Two p-adic logarithms: determinant bounds

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 p-adic analysis for linear forms in logarithms. The polynomial zero argument is the method proved in Two logarithms I: interpolation determinants. The weighted product formula is Places of number fields in extensions and the product formula, Theorem 5.2. The height convention and its power law are proved in 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 \(f\).

1. 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

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

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

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

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

2. Three integral determinant estimates

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

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

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

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

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. \(\square\)

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 \(0\le r_i\le R-1\),

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 \(K(R-1)L^2/8=NL(R-1)/8\). \(\square\)

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

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. \(\square\)

3. A close root and the analytic determinant

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

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

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

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 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

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

then

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

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

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

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

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

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

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

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

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 \(m\).

4. The arithmetic determinant and a parameter theorem

For the selected columns define

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

Proof. The polynomial

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

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

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

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). \(\square\)

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

If

then

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. \(\square\)

5. A bound with the height floor \(1/d\)

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

Define

Then

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

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

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

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

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 \(x\ge1\),

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

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

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

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,

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

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

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

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

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 \(a_1\ln p\ge1\),

In the injective case,

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

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 \(H\).

6. Coprime powers and a worked example

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

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

Use \(\ln A_1=\ln(5a)\), \(\ln A_2=\ln5\), and again \(b_1=b_2=p-1\). The exact identity

preserves the valuation, since \(c\ne p\).

In both cases the height choices are the stated \(\ln(5a),\ln(5b)\), and

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

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.

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

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

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

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. Figure program.

7. A deep-unit refinement

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

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

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

Then

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

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

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

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

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

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

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 \(z\ge G\),

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,

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

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

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

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

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

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 \(m\).

8. 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

No choice of complex logarithms is involved in this definition.

The complete finite-height count in 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

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

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,

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

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 Heights of algebraic numbers, Proposition 2.5, give

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 \(q^a\). \(\square\)

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 \(q\nmid a\).

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

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

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

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

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. \(\square\)

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

is consequently well defined.

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

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

9. 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 \(K\).

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

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

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

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 unselected archimedean weights is exactly \(\ell(d-f)/d\). Multiplying their bounds and the finite-place bounds gives

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

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

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 \(N\).

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

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

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. \(\square\)

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

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

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

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,

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 \(w_K[M_q:\mathbb Z^n]\).

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. \(\square\)

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

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

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. \(\square\)

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: height_counting_geometry.py.

10. 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.

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

Consequently,

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

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

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 Heights of algebraic numbers and 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 p-adic logarithms: power series, heights and the one-logarithm bound give

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 \(a<t\),

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

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 the same lesson, Section 5 gives

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

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

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

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 \(g_\gamma=|\langle\bar\gamma\rangle|\),

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.

11. 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.

A minor of minimum valuation

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

The derivations

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

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.

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

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.

Weighted relations with a sharp degree constant

We shall use the complete upper successive-minima inequality and subgroup-minor calculation in Modern estimates: Matveev and Waldschmidt. The following deduction uses the Euclidean radius \(1/\sqrt r\) in the weighted estimate.

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

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

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

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 \(K_{r,\nu}\le r^\nu\). \(\square\)

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 \(r\).

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

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

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

holds.

Proof. We use the fully proved stabilizer, jet and coordinate-degree arguments of the 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 \(J\),

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.

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

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 \((\prod_j\vartheta_j^{b_j})^m=\prod_i\gamma_i^{c_i}\).

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. \(\square\)

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

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: \(\sigma=\nu-1=0\).

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. Figure program.

12. 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

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

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

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

Suppose a positive \(\sigma\) satisfies

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

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

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,

Indeed 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

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

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

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

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.

13. 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.

Taylor coefficients at integral and fractional nodes

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

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

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

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.

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 \(j\ge1\),

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. \(\square\)

Independence of the shifted powers

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

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

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.

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. Figure program.

Affine changes of binomial jets

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

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.

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.

From prepared equations to the ordinary zero estimate

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

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

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 \(T\).

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

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

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

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.

14. Constructing the initial auxiliary coefficients

The prepared equations are a finite homogeneous linear system. The weighted Siegel lemma has a complete proof in 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

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 \(h_\infty(c)\le h_2(c)\).

