LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28

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A nonzero module over a Noetherian ring has a maximal element annihilator

Statement

Assume Dependent Choice.

Let $R$ be a Noetherian commutative ring and let $M$ be a nonzero left $R$ -module. Then there exists a nonzero element $m \in M$ such that ${Ann}_{R} (m)$ is maximal, under inclusion, among the annihilators of nonzero elements of $M$ .

Facts & Assumptions

Given: Dependent Choice, a Noetherian commutative ring $R$ , and a nonzero left $R$ -module $M$ .

[L1]

Assuming Dependent Choice, in a Noetherian commutative ring every nonempty set of ideals has a maximal member (A commutative ring is Noetherian exactly when every ideal is finitely generated, exactly when its ideals satisfy the ascending chain condition, and exactly when every nonempty set of ideals has a maximal member).

[L2]

For $m \in M$ , the annihilator is the set ${Ann}_{R} (m) = {r \in R : r m = 0}$ (Annihilators, torsion elements and the torsion subset of a module).

Proof

technique · direct

1.1L2givenchoosealgebra

For every $m \in M$ , the set ${Ann}_{R} (m)$ is an ideal: it contains $0$ , is closed under subtraction, and $a \in {Ann}_{R} (m)$ implies $r a \in {Ann}_{R} (m)$ for every $r \in R$ . Because $M \neq 0$ , choose $m_{0} \in M$ with $m_{0} \neq 0$ . Then $Σ = {{Ann}_{R} (m) : m \in M, m \neq 0}$ is a nonempty set of ideals of $R$ .

2.1L1step 1.1

By [L1], the nonempty set $Σ$ has a maximal member. Thus some nonzero $m \in M$ satisfies that ${Ann}_{R} (m)$ is maximal among the annihilators of nonzero elements of $M$ .

A nonzero module over a Noetherian ring has a maximal element annihilator

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources