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!


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