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

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Finite modules over Noetherian rings have finitely many associated primes

Statement

Let $R$ be a Noetherian commutative ring and let $M$ be a finitely generated left $R$ -module. Then ${Ass}_{R} (M)$ is a finite set.

Facts & Assumptions

Given: A Noetherian commutative ring $R$ and a finitely generated left $R$ -module $M$ .

[L1]

The module $M$ admits a prime filtration $0 = M_{0} \subset M_{1} \subset \dots \subset M_{n} = M$ with $M_{i} / M_{i - 1} ≅ R / p_{i}$ (Finite modules over Noetherian rings admit prime filtrations).

[L2]

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

Proof

technique · direct

1.1

Choose a prime filtration as in [L1]. If $n = 0$ , then $M = 0$ and ${Ass}_{R} (M) = \emptyset$ , which is finite.

L1given

1.2

Assume $n \geq 1$ . Applying [L2] to $0 ⟶ M_{n - 1} ⟶ M ⟶ M / M_{n - 1} ⟶ 0$ gives ${Ass}_{R} (M) \subseteq {Ass}_{R} (M_{n - 1}) \cup {Ass}_{R} (M / M_{n - 1}) .$ Since $M / M_{n - 1} ≅ R / p_{n}$ , the class of $1 + p_{n}$ has annihilator $p_{n}$ , so ${Ass}_{R} (M / M_{n - 1}) = {p_{n}}$ . Repeating this argument down the filtration yields ${Ass}_{R} (M) \subseteq {p_{1}, \dots, p_{n}} .$

L1L2algebra

2.1

The right-hand side of step 1.2 is finite, so ${Ass}_{R} (M)$ is finite.

step 1.1step 1.2∎

Depends on

Used by

Zero divisors on a module over a Noetherian ring are the union of its associated primes Theorem

Dependency tree · two levels

9 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., Theorem (17.21) (standard reference, not scraped)
The Stacks Project, Lemma 10.63.5 (standard reference, not scraped)