Let $R$ be a commutative Noetherian ring. Let $P,Q$ be some $R$-modules such that $-\otimes_R P $ and $ Hom_R(Q,-) $ are faithfully exact functors i.e., for any sequence of modules
$A \xrightarrow{f} B \xrightarrow{g} C$ , we have
$A \xrightarrow{f} B \xrightarrow{g} C$ is exact if and only if $A\otimes_RP \xrightarrow{id\otimes_f} B\otimes_R P \xrightarrow{id\otimes g} C\otimes_RP$ if and only if $Hom(Q,A) \xrightarrow{hom(f)} Hom(Q,B) \xrightarrow{hom(g)} Hom(Q,C)$ is exact. Also note that all this if and only if conditions are equivalent to saying : $P$ is faithfully flat and $Q$ is projective and $Hom(Q,X)\ne 0$ for every $R$-module $X$, and such $Q$ is also called faithfully projective.
My question is: Under the above conditions, when can we say that the functors $-\otimes_R P$ and $Hom_R(Q,-)$ are naturally isomorphic ? What are some examples when they are not isomorphic ? Is it true that if they are isomorphic, then $P\cong Q$ ?
Another, much more concrete question : Is it true that for every faithfully flat module $P$, there exists a faithfully projective module $Q$ such that the functors $-\otimes_R P$ and $Hom_R(Q,-)$ are naturally isomorphic ?
(NOTE: Of course if they are isomorphic, then $P\cong Hom_R(Q,R)$ )
Obviously , the functors $-\otimes_R P$ and $Hom_R(Q,-)$ are naturally isomorphic when $P\cong Q$ is free of finite rank. Apart from that I don't know any examples. Also, I would like to see some canonical counter-examples when they are not isomorphic .
Thanks
