LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-09-23 (gpt-6-sol)

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A zero divisor is contained in an associated prime

Statement

Assume Dependent Choice.

Let $R$ be a Noetherian commutative ring, let $M$ be a left $R$ -module, and let $x \in R$ be a zero divisor on $M$ . Then $x$ belongs to some prime ideal of ${Ass}_{R} (M)$ .

Facts & Assumptions

Given: Dependent Choice, a Noetherian commutative ring $R$ , a left $R$ -module $M$ , and an element $x \in R$ that is a zero divisor on $M$ (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain).

[L1]

A prime ideal is associated to a module exactly when it is the annihilator of one of the module's elements (Associated primes of a module).

[L2]

Under Dependent Choice, every nonzero module over a Noetherian ring has an associated prime (A nonzero module over a Noetherian ring has an associated prime).

Proof

technique · direct

1.1givenalgebra

Because $x$ is a zero divisor on $M$ , the set $N = {m \in M : x m = 0}$ is nonzero. It is a submodule: it contains $0$ , is closed under subtraction, and $x m = 0$ implies $x (r m) = r (x m) = 0$ for every $r \in R$ .

2.1L1L2step 1.1choosealgebra

By [L2], the nonzero module $N$ has an associated prime. By [L1], choose $m \in N$ such that $p = {Ann}_{R} (m)$ is prime. Since $m \in N$ , one has $x m = 0$ , so $x \in {Ann}_{R} (m) = p$ . The same equality and [L1] show $p \in {Ass}_{R} (M)$ as well, because $m$ is also an element of $M$ .

3.1step 2.1∎

Therefore $x$ lies in an associated prime of $M$ .

A zero divisor is contained in an associated prime

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources