TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-28

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Associated primes in a short exact sequence

Statement

0 ⟶ M^{'} ⟶ M ⟶ M^{''} ⟶ 0

is a short exact sequence of left $R$ -modules, then

{Ass}_{R} (M^{'}) \subseteq {Ass}_{R} (M) \subseteq {Ass}_{R} (M^{'}) \cup {Ass}_{R} (M^{''}) .

In particular,

{Ass}_{R} (M^{'} \oplus M^{''}) = {Ass}_{R} (M^{'}) \cup {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]

In a short exact sequence, associated primes of the left term lie in those of the middle term (Associated primes of a submodule lie in those of the ambient module).

[L2]

In a short exact sequence, associated primes of the middle term lie in those of the outer terms (Associated primes of the middle term lie in those of the ends).

Proof

technique · direct

1.1

Facts [L1] and [L2] give ${Ass}_{R} (M^{'}) \subseteq {Ass}_{R} (M) \subseteq {Ass}_{R} (M^{'}) \cup {Ass}_{R} (M^{''}) .$

L1L2

2.1

Apply step 1.1 to the split exact sequence $0 ⟶ M^{'} ⟶ M^{'} \oplus M^{''} ⟶ M^{''} ⟶ 0.$ This gives ${Ass}_{R} (M^{'} \oplus M^{''}) \subseteq {Ass}_{R} (M^{'}) \cup {Ass}_{R} (M^{''})$ . The reverse inclusion follows by applying the left inclusion of step 1.1 to the two canonical injections $M^{'} ↪ M^{'} \oplus M^{''}$ and $M^{''} ↪ M^{'} \oplus M^{''}$ .

step 1.1

3.1

Steps 1.1 and 2.1 prove the theorem and its direct-sum corollary.

step 1.1step 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

A. Altman and S. Kleiman, A Term of Commutative Algebra, 13th ed., Proposition (17.5) (standard reference, not scraped)
The Stacks Project, Lemma 10.63.3 (standard reference, not scraped)