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\section{The theorem on formal
functions}\label{the-theorem-on-formal-functions}

\emph{Written by GPT-6.1 Sol (OpenAI), in Codex, at Ultra effort,
October 2026; Appendix Z by Claude Opus 5.5 (Anthropic), October 2026.
Self-checked by the writing AI. Original text: public domain (CC0). See
the \href{licensing.html}{course notice} for authorship.}

A fiber remembers a family at one parameter. Its successive
infinitesimal neighborhoods remember powers of that parameter's maximal
ideal. Formal functions identifies the inverse limit of their cohomology
with the completion of a finite cohomology module. The individual
comparison maps can fail to be isomorphisms; what makes the theorem work
is a uniform bound on how long their errors persist.

We use \href{proper-morphisms-and-coherent-direct-images.md}{proper
coherent direct images},
\href{affine-cohomology-and-serres-criterion.md}{affine cohomology and
Serre's criterion}, and
\href{base-change-and-the-grothendieck-complex.md}{flat base change}.
\href{../AG-CA/AG-CA-19.html\#3-noetherian-exactness-tensoring-and-flatness}{Completion,
Theorems 3.1--3.3} supplies exact completion of finite modules over
Noetherian rings and faithful flatness when the completion ideal lies in
the Jacobson radical, including maximal-ideal completion of a Noetherian
local ring. The dimension consequences in Section 5 use
\href{../ag-etale-cohomology/cohomological-dimension-and-the-kunneth-formula.html\#7-the-bound-for-all-finite-type-schemes}{Cohomological
dimension and the Künneth formula, Lemma 7.1}, the full Noetherian
topological vanishing proof for arbitrary abelian sheaves. No flatness
assumption on the coherent sheaf is imposed in this lesson.

\subsection{0. A normality input in the algebraic Zariski Main
proof}\label{a-normality-input-in-the-algebraic-zariski-main-proof}

The quasi-finite input used in Section 5 has an algebraic Zariski Main
proof whose supporting coefficient argument requires polynomial
normality over an arbitrary normal domain. A proof for polynomial rings
over fields alone would not suffice. We supply the general statement
here. The valuation existence input is the written
\href{../AG-GS/prerequisites/AG-MO/valuation-rings-and-separatedness.html}{Valuation
rings and separatedness, Theorem 2.1}; it gives a valuation ring in a
specified field dominating a specified local subring, without a
Noetherian hypothesis or a discrete-value assumption.

\textbf{Lemma 0.1 (polynomial normality).} If \(R\) is an integrally
closed domain, then \(R[x_1,\ldots,x_n]\) is an integrally closed domain
for every finite \(n\).

\textbf{Proof.} Put \(F=\operatorname{Frac}R\). First we prove that
\(R\) is the intersection of all valuation subrings of \(F\) containing
\(R\). If \(z\in F\setminus R\), then \(z\) is not integral over \(R\).
In the ring \(R[z^{-1}]\), the element \(z^{-1}\) is not a unit: an
equation \[
1=z^{-1}(a_0+a_1z^{-1}+\cdots+a_mz^{-m})
\] would yield a monic polynomial equation for \(z\) over \(R\), after
multiplication by \(z^{m+1}\). Choose a maximal ideal containing
\(z^{-1}\) and localize at it. The earlier valuation theorem gives a
valuation ring \(V\subset F\) dominating this local ring. Thus
\(R\subset V\), while \(z^{-1}\) is in its maximal ideal and
\(z\notin V\). This separates every \(z\notin R\) from the intersection
and proves the assertion.

For such a valuation ring, let \(v:F^*\to\Gamma\) be its ordered value
group, with \(V=\{0\}\cup\{a:v(a)\geq0\}\). This group can be
constructed as \(F^*/V^*\), ordered by divisibility in \(V\). For a
nonzero polynomial \(h=\sum a_ix^i\in F[x]\), define \[
w(h)=\min_i v(a_i),
\] ignoring zero coefficients. We have \(w(hg)=w(h)+w(g)\): divide each
polynomial by a coefficient of least value. The resulting polynomials
have coefficients in \(V\) and nonzero reductions in
\((V/\mathfrak m_V)[x]\); their product has nonzero reduction because
this polynomial ring over a field is a domain. Also
\(w(h+g)\geq\min(w(h),w(g))\), with equality when the two values differ.
Consequently a sum with a unique term of strictly least value cannot
vanish.

Let \(u\in F(x)\) be integral over \(R[x]\). It is integral over
\(F[x]\), so \(u\in F[x]\). To justify this last assertion directly,
write \(u=a/b\) with coprime polynomials in the Euclidean domain
\(F[x]\). A monic integral equation implies that \(b\) divides \(a^m\).
Bézout's identity for \(a,b\) forces \(b\) to be a unit. Thus \(u\) is a
polynomial.

For every valuation ring \(V\) above, write its integral equation as \[
u^m+c_1(x)u^{m-1}+\cdots+c_m(x)=0,
\qquad c_i(x)\in R[x]\subset V[x].
\] If \(w(u)<0\), the term \(u^m\) has value \(m w(u)\), strictly less
than the value of every other nonzero term, since \(w(c_i)\geq0\). This
contradicts the least-value property. Hence every coefficient of \(u\)
belongs to \(V\). The intersection assertion puts all coefficients in
\(R\), proving that \(R[x]\) is integrally closed. Induction on the
number of variables proves the lemma. \(\square\)

This proof supplies the polynomial-normality step of the supporting
algebraic argument. Not all of the other inputs of that argument are
proved here.

\subsection{1. The completion map}\label{the-completion-map}

Let \(A\) be Noetherian, \(I\subset A\) an ideal,
\(f:X\to\operatorname{Spec}A\) proper, and \(F\) coherent. For
\(n\geq1\), put \[
X_n=X\times_A\operatorname{Spec}(A/I^n),\qquad F_n=F|_{X_n}.
\] The closed immersion \(i_n:X_n\hookrightarrow X\) has
\(i_{n*}F_n=F/I^nF\). Since closed immersion pushforward is exact and
has no positive direct images, we may compute \[
H^q(X_n,F_n)=H^q(X,F/I^nF).
\] Set \(M^q=H^q(X,F)\), a finite \(A\)-module by proper finiteness. The
quotient map on sheaves induces \[
M^q/I^nM^q\longrightarrow H^q(X_n,F_n),
\] because \(I^n\) annihilates the target. These maps are compatible
with reduction from \(n+1\) to \(n\). We seek an isomorphism from \[
\widehat{M^q}=\varprojlim_n M^q/I^nM^q
\] to the inverse limit of the right-hand terms. Completion here is with
respect to \(I\); neither \(A\) nor \(M^q\) is assumed complete in
advance.

The theorem is not a claim that every finite-level map is an
isomorphism. For example, the extension bundle on \(\mathbf P^1_{k[t]}\)
with class \(t\) has \(H^0(X,E)=0\), but has a nonzero section on the
fiber \(t=0\). We will return to its infinitesimal neighborhoods in
Exercise 7.6.

\subsection{2. All powers at once}\label{all-powers-at-once}

The \textbf{Rees algebra} is \[
R=\bigoplus_{n\geq0}I^nT^n\subset A[T],\qquad R_0=A.
\] If \(I=(a_1,\ldots,a_r)\), then \(R\) is generated by the degree-one
elements \(a_iT\), hence is a quotient of a polynomial algebra in
finitely many variables over \(A\). It is Noetherian. The symbol \(T\)
records degree; it need not itself belong to \(R\).

\textbf{Lemma 2.1 (graded cohomological finiteness).} For each
\(q\geq0\), the graded module \[
Q^q=\bigoplus_{n\geq0}H^q(X,I^nF)T^n
\] is finite over \(R\).

