The equivalence of three characterizations
of semisimple modules over a ring

The following standard definitions and results can be found for instance in Lang's Algebra (XVII,2), which includes the proof of the result from Bourbaki cited by Serre in the proof of Wedderburn's theorem. For that matter, Chapter XVII of Lang also contains a proof of Wedderburn's theorem itself, and Section 1 of that chapter describes matrices and linear maps in the non-commutative setting.

Let R be any ring. An R-module is said to be simple if it is not the zero module and has no submodules other than itself and {0}. If R is a k-algebra, then ``R-module'' becomes ``representation of R'' and ``simple'' becomes ``irreducible''. The following simple observation is a surprisingly powerful tool:

Schur's Lemma: A homomorphism between simple R-modules is either zero or an isomorphism.
Proof: The kernel and image are both submodules, etc.

Let E be any R-module. Then The following are equivalent:

1. E is the sum of a family of simple submodules.
2. E is the direct sum of a family of simple submodules.
3. Every submodule F of E is a ``direct summand'', that is, has a ``complementary'' submodule F' in E such that E is the direct sum of F and F'.
We shall say that an R-module satisfying these equivalent conditions is semisimple.

Proof: We shall show that 1 implies 2 implies 3 implies 1.

First, assume that there exists a family {Ei} of simple submodules of E of which E is the sum, that is, such that E is generated by the Ei as an R-module. Let {Ej} be a maximal subfamily for which the sum is direct, that is, such that the zero element of E cannot be written as a (finite) sum of nonzero elements from the Ej. [Such a subfamily exists by AC/Zorn; we won't need AC/Zorn as long as we're concerned only with finite-dimensional representations.] Let E' be this direct sum of the Ej. We claim that E' contains each Ei, and hence equals E. Indeed, the intersection of each Ei with E' is a submodule of Ei, so equals either {0} or Ei itself. But in the former case the subfamily is not maximal because it could be extended by Ei. Therefore the intersection equals Ei, which is thus contained in E' as claimed. We have thus shown that 1 implies 2.

To show that 2 implies 3, apply the same argument to a maximal subfamily {Ej} subject to the further condition that its direct sum E' intersect F in {0}. The same argument shows that E is then the direct sum of E' with F.

Finally, to go from 3 to 1, we show that E is the sum of all its simple submodules. Let that sum be F. By assumption, E is the direct sum of F with some other submodule F'. But this submodule then contains no simple submodules. But the only submodule of E that contains no simple submodule is {0}. To see this, let F' be any nonzero submodule of E, and v any nonzero element of F'. Using Zorn if necessary, find a maximal submodule M of Rv properly contained in Rv. (This corresponds to a maximal left ideal in the ring R/I, where I is the annihilator of v in R.) Now E is the direct sum of M with some submodule S. Intersecting with Rv, we obtain Rv as the direct sum of M with the intersection S' of S with Rv. Then S' is simple, since the direct sum of any submodule with M is a submodule of Rv containing M and thus equal to either M (when the submodule is {0}) or Rv (when it is all of S').