Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17
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.

Third isomorphism theorem for modules

Statement

If NLM, then L/N is a submodule of M/N and (M/N)/(L/N)M/L. See First isomorphism theorem for modules: M/kerfimf.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

For every module homomorphism f:MN, there is a module isomorphism M/kerf  imf, given by m+kerff(m). (First isomorphism theorem for modules: M/kerfimf).

[L2]

For NM, the additive cosets m+N form the quotient module M/N under the well-defined scalar action r(m+N):=rm+N. (Quotient module M/N with scalar multiplication on additive cosets).

[L3]

Let f:MP be a module homomorphism and let NM satisfy Nkerf. There is a unique module homomorphism fˉ:M/NP such that fˉ(m+N)=f(m), equivalently f=fˉπ. (A module homomorphism vanishing on N factors uniquely through M/N).

Proof

technique · direct
1.1

For NLM, send m+N to m+L.

L1L2L3givenalgebra
2.1

If m+N=m+N, then mmNL, so the images modulo L agree. The map is surjective, and its kernel consists exactly of the cosets m+N with mL, namely L/N; the first isomorphism theorem gives the displayed isomorphism.

step 1.1givenalgebra
3.1

The two coincident cases are admitted by NLM and hold. For N=L the submodule L/N is zero and the isomorphism reads (M/N)/0M/N=M/L; for L=M it is L/N=M/N and the isomorphism reads (M/N)/(M/N)=0M/M=0. This proves the stated claim.

step 2.1givenalgebra

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 27 results over 10 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources