TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-17

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Under Choice, the socle is the largest semisimple submodule

Statement

Assuming the Axiom of Choice, for every module $M$ , $Soc (M)$ is semisimple and contains every semisimple submodule of $M$ . See The socle as the sum of all simple submodules.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

For a left $R$ -module $M$ , the socle is $Soc (M) : = \sum {S \leq M : S is simple} .$ Concretely it is the submodule generated by the union of all simple submodules. If there are none, the sum is $0$ , so the definition is always meaningful. (The socle as the sum of all simple submodules).

[L2]

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).

[L3]

Assuming the Axiom of Choice, every submodule and every quotient of a semisimple module is semisimple. (Under Choice, submodules and quotients of semisimple modules are semisimple).

Proof

technique · direct

1.1L1L2L3givenalgebra

The sum of all simple submodules is semisimple by the sum characterization.

2.1step 1.1givenalgebra

Every semisimple submodule is itself a sum of simple submodules of the ambient module, hence lies in the socle.

3.1L1step 1.1step 2.1givenalgebra∎

If $M$ has no simple submodule, the defining sum of [L1] is empty, so $Soc (M) = 0$ ; the zero module is the empty direct sum and hence semisimple, and the only semisimple submodule of such an $M$ is $0$ itself, which it contains. This proves the stated claim.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

MIT 18.706, Lecture 2: Semisimple Modules, Socles, Artinian Rings, Wedderburn's Theorem (standard reference, not scraped)