LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 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 finite presentation reduces localised Hom to the finite free case

Statement

Let $R^{m} \overset{α}{\to} R^{n} \overset{β}{\to} M ⟶ 0$ be a finite presentation of an $R$ -module $M$ , and let $N$ be an $R$ -module. If the natural localisation maps for $R^{n}$ and $R^{m}$ are isomorphisms, then the natural localisation map $S^{- 1} ⁣ {Hom}_{R} (M, N) ⟶ {Hom}_{S^{- 1} R} (S^{- 1} M, S^{- 1} N)$ is an isomorphism as well.

Facts & Assumptions

Given: A commutative ring $R$ , a multiplicative subset $S \subseteq R$ , an $R$ -module $N$ , and a finite presentation $R^{m} \overset{α}{\to} R^{n} \overset{β}{\to} M \to 0$ .

[L1]

A finite presentation is an exact sequence $R^{m} \overset{α}{\to} R^{n} \overset{β}{\to} M \to 0$ , and equivalently $M ≅ R^{n} / im α$ (Finitely presented modules and finitely presented algebras).

[L2]

Localisation sends short exact sequences to short exact sequences (Localisation of modules is exact).

[L3]

The localisation map on Hom is natural in the source, with formula $θ_{X, N} (f / s) (x / u) = f (x) / (s u)$ (There is a natural localisation map on Hom).

Proof

technique · direct

1.1L1algebra

Because $β$ is surjective in [L1], a homomorphism $h : R^{n} \to N$ factors through $M$ exactly when $h$ kills $im α$ , equivalently when $h \circ α = 0$ . Thus ${Hom}_{R} (M, N)$ is the kernel of $α^{*} : {Hom}_{R} (R^{n}, N) \to {Hom}_{R} (R^{m}, N)$ , $h \mapsto h \circ α$ .

1.2L1L2algebra

By [L2], localising the presentation of [L1] gives an exact sequence $S^{- 1} R^{m} \overset{S^{- 1} α}{\to} S^{- 1} R^{n} \overset{S^{- 1} β}{\to} S^{- 1} M \to 0$ , so the same argument identifies ${Hom}_{S^{- 1} R} (S^{- 1} M, S^{- 1} N)$ with the kernel of $(S^{- 1} α)^{*}$ .

1.3L3algebra

Naturality from [L3] makes the square between $α^{*}$ and $(S^{- 1} α)^{*}$ commute.

2.1step 1.1step 1.2step 1.3algebra

If the two vertical maps in step 1.3 are isomorphisms, then they identify the kernel in step 1.1 with the kernel in step 1.2. Therefore the induced map on those kernels, namely $S^{- 1} ⁣ {Hom}_{R} (M, N) \to {Hom}_{S^{- 1} R} (S^{- 1} M, S^{- 1} N)$ , is an isomorphism.

3.1step 2.1∎

Step 2.1 is exactly the reduction from a finite presentation to the finite free case.

Depends on

Dependency tree · 2 levels

Sources

A. Altman and S. Kleiman, A Term of Commutative Algebra, 13th ed., Proposition 12.25