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Let $(A,B,P,Q,f,g)$ be a general Morita context (that is $A$,$B$ rings, ${}_AP_B$ and ${}_BQ_A$ and $f:Q\otimes_AP\rightarrow B$, $g:P\otimes_BQ \rightarrow A$ bimodule morphisms that satisfy the additional compatibility). In the particular case that $Q=P^*$ and $B=\mathrm{End}_A(P)$ it is (apparently well) known that if $P$ is an progenerator as an $A$ then the context gives rise to an equivalence ($f$ and $g$ are isomorphisms).

In the general case, if $P$ is a progenerator as an $A$-module and $Q$ a progenerator as a $B$-module, is this true as well? That is, are $f$ and $g$ isomorphisms? I can't seem to find a reference for this.

Thanks in advance

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    $\begingroup$ Could you edit the question to state clearly the conditions you want to assume? In the title you refer to $P$ and $Q$ being progenerators, but this doesn't seem to be what you're asking in the question. Also, progenerator for which ring or rings? $\endgroup$ – Jeremy Rickard Sep 28 '14 at 7:33
  • $\begingroup$ @Rickard I've edited the cuestion as you asked, stating more clearly the problem. $\endgroup$ – sjvega Sep 28 '14 at 13:36
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For arbitrary bimodules $P$ and $Q$ you get a Morita context with both $f=0$ and $g=0$, and you can certainly choose the bimodules to be progenerators for $A$ and $B$ respectively.

For the "well known fact" that you mention, I think you also need that $$f:P\otimes_BQ=P\otimes_B\operatorname{Hom}_A(P,A)\to A$$ is the natural map (i.e., the evaluation map $p\otimes\phi\mapsto\phi(p)$), and this condition doesn't make sense for general $Q$.

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  • $\begingroup$ Thanks! The fact that the null morphisms work as a Morita context completely slipped my mind. $\endgroup$ – sjvega Sep 28 '14 at 16:18

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