There are many ways I could proceed from here, all of which feel like a radical shift. But my goal was Artin-Wedderburn with some applications to representations and group rings, so probably the most important concept of this sequence of posts hasn’t been mentioned at all. This is the notion of being semisimple.
We’ll work from the definition that an R-module, M, is semisimple if every submodule is a direct summand. There are many equivalent ways of thinking of this.
First, note that a submodule of a semisimple module is semisimple. This just requires justifying that intersecting works nicely, and it does (a pretty straightforward exercise if you want to try it). An often useful equivalent condition for semisimple is that M is a direct sum of simple submodules.
The definitions I really want to get to are about rings (which is why I sort of breezed through that first part). A ring R is semisimple if it is a semisimple module over itself. But note that the submodules of R are just the left ideals, so R is semisimple iff every left ideal is a direct summand.
In fact, we have the following equivalent statements:
1) Every R-module is semisimple.
2) R is a semisimple ring.
3) R is a direct sum of a finite number of minimal left ideals.
Proof: is trivial. For
, we know that the simple submodules of R are the minimal left ideals of R, so
where
is minimal. So we just need this sum to be finite. But we know that
, a finite sum where
(reindex if you want rigor with indices since
isn’t necessarily a subset of
). So for any
, we have
. i.e.
. So
.
Now we said that a direct sum of minimal left ideals (simple submodules) was an equivalent definition of semisimple, so . So for
, let M be an R-module with R semisimple. Since any R-module is an epimorphic image of a free module, say
. But each
, so they are semisimple. Thus F is semisimple. But then M is a semisimple module.
With a view towards Artin-Wedderburn, I’ll present what is probably the most important example of a semisimple ring.
Let be a division ring. Then the claim is that
is semisimple. Let
. i.e. we have
as the matrix with zeros everywhere except the i-th column, which can be anything from the division ring. Certainly,
. And also each
is a left ideal. But why are they minimal?
Suppose some left ideal is properly contained in . Then there is some
. So take matrix
, we can easily form a matrix
such that
(since
is a division algebra) which contradicts L being a left ideal. Thus the
are minimal and hence
is a semisimple ring.
[...] This is equivalent to saying that the group ring is semisimple. [...]
By: Basics of group representation theory « Delta Epsilons on July 10, 2009
at 1:58 pm
Incidentally, I’m talking about basic representation theory over at our blog (http://deltaepsilons.wordpress.com/2009/07/10/basics-of-group-representation-theory/) as well…I’m trying to avoid overlapping with your posts, as much as possible, by stating theorems without proofs. The goal was to make sure we could refer back when we continue talking about somewhat more specialized subfields of representation theory. Quite a few of us have interests in this area of mathematics.
By: Akhil Mathew on July 10, 2009
at 2:11 pm
[...] already did the sufficient direction. So assume R is semisimple. Then where the are direct sums of isomorphic minimal left ideals [...]
By: Artin-Wedderburn « A Mind for Madness on July 21, 2009
at 9:30 pm