This is an exercise from Rotman, An Introduction to Homological Algebra, which I've been thinking now and then for a few days and I haven't solved it yet. I've decided to ask here because it is bugging me and I don't have any friends also studying this subject (in case it is of any use, I'm an undergraduate).
Let $R$ be a commutative noetherian ring. If $A$, $B$ are finitely generated $R$-modules, then $\operatorname{Hom}_R(A,B)$ is a finitely generated $R$-module.
Here's what I've thought as of yet (maybe the problem is that I haven't been trying to apply the useful theorems…):
First of all, since $R$ is noetherian, it suffices to inject $\operatorname{Hom}_R(A,B)$ into some finitely generated module (in noetherian rings, submodules of f.g. are f.g.).
Now, since $R$ is commutative (or since $R$ is noetherian), a finitely generated $R$-module is a quotient of $R^n$ for some $n$. Then let us write $A\simeq \frac{R^n}{I}$ and $B\simeq \frac{R^m}{J}$. Now we've got:
$\operatorname{Hom}_R(A,B)\simeq \operatorname{Hom}_R\left(\frac{R^n}{I},\frac{R^m}{J}\right)$.
If I had $\operatorname{Hom}_R\left(\frac{R^n}{I},\frac{R^m}{J}\right) \hookrightarrow \operatorname{Hom}_R(R^n, R^m)$ (this is what I tend to think is the wrong direction, but a interesting question-rising path nevertheless) then it would be over since $\operatorname{Hom}_R(R^n, R^m)\simeq R^{nm}$ which is finitely generated.
So I ask myself: can I see $\frac{R^n}{I}$ as a submodule of $R^n$? Because if I could, then doing the same thing for $R^m$ and passing it to the hom, then it would be over. Now, this would be true if the sequence $$0\to I\hookrightarrow R^n\to \frac{R^n}{I} \to 0$$ split. But this (of course) doesn't always happen. But if I somehow had the third module to be projective, or the first one to be injective, then it would happen.
But why would, for example, be $\frac{R^n}{I}$ be projective?
And this is where I got stuck. I'm sorry if this is overly detailed, but I remember reading that it is always good, when asking about a textbook question, to add what you've thought up to that moment. Also, even if my thoughts don't lead to the solution, I would like to know if they are correct!
Best Answer
You're nearly there: with your notations, $\mathrm{Hom}_R(A,B)$ injects in $\mathrm{Hom}_R(R^n,B) \simeq B^n$, and $B^n$ is finitely generated.
Indeed $\mathrm{Hom}_R(A,B) = \left\{f \in \mathrm{Hom}_R(R^n,B)\ |\ I \subset \ker f \right\}$.