Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 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.

Schur's lemma for simple modules

Statement

A nonzero homomorphism between simple modules is an isomorphism. Consequently the endomorphism ring of a simple module is a division ring. See Simple module: a nonzero module with no proper nonzero submodule.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

A left R-module M is simple if M≠0 and its only submodules are 0 and M. Equivalently, M has no proper nonzero submodule. (Simple module: a nonzero module with no proper nonzero submodule).

[L2]

For a left R-module M, define End⁡R(M):=Hom⁡R(M,M). Addition is pointwise and multiplication is composition, (fg)(m):=f(g(m)). The ring laws and the identity endomorphism are established in prop-endomorphisms-form-a-ring. (The endomorphism ring End⁡R(M) under addition and composition).

[L3]

For every left R-module M, pointwise addition and composition make End⁡R(M) a unital ring with identity id⁡M. (Module endomorphisms form a ring under pointwise addition and composition).

Proof

technique · direct
1.1L1L2L3givenalgebra

The kernel and image of a homomorphism between simple modules are each zero or whole.

2.1step 1.1givenalgebra

A nonzero homomorphism is therefore injective and surjective.

3.1step 2.1givenalgebra

Applied to a nonzero endomorphism, its inverse is linear, so the endomorphism ring is a division ring.

4.1step 2.1step 3.1givenalgebra∎

The excluded case is genuinely excluded rather than overlooked: the zero homomorphism between nonzero simple modules is not an isomorphism, which is why the hypothesis asks for a nonzero one, and it is the zero element of the endomorphism ring of step 3.1 rather than a non-invertible unit. Contrapositively, if two simple modules are not isomorphic then every homomorphism between them is zero. This proves the stated claim.

Depends on

Used by

Dependency tree · two levels

8 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