# An algebra which is a direct sum of simple sub-bimodules over a subalgebra

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.

• It is B that must be semisimple and I guess separable – Benjamin Steinberg Jun 13 at 16:33
• Ah ok, so to get this straight: From the assumptions above on $A$ and $B$, we can conclude that $B$ is a semisimple ring? – Boris Henriques Jun 13 at 16:35
• 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. – Benjamin Steinberg Jun 13 at 16:42
• Great! Please put this as an answer and I will accept it! – Boris Henriques Jun 13 at 16:51
• I probably won't have a chance to write a detailed answer until later. – Benjamin Steinberg Jun 13 at 17:06

@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.

• 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! – Boris Henriques Jun 13 at 21:50
• 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? – Benjamin Steinberg Jun 13 at 23:12
• I meant A projective as a left and right B-module. – Benjamin Steinberg Jun 14 at 0:35
• @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? – Benjamin Steinberg Jun 15 at 14:56
• 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. – Benjamin Steinberg Jun 15 at 15:00

No. Take any $$A$$ and take the ground field $$k=k1_A$$ as $$B$$.

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