TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17

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Equivalent characterizations of semisimple modules

Statement

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. See Semisimple modules as direct sums of simple modules.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

A left $R$ -module is semisimple when it is an internal direct sum of simple submodules, allowing the empty direct sum. Hence the zero module is semisimple. (Semisimple modules as direct sums of simple modules).

[L2]

Assuming the Axiom of Choice, every finitely generated nonzero module has a maximal proper submodule. (Under Choice, every finitely generated nonzero module has a maximal proper submodule).

[L3]

Assume the Axiom of Choice (def-axiom-of-choice). Let $(P, \leq)$ be a nonempty poset in which every chain has an upper bound. Then $P$ has a maximal element (def-maximal-element). Note the hypothesis asks only for an upper bound, not a least upper bound, and the conclusion asserts only that a maximal element exists, never that a greatest one does. (Zorn's lemma).

[L4]

Let $(M_{i})_{i \in I}$ be left $R$ -modules and $N$ a left $R$ -module. For every family of homomorphisms $f_{i} : M_{i} \to N$ , there is a unique homomorphism $f : ⨁_{i \in I} M_{i} ⟶ N$ such that $f \circ ȷ_{i} = f_{i}$ for every $i$ . It is given by $f ((m_{i})) = \sum_{i \in supp (m)} f_{i} (m_{i}) .$ For $I = \emptyset$ , this is the unique map $0 \to N$ . (Universal property of a direct sum of modules).

[L5]

The socle $Soc (M)$ is the sum of all simple submodules of $M$ . (The socle as the sum of all simple submodules).

Proof

technique · direct

1.1L1L2L3L4L5givenalgebra

A direct sum of simple submodules is plainly their sum. Conversely, if $M$ is a sum of simple submodules, Zorn's lemma applied to independent families of them gives a maximal direct sum $D$ ; if $D \neq M$ , a simple submodule not contained in $D$ meets $D$ trivially, contradicting maximality.

2.1L3step 1.1givenalgebra

Given $N \leq M$ and a direct-sum decomposition $M = ⨁_{i \in I} S_{i}$ into simples, use Zorn to choose a maximal sum $C = ⨁_{j \in J} S_{j}$ with $C \cap N = 0$ . If some $S_{i}$ were not contained in $N + C$ , simplicity would give $S_{i} \cap (N + C) = 0$ , so adjoining it would contradict maximality. Hence every $S_{i} \leq N + C$ , and therefore $M = N \oplus C$ .

3.1L2L5step 2.1givenalgebra∎

Conversely suppose every submodule has a complement. By [L5], choose $C$ with $M = Soc (M) \oplus C$ . If $C \neq 0$ , choose $0 \neq x \in C$ . The cyclic module $R x$ has a maximal proper submodule $K$ by [L2]. Let $D$ complement $K$ in $M$ . Then $R x = K \oplus (R x \cap D)$ , and $R x \cap D ≅ R x / K$ is a nonzero simple submodule of $C$ , contrary to $C \cap Soc (M) = 0$ . Hence $C = 0$ and $M$ is a sum of simples. The zero module is the empty sum.

Depends on

Used by

Dependency tree · two levels

16 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)