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Let $A$ be an infinite-dimensional noncommutative algebra over a field, let $B$ be an infinite-dimensional subalgebra of $A$, and let $A$ be a direct sum of projective simple $B$-sub-bimodules. Then can one conclude that $A$, or indeed $B$, is a semisimple ring?

EDIT: I should highlight that I am interested only in the case where $A$ and $B$ are infinite-dimensional algebras.

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  • $\begingroup$ It is B that must be semisimple and I guess separable $\endgroup$ – Benjamin Steinberg Jun 13 at 16:33
  • $\begingroup$ Ah ok, so to get this straight: From the assumptions above on $A$ and $B$, we can conclude that $B$ is a semisimple ring? $\endgroup$ – Boris Henriques Jun 13 at 16:35
  • $\begingroup$ I think so. B will be a semisimple bimodule and the simple summands are a subset of those appearing in A. Thus B is a projective B-B bimodule and hence separable. Any separable algebra over a field is semisimple. I'm assuming that in these bimodules the left and right actions of k agree which I guess is fine since this is true for the bimodule structure on A. $\endgroup$ – Benjamin Steinberg Jun 13 at 16:42
  • $\begingroup$ Great! Please put this as an answer and I will accept it! $\endgroup$ – Boris Henriques Jun 13 at 16:51
  • $\begingroup$ I probably won't have a chance to write a detailed answer until later. $\endgroup$ – Benjamin Steinberg Jun 13 at 17:06
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@BugsBunny answered the original version of the question. I'll answer the new version. The algebra $B$ must be finite dimensional and semisimple under these hypothesis, and even stronger, it must be separable meaning that it remains semisimple even under base extension.

Let $B^{e}=B\otimes_k B^{op}$ be the enveloping algebra. Note that (left) $B^e$-modules equal $B$-$B$-bimodules in which the left and right actions of $k$ coincide. In particular $A$ and $B$ are $B^e$-modules.

Recall that $B$ is separable over $k$ if $B$ is a projective left $B^e$-module. This is well known to be equivalent to $B$ being finite dimensional over $k$ and for each field extension $L/K$, the algebra $L\otimes_k B$ is semisimple. All these things can be found in Pierce's book.

Now under your assumption, $A$ is a direct sum of simple $B^e$-modules that are projective. Thus $A$ is a semisimple $B^e$-module and hence the same is true for its $B^e$-submodule $B$. Moreover, if $S$ is a simple $B^e$-submodule of $B$, then it must be nontrivial under the one of the projections of $A$ onto a simple $B^e$-summand and so $S$ is isomorphic to one of the simple $B^e$-summands in $A$ by Schur's lemma. Therefore, $B$ is a direct sum of projective $B^e$-modules and hence is projective. Thus $B$ is separable over $K$ and hence finite dimensional and semisimple (even after base change).

So your desired situation cannot occur if $B$ is infinite dimensional over $k$.

If you drop the projective hypothesis you could take $B$ a finite direct product of simple $k$-algebras at least one of which is infinite dimensional and take $A=B$ and $A$ will be a finite direct sum of simple $B$-bimodules. You can even make $B$ finitely presented as a $k$-algebra.

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  • $\begingroup$ Ok, so now I see the issue. I meant projective as a left module, and projective as a right module, not projective as a bimodule. But as I wrote it indicates projective as a bimodule. But it's too late to change to change it now, so thanks a lot for the answer! $\endgroup$ – Boris Henriques Jun 13 at 21:50
  • $\begingroup$ Maybe you should ask a question about the specific algebras you are interested in. You want A a direct sum of simple B-bimodules and B projective as a left and right B-module? $\endgroup$ – Benjamin Steinberg Jun 13 at 23:12
  • $\begingroup$ I meant A projective as a left and right B-module. $\endgroup$ – Benjamin Steinberg Jun 14 at 0:35
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    $\begingroup$ @BugsBunny I believe the issue with your counterexample is writing A as a direct sum of projective simple B-bimodules. I dont think your B is a projective B-bimodules. There is a big difference between projective as a one sided module and as a bimodule. If you take $B\otimes_{\mathbb Q} B$ it contains $\mathbb C\otimes_{\mathbb Q} \mathbb C$ as a summand which is some fairly huge and complicated commutative algebra which is uncountably dimension over $\mathbb C$. Is it semisimple? $\endgroup$ – Benjamin Steinberg Jun 15 at 14:56
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    $\begingroup$ The OP wasn't clear what kind of bimodules are allowed. Since he is taking about k-algebras I interpreted bimodules as $B^e$-modules. I am not convinced of you take $k=\mathbb Q$ and $B=\mathbb C=A$ that $B$ is a projective $\mathbb C\otimes_Q \mathbb C=B^e$-module. But since projectivity depends on the category of bimodules the OP should have been clearer. $\endgroup$ – Benjamin Steinberg Jun 15 at 15:00
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No. Take any $A$ and take the ground field $k=k1_A$ as $B$.

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    $\begingroup$ For latecomers, please note that this answered the original version of the question which has since changed. $\endgroup$ – Benjamin Steinberg Jun 13 at 20:43
  • $\begingroup$ It did not change that much :-)) $\endgroup$ – Bugs Bunny Jun 15 at 14:17

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