\textbf{Proof.} Form \(X_R=X\times_A\operatorname{Spec}R\), with affine
projection \(\pi:X_R\to X\). On \(X\), the graded sheaf
\(\bigoplus I^nF\) is a module over \(\mathcal O_X\otimes_A R\): a
degree-\(d\) element acts by multiplication into the degree shifted by
\(d\). It is a quotient of \(F\otimes_A R\), by the degreewise maps
\(F\otimes_A I^n\to I^nF\). The affine module/sheaf correspondence gives
a coherent sheaf \(G\) on \(X_R\) with \[
\pi_*G=\bigoplus_{n\geq0}I^nF.
\] Coherence follows because \(X_R\) is Noetherian and this sheaf is
locally a finite module, generated by a finite set of degree-zero
generators of \(F\).

The map \(X_R\to\operatorname{Spec}R\) is proper by base change. Proper
cohomology finiteness makes \(H^q(X_R,G)\) finite over \(R\). Affine
pushforward and Leray identify it with \(H^q(X,\pi_*G)\). Finally
cohomology of quasi-coherent sheaves on the quasi-compact,
quasi-separated scheme \(X\) commutes with filtered colimits, by the
third lesson. A direct sum is the filtered colimit of finite partial
sums, so \[
H^q(X,\pi_*G)=\bigoplus_{n\geq0}H^q(X,I^nF).
\] The identifications preserve multiplication and grading. This is the
asserted finite graded module. \(\square\)

We next extract two consequences. Write \[
J_n^q=\operatorname{im}\bigl(H^q(X,I^nF)\to M^q\bigr),
\qquad K_n^q=\ker\bigl(H^q(X,I^nF)\to M^q\bigr).
\] Both collections respect graded multiplication. Indeed inclusion of
ideal powers commutes with multiplying a section or cohomology class by
an element of \(I^d\). Thus \(\bigoplus J_n^qT^n\) is a finite graded
image module of \(Q^q\), and \(\bigoplus K_n^qT^n\) is a finite graded
submodule, using Noetherianness of \(R\).

\textbf{Lemma 2.2 (bounds on the image and kernel).} There are integers
\(c_J,c_K\geq0\) such that \[
I^nM^q\subset J_n^q\subset I^{n-c_J}M^q\quad(n\geq c_J),
\] and the transition map \(K_m^q\to K_n^q\) is zero whenever
\(m\geq n+c_K\).

\textbf{Proof.} Multiplication by an element of \(I^n\) factors as
\(F\to I^nF\to F\), proving the first inclusion. Choose finitely many
homogeneous generators of the image module, all in degrees at most
\(c_J\). The degree-\(n\) part is a sum of \(I^{n-d}J_d^q\) over
generator degrees \(d\leq c_J\). Each is contained in \(I^{n-c_J}M^q\),
proving the second inclusion. This is the cohomological analogue of an
Artin--Rees bound.

Choose homogeneous kernel generators \(k_\ell\in K_{d_\ell}^q\) with
\(d_\ell\leq c_K\). An element of \(K_m^q\) is a sum of products
\(a_\ell k_\ell\), with \(a_\ell\in I^{m-d_\ell}\). If \(m\geq n+c_K\),
each such coefficient is a sum of products \(ba'\), where \(b\in I^n\)
and \(a'\in I^{m-d_\ell-n}\). Under transition to \(H^q(X,I^nF)\), its
product with \(k_\ell\) is computed by first including \(I^{d_\ell}F\)
into \(F\), then multiplying by \(a'\), then by \(b\) into \(I^nF\). The
initial inclusion sends \(k_\ell\) to zero, by definition of the kernel.
Hence every product maps to zero. This proves the uniform transition
bound. \(\square\)

A system whose sufficiently distant transition maps to each fixed term
are zero is called \textbf{pro-zero}. Its individual terms may be
nonzero; its inverse limit is zero because each coordinate of a
compatible family is the image of a distant, necessarily zero,
transition. Lemma 2.2 supplies this stronger, bounded form of
disappearance for the kernel system.

\subsection{3. Stabilization of the thickened
cohomology}\label{stabilization-of-the-thickened-cohomology}

The short exact sequence \(0\to I^nF\to F\to F/I^nF\to0\) gives the
exact sequence \[
0\to M^q/J_n^q\longrightarrow H^q(X_n,F_n)\longrightarrow K_n^{q+1}\to0.
\tag{1}
\] It is a sequence of inverse systems. The left transitions are
surjective because \(J_m^q\subset J_n^q\) for \(m\geq n\). The right
system is pro-zero by Lemma 2.2 in degree \(q+1\).

\textbf{Proposition 3.1 (the Mittag--Leffler bound).} For fixed \(q\),
choose \(c\) as the kernel bound in degree \(q+1\). Then for
\(m\geq n+c\), \[
\operatorname{im}\bigl(H^q(X_m,F_m)\to H^q(X_n,F_n)\bigr)
=\operatorname{im}\bigl(M^q\to H^q(X_n,F_n)\bigr).
\] Consequently the thickened-cohomology system is Mittag--Leffler.

\textbf{Proof.} In diagram (1) for indices \(m,n\), the right transition
is zero. Therefore the middle transition lands in the left submodule
\(M^q/J_n^q\). It contains all of that submodule: the left transition
\(M^q/J_m^q\to M^q/J_n^q\) is onto, and these left modules inject into
the respective middle terms. Thus its image is exactly the claimed
submodule. The image is independent of all sufficiently large \(m\),
which is the Mittag--Leffler condition. \(\square\)

We can now take the limit without an unexplained exchange of limits and
cohomology. If \((u_n)\) is a compatible family of middle classes in
(1), its images in \(K_n^{q+1}\) form a compatible family in a pro-zero
system, so are all zero. Thus each \(u_n\) lies in the injective left
term, with a unique preimage. These preimages are automatically
compatible. We obtain the canonical identification \[
\varprojlim_n H^q(X_n,F_n)=\varprojlim_n M^q/J_n^q.
\tag{2}
\] This also proves why a transient class on a small thickening need not
define a class in the inverse limit. Compatibility with every higher
thickening is a real constraint.

\subsection{4. Global and stalk forms of formal
functions}\label{global-and-stalk-forms-of-formal-functions}

\textbf{Theorem 4.1 (formal functions).} Under the hypotheses of Section
1, the canonical maps give an isomorphism \[
\widehat{H^q(X,F)}\xrightarrow{\sim}\varprojlim_n H^q(X_n,F_n).
\] It is an isomorphism of \(\widehat A\)-modules and a homeomorphism
for their inverse-limit topologies.

\textbf{Proof.} Combine (2) with the inclusions from Lemma 2.2: \[
I^nM^q\subset J_n^q,\qquad J_{n+c_J}^q\subset I^nM^q.
\] They say that the two descending filtrations \((I^nM^q)\) and
\((J_n^q)\) are cofinal. The identification (2) also preserves the limit
topologies: each left term is a discrete submodule of the corresponding
middle term, and its unique coordinate preimage is continuous on that
submodule. The first inclusions give the canonical continuous map
between the inverse limits of the corresponding quotients. The second
give its inverse by taking the cofinal shifted indices \(n+c_J\). The
two composites are the usual transitions and hence induce the identity
on limits. Both maps are continuous for the quotient limit topologies,
proving the homeomorphism. The maps arise from \(A\)-linear maps at all
levels, where \(I^n\) annihilates the cohomology of \(F_n\), and so
preserve the induced \(\widehat A\)-action. \(\square\)

\textbf{Corollary 4.2 (the stalk form).} Let \(f:X\to S\) be proper,
\(S\) locally Noetherian, \(F\) coherent, and \(s\in S\). Put
\(A_s=\mathcal O_{S,s}\), with maximal ideal \(\mathfrak m_s\), and \[
X_n=X\times_S\operatorname{Spec}(A_s/\mathfrak m_s^n),\qquad F_n=F|_{X_n}.
\] Then \[
\widehat{(R^qf_*F)_s}\cong\varprojlim_n H^q(X_n,F_n)
\] as \(\widehat{A_s}\)-modules.

