# A-W Consequences

I said I’d do the uniqueness part of Artin-Wedderburn, but I’ve decided not to prove it. Here is the statement: Every left semisimple ring R is a direct product $R\cong M_{n_1}(\Delta_1)\times\cdots \times M_{n_m}(\Delta_m)$ where $\Delta_i$ are division rings (so far the same as before), and the numbers m, $n_i$, and the division rings $\Delta_i$ are uniquely determined by R.

The statement here is important since if we can figure one of those pieces of information out by some means, then we’ve completely figured it out, but I think the proof is rather unenlightening since it is just fiddling with simple components.

Let’s use this to write down the structure of kG where G is finite and k algebraically closed with characteristic not dividing $|G|$. This is due to Molien: then $kG\cong M_{n_1}(k)\times\cdots \times M_{n_m}(k)$.

By Maschke we know that kG is semisimple, and by Artin-Wedderburn, we get then that $kG\cong\prod M_{n_i}(\Delta_i)$. In fact, the proof of Artin-Wedderburn even tells us that $\Delta_i=End_{kG}(L_i)$ where $L_i$ is a minimal left ideal of $kG$. Thus, given some minimal left ideal L, it suffices to show that $\Delta=End_{kG}(L)\cong k$.

Note that $\Delta$ is a subspace of $kG$ as a vector space over k. Thus it is finite dimensional. Now we have both L and $\Delta$ as finite dimensional vector spaces (over k). Let $a\in k$, then this element acts on L by $u\mapsto au$. But $au=ua$, so $k\subset Z(\Delta)$. Choose $d\in \Delta$, then adjoin it to k: $k(d)$. Since this is commutative, and a subdivision ring, it is a field. i.e. $k(d)/k$ as a field extension is finite, and hence algebraic, so d is algebraic over k. But we assumed k algebraically closed, so $d\in k$. Thus $\Delta=k$, and we are done.

As a Corollary to this, we get that under the same hypotheses, $|G|=n_1^2+\cdots + n_m^2$. This is just counting dimensions under the isomorphism above, since $dim_k(kG)=|G|$ and $dim_k(M_{n_i})=n_i^2$. Note also that we can always take one of the $n_i$ to be 1, since we always have the trivial representation.

Let’s end today with an example to see how nice this is. Without needing to peek inside or know anything about representations of $S_3$, we know that $\mathbb{C}S_3\cong \mathbb{C}\times\mathbb{C}\times M_2(\mathbb{C})$, since the only way to write 6 as the sum of squares is 1+1+1+1+1+1, or 1+1+4, and the first one gives $\mathbb{C}^6$ which is abelian which can’t happen since $S_3$ is non-abelian. Thus it must be the second one.