LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-26

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.

A localised module fraction is zero exactly when one denominator kills its numerator

Statement

For $m \in M$ and $s \in S$ , $\frac{m}{s} = 0 in S^{- 1} M ⟺ u m = 0 for some u \in S .$

Facts & Assumptions

Given: A commutative ring $R$ , a multiplicative subset $S \subseteq R$ , a left $R$ -module $M$ , an element $m \in M$ , and an element $s \in S$ .

[L1]

In $S^{- 1} M$ , the equality $m / s = n / t$ means that $u (t m - s n) = 0$ for some $u \in S$ (Localisation of a module at a multiplicative subset).

Proof

technique · direct

1.1L1

If $m / s = 0 / 1$ , then [L1] gives $u (1 m - s 0) = u m = 0$ for some $u \in S$ .

1.2L1

If $u m = 0$ for some $u \in S$ , then $u (1 m - s 0) = 0$ , so [L1] gives $m / s = 0 / 1 = 0$ in $S^{- 1} M$ .

2.1step 1.1step 1.2∎

Steps 1.1 and 1.2 prove the equivalence.

Depends on

Localisation of a module at a multiplicative subset

Dependency tree · 2 levels

Sources

A. Altman and S. Kleiman, A Term of Commutative Algebra, 13th ed., (12.2)