1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ If I have not misunderstood, take $R$ to be a local Noetherian ring, $P$ its completion and $Q=R$. If $R$ is not complete, clearly $Q$ and $P$ are not isomorphic. $\endgroup$ – Mohan May 18 at 2:33
  • 1
    $\begingroup$ A simple example where the functors are not isomorphic is where $P$ and $Q$ are free of different nonzero finite ranks. $\endgroup$ – Jeremy Rickard May 18 at 10:34
  • 1
    $\begingroup$ It’s not exactly a duplicate, but I think that most of your questions are answered here: math.stackexchange.com/questions/2522672/… $\endgroup$ – Jeremy Rickard May 18 at 10:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.