd to obtain two positive
  coprime integers. With one nonzero coefficient, the one-base power law
  instead gives \((34/33)d\mathsf h_2/t+8dh(a)/t\). The first summand
  contains no height, so divide by the two-base product lower bound. In
  the second summand the active height cancels; use the singleton lower
  bound for the other base. Substitution of
  \(C_2=218295\mathrm e^4d^4\ell_d A/w\) gives the two fractions in
  (10.999), each with its stated field and torsion factors. Their sum is
  less than \(1/1000\) of the target.
\end{enumerate}

\textbf{Solution 88.} (a) If the second row coefficients are \(v_{lj}\),
the composed original coefficient of \(z_k\) is
\(\sum_{j,i}v_{lj}u_{ji}a_{ik}\). Its absolute value is at most
\(\sum_{j,i}|v_{lj}||u_{ji}||a_{ik}|\). Multiply by \(h(a_k)\) and sum
to obtain the required weight \(\sum_j|v_{lj}|s_j\). The actual residue
factor at the middle rank cancels in the product allowance. Denominator
clearing can introduce large new integer coefficients. It does not
identify their maximum or minimum nonzero absolute value with the
original \(R_{\max}\) or \(B_\dagger\). Lemma 10.221 instead keeps the
original Euler arguments and the original minimum-valuation coefficient
at every step.

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{1}
\item
  Write the constant as \[
  C_2(j)=\frac{c^{(2)}}P(a^{(2)}\mathrm eP)^jF(j)
  \frac{d^{j+2}\ell_d}{q^ut^3}\mathcal R_jA_j.
  \tag{10.1018}
  \] After multiplying by \(\Gamma_2^{n-r}\mathcal R_n/\mathcal R_r\),
  the factors \(\mathcal R_r\) cancel and the powers of d and P become
  exactly those at rank n.~The remaining exponential quotient is
  \((\mathrm e/q)^{n-r}\); the factorial quotient is \(F(r)/F(n)\).
  Finally \(A_r/A_n\le1\). Since \(F(j+1)/F(j)>\mathrm e\), the ratio is
  strictly below \(q^{r-n}\) when r\textless n.~At r=2 multiply this by
  \(3/2\), for the coefficient parameter, and \(100/123\), for the
  constant comparison. The largest resulting bound is
  \((3/2)(100/123)/2=25/41\), at n=3 and q=2.
\item
  Valuations at the rational primes 2,3,11 prove independence. The
  original full exponent lattice is saturated, so its q-primary index is
  one. For the pair \((6,11)\), a rational power product has equal
  integral valuations at 2 and 3 and an integral valuation at 11; its
  two rational exponents are therefore integers. This saturation index
  is also one. The original residues generate all of
  \(\mathbb F_5^\times\), whereas both 6 and 11 reduce to one. Including
  the rational torsion root \(-1\), the generated residue groups have
  orders four and two. Thus \(\delta(2,3,11)=1\), \(\delta(6,11)=2\).
\end{enumerate}

Here \(d=e=f=P=1,q=2,a^{(2)}=7\), and \(\Gamma_2=7\mathrm e^2/2>25\),
since \(\mathrm e>27/10\). Also \(\mathcal R_3\ge5\), while \[
\mathcal R_2=\max\{5/2,\mathrm e^2(\log5)^3/4\}
<\frac94\left(\frac{13}{8}\right)^3<10.
\tag{10.1019}
\] The latter upper bound exceeds 5/2 as well. The elementary bounds
\(\log2>2/3,\log3>1,\log6<2\) imply \[
\Gamma_2\frac{\mathcal R_3}{\mathcal R_2}
                 \log2\log3>\frac{25}{3}>\log6.
\tag{10.1020}
\] Multiplying by \(\log11\) proves (10.725). The individual weighted
conditions are equalities. Finally \(\Xi=66\) and \(\Xi-1=65=5\cdot13\),
so the valuation is one. This constructs an admissible rank-two
representation. It does not exclude a rank-one representation; no
least-rank assertion follows from that construction alone. \(\square\)

\textbf{Solution 89.} (a) If \(b_j=0\), its logarithmic entry is
\(\log(\omega(d)|b_n|/w_j)\), with a positive argument because
\(b_n\ne0\). In a varying coordinate it is still of the form
\(\log(A+D/u)\) used in (10.1023); D can vanish. The derivative on the
logarithmic branch is at least \(H_i-1\), and the branches meet
continuously. This proves the same monotonicity. Merely increasing the
product would give a bound with the old coefficient parameter; it would
not justify the smaller parameter computed from the new weights.

