Isomorphism between Homomorphism rings of rings and Homomorphism rings of localized rings I'm working on an assignment problem and I'm stuck.
Statement of Problem:
Let $R$ be a Noetherian ring, and $S\subset R$ a multiplicatively closed subset.  Show that if $M$ and $N$ are $R$-modules, with one of them finitely generated, then show that
$S^{-1}Hom_{R}(M,N)$ is isomorphic to $Hom_{S^{-1}R}(S^{-1}M, S^{-1}N)$
The "natural isomorphism" that I tried to define (that is, take an element $\frac{f}{s}$ in $S^{-1}Hom_{R}(M,N)$ and map it to the function $\phi$ defined by $\phi(\frac{m}{s}) = \frac{f(m)}{s}$) is not injective ($\phi$ "forgets" about the $s$ i started with in the denominator).
I'm actually really stumped on how to show this.  For starters, the hypothesis of one of $M$ and $N$ being finitely generated throws me off as I'm not sure how that will be involved.  Additionally, I'm not even sure if I should be trying to show it directly even.  On a previous assignment, I proved an isomorphism existed by constructing the mapping and showing it was an isomorphism.  After two pages, the solution was two lines using an exact sequence. :P
Any advice?
 A: Your putative natural isomorphism isn't well-defined. If I let $\theta$ denote your homomorphism, then observe these two examples:
$$ \theta(f/s)(m/s) = f(m)/s $$
$$ \theta(g/1)(n/1) = g(n)/1 $$
Now, if you let $f = sg$ and $m = sn$, the first equation simplifies to
$$ \theta(g/1)(n/1) = f(n) = s g(n)$$
What you're missing, I think, is the idea to have division work inversely to multiplication. $\theta(sf/1) = s \theta(f/1)$, so conversely, you want $\theta(f/s) = (1/s) \theta(f/1)$ 
It would be most natural to let your natural isomorphism have the property that
$$ \theta(f/1)(m/1) = f(m)/1$$
that is, $\theta$ just acts as the identity operator whenever it makes sense to think of it that way. Then, you just fill everything else in by multiplicativity:
$$ \theta(f/s)(m/s') = f(m)/(ss')$$
Hopefully you can take it from here. (Don't forget to verify that $\theta(f/s)$ is a well-defined homomorphism, and also that $\theta$ is a well-defined homomorphism!)
A: You need to define the map $\phi(m/s') = f(m)/ss'$ for it to be well defined (This observation was due to Hurkyl).
Okay here is one part:
Suppose $M$ is finitely generated and let $m_1,...,m_n$ be a finite set of generators of $M$. I will work with the map you defined.
Let $f/s \in S^{-1}Hom_R(M,N)$ such that $\phi(m/s')=f(m)/ss' = 0$ for all $m/s' \in S^{-1}M$. In particular $\phi(m_i/1) = 0 = f(m_i)/s$, for all $i \in \{1,...,n\}$. Thus, for all $i, \exists s_i \in S$ such that $s_if(m_i) = 0$ in $S^{-1}N$. Take $s''=s_1...s_n$. Then $s''f = 0$ in $Hom_R(M,N)$. Thus, $f/s = 0$ in $S^{-1}Hom_R(M,N)$. Thus your map is injective in this case.
Hope this gives you a hint as to how to proceed.
Added Later
Here is injectivity when $N$ is Noetherian. Let $N' = im(f)$. Then $N'$ is finitely generated with finite generators say $n_1,...,n_m$. Now, for all $m/s' \in S^{-1}M, \phi(m/s') = f(m)/ss' = 0$ means that for all $i$, $n_i/s = 0$. Hence, repeating the argument above, one can find an $s'' \in S$ such that $s''im(f) = 0$. But, this means that $s''f = 0$ in $Hom_R(M,N)$. So, $f/s$ is zero in $S^{-1}Hom_R(M,N)$, which proves injectivity again.
