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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 $k-Mod$.

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    $\begingroup$ 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
    Commented May 23, 2019 at 0:35
  • $\begingroup$ I have written algebra in the question, just not the $k$s. $\endgroup$
    – AMaths
    Commented May 24, 2019 at 7:36
  • $\begingroup$ Well, "algebra" doesn't always mean algebra over a field. $\endgroup$
    – Alex Kruckman
    Commented May 24, 2019 at 18:00
  • $\begingroup$ Technically Very true, in my mind it always is, Thank you for pointing that out! $\endgroup$
    – AMaths
    Commented May 24, 2019 at 19:23
  • $\begingroup$ I beleive marco's answer still holds thou $\endgroup$
    – AMaths
    Commented May 24, 2019 at 19:24

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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 $(r\cdot f)(x)=f(xr)$.

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  • $\begingroup$ Thank you Marco, I thought Hom(R,-) would work but couldn’t think of a good reason why. $\endgroup$
    – AMaths
    Commented May 21, 2019 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
    Commented May 21, 2019 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
    Commented May 21, 2019 at 21:30
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    $\begingroup$ I believe the answer to this question is in Prop 31 of projecteuclid.org/download/pdf_1/euclid.pjm/1102868049 $\endgroup$
    – AMaths
    Commented May 21, 2019 at 22:09
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    $\begingroup$ It is equivalent to R being finitely projective as a k module $\endgroup$
    – AMaths
    Commented May 21, 2019 at 22:09

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