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).


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