LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-28

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A maximal element annihilator is prime

Statement

Let $R$ be a commutative ring, let $M$ be a left $R$ -module, and let $m \in M$ be nonzero. If ${Ann}_{R} (m)$ is maximal among the annihilators of nonzero elements of $M$ , then ${Ann}_{R} (m)$ is a prime ideal.

Facts & Assumptions

Given: A commutative ring $R$ , a left $R$ -module $M$ , and a nonzero element $m \in M$ such that ${Ann}_{R} (m)$ is maximal among the annihilators of nonzero elements of $M$ .

[L1]

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

Proof

technique · direct

1.1L1givenalgebra

Put $I = {Ann}_{R} (m)$ . Since $m \neq 0$ , one has $1 \notin I$ , so $I$ is proper. Let $a b \in I$ and assume $b \notin I$ . Then $b m \neq 0$ . Also $I \subseteq {Ann}_{R} (b m)$ because every $r \in I$ satisfies $r (b m) = b (r m) = 0$ . If $r \in {Ann}_{R} (b m)$ , then $(r b) m = 0$ , so $r b \in I$ by [L1].

2.1step 1.1

The element $b m$ is nonzero, so maximality of $I$ forces ${Ann}_{R} (b m) = I$ . Since $a b m = 0$ , the element $a$ lies in ${Ann}_{R} (b m) = I$ . Therefore $a b \in I$ and $b \notin I$ imply $a \in I$ , so $I$ is prime.

3.1step 2.1∎

Thus ${Ann}_{R} (m)$ is a prime ideal.

Depends on

Annihilators, torsion elements and the torsion subset of a module

A maximal element annihilator is prime

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources