# Is this true: $\operatorname{Hom}_R(S,R) \otimes_S P \cong \operatorname{Hom}_R(P,R)$?

Let $$R \subset S$$ be a finite free ring extension where $$R$$ is a PID and $$S$$ a one-dimensional reduced ring. Let $$P$$ be a minimal prime ideal of $$S$$ such that $$S/P$$ is also finite free over $$R$$.

For every $$S$$-module $$M$$ we may consider $$\DeclareMathOperator{\Hom}{Hom}\Hom_R(M,R)$$ as an $$S$$-module via the multiplication into the argument.

My question is: Is it true that $$\Hom_R(P,R) \cong P \otimes_S \Hom_R(S,R) \quad \text{ as }S\text{-modules?}$$

To give a bit of a background: I originally wanted to show that

$$\Hom_R(S,R) \otimes_S S/P \cong \Hom_R(S/P,R)$$ as $$S/P$$-modules.

This is the commutative algebra translation of dualizing commutes with restriction to irreducible component where dualizing means applying $$\Hom_R(\cdot,R)$$ and regard it as an $$S$$-module. Of course, restricting to an irreducible component is locally given by tensoring with $$S/P$$.

To prove the original statement I proceeded as follows: We have the canonical sequence $$0 \to P \to S \to S/P \to 0$$ which is exact as a sequence of $$S$$-modules and as a sequence of $$R$$-modules. Since $$S/P$$ is also finite free over $$R$$, the sequence (of $$R$$-modules) splits and we obtain an isomorphism of $$R$$-modules $$S \cong P \oplus S/P$$. This provides $$\Hom_R(S,R) \cong \Hom_R(P,R) \oplus \Hom_R(S/P,R) \quad \text{ as }R\text{-modules.}$$ Then the sequence of $$R$$-modules $$0 \to \Hom_R(S/P,R) \to \Hom_R(S,R) \to \Hom_R(P,R) \to 0$$ is exact. But all involved maps are also $$S$$-linear and hence we obtain $$\Hom_R(S,R) \cong \Hom_R(P,R) \oplus \Hom_R(S/P,R) \quad \text{ as }S\text{-modules.}$$ Now we may tensor the above equality with $$S/P$$ to obtain $$\Hom_R(S,R) \otimes_{S} S/P \cong \underbrace{\Hom_R(P,R) \otimes_{S} S/P}_{(\star)} \oplus \Hom_R(S/P,R)\otimes_{S} S/P$$ where the latter of the RHS is isomorphic to $$\Hom_R(S/P,R)$$ regarded as an $$S/P$$-module. Thus now I hope that $$(\star) \cong 0$$. To show this, it is sufficient to prove that $$\Hom_R(P,R) \cong P \otimes_S \Hom_R(S,R)$$ as $$S$$-modules, hence my question.

What I tried: We can define a homomorphism of $$S$$-modules via $$P \otimes_S \Hom_R(S,R) \to \Hom_R(P,R),\quad p \otimes \varphi \mapsto \varphi \circ \psi_p$$ where $$\psi_p: S \to S$$, $$s \mapsto ps$$ is the multiplication by $$p$$ map. But I don't see how this is an isomorphism.

If you find a counter-example, I am also satisfied. Thank you in advance!