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

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

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.

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

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

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

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

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.

15. P-adic interpolation at integer nodes

We prove the analytic extrapolation estimates directly. The power-series algebra and its norm are those of Convergence and legitimate substitution, Lemmas 9.1–9.2. We use \(v_p(p)=1\) and \(v_p(0)=+\infty\). A 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 \(F_j(a)=F^{(j)}(a)/j!\).

Division and the Hermite remainder

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

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 \(a_s\).

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. \(\square\)

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

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

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

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

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

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). \(\square\)

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

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

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 \(v_p(F(\rho x))\ge\min\{\mu\theta,\Lambda\}\).

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 \(\mathbb Z_p\). \(\square\)

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. Figure program.

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.

The error from projecting a logarithmic curve

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

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

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 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).

Normalizing the prepared series

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

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

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

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

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.

16. 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.

Differentiating along the projected curve

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

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

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 \(h\ge0\),

Its scalar has valuation at least \(-h v_p(v(k))\).

For an indeterminate \(Z\), the defining product for \(\Delta\) gives

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. \(\square\)

Retaining the precision of each jet

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

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 \(0\le j\le\mu-1\),

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.

The input precision for extrapolation

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 \(x\in\mathbb Z_p\),

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

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

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.

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

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.

17. 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.

A nonzero prepared value

Use the height conventions of Section 14 and the product formula proved in Places of number fields in extensions and the product formula. The height identities for powers and inverses are proved in Heights of algebraic numbers.

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

If \(V\ne0\), then

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.

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

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

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. \(\square\)

Fractional nodes and the coefficient field

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

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

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.

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.

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. Figure program.

An explicit zero-forcing step

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

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

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

Proof. The first line of (10.120) supplies the analytic budget of Corollary 10.40. At every stated index its conclusion gives

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.

18. 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.

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

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

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

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 \([\mathrm e(1+(\Omega_\Sigma+r-1)/h)]^h\).

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

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. \(\square\)

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

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

A simpler consequence is

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

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. \(\square\)

The smaller nonzero-value budget

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

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

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 \(\mathcal A_I^*(X,T')\le\mathcal A_I(X,T')\).

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.

19. 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.

Separating the root monomials

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 \(\boldsymbol\rho\in\{0,\ldots,q-1\}^r\),

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.

The next prepared family

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

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

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

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

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

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 \(q=2\).

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. Figure program.

Root-of-unity phases and the coefficient field

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

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 \(P=p^a\), \(a\ge0\).

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.

20. 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.

The congruence and its phase classes

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:

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 \(K\).

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 \(s\). \(\square\)

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

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

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 \(\iota\in\Lambda\).

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

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

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

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

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

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 \(q\mid G_0\).

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. Figure program.

The analytic branch and arithmetic height

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

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

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.

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). \(\square\)

Constructing the local phase

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

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

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.

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

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. \(\square\)

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 \(K\).

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

If

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

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 \(T'=O\).

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

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.

21. 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.

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

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

The vector may be chosen with the separate-row budget

and hence with the mean-node budget

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

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

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

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 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.

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. Figure program.

22. 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.

The support-generated root field

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

Then

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.

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

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

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 \(e,f\). \(\square\)

A nonzero fractional prepared value

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,

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).

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

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.

Closing the first fractional-node extension

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

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

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

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.

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. Figure program.

23. 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.

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

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

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 \(s\),

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

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

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 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 \(\mathfrak H_\theta(-M)\).

Contraction and the original multiplicative bases

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

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

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

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.

The coefficient and nonzero-value budgets

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

and hence with

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

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. \(\square\)

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

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

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

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.

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. \(\square\)

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. Figure program.

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.

24. 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.

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

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

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

and therefore with

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

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 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. \(\square\)

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

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

Corollary 10.64 (an initial budget in the original heights). Under these hypotheses and \(N>M_*\), the initial coefficient vector may be chosen with

Replacing \(d_0\) everywhere by its upper bound \(d\) gives a valid further bound.

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.

Substitution at the numerical scale

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

Then a coefficient vector with every zero in Theorem 10.63 exists with

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 \(1+1/(2g+1)\).

Proof. Formula (10.175) gives

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

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.

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. Figure program.

25. 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.

Points in a translated closed box

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

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

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

The volume and translation rules used here are the same ones proved in the lattice argument of 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.

Retaining a single phase

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

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 \(B\Lambda\subset C-y_0\).

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 \(g=G_0,H=1\). \(\square\)

A certified surplus of prepared columns

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

Define the support-independent row upper bound

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

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.

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. Figure program.

26. 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.

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 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

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

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

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

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.

Lemma 10.69 (the saturation factor). The parameters above satisfy \(1\le\gamma\le q^\nu\).

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 \((h+x)(A+x)\ge hA\). \(\square\)

Lower bounds and the rounded equation count

Lemma 10.70 (lower budgets without lost floors). We have

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

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 \(\mathcal D\):

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,

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). \(\square\)

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

Then

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.

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 \(L>\widetilde D_0\),

Divide by \((2S+1)(1+\epsilon)T^r/r!\). Substitute (10.186) for \(\mathcal D\). The degree factors, widths and normalization factors cancel, leaving

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. \(\square\)

The height term with a real node budget

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

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

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. \(\square\)

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 Yu's free paper for further reading.

27. 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.

The degree available above an unramified root field

Lemma 10.73 (the torsion degree divided by ramification). With the notation of (10.183), we have

Proof. Put \(F=\mathbb Q(\zeta_{q^u})\subseteq K\). The cyclotomic Eisenstein proof in 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 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 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 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. \(\square\)

A lower volume and an upper coefficient count

Retain all real parameters in (10.183)–(10.186). Set

Lemma 10.74 (a uniform lower box volume). We have

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

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

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

Proof. By (10.180), \(|\Lambda|\le v+1\). Also

Use \(v\ge g_{61}\) to replace \(v+1\) by at most \((1+g_{61}^{-1})v\). The full saturation bound (10.55) gives

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

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

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

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.

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

Lemma 10.76 (a complete scalar row bound). The scalar bound in Lemma 10.63 may be taken to be

If \(\mathfrak v=1\), a maximizing index is \(u_*=0\). If \(\mathfrak v>1\), one is

More sharply, with \(R=s_0+2k\), the scalar bound may be taken to be

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

These steps form an initial interval, so an integer binary search finds a maximizing index, with no numerical root approximation.

Proof. For \(0\le a<k\), \(1\le\ell\le L\), write

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

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

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 \(u<t_0\),

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

Multiply by the clearing and Euler factors already proved. The ratio of the resulting expression at \(u+1\) to that at \(u\) is

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. \(\square\)

Corollary 10.77 (the normalized additive cost). If \(c_3q/t\le8\), the first factor in (10.198) contributes at most

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

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 \(\ln2>2/3\). \(\square\)

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

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. \(\square\)

Averaging the derivative orders

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

Let \(H(z)=-z\ln z-(1-z)\ln(1-z)\), with endpoint values zero, and put

The mean logarithm of the scalar row bounds is at most \(B_{\rm mean}\), and an integral initial kernel exists with

Moreover \(\overline u\le t_0/(r+1)\).

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

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. \(\square\)

A second way to clear the global rows

Corollary 10.80 (row scaling and an alternative mean bound). Define

In (10.206), \(B_{\rm mean}\) may be replaced by

One may also choose the smaller scalar bound separately at each row before averaging.

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 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

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. \(\square\)

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 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 Yu's free paper for further reading.

28. 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.

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

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.

Lemma 10.81 (the original principal-unit depth). If \(\beta\in L\) and \(v_p(\beta-1)>0\), then

The statement includes \(\beta^P=1\), with valuation infinity.

Proof. The valuation group of \(L\) is \(e^{-1}\mathbb Z\), by the complete degree-and-residue proof in Extensions of complete valued fields, Theorem 3.1. Thus \(v_p(\beta-1)\ge1/e\). The elementary power step in 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

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). \(\square\)

Coefficients and divided jets of the torus functions

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

Then \(F\in\mathcal A_{p^\theta}\), its Gauss norm is one, and for \(n\ge1\)

If \(v_p(z)\ge-\theta\), its divided jets satisfy

If \(\ell=0\), the positive-order coefficients and jets are zero. The constant coefficient and value have valuation zero.

