TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-17

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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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 $M \neq 0$ 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, $(f g) (m) : = f (g (m))$ . 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

1.1

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

L1L2L3givenalgebra

2.1

A nonzero homomorphism is therefore injective and surjective.

step 1.1givenalgebra

3.1

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

step 2.1givenalgebra

4.1

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.

step 2.1step 3.1givenalgebra∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 18 results over 15 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

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