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

How statement and proof provenance work

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 of a submodule lie in those of the ambient module

Statement

If $0 ⟶ M^{'} ⟶ M ⟶ M^{''} ⟶ 0$ is a short exact sequence of left $R$ -modules, then ${Ass}_{R} (M^{'}) \subseteq {Ass}_{R} (M) .$

Facts & Assumptions

Given: A commutative ring $R$ and a short exact sequence $0 \to M^{'} \to M \to M^{''} \to 0$ of left $R$ -modules.

[L1]

A prime $p$ is associated to a module exactly when $R / p$ embeds in that module (Associated primes are exactly primes of embedded cyclic residue modules).

[L2]

In a short exact sequence, the left map is injective (Exact sequences and short exact sequences of modules).

Proof

technique · direct

1.1L1L2

Let $p \in {Ass}_{R} (M^{'})$ . By [L1], there is an embedding $R / p ↪ M^{'}$ . Composing with the injective map $M^{'} ↪ M$ from [L2] gives an embedding $R / p ↪ M$ .

2.1L1step 1.1∎

Applying [L1] again, the embedding from step 1.1 shows that $p \in {Ass}_{R} (M)$ . Therefore ${Ass}_{R} (M^{'}) \subseteq {Ass}_{R} (M)$ .

Associated primes of a submodule lie in those of the ambient module

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources