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:



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



Adjoints of scalar extension and scalar coextension
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 $N\mapsto\text{Hom}\_S\left(\text{Hom}\_R(S,R),N\right)$.


