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Algebraic Zariski Main for Quasi-Finite Morphisms

1 · Prerequisites

2 · Summary

This page develops the algebraic form of Zariski's main theorem for quasi-finite morphisms of affine spectra, in the version proved in the Stacks Project's Section 10.123 and in Milne's Chapter 17, and isolates the choice dependence of its two global statements.

The first items set up the two notions the theorem needs for an arbitrary ring map $R \to S$ : the relative integral closure ${Int}_{R} (S)$ of the image of $R$ , a subring of $S$ that needs no injectivity, no reducedness and no Noetherian hypothesis, and quasi-finiteness at a prime $q$ , phrased through the fibre $S_{q} / p S_{q}$ over the contraction $p$ and equivalently through the schematic fibre $S \otimes_{R} κ (p)$ . The one-variable engine then follows: a root of a polynomial equation times its leading coefficient is integral, monic division corrects one variable while changing the algebra only at finitely many named elements, and the conductor of a finite map is controlled coefficient by coefficient. A strongly transcendental element is defined so that its powers satisfy no algebraic relation with a unit coefficient, and the two lemmas that use it show that strong transcendence survives passage to minimal-prime quotients and that a finite one-variable algebra generated by a strongly transcendental element is nowhere quasi-finite. Alongside these sits the local supplier that the polynomial algebra over an integrally closed domain is again integrally closed, proved without a Noetherian hypothesis; it is what makes the going-down step in the transcendence argument available. Quasi-finiteness is then shown to transfer through intermediate rings, both from a source prime and in the relative form needed later.

The main theorem is the local form of Zariski's main theorem: at a prime where a finite-type map is quasi-finite there is an element $g$ of the relative integral closure, avoiding that prime, such that $S_{g}^{'} ≅ S_{g}$ . The proof is a descent to a minimal prime, a division argument in one variable, and the conductor control assembled in the earlier items. Two global consequences are recorded: the quasi-finite locus is open, and a finite-type quasi-finite algebra admits a finite open factorization — a finite subalgebra $T$ of the relative integral closure with principal opens $D_{T} (g_{i})$ on which $T$ and $S$ become equal, and onto whose open union the contraction map is a homeomorphism. The final corollary restates this source-locally: every prime of a finite-type quasi-finite algebra has a principal open neighbourhood on which the algebra is a localisation of a finite algebra. The Axiom of Choice is declared in both global statements and in the local theorem. The local proof uses it through the nowhere-quasi-finiteness lemma; the global proofs also use compactness of the spectrum and turn a finite open cover by distinguished opens into a unit-ideal expression. The fibre computations and one-variable correction lemmas are choice-free. Two published proofs left "details omitted" — the case $n = 1$ of Theorem 10.123.12's reduction and part (3) of Lemma 10.123.14 — are written out here in full.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-09-27Open item page →

Integral elements subalgebra of an arbitrary ring map

Definition

Let $R \to S$ be a unital ring map of commutative rings, with $φ$ its underlying map, and let $φ (R) \subseteq S$ be its image. An element $s \in S$ is integral over the ring map $R \to S$ when $s$ is a root of a monic polynomial with coefficients in the image of $R$ (Integral elements over a commutative ring and algebraic integers): that is, when

$s^{d} + φ (a_{d - 1}) s^{d - 1} + \dots + φ (a_{1}) s + φ (a_{0}) = 0$

for some $d \geq 1$ and some $a_{0}, \dots, a_{d - 1} \in R$ . This is integrality of the element $s$ over $φ (R)$ . The ring map $R \to S$ is integral in the sense of Integral ring maps and integral extensions precisely when every element $s \in S$ satisfies such an equation.

Write

${Int}_{R} (S) = {s \in S : s is integral over R \to S}$

for this set. Then ${Int}_{R} (S)$ is a subring of $S$ (Subring: a subset containing $1_{R}$ and closed under addition, additive inverses and multiplication) containing $φ (R)$ , hence an $R$ -subalgebra of $S$ for the restricted structure map (Algebras over a commutative ring, central structure maps, and algebra homomorphisms); it is called the integral closure of the image of $R$ in $S$ , and in this page the notation $S^{'} \subseteq S$ always denotes this relative integral closure. The reason is Integral elements over a nonzero base ring form a subring: if $φ (R) \neq 0$ the published statement applies to the inclusion $φ (R) \subseteq S$ , whose integral elements over $φ (R)$ are exactly the elements listed above; if $φ (R) = 0$ then $1_{S} = φ (1_{R}) = 0$ , so $S = 0$ and ${Int}_{R} (S) = {0} = S$ is again a subring.

Conventions kept here. (i) No hypothesis of injectivity is imposed: the map $R \to S$ may have a kernel, and ${Int}_{R} (S)$ is a subring of $S$ containing the image, not of $R$ . (ii) No hypothesis excluding zero divisors or nilpotents is imposed, and all arguments below proceed in $S$ itself; in particular Integrality and integral closure commute with localisation may be applied through the structure map without assuming that $R \to S$ is injective. (iii) When $R \subseteq S$ are domains in the sense of Integral closure in an extension ring and integrally closed domains, this set is the integral closure of $R$ in $S$ , so the present definition specialises to the published one, which is not assumed here.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-sol)audited 2026-09-27Open item page →

Quasi-finiteness at a prime of a finite-type algebra

Definition

Let $R \to S$ be a ring map of finite type (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras), let $q \in Spec (S)$ be a prime, and let $p = q \cap R$ be its contraction to $R$ . Write $κ (p) = R_{p} / p R_{p} ≅ Frac (R / p)$ for the residue field ( $R_{p} / p R_{p} ≅ Frac (R / p)$ is the residue field at $p$ ).

The map $R \to S$ is quasi-finite at $q$ when the $κ (p)$ -algebra

$S_{q} / p S_{q}$

is finite over $κ (p)$ , that is, finitely generated as a $κ (p)$ -module, equivalently finite-dimensional over $κ (p)$ (Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis); the quotient is a $κ (p)$ -algebra because $p S_{q}$ is the extension of the ideal $p$ . The map $R \to S$ is quasi-finite when it is of finite type and quasi-finite at every prime of $S$ .

The fibre form. The fibre of $Spec (S) \to Spec (R)$ over $p$ is $Spec (S \otimes_{R} κ (p))$ (The tensor product $M \otimes_{R} N$ from the additive group underlying the free $Z$ -module on $M \times N$ , elementary tensors, and finite tensor sums). The prime $q$ determines a prime $\overline{q}$ of the fibre, and the local ring of the fibre at that prime is $S_{q} / p S_{q}$ . Indeed the quotient and tensor laws identify $S \otimes_{R} κ (p) ≅ S_{p} / p S_{p}$ (locally on the base use $M \otimes_{R} R / I ≅ M / I M$ naturally over the local ring $R_{p}$ and Localisation commutes with quotient rings: $S^{- 1} R / S^{- 1} I ≅ {\overset{ˉ}{S}}^{- 1} (R / I)$ ), and localising that $κ (p)$ -algebra at the prime $\overline{q}$ and quotienting by $p$ recovers $S_{q} / p S_{q}$ . Either description may be used as the definition: the two are related by these canonical identifications, and all items on this page use whatever form makes the step at hand shortest.

Conventions kept here. (i) The definition is phrased only at a prime of $S$ and never requires a chosen closed point, so it applies to nonreduced rings, to noninjective maps, and to primes of arbitrarily large residue field. (ii) A fibre may be empty or may have infinitely many primes; quasi-finiteness is a condition at one prime at a time, and the map is quasi-finite only when the condition holds at all primes of $S$ . (iii) This is distinct from the classical closed-point convention of Quasi-finite classical morphisms, which tests only closed-point fibres of classical varieties over an algebraically closed field; the algebraic notion above is the one used by the Zariski Main Theorem on this page.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-27Open item page →

The leading coefficient times a root is integral

Statement

Let $R \to S$ be a unital ring map of commutative rings and let $t \in S$ satisfy a relation

$φ (a_{0}) + φ (a_{1}) t + \dots + φ (a_{n}) t^{n} = 0$

with $n \geq 0$ and $a_{0}, \dots, a_{n} \in R$ . Then $φ (a_{n}) t$ is integral over $R$ (Integral ring maps and integral extensions). No hypothesis is imposed on $φ (a_{n})$ : it may be zero, a zero divisor, or a nilpotent, and the map $R \to S$ need not be injective.

Facts & Assumptions

Given: A unital ring map $φ : R \to S$ of commutative rings, an integer $n \geq 0$ , elements $a_{0}, \dots, a_{n} \in R$ , and an element $t \in S$ satisfying $φ (a_{0}) + φ (a_{1}) t + \dots + φ (a_{n}) t^{n} = 0$ .

[L1]

Let $A \to B$ be a homomorphism of commutative rings. An element $b \in B$ is integral over $A$ when it is a root of a monic polynomial in $A [X]$ (Integral elements over a commutative ring and algebraic integers).

[L2]

Let $f : A \to B$ be a homomorphism of commutative rings. The map $f$ is an integral ring map when every element of $B$ is integral over $A$ in the sense of Integral elements over a commutative ring and algebraic integers (Integral ring maps and integral extensions).

Proof

technique · direct

1.1

If $n = 0$ the given relation reads $φ (a_{0}) = 0$ , hence $φ (a_{0}) t = 0$ ; the element $0 \in S$ is a root of the monic polynomial $X$ , so $φ (a_{0}) t$ is integral over $R$ . This disposes of the case $n = 0$ and from here on we assume $n \geq 1$ .

givenL1

1.2

Set $y = φ (a_{n}) t \in S$ , so that $φ (a_{n})^{n} t^{n} = y^{n}$ , and for each $0 \leq i \leq n - 1$ the identity $φ (a_{n})^{n - 1} \cdot t^{i} = φ (a_{n})^{n - 1 - i} y^{i}$ holds in $S$ ; these are the ordinary power identities for a single element.

givenalgebra

2.1

Multiplying the given relation by $φ (a_{n})^{n - 1}$ and using step 1.2 in every summand gives the identity $y^{n} + \sum_{i = 0}^{n - 1} φ ⁣ (a_{i} a_{n}^{n - 1 - i}) y^{i} = 0$ in $S$ : the term $φ (a_{n})^{n - 1} φ (a_{i}) t^{i}$ equals $φ (a_{i} a_{n}^{n - 1 - i}) y^{i}$ , and the term of index $n$ becomes $y^{n}$ after multiplication, with coefficient $1$ .

step 1.2algebra

3.1

The displayed identity of step 2.1 is a vanishing statement for the monic polynomial $P (X) = X^{n} + \sum_{i = 0}^{n - 1} φ (c_{i}) X^{i}$ with coefficients $c_{i} = a_{i} a_{n}^{n - 1 - i} \in R$ : its leading coefficient is $1$ , so it is monic in the sense of [L1], and $P (y) = 0$ .

step 2.1algebra

4.1

Steps 1.2 and 3.1 exhibit $y = φ (a_{n}) t$ as a root of a monic polynomial with coefficients in $R$ , so $φ (a_{n}) t$ is integral over $R$ by [L1]. Together with step 1.1 (the case $n = 0$ ), this proves the statement for every $n \geq 0$ , including $a_{n} = 0$ and $t = 0$ , where $y = 0$ is integral and the identity of step 2.1 reads $0 = 0$ . ∎

step 1.1step 3.1L1L2

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-sol)audited 2026-09-27Open item page →

One-variable integral correction after leading-coefficient localization

Statement

Let $R$ be a commutative ring, let $φ : R [x] \to S$ be a unital ring map of commutative rings, and let $t \in S$ be integral over the image subring $φ (R [x]) \subseteq S$ (Integral elements over a commutative ring and algebraic integers, Integral ring maps and integral extensions). Let $p = a_{0} + a_{1} x + \dots + a_{k} x^{k} \in R [x]$ be a polynomial with

$t φ (p) \in Im (φ) .$

If $p$ is monic, then there exists $q \in R [x]$ such that $t - φ (q)$ is integral over $R$ .
In general there exist $q \in R [x]$ and an integer $n \geq 0$ such that $φ (a_{k})^{n} t - φ (q)$ is integral over $R$ .

No injectivity of $φ$ , no regularity of the leading coefficient $a_{k}$ , and no reducedness or domain hypothesis is imposed: $a_{k}$ may be $0$ , a zero divisor or a nilpotent, $t$ may be $0$ or a zero divisor, and the polynomial ring and all localisations are taken over the possibly nonreduced ring $R$ . Part 1 is the monic case of the one-variable integral correction; part 2 obtains it in general by inverting $a_{k}$ .

Facts & Assumptions

Given: A unital ring map $φ : R [x] \to S$ of commutative rings, an element $t \in S$ integral over the subring $φ (R [x]) \subseteq S$ , and a polynomial $p = a_{0} + a_{1} x + \dots + a_{k} x^{k} \in R [x]$ with $t φ (p) \in Im (φ)$ .

[L1]

An element $b$ of a commutative ring $B$ is integral over a subring $A$ exactly when it is a root of a monic polynomial in $A [X]$ (Integral elements over a commutative ring and algebraic integers).

[L2]

Let $R$ be a commutative ring and let $g \in R [x]$ be monic. For every $f \in R [x]$ there are unique $q, r \in R [x]$ with $f = q g + r$ and $r = 0$ or $\deg r < \deg g$ (Division by a monic polynomial over a commutative ring).

[L3]

Let $A \to B$ be a homomorphism of commutative rings, $S \subseteq A$ multiplicative and $b \in B$ . If $b$ is integral over $A$ then $b / 1$ is integral over $S^{- 1} A$ in $S^{- 1} B$ ; and if $b / 1$ is integral over $S^{- 1} A$ in $S^{- 1} B$ then some $s \in S$ makes $s b$ integral over $A$ (Integrality and integral closure commute with localisation).

[L4]

If $A \to B$ and $B \to C$ are integral ring maps of commutative rings then the composite $A \to C$ is integral (Integral extensions are transitive).

[L5]

Let $A \subseteq B$ be commutative rings with $A \neq 0$ and $b \in B$ . Then $b$ is integral over $A$ if and only if $A [b]$ is finitely generated as an $A$ -module (Integrality and finite-module characterizations for one element).

[L6]

Let $R$ be a commutative ring, $S \subseteq R$ multiplicative, and $r \in R$ . Then the image of $r$ in $S^{- 1} R$ is zero if and only if $u r = 0$ for some $u \in S$ ; and $S^{- 1} R$ is the zero ring if and only if $0 \in S$ (Equality, vanishing, and the kernel of the localisation map).

[L7]

For a commutative ring $R$ and $f \in R$ the powers of $f$ form a multiplicative subset and the principal localisation $R_{f}$ has elements $r / f^{n}$ ; $R_{0}$ is the zero ring (Principal localisation $R_{f} = {1, f, f^{2}, \dots}^{- 1} R$ ).

[L8]

Let $A \subseteq B$ be commutative rings with $A \neq 0$ . The elements of $B$ integral over $A$ form a subring of $B$ (Integral elements over a nonzero base ring form a subring).

Proof

technique · direct

1.1

We first dispose of the degenerate cases. If $R = 0$ then $R [x] = 0$ and $1_{S} = φ (0) = 0$ , so $S = 0$ , $t = 0$ and $q = 0$ satisfies both claims. If $S = 0$ then also $t = 0$ and $q = 0$ satisfies both claims. If $t$ is nilpotent, say $t^{N} = 0$ with $N \geq 1$ , then $t$ is a root of the monic polynomial $X^{N}$ , hence integral over $R$ by [L1], so $q = 0$ satisfies claim 1 and $(n, q) = (0, 0)$ satisfies claim 2. If $a_{k}$ is nilpotent, say $a_{k}^{N} = 0$ with $N \geq 1$ , then $φ (a_{k})^{N} t - φ (0) = 0$ is integral over $R$ by [L1], so $(n, q) = (N, 0)$ satisfies claim 2. We therefore assume $R \neq 0$ , $S \neq 0$ , $t$ not nilpotent, and in the treatment of claim 2 the element $a_{k}$ not nilpotent.

givenL1

1.2

Assume that $p$ is monic, so that claim 1 is at stake. Since $t φ (p) \in Im (φ)$ , choose $r \in R [x]$ with $t φ (p) = φ (r)$ . By [L2] applied to the monic divisor $p$ there are $q, r^{'} \in R [x]$ with $r = q p + r^{'}$ and $r^{'} = 0$ or $\deg r^{'} < \deg p$ . Set $t^{'} = t - φ (q) \in S$ . Then $t^{'} φ (p) = t φ (p) - φ (q) φ (p) = φ (r) - φ (q p) = φ (r^{'})$ . If $t^{'}$ is nilpotent, say $t^{' N} = 0$ for some $N \geq 1$ , then $t^{'}$ is integral over $R$ by the monic equation $X^{N} = 0$ , proving claim 1; below assume $t^{'}$ is not nilpotent.

givenL1L2construct

2.1

The element $t^{'} = t - φ (q)$ is integral over $φ (R [x])$ : the given element $t$ is integral over $φ (R [x])$ and $φ (q) \in φ (R [x])$ , while $1_{S} \neq 0$ makes the subring $φ (R [x]) \subseteq S$ nonzero, so [L8] applies.

step 1.2L8algebra

2.2

Write $p = x^{d} + p_{d - 1} x^{d - 1} + \dots + p_{0}$ with $d \geq 0$ (a monic constant is the case $d = 0$ ) and write $r^{'} = r_{d - 1}^{'} x^{d - 1} + \dots + r_{0}^{'}$ with $r_{i}^{'} = 0$ for $i \geq \deg r^{'}$ . In the principal localisation $S_{t^{'}}$ of [L7] the identity $t^{'} φ (p) = φ (r^{'})$ of step 1.2 becomes $φ (p) = t^{' - 1} φ (r^{'})$ , that is, the identity $φ (x)^{d} + \sum_{i = 0}^{d - 1} (φ (p_{i}) - t^{' - 1} φ (r_{i}^{'})) φ (x)^{i} = 0$ in $S_{t^{'}}$ . Its coefficients $φ (p_{i}) - t^{' - 1} φ (r_{i}^{'})$ lie in the subring $φ (R) [1 / t^{'}] \subseteq S_{t^{'}}$ and its leading coefficient is $1$ ; by [L1] the element $φ (x)$ is integral over $φ (R) [1 / t^{'}]$ .

step 1.2L1L7algebra

3.1

By [L5] applied to the nonzero ring $φ (R) [1 / t^{'}]$ and the integral element $φ (x)$ of step 2.2, the subalgebra $φ (R) [1 / t^{'}] [φ (x)]$ is a finitely generated $φ (R) [1 / t^{'}]$ -module, so every one of its elements is integral over $φ (R) [1 / t^{'}]$ by [L8]; equivalently the ring map $φ (R) [1 / t^{'}] \to φ (R) [1 / t^{'}] [φ (x)]$ is integral.

step 2.2L5L8algebra

4.1

Put $A = φ (R) [1 / t^{'}]$ and $B = A [φ (x)]$ inside $S_{t^{'}}$ . The element $t^{'}$ is integral over $B$ : its monic equation over $φ (R [x])$ from step 2.1 remains a monic equation over the larger subring $B$ . By step 3.1, $A \to B$ is integral. Since the integral elements over $B$ form a subring by [L8], the ring $B [t^{'}]$ is integral over $B$ ; transitivity [L4] makes $A \to B [t^{'}]$ integral. Therefore $t^{'} \in B [t^{'}]$ is integral over $A$ by [L1].

step 2.1step 3.1L1L4L8

5.1

By [L1] there are an integer $e \geq 1$ and coefficients $c_{0}, \dots, c_{e - 1} \in A = φ (R) [1 / t^{'}]$ with $t^{' e} + \sum_{i = 0}^{e - 1} c_{i} t^{' i} = 0$ . Each $c_{i}$ is a finite sum $\sum_{j = 1}^{m_{i}} φ (b_{i j}) / t^{' ℓ_{i j}}$ with $b_{i j} \in R$ and $ℓ_{i j} \geq 0$ , since $A$ is generated as a subring by $φ (R)$ and $t^{' - 1}$ . Choose $N \geq 0$ at least every exponent $ℓ_{i j}$ (take $N = 0$ if there are no summands). Multiplying the relation by $t^{' N}$ in $S_{t^{'}}$ gives $t^{' e + N} + \sum_{i = 0}^{e - 1} \sum_{j = 1}^{m_{i}} φ (b_{i j}) t^{' i + N - ℓ_{i j}} = 0,$ where every exponent $i + N - ℓ_{i j}$ is nonnegative.

step 4.1L1L7algebra

6.1

The identity of step 5.1 holds in $S_{t^{'}}$ , so by the kernel criterion [L6] applied to the localisation map $S \to S_{t^{'}}$ and the element $b = t^{' e + N} + \sum_{i = 0}^{e - 1} \sum_{j = 1}^{m_{i}} φ (b_{i j}) t^{' i + N - ℓ_{i j}} \in S$ there is $M \geq 0$ with $t^{' M} b = 0$ in $S$ . Hence $t^{' e + N + M} + \sum_{i = 0}^{e - 1} \sum_{j = 1}^{m_{i}} φ (b_{i j}) t^{' i + N - ℓ_{i j} + M} = 0$ in $S$ : a monic polynomial relation for $t^{'} = t - φ (q)$ with coefficients in $φ (R) \subseteq S$ . By [L1] the element $t - φ (q)$ is integral over $R$ , which is claim 1.

step 5.1L1L6algebra

7.1

Now let $p = \sum_{i = 0}^{k} a_{i} x^{i}$ be arbitrary with $t φ (p) \in Im (φ)$ , and assume first that $a_{k}$ is not nilpotent. Then the principal localisation $R^{'} = R_{a_{k}}$ of [L7] is nonzero by [L6], since its multiplicative set contains $1 = a_{k}^{0}$ while no power of $a_{k}$ is $0$ ; and $p^{'} = a_{k}^{- 1} p = x^{k} + \sum_{i = 0}^{k - 1} (a_{i} / a_{k}) x^{i} \in R^{'} [x]$ is monic. Let $S^{'} = S_{a_{k}}$ , let $φ^{'} : R^{'} [x] \to S^{'}$ be the localisation of $φ$ , and let $t^{'} \in S^{'}$ be the image of $t$ , so that $R^{'} [x] = R [x]_{a_{k}} = R_{a_{k}} [x]$ . The element $t^{'}$ is integral over $φ^{'} (R^{'} [x])$ , because the monic equation for $t$ over $φ (R [x])$ transports along the ring map induced by $φ$ ; and $t^{'} φ^{'} (p^{'}) = t φ (p) / a_{k}$ lies in $Im (φ^{'})$ , because $t φ (p) = φ (r)$ for some $r \in R [x]$ gives $t^{'} φ^{'} (p^{'}) = φ^{'} (r / a_{k})$ . Applying claim 1, already proved in steps 1.2-6.1 for the data $(R^{'}, S^{'}, φ^{'}, t^{'}, p^{'})$ , yields $q^{'} \in R^{'} [x]$ with $t^{'} - φ^{'} (q^{'})$ integral over $R^{'}$ .

step 6.1L1L6L7algebra

8.1

Write $q^{'}$ as a finite sum of terms $c / a_{k}^{m}$ with $c \in R [x]$ and $m \geq 0$ , and let $n$ be the maximum of the finitely many exponents $m$ occurring, with $n = 0$ when $q^{'} = 0$ . Then $a_{k}^{n} q^{'}$ is the image of some $q \in R [x]$ under $R [x] \to R^{'} [x]$ , namely $q = \sum_{j} c_{j} a_{k}^{n - m_{j}}$ for the finitely many summands $c_{j} / a_{k}^{m_{j}}$ of $q^{'}$ . With this $n$ and $q$ , the image in $S^{'}$ of $φ (a_{k})^{n} t - φ (q)$ is $a_{k}^{n} t^{'} - a_{k}^{n} φ^{'} (q^{'}) = a_{k}^{n} (t^{'} - φ^{'} (q^{'}))$ , and this is integral over $R^{'}$ because $t^{'} - φ^{'} (q^{'})$ is (step 7.1) and $a_{k}^{n} \in R^{'}$ .

step 7.1L1L7construct

9.1

Apply clause 2 of [L3] to the ring map $R \to S$ , the multiplicative subset ${a_{k}^{m} : m \geq 0} \subseteq R$ and the element $b = φ (a_{k})^{n} t - φ (q) \in S$ : its image in $S_{a_{k}} = S^{'}$ is integral over $R_{a_{k}} = R^{'}$ by step 8.1, so there exists $m \geq 0$ such that $a_{k}^{m} b$ is integral over $R$ . With $n^{'} = m + n \geq 0$ and $q^{''} = a_{k}^{m} q \in R [x]$ one has $φ (a_{k})^{n^{'}} t - φ (q^{''}) = a_{k}^{m} (φ (a_{k})^{n} t - φ (q))$ , which is integral over $R$ . This is claim 2 for arbitrary $p$ with $t φ (p) \in Im (φ)$ .

step 8.1L3algebra

10.1

Together, step 1.1 (the degenerate cases, including nilpotent $a_{k}$ and nilpotent $t$ ), step 6.1 (claim 1) and step 9.1 (claim 2) prove both assertions of the statement for every commutative ring $R$ , every unital $φ$ , every $t$ integral over $φ (R [x])$ and every $p$ with $t φ (p) \in Im (φ)$ ; this includes $p = 0$ , monic constants $p$ , the zero element $t = 0$ , and zero divisors or nilpotents among the $a_{i}$ . ∎

step 1.1step 6.1step 9.1given

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-09-27Open item page →

Strong transcendence over a subring

Definition

Let $R \subseteq S$ be an inclusion of commutative rings (Subring: a subset containing $1_{R}$ and closed under addition, additive inverses and multiplication) and let $x \in S$ . The element $x$ is strongly transcendental over $R$ when the implication

$u (a_{0} + a_{1} x + \dots + a_{k} x^{k}) = 0 ⟹ u a_{i} = 0 for every i \in {0, \dots, k}$

holds for every integer $k \geq 0$ , every multiplier $u \in S$ and all $a_{0}, a_{1}, \dots, a_{k} \in R$ ; the polynomial $a_{0} + a_{1} x + \dots + a_{k} x^{k}$ is the evaluation in $S$ of a polynomial over $R$ (The polynomial ring over a commutative ring as finitely supported coefficient sequences with convolution), so the condition is a statement about all polynomial relations that hold in $S$ between $x$ and the elements of $R$ .

Conventions kept here. (i) The multiplier $u$ is retained on purpose. It records annihilators: when $S$ has zero divisors the vanishing of a product $u \cdot P (x)$ does not force $P (x)$ to vanish, and it is the annihilator form above, not mere linear independence of the monomials, that is used in the minimal-prime argument of this page. (ii) No finiteness or noetherian hypothesis is imposed on $R$ , on $S$ or on the polynomial degree; $x$ has trivial annihilator in $S$ : applying the condition to $P (X) = X$ gives $u x = 0 \Rightarrow u = 0$ . The zero polynomial ( $k = 0$ , $a_{0} = 0$ ) is included. (iii) The condition is tied to the chosen pair $R \subseteq S$ and is not preserved by an arbitrary quotient: for $R = k$ a field $S = k [z] \times k [y]$ and $x = (z, y)$ , the element $x$ is strongly transcendental over $k$ (a multiplier $u = (u_{1}, u_{2})$ with $u \cdot (P (z), P (y)) = 0$ has $u_{1} P (z) = 0$ and $u_{2} P (y) = 0$ , so if $P = 0$ all coefficients vanish, whereas if $P \neq 0$ both $P (z)$ and $P (y)$ are nonzero, so the domain property forces $u_{1} = u_{2} = 0$ ; in either case $u a_{i} = 0$ for every $i$ ), yet in the quotient $S / q$ by the prime $q = (z - 1) k [z] \times k [y]$ the image of $x$ is $1 \in k$ , which is a root of the monic $X - 1$ and therefore not transcendental at all. The descent proved on this page is consequently formulated for minimal primes of reduced rings.

Domains. If $S$ is an integral domain (Zero divisor, and integral domain: a commutative ring with $1 \neq 0$ and no zero divisors), then strong transcendence of $x$ over $R$ is the same as saying that $x$ is transcendental over the fraction field $Frac (R)$ , viewed inside $Frac (S)$ (The field of fractions $Frac (D) = (D ∖ {0})^{- 1} D$ of an integral domain). Suppose first that $x$ is strongly transcendental over $R$ and let $P \in Frac (R) [X]$ be nonzero with $P (x) = 0$ ; clearing denominators gives $u \in R$ nonzero with $u P \in R [X]$ still nonzero, and evaluating gives $u \cdot (u P) (x) = u \cdot 0 = 0$ , so strong transcendence applied to the polynomial $u P$ and the multiplier $u$ forces $u \cdot (u p_{i}) = 0$ for every coefficient $p_{i}$ of $P$ ; since $S$ is a domain and $u \neq 0$ , this gives $p_{i} = 0$ for all $i$ , contradicting $P \neq 0$ . Conversely, if $x$ is transcendental over $Frac (R)$ and $u (a_{0} + \dots + a_{k} x^{k}) = 0$ with $u \neq 0$ , then cancelling the nonzero element $u$ in the domain $S$ exhibits the polynomial $a_{0} + \dots + a_{k} X^{k} \in R [X]$ as vanishing at $x$ ; if some $a_{i}$ were nonzero that polynomial would be a nonzero element of $Frac (R) [X]$ vanishing at $x$ , contradicting transcendence, so all $a_{i}$ vanish and $u a_{i} = 0$ holds; for $u = 0$ the conclusion is immediate. In this case the annihilator clause is automatic and the condition reduces to the classical notion. Nothing in the argument uses integrality; compare Integral elements over a commutative ring and algebraic integers, where the integral element is the opposite extreme, a root of a monic polynomial.

LemmaStatement: Literature-sourcedProof: AI-adaptedaudited 2026-09-27Open item page →

Strong transcendence descends to reduced minimal-prime quotients

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $R \subseteq S$ be an inclusion of reduced commutative rings (The nilradical and reduced rings), let $x \in S$ be strongly transcendental over $R$ (Strong transcendence over a subring), let $q \subseteq S$ be a minimal prime and let $p = R \cap q$ be its contraction to $R$ . Then the image of $x$ in the domain $S / q$ is strongly transcendental over the subring $R / p \subseteq S / q$ (The quotient ring $R / I$ with $(r + I) (s + I) = r s + I$ ).

The minimality of $q$ is essential: the local ring $S_{q}$ is then a field, which is what lets one clear a denominator outside $q$ , and the annihilator-sensitive form of strong transcendence is what makes the clearing argument work without assuming that $S$ is a domain.

Facts & Assumptions

Given: The Axiom of Choice; an inclusion of reduced commutative rings $R \subseteq S$ , an element $x \in S$ strongly transcendental over $R$ , a minimal prime $q \subseteq S$ and $p = R \cap q$ .

[L1]

For $R \subseteq S$ and $x \in S$ , strong transcendence of $x$ over $R$ means that $u (a_{0} + a_{1} x + \dots + a_{k} x^{k}) = 0$ with $u \in S$ and $a_{i} \in R$ implies $u a_{i} = 0$ for all $i$ (Strong transcendence over a subring).

[L2]

The nilradical $Nil (R)$ of a commutative ring is the radical of the zero ideal, so $x \in Nil (R)$ means $x^{n} = 0$ for some $n \geq 1$ , and $R$ is reduced when $Nil (R) = (0)$ (The nilradical and reduced rings).

[L3]

The nilradical of a commutative ring is the intersection of all of its prime ideals; this is the point where the Axiom of Choice is used, through the existence of primes avoiding a given element (The nilradical is the intersection of all prime ideals).

[L4]

For a commutative ring $R$ , a multiplicative subset $S \subseteq R$ and an ideal $I ⊴ R$ one has $S^{- 1} \sqrt{I} = \sqrt{S^{- 1} I}$ as ideals of $S^{- 1} R$ (Radicals commute with localization).

[L5]

For a commutative ring $R$ and a prime $p \in Spec (R)$ , contraction along $R \to R_{p}$ is an inclusion-preserving bijection from $Spec (R_{p})$ onto the set of primes $q \subseteq p$ of $R$ ; the inverse sends $q$ to $q R_{p}$ (Primes of a localization at a prime).

[L6]

For a multiplicative subset $S$ of a commutative ring $R$ and $r \in R$ , the image of $r$ in $S^{- 1} R$ is zero if and only if $u r = 0$ for some $u \in S$ (Equality, vanishing, and the kernel of the localisation map).

[L7]

For a prime $p \subseteq R$ the localisation $R_{p}$ is $(R ∖ p)^{- 1} R$ , with elements fractions $r / s$ for $s \notin p$ (Localisation at a prime ideal: $R_{p} = (R ∖ p)^{- 1} R$ ).

[L8]

For an ideal $I$ of a commutative ring $R$ the quotient ring $R / I$ has elements the cosets $r + I$ , and the canonical projection $R \to R / I$ , $π (r) = r + I$ , is a surjective ring homomorphism with kernel $I$ (The quotient ring $R / I$ with $(r + I) (s + I) = r s + I$ , The canonical projection $R \to R / I$ is a surjective ring homomorphism with kernel $I$ ).

[L9]

The Axiom of Choice is the assertion that every family of nonempty sets has a choice function; it is assumed in this item and used exactly through [L3] (The Axiom of Choice).

Proof

technique · direct

1.1

We work under [L9]. Set $\overline{R} = R / p$ and $\overline{S} = S / q$ , and let $\overline{x}$ denote the image of $x$ in $\overline{S}$ by [L8]. Since $p = R \cap q$ is the kernel of the composite $R \to S \to S / q$ , the map $\overline{R} \to \overline{S}$ induced by the inclusion is injective, so $\overline{R}$ is a subring of $\overline{S}$ ; both are domains, hence reduced by [L2].

givenL2L8

1.2

The localisation $S_{q}$ is reduced. Indeed, applying [L4] to the ideal $I = (0)$ of $S$ gives $S^{- 1} \sqrt{(0)} = \sqrt{S^{- 1} (0)}$ for the multiplicative subset $S ∖ q$ of [L7], that is, the extension of $Nil (S) = (0)$ is $Nil (S_{q})$ ; the extension of the zero ideal is zero, so $Nil (S_{q}) = (0)$ by [L2].

givenL2L4L7

1.3

By [L5] the primes of $S_{q}$ correspond bijectively to the primes of $S$ contained in $q$ . Since $q$ is a minimal prime, the only prime of $S$ contained in $q$ is $q$ itself, so $q S_{q}$ is the unique prime ideal of $S_{q}$ .

givenL5L7

2.1

By [L3] applied to the ring $S_{q}$ , its nilradical is the intersection of its prime ideals; by step 1.3 that intersection is the single ideal $q S_{q}$ , so $Nil (S_{q}) = q S_{q}$ . By step 1.2 the nilradical is zero, hence $q S_{q} = 0$ . In particular the image in $S_{q}$ of every element of $q$ is a member of $q S_{q} = 0$ , hence is zero.

step 1.2step 1.3L3

2.2

Now let $\overline{u} \in \overline{S}$ and ${\overline{a}}_{0}, \dots, {\overline{a}}_{k} \in \overline{R}$ satisfy $\overline{u} ({\overline{a}}_{0} + {\overline{a}}_{1} \overline{x} + \dots + {\overline{a}}_{k} {\overline{x}}^{k}) = 0$ in $\overline{S}$ . Choose preimages $u \in S$ and $a_{i} \in R$ under the quotient maps of [L8]; then $u (a_{0} + a_{1} x + \dots + a_{k} x^{k}) \in q$ , because its image in $\overline{S}$ is the left-hand side of the assumed relation.

givenstep 1.1L8algebra

3.1

By step 2.1 the image of the element $u (a_{0} + a_{1} x + \dots + a_{k} x^{k}) \in q$ in $S_{q}$ is zero, so the kernel criterion [L6] applied to the localisation $S \to S_{q}$ at the multiplicative subset $S ∖ q$ of [L7] provides $u^{'} \in S ∖ q$ with $u^{'} u (a_{0} + a_{1} x + \dots + a_{k} x^{k}) = 0$ in $S$ .

step 2.1step 2.2L6L7

4.1

The element $x$ is strongly transcendental over $R$ by hypothesis, so [L1] applied to the vanishing product of step 3.1 with multiplier $u^{'} u \in S$ gives $u^{'} u a_{i} = 0$ in $S$ for every $i$ .

givenstep 3.1L1

5.1

Since $u^{'} u a_{i} = 0 \in q$ and $u^{'} \notin q$ , the primality of $q$ gives $u a_{i} \in q$ for every $i$ . Passing to $\overline{S}$ by [L8], this says $\overline{u} {\overline{a}}_{i} = 0$ in $\overline{S}$ for every $i$ .

step 4.1L8algebra

6.1

Steps 2.2, 5.1 and the discussion of step 1.1 show: for every $k \geq 0$ , every multiplier $\overline{u} \in \overline{S}$ and all ${\overline{a}}_{i} \in \overline{R}$ , the vanishing $\overline{u} ({\overline{a}}_{0} + \dots + {\overline{a}}_{k} {\overline{x}}^{k}) = 0$ implies $\overline{u} {\overline{a}}_{i} = 0$ for all $i$ . By [L1] the image $\overline{x}$ is strongly transcendental over $\overline{R} = R / p$ inside $\overline{S} = S / q$ , which is the assertion; the Axiom of Choice was assumed in step 1.1 and used only through [L3] in step 2.1. ∎

step 1.1step 2.2step 5.1L1

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-sol)audited 2026-09-27Open item page →

Polynomial rings over normal domains are normal

Statement

Let $R$ be an integrally closed domain (Integral closure in an extension ring and integrally closed domains) with fraction field $K = Frac (R)$ (The field of fractions $Frac (D) = (D ∖ {0})^{- 1} D$ of an integral domain). Then the polynomial ring $R [x]$ (The polynomial ring over a commutative ring as finitely supported coefficient sequences with convolution) is integrally closed: every element of $Frac (R [x])$ that is integral over $R [x]$ (Integral elements over a commutative ring and algebraic integers) already lies in $R [x]$ .

No Noetherian hypothesis is imposed on $R$ : the proof reduces an arbitrary monic equation to a finitely generated $Z$ -subalgebra of $R$ and proves that this subalgebra is Noetherian by a direct finite-generator argument. The argument is choice-free.

Facts & Assumptions

Given: An integrally closed domain $R$ with fraction field $K = Frac (R)$ , and an element $z \in Frac (R [x])$ integral over $R [x]$ .

[L1]

A domain $A$ is integrally closed when every element of $Frac (A)$ integral over $A$ already lies in $A$ (Integral closure in an extension ring and integrally closed domains).

[L2]

If $D$ is an integral domain then $Frac (D) = (D ∖ {0})^{- 1} D$ , and its elements are fractions $a / b$ with $a, b \in D$ and $b \neq 0$ (The field of fractions $Frac (D) = (D ∖ {0})^{- 1} D$ of an integral domain).

[L3]

If $f : R \to A$ is a unital homomorphism of commutative rings and $f (s)$ is a unit of $A$ for every $s \in S$ , then there is a unique unital ring homomorphism $\tilde{f} : S^{- 1} R \to A$ with $\tilde{f} \circ λ_{S} = f$ , namely $\tilde{f} (r / s) = f (r) f (s)^{- 1}$ (Universal property of localisation: maps that invert $S$ factor uniquely through $S^{- 1} R$ ).

[L4]

If $R$ is an integral domain then $R [x]$ is an integral domain (A polynomial ring over an integral domain is an integral domain).

[L5]

For $0 \neq f = \sum_{i} a_{i} x^{i} \in R [x]$ the degree $\deg f$ is the largest $n$ with $a_{n} \neq 0$ and the leading coefficient is $lc (f) = a_{\deg f}$ ; the zero polynomial has no degree and no leading coefficient (Degree, leading coefficient and monic polynomial, with the zero polynomial having no degree).

[L6]

Let $R$ be a commutative ring and $f, g \in R [x]$ nonzero. If $f + g \neq 0$ then $\deg (f + g) \leq \max {\deg f, \deg g}$ ; the coefficient of $x^{\deg f + \deg g}$ in $f g$ is $lc (f) lc (g)$ (Degree inequalities for sums and products over a commutative ring).

[L7]

A domain has no zero divisors, so a product of nonzero elements is nonzero and cancellation of a nonzero factor is legitimate (Zero divisor, and integral domain: a commutative ring with $1 \neq 0$ and no zero divisors).

[L8]

Let $F$ be a field and $f, g \in F [x]$ not both zero. There are $A, B \in F [x]$ with $A f + B g = d$ , where $d$ is the monic greatest common divisor of $f, g$ , and $d$ divides both $f$ and $g$ (Bézout identity and the Euclidean algorithm for polynomials over a field).

[L9]

[L10]

A commutative ring is Noetherian exactly when every ideal of it is finitely generated (Left and right Noetherian rings, Noetherian modules: every submodule is finitely generated).

[L11]

In a commutative ring, $(S)$ consists of the finite sums $\sum_{i} r_{i} s_{i}$ with $r_{i} \in R$ and $s_{i} \in S$ , and the principal ideal $(a)$ equals $R a$ ; the empty sum is included and equals $0$ (In a commutative ring, $(S)$ consists of finite sums $\sum r_{i} s_{i}$ , and $(a) = R a$ ).

[L12]

Every subgroup of $(Z, +)$ equals $n Z = ⟨ n ⟩$ for exactly one natural number $n$ (Every subgroup of $(Z, +)$ is $⟨ n ⟩ = n Z$ for exactly one natural number $n$ ).

[L13]

For an ideal $I ⊴ R$ the canonical projection $π : R \to R / I$ , $π (r) = r + I$ , is a surjective ring homomorphism with kernel $I$ (The canonical projection $R \to R / I$ is a surjective ring homomorphism with kernel $I$ ).

[L14]

For $I ⊴ R$ , the maps $J \mapsto J / I$ and $K \mapsto π^{- 1} (K)$ are inverse inclusion-preserving bijections between the ideals $J$ of $R$ containing $I$ and the ideals $K$ of $R / I$ (Correspondence theorem: ideals of $R / I$ correspond to ideals of $R$ containing $I$ ).

[L15]

A subring $S \subseteq R$ contains $1_{R}$ and is closed under addition, negation and multiplication (Subring: a subset containing $1_{R}$ and closed under addition, additive inverses and multiplication).

Proof

technique · direct

1.1

We first prove the field case. Let $K$ be a field and let $z \in K (x) = Frac (K [x])$ be integral over $K [x]$ ; here $K [x]$ is a domain by [L4], so [L2] lets us write $z = f / g$ with $f, g \in K [x]$ and $g \neq 0$ . If $f = 0$ then $z = 0 \in K [x]$ . Otherwise $f, g$ are not both zero, so [L8] provides $A, B \in K [x]$ and the monic greatest common divisor $d = \gcd (f, g)$ with $d = A f + B g$ , $d ∣ f$ and $d ∣ g$ . Write $f = d f_{1}$ and $g = d g_{1}$ with $f_{1}, g_{1} \in K [x]$ . The polynomial $d$ is monic, hence nonzero, so cancellation in the domain $K [x]$ of [L4] gives $A f_{1} + B g_{1} = 1$ from $d (A f_{1} + B g_{1}) = d \cdot 1$ , and the same cancellation in $K (x)$ gives $z = f_{1} / g_{1}$ .

givenL2L4L7L8

1.2

Let $D$ be a domain with fraction field $L$ . Call an element $u \in L$ almost integral over $D$ when there is a nonzero $d \in D$ with $d u^{n} \in D$ for every integer $n \geq 0$ . Every element of $D$ is almost integral over $D$ with multiplier $1$ , the fraction field $L$ of $D$ being that of [L2] and $D$ being a domain in the sense of [L7]. This is a convention internal to the proof; the following steps establish the three properties of it that the argument uses.

givenL2L7construct

1.3

Every ideal $I$ of $Z$ is an additive subgroup of $(Z, +)$ , hence equals $n Z = (n)$ for some natural number $n$ by [L12] and [L11]. So every ideal of $Z$ is finitely generated and $Z$ is Noetherian by [L10].

