LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28

How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Associated primes localize forward

Statement

Let $R$ be a commutative ring, let $M$ be a left $R$ -module, let $S \subseteq R$ be multiplicative, and let $p \in {Ass}_{R} (M)$ with $p \cap S = \emptyset$ . Then $S^{- 1} p \in {Ass}_{S^{- 1} R} (S^{- 1} M) .$

Facts & Assumptions

Given: A commutative ring $R$ , a left $R$ -module $M$ , a multiplicative subset $S \subseteq R$ , and a prime ideal $p \in {Ass}_{R} (M)$ with $p \cap S = \emptyset$ .

[L1]

A prime ideal is associated to a module exactly when its residue module embeds in that module (Associated primes are exactly primes of embedded cyclic residue modules).

[L2]

Injective module maps remain injective after localisation (Injective module maps remain injective after localisation).

[L3]

Localisation commutes with quotient modules, so $S^{- 1} (R / p) ≅ (S^{- 1} R) / (S^{- 1} p)$ (Localisation commutes with quotient modules and arbitrary direct sums).

[L4]

If $p \cap S = \emptyset$ , then $S^{- 1} p$ is a prime ideal of $S^{- 1} R$ (Prime ideals of a localization are exactly the primes disjoint from the denominator set).

Proof

technique · direct

1.1L1L2

By [L1], there is an injective $R$ -module map $R / p ↪ M$ . Applying [L2] yields an injective $S^{- 1} R$ -module map $S^{- 1} (R / p) ↪ S^{- 1} M$ .

2.1L1L3L4step 1.1

By [L3], the source identifies with $(S^{- 1} R) / (S^{- 1} p)$ , and by [L4] the ideal $S^{- 1} p$ is prime. Therefore [L1], applied over the ring $S^{- 1} R$ , shows that $S^{- 1} p$ is associated to $S^{- 1} M$ .

Associated primes localize forward

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources