TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-09-23 (gpt-6-sol)

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Finite modules over Noetherian rings have finitely many associated primes

Statement

Assume the Axiom of Choice.

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: The Axiom of Choice, a Noetherian commutative ring $R$ and a finitely generated left $R$ -module $M$ (The Axiom of Choice).

[L1]

Under the stated Axiom of Choice, 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

1.1L1given

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

1.2L1L2algebra

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 ${Ass}_{R} (M) \subseteq {p_{1}, \dots, p_{n}} .$

2.1step 1.1step 1.2∎

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

Finite modules over Noetherian rings have finitely many associated primes

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources