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CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26
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Localisation commutes with finite intersections of submodules

Statement

Let N1,…,Nr be submodules of a left R-module M. Then S−1 ⁣(⋂i=1rNi)=⋂i=1rS−1Ni inside S−1M.

Facts & Assumptions

Given: A commutative ring R, a multiplicative subset S⊆R, a left R-module M, and submodules N1,…,Nr≤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
1.1L3algebra

Let δ:M→⨁i=1rM/Ni be the diagonal map δ(m)=(m+N1,…,m+Nr). By construction, ker⁡δ=⋂i=1rNi.

2.1L1L2step 1.1

By [L1], S−1(⋂iNi)≅ker⁡(S−1δ). By [L2], the codomain of S−1δ identifies with ⨁i(S−1M/S−1Ni), and under this identification S−1δ is the diagonal map S−1M→⨁i(S−1M/S−1Ni).

3.1step 2.1algebra

An element of S−1M lies in the kernel of that diagonal map exactly when its image in every quotient S−1M/S−1Ni is zero, that is, exactly when it lies in every submodule S−1Ni. Therefore ker⁡(S−1δ)=⋂iS−1Ni.

4.1step 2.1step 3.1∎

Combining steps 2.1 and 3.1 gives S−1(⋂iNi)=⋂iS−1Ni inside S−1M.

Depends on

Used by

Dependency tree · two levels

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Sources