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.

Residue phases and a coherent root choice

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

For a prime \(q\ne p\), the element

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

The choice may be made separately for every member of a finite family, without changing any of their \(q\)-th power identities.

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. \(\square\)

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, the figure program, and Yu's free paper, equations (2.1), (4.7) and (4.11), for further reading.

29. 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.

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

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

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),

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

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 \(J\).

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 \(\overline h>0\).

Lemma 10.85 (one budget for clearing and Euler factors). With \(\mathfrak v=\operatorname{lcm}(1,\ldots,k)\), put

Proof. The exact composition means in (10.204) give

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

Thus \(g_{91}\) is \(1+(1+3\ln\ln(3d))/g_0\) for \(d\ge2\). Assume in addition to Section 26 that

The added height condition is exactly \(R_{\max}\le dW(d)\mathrm e^h\). It is the coefficient-height condition associated with (10.218).

Theorem 10.86 (the original uniform mean Euler budget). Under these assumptions,

and

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

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

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

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

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 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

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). \(\square\)

The full initial coefficient-height bound

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

In particular

With \(d_0\) the actual row-field degree, its coefficients satisfy

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

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. \(\square\)

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. Figure program. Human-source context: Yu's free paper, equations (2.8), (4.23) and (5.6).

30. 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.

A centered basis with small generator heights

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:

Choose \(\theta'_j=\prod_i\theta_i^{U_{ij}}\). Then

for any root of unity \(\alpha_0\). This change preserves every normalized monomial, phase class, original support width and integer Euler argument after relabelling \(\lambda'=U^{-1}\lambda\).

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.

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

These are the original constants in Yu's equation (3.16); \(\mathrm e=\exp(1)\) is distinct from the ramification index.

Lemma 10.89 (the original omitted-product bound). For \(Z=S\mathcal D/d\) and every \(i=1,\ldots,r\),

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

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

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. \(\square\)

A uniform coefficient-field and initial height bound

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

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

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 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

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.

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. Figure program. Human-source context: Yu's free paper, equations (3.16), (3.23) and Lemma 4.2; Loher and Masser's free paper, Theorem 3, whose complete height-product proof is given in Section 9.

31. 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.

The actual projected slopes and the additive order

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

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:

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.

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

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.

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

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

Uniformly through total order \(T'\), its scalar bound is

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

At \(x=s/q^J\), \(s\in\mathbb Z\), all scalar denominators divide \(q^{\Xi_{I,J,u}}\), where

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

The field factor is retained even when these roots lie in the old completion.

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

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

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). \(\square\)

All three first integer extensions in the rational parameter family

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

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

The allowed coefficient family in (10.245), rather than one chosen pair \(b_1,b_2\), is covered by these comparisons.

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

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

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

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

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.

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.

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. Figure program. Human-source context: Yu's free paper, Lemma 5.2 and equations (5.25)–(5.38).

32. 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.

Hermite interpolation on nested intervals

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

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

Proof. For \(s\in E_m\), form the rational polynomial

It has value one at \(s\) and a zero of order \(\mu(t)\) at every other node \(t\). We claim

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

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

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

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

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

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

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

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.

The original first fractional range

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

This range uses the original unfloored \(S,T\), and the coefficient family is the whole family (10.245).

Proof. Theorem 10.93 and the initial construction give these nested integer data.

For output order \(T'=993090\), take

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

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

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

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

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.

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.

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. Figure program. Human-source context: Yu's free paper, Lemma 5.3.

33. 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.

Odd nodes inside nested full intervals

Lemma 10.96 (mixed nested interpolation). Let \(p\ne q\) be primes and \(1\le A\le R\) integers. Put

where \(\mu_0\ge\mu_1\ge\mu_2\ge1\). Define

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

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

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

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

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

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.

What each contraction preserves

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

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.

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,

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.

All stages with their original radii and orders

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

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 \(O_3\).

