TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26

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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 commutes with quotient modules and arbitrary direct sums

Statement

Let $R$ be a commutative ring and let $S \subseteq R$ be multiplicative.

For every submodule $N \leq M$ , there is a natural isomorphism $S^{- 1} (M / N) ≅ (S^{- 1} M) / (S^{- 1} N) .$
For every family $(M_{i})_{i \in I}$ of left $R$ -modules, there is a natural isomorphism $S^{- 1} ⁣ (⨁_{i \in I} M_{i}) ≅ ⨁_{i \in I} S^{- 1} M_{i} .$

Facts & Assumptions

Given: A commutative ring $R$ and a multiplicative subset $S \subseteq R$ .

[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]

Tensor products commute with arbitrary direct sums (Tensor products commute with arbitrary direct sums).

[L4]

A quotient module is the module of cosets $M / N$ (Quotient module $M / N$ with scalar multiplication on additive cosets).

Proof

technique · direct

1.1L1L2L4

For a submodule $N \leq M$ , the sequence $N \to M \to M / N \to 0$ is right exact, so [L2] and [L1] give a right-exact sequence $S^{- 1} N \to S^{- 1} M \to S^{- 1} (M / N) \to 0$ . Therefore $S^{- 1} (M / N)$ is the quotient of $S^{- 1} M$ by the image of $S^{- 1} N \to S^{- 1} M$ , namely by the submodule $S^{- 1} N$ .

1.2L1L3

For a family $(M_{i})_{i \in I}$ , [L3] and [L1] give $S^{- 1} (⨁_{i} M_{i}) ≅ (S^{- 1} R) \otimes_{R} (⨁_{i} M_{i}) ≅ ⨁_{i} ((S^{- 1} R) \otimes_{R} M_{i}) ≅ ⨁_{i} S^{- 1} M_{i}$ .

2.1step 1.1

Thus $S^{- 1} (M / N) ≅ (S^{- 1} M) / (S^{- 1} N)$ naturally in $M$ and $N$ .

3.1step 2.1step 1.2∎

Steps 2.1 and 1.2 prove the quotient and direct-sum claims.

Depends on

Used by

Dependency tree · two levels

19 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., Corollary 12.22 (standard reference, not scraped)
The Stacks Project, Section 10.9: Localization (standard reference, not scraped)