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

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Under Choice, submodules and quotients of semisimple modules are semisimple

Statement

Assuming the Axiom of Choice, every submodule and every quotient of a semisimple module is semisimple. See Equivalent characterizations of semisimple modules.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

Assuming the Axiom of Choice, for a module $M$ the following are equivalent: $M$ is a direct sum of simple submodules; $M$ is the sum of its simple submodules; and every submodule of $M$ has a complementary submodule. (Equivalent characterizations of semisimple modules).

[L2]

For submodules $L, N \leq M$ , there is a canonical isomorphism $L / (L \cap N) ≅ (L + N) / N .$ . (Second isomorphism theorem for modules).

Proof

technique · direct

1.1

A submodule inherits the complement property by intersecting a complement in the ambient module.

L1L2givenalgebra

2.1

For a quotient, complement the kernel and identify the quotient with that semisimple complement.

step 1.1givenalgebra

3.1

The extreme submodules are admitted and give nothing new: the zero submodule is the empty direct sum, hence semisimple, and its quotient $M / 0 ≅ M$ is semisimple by hypothesis; the whole submodule $M$ is semisimple by hypothesis and its quotient $M / M = 0$ is again the empty direct sum. This proves the stated claim.

step 1.1step 2.1givenalgebra∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 27 results over 7 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)