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LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck 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 mM be nonzero. If AnnR(m) is maximal among the annihilators of nonzero elements of M, then AnnR(m) is a prime ideal.

Facts & Assumptions

Given: A commutative ring R, a left R-module M, and a nonzero element mM such that AnnR(m) is maximal among the annihilators of nonzero elements of M.

[L1]

The annihilator of an element xM is AnnR(x)={rR:rx=0} (Annihilators, torsion elements and the torsion subset of a module).

Proof

technique · direct
1.1

Put I=AnnR(m). Since m0, one has 1I, so I is proper. Let abI and assume bI. Then bm0. Also IAnnR(bm) because every rI satisfies r(bm)=b(rm)=0. If rAnnR(bm), then (rb)m=0, so rbI by [L1].

L1givenalgebra
2.1

The element bm is nonzero, so maximality of I forces AnnR(bm)=I. Since abm=0, the element a lies in AnnR(bm)=I. Therefore abI and bI imply aI, so I is prime.

step 1.1
3.1

Thus AnnR(m) is a prime ideal.

step 2.1

Depends on

Used by

Dependency tree · two levels

4 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