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I'm working on building up a proof of Artin-Wedderburn theorem, given in some of the exercises of Dummit and Foote.

18.2.9 says that if $S$ is a simple, unital ring, satisfying the DCC on left ideals, then $S\cong M_n(\Delta)$ for a division ring $\Delta$. I have followed their suggestion and proved that $$ S^{op}\cong\operatorname{Hom}_S(L^n,L^n)\cong M_n(D) $$ where $L$ is a minimal nonzero left ideal of $S$, and $D=\operatorname{Hom}_S(L,L)$ is a division ring by Schur.

It says then that $S\cong M_n(D^{op})$, so $D^{op}=\Delta$ is the division ring. I think that $S\cong(S^{op})^{op}\cong M_n(D)^{op}$. Is it true that $M_n(D)^{op}\cong M_n(D^{op})$? I tried looking at the $2\times 2$ case, and it seems like $A\cdot B$ in $M_n(D)^{op}$ is the same as $AB^T$ in $M_n(D^{op})$, which doesn't look too good. Is there a better way to complete this?

Kristina Thai
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1 Answers1

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You're on the right track: $M_n(D^{op})\cong (M_n(D))^{op}$ via the transposition map. The only real challenge is checking if the map is multiplicative.

Let's use $\circ$ abusively to denote "opposite multiplication" in both rings, and let's use the convention that $T[a_{ij}]=[a_{ji}]=[\overline{a_{ij}}]$.

Take two matrices $[a_{ij}],[b_{ij}]\in M_n(D^{op})$. Then the $i,k$ entry of the product is $\sum_j a_{ij}\circ b_{jk}=\sum_jb_{jk}a_{ij}$. Therefore the $i,k$ entry of the transpose is $\sum_jb_{ji}a_{kj}$.

On the other hand, consider $T([a_{ij}])\circ T([b_{ij}])$. This is $[\overline{a_{ij}}]\circ[\overline{b_{ij}}]=[\overline{b_{ij}}][\overline{a_{ij}}]$ which has $i,k$ entry $\sum_j\overline{b_{ij}}\overline{a_{jk}}=\sum_jb_{ji}a_{kj}$. Notice this matches the other computation exactly.

rschwieb
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