5
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Suppose $R$ is a noncommutative ring. When could I reasonably expect $R^{op}\cong R$?

For instance I know group rings have a natural involution, i.e. if $SG$ is the group ring in question then $r\cdot g=g^{-1}\cdot r$. Does this imply $S[G]^{op}\cong S[G]$? If so, does this apply to any ring with a natural involution?

Finally, does $R^{op}\cong R$ imply all left ideals are right ideals (and vice versa)?

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  • $\begingroup$ @ThomasAndrews Sorry, I meant for the coefficient ring to be commutative $\endgroup$ – Sam Williams Aug 3 '17 at 16:51
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    $\begingroup$ In the last case, consider $R = M_2(A)$ for $A$ commtuative. Then the map $g \to g^t$ is an isomorphism $R \to R^{\text{op}}$, but $R$ has one-sided ideals; consider the set of matrices of the form $$g = \begin{pmatrix}* & * \\ 0 & 0\end{pmatrix},$$ for example. $\endgroup$ – anomaly Aug 3 '17 at 16:59
  • $\begingroup$ @anomaly Thanks for the counterexample. I was thinking of considering ideals as modules. Since any left module is a right module over its opposite ring, I expected left ideals to equal right ideals. Why exactly is this failing? $\endgroup$ – Sam Williams Aug 3 '17 at 17:04
  • $\begingroup$ @SamWilliams Thinking that this means all left ideals must all be right ideals is just non-sequitur. Yes, all of the left ideals correspond to right ideals via the isomorphism. And all left ideals that aren't right ideals correspond to right ideals that aren't left ideals. The example above shows precisely why. The nontrivial one-sided ideals form two disjoint sets, and they correspond to each other via the transposition map. $\endgroup$ – rschwieb Aug 3 '17 at 18:09
  • $\begingroup$ @SamWilliams: For one thing, the action of $R$ and $R^\text{op}$ are not the same. You can use an isomorphism $R\to R^{\text{op}}$ to transform a left $R$-module into a right $R$-module, but the induced map on modules is not trivial. $\endgroup$ – anomaly Aug 3 '17 at 18:20

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