Semisimplicity

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: $1\Rightarrow 2$ is trivial. For $2\Rightarrow 3$, we know that the simple submodules of R are the minimal left ideals of R, so $R=\oplus_{i\in I}L_i$ where $L_i$ is minimal. So we just need this sum to be finite. But we know that $1=x_1+\cdots +x_j$, a finite sum where $x_j\in L_j$ (reindex if you want rigor with indices since $\mathbb{N}$ isn’t necessarily a subset of $I$). So for any $r\in R$, we have $r=r1=rx_1+\cdots + rx_j$. i.e. $R\subset L_1\oplus\cdots \oplus L_j$. So $R=\oplus_{i=1}^j L_i$.

Now we said that a direct sum of minimal left ideals (simple submodules) was an equivalent definition of semisimple, so $3\Rightarrow 2$. So for $2\Rightarrow 1$, let M be an R-module with R semisimple. Since any R-module is an epimorphic image of a free module, say $F=\oplus Ra_i$. But each $Ra_i\cong R$, 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 $\Delta$ be a division ring. Then the claim is that $M_n(\Delta)$ is semisimple. Let $L_i=\{(0|\cdots | v | \cdots |0) : v\in \Delta^n\}$. i.e. we have $L_i$ as the matrix with zeros everywhere except the i-th column, which can be anything from the division ring. Certainly, $M_n(\Delta)=L_1\oplus\cdots\oplus L_n$. And also each $L_i$ is a left ideal. But why are they minimal?

Suppose some left ideal is properly contained in $L\subsetneq L_1$. Then there is some $V=(v | 0 \cdots 0)\in L_1\setminus L$. So take matrix $A\in L$, we can easily form a matrix $B\in M_n(\Delta)$ such that $BA=V$ (since $\Delta$ is a division algebra) which contradicts L being a left ideal. Thus the $L_j$ are minimal and hence $M_n(\Delta)$ is a semisimple ring.