CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26

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Localisation commutes with finite intersections of submodules

Statement

Let $N_{1}, \dots, N_{r}$ be submodules of a left $R$ -module $M$ . Then

S^{- 1} ⁣ (⋂_{i = 1}^{r} N_{i}) = ⋂_{i = 1}^{r} S^{- 1} N_{i}

inside $S^{- 1} M$ .

Facts & Assumptions

Given: A commutative ring $R$ , a multiplicative subset $S \subseteq R$ , a left $R$ -module $M$ , and submodules $N_{1}, \dots, N_{r} \leq M$ .

[L1]

Localisation identifies kernels with the kernels of localised maps (Localisation commutes with kernels images and cokernels).

[L2]

Localisation commutes with quotient modules and finite direct sums (Localisation commutes with quotient modules and arbitrary direct sums).

[L3]

The direct sum has its universal diagonal map into a family of targets (Universal property of a direct sum of modules, The direct sum of an indexed family of modules).

Proof

technique · direct

1.1

Let $δ : M \to ⨁_{i = 1}^{r} M / N_{i}$ be the diagonal map $δ (m) = (m + N_{1}, \dots, m + N_{r})$ . By construction, $\ker δ = ⋂_{i = 1}^{r} N_{i}$ .

L3algebra

2.1

By [L1], $S^{- 1} (⋂_{i} N_{i}) ≅ \ker (S^{- 1} δ)$ . By [L2], the codomain of $S^{- 1} δ$ identifies with $⨁_{i} (S^{- 1} M / S^{- 1} N_{i})$ , and under this identification $S^{- 1} δ$ is the diagonal map $S^{- 1} M \to ⨁_{i} (S^{- 1} M / S^{- 1} N_{i})$ .

L1L2step 1.1

3.1

An element of $S^{- 1} M$ lies in the kernel of that diagonal map exactly when its image in every quotient $S^{- 1} M / S^{- 1} N_{i}$ is zero, that is, exactly when it lies in every submodule $S^{- 1} N_{i}$ . Therefore $\ker (S^{- 1} δ) = ⋂_{i} S^{- 1} N_{i}$ .

step 2.1algebra

4.1

Combining steps 2.1 and 3.1 gives $S^{- 1} (⋂_{i} N_{i}) = ⋂_{i} S^{- 1} N_{i}$ inside $S^{- 1} M$ .

step 2.1step 3.1∎

Depends on

Used by

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Dependency tree · two levels

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Sources

A. Altman and S. Kleiman, A Term of Commutative Algebra, 13th ed., Section 12 (standard reference, not scraped)