\textbf{Proof.} The canonical map \(\operatorname{Spec}A_s\to S\) is
flat: on an affine neighborhood it is localization. Flat base change and
the affine description of quasi-coherent direct images identify the
stalk with
\(H^q(X\times_S\operatorname{Spec}A_s,F|_{X\times_S\operatorname{Spec}A_s})\).
The local ring is Noetherian, the changed morphism proper, and the
changed sheaf coherent. Apply Theorem 4.1 with ideal \(\mathfrak m_s\).
Its thickenings are precisely the displayed \(X_n\), since their
parameter maps factor through \(\operatorname{Spec}A_s\). \(\square\)

\subsection{5. Dimension and finiteness
consequences}\label{dimension-and-finiteness-consequences}

\textbf{Theorem 5.1 (vanishing above a fiber's dimension).} For the
morphism of Corollary 4.2, \[
(R^qf_*F)_s=0\quad\text{if }q>\dim X_s.
\]

\textbf{Proof.} Each \(X_n\) has the same underlying topological space
as \(X_s\): the extra ideal is nilpotent and therefore belongs to every
prime. That space is Noetherian, of finite dimension \(d\), because
\(X_s\) is of finite type over a field. The Noetherian topological
vanishing theorem,
\href{../ag-etale-cohomology/cohomological-dimension-and-the-kunneth-formula.html\#7-the-bound-for-all-finite-type-schemes}{Cohomological
dimension and the Künneth formula, Lemma 7.1}, gives \(H^q(X_n,F_n)=0\)
for \(q>d\). Formal functions makes the completed stalk zero. The
uncompleted stalk is finite by proper coherence. For a finite module
over a Noetherian local ring, completion preserves the residue-field
quotient; zero completion implies \(M/\mathfrak mM=0\), and Nakayama
implies \(M=0\). This proves the assertion. An empty fiber gives empty
thickenings and the same conclusion in every degree. \(\square\)

The vanishing is a statement about the whole stalk, stronger than a
statement about its tensor with the residue field. In particular, if all
fiber dimensions are at most \(d\), then \(R^qf_*F=0\) for \(q>d\),
without a flatness assumption.

\textbf{Theorem 5.2 (proper with finite fibers is finite).} A proper
morphism over a locally Noetherian base whose fibers have finitely many
points is finite.

\textbf{Proof.} Its fibers are finite type over fields. Such schemes
with finitely many points are zero-dimensional by Noether normalization.
Indeed an affine chart admits a finite integral surjection onto affine
space of its dimension; a positive-dimensional affine space over any
field has infinitely many points, so a finite-point chart cannot have
positive dimension. Their zero-dimensional coordinate rings are
finite-dimensional by Noether normalization. Thus Theorem 5.1 makes
\(R^1f_*G=0\) for every coherent \(G\).

Work over an affine Noetherian neighborhood \(S=\operatorname{Spec}A\).
Its inverse image \(X\) is Noetherian, separated and quasi-compact. Any
finite type ideal \(J\subset\mathcal O_X\) is coherent, and Leray over
the affine base gives \[
H^1(X,J)=\Gamma(S,R^1f_*J)=0.
\] Here affine vanishing eliminates the other Leray terms, since the
direct images are quasi-coherent. The finite-ideal version of Serre's
affine criterion from the third lesson now implies that \(X\) is affine.
Its function ring \(H^0(X,\mathcal O_X)\) is a finite \(A\)-module by
proper cohomology finiteness. Hence \(X\to\operatorname{Spec}A\) is
finite. This is local on the base and proves the theorem. \(\square\)

A quasi-finite proper morphism over a locally Noetherian base is
therefore finite. The existing earlier \emph{Zariski's Main Theorem}
lesson in \emph{Morphisms of schemes} gives another route; the present
route uses formal functions and the cohomological affine criterion.

\textbf{Corollary 5.3 (a finite fiber has a finite neighborhood).} If
\(f:X\to S\) is proper, \(S\) is locally Noetherian, and \(X_s\) is a
finite set, then \(f\) is finite over an open neighborhood of \(s\).

\textbf{Proof.} The quasi-finite locus is open by the existing earlier
\href{../AG-MO/AG-MO-12.html\#3-from-local-equalities-to-finite-affine-completions}{\emph{Zariski's
Main Theorem}, Theorem 3.1}. It contains the entire finite fiber: the
pointwise finite-fiber test is proved in \emph{Quasi-finite morphisms
and Chevalley}, Theorem 1.1. Its closed complement has closed image in
\(S\), by properness, and that image misses \(s\). Remove the image. The
restricted map has only finite fibers, and Theorem 5.2 makes it finite.
\(\square\)

\textbf{Lemma 5.4 (cartesian nilpotent thickenings).} Let
\(Y_0\hookrightarrow Y\) be defined by an ideal \(J\) with \(J^N=0\),
let \(f:X\to Y\), and put \(X_0=X\times_Y Y_0\). If \(f_0:X_0\to Y_0\)
is of finite type, a closed immersion, or proper, respectively, then so
is \(f\).

\textbf{Proof.} First a nilpotent thickening \(T\) of an affine scheme
\(T_0\) is affine. Their topological spaces coincide, so \(T\) is
quasi-compact and quasi-separated. For every quasi-coherent \(M\) on
\(T\), its finite filtration by powers of the nilpotent ideal has layers
which are quasi-coherent on \(T_0\). Affine acyclicity and the
cohomology sequences give \(H^1(T,M)=0\). The Serre affine criterion
proved in the affine-cohomology lesson now gives affineness.

Work over an affine of \(Y\), and cover \(X_0\) by finitely many affines
if \(f_0\) is of finite type. The corresponding open thickenings in
\(X\) are affine by the preceding argument. On one of them, write the
ring map \(B\to C\). If \(C/JC\) is generated over \(B/J\) by finitely
many elements, lift them and let \(D\subset C\) be the \(B\)-subalgebra
they generate. Then \(C=D+JC\); iteration gives \(C=D+J^NC=D\). This
proves finite type. If \(f_0\) is a closed immersion, \(X_0\) over this
affine is affine, hence \(X\) is affine. Surjectivity modulo \(J\) gives
\(C=\operatorname{im}(B)+JC\); the same iteration makes \(B\to C\)
surjective, proving closed immersion. These affine conclusions are local
on the target.

If \(f_0\) is proper, finite type was just proved. Its diagonal is a
closed immersion; the diagonal of \(f_0\) is the cartesian
nilpotent-base restriction of the diagonal of \(f\), so the
closed-immersion result makes \(f\) separated. Finally, after every base
change, both the source and target have the same underlying spaces as
their nilpotent-base restrictions. Images of closed subsets are
therefore closed, because the base change of \(f_0\) is closed. This
proves universal closedness and hence properness. \(\square\)

\subsection{6. The blow-up of a plane
point}\label{the-blow-up-of-a-plane-point}

Let \(b:X\to\mathbf A^2_k\) be the blow-up of the origin and
\(E\cong\mathbf P^1_k\) its exceptional curve. Its two charts are \[
U=\operatorname{Spec}k[x,v],\quad y=xv,
\qquad V=\operatorname{Spec}k[u,y],\quad x=uy.
\] They identify the blow-up with the closed incidence scheme in
\(\mathbf A^2\times\mathbf P^1\), so \(b\) is projective and proper. Off
the origin it is an isomorphism. The pulled-back origin ideal
\(\mathfrak m=(x,y)\) is generated by \(x\) on \(U\) and by \(y\) on
\(V\), and is the invertible ideal \(J=\mathcal O_X(-E)\). On the
overlap \(y=vx\), so its conormal restriction is \(\mathcal O_E(1)\).
Consequently \[
J^j/J^{j+1}\cong\mathcal O_E(j)\quad(j\geq0).
\] The positive sign is essential: the normal bundle of \(E\) is
\(\mathcal O_E(-1)\), whereas the successive ideal layers use its dual.

