Localization and dual commute

Question

Let $A$ be a noetherian ring, and let $S\subset A$ a multiplicative set. Given a finitely generated $A$-module $M,$ let us denote with $W^\wedge=\operatorname{Hom}(W,A)$ the dual module. Is it true that $$S^{-1}W^\wedge\simeq\left(S^{-1}W\right)^\wedge$$

Answer 1

If $M$ is a finitely presented $R$-module, $N$ any $R$-module, and $S$ a multiplicatively closed subset of $R$, then $\operatorname{Hom}$ commutes with localization in the sense that $$ S^{-1} \operatorname{Hom}_R(M, N) \cong \operatorname{Hom}_{S^{-1}R}(S^{-1}M, S^{-1}N). $$

See this question for a proof.

In your notation, since $R$ is Noetherian and $W$ finitely generated, it follows that $W$ is finitely presented and we have the isomorphism $$ S^{-1} \operatorname{Hom}_R(W, R) \cong \operatorname{Hom}_{S^{-1}R}(S^{-1}W, S^{-1}R). $$

In other words, $$ S^{-1} (W^\wedge) \cong (S^{-1}W)^\wedge. $$