L10L11L12

1.4

Now let $C$ be a commutative ring in which every ideal is finitely generated, and let $I \subseteq C [x]$ be an ideal. For $n \geq 0$ let $S_{n}$ be the set consisting of $0$ and of the leading coefficients of all elements of $I$ of degree $\leq n$ [L5], and put $J_{n} = (S_{n}) \subseteq C$ [L11]; then $J_{0} \subseteq J_{1} \subseteq \dots$ , since $S_{n} \subseteq S_{n + 1}$ , and $J = ⋃_{n \geq 0} J_{n}$ is an ideal of $C$ . By hypothesis $J = (g_{1}, \dots, g_{m})$ for finitely many $g_{i} \in C$ [L11], and each $g_{i} \in J_{n_{i}}$ for some $n_{i}$ because $J$ is the union of the $J_{n}$ ; with $N = \max {n_{1}, \dots, n_{m}}$ this gives $J = J_{N}$ . For each $n \leq N$ the ideal $J_{n}$ is finitely generated and is generated by $S_{n}$ , so finitely many elements of $S_{n}$ , say the leading coefficients of polynomials $p_{n, 1}, \dots, p_{n, k_{n}} \in I$ of degree $\leq n$ , generate $J_{n}$ [L11]. Let $W$ be the finite set of all these polynomials for $n \leq N$ .

givenL5L11construct

1.5

Let $C$ be a commutative ring in which every ideal is finitely generated, let $I \subseteq C$ be an ideal and let $K \subseteq C / I$ be an ideal. By [L14] the preimage $J = π^{- 1} (K)$ is an ideal of $C$ containing $I$ and $J / I = K$ , and $J$ is finitely generated, say $J = (c_{1}, \dots, c_{k})$ ; by [L13] the projection $π$ is surjective, so $K = π (J)$ is generated by the images $π (c_{1}), \dots, π (c_{k})$ [L11]. Hence every ideal of $C / I$ is finitely generated.

givenL11L13L14

2.1

Let $z^{n} + c_{n - 1} z^{n - 1} + \dots + c_{0} = 0$ with $n \geq 1$ and $c_{j} \in K [x]$ be a monic equation for $z$ over $K [x]$ , which exists because $z$ is integral over $K [x]$ . Substituting $z = f_{1} / g_{1}$ and multiplying by $g_{1}^{n}$ gives $f_{1}^{n} = - (c_{n - 1} f_{1}^{n - 1} g_{1} + \dots + c_{0} g_{1}^{n})$ , so $f_{1}^{n} \in (g_{1})$ by [L11]. Raising $A f_{1} + B g_{1} = 1$ to the $n$ -th power and expanding by the binomial theorem, every term of the expansion contains a factor $f_{1}^{n}$ or a factor $g_{1}$ , so $1 = A^{'} f_{1}^{n} + B^{'} g_{1}$ for suitable $A^{'}, B^{'} \in K [x]$ ; by [L11] both summands lie in the ideal $(g_{1})$ , so $1 \in (g_{1})$ , say $g_{1} w = 1$ with $w \in K [x]$ . Then $z = f_{1} / g_{1} = f_{1} w \in K [x]$ . Hence every element of $K (x)$ integral over $K [x]$ lies in $K [x]$ : for every field $K$ the ring $K [x]$ is integrally closed in its fraction field.

step 1.1L11algebra

2.2

Let $u, v \in L$ be almost integral over $D$ with multipliers $d, e \in D ∖ {0}$ . Then $d e \neq 0$ by [L7], and for every $n \geq 0$ one has $(d e) (u v)^{n} = (d u^{n}) (e v^{n}) \in D$ and $(d e) (u + v)^{n} = \sum_{j = 0}^{n} (\binom{n}{j}) (d u^{j}) (e v^{n - j}) \in D$ , each summand being a product of two elements of $D$ . So the almost integral elements of $L$ form a subring of $L$ containing $D$ .

step 1.2L7algebra

2.3

Let $u \in L$ be integral over $D$ . By [L9] the ring $D [u]$ is a finitely generated $D$ -module; choose generators $h_{1}, \dots, h_{N} \in D [u]$ and write $h_{i} = a_{i} / d_{i}$ with $a_{i}, d_{i} \in D$ and $d_{i} \neq 0$ by [L2]. Then $d = d_{1} \dots d_{N} \neq 0$ by [L7] and $d h_{i} = a_{i} \prod_{j \neq i} d_{j} \in D$ for every $i$ , so $d \cdot D [u] \subseteq D$ ; in particular $d u^{n} \in D$ for all $n \geq 0$ . Hence every element of $L$ integral over $D$ is almost integral over $D$ .

step 1.2L2L7L9

2.4

Suppose every ideal of $D$ is finitely generated and $u \in L$ is almost integral over $D$ with multiplier $d \neq 0$ . Then $D [u] \subseteq d^{- 1} D = {c / d : c \in D}$ , since $d u^{n} \in D$ for every $n \geq 0$ and $d^{- 1} D$ is a $D$ -submodule of $L$ . Multiplication by $d$ is an injective $D$ -module map $D [u] \to D$ , because $L$ is a field and $d \neq 0$ , and its image $d D [u]$ is an ideal of $D$ ; by hypothesis $d D [u] = (y_{1}, \dots, y_{k})$ for some $y_{j} \in D$ [L11]. Then $D [u] = (d^{- 1} y_{1}, \dots, d^{- 1} y_{k})$ inside $L$ : each $d^{- 1} y_{j}$ lies in $D [u]$ because $y_{j} = d x_{j}$ for some $x_{j} \in D [u]$ and cancellation in $L$ gives $d^{- 1} y_{j} = x_{j}$ , while for $x \in D [u]$ the identity $d x = \sum_{j} c_{j} y_{j}$ with $c_{j} \in D$ gives $x = \sum_{j} c_{j} (d^{- 1} y_{j})$ . So $D [u]$ is a finitely generated $D$ -module and [L9] makes $u$ integral over $D$ . Consequently, over a domain in which every ideal is finitely generated, almost integral and integral elements of the fraction field coincide.

step 1.2L7L9L11

2.5

We show that the finite set $W$ generates $I$ . Let $f \in I$ be nonzero of degree $m$ and leading coefficient $b \neq 0$ [L5], and put $m^{*} = \min {m, N}$ . The leading coefficient $b$ lies in $J_{m}$ , and $J_{m} = J_{m^{*}}$ : for $m \leq N$ this is $m^{*} = m$ , and for $m > N$ it holds because $J_{N} \subseteq J_{m} \subseteq J = J_{N}$ . So $b = \sum_{j} r_{j} lc (p_{m^{*}, j})$ with $r_{j} \in C$ [L11]. Every $p_{m^{*}, j}$ has degree $\leq m^{*} \leq m$ , and the polynomial $f - \sum_{j} r_{j} x^{m - \deg p_{m^{*}, j}} p_{m^{*}, j}$ lies in $I$ , is congruent to $f$ modulo $(W)$ , and has degree $< m$ by [L6]. Induction on $m$ therefore exhibits every element of $I$ as an element of $(W)$ ; hence $I$ is finitely generated, and every ideal of $C [x]$ is finitely generated.

step 1.4L5L6L11algebra

2.6

Let $C$ be a commutative ring in which every ideal is finitely generated, let $A$ be a commutative $C$ -algebra and let $t_{1}, \dots, t_{d} \in A$ . The $C$ -subalgebra of $A$ generated by $t_{1}, \dots, t_{d}$ is the image of the evaluation homomorphism $C [x_{1}, \dots, x_{d}] \to A$ with $x_{i} \mapsto t_{i}$ , hence is a quotient of $C [x_{1}, \dots, x_{d}]$ ; iterating step 1.4 and applying step 1.5, every ideal of it is finitely generated. In particular, taking $C = Z$ and using step 1.3, every $Z$ -subalgebra of a commutative ring that is generated by finitely many elements is Noetherian by [L10].

step 1.3step 1.4step 1.5L10L15algebra

3.1

Let $D$ be a domain with fraction field $K_{0}$ and let $f = α_{0} + α_{1} x + \dots + α_{r} x^{r} \in K_{0} [x]$ with $α_{r} \neq 0$ be almost integral over $D [x]$ , with multiplier $h = b_{0} + b_{1} x + \dots + b_{s} x^{s} \in D [x] ∖ {0}$ and $b_{s} \neq 0$ . For every $n \geq 0$ the coefficient of $x^{r n + s}$ in the product $h f^{n}$ is $b_{s} α_{r}^{n}$ by [L6], and $h f^{n} \in D [x]$ , so $b_{s} α_{r}^{n} \in D$ ; thus $α_{r}$ is almost integral over $D$ with multiplier $b_{s}$ . Then $α_{r} x^{r}$ is almost integral over $D [x]$ with the same multiplier, because $b_{s} (α_{r} x^{r})^{n} = (b_{s} α_{r}^{n}) x^{r n} \in D [x]$ , and by step 2.2 the difference $f - α_{r} x^{r} \in K_{0} [x]$ , which has degree $< r$ , is almost integral over $D [x]$ as well. Induction on $r$ and on the degree of $f$ therefore shows that every coefficient $α_{i}$ of $f$ is almost integral over $D$ .

step 2.2L4L5L6algebra

3.2

Now return to the given data: $R$ is an integrally closed domain with fraction field $K$ [L1], and $z \in Frac (R [x])$ is integral over $R [x]$ . The rings $R [x] \subseteq K [x]$ are domains by [L4], $K (x) = Frac (K [x])$ is a field, and the inclusion $R [x] \to K (x)$ carries every nonzero element of $R [x]$ to a unit, so by [L3] it extends uniquely to a unital ring homomorphism $Frac (R [x]) \to K (x)$ , $f / g \mapsto f / g$ ; this map is injective because its restriction to the domain $R [x]$ is injective. We therefore regard $z$ as an element of $K (x)$ . The monic equation for $z$ over $R [x]$ is in particular a monic equation over $K [x]$ , so $z$ is integral over $K [x]$ , and step 2.1 applied to the field $K$ gives $z \in K [x]$ . Write $z = α_{0} + α_{1} x + \dots + α_{r} x^{r}$ with $α_{i} \in K$ .

givenstep 2.1L1L2L3L4

3.3

Choose a monic equation $z^{n} + c_{n - 1} z^{n - 1} + \dots + c_{0} = 0$ with all $c_{j} \in R [x]$ , and write each $α_{i} = a_{i} / b_{i}$ with $a_{i}, b_{i} \in R$ and $b_{i} \neq 0$ by [L2]. Let $R_{0} \subseteq R$ be the $Z$ -subalgebra generated by the finitely many coefficients of the polynomials $c_{j}$ together with all the elements $a_{i}, b_{i}$ . Then $R_{0}$ is a subring of $R$ containing $1$ , hence a domain by [L15] and [L7], and every ideal of $R_{0}$ is finitely generated by step 2.6 applied with $C = Z$ and step 1.3. Moreover $α_{i} = a_{i} / b_{i}$ with $a_{i}, b_{i} \in R_{0}$ and $b_{i} \neq 0$ , so $α_{i} \in K_{0} : = Frac (R_{0}) \subseteq K$ [L2], and $z \in K_{0} [x]$ ; since all coefficients of the $c_{j}$ lie in $R_{0}$ , the displayed monic equation has coefficients in $R_{0} [x]$ . Thus $z \in K_{0} [x]$ is integral over $R_{0} [x]$ .

givenstep 1.3step 2.6L2L7L15

4.1

Let $D$ be a domain in which every ideal is finitely generated, put $K_{0} = Frac (D)$ , and let $f \in K_{0} [x]$ be integral over $D [x]$ . Then every coefficient of $f$ is integral over $D$ : the ring $D [x]$ is a domain by [L4] and $f$ lies in its fraction field $K_{0} (x) = Frac (K_{0} [x])$ , so step 2.3 makes $f$ almost integral over $D [x]$ ; step 3.1 makes each coefficient of $f$ almost integral over $D$ ; and step 2.4 turns almost integrality over $D$ into integrality over $D$ .

step 2.3step 2.4step 3.1L4

5.1

By step 4.1 applied to the domain $R_{0}$ , in which every ideal is finitely generated, each coefficient $α_{i}$ of $z$ is integral over $R_{0}$ . The monic polynomial over $R_{0} \subseteq R$ witnessing this has coefficients in $R$ , so each $α_{i}$ is integral over $R$ ; since $R$ is integrally closed in $K = Frac (R)$ [L1], each $α_{i}$ lies in $R$ .

step 4.1step 3.3L1

6.1

Consequently $z = α_{0} + α_{1} x + \dots + α_{r} x^{r} \in R [x]$ . Since the homomorphism of step 3.2 is injective and restricts to the identity on $R [x]$ , every element of $Frac (R [x])$ integral over $R [x]$ is already an element of $R [x]$ : the polynomial ring $R [x]$ is an integrally closed domain by [L1] and [L4], for every integrally closed domain $R$ and with no Noetherian hypothesis on $R$ . ∎

step 2.1step 3.2step 5.1L1L4

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-sol)audited 2026-09-27Open item page →

Quasi-finite local fibres transfer through quotients and intermediate rings

Statement

Let $R \to S$ be a ring map of finite type that is quasi-finite at the prime $q \in Spec (S)$ , and put $p = q \cap R$ . Then the following hold.

For every intermediate $R$ -subalgebra $T$ , that is $Im (R) \subseteq T \subseteq S$ , with $r = q \cap T$ , the map $T \to S$ is of finite type and is quasi-finite at $q$ .
Let $T$ be an intermediate $R$ -subalgebra that is of finite type over $R$ , let $u \in T ∖ q$ and suppose that $T_{u} = S_{u}$ as subrings of $S_{u}$ , with $r = q \cap T$ . Then $R \to T$ is quasi-finite at $r$ .
Let $R \to R^{'}$ be an arbitrary ring map, put $S^{'} = S \otimes_{R} R^{'}$ and let $q^{'} \in Spec (S^{'})$ be a prime of $S^{'}$ that lies over $q$ , i.e. $q^{'} \cap S = q$ . Then $R^{'} \to S^{'}$ is of finite type and quasi-finite at $q^{'}$ .
Let $J \subseteq q$ be an ideal of $S$ , put $\overset{ˉ}{S} = S / J$ and let $\overset{ˉ}{q} = q / J$ be the image of $q$ . Then $R \to \overset{ˉ}{S}$ is of finite type and quasi-finite at $\overset{ˉ}{q}$ . Consequently, if a quotient $\overset{ˉ}{S} = S / J$ with $J \subseteq q$ is not quasi-finite over $R$ at the image of $q$ , then $R \to S$ is not quasi-finite at $q$ .

The transfers of (3) and (4) are the ones used later on this page to move quasi-finiteness between a finite-type algebra and its quotients and base changes; no Noetherian hypothesis is imposed anywhere.

Facts & Assumptions

Given: A finite-type ring map $R \to S$ , a prime $q \in Spec (S)$ with contraction $p = q \cap R$ , and the hypothesis that $R \to S$ is quasi-finite at $q$ .

[L1]

The map $R \to S$ is quasi-finite at $q$ when the $κ (p)$ -algebra $S_{q} / p S_{q}$ is finite over $κ (p)$ , that is, finitely generated as a $κ (p)$ -module, equivalently finite-dimensional over $κ (p)$ ; the map is quasi-finite when it is of finite type and quasi-finite at every prime of $S$ (Quasi-finiteness at a prime of a finite-type algebra).

[L2]

An $R$ -algebra $A$ is of finite type over $R$ when $A = R [a_{1}, \dots, a_{n}]$ for some $n \geq 0$ and some elements $a_{i} \in A$ ; equivalently $A$ is isomorphic as an $R$ -algebra to a quotient $R [x_{1}, \dots, x_{n}] / a$ (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras).

[L3]

For a multiplicative subset $T \subseteq A$ of a commutative ring, $T^{- 1} A$ consists of the classes of pairs $(a, t)$ , written $a / t$ , with $a / t = a^{'} / t^{'}$ if and only if $v (a t^{'} - a^{'} t) = 0$ for some $v \in T$ ; every $t \in T$ maps to a unit of $T^{- 1} A$ (Multiplicative subsets and the localisation $S^{- 1} R$ as equivalence classes of fractions).

[L4]

For an ideal $I$ of a commutative ring $A$ and a multiplicative subset $T \subseteq A$ there is a canonical isomorphism $(T^{- 1} A) / (T^{- 1} I) ≅ {\overset{ˉ}{T}}^{- 1} (A / I)$ , where $\overset{ˉ}{T}$ is the image of $T$ in $A / I$ (Localisation commutes with quotient rings: $S^{- 1} R / S^{- 1} I ≅ {\overset{ˉ}{S}}^{- 1} (R / I)$ ).

[L5]

For an ideal $I ⊴ A$ and an $A$ -module $M$ there is a natural isomorphism $M \otimes_{A} (A / I) ≅ M / I M$ ( $M \otimes_{R} R / I ≅ M / I M$ naturally).

[L6]

Contraction along the quotient map $A \to A / I$ is an inclusion-preserving bijection from $Spec (A / I)$ onto the primes of $A$ containing $I$ , with inverse $p \mapsto p / I$ (Prime ideals of a quotient ring are exactly the prime ideals containing the ideal).

[L7]

If $f : A \to B$ is a unital homomorphism of commutative rings and $f (t)$ is a unit of $B$ for every $t \in T$ , then there is a unique unital ring homomorphism $\tilde{f} : T^{- 1} A \to B$ with $\tilde{f} \circ λ_{T} = f$ (Universal property of localisation: maps that invert $S$ factor uniquely through $S^{- 1} R$ ).

[L8]

For a domain $D$ the field of fractions is $Frac (D) = (D ∖ {0})^{- 1} D$ , with elements fractions $a / b$ for $a, b \in D$ , $b \neq 0$ (The field of fractions $Frac (D) = (D ∖ {0})^{- 1} D$ of an integral domain).

[L9]

If $M$ is a multiplicative subset of a commutative ring $A$ and $A \to C$ is a ring map, then $(M^{- 1} A) \otimes_{A} C$ is canonically isomorphic to the localization of $C$ at the image of $M$ (Presentations and localization under base extension).

Proof

technique · direct

1.1

Let $T$ be an intermediate $R$ -subalgebra, so that the given map factors as $R \to T \to S$ , and let $r = q \cap T$ . By [L2] the $R$ -algebra $S$ is generated by finitely many elements $s_{1}, \dots, s_{n} \in S$ ; these same elements generate $S$ as a $T$ -algebra, so $T \to S$ is of finite type by [L2]. Moreover $r$ is a prime of $T$ and $p = r \cap R$ , since $p = q \cap R = (q \cap T) \cap R$ .

givenL2

1.2

Now assume in addition that $T$ is of finite type over $R$ , that $u \in T ∖ q$ , and that $T_{u} = S_{u}$ ; set $r = q \cap T$ , so that $u \notin r$ and $p = r \cap R$ . Both $T_{r}$ and $S_{q}$ are localisations of the common ring $A : = T_{u} = S_{u}$ : the former is $A$ localised at the multiplicative subset generated by the image of $T ∖ r$ , the latter at the multiplicative subset generated by the image of $S ∖ q$ , and every element of $S_{u}$ is a fraction $a / u^{k}$ with $a \in T$ by [L3].

givenL3

1.3

Now let $R \to R^{'}$ be a ring map, put $S^{'} = S \otimes_{R} R^{'}$ and let $q^{'} \in Spec (S^{'})$ lie over $q$ ; set $p^{'} = q^{'} \cap R^{'}$ , so that $p^{'} \cap R = q^{'} \cap R = q \cap R = p$ . By [L2] write $S = R [s_{1}, \dots, s_{n}]$ ; then $S^{'}$ is generated as an $R^{'}$ -algebra by the images of $s_{1}, \dots, s_{n}$ , because $S$ is a quotient of a polynomial ring $R [x_{1}, \dots, x_{n}]$ and tensoring the quotient presentation with $R^{'}$ over $R$ gives a quotient presentation of $S^{'}$ over $R^{'}$ by [L5]. In particular $R^{'} \to S^{'}$ is of finite type by [L2].

givenL2L5

1.4

Put $E = S_{q} / p S_{q}$ . By hypothesis and [L1], $E$ is finite-dimensional over $κ (p)$ ; choose a basis $x_{1}, \dots, x_{d}$ .

givenL1

1.5

Finally let $J \subseteq q$ be an ideal of $S$ , put $\overset{ˉ}{S} = S / J$ and let $\overset{ˉ}{q} = q / J$ . By [L6] the ideal $\overset{ˉ}{q}$ is a prime of $\overset{ˉ}{S}$ with $\overset{ˉ}{q} \cap R = p$ ; $\overset{ˉ}{S}$ is a quotient of the finite-type $R$ -algebra $S$ , hence of finite type over $R$ by [L2]. By [L4] there is a canonical isomorphism ${\overset{ˉ}{S}}_{\overset{ˉ}{q}} ≅ S_{q} / J S_{q}$ identifying the extensions of $p$ , so ${\overset{ˉ}{S}}_{\overset{ˉ}{q}} / p {\overset{ˉ}{S}}_{\overset{ˉ}{q}} ≅ S_{q} / (p S_{q} + J S_{q})$ is a quotient of $S_{q} / p S_{q}$ .

givenL2L4L6

2.1

The inclusion $p \subseteq r$ gives $p S_{q} \subseteq r S_{q}$ , so the quotient map $S_{q} / p S_{q} \to S_{q} / r S_{q}$ is a surjective $κ (p)$ -algebra homomorphism. By hypothesis and [L1] the algebra $S_{q} / p S_{q}$ is finite over $κ (p)$ , hence so is its quotient $S_{q} / r S_{q}$ .

givenstep 1.1L1

2.2

The homomorphism $T_{r} \to S_{q}$ induced by the inclusion $T \subseteq S$ is injective. Indeed, let $a \in T$ and $s \in T ∖ r$ with $a / s$ mapping to $0$ in $S_{q}$ ; by [L3] there is $σ \in S ∖ q$ with $σ a = 0$ in $S$ . By [L3] again write $σ = t / u^{k}$ with $t \in T$ , $k \geq 0$ . If $t$ lay in $q$ , then $σ = t / u^{k}$ would lie in the prime $q S_{u}$ of $S_{u}$ , contradicting $σ \notin q$ ; hence $t \notin q$ , so $t \in T ∖ r$ . The vanishing $σ a = 0$ in $S_{u}$ means $u^{m} t a = 0$ in $S$ for some $m \geq 0$ by [L3]; this element lies in $T$ , and $u^{m} t \in T ∖ r$ is inverted in $T_{r}$ , so $a / s = 0$ .

givenstep 1.2L3

2.3

The composite $S \to S^{'} \to S_{q^{'}}^{'}$ inverts every element of $S ∖ q$ , because such an element lies outside $q^{'} \cap S = q$ and hence outside $q^{'}$ . It also kills $p$ in the quotient by $p^{'} S_{q^{'}}^{'}$ . Thus the map factors through a $κ (p)$ -algebra homomorphism $E \to E^{'} : = S_{q^{'}}^{'} / p^{'} S_{q^{'}}^{'}$ , and the residue-field map $κ (p) \to κ (p^{'})$ is induced by $R \to R^{'}$ .

givenstep 1.3step 1.4L1L7

2.4

The quotient of step 1.5 is therefore finite over $κ (p) = κ (\overset{ˉ}{q} \cap R)$ by hypothesis and [L1], so $R \to \overset{ˉ}{S}$ is quasi-finite at $\overset{ˉ}{q}$ by [L1]. Contrapositively, if $J \subseteq q$ and the quotient map $R \to S / J$ fails to be quasi-finite at $q / J$ , then $R \to S$ is not quasi-finite at $q$ .

step 1.5L1

3.1

The ring $T / r$ is a domain and the composite $T / r \to S_{q} / r S_{q}$ is injective: an element of $T$ has vanishing image in $S_{q} / r S_{q}$ exactly when it lies in $q$ , and $q \cap T = r$ . Every class $t + r$ with $t \in T ∖ r$ lies outside $q$ , hence is a unit of $S_{q}$ and therefore a unit of the quotient $S_{q} / r S_{q}$ ; by [L7] and [L8] the field of fractions $κ (r) = Frac (T / r)$ therefore embeds in $S_{q} / r S_{q}$ as a subring containing the image of $κ (p)$ .

