Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck 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.

Addition of localised module fractions is independent of representatives

Statement

If m/s=m′/s′ and n/t=n′/t′ in S−1M, then tm+snst=t′m′+s′n′s′t′. So the addition formula of Localisation of a module at a multiplicative subset is independent of the chosen representatives.

Facts & Assumptions

Given: A commutative ring R, a multiplicative subset S⊆R, a left R-module M, and equalities m/s=m′/s′ and n/t=n′/t′ in S−1M.

[L1]

In S−1M, the equality x/u=y/v means that q(vx−uy)=0 for some q∈S, and addition is defined by (x/u)+(y/v)=(vx+uy)/(uv) (Localisation of a module at a multiplicative subset).

[L2]

The relation defining S−1M is an equivalence relation (The module-fraction relation is an equivalence relation).

Proof

technique · direct
1.1givenL1L2choose

By [L1], choose u,v∈S with u(s′m−sm′)=0 and v(t′n−tn′)=0.

2.1step 1.1algebra

Multiplying the first equality by vtt′ and the second by uss′ and then adding gives uv(s′t′(tm+sn)−st(t′m′+s′n′))=0.

3.1step 2.1L1∎

Step 2.1 is exactly the relation witnessing (tm+sn)/(st)=(t′m′+s′n′)/(s′t′), so the addition formula is independent of representatives.

Depends on

Used by

Cited to discharge well-definedness by Localisation of a module at a multiplicative subset.

Dependency tree · two levels

4 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