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.

Third isomorphism theorem for modules

Statement

If $N \leq L \leq M$ , then $L / N$ is a submodule of $M / N$ and $(M / N) / (L / N) ≅ M / L .$ 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 $N \leq M$ , 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).

[L3]

Let $f : M \to P$ be a module homomorphism and let $N \leq M$ satisfy $N \subseteq \ker f$ . There is a unique module homomorphism $\overset{ˉ}{f} : M / N ⟶ P$ such that $\overset{ˉ}{f} (m + N) = f (m)$ , equivalently $f = \overset{ˉ}{f} \circ π$ . (A module homomorphism vanishing on $N$ factors uniquely through $M / N$ ).

Proof

technique · direct

1.1L1L2L3givenalgebra

For $N \leq L \leq M$ , send $m + N$ to $m + L$ .

2.1step 1.1givenalgebra

If $m + N = m^{'} + N$ , then $m - m^{'} \in N \leq L$ , so the images modulo $L$ agree. The map is surjective, and its kernel consists exactly of the cosets $m + N$ with $m \in L$ , namely $L / N$ ; the first isomorphism theorem gives the displayed isomorphism.

3.1step 2.1givenalgebra∎

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

Depends on

Used by

Dependency tree · two levels

10 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)