LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 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.

Injective module maps remain injective after localisation

Statement

Let $f : M^{'} \to M$ be an injective $R$ -module homomorphism. Then the induced map

S^{- 1} f : S^{- 1} M^{'} ⟶ S^{- 1} M, (m^{'} / s) ⟼ f (m^{'}) / s,

is injective.

Facts & Assumptions

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

[L1]

A localised fraction is zero exactly when one element of $S$ kills its numerator (A localised module fraction is zero exactly when one denominator kills its numerator).

[L2]

A module homomorphism preserves scalar multiplication, so $f (r m^{'}) = r f (m^{'})$ for every $r \in R$ and $m^{'} \in M^{'}$ (Module homomorphism and isomorphism, kernel, image and cokernel).

Proof

technique · direct

1.1

Suppose $(S^{- 1} f) (m^{'} / s) = 0$ . Then $f (m^{'}) / s = 0$ , so [L1] gives $u f (m^{'}) = 0$ for some $u \in S$ .

givenL1

2.1

By [L2], $u f (m^{'}) = f (u m^{'})$ , so injectivity of $f$ gives $u m^{'} = 0$ .

step 1.1L2

3.1

Applying [L1] again, step 2.1 gives $m^{'} / s = 0$ in $S^{- 1} M^{'}$ . Hence $S^{- 1} f$ is injective.

step 2.1L1∎

Depends on

Used by

Localisation of modules is exact Theorem

Dependency tree · two levels

6 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)