Alphabeta Math
LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck 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.

Surjective module maps remain surjective after localisation

Statement

Let f:M→M′′ be a surjective R-module homomorphism. Then the induced map S−1f:S−1M⟶S−1M′′,(m/s)⟼f(m)/s, is surjective.

Facts & Assumptions

Given: A commutative ring R, a multiplicative subset S⊆R, left R-modules M,M′′, and a surjective R-module homomorphism f:M→M′′.

[L1]

A module homomorphism preserves scalar multiplication (Module homomorphism and isomorphism, kernel, image and cokernel).

[L2]

Elements of S−1M′′ are fractions m′′/s with m′′∈M′′ and s∈S (Localisation of a module at a multiplicative subset).

Proof

technique · direct
1.1givenL2choose

Let m′′/s∈S−1M′′. Since f is surjective, choose m∈M with f(m)=m′′.

2.1step 1.1L1

Then (S−1f)(m/s)=f(m)/s=m′′/s, so m′′/s lies in the image of S−1f.

3.1step 2.1L2∎

Since every element of S−1M′′ is hit, S−1f is surjective.

Depends on

Used by

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