The thickened fiber \(X_n\) is defined by \(J^n\). For \(n\geq2\), \[
0\to\mathcal O_E(n-1)\to\mathcal O_{X_n}\to\mathcal O_{X_{n-1}}\to0.
\] Projective-line cohomology gives \(H^1(E,\mathcal O_E(j))=0\) for
every \(j\geq0\). The long exact sequences, starting with \(X_1=E\),
give \(H^1(X_n,\mathcal O_{X_n})=0\) for all \(n\). By formal functions
and Nakayama, \((R^1b_*\mathcal O_X)_0=0\). Away from zero the morphism
is an isomorphism, so \[
R^1b_*\mathcal O_X=0.
\] All still higher direct images vanish by the fiber-dimension bound,
because the largest fiber dimension is one.

We can also see the completed degree-zero comparison explicitly. The
canonical polynomial map \[
k[x,y]/(x,y)^n\longrightarrow H^0(X_n,\mathcal O_{X_n})
\] is an isomorphism. For \(n=1\) both sides are \(k\). At the induction
step, its map on the leftmost ideal layers is \[
(x,y)^{n-1}/(x,y)^n\xrightarrow{\sim}H^0(E,\mathcal O_E(n-1)),
\] the monomial basis identification from projective-space cohomology.
Both the polynomial quotient sequence and the section sequence are short
exact, the latter because the layer has zero \(H^1\). Induction proves
the isomorphism. Their limits give \(k[[x,y]]\). The unit map from the
base local ring to \((b_*\mathcal O_X)_0\) becomes an isomorphism after
completion. Exactness and faithful flatness of completion for finite
modules make it an isomorphism before completion; off the origin the
same holds directly. Thus \(b_*\mathcal O_X=\mathcal O_{\mathbf A^2}\),
as well as higher vanishing.

\subsection{7. Exercises with solutions}\label{exercises-with-solutions}

\textbf{Exercise 7.1 (easy: completed constants).} State and prove the
completed-stalk formula for \(f_*\mathcal O_X\) at a point of a locally
Noetherian base.

\textbf{Solution.} For proper \(f\), Corollary 4.2 in degree zero gives
\(\widehat{(f_*\mathcal O_X)_s}=\varprojlim_n H^0(X_n,\mathcal O_{X_n})\),
with \(X_n\) defined over \(\mathcal O_{S,s}/\mathfrak m_s^n\).
Localization is flat, so identifies the stalk with degree-zero
cohomology over the local ring; global formal functions for that ring
and ideal gives the formula. All maps preserve multiplication, since
they are restriction maps of structure sheaves. Hence it is an
isomorphism of complete algebras, not only of modules.

\textbf{Exercise 7.2 (medium: the exceptional layers).} Compute the
ideal layers of the exceptional curve in the plane blow-up and deduce
\(R^1b_*\mathcal O_X=0\).

\textbf{Solution.} The local ideal frames are \(x\) and \(y\), related
by \(y=vx\); their restriction has the transition of
\(\mathcal O_{\mathbf P^1}(1)\). Since the ideal is invertible, its
\(j\)-th layer is its restricted \(j\)-th tensor power,
\(\mathcal O_{\mathbf P^1}(j)\). Their \(H^1\) groups vanish for
\(j\geq0\), so induction on the exact thickening sequences gives zero
\(H^1\) on every \(X_n\). Formal functions gives zero completed stalk at
the origin, and finite-module Nakayama gives zero stalk. Off the origin
the map is an isomorphism, proving the global assertion.

\textbf{Exercise 7.3 (medium: finite fibers).} Derive finiteness of a
proper morphism with finite fibers using only formal functions, affine
cohomology and Serre's criterion.

\textbf{Solution.} Every fiber and thickening has dimension zero, so its
positive cohomology is zero. Formal functions and Nakayama annihilate
\(R^1f_*J\) for every coherent ideal. Over an affine Noetherian base,
Leray and affine vanishing give \(H^1(X,J)=0\). The finite-ideal affine
criterion makes \(X\) affine. Proper finiteness makes its coordinate
algebra finite over the base algebra, which is exactly finiteness of the
morphism. This also covers empty fibers.

\textbf{Exercise 7.4 (medium: a uniform dimension bound).} If
\(\dim X_s\leq d\) for every \(s\), prove \(R^qf_*F=0\) for \(q>d\). Can
\(\dim X\) replace the fiber calculation in the proof without further
assumptions?

\textbf{Solution.} Theorem 5.1 kills every stalk in the stated degrees,
hence kills the sheaf. The proof works on the Noetherian topology of
each thickened fiber and does not require a dimension bound on the total
space. A locally Noetherian base can have unbounded or infinite Krull
dimension, so a total-space bound might be unavailable and would in any
event miss the sharper relative assertion.

\textbf{Exercise 7.5 (hard: the inverse-limit step).} Prove the
stabilization and limit claims from (1), without assuming the middle
transitions are surjective.

\textbf{Solution.} The right transition \(K_m^{q+1}\to K_n^{q+1}\) is
zero for \(m\geq n+c\), so the middle image lies in \(M^q/J_n^q\). The
left transition is onto, so the middle image contains this entire
submodule. Equality proves Mittag--Leffler stabilization. A compatible
family in the middle maps to a compatible family in the pro-zero right
system. Each of its right coordinates is the image of a sufficiently
distant zero transition, so is zero. Each middle coordinate therefore
has a unique preimage in the injective left term, and uniqueness forces
compatibility of those preimages. Thus the two limits agree. Finally
\(I^nM^q\subset J_n^q\) and \(J_{n+c_J}^q\subset I^nM^q\) identify the
left limit continuously with the adic completion. Every assertion is
justified without finite-level surjectivity of the middle system.

\textbf{Exercise 7.6 (challenging: transient infinitesimal sections).}
For the extension with class \(t\) on \(\mathbf P^1_{k[t]}\), compute
\(H^0\) on every thickening \(B_n=k[t]/(t^n)\), its transition maps, and
its inverse limit.

\textbf{Solution.} Its Grothendieck complex is \([A\xrightarrow{t}A]\)
in degrees zero and one. Over \(B_n\), the kernel of multiplication by
\(t\) is \(k\,t^{n-1}\), including \(k\) when \(n=1\). Reduction
\(B_{n+1}\to B_n\) sends its kernel generator \(t^n\) to zero. Thus all
adjacent transitions on degree-zero cohomology are zero, and its inverse
limit is zero. This agrees with the completion of \(H^0(X,E)=0\),
although every finite thickening has a nonzero section. In degree one,
the cokernel is \(k\) at every level and the transitions are identities;
its limit agrees with the completion of \(A/(t)\). This exhibits both
the transient error and the persistent cohomology in one family.

\subsection{8. Ampleness near a fiber}\label{ampleness-near-a-fiber}

\textbf{Theorem 8.1.} If \(f:X\to S\) is proper with \(S\) locally
Noetherian and \(L|_{X_s}\) is ample, then \(L\) is relatively ample
over an open neighborhood of \(s\).

