I'm in the following situation. I have a self-injective finite-dimensional basic algebra $\Lambda$ (hence Frobenius) over a perfect field and two finite-dimensional invertible $\Lambda$-bimodules $M$ and $N$ which are isomorphic in the stable category of $\Lambda$-bimodules, i.e. there are finite-dimensional projective $\Lambda$-bimodules $P$ and $Q$ such that $M\oplus P\cong N\oplus Q$. I wish this implied that they were strictly isomorphic as $\Lambda$-bimodules $M\cong N$, but I'm not totally sure whether this is true, nor am I able to prove it. Is this known? Are there maybe counterexamples?
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asked Jul 26, 2018 at 11:21
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3$\begingroup$ Selfinjective does not imply Frobenius (but the other direction holds). Can you give the definition of invertible? $\endgroup$– MareJul 26, 2018 at 11:28
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$\begingroup$ @Mare, OK, I've added basic, my question is Morita invariant anyway. Invertible means from the perspective of the tensor product over $\Lambda$. A $\Lambda$-bimodule is invertible if there is another one such that the two obvious tensor products are isomorphic to $\Lambda$. $\endgroup$– Fernando MuroJul 26, 2018 at 14:46
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Here is an easy counterexample: let $\Lambda =F\oplus F$ (sum of 2 copies of the base field). Then any $\Lambda-$bimodule is projective, so everything is isomorphic in the stable category. However non-isomorphic invertible bimodules do exist: take $M=\Lambda$ and $N=M$ but with the right action twisted by an automorphism of $\Lambda$ permuting two copies of $F$. Thus some extra assumptions are needed..
answered Jul 26, 2018 at 18:20
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$\begingroup$ Thanks! You're right, I'll probably have to ask $\Lambda$ to be irreducible. In this case the proof is easy. $\endgroup$ Jul 26, 2018 at 20:43