LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Associated primes of the middle term lie in those of the ends

Statement

0 ⟶ M^{'} ⟶ M ⟶ M^{''} ⟶ 0

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

{Ass}_{R} (M) \subseteq {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, the image of the left map equals the kernel of the right map (Exact sequences and short exact sequences of modules).

Proof

technique · direct

1.1

Let $p \in {Ass}_{R} (M)$ , and choose $m \in M$ with ${Ann}_{R} (m) = p$ . If there exists $g \notin p$ with $g m \in M^{'}$ , then $g m \neq 0$ . Also $p \subseteq {Ann}_{R} (g m)$ , and if $a g m = 0$ then $a g \in p$ ; since $p$ is prime and $g \notin p$ , this gives $a \in p$ . Hence ${Ann}_{R} (g m) = p$ , so $p \in {Ass}_{R} (M^{'})$ .

givenalgebra

2.1

If no such $g$ exists, let $\overset{ˉ}{m}$ be the image of $m$ in $M^{''}$ . Then $\overset{ˉ}{m} \neq 0$ , for otherwise $m \in M^{'}$ and $g = 1$ would contradict the assumption. Every $a \in p$ kills $m$ , hence kills $\overset{ˉ}{m}$ . Conversely, if $a \overset{ˉ}{m} = 0$ , then $a m \in M^{'}$ . By the standing assumption, this forces $a \in p$ . Therefore ${Ann}_{R} (\overset{ˉ}{m}) = p$ , so $p \in {Ass}_{R} (M^{''})$ .

L1step 1.1algebra

3.1

Steps 1.1 and 2.1 show that every associated prime of $M$ lies in ${Ass}_{R} (M^{'}) \cup {Ass}_{R} (M^{''})$ .

step 1.1step 2.1∎

Depends on

Exact sequences and short exact sequences of modules

Used by

Associated primes in a short exact sequence Theorem

Dependency tree · two levels

3 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)