Let $A$ be a ring and $S$ a multiplicative subset of $A$ such that $1 \in S$. Let $G$ be the forgetful functor from $Mod_{S^{-1}A} \rightarrow Mod_A$. Taking an $S^{-1}A$-module N and consider it as an $A$-module. Let $F: Mod_A \rightarrow Mod_{S^{-1}A} $ be its adjoint functor. It takes an $A$-module $M$ to $S^{-1}M$.
I want to show that $Hom_{S^{-1}A}(FM, N)$ is isomorphic to $Hom_A(M,GN)$ (my goal is to show that $F$ and $G$ are indeed adjoints). Could someone please explain me how to show this isomorphism? Thanks!
There is a natural morphism $\alpha_M\colon M\to S^{-1}M$ (of $A$-modules) defined by $\alpha_M(x)=\frac{x}{1}$; given $f\in\operatorname{Hom}_{S^{-1}A}(S^{-1}M,N)$, consider $f\circ \alpha_M\colon M\to N$, which is a morphism of $A$-modules.
This gives a map $$ \operatorname{Hom}_{S^{-1}A}(FM,N)\to \operatorname{Hom}_{A}(M,GN) $$
Next, if $g\colon M\to GN=N$ is a morphism of $A$-modules, define $$ \hat{g}\colon S^{-1}M\to N $$ by $$ \hat{g}\left(\frac{x}{s}\right)=\frac{g(x)}{s}\qquad (x\in M, s\in S) $$ The only problem is proving that $\hat{g}$ is well defined. If $x/s=y/t$, then there exists $u\in S$ such that $u(tx-sy)=0$; then $$ 0=g(utx-usy)=u(t(g(x)-sg(y)) $$ which amounts to saying that $$ \frac{g(x)}{s}=\frac{g(y)}{t}. $$ Proving that $\hat{g}$ is a morphism of $S^{-1}A$-modules is easy. And it's easy also to see that $\hat{g}\circ\alpha_M=g$. Can you show that $\widehat{f\circ\alpha_M}=f$?
When you'll learn about tensor product, it will be easier.