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

How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Localisation commutes with kernels images and cokernels

Statement

For every $R$ -module homomorphism $f : M \to N$ , localisation identifies $S^{- 1} (\ker f) ≅ \ker (S^{- 1} f), S^{- 1} (im f) ≅ im (S^{- 1} f), S^{- 1} (coker f) ≅ coker (S^{- 1} f) .$

Facts & Assumptions

Given: A commutative ring $R$ , a multiplicative subset $S \subseteq R$ , left $R$ -modules $M, N$ , and an $R$ -module homomorphism $f : M \to N$ .

[L1]

Localisation sends short exact sequences to short exact sequences (Localisation of modules is exact).

[L2]

The kernel, image, and cokernel of $f$ are the standard submodule and quotient constructions associated to $f$ (Module homomorphism and isomorphism, kernel, image and cokernel).

Proof

technique · direct

1.1L1L2

The standard short exact sequences $0 \to \ker f \to M \to im f \to 0$ and $0 \to im f \to N \to coker f \to 0$ localise, by [L1], to short exact sequences $0 \to S^{- 1} (\ker f) \to S^{- 1} M \to S^{- 1} (im f) \to 0$ and $0 \to S^{- 1} (im f) \to S^{- 1} N \to S^{- 1} (coker f) \to 0$ .

2.1step 1.1L2

In the first localised sequence, the middle map is $S^{- 1} f$ restricted to $S^{- 1} M$ , so its kernel is exactly $S^{- 1} (\ker f)$ and its image is exactly the embedded copy of $S^{- 1} (im f)$ .

3.1step 2.1L2

In the second localised sequence, the quotient by the image of $S^{- 1} f$ is therefore $S^{- 1} (coker f)$ , so $S^{- 1} (coker f) ≅ coker (S^{- 1} f)$ .

4.1step 2.1step 3.1∎

Steps 2.1 and 3.1 prove the kernel, image, and cokernel identifications.

Depends on

Used by

Dependency tree · two levels

9 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., Theorem 12.20 (standard reference, not scraped)