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{1}
\item
  For odd primes, insert
  \(\max\{p/\delta(\boldsymbol a),\mathrm e^nt^{n+1}/n^n\}
  <(n+1)p\) into (10.642). The factors \(q^u=2\) and the leading 2
  cancel. Divide by (10.1024) and use (10.1026). The remaining
  coefficient is \((c^{(2)}/P)/(256\mathrm e)\), the exponential factor
  is \((7P/256)^n\), and the remaining rank factor is
  \((n+1)^3(A/t)/(n^{3/2}\log^2(2n))\). At primes at least five use
  \(c^{(2)}/P\le2228/3,A/t\le n+2\); at three use
  \(c^{(2)}/P=216,A/t\le2(n+2)\). These give respectively
  \(10n^3(7/256)^n\) and \(6n^3(21/256)^n\). Their n=2 values are
  245/4096 and 1323/4096, and their successive ratios are at most
  189/2048 and 567/2048. At two, retain \(d=2,f=2,P=4,q^u=3\) and the
  maximum bound 6. The exact constant before the factorial estimate is
  \(203(104\mathrm e)^n/(16\mathrm e)^{2(n+1)}\). The second inequality
  of (10.1026) leaves \(203(13/32)^n/(256\mathrm e)\) times
  \((n+1)^2A/(tn^{3/2}\log^2(2n))\). The estimates in the proof give the
  scalar factor 345303/262144, which is less than 4/3. Thus the final
  envelope is \((4/3)n^2(13/32)^n\), with initial value 169/192 and
  ratio at most 117/128. Every starting value and ratio is below one;
  induction, rather than a finite numerical sample, proves all ranks.
\item
  Put \(r=5^8=390625\). At \(\varepsilon=1/2\), \(L=51+\log500\); the
  two branches are \((\log3)(51+\log500)\) and 390625/2, and
  \(w_2^*=390625/(2(51+\log500))\). This is the larger weight:
  \(\log500<7\), \(\log3<2\), and \(390625/2>116\) suffice. At
  \(\varepsilon=1/r\), \(L=51+\log(250r)\), the second branch is one,
  and \(w_2^*=\log3\), since \(1/L<1<\log3\). To justify the rough
  estimates just used, \(\mathrm e^7>(8/3)^7>500\) and
  \(\mathrm e^2>(8/3)^2>3\). The congruences in the worked example give
  valuation exactly one. For a dependent list the application of Theorem
  10.223 is unavailable; padding heights changes no multiplicative
  relations. In particular it cannot turn (1,3,3) into an independent
  list or bound new coefficients arising from eliminating a relation.
  Those are additional proof obligations.
\end{enumerate}

\textbf{Solution 90.} (a) The valuation vectors at two and three are
(2,-2) and (3,-3). Their primitive integral direction is (1,-1), so take
\(\beta=2/3\), with exponents (2,3). For (3,-2) the combined exponent is
zero and the product is one; it is excluded. For (2,-1) the combined
exponent is one and the product is 2/3. At two, \((-3)9=-27\), and
\(-27-1=-28\) has valuation two. The primitive positive generator is 3,
its exponent is three, and the sign is negative. Squaring gives \(3^6\).
Here \(v_2(3^2-1)=3\), so the power identity gives \(v_2(3^6-1)=3\),
consistent with the inequality in (10.1043).

