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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-09-09 (gpt-6-astra)
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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Equivalent module-theoretic characterizations of semisimple rings

Statement

Assuming the Axiom of Choice, for a unital ring R the following are equivalent: RR is semisimple; every left R-module is semisimple; every short exact sequence of left R-modules splits; and every left R-module is projective. See A semisimple ring as a ring whose left regular module is semisimple.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

A unital ring R is semisimple when its left regular module RR is semisimple. This is a left-module definition and uses no Jacobson radical. For the zero ring, the regular module is zero and hence semisimple; the Wedderburn-Artin theorem below is stated for nonzero rings. (A semisimple ring as a ring whose left regular module is semisimple).

[L2]

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).

[L3]

For every left R-module M, the free module R(M) on its underlying set admits a canonical surjection εM:R(M)→M, determined by εM(em)=m. Consequently M≅R(M)/ker⁡εM. (Every module is a quotient of a free module).

[L4]

In a short exact sequence 0→A→iB→pC→0, a section of p is a homomorphism s:C→B with p∘s=id⁡C, and a retraction of i is a homomorphism r:B→A with r∘i=id⁡A. (Split short exact sequences, sections, and retractions).

[L5]

For a short exact sequence 0→A→iB→pC→0, the following are equivalent: 1. p has a section s:C→B; 2. i has a retraction r:B→A; 3. there is an isomorphism Φ:A⊕C→B with Φ(a,0)=i(a) and p(Φ(a,c))=c. (The splitting lemma for short exact sequences of modules).

[L6]

A left R-module P is projective if it has the lifting property for epimorphisms: whenever q:E→M is a surjective module homomorphism and f:P→M is a module homomorphism, there exists a module homomorphism f~:P→E such that q∘f~=f (def-module-homomorphism-kernel-image-and-cokernel, def-injection-surjection-bijection). (Projective modules and the lifting property).

[L7]

For a left R-module P, assertions 1 to 3 below are equivalent without choice. Under the Axiom of Choice, they are also equivalent to assertion 4: 1. P is projective; 2. every short exact sequence 0→K→E→P→0 splits; 3. Hom⁡R(P,−) takes every short exact sequence to a short exact sequence; 4. P is a direct summand of a free module. (Equivalent characterizations of projective modules).

[L8]

Assuming the Axiom of Choice, a module is semisimple if and only if every submodule has a complementary submodule. (Equivalent characterizations of semisimple modules).

[L9]

Assume The Axiom of Choice. Its actual use here is the AC-qualified semisimple complement and submodule/quotient results [L8] and [L2], whose proofs use Zorn's lemma. The canonical free cover [L3] and the splitting/projectivity equivalence in clauses 1–2 of [L7] are choice-free.

Proof

technique · direct
1.1L1L2L3L9givenalgebra

If RR is semisimple, fix a decomposition R=⨁j∈JSj into simple left submodules. In the free module R(X), put a copy of each Sj in each coordinate x∈X. These simple submodules sum directly: any element has finite coordinate support and in each coordinate a finite decomposition in the Sj, uniquely. Thus R(X) is semisimple. Every left module is a quotient of such a free module by [L3], hence is semisimple by the AC-qualified [L2].

2.1L5L7L8L9step 1.1givenalgebra

If every left module is semisimple, [L8] under [L9] gives every submodule a complement, and [L5] makes every short exact sequence split. If every short exact sequence splits, clauses 1–2 of [L7] make every module projective. If every module is projective, apply those same clauses to every quotient map M→M/U; its splitting gives a complement to U by [L5], so [L8] under [L9] makes every module semisimple.

3.1step 2.1givenalgebra∎

Applying the universal module condition to the left regular module recovers the first condition, and every clause is left-handed as asserted. This proves the stated claim.

Remarks

These are left-module characterizations. Applying them to right modules requires a separately justified opposite-ring or left/right semisimplicity interface; injectivity is not an additional conclusion of this statement.

Depends on

Used by

Dependency tree · two levels

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Sources