\textbf{Proof.} Work first over \(A=\mathcal O_{S,s}\), with maximal
ideal \(\mathfrak m\). Put \(X_n=V(\mathfrak m^n\mathcal O_X)\). The
layers \(D_n=\mathfrak m^n\mathcal O_X/\mathfrak m^{n+1}\mathcal O_X\)
form a finite graded module over
\(\mathcal O_{X_1}\otimes_{A/\mathfrak m}\operatorname{gr}_{\mathfrak m}A\):
multiplication gives a surjection from that ring onto their direct sum,
generated in degree zero. The associated graded ring is Noetherian and
finite type. Apply the Serre vanishing theorem to this finite module on
the base change of the proper fiber \(X_1\), with the pulled-back ample
\(L_1\). Affine pushforward and the finite affine-cover complex split
its cohomology by graded degree. Thus one \(d_0\) gives
\(H^1(X_1,D_n\otimes L_1^d)=0\) for every \(n\) and every \(d\ge d_0\).
The exact thickening sequences consequently lift each section of
\(L_1^d\) successively to all \(X_n\). Formal functions identifies the
compatible lift with an element of the completion of the finite module
\(H^0(X,L^d)\). Its reduction modulo \(\mathfrak m\) comes from that
module itself, so the original section lifts to an actual section on
\(X\).

Choose \(d\ge d_0\) for which finitely many sections embed the fiber in
projective space, using the ample embedding lemma in the Serre lesson.
Lift them as above. The local-base cohomology localization proved in the
affine lesson represents the finite collection over an affine
neighborhood of \(s\). Their evaluation cokernel is coherent; its proper
closed support has closed image missing \(s\). Shrink away from that
image. They now generate \(L^d\) and define a morphism
\(h:X\to\mathbf P^r_S\), proper by the closed-graph factorization, whose
fiber at \(s\) is a closed immersion. The quasi-finite open of \(h\)
contains all of \(X_s\). Its closed complement has closed image in
\(S\), since \(X/S\) is proper. Shrinking once more makes \(h\)
quasi-finite everywhere. Theorem 5.2 makes it finite, checked on affine
opens of its target. The inverse images of standard affine projective
charts are now affine; these are exactly the section opens for the
pulled-back coordinate sections of \(L^d=h^*\mathcal O(1)\). They yield
an affine basis after allowing all powers of the line bundle. Indeed, on
one such affine inverse image \(X_{s_i}\), take any principal open
\(D(a)\). Lemma 1.1 of the Serre lesson extends \(a\) to a section \(t\)
of a power of \(L^d\), with \(a=t/s_i^b\) on that chart. Multiplying
\(t\) once more by \(s_i\) forces its nonvanishing open to be contained
in that chart, and that open is exactly \(D(a)\). These principal opens
form a basis. Hence \(L^d\) is relatively ample. The positive-power
criterion proved in the Serre lesson makes \(L\) relatively ample.
\(\square\)

\subsection{References}\label{references}

\begin{itemize}
\tightlist
\item
  \textbf{{[}Stacks{]}} The Stacks project authors, \emph{The Stacks
  project}, in its AI Integrated Stacks Project edition: graded powers
  \href{https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/coherent.html\#coherent-lemma-cohomology-powers-ideal-times-F}{Tag
  02O8}; filtration consequences
  \href{https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/coherent.html\#coherent-lemma-cohomology-powers-ideal-application}{Tag
  02OA}; Mittag--Leffler
  \href{https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/coherent.html\#coherent-lemma-ML-cohomology-powers-ideal}{Tag
  02OB}; formal functions
  \href{https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/coherent.html\#coherent-theorem-formal-functions}{Tag
  02OC}; stalk form
  \href{https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/coherent.html\#coherent-lemma-formal-functions-stalk}{Tag
  02OD}.
\item
  Dimension bound
  \href{https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/coherent.html\#coherent-lemma-higher-direct-images-zero-above-dimension-fibre}{Tag
  02V7}, using the open topological theorem
  \href{https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/cohomology.html\#cohomology-proposition-vanishing-Noetherian}{Tag
  02UZ}; finiteness
  \href{https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/coherent.html\#coherent-lemma-characterize-finite}{Tag
  02OG}. The additional ampleness proof is
  \href{https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/coherent.html\#coherent-lemma-ample-on-fibre}{Tag
  0D2M} and
  \href{https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/coherent.html\#coherent-lemma-ample-in-neighbourhood}{Tag
  0D2N}.
\item
  The linked Stacks Project texts are under GNU FDL 1.2. Sections 2--5
  prove the assigned formal-functions, vanishing and finiteness results;
  the completion algebra belongs to the named prerequisite, and Section
  8 proves the additional ampleness consequence.
\end{itemize}

\subsection{Appendix Z. Conductor coefficients and strong
transcendence}\label{appendix-z.-conductor-coefficients-and-strong-transcendence}

Lemma 2.1 of \href{../AG-MO/AG-MO-12.html}{Zariski's Main Theorem},
Section 2, treats an algebra that is finite over one generator. It needs
two facts about a finite algebra over a polynomial ring: the
coefficients of a polynomial that lands in the radical of the conductor
land there one by one, and a strongly transcendental element rules out
quasi-finite points. This appendix proves both, together with the lemmas
they rest on. Rings are commutative with \(1\), ring maps preserve
\(1\), and ``finite'' means finite as a module. We write \(R[X]\) for a
polynomial ring and \(R[x]\subset S\) for the subalgebra generated by an
element \(x\in S\). No ring is assumed Noetherian; reducedness and
integrality hypotheses are stated where they are used.

We rely on these earlier results. From
\href{../AG-CA/AG-CA-05.html}{Integral extensions: lying over, going up
and going down}: the determinant test and the closure and transitivity
of integrality (Theorems 1.1--1.2, Proposition 1.3), lying over (Theorem
3.2) and going down over a normal domain (Theorem 4.3). From
\href{../AG-CA/AG-CA-02.html}{Localization, local properties and
support}, Section 1 and Theorems 3.1--3.2: when a fraction is zero, the
correspondence of primes, and fibres as spectra of residue-field base
changes. From \href{../AG-CA/AG-CA-01.html}{Spectra of rings}, Theorem
1.2, Theorem 5.2 and Corollary 5.3: the nilradical as the intersection
of primes, and open-and-closed subsets as product decompositions. From
\href{../AG-CA/AG-CA-03.html}{Noetherian and Artinian rings}, Lemma 4.1
and Theorem 4.2: spectra of Artinian rings. A prime of a finite type
algebra is a quasi-finite point exactly when its residue field and its
local fibre algebra are finite over the residue field below; this
pointwise criterion is \href{../AG-MO/AG-MO-06.html}{Quasi-finite
morphisms and Chevalley's theorem, Theorem 1.1}. Polynomial rings over a
normal domain are normal by
\href{the-theorem-on-formal-functions.html\#0-a-normality-input-in-the-algebraic-zariski-main-proof}{Lemma
0.1 above}.

\subsubsection{Z.1. Integral closure and
localization}\label{z.1.-integral-closure-and-localization}

\textbf{Lemma Z.1.} For a ring map \(A\to B\) and a multiplicative
subset \(T\subset A\), write \(B^{\mathrm{int}}\subset B\) for the
subring of elements integral over \(A\). Inside \(T^{-1}B\), the
integral closure of \(T^{-1}A\) equals \(T^{-1}B^{\mathrm{int}}\).

\textbf{Proof.} Localization is exact, so \(T^{-1}B^{\mathrm{int}}\) is
a subring of \(T^{-1}B\). If \(b\) satisfies
\(b^d+a_1b^{d-1}+\cdots+a_d=0\) with \(a_i\in A\), then dividing by
\(s^d\), for \(s\in T\), gives the monic equation
\((b/s)^d+(a_1/s)(b/s)^{d-1}+\cdots+a_d/s^d=0\) over \(T^{-1}A\). So
\(T^{-1}B^{\mathrm{int}}\) is integral over \(T^{-1}A\).