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{1}
\item
  The exact coefficient transfer gives \(|M|<2nB_n\Omega r\). Since
  \(r\le U/\varepsilon\), its logarithm is less than
  \(\log(2n)+\log B_n+\log(\varepsilon^{-1})
   +\log\Omega+\log U\). The first three terms together are less than L,
  which is at most U; each remaining logarithm is less than U. This
  proves the strict factor three. The generator height is at most
  \(\Omega\), and \(\Omega<U/((n+1)\log p)\). After adding the height
  term and \(\log2<U\), (10.1043) is bounded by the first line of
  (10.1045). Since \(2/(n+1)\le2/3\) and \(\log p<p-1\), its bracket is
  less than \((14/3)(p-1)<14p/3\). Rationality supplies an integral
  primitive valuation vector representing an actual positive rational
  number, the lower height bound \(h(\beta)\ge\log2\), and the
  elementary residue and local-height bounds in the rational field.
  These assertions cannot be transferred unchanged to arbitrary number
  fields.
\item
  Modulo four, \(3\cdot5^k+5\) is zero. At k=1 it is 20; for k greater
  than one its quotient by five is \(3\cdot5^{k-1}+1\), prime to five.
  Therefore \(v_5(M/4)=1\), and the power identity for \(2^4=16\) gives
  \(v_5(2^M-1)=1+1=2\). At \(\varepsilon=1/2\), the two branches
  defining U are \(\Omega(92+\log(2500\Omega))\) and \(5^k/2\); U is
  their maximum. At \(\varepsilon=5^{-k}\), they are
  \(\Omega(92+\log(1250\Omega5^k))\) and one. The first is larger, since
  \(\Omega=(\log8)(\log4)>1\) and its logarithmic factor exceeds 92. Two
  rational unit parts have rank at most two. Rank two is independence,
  and lower ranks are covered by Theorem 10.228. The three bases (2,3,6)
  at five have rank two: their first two valuation vectors are
  independent and the third is their sum. This is an intermediate
  dependent rank, which neither Theorem 10.226 nor Theorem 10.228
  covers. Theorem 10.232 treats it.
\end{enumerate}

\textbf{Solution 91.} (a) At primes two and three the first two
valuation vectors are the standard basis, while the last is their sum.
Take \(d_0=1,D=2\). Cramer's numerators give exponents \(M_1=2(2+1)=6\)
and \(M_2=2(1+1)=4\). The original product is \(2^2\cdot3\cdot6=72\),
and its square is \(2^6 3^4=5184\). Both \(72-1=71\) and \(5184-1=5183\)
are prime to five, so both valuations are zero. The third base is
redundant, so Proposition 10.231 applies directly.

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{1}
\item
  The first two valuation vectors at three are 1 and 3. Take
  \((k_1,k_2)=(-3,1)\), whose product is \((-3)^{-3}27=-1\). Eliminate
  the second index. Doubling this primitive relation gives
  \(\kappa=(-6,2,0)\) and \(\kappa_x=2\). For coefficients (1,1,1), the
  surviving coefficients are (8,2), and the exact identity is
  \((-405)^2=(-3)^8 5^2=164025\). We have \(v_2(-405-1)=1\), whereas
  \(v_2(164025-1)=v_2(8\cdot20503)=3\). The minimum coefficient
  valuation increases from zero to one, with the distinguished
  coefficient still attaining it. For original coefficients (2,2,2), the
  product is 164025 and the surviving coefficients are (16,4). Its
  square is \((-3)^{16}5^4\). The original difference has valuation
  three; the square has valuation four, since 164025+1 is twice an odd
  number. The minimum coefficient valuation increases from one to two.
  Thus both examples satisfy the factored-power inequality and the
  required minimum condition, including the common factor at two.
\item
  Substitute \(B_n'=\kappa_xB_n\),
  \(\varepsilon'=\kappa_x\varepsilon/(2K)\), and \(V'=V/w_x\) into the
  definition of T for length n-1. The factors \(\kappa_x\) cancel,
  leaving a ratio \((2K/w_x)\exp(D_{n-1}-D_n)/p\) to the original T.
  Taking logarithms gives (10.1062). Since \(\log(2K)<L\), its value is
  less than 2L. The other branch of U is preserved exactly. At two every
  doubled circuit coefficient is divisible by two. If the chosen
  primitive coefficient is odd, the pivot has valuation exactly one.
  Every transferred coefficient is divisible by \(2^{v_2(b_n)+1}\), and
  the distinguished one has exactly that valuation. Hence its minimum
  condition persists. One must track this common factor rather than
  incorrectly claiming that the pivot itself remains odd.
\item
  Either the last vector is in the span of the preceding ones, or
  adjoining it raises rank by one. In the latter case, dependence of the
  whole list forces dependence among those preceding vectors. This is
  exactly the circuit needed for induction, and its support avoids the
  distinguished coefficient. For rational numbers the prime-valuation
  map has only signs as its kernel, and bounds each coordinate by the
  logarithmic height divided by \(\log2\). In a number field this kernel
  can contain units of infinite order. For example, in
  \(\mathbb Q(\sqrt2)\), the number \(1+\sqrt2\) and its inverse
  \(\sqrt2-1\) are algebraic integers, as the monic equations
  \(X^2-2X-1=0\) and \(X^2+2X-1=0\) show. Their product is one, so their
  finite-prime valuations vanish; the powers of \(1+\sqrt2>1\) are
  distinct. Additional archimedean and unit information is therefore
  needed in general.
\end{enumerate}