givenstep 1.1step 2.1L7L8

3.2

The homomorphism $T_{r} \to S_{q}$ of step 2.2 is also surjective. An element of $S_{q}$ is a fraction $σ / τ$ with $σ, τ \in S$ and $τ \notin q$ ; since $S_{u} = T_{u}$ as subrings of $S_{u}$ , and $u \notin q$ , we may write $σ = t / u^{k}$ and $τ = t^{'} / u^{l}$ with $t, t^{'} \in T$ by [L3]. Then $t^{'} \notin q$ , hence $t^{'} \in T ∖ r$ , and $σ / τ = t u^{l} / (t^{'} u^{k})$ with numerator $t u^{l} \in T$ and denominator $t^{'} u^{k} \in T ∖ r$ , so $σ / τ$ is the image of an element of $T_{r}$ .

givenstep 1.2L3

3.3

Let $F = S \otimes_{R} κ (p)$ and let $\overset{ˉ}{q}$ be the prime of this fiber induced by $q$ , so $E ≅ F_{\overset{ˉ}{q}}$ . The prime $q^{'}$ induces a prime ${\overset{ˉ}{q}}^{'}$ of $F \otimes_{κ (p)} κ (p^{'})$ lying over $\overset{ˉ}{q}$ . By [L9], localization commutes with this scalar extension; localizing further at ${\overset{ˉ}{q}}^{'}$ gives the canonical isomorphism $E^{'} ≅ (E \otimes_{κ (p)} κ (p^{'}))_{{\tilde{q}}^{'}},$ where ${\tilde{q}}^{'}$ is the corresponding prime after the first localization.

givenstep 2.3L9

4.1

Since $κ (r)$ is a $κ (p)$ -subspace of the finite-dimensional $κ (p)$ -vector space $S_{q} / r S_{q}$ of step 2.1, the field extension $κ (r) / κ (p)$ is finite. The algebra $S_{q} / r S_{q}$ of step 2.1 is a module over the field $κ (r)$ by step 3.1, and a $κ (r)$ -linearly independent subset of it is $κ (p)$ -linearly independent, so $S_{q} / r S_{q}$ is finite-dimensional over $κ (r)$ ; by [L1] the map $T \to S$ is quasi-finite at $q$ , which is assertion (1).

step 2.1step 3.1L1

4.2

By steps 2.2 and 3.2 the inclusion induces an isomorphism $T_{r} ≅ S_{q}$ ; it carries $p T_{r}$ onto $p S_{q}$ because it is an isomorphism of $R$ -algebras. Hence $T_{r} / p T_{r} ≅ S_{q} / p S_{q}$ is finite over $κ (p)$ by hypothesis and [L1], and since $p = r \cap R$ this says by [L1] that $R \to T$ is quasi-finite at $r$ , which is assertion (2).

givenstep 2.2step 3.2L1

4.3

The algebra $E \otimes_{κ (p)} κ (p^{'})$ is finite-dimensional over $κ (p^{'})$ , since the images of the basis in step 1.4 span it. Any localization $A_{r}$ of a finite-dimensional algebra $A$ over a field at a prime $r$ is finite-dimensional: for $s \notin r$ , the descending chain of vector subspaces $(s^{n}) \subseteq A$ stabilizes, so for some $N \geq 0$ and $a \in A$ one has $s^{N} = s^{N + 1} a$ ; in $A_{r}$ this gives $1 = s a$ , so the inverse of every denominator is already in the image of $A$ . Hence $A \to A_{r}$ is surjective and its target is finite-dimensional. Applying this to the localization in step 3.3 shows $E^{'}$ is finite-dimensional over $κ (p^{'})$ .

step 1.4step 3.3algebra

5.1

Thus the $κ (p^{'})$ -algebra $S_{q^{'}}^{'} / p^{'} S_{q^{'}}^{'}$ is generated as a $κ (p^{'})$ -module by $y_{1}, \dots, y_{d}$ , hence is finite over $κ (p^{'})$ by [L1]; that is, $R^{'} \to S^{'}$ is quasi-finite at $q^{'}$ by [L1], which is assertion (3).

step 2.3step 4.3L1

6.1

Assertion (1) is step 4.1, assertion (2) is step 4.2, assertion (3) is step 5.1 and assertion (4) with its contrapositive form is step 2.4; all four reduce to the single finite-dimensionality condition of [L1] at the relevant prime, and no Noetherian hypothesis and no form of the Axiom of Choice was used. ∎

step 4.1step 4.2step 5.1step 2.4L1

LemmaStatement: Literature-sourcedProof: AI-adaptedaudited 2026-09-27Open item page →

Finite algebras over a strongly transcendental variable are nowhere quasi-finite

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $R \subseteq S$ be an inclusion of reduced commutative rings (The nilradical and reduced rings), let $x \in S$ be strongly transcendental over $R$ (Strong transcendence over a subring), and suppose that $S$ is module-finite over the $R$ -subalgebra $R [x] \subseteq S$ generated by $x$ (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras). Then $R \to S$ is a ring map of finite type that is quasi-finite (Quasi-finiteness at a prime of a finite-type algebra) at no prime of $S$ .

The hypotheses are exactly those of the one-variable case of Zariski's main theorem with the roles of the variable and the finite extension separated: $x$ contributes a transcendental direction, and the finiteness of $S$ over $R [x]$ prevents a local fibre from being finite-dimensional over its base residue field. A local fibre can nevertheless have Krull dimension zero: for $R = k$ , $S = k [x]$ and $q = (0)$ it is $k (x)$ , which has infinite dimension over $k$ . In the normal-domain part of the proof, finite residue degree forces a strict prime chain in the local fibre; normalization and minimal-prime descent then give the general result. The Axiom of Choice is used exactly through going down, lying over, the minimal-prime descent of strong transcendence and the existence of a minimal prime below a given prime.

Facts & Assumptions

Given: The Axiom of Choice; an inclusion of reduced commutative rings $R \subseteq S$ ; an element $x \in S$ strongly transcendental over $R$ ; and the hypothesis that $S$ is module-finite over $R [x]$ .

[L1]

The map $R \to S$ is quasi-finite at $q \in Spec (S)$ when $S_{q} / p S_{q}$ is finite over $κ (p)$ for $p = q \cap R$ , that is, finitely generated as a $κ (p)$ -module, equivalently finite-dimensional over $κ (p)$ (Quasi-finiteness at a prime of a finite-type algebra).

[L2]

An $R$ -algebra $A$ is of finite type over $R$ when $A = R [a_{1}, \dots, a_{n}]$ for some $n \geq 0$ and some $a_{i} \in A$ , and $A$ is module-finite over $R$ when $A$ is finitely generated as an $R$ -module (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras).

[L3]

For a prime $p$ of a commutative ring $A$ there is a canonical field isomorphism $A_{p} / p A_{p} ≅ Frac (A / p)$ ; this quotient is the residue field of the local ring $A_{p}$ ( $R_{p} / p R_{p} ≅ Frac (R / p)$ is the residue field at $p$ ).

[L4]

For a commutative ring $A$ and an ideal $P ⊴ A$ , the quotient $A / P$ is an integral domain if and only if $P$ is a prime ideal ( $R / P$ is an integral domain if and only if $P$ is a prime ideal).

[L5]

For $R \subseteq S$ and $x \in S$ , strong transcendence of $x$ over $R$ is the annihilator-sensitive condition that $u (a_{0} + a_{1} x + \dots + a_{k} x^{k}) = 0$ with $u \in S$ and $a_{i} \in R$ implies $u a_{i} = 0$ for all $i$ ; when $S$ is a domain this is the same as saying that $x$ is transcendental over the fraction field of $R$ (Strong transcendence over a subring).

[L6]

The evaluation homomorphism ${ev}_{φ, s} : R [x] \to S$ extending a unital ring homomorphism $φ : R \to S$ and sending $x$ to $s$ is unique; in particular the evaluation of a polynomial of $R [x]$ at the indeterminate $x$ is that same polynomial (Universal property of $R [x]$ : a coefficient homomorphism and the image of $x$ determine a unique ring homomorphism).

[L7]

If $R$ is an integral domain then $R [x]$ is an integral domain (A polynomial ring over an integral domain is an integral domain).

[L8]

For an integrally closed domain $R$ , the polynomial ring $R [x]$ is integrally closed, with no Noetherian hypothesis (Polynomial rings over normal domains are normal).

[L9]

For a domain $A$ and a homomorphism $A \to B$ , the integral closure of $A$ in $B$ is the set of elements of $B$ integral over $A$ ; and $A$ is integrally closed when every element of $Frac (A)$ integral over $A$ already lies in $A$ . The integral closure of a domain in a field extension of its fraction field is itself an integrally closed domain (Integral closure in an extension ring and integrally closed domains, The integral closure of a domain in a field extension is integrally closed).

[L10]

Let $A \subseteq B$ be commutative rings with $A \neq 0$ and $b \in B$ . Then $b$ is integral over $A$ if and only if $A [b]$ is finitely generated as an $A$ -module; in particular a module-finite extension is integral (Integrality and finite-module characterizations for one element).

[L11]

Assume the Axiom of Choice. Let $A \subseteq B$ be an integral extension of domains with $A$ integrally closed; if $p_{0} \subseteq p_{1}$ are primes of $A$ and $q_{1}$ is a prime of $B$ with $q_{1} \cap A = p_{1}$ , then there is a prime $q_{0} \subseteq q_{1}$ of $B$ with $q_{0} \cap A = p_{0}$ (Going down holds for integral extensions over integrally closed domains).

[L12]

If $f : A \to B$ is a unital homomorphism and $f (t)$ is a unit of $B$ for every $t \in T$ , then there is a unique unital homomorphism $T^{- 1} A \to B$ with $\tilde{f} \circ λ_{T} = f$ ; in particular an injection of domains extends canonically to their fraction fields (Universal property of localisation: maps that invert $S$ factor uniquely through $S^{- 1} R$ ).

[L13]

For a unital ring map $R \to S$ , the set ${Int}_{R} (S)$ of elements of $S$ integral over the map is a subring of $S$ containing the image of $R$ , hence an $R$ -subalgebra (Integral elements subalgebra of an arbitrary ring map).

[L14]

Let $A \subseteq B$ be commutative rings and let $b_{1}, \dots, b_{n} \in B$ be integral over $A$ ; then the $A$ -subalgebra $A [b_{1}, \dots, b_{n}]$ is module-finite over $A$ (A subalgebra generated by finitely many integral elements is module-finite).

[L17]

Assume the Axiom of Choice. Let $f : A \to B$ be an integral ring map and let $p \in Spec (A)$ with $\ker f \subseteq p$ . Then there is $q \in Spec (B)$ with $f^{- 1} (q) = p$ (Lying over for integral ring maps).

[L18]

Contraction along $A \to A / I$ is an inclusion-preserving bijection from $Spec (A / I)$ onto the primes of $A$ containing $I$ , with inverse $p \mapsto p / I$ (Prime ideals of a quotient ring are exactly the prime ideals containing the ideal).

[L19]

For a unital ring $A$ , a right $A$ -module $M$ and a left $A$ -module $N$ , the tensor product $M \otimes_{A} N$ is the quotient of the free abelian group on $M \times N$ by the balanced relations, with elementary tensors $m \otimes n$ (The tensor product $M \otimes_{R} N$ from the additive group underlying the free $Z$ -module on $M \times N$ , elementary tensors, and finite tensor sums).

[L20]

Quasi-finiteness transfers: for a finite-type map $R \to S$ quasi-finite at $q$ and any ring map $R \to R^{'}$ , the base-changed map $R^{'} \to S \otimes_{R} R^{'}$ is quasi-finite at every prime of $S \otimes_{R} R^{'}$ lying over $q$ ; and for an ideal $J \subseteq q$ the quotient map $R \to S / J$ is quasi-finite at the image of $q$ (Quasi-finite local fibres transfer through quotients and intermediate rings).

[L21]

Assume the Axiom of Choice. If $R \subseteq S$ are reduced rings, $x \in S$ is strongly transcendental over $R$ , $q$ is a minimal prime of $S$ and $p = R \cap q$ , then the image of $x$ in $S / q$ is strongly transcendental over $R / p$ (Strong transcendence descends to reduced minimal-prime quotients).

[L22]

Contraction along $A \to A_{p}$ induces an inclusion-preserving bijection from $Spec (A_{p})$ onto the set of primes $q \subseteq p$ of $A$ (Primes of a localization at a prime).

[L23]

Assume the Axiom of Choice. For a commutative ring $A$ and a proper ideal $I ⊴ A$ there is a prime ideal of $A$ containing $I$ that is minimal among the primes containing $I$ (Minimal primes over a proper ideal exist).

[L24]

For a multiplicative subset $T$ of a commutative ring $A$ and $r \in A$ one has $r / s = 0$ in $T^{- 1} A$ if and only if $u r = 0$ for some $u \in T$ ; moreover $T^{- 1} A$ is the zero ring if and only if $0 \in T$ (Equality, vanishing, and the kernel of the localisation map).

[L25]

For a prime $p$ of $A$ the localisation is $A_{p} = (A ∖ p)^{- 1} A$ , with elements fractions $r / s$ for $s \notin p$ (Localisation at a prime ideal: $R_{p} = (R ∖ p)^{- 1} R$ ).

[L26]

The Axiom of Choice is the assertion that every family of nonempty sets has a choice function; it is assumed here and used exactly through [L11], [L17], [L21] and [L23] (The Axiom of Choice).

Proof

technique · direct

1.1

By the hypothesis of module-finiteness there are $t_{1}, \dots, t_{m} \in S$ with $S = R [x] t_{1} + \dots + R [x] t_{m}$ . Hence $S = R [x, t_{1}, \dots, t_{m}]$ is generated as an $R$ -algebra by the finitely many elements $x, t_{1}, \dots, t_{m}$ and is of finite type over $R$ by [L2]; consequently the notion of quasi-finiteness of [L1] is defined for $R \to S$ , and it remains to show that no prime of $S$ satisfies it.

givenL2

1.2

We first treat the case in which $R$ and $S$ are domains; here, by [L5], the strong transcendence of $x$ over $R$ says exactly that $x$ is transcendental over $K = Frac (R)$ . Let $q \in Spec (S)$ , put $p = q \cap R$ and $r = q \cap R [x]$ , and suppose for contradiction that $R \to S$ is quasi-finite at $q$ .

givenL5

1.3

Now let $R \subseteq S$ be reduced and let $q \in Spec (S)$ . By [L25] the localisation $S_{q} = (S ∖ q)^{- 1} S$ is nonzero, because $0 \in q$ excludes $0 \in S ∖ q$ and $S_{q} = 0$ would force $0 \in S ∖ q$ by [L24]; so the zero ideal is a proper ideal of $S_{q}$ , and by [L23] there is a prime $P$ of $S_{q}$ minimal over $(0)$ . By [L22] every prime of $S_{q}$ is the extension of a prime of $S$ contained in $q$ , so $P = q_{0} S_{q}$ for the prime $q_{0} = P \cap S \subseteq q$ , and the inclusion-preserving bijection shows that $q_{0}$ is a minimal prime of $S$ .

givenL22L23L24L25

2.1

First consider the normal-domain case, so assume here that $R$ is a domain and integrally closed in $K = Frac (R)$ . Then $R [x]$ is an integral domain by [L7] and is integrally closed by [L8] and [L9]. Moreover $R [x] \subseteq S$ is an integral extension: every element of the module-finite $R [x]$ -algebra $S$ is integral over $R [x]$ by [L10].

givenstep 1.1L7L8L9L10

2.2

By [L1] the algebra $S_{q} / p S_{q}$ is finite over $κ (p)$ , and by [L3] the residue field $κ (q)$ is $S_{q} / q S_{q}$ , which is a quotient of $S_{q} / p S_{q}$ because $p \subseteq q$ . Hence $κ (q)$ is finite over $κ (p)$ .

givenstep 1.2L1L3

2.3

For the general domain case drop the normality of $R$ : let $\overline{R}$ be the integral closure of $R$ in $K = Frac (R)$ as in [L9], which is a subring of $K$ by [L13] and is integrally closed by [L9]; put $L = Frac (S)$ and let $\overline{S} = \overline{R} [S]$ be the subring of $L$ generated by $\overline{R}$ and $S$ , a domain because it is a subring of the field $L$ .

givenstep 1.2L9L13

2.4

By [L21] the image $\overset{ˉ}{x}$ of $x$ in $S / q_{0}$ is strongly transcendental over $R / p_{0}$ , where $p_{0} = q_{0} \cap R$ . Both $R / p_{0}$ and $S / q_{0}$ are domains by [L4], so by [L5] the element $\overset{ˉ}{x}$ is transcendental over $κ (p_{0}) = Frac (R / p_{0})$ .

givenstep 1.3L4L5L21

2.5

The quotient $S / q_{0}$ is module-finite over $(R / p_{0}) [\overset{ˉ}{x}]$ : it is the image of the module-finite $R [x]$ -algebra $S$ under the quotient map, and the image of $R [x]$ is the subalgebra generated by $\overset{ˉ}{x}$ over $R / p_{0}$ .

givenstep 1.1step 1.3

3.1

The kernel of $R [x] \to S / q$ is exactly $r$ , so $R [x] / r$ is a subring of the domain $S / q$ by [L4]; by [L3] the residue fields are $κ (r) = Frac (R [x] / r)$ and $κ (q) = Frac (S / q)$ , and by [L12] the inclusion of domains extends to an injection $κ (r) \to κ (q)$ of fields carrying the image of $κ (p)$ onto a subfield contained in $κ (r)$ . Since a $κ (p)$ -linearly independent subset of $κ (r)$ is also $κ (p)$ -linearly independent in $κ (q)$ , step 2.2 shows that $κ (r)$ is finite over $κ (p)$ .

givenstep 1.2step 2.2L3L4L12

3.2

Since $R \subseteq \overline{R} \subseteq K$ and $K = Frac (R)$ , their fraction fields satisfy $K \subseteq Frac (\overline{R}) \subseteq K$ , hence $Frac (\overline{R}) = K$ . The element $x$ is transcendental over $K$ by [L5]: if a nonzero polynomial over $K$ vanished at $x$ , clearing its finitely many denominators would give a nonzero $h \in R [X]$ with $h (x) = 0$ , and strong transcendence with multiplier $1$ would force every coefficient of $h$ to be zero, a contradiction. Hence $x$ is transcendental over $Frac (\overline{R})$ as well.

givenstep 2.3L5algebra

3.3

The algebra $\overline{S}$ is integral over $S$ : by [L13] the set ${Int}_{S} (\overline{S})$ is a subring of $\overline{S}$ containing $S$ , and it contains every element of $\overline{R}$ , which is integral over $R$ by [L9] and hence over $S$ by the same monic equation; being a subring containing $S$ and $\overline{R}$ , it contains the subring they generate, namely $\overline{S}$ , so $\overline{S} = {Int}_{S} (\overline{S})$ is integral over $S$ .

givenstep 2.3L9L13algebra

3.4

The algebra $\overline{S}$ is module-finite over $\overline{R} [x]$ : from $S = R [x] t_{1} + \dots + R [x] t_{m}$ of step 1.1 we get $\overline{S} = \overline{R} [x] [t_{1}, \dots, t_{m}]$ , and each $t_{j}$ is integral over $R [x]$ by module-finiteness and [L10], hence over $\overline{R} [x] \supseteq R [x]$ by the same monic equation, so [L14] makes $\overline{S}$ module-finite over $\overline{R} [x]$ .

givenstep 1.1step 2.1step 2.3L14algebra

4.1

Consequently $r \neq p R [x]$ : if equality held, then $R [x] / r = (R / p) [x]$ and $κ (r) = κ (p) (\overset{ˉ}{x})$ where $\overset{ˉ}{x}$ is the class of $x$ . Then $\overset{ˉ}{x}$ is transcendental over $κ (p)$ : if $0 \neq h \in κ (p) [X]$ satisfied $h (\overset{ˉ}{x}) = 0$ , then writing $h = d^{- 1} g$ with $0 \neq d \in R / p$ and $g \in (R / p) [X]$ would give $g (\overset{ˉ}{x}) = 0$ , that is $g = 0$ because evaluation at the indeterminate is the identity by [L6], contradicting $g \neq 0$ in the domain $(R / p) [x]$ by [L7]. But then the powers $1, \overset{ˉ}{x}, \dots, {\overset{ˉ}{x}}^{d}$ for $d = \dim_{κ (p)} κ (r) < \infty$ would be $κ (p)$ -linearly dependent, yielding a nonzero polynomial relation over $κ (p)$ of degree $\leq d$ satisfied by $\overset{ˉ}{x}$ , a contradiction.

givenstep 3.1L6L7

5.1

Since $p R [x] ⊊ r$ and $q$ lies over $r$ along the integral extension $R [x] \subseteq S$ of step 2.1, going down [L11] provides a prime $q^{'} \subseteq q$ of $S$ with $q^{'} \cap R [x] = p R [x]$ ; then $q^{'} \neq q$ , and $q^{'} \cap R = (q^{'} \cap R [x]) \cap R = p R [x] \cap R = p$ .

givenstep 2.1step 4.1L11

6.1

The prime ideals $q^{'} S_{q} \subseteq q S_{q}$ of $S_{q}$ are distinct and both contain $p S_{q}$ by step 5.1, so $A : = S_{q} / p S_{q}$ carries a strict chain of primes; $A$ is finite-dimensional over $κ (p)$ by [L1]. But every prime $P$ of a finite-dimensional algebra $A$ over a field $κ (p)$ is maximal: $A / P$ is a domain by [L4], finite-dimensional over $κ (p)$ , so for $0 \neq u \in A / P$ the $κ (p)$ -linear map $a \mapsto u a$ is injective by [L4] and hence surjective by finite dimension, which makes $u$ a unit and $A / P$ a field. This contradiction shows that, when $R, S$ are domains and $R$ is integrally closed, $R \to S$ is quasi-finite at no prime of $S$ , and then also when $R, S$ are any domains, as the next steps show.

givenstep 5.1L1L4

7.1

Applying the result of step 6.1 to the domains $\overline{R} \subseteq \overline{S}$ , where $\overline{R}$ is integrally closed by step 2.3 and $x$ is transcendental over $Frac (\overline{R})$ by step 3.2 while $\overline{S}$ is module-finite over $\overline{R} [x]$ by step 3.4, we obtain: $\overline{R} \to \overline{S}$ is quasi-finite at no prime of $\overline{S}$ .

