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What is an example of a non-free finitely generated $R$-bimodule $M$ satisfying

i) $M$ is projective as both a left and right $R$-module

ii) the right dual $\mathrm{Hom}_R(M,R)$ and the left dual $_R\mathrm{Hom}(M,R)$ are isomorphic as bimodules,

where $R$ is a noncommutative unital algebra defined over a field $k$ with non-zero characteristic.

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  • $\begingroup$ Why does $M=R$ not work? $\endgroup$ Feb 26 '20 at 23:17
  • $\begingroup$ I guess I want a non-trivial example. $\endgroup$ Feb 26 '20 at 23:33
  • $\begingroup$ You mean that $M$ is not free? $\endgroup$ Feb 26 '20 at 23:40
  • $\begingroup$ yes, I have put this in the question $\endgroup$ Feb 26 '20 at 23:42
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    $\begingroup$ Just a minor rant about terminology. $M=R$ is not free as a bimodule. Also, a bimodule that is projective on the left and on the right is not necessarily a projective bimodule. $\endgroup$ Feb 27 '20 at 9:49
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Take $G$ a nonabelian finite group of size coprime to $p$, and $k$ a field of characteristic $p>0$. Then $k[G]$ is semisimple (Maschke's theorem), so every module on left or right is projective.

In particular, take $\underline{k}$ to be the trivial representation of $G$. So this is a bimodule projective on both sides, and $\mathrm{Hom}_G(\underline{k}, k[G])={_G\mathrm{Hom}}(\underline{k}, k[G])=\underline{k}$, as bimodules (both of these being naturally the same subspace of $k[G]$, i.e. $\langle\sum_{g \in G}g\rangle$).

... This satisfies all the criteria of the question; somehow, however, this still does not feel as if it should count for a non-trivial example.

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    $\begingroup$ In fact, for any group algebra $k[G]$ of a finite group, semisimple or not, and for any bimodule $M$, the left dual and right dual are isomorphic. And there are plenty of natural bimodules that are projective on both sides: e.g., $k[G]\otimes_{k[H]}k[G]$ for a subgroup $H\leq G$. $\endgroup$ Feb 27 '20 at 9:49

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