\textbf{Solution 92.} (a) The valuation vectors at two and three are
\((1,0),(0,1),(2,0)\), so the greedy basis is \((2,3)\). It is already
saturated over the rationals: a rational root of an integral combination
has integral valuations at both primes. The residue of two generates all
four nonzero classes at five, so \(\delta_\beta=1\). Here P=1, t=log5,
and \(L_3=3/(25\log5)<2/25<\log2\). Therefore
\(\Omega_3=(\log2)(\log3)(\log4)\), \(H=\log5\), and \(A_3=4+\log4\).
The second entry in \(M_3\) is \(\mathrm e^3(\log5)^4/27<5\), so
\(M_3=5\). The strict inequality follows from the positive-series
enclosures in (10.1069); the exact calculation gives a positive rational
lower bound for \(135-\mathrm e^3(\log5)^4\). The product is96 and
\(v_5(96-1)=v_5(95)=1\). The circuit \(4\cdot2^{-2}=1\) deletes4 with
coefficient one, leaving exponents \((1+2\cdot2,1)=(5,1)\) on \((2,3)\);
their product is again96. This example lies in the small-coefficient
branch: \(B/\log B<3\), whereas (10.1071) gives \(W_3>200\mathrm e^3\).

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{1}
\item
  The first base is torsion of order two, and the greedy basis is still
  \((2,3)\). The original product is \(-24\), with \(v_5(-24-1)=2\).
  Squaring deletes the torsion base and yields \(2^2 3^2 4^2=576\); its
  distance from one is575=25·23, also of valuation two. Thus the
  factored-power inequality is equality here. The old modified product
  is \(L_4(\log2)(\log3)(\log4)\), because all three positive-base
  heights exceed \(L_4=4/(25\log5)\). The new product omits that factor,
  and the selected basis and its residue index remain exactly the same.
  A bound B=3 becomes a valid coefficient bound \(B^2=9\), although the
  actual new coefficients are only two.
\item
  Before \(t_0=(n/\mathrm e)^{n/(n+1)}\), the comparison quotient is
  \(t^j\); after that point it is \(n^n\mathrm e^{-n}t^{j-n-1}\). The
  first increases and the second is nonincreasing for \(j\le n+1\),
  proving the bound at their common endpoint. The ratio comparison
  following (10.1065) proves the stated strict upper bound. At two, P is
  at least4. A primitive cube root has a nontrivial reduction of order
  three, so \(3\mid2^f-1\) and f is at least two. Hence
  \(Pt\ge4\log4>16/3\). Together with \(\mathrm e>8/3\), this gives
  \((4\mathrm ePt)^{-1}<9/512\). Finally the exponential series through
  degree six is1957/720 and the remaining terms are positive. Direct
  rational arithmetic gives \(13(1957/720)^2-96-9/512=25049/1036800>0\).
\item
  If the maximal-height base is the deleted base, normalize the
  coefficient bound by \(BdH_n(a_x)/8\); the two ratios
  \(h(a_x)/H_n(a_x)\) are at most one. If it is a surviving selected
  base, cancel its height before applying the independent product bound:
  the required independent list has length j-1, and its contribution
  loses a factor \(\mathrm e\). If it is a surviving nonbasis member,
  cancel its actual height against its modified height and use one fewer
  floor. This changes the powers to \(\varkappa^{s-1}/P^{r+1}\) and
  \(t^{s+1}/X_n^{s-1}\). There are n-r nonbasis indices in total. One is
  the deleted index and another is this surviving maximal-height index,
  so n-r is at least two and s=n-r-1 is at least one. These are exactly
  the three normalizations bounded by (10.1081).
\end{enumerate}

\textbf{Solution 93.} The first polynomial is \(X^2-14X-49\), with roots
\(7(1\pm\sqrt2)\). Both absolute values exceed one, since
\(\sqrt2>8/7\), so the logarithmic height is \(\tfrac12\log49=\log7\).
The second primitive polynomial is \(49X^2-84X+34\). Its roots are
\((6\pm\sqrt2)/7\); the positive-sign root is greater than one and the
other is between zero and one. Its height is therefore
\(\tfrac12\log(7(6+\sqrt2))\). The polynomials are irreducible since
their roots generate the quadratic field, and their coefficients have
greatest common divisor one.