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

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

For the next fractional step use Lemma 10.96 with \(q=2\), \(A=A_I\), \(R=R_I\), and

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

Set \(r=X/2^I+99\), \(D=D_{\mathrm a}\), and \(V=3^{50}\). The pointwise scalar bound in Theorem 10.92 is

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

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 \((U+G_0)/Z\ge7.294535093395\).

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. \(\square\)

The last support has one exponent

Corollary 10.99 (a complete valuation bound for the stated family). For every pair of nonzero integers \(b_1,b_2\) satisfying (10.245),

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

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.

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. Figure program. Human-source context: Yu's free paper, Lemmas5.3–5.4 and the support passage in Section5.

34. 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:

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

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

Theorem 10.100 (all geometric-stop degree inequalities). Under these hypotheses, the allocations satisfy (10.67), and

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

They satisfy the uniform strict bounds

Thus every allocation and degree hypothesis of Theorem 10.25 holds, for every original rank, field and prime parameter case.

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.

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

Since \(r!\le r^r\), one has \(F_r\ge(1+1/r)^r>2\). The definitions (10.185) consequently imply

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

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

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,

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

Similarly the additive degree divided by the real output order satisfies

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 \(\mathsf D_0>2(\overline T+r)>T_r+r\).

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

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

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

All these subfamilies are independent, so Theorem 10.18 applies separately to each one:

Furthermore \(E_I/q^\nu>\exp(3A)\), and

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

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

Since \(\delta\ge1\), \(t>2/3\), \(p^f=\exp t\le\exp A\), and \(r\ge2\), \(M<\exp A(3/2)^r\). The identities

retain the saturation factors before they are bounded. Cancellation of \(\mathcal D\) yields

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

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. \(\square\)

The resulting bases remain in the original field

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

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 \(c_i\),

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.

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. Figure program. Human-source context: Yu's free paper, Section 6.

35. 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

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).

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 \(n\ge r\),

This preserves the original residue-index-dependent allowance, including the different definition at rank one.

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,

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

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

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

Combine this with \(t^mM_m\le p^ft\le\exp(A)A\), \(\mathrm e<3\), and \(\eta>18/25\). It follows that

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

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

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). \(\square\)

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

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 \(B_j\ne0\).

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.

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

Thus every required independence, representation and weighted-height condition survives. Finally Theorem 10.102 and (10.277) give

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. \(\square\)

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

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.

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. Figure program. Human-source context: Yu's free paper, (2.10) and (6.18)–(6.19).

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.

36. 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).

Every terminating decimal in this table denotes an exact rational number. Put

Coupled rounding and coefficient-count bounds

Lemma 10.104 (uniform lower budgets in every original case). For every integer \(r\ge2\), the table and (10.183)–(10.186) give

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

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

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

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

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 \(z\ge0\),

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. \(\square\)

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

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

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 \(\mathcal D>b_r(A+x)t>5b_r\),

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

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 \(Z=S\mathcal D/d\). \(\square\)

The original height parameter

For the next average take the original

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.

Lemma 10.106 (the general Euler logarithm constant). For these original choices, \(g_{91}\) of (10.222) satisfies

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 \(z=y+\ln3\),

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 \(\ln6<9/5\). \(\square\)

Keeping the zero rows in the simplex

Put \(t_0=\lfloor T\rfloor\), and now retain 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 \(t_0\).

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

where \(s_0=\lfloor S\rfloor\), \(H_1=h+x\). In fact

Proof. Permutation of the \(r+1\) composition coordinates is a bijection. Their means are therefore exactly \(t_0/(r+1)\). Consequently

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

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

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. \(\square\)

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

Here \(g_{12}=0\) when \(d=1\); in every case one may further use

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

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. \(\square\)

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.

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

If the actual value \(V\in F\) is nonzero, then

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 \(F,d_F,e_F,f_F\).

Proof. Multiply the whole algebraic value, at this one fixed additive order, by the common nonzero rational scalar

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

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. \(\square\)

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. Figure program. Human-source background: Yu's free paper, equations (1.11), (3.12)–(3.18) and the initial row system in Section 4.

