Example of the semisimple ring $R$ but $R^{{\rm op}}$ is not.

Question

Is there any example of this kind of rings? i don't have any imagine of this rings, if they are exist!

egreg · Accepted Answer · 2015-02-22T16:52:29.900

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By the Wedderburn-Artin theorem, a semisimple ring is a product of full matrix rings over division rings. Conversely, a product of simple artinian rings (like full matrix rings over division rings) is semisimple. This is left-right symmetric.

edited Feb 22 '15 at 16:52

answered Feb 21 '15 at 14:29

egreg

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@ egreg: thanks for your complete answer. – Feb 21 '15 at 14:46
Just a nitpick, but "left-right symmetric" is the phrase you're looking for. Great answer, though. – rschwieb Feb 22 '15 at 14:51
@rschwieb The property of being semisimple is left-right invariant. – egreg Feb 22 '15 at 14:54
@egreg I know what you mean, but you will find that it is hard to find that phrase the way you are using it, but it is very easy to find the one I'm using. Although I guess one can make an argument for "invariant," it is still not as natural or standard of a choice as "symmetric" – rschwieb Feb 22 '15 at 16:51

score 0 · Answer 2 · answered Feb 21 '15 at 14:21

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No. Every semisimple ring is isomorphic to its opposite ring, and this fact is a corollary of Artin-Wedderburn theorem. So if $R$ is semisimple, then $R^{op}$ is semisimple.

answered Feb 21 '15 at 14:21

Crostul

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Isomorphic? Perhaps you're thinking to commutative rings. – egreg Feb 21 '15 at 14:26
No. If you have a matrix ring $R=M(n \times n, D)$, then transposition of matrices is an isomorphism between $R$ and $R^{op}$. The same holds for products of such rings, and all semisimple rings are of this form. – Crostul Feb 21 '15 at 14:29
@Crostul Even if $K$ is a division ring not a field? In particular, does a division ring necessarily have an involution? – egreg Feb 21 '15 at 14:30
Mh. I never checked it: does $(AB)^T=B^TA^T$ fail in the non commutative case? – Crostul Feb 21 '15 at 14:32
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@Crostul Try it. ;-) It's easy: use $n=1$. – egreg Feb 21 '15 at 14:33
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@Crostul Sorry, I have to downvote this. The idea is good, of course, but the statement that a semisimple ring is isomorphic to its opposite ring is false. – egreg Feb 22 '15 at 14:56
@Crostul I had to downvote for the same reason. There are even division rings which aren't isomorphic to their opposite rings.. It's also an important lesson that transposition in $M_n(R)$ isn't an anti-homomorphism when $R$ is not commutative. – rschwieb Feb 23 '15 at 13:08

Example of the semisimple ring $R$ but $R^{{\rm op}}$ is not.

2 Answers2