step 6.1step 2.3step 3.2step 3.4

8.1

Now suppose that $R \to S$ is quasi-finite at a prime $q \in Spec (S)$ . By [L17] applied to the integral extension $S \subseteq \overline{S}$ of step 3.3 there is $\overline{q} \in Spec (\overline{S})$ with $\overline{q} \cap S = q$ . The $R$ -algebra homomorphism $S \otimes_{R} \overline{R} \to \overline{S}$ of [L19] sending $s \otimes r$ to $s r$ is surjective because its image is a subring of $\overline{S}$ containing the images of $S$ and of $\overline{R}$ ; with $J$ its kernel we have $\overline{S} ≅ (S \otimes_{R} \overline{R}) / J$ , and by [L18] the prime $\overline{q}$ corresponds to a prime $\tilde{q}$ of $S \otimes_{R} \overline{R}$ mapping to $q$ in $S$ . By [L20] the base change $\overline{R} \to S \otimes_{R} \overline{R}$ is quasi-finite at $\tilde{q}$ , and by [L20] again, applied to the quotient $\overline{S}$ , the map $\overline{R} \to \overline{S}$ is quasi-finite at $\overline{q}$ , contradicting step 7.1. Hence in the domain case $R \to S$ is quasi-finite at no prime of $S$ .

givenstep 3.3step 7.1L17L18L19L20

9.1

By step 8.1 applied to the domains $R / p_{0} \subseteq S / q_{0}$ with the element $\overset{ˉ}{x}$ of step 2.4, the map $R / p_{0} \to S / q_{0}$ is quasi-finite at no prime of $S / q_{0}$ ; in particular it is not quasi-finite at the prime $q / q_{0}$ of $S / q_{0}$ , which exists by [L18] because $q_{0} \subseteq q$ and $q_{0}$ is prime.

givenstep 8.1step 2.4step 2.5L18

10.1

If $R \to S$ were quasi-finite at $q$ , then by [L20] applied to the ideal $q_{0} \subseteq q$ the quotient map $R \to S / q_{0}$ would be quasi-finite at $q / q_{0}$ , contradicting step 9.1: the actions factor through $R / p_{0}$ , and $κ (p / p_{0}) = κ (p)$ , so these two base-ring descriptions give the same local fibre. Hence $R \to S$ is not quasi-finite at $q$ ; since $q \in Spec (S)$ was arbitrary, $R \to S$ is quasi-finite at no prime of $S$ .

givenstep 9.1L20

11.1

Steps 1.1 and 10.1 prove the Statement: $R \to S$ is of finite type whenever $S$ is module-finite over $R [x]$ , and it is quasi-finite at no prime of $S$ for reduced $R \subseteq S$ and strongly transcendental $x$ . The Axiom of Choice was assumed in the Statement and used exactly through [L11] in step 5.1, [L17] in step 8.1, [L21] in step 2.4 and [L23] in step 1.3; all other steps are choice-free. ∎

step 1.1step 10.1L11L17L21L23L26

LemmaStatement: Literature-sourcedProof: AI-adaptedaudited 2026-09-27Open item page →

A quasi-finite one-generator quotient is locally its integral closure

Statement

Let $R$ be a commutative ring, let $I ⊴ R [x]$ be an ideal, put $S = R [x] / I$ and let $q \in Spec (S)$ . Assume that $R \to S$ is quasi-finite at $q$ (Quasi-finiteness at a prime of a finite-type algebra) and let $S^{'} \subseteq S$ be the integral closure of the image of $R$ in $S$ (Integral elements subalgebra of an arbitrary ring map). Then there is an element $g \in S^{'} ∖ q$ such that the canonical homomorphism $S_{g}^{'} \to S_{g}$ of localisations at $g$ (Multiplicative subsets and the localisation $S^{- 1} R$ as equivalence classes of fractions) is an isomorphism.

Thus, at a quasi-finite prime, a one-generator quotient $R [x] / I$ already agrees with its relative integral closure after inverting a single element of that closure. No regularity, Noetherian or finiteness hypothesis beyond the finite type built into $R [x] / I$ is used, and $R \to S$ need not be injective.

Facts & Assumptions

Given: A commutative ring $R$ , an ideal $I ⊴ R [x]$ , the quotient $S = R [x] / I$ with quotient map $φ : R [x] \to S$ , a prime $q \in Spec (S)$ with $p = q \cap R$ , the hypothesis that $R \to S$ is quasi-finite at $q$ , and the relative integral closure $S^{'} = {Int}_{R} (S) \subseteq S$ of the image of $R$ in $S$ .

[L1]

The map $R \to S$ is quasi-finite at $q$ when $S_{q} / p S_{q}$ is finite over $κ (p)$ , equivalently finite-dimensional over $κ (p)$ (Quasi-finiteness at a prime of a finite-type algebra).

[L2]

$S = R [x] / I$ is generated as an $R$ -algebra by the class of $x$ , hence is of finite type over $R$ ; a quotient of a polynomial ring by an ideal is of finite type by definition (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras).

[L3]

For a multiplicative subset $T$ of a commutative ring $A$ , the localisation $T^{- 1} A$ consists of fractions $a / t$ with the usual arithmetic, every $t \in T$ becoming a unit; for an element $g \in A$ the notation $A_{g}$ means the localisation at the multiplicative set generated by $g$ (Multiplicative subsets and the localisation $S^{- 1} R$ as equivalence classes of fractions).

[L4]

For a unital ring map $R \to S$ , the set ${Int}_{R} (S)$ of elements of $S$ integral over the map is a subring of $S$ containing the image of $R$ (Integral elements subalgebra of an arbitrary ring map).

[L5]

If $t \in S$ satisfies a relation $a_{0} + a_{1} t + \dots + a_{n} t^{n} = 0$ with $a_{j}$ in a ring $A \subseteq S$ , then $a_{n} t$ is integral over $A$ , even when $a_{n}$ is a zero divisor (The leading coefficient times a root is integral).

[L6]

If $A \to B$ and $B \to C$ are integral ring maps, then the composite $A \to C$ is integral (Integral extensions are transitive).

[L7]

For an ideal $J$ of a commutative ring $A$ and a multiplicative subset $T \subseteq A$ there is a canonical isomorphism $(T^{- 1} A) / (T^{- 1} J) ≅ {\overset{ˉ}{T}}^{- 1} (A / J)$ , where $\overset{ˉ}{T}$ is the image of $T$ in $A / J$ (Localisation commutes with quotient rings: $S^{- 1} R / S^{- 1} I ≅ {\overset{ˉ}{S}}^{- 1} (R / I)$ ).

[L8]

For an ideal $J$ of a commutative ring $A$ , the kernel of the canonical map $A [x] \to (A / J) [x]$ is the ideal $J A [x]$ of polynomials with coefficients in $J$ , so $A [x] / J A [x] ≅ (A / J) [x]$ (First isomorphism theorem for rings: $R / \ker f ≅ im f$ ).

[L9]

The evaluation homomorphism $R [x] \to S$ extending a fixed unital map on constants and sending $x$ to a prescribed element is unique; in particular, evaluation of a polynomial of $A [x]$ at the indeterminate is that same polynomial, so a polynomial of $A [x]$ all of whose coefficients lie in an additive subgroup vanishes as an element of $A [x]$ only if all its coefficients do (Universal property of $R [x]$ : a coefficient homomorphism and the image of $x$ determine a unique ring homomorphism).

[L10]

If $D$ is an integral domain then $D [x]$ is an integral domain (A polynomial ring over an integral domain is an integral domain).

[L11]

For a multiplicative subset $T$ of a commutative ring $A$ one has $a / t = 0$ in $T^{- 1} A$ if and only if $u a = 0$ for some $u \in T$ , and $a / t = a^{'} / t^{'}$ if and only if $u (a t^{'} - a^{'} t) = 0$ for some $u \in T$ ; consequently the localisation map is injective on a domain (Equality, vanishing, and the kernel of the localisation map).

Proof

technique · direct

1.1

We first produce a relation over $S^{'}$ of positive degree with a coefficient outside $q$ . Suppose that every element of $I$ had all its coefficients in $p$ , that is $I \subseteq p R [x]$ . Then by [L8] the quotient $S / p S ≅ R [x] / (I + p R [x]) = R [x] / p R [x] ≅ (R / p) [x]$ is a polynomial ring over the domain $R / p$ ; writing $\overset{ˉ}{q}$ for the image of $q$ in $S / p S$ and using [L7] with $T = S ∖ q$ , we get $S_{q} / p S_{q} ≅ (S / p S)_{\overset{ˉ}{q}}$ , a localisation of $(R / p) [x]$ .

givenL2L7L8

1.2

Such a localisation is infinite-dimensional over $κ (p) = Frac (R / p)$ . Indeed the powers $1, \overset{ˉ}{x}, {\overset{ˉ}{x}}^{2}, \dots$ of the class of $x$ in $(R / p) [x]$ are $κ (p)$ -linearly independent: a relation $\sum_{i} c_{i} {\overset{ˉ}{x}}^{i} = 0$ with $c_{i} \in κ (p)$ becomes, after multiplying by a common nonzero denominator $d \in R / p$ , a relation $\sum_{i} (d c_{i}) {\overset{ˉ}{x}}^{i} = 0$ with coefficients in the domain $R / p$ and not all zero, which by [L9] is a nonzero polynomial of $(R / p) [x]$ equal to zero — impossible by [L10] unless all coefficients vanish; and this linear independence is preserved in the localisation because the localisation map is injective on the domain $(R / p) [x]$ by [L11].

givenL9L10L11

2.1

Therefore $I \subseteq̸ p R [x]$ : otherwise steps 1.1 and 1.2 would exhibit $S_{q} / p S_{q}$ as an infinite-dimensional $κ (p)$ -algebra, contradicting the quasi-finiteness hypothesis through [L1]. So there are $n \geq 0$ and $a_{0}, \dots, a_{n} \in R$ with $f = \sum_{j} a_{j} x^{j} \in I$ and $a_{j} \notin p$ for at least one $j$ ; applying $φ$ and writing $b_{j} = φ (a_{j})$ (the element of the relative integral closure $S^{'}$ given by the image of $R$ ) gives a relation $\sum_{j = 0}^{n} b_{j} φ (x)^{j} = 0$ in $S$ with some $b_{j} \notin q$ , because $a_{j} \in p$ if and only if $φ (a_{j}) \in q$ .

givenstep 1.1step 1.2L1L4

3.1

Among all relations $\sum_{j = 0}^{m} b_{j} φ (x)^{j} = 0$ in $S$ with all $b_{j} \in S^{'}$ and at least one $b_{j} \notin q$ , choose one of minimal degree $m$ , which exists by step 2.1; then $m \geq 1$ , because a relation of degree $0$ reads $b_{0} = 0$ with $b_{0} \in S^{'}$ , and $b_{0} \notin q$ would be impossible. We show that its leading coefficient satisfies $b_{m} \notin q$ .

givenstep 2.1

4.1

Assume first that $b_{m} \notin q$ . By [L5] applied to the relation of step 3.1 with $A = S^{'}$ and $t = φ (x)$ , the element $b_{m} φ (x)$ is integral over $S^{'}$ ; hence the inclusion $S^{'} \to S^{'} [b_{m} φ (x)]$ is an integral ring map, and since $S^{'} = {Int}_{R} (S)$ consists of elements integral over $R$ by [L4], the composite $R \to S^{'} [b_{m} φ (x)]$ is integral by [L6]. So $b_{m} φ (x)$ is integral over $R$ , that is $b_{m} φ (x) \in {Int}_{R} (S) = S^{'}$ .

givenstep 3.1L4L5L6

5.1

Assume instead that $b_{m} \in q$ . The argument of step 4.1 still gives $b_{m} φ (x) \in S^{'}$ . Set $b_{m - 1}^{'} = b_{m} φ (x) + b_{m - 1} \in S^{'}$ . Then $\sum_{j = 0}^{m - 2} b_{j} φ (x)^{j} + b_{m - 1}^{'} φ (x)^{m - 1} = 0$ by step 3.1, and $b_{m - 1}^{'} - b_{m - 1} = b_{m} φ (x) \in q$ because $b_{m} \in q$ ; so $b_{m - 1}^{'} \notin q$ if $b_{m - 1} \notin q$ , and otherwise one of $b_{m - 2}, \dots, b_{0}$ lies outside $q$ by step 3.1. Either way we have a relation of degree $m - 1$ with all coefficients in $S^{'}$ and at least one coefficient outside $q$ , contradicting the choice of the minimal degree $m$ . Hence the case $b_{m} \in q$ cannot occur.

givenstep 3.1step 4.1

6.1

By steps 3.1 and 5.1 the minimal relation of step 3.1 has $b_{m} \in S^{'} ∖ q$ , and $b_{m} φ (x) \in S^{'}$ by step 4.1. Put $g = b_{m}$ . In the localisation $S_{g}^{'}$ the element $g$ is a unit by [L3], so $φ (x) = (b_{m} φ (x)) / b_{m} \in S_{g}^{'}$ ; since $S$ is generated as an $S^{'}$ -algebra by $φ (x)$ by [L2], the subring $S_{g}^{'} \subseteq S_{g}$ already contains a generating set of $S_{g}$ over $S_{g}^{'}$ and hence equals $S_{g}$ .

givenstep 4.1step 5.1L2L3

7.1

The canonical homomorphism $S_{g}^{'} \to S_{g}$ is injective, so it is an isomorphism: if a fraction $a / g^{k}$ with $a \in S^{'}$ maps to zero in $S_{g}$ , then by [L11] there is $l \geq 0$ with $g^{l} a = 0$ in $S$ ; since $a, g \in S^{'}$ , the product $g^{l} a$ lies in $S^{'}$ , and [L11] again gives $a / g^{k} = 0$ in $S_{g}^{'}$ .

givenstep 6.1L11

8.1

Steps 2.1, 6.1 and 7.1 produce $g \in S^{'} ∖ q$ with $S_{g}^{'} \to S_{g}$ an isomorphism, which is the assertion. The argument is choice-free: the only selection is that of a relation of minimal degree in step 3.1, and the minimal-degree argument uses only that degrees are natural numbers. ∎

step 2.1step 6.1step 7.1

LemmaStatement: Literature-sourcedProof: AI-adaptedaudited 2026-09-27Open item page →

Conductor radical detects every polynomial coefficient

Statement

Let $φ : R [x] \to S$ be a unital ring map of commutative rings that is finite, that is, $S$ is finitely generated as an $R [x]$ -module (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras), and let $J = {g \in S : g S \subseteq Im (φ)}$ be the conductor of the image of $φ$ . Assume the following integral-image condition: every element of $S$ integral over the ring map $R \to S$ in the sense of Integral elements subalgebra of an arbitrary ring map lies in $Im (φ)$ .

Then $J$ is an ideal of $S$ (The radical of an ideal) and the following hold, for $u \in S$ and $P = a_{0} + a_{1} x + \dots + a_{k} x^{k} \in R [x]$ with $k \geq 0$ (The polynomial ring over a commutative ring as finitely supported coefficient sequences with convolution):

if $u φ (P) \in J$ , then $u φ (a_{k})^{m} \in J$ for some integer $m \geq 0$ ;
if $u φ (P) \in \sqrt{J}$ , then $u φ (a_{i}) \in \sqrt{J}$ for every $i \in {0, \dots, k}$ .

No hypothesis is imposed on the leading coefficient: it may be zero, a zero divisor or a nilpotent, and $φ$ need not be injective.

Facts & Assumptions

Given: A finite unital ring map $φ : R [x] \to S$ such that every element of $S$ integral over $R$ lies in $Im (φ)$ , the conductor $J = {g \in S : g S \subseteq Im (φ)}$ , an element $u \in S$ and a polynomial $P = a_{0} + a_{1} x + \dots + a_{k} x^{k} \in R [x]$ with $k \geq 0$ .

[L1]

For a unital ring map $R [x] \to S$ the set of elements of $S$ integral over the map is a subring of $S$ containing the image of $R [x]$ ; an element of $S$ integral over $R$ is in particular integral over the ring map $R [x] \to S$ , and the separate integral-image condition in the Statement places each $R$ -integral element in $Im (φ)$ (Integral elements subalgebra of an arbitrary ring map).

[L2]

A commutative algebra is module-finite over its base ring when it is finitely generated as a module; in particular the finite map $φ$ exhibits $S$ as generated by finitely many elements $t_{1}, \dots, t_{n}$ as an $R [x]$ -module (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras).

[L3]

Let $A \subseteq B$ be commutative rings with $A \neq 0$ and $b \in B$ . Then $b$ is integral over $A$ if and only if $A [b]$ is finitely generated as an $A$ -module; for that reason every element of a module-finite $A$ -algebra $B$ is integral over $A$ (Integrality and finite-module characterizations for one element).

[L4]

Let $φ : R [x] \to S$ be a unital ring map, let $t \in S$ be integral over $φ (R [x])$ and let $p \in R [x]$ , $p = c_{0} + \dots + c_{k} x^{k}$ , satisfy $t φ (p) \in Im (φ)$ . Then there are $q \in R [x]$ and $n \geq 0$ with $φ (c_{k})^{n} t - φ (q)$ integral over $R$ (One-variable integral correction after leading-coefficient localization).

[L5]

For an ideal $I$ of a commutative ring the radical is $\sqrt{I} = {x : x^{n} \in I for some n \geq 1}$ , and $\sqrt{I}$ is an ideal containing $I$ (The radical of an ideal, The radical of an ideal is an ideal).

[L6]

Let $R$ be a commutative ring and $f, g \in R [x]$ nonzero. Then the coefficient of $x^{\deg f + \deg g}$ in $f g$ is $lc (f) lc (g)$ , and the coefficients of a product of polynomials are polynomial expressions in the coefficients of the factors (Degree inequalities for sums and products over a commutative ring).

Proof

technique · direct

1.1

The conductor $J$ is an ideal of $S$ . It contains $0$ because $0 \cdot S = 0 \subseteq Im (φ)$ ; it is closed under addition, since $(g + g^{'}) S \subseteq g S + g^{'} S \subseteq Im (φ)$ for $g, g^{'} \in J$ ; and it is closed under multiplication by $S$ , since $(s g) S = g (s S) \subseteq g S \subseteq Im (φ)$ for $g \in J$ and $s \in S$ . If $S = 0$ , then $J = S = {0}$ and both assertions of the Statement hold trivially; hence we may assume from now on that $S \neq 0$ and, by [L2], that $S = R [x] t_{1} + \dots + R [x] t_{n}$ for finitely many $t_{j} \in S$ .

givenL2

2.1

For $g \in S$ one has $g \in J$ if and only if $g t_{j} \in Im (φ)$ for all $j$ : one direction is immediate from $t_{j} \in S$ , and conversely $g S \subseteq \sum_{j} g R [x] t_{j} \subseteq Im (φ)$ because $g t_{j} \in Im (φ)$ for the finitely many module generators $t_{j}$ of [L2] and $Im (φ)$ is closed under multiplication by elements of $φ (R [x])$ .

givenstep 1.1L2

3.1

Assume now that $u φ (P) \in J$ with $u \in S$ and $P = \sum_{i = 0}^{k} a_{i} x^{i}$ . For each $j$ the element $u t_{j} \in S$ is integral over $R [x]$ by [L3], because $S$ is module-finite over $R [x]$ by [L2]; and $u φ (P) t_{j} \in Im (φ)$ by step 2.1. Applying [L4] with $t = u t_{j}$ and $p = P$ provides $q_{j} \in R [x]$ and an integer $n_{j} \geq 0$ such that $φ (a_{k})^{n_{j}} u t_{j} - φ (q_{j})$ is integral over $R$ ; this element lies in $Im (φ)$ by the integral-image condition of the Statement. Hence $φ (a_{k})^{n_{j}} u t_{j} \in Im (φ)$ for every $j$ .

givenstep 1.1step 2.1L1L2L3L4

4.1

Put $m = \max_{j} n_{j} \geq 0$ . For every $j$ , $u φ (a_{k})^{m} t_{j} = φ (a_{k})^{m - n_{j}} (φ (a_{k})^{n_{j}} u t_{j}) \in Im (φ)$ , because $Im (φ)$ is closed under multiplication by elements of $φ (R [x])$ and $m - n_{j} \geq 0$ . By the criterion of step 2.1 this says $u φ (a_{k})^{m} \in J$ , which is assertion 1.

givenstep 2.1step 3.1L2

5.1

Now assume $u φ (P) \in \sqrt{J}$ . By [L5] there is $N \geq 1$ with $(u φ (P))^{N} \in J$ . Since $φ$ is a ring homomorphism, $(u φ (P))^{N} = u^{N} φ (P^{N})$ . The coefficient of $x^{N k}$ in $P^{N}$ is $a_{k}^{N}$ by polynomial multiplication, even if this coefficient is zero and the degree is less than $N k$ . Applying assertion 1 in the form of step 4.1 to $u^{N}$ and $P^{N}$ gives $m \geq 0$ with $u^{N} φ (a_{k})^{N m} \in J$ . If $m \geq 1$ , multiply by $u^{N (m - 1)}$ to get $(u φ (a_{k}))^{N m} \in J$ ; if $m = 0$ , multiply $u^{N} \in J$ by $φ (a_{k})^{N}$ to get $(u φ (a_{k}))^{N} \in J$ . In either case a positive power of $u φ (a_{k})$ lies in $J$ , so $u φ (a_{k}) \in \sqrt{J}$ by [L5].

givenstep 4.1L5L6

6.1

Since $\sqrt{J}$ is an ideal of $S$ by [L5], the element $u φ (a_{k} x^{k}) = u φ (a_{k}) φ (x)^{k}$ lies in $\sqrt{J}$ , and therefore $u φ (P - a_{k} x^{k}) = u φ (P) - u φ (a_{k} x^{k}) \in \sqrt{J}$ ; here $P - a_{k} x^{k} = a_{0} + a_{1} x + \dots + a_{k - 1} x^{k - 1}$ has degree $\leq k - 1$ . Induction on $k \geq 0$ now gives $u φ (a_{i}) \in \sqrt{J}$ for every $i \in {0, \dots, k}$ : the case $k = 0$ is the hypothesis itself, and the induction step is the reduction just performed together with step 5.1.

givenstep 5.1L5

7.1

Steps 4.1 and 6.1 prove assertions 1 and 2 for every $u \in S$ and every $P \in R [x]$ ; the proof uses only finite choice (the finitely many module generators $t_{j}$ and the finite maximum $m$ ) and is therefore choice-free. ∎

step 4.1step 6.1

TheoremStatement: Literature-sourcedProof: AI-adaptedaudited 2026-09-27Open item page →

Algebraic Zariski Main localization at a quasi-finite prime

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $R \to S$ be a ring map of finite type (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras), let $S^{'} \subseteq S$ be the integral closure of the image of $R$ in $S$ (Integral elements subalgebra of an arbitrary ring map), and let $q \in Spec (S)$ be a prime at which $R \to S$ is quasi-finite (Quasi-finiteness at a prime of a finite-type algebra). Then there exists $g \in S^{'}$ with $g \notin q$ such that the inclusion $S^{'} \to S$ induces an isomorphism $S_{g}^{'} \overset{≅}{\to} S_{g}$ of localisations at $g$ (Multiplicative subsets and the localisation $S^{- 1} R$ as equivalence classes of fractions).