Conversely, suppose \(z=b/s\) satisfies
\(z^d+\sum_{i=1}^d(a_i/t_i)z^{d-i}=0\) with \(a_i\in A\) and
\(t_i\in T\). Put \(t=t_1\cdots t_d\) and \(y=tb\). Multiplying the
equation by \((ts)^d\) turns it into

\[
g(y)=y^d+\sum_{i=1}^dc_iy^{d-i}=0\ \text{ in }T^{-1}B,\qquad c_i=a_i\,\frac{t}{t_i}\,t^{\,i-1}s^i\in A .
\]

So some \(v\in T\) has \(vg(y)=0\) in \(B\), and then
\(0=v^dg(y)=(vy)^d+\sum_iv^ic_i(vy)^{d-i}\). Hence
\(vy\in B^{\mathrm{int}}\), and \(z=vy/(vts)\) lies in
\(T^{-1}B^{\mathrm{int}}\). Zero divisors cause no difficulty here,
since the factor \(v\) has been absorbed into the element. \(\square\)

\subsubsection{Z.2. Removing a monic
denominator}\label{z.2.-removing-a-monic-denominator}

\textbf{Lemma Z.2.} Suppose \(\varphi:R[X]\to S\) is a ring map and
\(t\in S\) is integral over \(R[X]\). If
\(t\varphi(p)\in\operatorname{im}\varphi\) for a monic \(p\in R[X]\),
then \(t-\varphi(q)\) is integral over \(R\) for some \(q\in R[X]\).

\textbf{Proof.} Write \(t\varphi(p)=\varphi(r)\). Long division by the
monic \(p\) works over any ring and gives \(r=qp+r'\) with
\(\deg r'<\deg p\). Put \(\tau=t-\varphi(q)\); then
\(\tau\varphi(p)=\varphi(r')\), and \(\tau\) is integral over \(R[X]\)
because integral elements form a subring. If \(\deg p=0\), then \(p=1\),
\(r'=0\) and \(\tau=0\). Let \(\deg p=n\ge1\) and pass to \(S_\tau\),
where \(\tau\) is a unit. There \(x=\varphi(X)\) satisfies

\[
p(x)-\tau^{-1}r'(x)=0,
\]

a monic equation of degree \(n\) over the subring
\(D=\varphi(R)[\tau^{-1}]\), because \(\deg r'<n\). So \(D[x]\) is
integral over \(D\). The image of \(\tau\) in \(S_\tau\) is integral
over the image of \(R[X]\), which lies in \(D[x]\), and by transitivity
\(\tau\) is integral over \(D\). Write such an equation as
\(\tau^m+\sum_{i<m}d_i\tau^i=0\) with
\(d_i=\sum_{j=0}^{N}\varphi(b_{ij})\tau^{-j}\), \(b_{ij}\in R\).
Multiplying by \(\tau^N\) gives an equation in \(S_\tau\) with only
nonnegative powers of \(\tau\), and multiplying by a further power
\(\tau^M\) makes it hold in \(S\):

\[
\tau^{m+N+M}+\sum_{i<m}\sum_{j=0}^{N}\varphi(b_{ij})\,\tau^{\,i+N+M-j}=0 .
\]

All exponents in the sum are smaller than \(m+N+M\), so this is a monic
equation for \(\tau\) over \(R\). If \(\tau\) is nilpotent, then
\(S_\tau=0\) and \(\tau\) is integral over \(R\) anyway. \(\square\)

\subsubsection{Z.3. Clearing a leading
coefficient}\label{z.3.-clearing-a-leading-coefficient}

\textbf{Lemma Z.3.} Suppose again that \(\varphi:R[X]\to S\) is a ring
map and \(t\in S\) is integral over \(R[X]\), and let

\[
p=\sum_{i=0}^ka_iX^i\in R[X]
\]

with \(t\varphi(p)\in\operatorname{im}\varphi\). For some \(n\ge0\) and
\(q\in R[X]\), the element \(\varphi(a_k)^nt-\varphi(q)\) is integral
over \(R\).

\textbf{Proof.} Put \(a=a_k\). If \(a\) is nilpotent, \(a^n=0\) for some
\(n\) and \(q=0\) works. Otherwise invert \(a\) in \(R\) and
\(\varphi(a)\) in \(S\). Over \(R_a\) the polynomial \(a^{-1}p\) is
monic, and the images of \(t\) and of \(t\varphi(a^{-1}p)\) satisfy the
hypotheses of Lemma Z.2 for \(R_a[X]\to S_{\varphi(a)}\). That lemma
gives \(q'\in R_a[X]\) with \(t-\varphi(q')\) integral over \(R_a\).
Write \(a^eq'=q_0\) with \(q_0\in R[X]\). The image of
\(\delta=\varphi(a)^et-\varphi(q_0)\) in \(S_{\varphi(a)}\) is
\(\varphi(a)^e\) times an integral element, so it is integral over
\(R_a\). By Lemma Z.1 for \(R\to S\) and the powers of \(a\), it equals
\(c/\varphi(a)^j\) for some \(c\in S\) integral over \(R\). Equality of
fractions gives \(\ell\ge0\) with
\(\varphi(a)^{\ell}(\varphi(a)^j\delta-c)=0\) in \(S\). Then
\(\varphi(a)^{j+\ell}\delta=\varphi(a)^\ell c\) is integral over \(R\),
which is the assertion with \(n=e+j+\ell\) and \(q=a^{j+\ell}q_0\).
\(\square\)

\subsubsection{Z.4. The conductor setup}\label{z.4.-the-conductor-setup}

\textbf{Situation Z.4.} Let \(\varphi:R[X]\to S\) be a finite ring map.
Put \(A=\varphi(R)\), \(x=\varphi(X)\) and \(C=A[x]=\varphi(R[X])\).
Assume that \(A\) is integrally closed in \(S\), that is, every element
of \(S\) integral over \(A\) lies in \(A\); \(A\) need not be a domain.
The \textbf{conductor} of \(C\subset S\) is

\[
J=\{g\in S:\ gS\subset C\}.
\]

It is an ideal of \(S\) and is contained in \(C\), since \(g=g\cdot1\).
Fix \(C\)-module generators \(t_1,\ldots,t_r\) of \(S\). Then \(g\in J\)
if and only if \(gt_i\in C\) for every \(i\), because every element of
\(gS\) is a \(C\)-combination of the \(gt_i\). Nothing here excludes
\(S=0\).

\subsubsection{Z.5. The leading coefficient enters the
conductor}\label{z.5.-the-leading-coefficient-enters-the-conductor}

\textbf{Lemma Z.5.} Assume Situation Z.4, and take \(u\in S\) and
\(p=\sum_{i=0}^ka_iX^i\in R[X]\). If \(u\varphi(p)\in J\), then
\(u\varphi(a_k)^m\in J\) for some \(m\ge0\).

\textbf{Proof.} Each product \(ut_i\) is integral over \(R[X]\): \(S\)
is a finite \(R[X]\)-algebra, and finite algebras are integral by the
determinant test. Also
\((ut_i)\varphi(p)=(u\varphi(p))t_i\in C=\operatorname{im}\varphi\).
Lemma Z.3 gives \(n_i\ge0\) and \(q_i\in R[X]\) with
\(\varphi(a_k)^{n_i}ut_i-\varphi(q_i)\) integral over \(R\), that is,
over \(A\). By the hypothesis on \(A\) this element lies in
\(A\subset C\), so \(\varphi(a_k)^{n_i}ut_i\in C\). For \(m=\max_in_i\),
multiplying by further powers of \(\varphi(a_k)\in C\) gives
\(u\varphi(a_k)^mt_i\in C\) for all \(i\), so \(u\varphi(a_k)^m\in J\).
\(\square\)

\subsubsection{Z.6. Every coefficient enters the
radical}\label{z.6.-every-coefficient-enters-the-radical}

\textbf{Lemma Z.6.} Assume Situation Z.4, with \(u\in S\) and
\(p=\sum_{i=0}^ka_iX^i\in R[X]\). If \(u\varphi(p)\in\sqrt J\), then
\(u\varphi(a_i)\in\sqrt J\) for every \(i\).

\textbf{Proof.} We induct on \(k\). Choose \(N\ge1\) with
\(u^N\varphi(p)^N\in J\). The polynomial \(p^N\), written with top index
\(kN\), has coefficient \(a_k^N\) there, so Lemma Z.5 gives \(m\ge1\)
with \(u^N\varphi(a_k)^{Nm}\in J\); enlarging \(m\) keeps this true, as
\(J\) is an ideal. It follows that
\((u\varphi(a_k))^{Nm}=u^{N(m-1)}\cdot u^N\varphi(a_k)^{Nm}\in J\), so
\(u\varphi(a_k)\in\sqrt J\). Hence
\(u\varphi(p-a_kX^k)=u\varphi(p)-u\varphi(a_k)x^k\in\sqrt J\), and the
induction hypothesis applies to \(p-a_kX^k\), whose top index is
\(k-1\). For \(k=0\) the first step is the whole proof. \(\square\)

\subsubsection{Z.7. Strong transcendence and the conductor
quotient}\label{z.7.-strong-transcendence-and-the-conductor-quotient}

\textbf{Definition Z.7.} Let \(R\subset S\) be rings. An element
\(x\in S\) is \textbf{strongly transcendental over \(R\)} if, whenever
\(u\in S\) and \(a_0,\ldots,a_k\in R\) satisfy \(u\sum_ia_ix^i=0\),
every product \(ua_i\) is zero. For domains \(R\subset S\) this is
ordinary transcendence: \(u=1\) gives transcendence, and in a domain a
nonzero \(u\) can be cancelled.

In Situation Z.4, let

\[
S_0=S/\sqrt J,\qquad A_0=A/(A\cap\sqrt J),
\]

with \(x_0\) the class of \(x\) in \(S_0\). Then \(A_0\subset S_0\) are
reduced rings, \(S_0\) is a finite \(A_0[x_0]\)-module, and over \(A_0\)
the element \(x_0\) is strongly transcendental.

\textbf{Proof.} The map \(A_0\to S_0\) is injective by construction. The
ideal \(\sqrt J\) is radical, so \(S_0\) and its subring \(A_0\) are
reduced. The images of \(t_1,\ldots,t_r\) generate \(S_0\) over
\(A_0[x_0]\). Now let \(u_0\sum_i\bar a_ix_0^i=0\) in \(S_0\), with
\(u_0\in S_0\) and \(\bar a_i\in A_0\). Choose \(u\in S\) lifting
\(u_0\) and \(r_i\in R\) with \(\varphi(r_i)\) lifting \(\bar a_i\).
Then \(u\varphi(\sum_ir_iX^i)\in\sqrt J\), and Lemma Z.6 gives
\(u\varphi(r_i)\in\sqrt J\) for every \(i\), that is, \(u_0\bar a_i=0\).
\(\square\)

\subsubsection{Z.8. Quasi-finiteness in a four-ring
diagram}\label{z.8.-quasi-finiteness-in-a-four-ring-diagram}

\textbf{Lemma Z.8.} Let

\[
\begin{array}{ccc}
R&\longrightarrow&S\\
\downarrow&&\downarrow\\
R'&\longrightarrow&S'
\end{array}
\]

be a commutative square of rings in which \(R\to S\) is of finite type
and \(S\otimes_RR'\to S'\) is surjective. Let \(\mathfrak q'\subset S'\)
be a prime, with contractions \(\mathfrak q\subset S\),
\(\mathfrak p'\subset R'\) and \(\mathfrak p\subset R\).
Quasi-finiteness of \(R\to S\) at \(\mathfrak q\) implies
quasi-finiteness of \(R'\to S'\) at \(\mathfrak q'\).

\textbf{Proof.} Since \(S\otimes_RR'\) is of finite type over \(R'\) and
maps onto \(S'\), the map \(R'\to S'\) is of finite type. Let
\(k=\kappa(\mathfrak p)\), \(k'=\kappa(\mathfrak p')\), and let
\(F=S\otimes_Rk\) and \(F'=S'\otimes_{R'}k'\) be the two fibre rings. By
the pointwise criterion, the point \(\mathfrak q\) of
\(\operatorname{Spec}F\) is closed as well as isolated. So
\(\{\mathfrak q\}\) is open and closed, and \(F=F_1\times F_2\) with
\(\operatorname{Spec}F_1=\{\mathfrak q\}\). The ring \(F_1\) has a
single prime ideal, so it equals its localization there, which is the
local fibre algebra at \(\mathfrak q\). The pointwise criterion makes it
finite over \(k\).

Tensoring the surjection \(S\otimes_RR'\to S'\) with \(k'\) over \(R'\)
gives a surjection \(F\otimes_kk'\to F'\). The image of the idempotent
of \(F_1\) splits \(F'=F_1'\times F_2'\), where \(F_1'\) is a quotient
of \(F_1\otimes_kk'\) and hence finite over \(k'\). The map
\(F\to F'\to\kappa(\mathfrak q')\) factors through
\(\kappa(\mathfrak q)\), because \(\mathfrak q'\) contracts to
\(\mathfrak q\). So it sends the idempotent of \(F_1\) to \(1\), and
\(\mathfrak q'\) lies in \(\operatorname{Spec}F_1'\). A finite algebra
over a field is Artinian, since its ideals are subspaces, so it has
finitely many primes, all maximal, and its spectrum is discrete. Thus
\(\mathfrak q'\) is isolated in the open and closed subset
\(\operatorname{Spec}F_1'\) of \(\operatorname{Spec}F'\), and so in its
fibre. \(\square\)

\subsubsection{Z.9. Passage to a minimal
prime}\label{z.9.-passage-to-a-minimal-prime}

\textbf{Lemma Z.9.} Suppose \(R\subset S\) are reduced and \(x\in S\) is
strongly transcendental over \(R\). For a minimal prime \(\mathfrak q\)
of \(S\) with \(\mathfrak p=R\cap\mathfrak q\), the class of \(x\) in
\(S/\mathfrak q\) is strongly transcendental over \(R/\mathfrak p\).

\textbf{Proof.} First, \(\mathfrak qS_{\mathfrak q}=0\). The ring
\(S_{\mathfrak q}\) is reduced: if \((y/s)^n=0\), then \(wy^n=0\) for
some \(w\notin\mathfrak q\), so \((wy)^n=0\), \(wy=0\) by reducedness of
\(S\), and \(y/s=0\). Its primes correspond to the primes of \(S\)
inside \(\mathfrak q\), and by minimality only
\(\mathfrak qS_{\mathfrak q}\) remains. The nilradical is the
intersection of all primes, so \(\mathfrak qS_{\mathfrak q}\) is the
nilradical, which is zero.

