TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck 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.

Second isomorphism theorem for modules

Statement

For submodules $L, N \leq M$ , there is a canonical isomorphism $L / (L \cap N) ≅ (L + N) / N .$ See First isomorphism theorem for modules: $M / \ker f ≅ im f$ .

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

For every module homomorphism $f : M \to N$ , there is a module isomorphism $M / \ker f ≅ im f,$ given by $m + \ker f \mapsto f (m)$ . (First isomorphism theorem for modules: $M / \ker f ≅ im f$ ).

[L2]

For submodules $A, B \leq M$ , the intersection $A \cap B$ and the sum $A + B : = {a + b : a \in A, b \in B}$ are submodules of $M$ . (The one-step submodule criterion; intersections and sums of submodules are submodules).

[L3]

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

Proof

technique · direct

1.1L1L2L3givenalgebra

For submodules $L, N \leq M$ , map $L \to (L + N) / N$ .

2.1step 1.1givenalgebra

Its kernel is $L \cap N$ , and it is surjective by the definition of $L + N$ ; the first isomorphism theorem gives $L / (L \cap N) ≅ (L + N) / N$ .

3.1step 2.1givenalgebra∎

The coincident and zero cases are admitted and give equalities rather than exceptions. For $L = 0$ both sides are $0$ , since $0 / (0 \cap N) = 0$ and $(0 + N) / N = N / N = 0$ ; for $N = 0$ both sides are $L$ ; and for $L = N$ both sides are $0$ . This proves the stated claim.

Depends on

Used by

Dependency tree · two levels

11 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

William Crawley-Boevey, Noncommutative Algebra, Chapter 1 Sections 1.1-1.9 (standard reference, not scraped)