Show that this mapping between localized modules is an isomorphism 
Let $R$ be a ring. Let $M$ and $N$ be $R$-modules where $M$ is finitely presented. Then for every multliplicative set $S \subset R$ the canonical mapping
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~Hom_R(M,N) \otimes_R R_S \to Hom_{R_S}(M_s,N_s)$
   is an isomorphism between $R_S$ modules.   

I found this in a texbook and the proof just says 'diagram chase'. I am not even sure what diagram I am supposed to chase. Could someone elaborate this some more?
 A: Let
$$R^{\oplus m} \longrightarrow R^{\oplus n} \longrightarrow M \longrightarrow 0$$
be a finite presentation for $M$. Note that $\operatorname{Hom}_R(-,N) \otimes_R R_S \to \operatorname{Hom}_{R_S}(-_S,N_S)$ is a morphism of contravariant, left-exact functors. We therefore have the commutative diagram
$$\require{AMScd}
\begin{CD}
  0 @>>> \operatorname{Hom}_R(M,N) \otimes_R R_S @>>> \operatorname{Hom}_R(R^{\oplus n},N) \otimes_R R_S @>>> \operatorname{Hom}_R(R^{\oplus m},N) \otimes_R R_S\\
  @. @VVV @VVV @VVV\\
  0 @>>> \operatorname{Hom}_{R_S}(M_S,N_S) @>>> \operatorname{Hom}_{R_S}(R_S^{\oplus n},N_S) @>>> \operatorname{Hom}_{R_S}(R_S^{\oplus m},N_S)
\end{CD}$$
with exact rows.
Now by the universal property of kernels (or just a diagram chase), it suffices to show that the middle and right vertical arrows are isomorphisms. This is clear since
$$\operatorname{Hom}_R(R^{\oplus l},N) \otimes_R R_S \simeq N^{\oplus l} \otimes_R R_S \simeq N^{\oplus l}_S \simeq \operatorname{Hom}_{R_S}(R_S^{\oplus l},N_S)$$
for $l \in \{m,n\}$ (although you have to make sure that this composition is the arrow in the commutative diagram above).
Remark. You only really need that localization $-_S$ is exact in this argument. For a reference, see Bourbaki's Commutative Algebra, Chapter 1, §2, Proposition 11.
