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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∈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∈M such that Ann⁡R(m) is maximal among the annihilators of nonzero elements of M.

[L1]

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

Proof

technique · direct
1.1L1givenalgebra

Put I=Ann⁡R(m). Since m≠0, one has 1∉I, so I is proper. Let ab∈I and assume b∉I. Then bm≠0. Also I⊆Ann⁡R(bm) because every r∈I satisfies r(bm)=b(rm)=0. If r∈Ann⁡R(bm), then (rb)m=0, so rb∈I by [L1].

2.1step 1.1

The element bm is nonzero, so maximality of I forces Ann⁡R(bm)=I. Since abm=0, the element a lies in Ann⁡R(bm)=I. Therefore ab∈I and b∉I imply a∈I, so I is prime.

3.1step 2.1∎

Thus Ann⁡R(m) is a prime ideal.

Depends on

Used by

Dependency tree · two levels

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Sources