# Artin-Wedderburn

I’m back after a brief hiatus in which I worked through a set of problems on Lie derivatives, flows, and vector fields. At the end of the day, I just never seemed to muster the strength to look at algebra. Here goes.

We need a lemma first (I know. If I carefully had planned this, it would have been taken care of already). Lemma: If R is considered as a left module over itself, then $R^{op}\cong End_R(R)$. The natural map to check is $\phi: End_R(R)\to R^{op}$ by $\phi(f)=f(1)$. It is just routine checking that this works. We are in $R^{op}$ since multiplication gets reversed: $\phi(fg)=(f\circ g)(1)=f(g(1)\cdot 1)=g(1)f(1)=\phi(g)\phi(f)$.

Artin-Wedderburn: A ring R is semisimple iff R is isomorphic to a direct product of matrix rings over division rings.

We already did the sufficient direction. So assume R is semisimple. Then $R=B_1\oplus \cdots \oplus B_m$ where the $B_i$ are direct sums of isomorphic minimal left ideals (decompose into all minimal left ideals $L_{j}$, and then group isomorphic ones into the $B_j$). By our above lemma $R^{op}\cong End_R(R)$. As a consequence of Schur’s Lemma, $Hom_R(B_i, B_j)=\{0\}$ when $i\neq j$.

Thus we now have $R^{op}\cong End_R(R)\cong End_R(B_1)\times \cdots\times End_R(B_m)$. But the $B_j$ can be decomposed into the isomorphic minimal left ideals and we get $End_R(B_i)\cong M_{n_i}(End_R(L_i))$.

But by Schur again $End_R(L_i)$ is some division ring $\Delta_i$ and hence $R^{op}\cong M_{n_1}(\Delta_1)\times\cdots\times M_{n_m}(\Delta_m)$. So $R\cong \left(M_{n_1}(\Delta_1)\right)^{op}\times\cdots \times \left(M_{n_m}(\Delta_m)\right)^{op}$.

Note that $\left(M_{n_i}(\Delta)\right)^{op}\cong M_{n_i}(\Delta_i^{op})$ and that $\Delta_i^{op}$ is also a division ring (now just rename these division rings) to get that $R\cong M_{n_1}(\Delta_1)\times\cdots\times M_{n_m}(\Delta_m)$ where the $\Delta_j$ are division rings.

We immediately get some nice corollaries. One is that a ring is left semisimple iff it is right semisimple, since a ring is left semisimple iff its opposite ring is right semisimple. Another is that a commutative ring is semisimple iff it is a product of fields.

Next time I’ll do the “uniqueness” part and start on some of the group ring and representation theory consequences (of which there are many).