Write a lift of 3 as \(3+7k\). Modulo 49, its square minus two is
\(7+42k\). Thus \(1+6k\equiv0\pmod7\), giving \(k\equiv1\) and the root
10 modulo 49. For \(\Xi=8+7\sqrt2\), the residue of \(\Xi-1\) is 28
modulo 49. It is divisible by seven but not 49, so its prime-ideal
valuation is exactly one.

At n=2 the first envelope is less than \[
\frac{3627}{1024}\left(\frac{21}{128}\right)^2\frac{135}{14}
=\frac{30847635}{33554432}<1.
\] Its positive margin is \(2706797/33554432\). The successive ratio is
less than \(7/18\), as proved in (10.1092); induction therefore bounds
every later rank too. A finite plot alone would not establish that
quantifier.

For the final list, \(b_2=1\) has minimum seven-adic valuation zero,
while \(v_7(b_1)=+\infty\). The total valuation is zero and the product
is \(1+\sqrt2\ne1\). However, its normalized bases are \((1,1+\sqrt2)\),
a dependent list. The independence hypothesis in Theorem 10.239 fails.
This example neither violates that theorem nor supplies a proof for the
dependent case.

\textbf{Solution 94.} (a) Conjugation sends \(\sqrt2\) to \(-\sqrt2\),
giving norms \(-1,34,9\). If \(a^r c^s3^t=1\), taking the product with
its conjugate and then the valuations at seventeen and three gives
\(s=t=0\). Since \(a>1\) in the chosen real embedding, \(r=0\). A
torsion relation reduces to this case after raising to its order, so the
same list is independent modulo torsion. The first relation has
\(k_x=1\) at x=3 and \(\kappa=1\); (10.1116) gives \((2,2,6)\). As
\(c_3(4,2)/c_3(3,2)=16\), the new bounds are \(B'=12\), \(B_n'=6\), and
\(\varepsilon'=1/64\). Thus

\[
\frac{\varepsilon'B'}{B_n'c_3(3,2)}
=\frac{1}{32c_3(3,2)}
=\frac{1}{2c_3(4,2)}
=\frac{\varepsilon B}{B_nc_3(4,2)}.
\]

The universal proof uses \(\kappa=AR\); any smaller verified relation
bound also satisfies its comparisons. Here the explicit relation
supplies that smaller bound.

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{1}
\item
  Now \(k_x=7\), and the reduced coefficients are \((14,14,42)\). Take
  \(B'=84\), \(B_n'=42\), and again \(\varepsilon'=1/64\). The ratio
  \(B'/B_n'\) is still two, so the displayed equality is unchanged. All
  three new coefficients have seven-adic valuation one, including the
  distinguished coefficient. The product of the reduced bases is the
  seventh power of the original product; its difference from one has
  valuation at least that of the original difference. The proof requires
  only that \(k_x\) have minimum valuation among the relation
  coefficients. Its valuation can be positive.
\item
  Write the first lift as \(3+7z\). Modulo 49 its square minus two is
  \(7(1+6z)\), so z=1 gives the root 10. Next write \(10+49z\). Modulo
  343 its square minus two is \(49(2+20z)\), so z=2 gives 108. Now
  \(ac=8+7\sqrt2\equiv78\pmod{343}\), and
\end{enumerate}

\[
3^6\cdot78^2-1\equiv245=5\cdot49\pmod{343}.
\]

The valuation is therefore exactly two. For the alternate coefficient
vector, the relation with \(k_x=1\) gives \((2,0,6)\). Its last
coefficient has valuation zero, equal to the minimum of zero, infinity
and zero. This nonidentity product is allowed by Theorem 10.242; the
zero coefficient does not create a new dependence assumption.

\textbf{Solution 95.} Let \(z=N_{L/K}(\beta)\), so \(z^T=A^2\). Then
\(\theta=z^{(T+1)/2}/A\) belongs to \(K\) and satisfies \(\theta^T=A\).
Applying this to every vector in the 3-primary saturation proves
equality of the two lattices and allows the roots representing a
saturated basis to be chosen in \(K\).

For odd \(f\), these roots reduce to a group of order \(g\mid N-1\). The
3-primary root-of-unity group has order \(m\mid N+1\). The orders are
coprime, so the combined group has order \(mg\), and
\(\delta_L=(N^2-1)/(mg)\). Thus
\(N^2/(m\delta_L)=N^2g/(N^2-1)\le N^2/(N+1)<N\). The factor \(m\) is
essential to this calculation.

