LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedaudited 2026-09-27

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

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

Depends on

Used by

Algebraic Zariski Main localization at a quasi-finite prime Theorem

Dependency tree · two levels

29 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

The Stacks Project, Commutative Algebra, Section 10.123, Situation 10.123.4 and Lemmas 10.123.5 and 10.123.6 with their proofs (standard reference, not scraped)
J. S. Milne, A Primer of Commutative Algebra, version 4.03, Section 17 (standard reference, not scraped)