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

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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.

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Associated primes commute with localization for finite modules

Statement

Let $R$ be a Noetherian commutative ring, let $M$ be a finitely generated left $R$ -module, and let $S \subseteq R$ be multiplicative. Then

{Ass}_{S^{- 1} R} (S^{- 1} M) = {S^{- 1} p : p \in {Ass}_{R} (M), p \cap S = \emptyset} .

Facts & Assumptions

Given: A Noetherian commutative ring $R$ , a finitely generated left $R$ -module $M$ , and a multiplicative subset $S \subseteq R$ .

[L1]

Associated primes of $M$ disjoint from $S$ localize to associated primes of $S^{- 1} M$ (Associated primes localize forward).

[L2]

Every associated prime of $S^{- 1} M$ is the localization of an associated prime of $M$ disjoint from $S$ (Associated primes of a localized finite module come from upstairs).

Proof

technique · direct

1.1

If $p \in {Ass}_{R} (M)$ and $p \cap S = \emptyset$ , then [L1] gives $S^{- 1} p \in {Ass}_{S^{- 1} R} (S^{- 1} M)$ .

1.2

Conversely, if $q \in {Ass}_{S^{- 1} R} (S^{- 1} M)$ , then [L2] produces $p \in {Ass}_{R} (M)$ with $p \cap S = \emptyset$ and $q = S^{- 1} p$ .

2.1

Steps 1.1 and 1.2 prove the stated equality of associated-prime sets.

step 1.1step 1.2∎

Depends on

Used by

Localisation of a primary submodule either stays primary or becomes the whole module Lemma

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., Proposition (17.10) (standard reference, not scraped)
The Stacks Project, Lemma 10.63.16 (standard reference, not scraped)