This is the local form of Zariski's main theorem: at a quasi-finite prime a finite-type algebra is, after inverting one element of its relative integral closure, a principal localisation of that closure. The proof is an induction on the least number of elements over which $S$ becomes module-finite over a polynomial extension of $R$ , and its one-variable step is the conductor argument supplied by Conductor radical detects every polynomial coefficient. The Axiom of Choice is used exactly once, in the nowhere-quasi-finiteness lemma Finite algebras over a strongly transcendental variable are nowhere quasi-finite (going down and lying over); all other steps make finitely many choices only.

Facts & Assumptions

Given: A unital ring map $R \to S$ of finite type that is quasi-finite at a prime $q \in Spec (S)$ , with contraction $p = q \cap R$ , together with the relative integral closure $S^{'} = {Int}_{R} (S) \subseteq S$ of the image of $R$ in $S$ ; the Axiom of Choice is assumed throughout.

[L1]

A finite type map $R \to S$ is quasi-finite at $q$ when the $κ (p)$ -algebra $S_{q} / p S_{q}$ is finite over $κ (p)$ , that is finitely generated as a module, equivalently finite-dimensional over $κ (p)$ (Quasi-finiteness at a prime of a finite-type algebra).

[L2]

For a unital ring map $R \to S$ the relative integral closure ${Int}_{R} (S)$ is the set of elements of $S$ integral over the map; it is a subring of $S$ containing the image of $R$ , hence an $R$ -subalgebra, and it is exactly the integral closure of the image of $R$ in $S$ (Integral elements subalgebra of an arbitrary ring map).

[L3]

An $R$ -algebra $A$ is of finite type over $R$ when $A = R [a_{1}, \dots, a_{n}]$ for finitely many $a_{i} \in A$ , equivalently when $A$ is isomorphic to a quotient $R [x_{1}, \dots, x_{n}] / a$ ; it is module-finite over $R$ when it is finitely generated as an $R$ -module (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras).

[L4]

Let $R \to S$ be of finite type and quasi-finite at $q$ . Then (1) for every intermediate $R$ -subalgebra $T$ , $Im (R) \subseteq T \subseteq S$ , the map $T \to S$ is of finite type and quasi-finite at $q$ ; (2) if such a $T$ is of finite type over $R$ , if $u \in T ∖ q$ and $T_{u} = S_{u}$ as subrings of $S_{u}$ , then $R \to T$ is quasi-finite at $q \cap T$ ; and (4) for an ideal $J \subseteq q$ the quotient $R \to S / J$ is of finite type and quasi-finite at $q / J$ (Quasi-finite local fibres transfer through quotients and intermediate rings).

[L5]

Let $φ : R [x] \to S$ be a finite ring map such that every element of $S$ integral over $R$ lies in $Im (φ)$ , and put $J = {g \in S : g S \subseteq Im (φ)}$ . Then $J$ is an ideal of $S$ , and for all $u \in S$ and $P = a_{0} + a_{1} x + \dots + a_{k} x^{k} \in R [x]$ : if $u φ (P) \in J$ then $u φ (a_{k})^{m} \in J$ for some $m \geq 0$ , and if $u φ (P) \in \sqrt{J}$ then $u φ (a_{i}) \in \sqrt{J}$ for every $i$ (Conductor radical detects every polynomial coefficient).

[L6]

Assume the Axiom of Choice. If $R \subseteq S$ are reduced rings, $x \in S$ is strongly transcendental over $R$ and $S$ is module-finite over $R [x]$ , then $R \to S$ is a finite type ring map that is quasi-finite at no prime of $S$ (Finite algebras over a strongly transcendental variable are nowhere quasi-finite).

[L7]

Let $R$ be a commutative ring, $I ⊴ R [x]$ an ideal, $S = R [x] / I$ , $q \in Spec (S)$ , and assume that $R \to S$ is quasi-finite at $q$ . Then the integral closure $S^{'}$ of the image of $R$ in $S$ contains an element $g \notin q$ with $S_{g}^{'} \to S_{g}$ an isomorphism (A quasi-finite one-generator quotient is locally its integral closure).

[L8]

For an inclusion $R \subseteq S$ and $x \in S$ , the element $x$ is strongly transcendental over $R$ when $u (a_{0} + a_{1} x + \dots + a_{k} x^{k}) = 0$ with $u \in S$ and $a_{i} \in R$ implies $u a_{i} = 0$ for every $i$ (Strong transcendence over a subring).

[L9]

If $A \to B$ and $B \to C$ are integral ring maps then the composite $A \to C$ is integral (Integral extensions are transitive).

[L10]

Let $A \subseteq B$ be commutative rings with $A \neq 0$ and $b \in B$ : $b$ is integral over $A$ if and only if $A [b]$ is finitely generated as an $A$ -module; in particular a module-finite extension is integral (Integrality and finite-module characterizations for one element).

[L11]

If $b_{1}, \dots, b_{n} \in B$ are integral over a subring $A \subseteq B$ , then the $A$ -subalgebra $A [b_{1}, \dots, b_{n}]$ is module-finite over $A$ (A subalgebra generated by finitely many integral elements is module-finite).

[L12]

The Axiom of Choice is the statement that every family of nonempty sets has a choice function (The Axiom of Choice).

[L13]

An injective module homomorphism remains injective after localisation: the induced map on localised modules is injective (Injective module maps remain injective after localisation).

[L14]

For multiplicative subsets $S, T \subseteq R$ with image $\overset{ˉ}{T}$ in $S^{- 1} R$ and $U$ the multiplicative subset generated by $S \cup T$ there is a unique $R$ -algebra isomorphism ${\overset{ˉ}{T}}^{- 1} (S^{- 1} R) ≅ U^{- 1} R$ ; in particular $(R_{f})_{g} ≅ R_{f g}$ for $f, g \in R$ (Localising twice is localising once at the multiplicative set generated by both denominator sets).

[L15]

For $f$ in a commutative ring $R$ the principal localisation is $R_{f} = S_{f}^{- 1} R$ with $S_{f} = {1, f, f^{2}, \dots}$ ; in particular $R_{1}$ is canonically isomorphic to $R$ (Principal localisation $R_{f} = {1, f, f^{2}, \dots}^{- 1} R$ ).

[L16]

A proper ideal $P ⊊ R$ is prime when $a b \in P$ implies $a \in P$ or $b \in P$ (Prime ideals and maximal ideals in a commutative ring).

[L17]

For an ideal $I ⊴ R$ the radical is $\sqrt{I} = {x \in R : x^{n} \in I for some integer n \geq 1}$ (The radical of an ideal).

[L18]

$\sqrt{I}$ is an ideal of $R$ containing $I$ (The radical of an ideal is an ideal).

[L19]

The ring $R$ is reduced when the only nilpotent element of $R$ is $0$ , that is when $Nil (R) = (0)$ for the radical of the zero ideal (The nilradical and reduced rings).

[L20]

For an ideal $I ⊴ R$ contraction along the quotient map $R \to R / I$ is an inclusion-preserving bijection $Spec (R / I) \to V (I)$ , whose inverse sends $p \supseteq I$ to $p / I$ (Prime ideals of a quotient ring are exactly the prime ideals containing the ideal).

[L21]

For a prime ideal $p \subset R$ the localisation of $R$ at $p$ is $R_{p} = (R ∖ p)^{- 1} R$ , with elements fractions $r / s$ where $s \notin p$ (Localisation at a prime ideal: $R_{p} = (R ∖ p)^{- 1} R$ ).

Proof

technique · direct

1.1

We assume the Axiom of Choice throughout; it is recorded in [L12] and will be used exactly once below, through [L6] in the nowhere-quasi-finiteness argument. By [L3] the finite type $R$ -algebra $S$ has the form $S = R [x_{1}, \dots, x_{n}]$ for some $n \geq 0$ and some $x_{1}, \dots, x_{n} \in S$ ; for these elements $S$ is module-finite over $R [x_{1}, \dots, x_{n}] = S$ , generated as a module over itself by $1$ . We prove by induction on $n \geq 0$ the statement $P (n)$ : for every unital ring map $A \to B$ of finite type that is quasi-finite at a prime $q_{B} \in Spec (B)$ in the sense of [L1], and every $y_{1}, \dots, y_{n} \in B$ such that $B$ is module-finite over the subalgebra $A [y_{1}, \dots, y_{n}]$ of [L3], there exists $g$ in the relative integral closure $B^{'}$ of the image of $A$ in $B$ with $g \notin q_{B}$ and $B_{g}^{'} \to B_{g}$ an isomorphism. The Theorem is the case $A = R$ , $B = S$ , $q_{B} = q$ and $y_{i} = x_{i}$ , and it is exactly the assertion to be proved.

givenL1L3L12

1.2

We begin with the case $n = 0$ of $P (0)$ : here $B$ is module-finite over $A [y_{1}, \dots, y_{0}] = A$ , so every element of $B$ is integral over $A$ by [L10]; hence $B \subseteq B^{'}$ , while $B^{'} \subseteq B$ by [L2], so that $B^{'} = B$ . Take $g = 1$ : the prime $q_{B}$ is proper by [L16], so $1 \notin q_{B}$ , and $B_{1}^{'} = B^{'} = B = B_{1}$ by [L15], an isomorphism of localisations. Thus $P (0)$ holds.

givenL2L10L15L16

2.1

Now suppose $n = 1$ , write $y = y_{1}$ and put $T : = B^{'}$ , the integral closure of the image of $A$ in $B$ , which is an intermediate $A$ -subalgebra by [L2]; also $B$ is module-finite over $T [y]$ because $A [y] \subseteq T [y]$ . By [L4] the map $T \to B$ is of finite type and quasi-finite at $q_{B}$ , and $T$ is integrally closed in $B$ : if $b \in B$ is integral over $T$ , then $b$ is integral over $A$ by [L9], since every element of $T$ is integral over $A$ by [L2], so $b \in B^{'} = T$ . Consequently, for $P (1)$ it suffices to prove the following special case. $(†)$ If $R$ is a ring integrally closed in a ring $S$ , if $R \to S$ is of finite type and quasi-finite at $q \in Spec (S)$ , and if $S$ is module-finite over $R [x]$ for one element $x \in S$ , then there is $g \in R ∖ q$ with the canonical map $R_{g} \to S_{g}$ an isomorphism. Indeed, applying $(†)$ to $T \to B$ with the element $y$ and the prime $q_{B}$ produces $g \in T ∖ q_{B} = B^{'} ∖ q_{B}$ with $T_{g} ≅ B_{g}$ , which is the conclusion of $P (1)$ because $T = B^{'}$ .

givenstep 1.1L2L4L9

2.2

Now let $n \geq 2$ and assume $P (m)$ for all $m < n$ ; suppose $S$ is module-finite over $R [x_{1}, \dots, x_{n}]$ and $R \to S$ is of finite type and quasi-finite at $q$ . Put $R^{'} : = {Int}_{R [x_{1}, \dots, x_{n - 1}]} (S)$ , the integral closure of $R [x_{1}, \dots, x_{n - 1}]$ in $S$ by [L2], an intermediate $R$ -subalgebra containing $x_{1}, \dots, x_{n - 1}$ . By [L4] the map $R^{'} \to S$ is of finite type and quasi-finite at $q$ , and $S$ is module-finite over $R^{'} [x_{n}]$ because $R [x_{1}, \dots, x_{n}] \subseteq R^{'} [x_{n}]$ .

givenstep 1.1L2L4

3.1

We now prove $(†)$ , so we assume $R = S^{'} = {Int}_{R} (S)$ , that $R \to S$ is of finite type and quasi-finite at $q$ , and that $S$ is module-finite over $R [x]$ for some $x \in S$ . Let $φ : R [x] \to S$ be the $R$ -algebra homomorphism with $φ (x) = x$ , put $I = \ker φ$ and $A = Im (φ) = R [x] / I$ , so that $R = φ (R) \subseteq A \subseteq S$ ; the map $φ$ is finite, that is, $S$ is a finitely generated $R [x]$ -module. Let $J = {g \in S : g S \subseteq A}$ be the conductor of $A$ in $S$ . Every element of $S$ integral over $R$ lies in $R \subseteq A$ because $R = S^{'}$ , so the hypothesis of [L5] is satisfied and $J$ is an ideal of $S$ .

givenstep 2.1L2L5

3.2

The ring $R^{'}$ is integrally closed in $S$ : if $s \in S$ is integral over $R^{'}$ , then $s$ is integral over $R [x_{1}, \dots, x_{n - 1}]$ by [L9], because every element of $R^{'}$ is integral over $R [x_{1}, \dots, x_{n - 1}]$ by [L2]; hence $s \in R^{'}$ .

givenstep 2.2L2L9

4.1

The quotient $\overset{ˉ}{S} : = S / \sqrt{J}$ is reduced, and so is its subring $\overset{ˉ}{R} : = R / (R \cap \sqrt{J})$ : if $s \in S$ has nilpotent image in $\overset{ˉ}{S}$ , say $s^{m} \in \sqrt{J}$ for some $m \geq 1$ , then by [L17] there is $k \geq 1$ with $s^{m k} = (s^{m})^{k} \in J$ , so $s \in \sqrt{J}$ by [L17] again, that is, the image of $s$ in $\overset{ˉ}{S}$ is zero; thus $\overset{ˉ}{S}$ has no nonzero nilpotent and is reduced by [L19].

givenstep 3.1L17L19

5.1

The image $\overset{ˉ}{x}$ of $x$ in $\overset{ˉ}{S}$ is strongly transcendental over $\overset{ˉ}{R}$ in the sense of [L8], and $\overset{ˉ}{S}$ is module-finite over $\overset{ˉ}{R} [\overset{ˉ}{x}]$ . For the first assertion, let $\overset{ˉ}{u} \in \overset{ˉ}{S}$ and $\overset{ˉ}{P} = \sum_{i} {\overset{ˉ}{a}}_{i} z^{i} \in \overset{ˉ}{R} [z]$ satisfy $\overset{ˉ}{u} \overset{ˉ}{P} (\overset{ˉ}{x}) = 0$ , and lift $\overset{ˉ}{u}$ , $\overset{ˉ}{P}$ to $u \in S$ , $P = \sum_{i} a_{i} z^{i} \in R [z]$ ; then $u φ (P) \in \sqrt{J}$ , so [L5] gives $u φ (a_{i}) \in \sqrt{J}$ for every $i$ , which says $\overset{ˉ}{u} {\overset{ˉ}{a}}_{i} = 0$ in $\overset{ˉ}{S}$ for every $i$ , exactly the condition of [L8]. Here $\overset{ˉ}{R} \to \overset{ˉ}{S}$ is injective, since an element of $R$ lies in $R \cap \sqrt{J}$ precisely when its image in $\overset{ˉ}{S}$ vanishes. For the second assertion, the image of $R [x]$ in $\overset{ˉ}{S}$ is $\overset{ˉ}{R} [\overset{ˉ}{x}]$ and the images of finitely many $R [x]$ -module generators of $S$ generate $\overset{ˉ}{S}$ over that subring, so $\overset{ˉ}{S}$ is module-finite over $\overset{ˉ}{R} [\overset{ˉ}{x}]$ by [L3].

givenstep 3.1step 4.1L3L5L8

6.1

By [L6], whose hypothesis is exactly the combination of steps 4.1 and 5.1, the map $\overset{ˉ}{R} \to \overset{ˉ}{S}$ is of finite type and quasi-finite at no prime of $\overset{ˉ}{S}$ . This is the only step of the proof that uses the Axiom of Choice, through the going down and lying over arguments inside [L6].

givenstep 4.1step 5.1L6L12

7.1

We deduce that $J \subseteq̸ q$ . Suppose instead that $\sqrt{J} \subseteq q$ ; then $\overset{ˉ}{q} : = q / \sqrt{J}$ is a prime of $\overset{ˉ}{S}$ , and [L4] applied to the finite type map $R \to S$ and the ideal $\sqrt{J} \subseteq q$ shows that $R \to \overset{ˉ}{S}$ is of finite type and quasi-finite at $\overset{ˉ}{q}$ . Applying [L4] once more with the intermediate $R$ -subalgebra $\overset{ˉ}{R} = Im (R \to \overset{ˉ}{S}) \subseteq \overset{ˉ}{S}$ gives that $\overset{ˉ}{R} \to \overset{ˉ}{S}$ is quasi-finite at $\overset{ˉ}{q}$ , contradicting step 6.1. Since the primes containing $J$ are precisely those containing $\sqrt{J}$ , by [L18], the assumption $J \subseteq q$ is therefore impossible, and we may choose $s \in J ∖ q$ .

givenstep 6.1L4L18

8.1

Because $J = {g \in S : g S \subseteq A} \subseteq A$ , the element $s$ lies in $A = Im (φ)$ , so $s = φ (f)$ for some $f \in R [x]$ ; and $f \notin q_{0} : = φ^{- 1} (q)$ because $φ (f) = s \notin q$ . Put $q^{'} : = q_{0} / I$ . By [L20] the ideal $q^{'}$ is a prime of $A = R [x] / I$ with $q^{'} \cap R = q_{0} \cap R = p$ , and $q^{'} = q \cap A$ .

givenstep 3.1step 7.1L20

9.1

The canonical $R$ -algebra homomorphism $A_{f} \to S_{s}$ is an isomorphism: it is injective as the localisation of the injective map $A \to S$ by [L13], and it is surjective because $s S \subseteq A$ . Indeed, for $σ \in S$ the element $s σ$ lies in $A$ , say $s σ = φ (r)$ with $r \in R [x]$ , so $σ / 1 = (s σ) / s = φ (r) / s$ is the image of $φ (r) / f$ (note $φ (f) = s$ ).

givenstep 3.1step 7.1step 8.1L13

10.1

The map $R \to A$ is quasi-finite at $q^{'}$ by [L4]: the ring $A$ is an intermediate $R$ -subalgebra of $S$ that is of finite type over $R$ as a quotient of $R [x]$ by [L3]; the element $f$ lies in $A ∖ q^{'}$ because $φ (f) = s \notin q$ ; and step 9.1 says that $A_{f} = S_{f}$ as subrings of $S_{f}$ , with $q^{'} = q \cap A$ by step 8.1.

givenstep 8.1step 9.1L3L4

11.1

The integral closure of the image of $R$ in $A$ is $R$ itself: every element of $A$ that is integral over $R$ is an element of $S$ integral over $R$ , hence lies in $S^{'} = R$ because $R$ is integrally closed in $S$ ; and $R \subseteq A$ . Since $A = R [x] / I$ and $R \to A$ is quasi-finite at $q^{'}$ by step 10.1, [L7] provides $h \in R ∖ q^{'}$ such that the canonical map $R_{h} \to A_{h}$ is an isomorphism.

givenstep 10.1L2L7

12.1

Inside $A_{h} = R_{h}$ the element $f \in A$ has the form $f = r / h^{k}$ for some $r \in R$ and some $k \geq 0$ . We claim $r \notin q$ . Otherwise $r \in q \cap A = q^{'}$ ; localising $A$ at the prime $q^{'}$ by [L21], the equation $f h^{k} = r$ would exhibit $f h^{k}$ as an element of the maximal ideal $q^{'} A_{q^{'}}$ , although $f \notin q^{'}$ and $h \notin q^{'}$ make $f h^{k}$ a unit of $A_{q^{'}}$ .

givenstep 8.1step 10.1step 11.1L21

13.1

Set $g : = r h \in R$ ; then $g \notin q$ because $r \notin q$ , $h \notin q$ and $q$ is prime by [L16]. We show that the canonical map $R_{g} \to S_{g}$ is an isomorphism. First $A_{f h} ≅ S_{s h}$ : localising the isomorphism $A_{f} ≅ S_{s}$ of step 9.1 at the common element $h$ gives $(A_{f})_{h} ≅ (S_{s})_{h}$ , and $(A_{f})_{h} ≅ A_{f h}$ , $(S_{s})_{h} ≅ S_{s h}$ by [L14]. Second $A_{f h} ≅ R_{r h} = R_{g}$ : by step 11.1 we have $A_{h} ≅ R_{h}$ , so $A_{f h} ≅ (R_{h})_{f}$ by [L14], and inside $R_{h}$ the elements $f$ and $r$ differ by the unit $h^{- k}$ , so $(R_{h})_{f} ≅ (R_{h})_{r} ≅ R_{h r}$ by [L14]. Third $S_{s h} ≅ S_{r h} = S_{g}$ : the relation $f h^{k} = r$ holds in $A_{h}$ and hence in $S_{h}$ , and applying the $R$ -algebra map $φ$ gives $s h^{k} = φ (f h^{k}) = φ (r) = r$ there; so $s$ and $r$ also differ by the unit $h^{- k}$ in $S_{h}$ , whence $(S_{h})_{s} ≅ (S_{h})_{r}$ , that is $S_{s h} ≅ S_{r h} = S_{g}$ by [L14]. Composing these isomorphisms gives an isomorphism $R_{g} \to S_{g}$ ; every map in the chain is the canonical localisation of one of the inclusions $R \subseteq A \subseteq S$ , so the composite is the map induced by the inclusion $R \subseteq S$ . This proves $(†)$ .

givenstep 9.1step 11.1step 12.1L14L16

14.1

Steps 3.1 to 13.1 prove $(†)$ , and step 2.1 reduces $P (1)$ to $(†)$ . Hence $P (1)$ holds: for a finite type map $A \to B$ quasi-finite at $q_{B}$ with $B$ module-finite over $A [y_{1}]$ , there is $g \in B^{'} ∖ q_{B}$ with $B_{g}^{'} ≅ B_{g}$ .

step 2.1step 13.1

14.2

By $(†)$ , that is by step 13.1 applied under the hypotheses verified in steps 2.2 and 3.2, there is $g^{'} \in R^{'} ∖ q$ with $(R^{'})_{g^{'}} ≅ S_{g^{'}}$ via the inclusion.

