This question is related to this other question I have asked some time ago. Let $R$ and $S$ be two rings and let $\phi:R\to S$ be a ring homomorphism.

It is well-known that $\phi$ is an epimorphism in the category of rings if and only if the counit of the adjunction $(-\otimes_R{}_RS_S,-\otimes_S{}_SS_R)$ (where the structures of left or right $R$-module on $S$ are induced using $\phi$), is a natural isomorphism. I usually denote the adjunction described above by $(\phi_!,\phi^*)$, here $\phi^*:Mod(S)\to Mod(R)$ is a restriction of scalars, while $\phi_!:Mod(R)\to Mod(S)$ is extension of scalars.

Now, since $\phi_!$ is left adjoint, we can consider its restriction to the respective categories of finitely presented modules, obtaining a new functor $\Phi:=(\phi_!)_{mod(R)}:mod(R)\to mod(S)$, between the two preadditive categories $mod (R) $ and $mod (S)$. Of course, this new functor induces extension and restriction of scalars: $$\Phi_!:((mod (R))^{op},Ab)\rightleftarrows ((mod (S))^{op},Ab):\Phi^*$$ and we can ask under which conditions the counit $\Phi_!\Phi^*\to id$ is an isomorphism. This is probably the same as saying that $\Phi$ is an epimorphism in the category of preadditive small categories (with additive functors as morphisms).

My question is: Can we characterize those $\phi$ for which $\Phi_!\Phi^*\to id$ is an isomorphism? Is this equivalent to $\phi_!\phi^*\to id$ being an isomorphism, so equivalent to say that $\phi$ is a ring epimorphism, or is it a genuinely stronger condition?

  • $\begingroup$ This might be a silly/easy question, but if $\phi:R\to S$ is a ring epimorphism and $M$ is a finitely presented $S$-module, do you know whether there is always a finitely presented $R$-module $N$ such that $M\cong N\otimes_RS$ (or at least $M$ is a direct summand of $N\otimes_RS$)? It's true when $\phi$ is a surjection. $\endgroup$ – Jeremy Rickard Jul 21 '15 at 13:46
  • $\begingroup$ So you are asking whether $\Phi$ in my question is surjective on objects, provided $\phi$ is an epimorphism. I do not know if this is true but I do not see any easy counterexample right now. $\endgroup$ – Simone Virili Jul 21 '15 at 18:42
  • $\begingroup$ Yes. I wondered if there might be some functor on f.p. $S$-modules that was zero on $N\otimes_RS$ for all f.p. $R$-modules $N$. If that were the case, then it would show that $\Phi_!\Phi^*\to\text{id}$ need not be an isomorphism. $\endgroup$ – Jeremy Rickard Jul 21 '15 at 19:32
  • $\begingroup$ It seems that the answer to my question is "yes" (if it had been "no" then it would have implied the existence of examples where $\Phi_!\Phi^*\to\text{id}$ is not an isomorphism). $\endgroup$ – Jeremy Rickard Jul 22 '15 at 10:23
  • $\begingroup$ Your question of whether $\Phi_!\Phi^*\to\text{id}$ must be an isomorphism is equivalent to asking whether $\Phi^*$ must be fully faithful. It follows from my last comment that it's faithful. $\endgroup$ – Jeremy Rickard Jul 22 '15 at 10:24

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