LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck 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.

There is a natural localisation map on Hom

Statement

Let $R$ be a commutative ring, let $S \subseteq R$ be multiplicative, and let $M, N$ be left $R$ -modules. There is a natural $S^{- 1} R$ -linear map $θ_{M, N} : S^{- 1} ⁣ {Hom}_{R} (M, N) ⟶ {Hom}_{S^{- 1} R} (S^{- 1} M, S^{- 1} N)$ given by $θ_{M, N} (f / s) (m / u) = f (m) / (s u) .$

Facts & Assumptions

Given: A commutative ring $R$ , a multiplicative subset $S \subseteq R$ , and left $R$ -modules $M, N$ .

[L1]

For a commutative ring $A$ and $A$ -modules $X, Y$ , the group ${Hom}_{A} (X, Y)$ is an $A$ -module under pointwise scalar multiplication (Over a commutative ring the homomorphism group ${Hom}_{R} (M, N)$ is an $R$ -module).

[L2]

The localisation map of a module is universal for maps into modules over the localised ring (Universal property of localisation for modules).

Proof

technique · direct

1.1L2construct

For each $R$ -linear map $f : M \to N$ , apply [L2] to the composite $M \overset{f}{\to} N \to S^{- 1} N$ to obtain a unique $S^{- 1} R$ -linear map $S^{- 1} f : S^{- 1} M \to S^{- 1} N$ with $(S^{- 1} f) (m / u) = f (m) / u$ .

2.1step 1.1L1algebra

The assignment $f \mapsto S^{- 1} f$ is $R$ -linear because for $f, g \in {Hom}_{R} (M, N)$ , $r \in R$ , and $m / u \in S^{- 1} M$ one has $(S^{- 1} (f + g)) (m / u) = (f (m) + g (m)) / u$ and $(S^{- 1} (r f)) (m / u) = r f (m) / u = (r / 1) (f (m) / u)$ .

3.1step 2.1L1L2construct

By [L1], ${Hom}_{R} (M, N)$ is an $R$ -module, while ${Hom}_{S^{- 1} R} (S^{- 1} M, S^{- 1} N)$ is an $S^{- 1} R$ -module; therefore [L2] applied to the $R$ -linear map of step 2.1 gives a unique $S^{- 1} R$ -linear map $θ_{M, N}$ with $θ_{M, N} (f / 1) = S^{- 1} f$ .

4.1step 3.1algebra

Since $θ_{M, N}$ is $S^{- 1} R$ -linear, $θ_{M, N} (f / s) = (1 / s) θ_{M, N} (f / 1)$ , so for $m / u \in S^{- 1} M$ one gets $θ_{M, N} (f / s) (m / u) = (1 / s) (f (m) / u) = f (m) / (s u)$ .

5.1step 3.1step 4.1∎

Steps 3.1 and 4.1 produce the stated natural $S^{- 1} R$ -linear map.

Depends on

Used by

Dependency tree · two levels

10 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

A. Altman and S. Kleiman, A Term of Commutative Algebra, 13th ed., Proposition 12.25 (standard reference, not scraped)