MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $S$ and $R$ be two (not necessarily commutative) $k$-algebras for $k$ a field. If I have a $S$-$R$ bimodule $_SM_R$, I can form the functor $_SM_R\otimes_R (-):R\text{Mod} \rightarrow S\text{Mod}$. Similarly, I can form $(-)\otimes_S {_SM_R}$ to get a functor from $\text{Mod}S \rightarrow \text{Mod}R$. If there are left adjoints to these tensor product functors, I want to show that $M$ must be projective and finitely generated over $R$ and $S$ respectively. Does anyone know any sources that might contain such an argument?

share|cite|improve this question
Depending on what you want this for, this may be circular, but: by Eilenberg-Watts the left adjoint must itself be tensor product with some bimodule, so your functor is also a Hom functor, which ought to imply that your bimodule is dualizable. – Qiaochu Yuan Feb 13 '14 at 22:21
Hi, thanks for the reply. I realize that the bimodule must be dualizable; I'm actually using this to extract the criterion for when an algebra A in the (infinity-) 2-category of Algebras Bimodules and Intertwiners is fully dualizable. It should be that it is projective over the enveloping algebra (=> separable <=> semisimple if over a field of char 0). – Geoffrey Feb 13 '14 at 22:42
up vote 1 down vote accepted

I believe I have worked out an argument: We want the functor $F:R\text{Mod}\rightarrow S\text{Mod}$ by $_S M_R \otimes_R (-)$. By tensor-hom duality, this functor is always a left adjoint so in particular it is right exact. If this functor is also a right adjoint, then it must be left exact which means that $M$ is flat as an $S$-module. In the case that I'm interested in, $R$ is the algebra $A^e$ and $M$ is $A$ as a $k-A^e$ bimodule where $k$ is some ground field. $A$ being flat over $A^e$ as a right module is equivalent to $A$ being separable as a $k$-algebra.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.