Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26
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Localisation commutes with quotient modules and arbitrary direct sums

Statement

Let R be a commutative ring and let SR be multiplicative.

  1. For every submodule NM, there is a natural isomorphism
S1(M/N)(S1M)/(S1N).
  1. For every family (Mi)iI of left R-modules, there is a natural isomorphism
S1 ⁣(iIMi)iIS1Mi.

Facts & Assumptions

Given: A commutative ring R and a multiplicative subset SR.

[L1]

Localisation is naturally (S1R)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
1.1

For a submodule NM, the sequence NMM/N0 is right exact, so [L2] and [L1] give a right-exact sequence S1NS1MS1(M/N)0. Therefore S1(M/N) is the quotient of S1M by the image of S1NS1M, namely by the submodule S1N.

L1L2L4
1.2

For a family (Mi)iI, [L3] and [L1] give S1(iMi)(S1R)R(iMi)i((S1R)RMi)iS1Mi.

L1L3
2.1

Thus S1(M/N)(S1M)/(S1N) naturally in M and N.

step 1.1
3.1

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

step 2.1step 1.2

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