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Let $h\colon R\rightarrow S$ be a morphism of commutative rings. We consider the following functors (I am aware that the notations may be different in other contexts):

  1. $h^*$: Scalar extension by means of $h$, i.e. $h^*(M)=M\otimes_RS$;

  2. $h_*$: Scalar restriction by means of $h$;

  3. $\widetilde{h}$: Scalar coextension by means of $h$, i.e., $\widetilde{h}(M)={\rm Hom}_R(S,M)$.

These functor form the adjoint pairs $(h^*,h_*)$ and $(h_*,\widetilde{h})$. Moreover, using some general nonsense one can show that the following statements are equivalent: (i) $h^*$ has a left adjoint; (ii) $\widetilde{h}$ has a right adjoint; (iii) The $R$-module $h_*(S)$ is projective and of finite type.

So, suppose that (i)-(iii) are fulfilled. My question is then:

What are the left adjoint of $h^*$ and the right adjoint of $\widetilde{h}$?

ADDENDUM: It is a result of Morita (Theorem 4.1 in K. Morita, Adjoint pairs of functors and Frobenius extensions, Sci. Rep. Tokyo Kyoiku Daigaku Sect. A 9 (1965), 40--71) that (still under the above conditions (i)-(iii)) $$h^*\cong\widetilde{h}\quad\Longleftrightarrow\quad\widetilde{h}(R)\cong S.$$ Therefore, it remains to consider situations where $\widetilde{h}(R)$ and $S$ are not isomorphic.

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  • $\begingroup$ Although I am only interested in the commutative case, the question can naturally be studied in a non-commutative setting. So, I have added the tag ra.rings-and-algebras. $\endgroup$
    – Fred Rohrer
    Commented Jun 4, 2018 at 7:12
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    $\begingroup$ I think $\widetilde{h}(M)$ should be $\text{Hom}_R(S,M)$, not $\text{Hom}_R(M,S)$? $\endgroup$
    – Jeremy Rickard
    Commented Jun 4, 2018 at 8:39

1 Answer 1

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If $X$ is an $R$-module, there is a natural map $M\otimes_RX\to\text{Hom}_R\left(\text{Hom}_R(X,R),M\right)$ given by $m\otimes x\mapsto[\varphi\mapsto m\varphi(x)]$ that is easily checked to be an isomorphism when $X=R$, and hence (by additivity) when $X$ is a finitely generated projective.

So assuming (i)-(iii), $h^*(M)=M\otimes_RS\cong\text{Hom}_R\left(\text{Hom}_R(S,R),M\right)$. This is an isomorphism of $S$-modules, since for $x,s\in S$ and $m\in M$, $m\otimes xs\mapsto[\varphi\mapsto m\varphi(xs)$. Thus $h^*$ has left adjoint $N\mapsto N\otimes_S\text{Hom}_R(S,R)$.

Similarly, there is a natural map $M\otimes_R\text{Hom}_R(X,R)\to\text{Hom}_R(X,M)$ given by $m\otimes\vartheta\mapsto[x\mapsto m\vartheta(x)]$ that is an isomorphism for $X=R$ and hence for $X$ a finitely generated projective.

So assuming (i)-(iii), $\widetilde{h}(M)=\text{Hom}_R(S,M)\cong M\otimes_R\text{Hom}_R(S,R)$. Again, this is an isomorphism of $S$-modules, since for $m\in M$, $\vartheta\in\text{Hom}_R(S,R)$ and $x,s\in S$, $(\vartheta s)(x)=\vartheta(sx)$, so the isomorphism (in the reverse direction) maps $m\otimes(\vartheta s)\mapsto [x\mapsto m\vartheta(sx)]$. Thus $\widetilde{h}$ has right adjoint $N\mapsto\text{Hom}_S\left(\text{Hom}_R(S,R),N\right)$.

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  • $\begingroup$ Dear Jeremy, thank you very much for your answer. $\endgroup$
    – Fred Rohrer
    Commented Jun 4, 2018 at 9:21
  • $\begingroup$ Hm, going through the details this is not completely clear to me. For example, the second paragraph seems to show only that $h_*(\widetilde{h}(R))\otimes_R\bullet$ is left adjoint to $h_*(h^*(\bullet))$. May I ask you to give a bit more details (for example, insert $h_*$ everywhere it occurs)? Or do you know a reference where this is completely spelled out? $\endgroup$
    – Fred Rohrer
    Commented Jun 7, 2018 at 7:41
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    $\begingroup$ @FredRohrer By naturality,(if you take my $X$ to be an $R$-$T$-bimodule for some ring $T$ then the maps in the first and third paragraphs are $T$-module homomorphisms: then take $X=S$ considered as an $R$-$S$-bimodule), the isomorphisms commute with the action of $S$ on itself and so are $S$-module isomorphisms, not just $R$-module isomorphisms. I'll add some details when I have time. $\endgroup$
    – Jeremy Rickard
    Commented Jun 7, 2018 at 7:59
  • $\begingroup$ Dear Jeremy, thanks for your explanations. Unfortunately, I am still confused. How do you conclude at the end of the second paragraph that $h^*$ has a left adjoint? I think the Hom-tensor adjunction in this context is (for an arbitrary $S$-module $M$) given by $(M\otimes_R\bullet,{\rm Hom}_R(h_*(M),h_*(\bullet)))$ (where the first functor goes from $R$-modules to $S$-modules). $\endgroup$
    – Fred Rohrer
    Commented Jun 7, 2018 at 18:16
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    $\begingroup$ @FredRohrer For any $S$-$R$-bimodule $T$, the functor $-\otimes_ST$ from $S$-modules to $R$-modules is left adjoint to the functor $\text{Hom}_R(T,-)$ from $R$-modules to $S$-modules. Here $T=\text{Hom}_R(S,R)$. $\endgroup$
    – Jeremy Rickard
    Commented Jun 7, 2018 at 19:01

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