step 13.1step 2.2step 3.2

15.1

The localisation $S_{g^{'}}$ is a finitely generated $R$ -algebra: $S$ is generated as an $R$ -algebra by finitely many elements by [L3], and adjoining $1 / g^{'}$ exhibits $S_{g^{'}}$ as generated by those elements together with $1 / g^{'}$ by [L3]. Choose $z_{1}, \dots, z_{M} \in S_{g^{'}}$ generating $S_{g^{'}}$ over $R$ ; by step 14.2 each $z_{j}$ lies in $(R^{'})_{g^{'}}$ , so $z_{j} = y_{j} / (g^{'})^{n_{j}}$ for some $y_{j} \in R^{'}$ and $n_{j} \geq 0$ . Put $R^{''} : = R [x_{1}, \dots, x_{n - 1}, y_{1}, \dots, y_{M}, g^{'}] \subseteq R^{'}$ , an $R$ -subalgebra that is of finite type over $R$ by [L3]. Then $(R^{''})_{g^{'}} = S_{g^{'}}$ : the inclusion $(R^{''})_{g^{'}} \subseteq (R^{'})_{g^{'}} = S_{g^{'}}$ is clear, while each $z_{j} = y_{j} / (g^{'})^{n_{j}}$ lies in $(R^{''})_{g^{'}}$ , so $S_{g^{'}} = R [z_{1}, \dots, z_{M}] \subseteq (R^{''})_{g^{'}}$ .

givenstep 14.2L3

16.1

The algebra $R^{''}$ is module-finite over $R [x_{1}, \dots, x_{n - 1}]$ : each of $y_{1}, \dots, y_{M}, g^{'}$ lies in $R^{'}$ and is therefore integral over $R [x_{1}, \dots, x_{n - 1}]$ by [L2], so [L11] applies to $R^{''} = R [x_{1}, \dots, x_{n - 1}] [y_{1}, \dots, y_{M}, g^{'}]$ .

givenstep 15.1L2L11

17.1

By [L4] applied to the intermediate subalgebra $R^{''}$ , the element $g^{'} \in R^{''} ∖ q$ and the equality $(R^{''})_{g^{'}} = S_{g^{'}}$ of step 15.1, the map $R \to R^{''}$ is quasi-finite at $q^{''} : = q \cap R^{''}$ ; it is of finite type by step 15.1. Since $R^{''}$ is module-finite over $R [x_{1}, \dots, x_{n - 1}]$ by step 16.1, the induction hypothesis $P (n - 1)$ applies to the map $R \to R^{''}$ , the prime $q^{''}$ and the elements $x_{1}, \dots, x_{n - 1} \in R^{''}$ : there is $g^{''} \in R^{'''} ∖ q^{''}$ with $(R^{'''})_{g^{''}} ≅ (R^{''})_{g^{''}}$ via the inclusion, where $R^{'''} = {Int}_{R} (R^{''}) \subseteq S^{'}$ is the integral closure of the image of $R$ in $R^{''}$ .

givenstep 15.1step 16.1step 14.1L2L4

18.1

The image of $g^{'}$ in $(R^{''})_{g^{''}} = (R^{'''})_{g^{''}}$ has the form $g^{'''} / (g^{''})^{m}$ for some $g^{'''} \in R^{'''}$ and some $m \geq 0$ , since the elements of that localisation are fractions with numerator in $R^{'''}$ by [L21]. If $g^{'''} \in q$ , then $g^{'''} / (g^{''})^{m}$ would lie in the prime $q (R^{'''})_{g^{''}}$ of $(R^{'''})_{g^{''}}$ , because $g^{''} \notin q$ ; but this element is the image of $g^{'} \in R^{''} ∖ q$ under $R^{''} \to (R^{''})_{g^{''}}$ , and $g^{''} \notin q^{''} = q \cap R^{''}$ forces that image to lie outside $q (R^{''})_{g^{''}}$ . Hence $g^{'''} \notin q$ , and since also $g^{''} \notin q$ and $q$ is prime by [L16], the product $g : = g^{''} g^{'''} \in R^{'''} \subseteq S^{'}$ satisfies $g \notin q$ .

givenstep 17.1L16L21

19.1

Finally $(R^{'''})_{g} ≅ S_{g}$ . By [L14], $(R^{'''})_{g} = ((R^{'''})_{g^{''}})_{g^{'''}} ≅ ((R^{''})_{g^{''}})_{g^{'''}}$ , and inside $(R^{''})_{g^{''}}$ the element $g^{'''}$ differs from $g^{'}$ by the unit $(g^{''})^{- m}$ : indeed step 18.1 says that the image of $g^{'}$ equals $g^{'''} / (g^{''})^{m}$ , that is $g^{'''} = g^{'} (g^{''})^{m}$ . Hence inverting $g^{'''}$ is the same as inverting $g^{'}$ , and $((R^{''})_{g^{''}})_{g^{'''}} ≅ ((R^{''})_{g^{''}})_{g^{'}} = (R^{''})_{g^{''} g^{'}}$ by [L14]. By step 15.1, $(R^{''})_{g^{'}} = S_{g^{'}}$ , so $(R^{''})_{g^{''} g^{'}} = (S_{g^{'}})_{g^{''}} = S_{g^{''} g^{'}}$ by [L14]. The same relation $g^{'} = g^{'''} / (g^{''})^{m}$ , read in $S_{g^{''}}$ through the injective map $R^{''} \to S$ and its localisation, shows that $g^{'}$ and $g^{'''}$ differ by a unit of $S_{g^{''}}$ , so also $S_{g^{''} g^{'}} ≅ S_{g^{''} g^{'''}} = S_{g}$ by [L14]. Thus $(R^{'''})_{g} ≅ S_{g}$ via the canonical maps. As $R^{'''} \subseteq S^{'} \subseteq S$ , the inclusion $S^{'} \to S$ becomes an isomorphism after inverting $g$ , since $S_{g}^{'}$ lies between the subrings $(R^{'''})_{g}$ and $S_{g}$ , which coincide. This is the conclusion of $P (n)$ .

givenstep 15.1step 18.1L14

20.1

Steps 1.2, 14.1 and 15.1 to 19.1 establish $P (m)$ for every $m \geq 0$ by induction. Applying $P (n)$ to the originally given map $R \to S$ , the prime $q$ and the elements $x_{1}, \dots, x_{n}$ of step 1.1 produces $g \in S^{'} ∖ q$ such that $S^{'} \to S$ induces an isomorphism $S_{g}^{'} ≅ S_{g}$ , which is the assertion of the Theorem. The Axiom of Choice was used only in step 6.1 through [L6]; every other step selected only finitely many elements (the generators $x_{i}$ , the element $s$ , the finite lists $y_{j}$ and $z_{j}$ , the elements $r$ and $g$ ), so no use of the axiom is hidden elsewhere. ∎

step 1.1step 1.2step 14.1step 19.1L12

CorollaryStatement: Literature-sourcedProof: AI-adaptedaudited 2026-09-27Open item page →

The quasi-finite locus of a finite-type algebra is open

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $R \to S$ be a ring map of finite type (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras). Then the set of primes

$U = {q \in Spec (S) : R \to S is quasi-finite at q}$

is open in $Spec (S)$ (Quasi-finiteness at a prime of a finite-type algebra, Principal distinguished subsets of the prime spectrum, The vanishing sets define the Zariski topology on the prime spectrum).

At every point of $U$ the local theorem Algebraic Zariski Main localization at a quasi-finite prime supplies an element of the relative integral closure at which the algebra becomes a principal localisation of that closure. Because the algebra is of finite type, the closure may be replaced there by the finite subalgebra it generates, and such a finite algebra is quasi-finite over the base at every one of its primes; this is the content of the proof below, which is where the Axiom of Choice enters, through the local theorem.

Facts & Assumptions

Given: A unital ring map $R \to S$ of finite type, the relative integral closure $S^{'} = {Int}_{R} (S) \subseteq S$ of the image of $R$ in $S$ , and the Axiom of Choice, assumed throughout.

[L1]

The map $R \to S$ is quasi-finite at $q$ when the $κ (p)$ -algebra $S_{q} / p S_{q}$ is finite over $κ (p)$ , that is, finitely generated as a $κ (p)$ -module, equivalently finite-dimensional over $κ (p)$ (Quasi-finiteness at a prime of a finite-type algebra).

[L2]

An $R$ -algebra $A$ is of finite type over $R$ when $A = R [a_{1}, \dots, a_{n}]$ for finitely many elements, equivalently a quotient of a polynomial ring, and it is module-finite over $R$ when it is finitely generated as an $R$ -module (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras).

[L3]

Assume the Axiom of Choice. For a finite type map $R \to S$ that is quasi-finite at $q \in Spec (S)$ there is $g \in S^{'} ∖ q$ , where $S^{'}$ is the integral closure of the image of $R$ in $S$ , such that the inclusion induces an isomorphism $S_{g}^{'} ≅ S_{g}$ of localisations (Algebraic Zariski Main localization at a quasi-finite prime).

[L4]

[L5]

If $A \subseteq B$ is a subring and $b_{1}, \dots, b_{n} \in B$ are integral over $A$ , then the $A$ -subalgebra $A [b_{1}, \dots, b_{n}]$ is module-finite over $A$ (A subalgebra generated by finitely many integral elements is module-finite).

[L6]

For multiplicative subsets $S, T \subseteq R$ with image $\overset{ˉ}{T}$ in $S^{- 1} R$ and $U$ the multiplicative subset generated by $S \cup T$ there is a unique $R$ -algebra isomorphism ${\overset{ˉ}{T}}^{- 1} (S^{- 1} R) ≅ U^{- 1} R$ ; in particular $(R_{f})_{g} ≅ R_{f g}$ (Localising twice is localising once at the multiplicative set generated by both denominator sets).

[L7]

For an ideal $I$ of a commutative ring $R$ and a multiplicative subset $S \subseteq R$ with image $\overset{ˉ}{S}$ in $R / I$ there is a canonical isomorphism $(S^{- 1} R) / (S^{- 1} I) ≅ {\overset{ˉ}{S}}^{- 1} (R / I)$ (Localisation commutes with quotient rings: $S^{- 1} R / S^{- 1} I ≅ {\overset{ˉ}{S}}^{- 1} (R / I)$ ).

[L8]

The localisation $S^{- 1} R$ consists of the classes $r / s$ with $r \in R$ , $s \in S$ , and every $s \in S$ maps to a unit, with $(s / 1)^{- 1} = 1 / s$ (Multiplicative subsets and the localisation $S^{- 1} R$ as equivalence classes of fractions).

[L9]

For a prime ideal $p$ of a commutative ring $R$ there is a canonical field isomorphism $R_{p} / p R_{p} ≅ Frac (R / p)$ , the residue field $κ (p)$ ( $R_{p} / p R_{p} ≅ Frac (R / p)$ is the residue field at $p$ ).

[L10]

The Axiom of Choice (AC) is the statement that every family of nonempty sets has a choice function (The Axiom of Choice).

[L11]

For a multiplicative subset $S \subseteq R$ contraction along the localisation map induces an inclusion-preserving bijection $Spec (S^{- 1} R) \to {p \in Spec (R) : p \cap S = \emptyset}$ , with inverse $p \mapsto S^{- 1} p$ (Prime ideals of a localization are exactly the primes disjoint from the denominator set).

[L12]

For $f$ in a commutative ring $R$ the principal localisation is $R_{f} = S_{f}^{- 1} R$ with $S_{f} = {1, f, f^{2}, \dots}$ , and its elements may be written $r / f^{n}$ ; in particular $R_{1}$ is canonically isomorphic to $R$ (Principal localisation $R_{f} = {1, f, f^{2}, \dots}^{- 1} R$ ).

[L13]

For $f \in R$ the principal distinguished subset is $D (f) = {p \in Spec (R) : f \notin p}$ , the complement of $V ((f))$ inside $Spec (R)$ (Principal distinguished subsets of the prime spectrum).

[L14]

The subsets $V (I)$ , as $I$ ranges over the ideals of $R$ , contain $Spec (R)$ and $\emptyset$ , are closed under arbitrary intersections and finite unions, and define a topology on $Spec (R)$ (The vanishing sets define the Zariski topology on the prime spectrum).

Proof

technique · direct

1.1

We assume the Axiom of Choice as recorded in [L10]. By [L2] the finite type $R$ -algebra $S$ has the form $S = R [s_{1}, \dots, s_{n}]$ for some $n \geq 0$ and elements $s_{i} \in S$ , and $S^{'} = {Int}_{R} (S)$ is an $R$ -subalgebra of $S$ containing the image of $R$ by [L4]. We must show that the set $U$ of primes $q$ at which $R \to S$ is quasi-finite, in the sense of [L1], is open in the topology of [L14].

givenL1L2L4L10L14

1.2

We record a general fact about module-finite algebras. Let $B$ be module-finite over a ring $C$ and let $p \in Spec (C)$ ; choose $b_{1}, \dots, b_{m} \in B$ generating $B$ as a $C$ -module. Then the images of $b_{1}, \dots, b_{m}$ generate $B / p B$ as a $C / p$ -module, and therefore generate the $C / p$ -algebra $(B / p B) \otimes_{C / p} κ (p)$ as a module over $κ (p) = Frac (C / p)$ by [L9]: localising a module at a multiplicative set multiplies its generating set by scalars and never enlarges it. Hence $(B / p B) \otimes_{C / p} κ (p)$ is a finite-dimensional $κ (p)$ -algebra.

givenL2L9

1.3

We also record a general fact about localising finite-dimensional algebras. Let $C$ be a finite-dimensional algebra over a field $K$ and let $W \subseteq C$ be a multiplicative subset. For every $u \in W$ the ideals $(u) \supseteq (u^{2}) \supseteq (u^{3}) \supseteq \dots$ of $C$ are $K$ -subspaces of the finite-dimensional $K$ -vector space $C$ , so the chain stabilises: there is $N \geq 1$ with $(u^{N}) = (u^{N + 1})$ , hence $u^{N} = u^{N + 1} x$ for some $x \in C$ and therefore $u^{N} (1 - u x) = 0$ in $C$ . Since $u$ is a unit of $W^{- 1} C$ with inverse $1 / u$ by [L8], the image of $u^{N}$ is invertible in $W^{- 1} C$ , so the image of $1 - u x$ is zero there, that is $x = 1 / u$ in $W^{- 1} C$ . Consequently every element of $W^{- 1} C$ is of the form $c \cdot x$ with $c \in C$ and hence lies in the image of the localisation map $C \to W^{- 1} C$ , which is therefore surjective as a $K$ -linear map; so $\dim_{K} W^{- 1} C \leq \dim_{K} C$ and $W^{- 1} C$ is finite-dimensional over $K$ .

givenL8

2.1

Let $q \in U$ and put $p = q \cap R$ . By [L3] there is $g \in S^{'} ∖ q$ such that the inclusion $S^{'} \to S$ induces an isomorphism $S_{g}^{'} \to S_{g}$ of localisations; we fix such a $g$ and view $S_{g}^{'} = S_{g}$ as subrings of the common ring $S_{g}$ .

givenstep 1.1L3

2.2

The localisation $S_{g}$ is a finitely generated $R$ -algebra: its elements are the fractions with numerator in $S$ and denominator a power of $g$ by [L12], so the images of $s_{1}, \dots, s_{n}$ together with $1 / g$ generate $S_{g}$ over $R$ by [L2]. Hence there are finitely many elements $z_{1}, \dots, z_{N} \in S_{g}$ generating $S_{g}$ as an $R$ -algebra; taking them to be the explicit generators just listed.

givenstep 1.1L2L12

3.1

Each $z_{j}$ lies in $S_{g}^{'} = S_{g}$ , so by [L12] and [L8] it can be written $z_{j} = y_{j} / g^{m_{j}}$ with $y_{j} \in S^{'}$ and some exponent $m_{j} \geq 0$ . Put $T : = R [g, y_{1}, \dots, y_{N}] \subseteq S^{'}$ . Then $T_{g} = S_{g}$ : the inclusion $T \subseteq S^{'}$ gives $T_{g} \subseteq S_{g}^{'} = S_{g}$ , while each generator $z_{j} = y_{j} / g^{m_{j}}$ of $S_{g}$ over $R$ lies in $T_{g}$ , so $S_{g} \subseteq T_{g}$ . Also $T$ is an $R$ -subalgebra of $S$ containing the image of $R$ .

givenstep 2.1step 2.2L2L8L12

4.1

Every generator $g, y_{1}, \dots, y_{N}$ of $T$ lies in $S^{'}$ and is therefore integral over $R$ by [L4]. Writing $A$ for the image of $R$ in $S$ , the elements $g, y_{1}, \dots, y_{N}$ are integral over the subring $A \subseteq S$ , so [L5] shows that $T = A [g, y_{1}, \dots, y_{N}]$ is module-finite over $A$ ; the same finite list generates $T$ as an $R$ -module, so $T$ is module-finite over $R$ , and $T$ is of finite type over $R$ by [L2].

givenstep 3.1L2L4L5

4.2

Now let $q^{'} \in Spec (S)$ with $g \notin q^{'}$ , and put $q^{''} : = q^{'} \cap T$ , $p^{'} : = q^{'} \cap R = q^{''} \cap R$ and $T_{g}$ -prime $q^{''} T_{g} \subseteq T_{g}$ , $q^{'} S_{g} \subseteq S_{g}$ . The primes $q^{''} T_{g}$ and $q^{'} S_{g}$ correspond under the $R$ -algebra isomorphism $T_{g} ≅ S_{g}$ of step 3.1, because contraction along $T \to T_{g}$ and $S \to S_{g}$ recovers $q^{''}$ and $q^{'}$ by [L11] and $q^{'} \cap T = q^{''}$ .

givenstep 3.1L11

5.1

Consequently, for the module-finite $R$ -algebra $T$ of step 4.1 and any prime $r \in Spec (T)$ with contraction $p = r \cap R$ , the algebra $T_{r} / p T_{r}$ is finite-dimensional over $κ (p)$ . Indeed, $p T \subseteq r$ , so [L7] gives $T_{r} / p T_{r} ≅ (T / p T)_{\overset{ˉ}{r}}$ for the prime $\overset{ˉ}{r} = r / p T$ ; put $C : = (T / p T) \otimes_{R / p} κ (p)$ , which is finite-dimensional over $κ (p)$ by step 1.2, and let $S_{0}$ be the image of $(R / p) ∖ {0}$ in $T / p T$ and $\overset{ˉ}{U}$ the image of $(T / p T) ∖ \overset{ˉ}{r}$ . Since $\overset{ˉ}{r} \cap (R / p) = 0$ , we have $S_{0} \subseteq (T / p T) ∖ \overset{ˉ}{r}$ ; applying [L6] to the multiplicative subsets $S_{0}$ and $(T / p T) ∖ \overset{ˉ}{r}$ of the ring $T / p T$ shows that $(T / p T)_{\overset{ˉ}{r}} ≅ {\overset{ˉ}{U}}^{- 1} C$ , a localisation of the finite-dimensional $κ (p)$ -algebra $C$ at a multiplicative subset, which is finite-dimensional over $κ (p)$ by step 1.3.

givenstep 4.1step 1.2step 1.3L6L7L9

5.2

Hence the canonical $R$ -algebra homomorphism $T_{q^{''}} \to S_{q^{'}}$ is an isomorphism: both rings are localisations of the common ring $T_{g} = S_{g}$ at the primes $q^{''} T_{g}$ and $q^{'} S_{g}$ of step 4.2, and [L6] exhibits each of them as a localisation at the multiplicative subset of the common ring generated by ${g^{n}}$ together with the elements outside that prime.

givenstep 4.2L6L12

6.1

Therefore $T_{q^{''}} / p^{'} T_{q^{''}} ≅ S_{q^{'}} / p^{'} S_{q^{'}}$ is finite-dimensional over $κ (p^{'})$ by step 5.1, and since $S$ is of finite type over $R$ by step 1.1 and $p^{'} = q^{'} \cap R$ , the map $R \to S$ is quasi-finite at $q^{'}$ by [L1]. Thus $D_{S} (g) \subseteq U$ .

givenstep 5.1step 5.2L1

7.1

It remains to draw the topological conclusion. By [L13] and [L14] each set $D_{S} (g) = Spec (S) ∖ V ((g))$ is the complement of a closed subset, hence open in the topology of [L14]. Step 2.1 attaches to every $q \in U$ an element $g \in S$ with $q \in D_{S} (g)$ , and step 6.1 shows $D_{S} (g) \subseteq U$ ; thus every point of $U$ has an open neighbourhood contained in $U$ , which is the defining property of an open subset, so $U$ is open (no selection from infinitely many points is needed, the argument being applied to one prime at a time).

givenstep 2.1step 6.1L13L14

8.1

The proof used the Axiom of Choice only in step 2.1, through [L3]; every other step selected finitely many elements (the generators $s_{i}$ , the generators $z_{j}$ , the numerators $y_{j}$ , the module generators $b_{i}$ , the exponent $N$ and the elements $u, x$ of step 1.3). ∎

givenstep 1.1step 2.1L10

TheoremStatement: Literature-sourcedProof: AI-adaptedaudited 2026-09-27Open item page →

A quasi-finite algebra factors openly through a finite algebra

Statement

There are a finite $R$ -subalgebra $T \subseteq S^{'}$ that is module-finite over $R$ and finitely many elements $g_{1}, \dots, g_{n} \in T$ such that $U : = D_{T} (g_{1}) \cup \dots \cup D_{T} (g_{n})$ is an open subset of $Spec (T)$ (Principal distinguished subsets of the prime spectrum), the contraction map $Spec (S) \to Spec (T)$ (The prime-spectrum construction is a contravariant functor to topological spaces) is a homeomorphism onto $U$ , and the inclusion $T \to S$ induces an isomorphism $T_{g_{i}} \to S_{g_{i}}$ of principal localisations (Principal localisation $R_{f} = {1, f, f^{2}, \dots}^{- 1} R$ ) for every $i$ .
For every $g \in T$ with $D_{T} (g) \subseteq U$ the inclusion induces an isomorphism $T_{g} \to S_{g}$ .

Thus a quasi-finite finite-type algebra is, after replacing the base by a finite subalgebra of the relative integral closure, an open piece of that finite algebra: locally on the source it is a principal localisation of a finite algebra, and globally on the source the map is a homeomorphism onto an open subset. The proof is a finite-principal-open patching, with the Axiom of Choice used for the compactness of the spectrum and for turning a finite open cover by distinguished opens into a unit-ideal expression; the published Stacks proof of the finite-algebra part is the phrase "Details omitted", which is spelled out here.

Facts & Assumptions

Given: A ring map $R \to S$ of finite type that is quasi-finite at every prime of $S$ , the relative integral closure $S^{'} = {Int}_{R} (S) \subseteq S$ of the image of $R$ in $S$ , and the Axiom of Choice, assumed throughout.

[L1]

The map $R \to S$ is quasi-finite at $q$ when the $κ (p)$ -algebra $S_{q} / p S_{q}$ is finite over $κ (p)$ , and $R \to S$ is quasi-finite when it is of finite type and quasi-finite at every prime of $S$ (Quasi-finiteness at a prime of a finite-type algebra).

[L2]

An $R$ -algebra $A$ is of finite type over $R$ when $A = R [a_{1}, \dots, a_{n}]$ for finitely many elements, equivalently a quotient of a polynomial ring, and module-finite over $R$ when it is finitely generated as an $R$ -module (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras).

