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**:

- E is the sum of a family of simple submodules.
- E is the direct sum of a family of simple submodules.
- 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'.

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

First, assume that there exists a family {E_{i}}
of simple submodules of E of which E is the sum, that is,
such that E is generated by the E_{i} as an R-module.
Let {E_{j}} 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 E_{j}.
[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 E_{j}. We claim that E'
contains each E_{i}, and hence equals E. Indeed, the
intersection of each E_{i} with E' is a submodule
of E_{i}, so equals either {0} or E_{i} itself.
But in the former case the subfamily is not maximal because it could be
extended by E_{i}. Therefore the intersection equals
E_{i}, 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 {E_{j}} 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').