CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17

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Integral elements over a nonzero base ring form a subring

Statement

Let $A \subseteq B$ be commutative rings with $A \neq 0$ . The elements of $B$ integral over $A$ form a subring of $B$ . See Integrality and finite-module characterizations for one element.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

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

[L2]

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 $A [X]$ . The extension is integral when every element is integral. An algebraic integer is a complex number integral over $Z$ . (Integral elements over a commutative ring and algebraic integers).

Proof

technique · direct

1.1L1L2givenalgebra

If $x, y$ are integral, $A [x]$ is finite over $A$ , and the monic equation for $y$ over $A$ is also one over $A [x]$ ; multiplying finite generating sets shows that $A [x, y]$ is finite over $A$ .

2.1step 1.1givenalgebra

For each $z \in {x + y, x - y, x y}$ , the $A [z]$ -module $A [x, y]$ is faithful and finite over $A$ , so the finite-module criterion makes $z$ integral.

3.1L2step 2.1givenalgebra∎

The elements $0$ and $1$ are roots of the monic polynomials $X$ and $X - 1$ , respectively; step 2.1 supplies closure under addition, multiplication, and additive inverses, including coincident elements. Thus the integral elements form a unital subring of $B$ .

Depends on

Used by

Dependency tree · two levels

7 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

Eloisa Grifo, Commutative Algebra I, Section 1.4 (standard reference, not scraped)