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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17
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Simple modules over a product of matrix rings over division rings

Statement

Let r≥1, let every ni≥1, let every Di be a division ring, and put R=∏i=1rMni(Di). Then every simple left R-module is supported on exactly one factor and is isomorphic to that factor's column module Dini. These column modules give all simple left R-module isomorphism classes, with one class for each factor. See Wedderburn–Artin theorem for semisimple rings.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

A nonzero homomorphism between simple modules is an isomorphism. Consequently the endomorphism ring of a simple module is a division ring. (Schur's lemma for simple modules).

[L2]

If D is a division ring and n≥1, then the left regular module of Mn(D) is the direct sum of its simple column ideals Mn(D)ejj≅Dn. (Matrix rings over division rings are semisimple).

Proof

technique · direct
1.1givenalgebra

Write ci∈R for the central idempotent that is 1 in factor i and 0 elsewhere. For a simple left R-module S, every ciS is a submodule and S=∑iciS. Hence some ciS is nonzero and therefore equals S; then cjS=cjciS=0 for j≠i. Thus S is supported on exactly one factor Ai=Mni(Di).

2.1L1L2step 1.1givenalgebra

Choose 0≠s∈S. The map Ai→S, a↦as, is surjective because its image is a nonzero submodule. By [L2], Ai is a direct sum of simple column ideals Cj≅Dini. At least one restriction Cj→S is nonzero, so [L1] makes it an isomorphism. Hence S≅Dini.

3.1L2step 1.1step 2.1givenalgebra∎

Conversely each column module is simple by [L2]. Modules supported on different factors cannot be isomorphic, because the corresponding ci acts as the identity on one and as zero on the other. For a fixed factor all column ideals are isomorphic to Dini by [L2]. This proves the classification, including the one-factor case r=1.

Depends on

Used by

Dependency tree · two levels

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Sources