37. 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

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 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.

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

The symbols \(v_r\) are weights; they are distinct from the integer \(v(k)\). Put

As usual, \(0!=1\).

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

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. \(\square\)

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.

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 \(1\le y<a_s\), \(s=2,3,4\).

Lemma 10.111 (exact finite certificate). The cutoffs and every interval in the table have the asserted property. Set

With \(Q=10^{12}\), \(N=10^6\), \(U=27/26\), \(L=25/26\), and \(\varepsilon=9/40000\), the following rational inequalities hold:

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 \(U<\psi(113)/113<107/103\).

Proof and reproducible finite computation. The full standard-library program 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

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:

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. \(\square\)

Continuing the bounds to every real argument

Theorem 10.112 (a sharp elementary lcm bound). For every real \(x>0\),

Consequently (10.299) holds at every positive integer \(k\).

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:

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 \(\ln v(k)=\psi(k)\). \(\square\)

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

The individual clearing factor also satisfies \(u\ln v(k)<(107/103)uk\) whenever \(u>0\), and is zero when \(u=0\).

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. \(\square\)

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. Figure program. Human method and interval choices: Costa Pereira's free paper, pages 321–323; all required identities, finite checks, error bounds and induction are given above.

38. 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 \(g_9=\max\{g_{91},107/103\}\).

The exact sharp regime for the Euler constant

Lemma 10.114. For every original case and every \(r\ge2\),

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). \(\square\)

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.

Theorem 10.115 (an individual clearing/Euler bound). For every pair of nonnegative integers \(u,h_{\mathrm E}\),

In the original parameters the remainder satisfies

The symbol \(\mathrm e\) in this formula is \(\exp(1)\), and \(P=p^\kappa\) is the original prime power.

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

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:

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. \(\square\)

Retaining the factorial and the actual field

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

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

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).

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. \(\square\)

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. Figure program. Human background: 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.

39. 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.

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

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.

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

If \(u>D_{\mathrm a}\), both prepared functions vanish identically.

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

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.

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

The minimum is over integers. It computes a lower-bound envelope for the jets; it does not assert that a jet attains that valuation.

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 \(|\boldsymbol t|+j\le T\),

Here \(F_u,G_u\) are exactly (10.238), and the subscript \(j\) is ordinary divided differentiation. In particular, with

one may use the linear loss \(C_{\mathrm{sharp}}\) instead of \(C_0\). Formula (10.316) can be stronger than this linear consequence.

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

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

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. \(\square\)

Full, nested and q-deleted interpolation

For an integer maximum multiplicity \(M\ge1\) and separation bound \(B\ge0\), put

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.

Theorem 10.119 (interpolation at the retained jet precision). Use any one of the following proved node systems:

• The full nested intervals of Lemma 10.94. Set \(N_*\) as in (10.247), \(M=\mu_0\), and \(E=0\).
• The mixed q-deleted and full intervals of Lemma 10.96. Set \(N_*\) as in (10.251), \(M=\mu_0\), and \(E=\mu_0-\mu_1\).
• 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 \(E=M\).

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 \(x\in\mathbb Z_p\subset\mathbb Q_p\),

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

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.

The comparison with an actual algebraic value

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

In particular the sufficient input budget

and the strict comparison \(N_*\theta>\mathcal A_u(x)\) imply that zero.

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.

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. Figure program. Human-source context: the original ordinary-jet extrapolation in Yu's freely accessible paper, Section 5, equations (5.25)–(5.32); the order-dependent refinement is proved above.

40. 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.

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

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.

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.

Coupled field bounds

Lemma 10.121 (surplus and depth in the same field). Write

The following bounds hold simultaneously. In the third column, the lower bound for \(Y\) is the displayed number divided by \(1+10^{-6}\).

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 \(\Omega_{\mathrm n}\le3/2\). \(\square\)

The additive polynomial and the contracted denominator

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

At \(I=0\) there is no q-denominator. At \(I\ge1,j\ge1\), its full denominator cost and the torus height obey

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

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 \(\exp(-3D)+3\lambda c_2\ln q/(c_4q^{j+1}\ln Q)\).