Given any algebra $R,$ when does the forgetful functor $R\text{-}Mod \rightarrow Vec$ have a right adjoint? Does this imply any finiteness conditions on R? Is there a book/paper discussing this?

I've assumed $R$ is $k$ algebra where $k$ is a field. but if $k$ is not a field, and just a commutative ring then Marco's answer should hold up still with replacing $Vec$ by $k-Mod$.

  • 1
    $\begingroup$ I don't understand why this forgetful functor exists, e.g. if $R=\mathbb{Z}$. Are you assuming $R$ is an algebra over some field $k$? And Vec is the category of vector spaces over $k$? $\endgroup$ – Alex Kruckman May 23 '19 at 0:35
  • $\begingroup$ I have written algebra in the question, just not the $k$s. $\endgroup$ – AMaths May 24 '19 at 7:36
  • $\begingroup$ Well, "algebra" doesn't always mean algebra over a field. $\endgroup$ – Alex Kruckman May 24 '19 at 18:00
  • $\begingroup$ Technically Very true, in my mind it always is, Thank you for pointing that out! $\endgroup$ – AMaths May 24 '19 at 19:23
  • $\begingroup$ I beleive marco's answer still holds thou $\endgroup$ – AMaths May 24 '19 at 19:24

Hom is right adjoint to tensor product. Have you tried to write the forgetfull functor as $F(M)=R\otimes_R M$?

I didn't check the details, but I think you can do something like

$$Hom_k(Forget(M),W)=Hom_k(M,W)\cong Hom_k(R\otimes_RM,W)\cong Hom_R(M,Hom_k(R,W))$$ where $M$ is an $R$-module, $W$ is a vector space, and the $R$-structure on $Hom(R,W)$ is from the right structure of $R$, that is $(r\cdot f)(x)=f(xr)$.

| cite | improve this answer | |
  • $\begingroup$ Thank you Marco, I thought Hom(R,-) would work but couldn’t think of a good reason why. $\endgroup$ – AMaths May 21 '19 at 20:33
  • $\begingroup$ Are you aware of a similar construction for the forgetful functor from $Forg : R-Comod \rightarrow Vec$ having a LEFT adjoint? $\endgroup$ – AMaths May 21 '19 at 20:37
  • $\begingroup$ mmm.. that seems more dificult. Now $R$ is a coalgebra, if $M$ is a (right) comodule then $M$ is a (left) $R^*$-module, and $\endgroup$ – Marco Farinati May 21 '19 at 21:30
  • 2
    $\begingroup$ I believe the answer to this question is in Prop 31 of projecteuclid.org/download/pdf_1/euclid.pjm/1102868049 $\endgroup$ – AMaths May 21 '19 at 22:09
  • 2
    $\begingroup$ It is equivalent to R being finitely projective as a k module $\endgroup$ – AMaths May 21 '19 at 22:09

Your Answer

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

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