LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-sol)audited 2026-09-27

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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

Depends on

Used by

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

Dependency tree · two levels

5 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, Lemma 10.123.1 (standard reference, not scraped)
J. S. Milne, A Primer of Commutative Algebra, version 4.03, Section 17 (standard reference, not scraped)