TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 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.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Universal property of localisation for modules

Statement

Let $R$ be a commutative ring, let $S \subseteq R$ be multiplicative, let $M$ be a left $R$ -module, and let $N$ be an $S^{- 1} R$ -module. Every $R$ -linear map $f : M \to N$ factors uniquely through the localisation map $λ_{M} : M \to S^{- 1} M$ by an $S^{- 1} R$ -linear map $\tilde{f} : S^{- 1} M ⟶ N, \tilde{f} (m / s) = (1 / s) f (m) .$

Facts & Assumptions

Given: A commutative ring $R$ , a multiplicative subset $S \subseteq R$ , a left $R$ -module $M$ , an $S^{- 1} R$ -module $N$ , and an $R$ -linear map $f : M \to N$ .

[L1]

In $S^{- 1} M$ , the localisation map is $λ_{M} (m) = m / 1$ , addition is $(m / s) + (n / t) = (t m + s n) / (s t)$ , and the scalar action is $(a / u) (m / s) = a m / (u s)$ (Localisation of a module at a multiplicative subset).

[L2]

The addition formula of $S^{- 1} M$ is independent of representatives (Addition of localised module fractions is independent of representatives).

[L3]

The scalar action of $S^{- 1} R$ on $S^{- 1} M$ is independent of representatives (The localised scalar action is independent of representatives).

[L4]

In $S^{- 1} R$ , every $s / 1$ with $s \in S$ is a unit with inverse $1 / s$ (The localisation relation is an equivalence relation and fraction arithmetic is well defined).

Proof

technique · direct

1.1L1L4givenalgebra

Define $\tilde{f} (m / s) : = (1 / s) f (m)$ . If $m / s = m^{'} / s^{'}$ , choose $u \in S$ with $u (s^{'} m - s m^{'}) = 0$ ; applying $f$ gives $(u / 1) ((s^{'} / 1) f (m) - (s / 1) f (m^{'})) = 0$ , and multiplying by the units $(u / 1)^{- 1} (s / 1)^{- 1} (s^{'} / 1)^{- 1}$ from [L4] gives $(1 / s) f (m) = (1 / s^{'}) f (m^{'})$ .

2.1step 1.1L1L2L3algebra

For $m / s, n / t \in S^{- 1} M$ , $\tilde{f} ((m / s) + (n / t)) = (1 / s t) f (t m + s n) = (1 / s) f (m) + (1 / t) f (n) = \tilde{f} (m / s) + \tilde{f} (n / t)$ , and for $a / u \in S^{- 1} R$ one has $\tilde{f} ((a / u) (m / s)) = (1 / u s) f (a m) = (a / u) \tilde{f} (m / s)$ .

2.2step 1.1L1

For every $m \in M$ , $\tilde{f} (λ_{M} (m)) = \tilde{f} (m / 1) = f (m)$ .

3.1L1L4step 2.2

If $g : S^{- 1} M \to N$ is $S^{- 1} R$ -linear and $g λ_{M} = f$ , then for every $m / s$ one has $g (m / s) = g ((1 / s) (m / 1)) = (1 / s) g (m / 1) = (1 / s) f (m) = \tilde{f} (m / s)$ , so $g = \tilde{f}$ .

4.1step 1.1step 2.1step 2.2step 3.1∎

Steps 1.1, 2.1, 2.2, and 3.1 prove the stated unique $S^{- 1} R$ -linear factorisation.

Depends on

Used by

Module sheaf on an affine scheme Definition
Kähler differentials commute with localization Lemma
There is a natural localisation map on Hom Lemma
Localisation of modules is extension of scalars Theorem

Dependency tree · two levels

7 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., Section 12 (standard reference, not scraped)
J. S. Milne, A Primer of Commutative Algebra, v4.03, Proposition 5.10 (standard reference, not scraped)