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LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28
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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Associated primes are exactly primes of embedded cyclic residue modules

Statement

Let R be a commutative ring, let M be a left R-module, and let p be a prime ideal of R. Then p∈Ass⁡R(M) if and only if there exists an injective R-module homomorphism R/p↪M.

Facts & Assumptions

Given: A commutative ring R, a left R-module M, and a prime ideal p⊆R.

[L1]

A prime ideal is associated to M exactly when it is the annihilator of some element of M (Associated primes of a module).

[L2]

For any m∈M, the cyclic submodule Rm is naturally isomorphic to R/Ann⁡R(m) (A cyclic submodule is a residue module by its annihilator).

Proof

technique · direct
1.1L1L2choose

Assume p∈Ass⁡R(M). By [L1], choose m∈M with Ann⁡R(m)=p. Then [L2] gives Rm≅R/p, and the inclusion Rm↪M composes with this isomorphism to give an embedding R/p↪M.

1.2L1givenalgebra

Conversely, let j:R/p↪M be injective, and put m=j(1+p). Every r∈p kills 1+p, so rm=0. If rm=0, then j(r+p)=0, and injectivity gives r+p=0, hence r∈p. Therefore Ann⁡R(m)=p, so p∈Ass⁡R(M) by [L1].

2.1step 1.1step 1.2∎

Steps 1.1 and 1.2 prove the equivalence.

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