[L3]

Assume the Axiom of Choice. For a finite type map $R \to S$ quasi-finite at $q \in Spec (S)$ there is $g \in S^{'} ∖ q$ with $S^{'} \to S$ inducing an isomorphism $S_{g}^{'} ≅ S_{g}$ (Algebraic Zariski Main localization at a quasi-finite prime).

[L4]

For a unital ring map $R \to S$ the relative integral closure ${Int}_{R} (S)$ is the set of elements of $S$ integral over the map, an $R$ -subalgebra of $S$ containing the image of $R$ and equal to the integral closure of that image (Integral elements subalgebra of an arbitrary ring map).

[L5]

If $A \subseteq B$ is a subring and $b_{1}, \dots, b_{n} \in B$ are integral over $A$ , then $A [b_{1}, \dots, b_{n}]$ is module-finite over $A$ (A subalgebra generated by finitely many integral elements is module-finite).

[L6]

For multiplicative subsets $S, T \subseteq R$ with image $\overset{ˉ}{T}$ in $S^{- 1} R$ and $U$ generated by $S \cup T$ there is a unique $R$ -algebra isomorphism ${\overset{ˉ}{T}}^{- 1} (S^{- 1} R) ≅ U^{- 1} R$ ; in particular $(R_{f})_{g} ≅ R_{f g}$ (Localising twice is localising once at the multiplicative set generated by both denominator sets).

[L7]

Assume the Axiom of Choice. For every commutative ring $R$ the space $Spec (R)$ is compact (The prime spectrum is compact in the library's non-Hausdorff sense).

[L8]

Assume the Axiom of Choice. If a family of elements $f_{λ}$ of $R$ satisfies $Spec (R) = ⋃_{λ} D (f_{λ})$ , then the ideal generated by the family is the unit ideal $R$ (A distinguished-open cover of the spectrum forces the covering ideal to be the unit ideal).

[L9]

For a multiplicative subset $S \subseteq R$ the contraction map along $R \to S^{- 1} R$ is a homeomorphism from $Spec (S^{- 1} R)$ onto ${p \in Spec (R) : p \cap S = \emptyset}$ (The spectrum of a localisation is the subspace of primes disjoint from the denominator set).

[L10]

For $f \in R$ the principal distinguished subset is $D (f) = {p : f \notin p}$ , the complement of $V ((f))$ (Principal distinguished subsets of the prime spectrum).

[L11]

The subsets $V (I)$ of $Spec (R)$ , as $I$ ranges over the ideals of $R$ , contain $Spec (R)$ and $\emptyset$ , are closed under arbitrary intersections and finite unions, and define a topology on $Spec (R)$ (The vanishing sets define the Zariski topology on the prime spectrum).

[L12]

For every ring homomorphism $φ : R \to A$ contraction defines a continuous map $Spec (A) \to Spec (R)$ , and these maps compose contravariantly (The prime-spectrum construction is a contravariant functor to topological spaces).

[L13]

For a multiplicative subset $S \subseteq R$ contraction along $R \to S^{- 1} R$ is an inclusion-preserving bijection onto the primes disjoint from $S$ , with inverse $p \mapsto S^{- 1} p$ (Prime ideals of a localization are exactly the primes disjoint from the denominator set).

[L14]

For $f$ in a commutative ring $R$ the principal localisation is $R_{f} = S_{f}^{- 1} R$ with $S_{f} = {1, f, f^{2}, \dots}$ , and its elements may be written $r / f^{n}$ (Principal localisation $R_{f} = {1, f, f^{2}, \dots}^{- 1} R$ ).

[L15]

Localisation preserves injectivity: if $f : M^{'} \to M$ is an injective $R$ -module homomorphism and $S \subseteq R$ is multiplicative, then $S^{- 1} f : S^{- 1} M^{'} \to S^{- 1} M$ is injective (Injective module maps remain injective after localisation).

[L16]

The Axiom of Choice (AC) is the statement that every family of nonempty sets has a choice function (The Axiom of Choice).

Proof

technique · direct

1.1

We assume the Axiom of Choice as recorded in [L16]. By [L2] the finite type $R$ -algebra $S$ has the form $S = R [s_{1}, \dots, s_{m}]$ , and $S^{'} = {Int}_{R} (S)$ is an $R$ -subalgebra of $S$ containing the image of $R$ by [L4]. The hypothesis is that $R \to S$ is quasi-finite at every prime of $S$ in the sense of [L1].

givenL1L2L4L16

2.1

For each prime $q \in Spec (S)$ the local theorem [L3] applies and produces an element $g_{q} \in S^{'} ∖ q$ such that the inclusion $S^{'} \to S$ induces an isomorphism $S_{g_{q}}^{'} \to S_{g_{q}}$ . Since $q \in D_{S} (g_{q}) = {r : g_{q} \notin r}$ , and each $D_{S} (g_{q})$ is open in the topology of [L11] because it is the complement of $V ((g_{q}))$ by [L10], the family ${D_{S} (g_{q})}_{q}$ is an open cover of $Spec (S)$ .

givenstep 1.1L3L10L11

3.1

By compactness [L7] there are finitely many elements $g_{1}, \dots, g_{n} \in S^{'}$ with $Spec (S) = D_{S} (g_{1}) \cup \dots \cup D_{S} (g_{n}),$ and then the ideal generated by $g_{1}, \dots, g_{n}$ is the unit ideal of $S$ by [L8].

givenstep 2.1L7L8

3.2

For each $i$ the localisation $S_{g_{i}}$ is a finitely generated $R$ -algebra: by [L2] the images of $s_{1}, \dots, s_{m}$ together with the inverse of (the image of) $g_{i}$ generate it over $R$ , and $1 / g_{i}$ is an element of $S_{g_{i}}$ by [L14]. Hence for each $i$ there are finitely many elements $z_{i 1}, \dots, z_{i N_{i}} \in S_{g_{i}}$ generating $S_{g_{i}}$ as an $R$ -algebra; using the isomorphism $S_{g_{i}}^{'} ≅ S_{g_{i}}$ of step 2.1 and [L14] we may write $z_{i j} = y_{i j} / g_{i}^{m_{i j}} with y_{i j} \in S^{'}, m_{i j} \geq 0.$ Put $T : = R [g_{1}, \dots, g_{n}, y_{i j} (1 \leq i \leq n, 1 \leq j \leq N_{i})] \subseteq S^{'} .$ Then $T_{g_{i}} = S_{g_{i}}$ for every $i$ : the inclusion $T \subseteq S^{'}$ gives $T_{g_{i}} \subseteq S_{g_{i}}^{'} = S_{g_{i}}$ , while every generator $z_{i j} = y_{i j} / g_{i}^{m_{i j}}$ of $S_{g_{i}}$ over $R$ lies in $T_{g_{i}}$ , so $S_{g_{i}} \subseteq T_{g_{i}}$ .

givenstep 2.1L2L14

4.1

Every generator $g_{i}$ , $y_{i j}$ of $T$ lies in $S^{'}$ and hence is integral over $R$ by [L4]. Writing $A$ for the image of $R$ in $S$ , [L5] shows that $T = A [g_{i}, y_{i j}]$ is module-finite over $A$ , and the same finite list generates $T$ as an $R$ -module, so $T$ is module-finite over $R$ ; in particular $T$ is of finite type over $R$ by [L2], and it is a subalgebra of $S^{'}$ by step 3.2.

givenstep 3.2L2L4L5

4.2

Let $φ : Spec (S) \to Spec (T)$ be the contraction map along the inclusion $T \subseteq S$ , which is continuous by [L12]. For $h \in T \subseteq S$ one has $φ^{- 1} (D_{T} (h)) = D_{S} (h)$ , because a prime $r \in Spec (S)$ contracts to a prime containing $h$ exactly when $r$ itself contains $h$ . Consequently $φ^{- 1} (U) = ⋃_{i} φ^{- 1} (D_{T} (g_{i})) = ⋃_{i} D_{S} (g_{i}) = Spec (S)$ for $U : = D_{T} (g_{1}) \cup \dots \cup D_{T} (g_{n})$ , so $φ$ maps $Spec (S)$ into $U$ ; and $U$ is open in $Spec (T)$ by [L10] and [L11].

givenstep 3.1L10L11L12

4.3

For each $i$ the restriction of $φ$ to $D_{S} (g_{i})$ is a homeomorphism onto $D_{T} (g_{i})$ . Indeed [L9] identifies $D_{S} (g_{i})$ with $Spec (S_{g_{i}})$ and $D_{T} (g_{i})$ with $Spec (T_{g_{i}})$ through the contraction maps of the localisations, and the ring isomorphism $T_{g_{i}} ≅ S_{g_{i}}$ of step 3.2 induces a homeomorphism $Spec (S_{g_{i}}) \to Spec (T_{g_{i}})$ ; by [L13] the inverse of each of these identifications is the extension of a prime, and contracting a localised prime back to $T$ recovers the contraction of the original prime, so the composite is exactly the restriction of $φ$ .

givenstep 3.2L9L12L13

4.4

For every $i$ the inclusion induces an isomorphism $T_{g g_{i}} \to S_{g g_{i}}$ : localising the isomorphism $T_{g_{i}} \to S_{g_{i}}$ of step 3.2 at the element $g$ , and rewriting $(T_{g_{i}})_{g} ≅ T_{g_{i} g}$ and $(S_{g_{i}})_{g} ≅ S_{g_{i} g}$ by [L6], gives the claim.

givenstep 3.2L6

5.1

Now let $g \in T$ with $D_{T} (g) \subseteq U$ , where the subalgebra $T$ , the elements $g_{1}, \dots, g_{n}$ and the open set $U$ are the ones constructed in steps 3.2 and 4.2, and first record that the images of $g_{1}, \dots, g_{n}$ generate the unit ideal of $T_{g}$ . By [L9] applied to the principal localisation $T \to T_{g}$ the spectrum $Spec (T_{g})$ is identified with $D_{T} (g)$ , and under this identification the subset $D_{T_{g}} (g_{i} / 1)$ corresponds to $D_{T} (g) \cap D_{T} (g_{i})$ : a prime $p \in D_{T} (g)$ contains $g_{i}$ exactly when the corresponding prime of $T_{g}$ contains $g_{i} / 1$ by [L13] and [L14]. Hence the inclusion $D_{T} (g) \subseteq ⋃_{i} D_{T} (g_{i})$ gives $Spec (T_{g}) = ⋃_{i} D_{T_{g}} (g_{i} / 1)$ , and [L8] applied to the ring $T_{g}$ and the finite family $g_{i} / 1$ shows that the ideal generated by the images of $g_{1}, \dots, g_{n}$ in $T_{g}$ is the unit ideal, say $1 = \sum_{i} a_{i} (g_{i} / 1)$ with $a_{i} \in T_{g}$ .

givenstep 3.2step 4.2L8L9L13L14

5.2

The map $φ$ is injective and has image $U$ . For injectivity, let $q_{1}, q_{2} \in Spec (S)$ with $φ (q_{1}) = φ (q_{2}) = : p \in U$ ; choose $i$ with $p \in D_{T} (g_{i})$ . Since $g_{i} \in T$ and $g_{i} \notin p = q_{j} \cap T$ , we get $g_{i} \notin q_{j}$ for $j = 1, 2$ , so both primes lie in $D_{S} (g_{i})$ , where $φ$ is injective by step 4.3. For surjectivity onto $U$ , let $p \in U$ and choose $i$ with $p \in D_{T} (g_{i})$ ; the prime $p$ corresponds under [L9] to a prime of $T_{g_{i}}$ , which we transport across the isomorphism $T_{g_{i}} ≅ S_{g_{i}}$ of step 3.2 to a prime of $S_{g_{i}}$ , and its contraction to $S$ by [L9] is a prime $q \in D_{S} (g_{i})$ with $φ (q) = p$ by step 4.3.

givenstep 4.2step 4.3L9

6.1

The map $T_{g} \to S_{g}$ induced by the inclusion is an isomorphism. It is injective: $T \to S$ is injective with $S$ viewed as a $T$ -module, so [L15] applies to the multiplicative subset ${1, g, g^{2}, \dots}$ of $T$ . For surjectivity let $s \in S_{g}$ . By step 4.4 the localisation $(T_{g})_{g_{i}} \to (S_{g})_{g_{i}}$ is an isomorphism for every $i$ , so for each $i$ there are $m_{i} \geq 0$ and $t_{i} \in T_{g}$ with $g_{i}^{m_{i}} s = t_{i}$ . Put $M : = m_{1} + \dots + m_{n}$ and expand $1 = (\sum_{i} a_{i} (g_{i} / 1))^{M}$ from step 5.1: every monomial in the expansion has the form $(product of a ’s) \cdot \prod_{i} g_{i}^{ν_{i}}$ with $ν_{1} + \dots + ν_{n} = M$ , hence some $ν_{i} \geq m_{i}$ and the monomial is divisible by $g_{i}^{m_{i}}$ ; consequently $1 \in (g_{1}^{m_{1}}, \dots, g_{n}^{m_{n}})$ in $T_{g}$ , say $1 = \sum_{i} c_{i} g_{i}^{m_{i}}$ with $c_{i} \in T_{g}$ . Then $s = \sum_{i} c_{i} g_{i}^{m_{i}} s = \sum_{i} c_{i} t_{i}$ lies in $T_{g}$ . Hence $T_{g} \to S_{g}$ is bijective, and being a ring homomorphism induced by the inclusion it is an isomorphism.

givenstep 5.1step 4.4L6L15

6.2

The restricted map $φ : Spec (S) \to U$ is a homeomorphism. It is continuous by step 4.2 and bijective by step 5.2. To see that it is open, let $W \subseteq Spec (S)$ be open and write $W = ⋃_{i} (W \cap D_{S} (g_{i}))$ using that the $D_{S} (g_{i})$ cover $Spec (S)$ by step 3.1; each $W \cap D_{S} (g_{i})$ is open in $D_{S} (g_{i})$ and therefore has image $φ (W \cap D_{S} (g_{i}))$ open in $D_{T} (g_{i})$ by step 4.3, hence open in $U$ because $D_{T} (g_{i}) \subseteq U$ is open in $Spec (T)$ . Thus $φ (W) = ⋃_{i} φ (W \cap D_{S} (g_{i}))$ is a union of subsets open in $U$ and is open in $U$ . A continuous, open bijection onto $U$ is a homeomorphism.

givenstep 3.1step 4.2step 4.3step 5.2

7.1

Part 1 is now established by the objects constructed above: $T$ is a finite $R$ -subalgebra of $S^{'}$ by step 4.1, the set $U = D_{T} (g_{1}) \cup \dots \cup D_{T} (g_{n})$ is open by step 4.2, the contraction map $φ : Spec (S) \to Spec (T)$ is a homeomorphism onto $U$ by step 6.2, and $T_{g_{i}} ≅ S_{g_{i}}$ for every $i$ by step 3.2. Part 2 is step 6.1.

givenstep 3.2step 4.1step 4.2step 6.2step 6.1

8.1

The Axiom of Choice was used in step 2.1 through [L3], in step 3.1 through the compactness of $Spec (S)$ [L7] and the cover-to-unit-ideal statement [L8], and in step 5.1 through [L8] again. Every other selection was finite: the generators $s_{i}$ of step 1.1, the finitely many $g_{i}$ of step 3.1, the generators $z_{i j}$ and numerators $y_{i j}$ of step 3.2, the index $i$ in steps 5.2 and 5.1 and the exponents $m_{i}$ of step 6.1. ∎

givenstep 2.1step 3.1step 5.1L3L7L8L16

CorollaryStatement: Literature-sourcedProof: AI-adaptedaudited 2026-09-27Open item page →

Quasi-finite algebras are source locally localizations of finite algebras

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $R \to S$ be a ring map of finite type (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras) that is quasi-finite at every prime of $S$ (Quasi-finiteness at a prime of a finite-type algebra), and let $S^{'} \subseteq S$ be the integral closure of the image of $R$ in $S$ (Integral elements subalgebra of an arbitrary ring map). Then for every prime $q \in Spec (S)$ there are a finite $R$ -subalgebra $T \subseteq S^{'}$ , module-finite over $R$ , and an element $g \in T$ with $g \notin q$ such that the inclusion $T \to S$ induces an isomorphism $T_{g} \overset{≅}{\to} S_{g}$ of principal localisations (Principal localisation $R_{f} = {1, f, f^{2}, \dots}^{- 1} R$ ).

In other words, on the source every point of a quasi-finite finite-type algebra has an open neighbourhood, the principal open $D_{S} (g)$ , on which the algebra is a principal localisation of a finite algebra, and this already happens inside the relative integral closure. The element $g$ lies in the finite intermediate algebra $T$ and need not be the image of an element of $R$ : the corollary is a statement about the source, and it makes no base-principal or globally finite claim. The Axiom of Choice is inherited from the factorization theorem A quasi-finite algebra factors openly through a finite algebra used in the proof, whose finite cover it reuses.

Facts & Assumptions

[L1]

The map $R \to S$ is quasi-finite when it is of finite type and quasi-finite at every prime of $S$ , where quasi-finiteness at $q$ is finiteness of $S_{q} / p S_{q}$ over $κ (p)$ (Quasi-finiteness at a prime of a finite-type algebra).

[L2]

An $R$ -algebra $A$ is of finite type over $R$ when $A = R [a_{1}, \dots, a_{n}]$ for finitely many elements, and module-finite over $R$ when it is finitely generated as an $R$ -module (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras).

[L3]

For a unital ring map $R \to S$ the relative integral closure ${Int}_{R} (S)$ is an $R$ -subalgebra of $S$ containing the image of $R$ , namely the set of elements integral over the map (Integral elements subalgebra of an arbitrary ring map).

[L4]

Assume the Axiom of Choice. If $R \to S$ is of finite type and quasi-finite at every prime of $S$ and $S^{'}$ is the integral closure of the image of $R$ in $S$ , then there are a finite $R$ -subalgebra $T \subseteq S^{'}$ , module-finite over $R$ , and finitely many elements $g_{1}, \dots, g_{n} \in T$ such that $U = D_{T} (g_{1}) \cup \dots \cup D_{T} (g_{n})$ is open in $Spec (T)$ , the contraction map $Spec (S) \to Spec (T)$ is a homeomorphism onto $U$ , $T_{g_{i}} ≅ S_{g_{i}}$ for every $i$ , and for every $g \in T$ with $D_{T} (g) \subseteq U$ the inclusion induces an isomorphism $T_{g} \to S_{g}$ (A quasi-finite algebra factors openly through a finite algebra).

[L5]

[L6]

The Axiom of Choice (AC) is the statement that every family of nonempty sets has a choice function (The Axiom of Choice).

Proof

technique · direct

1.1

We assume the Axiom of Choice as recorded in [L6]. The hypothesis on $R \to S$ together with [L2] says that $S$ is a finitely generated $R$ -algebra and that the quasi-finiteness condition of [L1] holds at every prime of $S$ , so the factorization theorem [L4] applies; the relative integral closure $S^{'}$ is an $R$ -subalgebra of $S$ by [L3].

givenL1L2L3L4L6

2.1

Fix $q \in Spec (S)$ . By step 1.1 and [L4] there are a finite $R$ -subalgebra $T \subseteq S^{'}$ , module-finite over $R$ , and finitely many elements $g_{1}, \dots, g_{n} \in T$ such that $U = D_{T} (g_{1}) \cup \dots \cup D_{T} (g_{n})$ is an open subset of $Spec (T)$ onto which the contraction map $Spec (S) \to Spec (T)$ is a homeomorphism, and $T_{g_{i}} ≅ S_{g_{i}}$ for every $i$ .

givenstep 1.1L4

3.1

The image $p : = q \cap T$ of $q$ under the contraction map lies in the image of that homeomorphism, namely $U$ ; hence $p \in D_{T} (g_{i})$ for some index $i$ , that is $g_{i} \notin p = q \cap T$ . Since $g_{i} \in T \subseteq S$ , this means $g_{i} \notin q$ .

givenstep 2.1L4

4.1

For this index $i$ the inclusion $T \subseteq S$ induces an isomorphism $T_{g_{i}} \to S_{g_{i}}$ . Indeed $D_{T} (g_{i}) \subseteq U$ because $U$ is the union of the $D_{T} (g_{j})$ , and $T_{g_{i}} ≅ S_{g_{i}}$ is also one of the conclusions of step 2.1; either form of the factorization theorem gives the isomorphism, the first by its local form and the second by its cover statement.

givenstep 2.1step 3.1L4

5.1

Taking $g : = g_{i} \in T$ we have produced a finite $R$ -subalgebra $T \subseteq S^{'}$ that is module-finite over $R$ and an element $g \in T$ with $g \notin q$ such that $T_{g} ≅ S_{g}$ ; by [L5] these are principal localisations, so the source principal open $D_{S} (g)$ has algebra $S_{g} ≅ T_{g}$ over $R$ . No step uses that $g$ lies in the image of $R$ , and none asserts finiteness of $S$ or of $T_{g}$ over $R$ beyond the module-finiteness of $T$ , so no base-principal or globally finite statement is claimed.

givenstep 1.1step 3.1step 4.1L5

6.1

The Axiom of Choice was used only in step 2.1, through the factorization theorem [L4]; the only other selections are the single index $i$ of step 3.1 and the single element $g$ of step 5.1. This proves the corollary. ∎

givenstep 2.1step 5.1L4L6

5 · Examples, counterexamples and false statements

None yet.

Algebraic Zariski Main for Quasi-Finite Morphisms

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

Integral elements subalgebra of an arbitrary ring map

Definition

Quasi-finiteness at a prime of a finite-type algebra

Definition

The leading coefficient times a root is integral

Statement

Facts & Assumptions

Proof

One-variable integral correction after leading-coefficient localization

Statement

Facts & Assumptions

Proof

Strong transcendence over a subring

Definition

Strong transcendence descends to reduced minimal-prime quotients

Statement

Facts & Assumptions

Proof

Polynomial rings over normal domains are normal

Statement

Facts & Assumptions

Proof

Quasi-finite local fibres transfer through quotients and intermediate rings

Statement

Facts & Assumptions

Proof

Finite algebras over a strongly transcendental variable are nowhere quasi-finite

Statement

Facts & Assumptions

Proof

A quasi-finite one-generator quotient is locally its integral closure

Statement

Facts & Assumptions

Proof

Conductor radical detects every polynomial coefficient

Statement

Facts & Assumptions

Proof

Algebraic Zariski Main localization at a quasi-finite prime

Statement

Facts & Assumptions

Proof

The quasi-finite locus of a finite-type algebra is open

Statement

Facts & Assumptions

Proof

A quasi-finite algebra factors openly through a finite algebra

Statement

Facts & Assumptions

Proof

Quasi-finite algebras are source locally localizations of finite algebras

Statement

Facts & Assumptions

Proof

5 · Examples, counterexamples and false statements

Sources