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 lie in the support

Statement

Let $R$ be a commutative ring and let $M$ be a left $R$ -module. Then ${Ass}_{R} (M) \subseteq {Supp}_{R} (M) .$

Facts & Assumptions

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

[L1]

A prime ideal $p$ is associated to $M$ exactly when $p = {Ann}_{R} (m)$ for some $m \in M$ (Associated primes of a module).

[L2]

A prime ideal lies in ${Supp}_{R} (M)$ exactly when the localization $M_{p}$ is nonzero (Support of a module).

[L3]

The module localization $M_{p}$ consists of fractions $m / s$ with $s \in R ∖ p$ , and $m / 1 = 0$ exactly when some $s \notin p$ kills $m$ (Localisation of a module at a multiplicative subset, A localised module fraction is zero exactly when one denominator kills its numerator, Localisation at a prime ideal: $R_{p} = (R ∖ p)^{- 1} R$ ).

Proof

technique · direct

1.1L1L3choosealgebra

Let $p \in {Ass}_{R} (M)$ . By [L1], choose $m \in M$ with ${Ann}_{R} (m) = p$ . If $m / 1 = 0$ in $M_{p}$ , then by the definition of localization there exists $s \notin p$ with $s m = 0$ . Hence $s \in {Ann}_{R} (m) = p$ , a contradiction. So $m / 1 \neq 0$ in $M_{p}$ .

2.1L2step 1.1∎

By [L2], step 1.1 shows $p \in {Supp}_{R} (M)$ . Thus ${Ass}_{R} (M) \subseteq {Supp}_{R} (M)$ .

Associated primes lie in the support

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources