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?

$\begingroup$ Depending on what you want this for, this may be circular, but: by EilenbergWatts 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. $\endgroup$– Qiaochu YuanFeb 13, 2014 at 22:21

$\begingroup$ 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) 2category 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). $\endgroup$– GeoffreyFeb 13, 2014 at 22:42
1 Answer
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 tensorhom 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 $kA^e$ 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.