Question. Let $R$ be a ring, $\mathfrak{p}$ a prime, $M$ a finitely-generated $R$-module, and $N$ any $R$-module. Is the natural map $$\textrm{Hom}_R(M, N)_\mathfrak{p} \to \textrm{Hom}_{R_\mathfrak{p}}(M_\mathfrak{p}, N_\mathfrak{p})$$ an isomorphism (of $R_\mathfrak{p}$-modules)?
I can prove this in the case when $M$ is finitely presented: indeed, let $S$ be any flat $R$-algebra, and let $$R^m \to R^n \to M \to 0$$ be a right-exact sequence; tensoring with $S$ gives another right-exact sequence $$S^m \to S^n \to M \mathbin{\otimes_R} S \to 0$$ and applying hom functors, we get left-exact sequences $$0 \to \textrm{Hom}_R(M, N) \to N^n \to N^m$$ $$0 \to \textrm{Hom}_S(M \mathbin{\otimes_R} S, N \mathbin{\otimes_R} S) \to (N \mathbin{\otimes_R} S)^n \to (N \mathbin{\otimes_R} S)^m$$ and $S$ is flat, so tensoring the first sequence yields $$0 \to \textrm{Hom}_R(M, N) \mathbin{\otimes_R} S \to (N \mathbin{\otimes_R} S)^n \to (N \mathbin{\otimes_R} S)^m$$ but extending the sequences by $0$ to the left, and putting in vertical maps between the last two, we conclude that $\textrm{Hom}_R(M, N) \mathbin{\otimes_R} S \cong \textrm{Hom}_S(M \mathbin{\otimes_R} S, N \mathbin{\otimes_R} S)$ by the five lemma (modulo checking commutativity of diagrams). This is essentially the proof Eisenbud gives [Commutative Algebra, Prop. 2.10].
The trouble with extending it to a proof for finitely-generated modules is that we have to replace $R^m$ with a potentially arbitrary submodule $K$ of $R^n$, and that may not be free or even finitely-generated without some additional assumptions on $R$. I can't see an abstract nonsense proof of the claim, but I admit I haven't tried a bare-hands proof. However, is the claim even true?