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

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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.
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Localisation of modules is exact

Statement

If $0 ⟶ M^{'} \overset{f}{\to} M \overset{g}{\to} M^{''} ⟶ 0$ is a short exact sequence of $R$ -modules, then $0 ⟶ S^{- 1} M^{'} \overset{S^{- 1} f}{\to} S^{- 1} M \overset{S^{- 1} g}{\to} S^{- 1} M^{''} ⟶ 0$ is a short exact sequence of $S^{- 1} R$ -modules.

Facts & Assumptions

Given: A commutative ring $R$ , a multiplicative subset $S \subseteq R$ , and a short exact sequence $0 \to M^{'} \overset{f}{\to} M \overset{g}{\to} M^{''} \to 0$ .

[L1]

Localisation is naturally $(S^{- 1} R) \otimes_{R} -$ (Localisation of modules is extension of scalars).

[L2]

Tensoring a right-exact sequence with a fixed module preserves right exactness (Tensoring is right exact).

[L3]

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

[L4]

A short exact sequence is exact, with left map injective and right map surjective (Exact sequences and short exact sequences of modules).

Proof

technique · direct

1.1L1L2L4

By [L4], the tail $M^{'} \overset{f}{\to} M \overset{g}{\to} M^{''} \to 0$ is exact. Using [L1] to identify localisation with tensor product, [L2] gives an exact sequence $S^{- 1} M^{'} \overset{S^{- 1} f}{\to} S^{- 1} M \overset{S^{- 1} g}{\to} S^{- 1} M^{''} \to 0$ .

1.2L3L4

The map $f$ is injective by [L4], so [L3] makes $S^{- 1} f$ injective.

2.1step 1.1step 1.2∎

Steps 1.1 and 1.2 show that the localised sequence is short exact.

Localisation of modules is exact

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources