Let $A,B$ be two rings such that their categories of bimodules are equivalent: $$A\mathsf{-Bimod} \simeq B\mathsf{-Bimod}$$ What can we say about $A$ and $B$? Are they isomorphic? Are they Morita-equivalent? Probably this has been studied in the literature, so this is primarily a reference request.
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5This is equivalent to the condition that $A \otimes A^{op}$ and $B \otimes B^{op}$ are Morita equivalent. I think a more interesting question is to ask that the equivalence respect monoidal structures. – Qiaochu Yuan Jan 19 '16 at 20:54
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Yes (to both). But I would like to ask this question first :). – Martin Brandenburg Jan 19 '16 at 21:02
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Are bimodules sufficiently different from modules to deserve a tag? An $(A,B)$-bimodule is merely an $A \otimes B^{op}$-module after all... – Najib Idrissi Jan 21 '16 at 08:33
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1The tensor product of $A$-bimodules is not the tensor product over $A \otimes A^{op}$. The tensor product of free $A$-bimodules is usually not free. – Martin Brandenburg Jan 21 '16 at 08:36
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By the way, there are tags for p-groups, finite groups, cyclic groups, ... – Martin Brandenburg Jan 21 '16 at 09:14
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@Najib: Yes. Modules just form a category, but bimodules form the 1-morphisms in a 2-category. – Qiaochu Yuan Jan 22 '16 at 05:48
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Let me work over a field $k$ and talk about bimodules over $k$ and $k$-linear categories (so we ask for an equivalence respecting the $k$-linear structure). Then it does not follow that $A$ and $B$ are Morita equivalent (or isomorphic) over $k$.
The given condition is equivalent to the condition that $A \otimes_k A^{op}$ and $B \otimes_k B^{op}$ are Morita equivalent. If $A, B$ are both central simple algebras over $k$ then both of these algebras are Morita equivalent to $k$, but $A$ and $B$ themselves are Morita equivalent iff they represent the same class in the Brauer group $\text{Br}(k)$. So, for example, we can take the two representatives $\mathbb{R}, \mathbb{H}$ of the Brauer group $\text{Br}(\mathbb{R})$.
On the other hand, if $A$ and $B$ are Morita equivalent, then $\text{Bimod}(A)$ and $\text{Bimod}(B)$ are equivalent even monoidally, because they can be described as the monoidal categories of $k$-linear cocontinuous endofunctors of $\text{Mod}(A)$ and $\text{Mod}(B)$ respectively by Eilenberg-Watts.
Qiaochu Yuan
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3You can get a counterexample over $\mathbb{Z}$ by looking at central simple algebras over $\mathbb{Q}$, since $\otimes_\mathbb{Q}$ coincides with $\otimes_\mathbb{Z}$ for $\mathbb{Q}$-algebras. – Eric Wofsey Jan 20 '16 at 04:03
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In case the field is algebraically closed(or more generally both algebras are split over the field), the result might be true. At least in characteristic 0. – Mare Jan 20 '16 at 12:14
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1If $A$ is a central simple algebra over $k$, isn't the equivalence $U\mapsto U\otimes_kA$ from $\text{Mod}(k)$ to $\text{Bimod}(A)$ always a monoidal equivalence? – Jeremy Rickard Jan 21 '16 at 11:39
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