Direct cancellation of factorials gives \[
\frac{T_{n+1}}{T_n}
=\left(1+\frac1{n+1}\right)^{n+1}
  \frac{(n+2)\sqrt n}{(n+1)^{3/2}}.
\] The first factor is less than \(\mathrm e\). The logarithmic
derivative of the second is \((2-n)/(2n(n+1)(n+2))\), so its maximum for
\(n\ge2\) occurs at two. Its square there is \(32/27<121/100\). This
proves (10.1128). In the rational row of (10.1129), the endpoint is
bounded by \[
\frac{406}{1024}
\left(\frac{13}{16}\right)^2
\left(\frac38\right)^4\frac{135}{14}\frac{81}{8}
=\frac{4340987235}{8589934592}<1.
\] The positive margin is \(4248947357/8589934592\). The successive
ratio is less than \(143/160<1\), so this endpoint controls every larger
rank in that row. The other rows use the smaller endpoints and ratios
displayed in (10.1129).

For the rational example, \(d=e=f=1\). The valuations of \((6,11/2,33)\)
at two are \((1,-1,0)\), so (10.1087) gives
\(\boldsymbol\beta=(3,11,33)\). This list has rank two: prime valuations
prove that three and eleven are independent, and their product is
thirty-three. The original product is \(33^2\), hence \[
\Xi-1=33^2-1=1088=2^6\cdot17,\qquad v_2(\Xi-1)=6.
\] The minimum coefficient valuation is zero. The weights
\((\log6,\log11,\log33)\) are the original rational heights and exceed
\(Q^{-1}\), where \(Q=16\mathrm e^2\). Theorem 10.246 therefore applies
with the stated bounds. Neither the nonunit original inputs nor the
dependence of their normalized units requires a change to the field
degree or height floor.

\subsection{References}\label{references}

\begin{itemize}
\item
  Yann Bugeaud and Michel Laurent, \emph{Minoration effective de la
  distance p-adique entre puissances de nombres algébriques}, Journal of
  Number Theory 61 (1996), 311--342.
  \href{https://irma.math.unistra.fr/~bugeaud/travaux/logpadicdef.ps}{Author-hosted
  full text}.
\item
  Tomohiro Yamada, \emph{A note on the paper by Bugeaud and Laurent
  ``Minoration effective de la distance p-adique entre puissances de
  nombres algébriques''}, arXiv:math/0607072, version 3 (2007).
  \href{https://arxiv.org/pdf/math/0607072}{Free full text}.
\item
  Yann Bugeaud, \emph{Linear forms in p-adic logarithms and the
  Diophantine equation \((x^n-1)/(x-1)=y^q\)}, Mathematical Proceedings
  of the Cambridge Philosophical Society 127 (1999), 373--381.
  \href{https://irma.math.unistra.fr/~bugeaud/travaux/shopuissdef.ps}{Author-hosted
  full text}.
\item
  Kunrui Yu, \emph{p-adic logarithmic forms and a problem of Erdős},
  Acta Mathematica 211 (2013), 315--382.
  \href{https://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6782-11511_2013_Article_106.pdf}{Free
  university-hosted full text}.
\item
  Thomas Loher and David Masser, \emph{Uniformly counting points of
  bounded height}, Acta Arithmetica 111 (2004), 277--297.
  \href{https://www.impan.pl/shop/en/publication/transaction/download/product/82907}{Free
  publisher full text}.
\item
  Kunrui Yu, \emph{Linear forms in p-adic logarithms II}, Compositio
  Mathematica 74 (1990), 15--113.
  \href{https://www.numdam.org/item/CM_1990__74_1_15_0.pdf}{Free Numdam
  full text}.
\item
  N. Costa Pereira, \emph{Elementary estimates for the Chebyshev
  function \(\psi(x)\) and for the Möbius function \(M(x)\)}, Acta
  Arithmetica 52 (1989), 307--337.
  \href{https://www.impan.pl/shop/en/publication/transaction/download/product/106154}{Free
  publisher full text}.
\item
  Kunrui Yu, \emph{p-adic logarithmic forms and group varieties II},
  Acta Arithmetica 89 (1999), 337--378.
  \href{https://matwbn.icm.edu.pl/ksiazki/aa/aa89/aa8944.pdf}{Free
  publisher full text}.
\end{itemize}

\end{document}