Let \(u\in S\) and \(a_i\in R\) with \(u\sum_ia_ix^i\in\mathfrak q\).
This element vanishes in \(S_{\mathfrak q}\), so \(wu\sum_ia_ix^i=0\)
for some \(w\notin\mathfrak q\). Strong transcendence gives \(wua_i=0\),
and since \(w\notin\mathfrak q\), each \(ua_i\) lies in \(\mathfrak q\).
This is the required property for the images in \(S/\mathfrak q\), and
\(R/\mathfrak p\to S/\mathfrak q\) is injective by the choice of
\(\mathfrak p\). \(\square\)

\subsubsection{Z.10. Transcendence excludes quasi-finite points for
domains}\label{z.10.-transcendence-excludes-quasi-finite-points-for-domains}

\textbf{Lemma Z.10.} Let \(x\in S\) be transcendental over \(R\), where
\(R\subset S\) are domains and \(S\) is a finite \(R[x]\)-module. Then
\(R\to S\) has no quasi-finite point.

\textbf{Proof.} Generators of \(S\) as an \(R[x]\)-module, together with
\(x\), generate \(S\) as an \(R\)-algebra, so \(R\to S\) is of finite
type.

\emph{Case 1: \(R\) is normal.} Then \(R[x]\cong R[X]\) is a normal
domain by Lemma 0.1. Let \(\mathfrak q\subset S\) be a prime, with
\(\mathfrak p=R\cap\mathfrak q\) and
\(\mathfrak r=R[x]\cap\mathfrak q\), and suppose \(R\to S\) is
quasi-finite at \(\mathfrak q\). Then \(\kappa(\mathfrak q)\) is finite
over \(\kappa(\mathfrak p)\), and so is the intermediate field
\(\kappa(\mathfrak r)\). The prime \(\mathfrak pR[x]\) has residue field
\(\kappa(\mathfrak p)(X)\), which is not finite over
\(\kappa(\mathfrak p)\); hence \(\mathfrak pR[x]\subsetneq\mathfrak r\).
The extension \(R[x]\subset S\) is integral, \(S\) is a domain and
\(R[x]\) is normal, so going down gives a prime
\(\mathfrak q_0\subset\mathfrak q\) with
\(R[x]\cap\mathfrak q_0=\mathfrak pR[x]\). Then
\(\mathfrak q_0\ne\mathfrak q\), while
\(R\cap\mathfrak q_0=\mathfrak p\). So \(\mathfrak q_0\) lies in the
fibre over \(\mathfrak p\) and specializes to \(\mathfrak q\). Every
open set containing \(\mathfrak q\) contains its generization
\(\mathfrak q_0\), so \(\mathfrak q\) is not isolated in the fibre, a
contradiction.

\emph{Case 2: general \(R\).} Let \(K\) and \(L\) be the fraction fields
of \(R\) and \(S\), let \(R'\subset K\) be the integral closure of
\(R\), and let \(S'\) be the subring of \(L\) generated by \(S\)
together with \(R'\). The domain \(R'\) is normal with fraction field
\(K\): if an element of \(K\) satisfies a monic equation over \(R'\),
transitivity makes it integral over \(R\), so it lies in \(R'\). The
element \(x\) is transcendental over \(R'\), since clearing denominators
turns a relation over \(R'\) into one over \(R\). The map
\(S\otimes_RR'\to S'\) is surjective, and the images of \(R[x]\)-module
generators of \(S\) generate \(S'\) over \(R'[x]\). So \(R'\subset S'\)
satisfy the hypotheses of Case 1. The ring \(S'\) is generated over
\(S\) by elements of \(R'\), which are integral over \(R\subset S\), so
\(S\subset S'\) is integral. By lying over, a given prime
\(\mathfrak q\subset S\) is the contraction of some prime
\(\mathfrak q'\subset S'\). If \(R\to S\) were quasi-finite at
\(\mathfrak q\), Lemma Z.8 for the square \(R,S,R',S'\) would make
\(R'\to S'\) quasi-finite at \(\mathfrak q'\), contradicting Case 1.
Neither finiteness of \(R'\) over \(R\) nor Noetherianity is used.
\(\square\)

\subsubsection{Z.11. Strong transcendence excludes quasi-finite
points}\label{z.11.-strong-transcendence-excludes-quasi-finite-points}

\textbf{Lemma Z.11.} Let \(R\subset S\) be reduced, with \(x\in S\)
strongly transcendental over \(R\) and \(S\) a finite \(R[x]\)-module.
Then \(R\to S\) has no quasi-finite point.

\textbf{Proof.} As before, \(R\to S\) is of finite type. Let
\(\mathfrak q\subset S\) be a prime. It contains a minimal prime
\(\mathfrak q_0\) of \(S\), by Zorn's lemma for the primes inside
\(\mathfrak q\) ordered by reverse inclusion. The hypothesis of Zorn's
lemma holds because the intersection of a chain of primes is prime: if
\(ab\) lies in it while \(a\) and \(b\) do not, choose members of the
chain missing \(a\) and missing \(b\). The smaller of the two misses
both, yet contains \(ab\).

Let \(\mathfrak p_0=R\cap\mathfrak q_0\), \(R_0=R/\mathfrak p_0\) and
\(S_0=S/\mathfrak q_0\). These are domains with \(R_0\subset S_0\). By
Lemma Z.9 the image \(x_0\) of \(x\) is strongly transcendental over
\(R_0\), hence transcendental. The ring \(S_0\) is finite over
\(R_0[x_0]\). By Lemma Z.10, \(R_0\to S_0\) is not quasi-finite at
\(\mathfrak q/\mathfrak q_0\). Lemma Z.8 applies to the square
\(R,S,R_0,S_0\), since \(S\otimes_RR_0=S/\mathfrak p_0S\) maps onto
\(S/\mathfrak q_0\). So \(R\to S\) is not quasi-finite at
\(\mathfrak q\). There may be infinitely many minimal primes; the
argument uses only one. \(\square\)

\subsubsection{Z.12. The conductor is avoided at a quasi-finite
point}\label{z.12.-the-conductor-is-avoided-at-a-quasi-finite-point}

\textbf{Corollary Z.12.} Assume Situation Z.4. If \(\mathfrak q\) is a
prime of \(S\) at which \(A\to S\) is quasi-finite, the conductor \(J\)
is not contained in \(\mathfrak q\).

\textbf{Proof.} Suppose \(J\subset\mathfrak q\), so
\(\sqrt J\subset\mathfrak q\). Z.7 and Lemma Z.11, applied to
\(A_0\subset S_0\), which are reduced, and to \(x_0\), which is strongly
transcendental over \(A_0\), show that \(A_0\to S_0\) is not
quasi-finite at \(\mathfrak q/\sqrt J\). But \(A\to S\) is of finite
type, \(S\) being finite over \(A[x]\), and Lemma Z.8 applies to the
square \(A,S,A_0,S_0\), since \(S\otimes_AA_0=S/(A\cap\sqrt J)S\) maps
onto \(S/\sqrt J\). That lemma carries quasi-finiteness of \(A\to S\) at
\(\mathfrak q\) to quasi-finiteness of \(A_0\to S_0\) at
\(\mathfrak q/\sqrt J\), a contradiction. \(\square\)

For Lemma 2.1 of \href{../AG-MO/AG-MO-12.html}{Zariski's Main Theorem},
Corollary Z.12 provides the conductor element outside the given prime.
The two supporting facts behind it are Lemma Z.6, on coefficients, and
Lemma Z.11, on reduced rings with a strongly transcendental element. The
monogenic argument and the main induction are in that lesson.

The statements of this appendix correspond to the Stacks Project,
\emph{Algebra}, Tags 0307, 00PT, 00PV, 00PW, 00PX, 00PY, 00PZ (strong
transcendence), 00Q0, 00Q1, 00Q2 and 00PN, in the AI Integrated Stacks
Project edition pinned at
\href{https://github.com/KokunoYumeto/unofficial-stacks-project-ai-drafts/blob/565b10e987aba5969b21145a0833f42d69f96790/algebra.tex}{revision
565b10e9}.
\end{document}
