# Hom of finitely generated modules over a noetherian ring

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!

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The argument "Now, since R is commutative (or since R is noetherian), a finitely generated R-module is a quotient of R^n for some n." is quite misleading / wrong. – Martin Brandenburg Jul 17 '13 at 8:52

## 1 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 $\mathrm{Hom}_R(A,B) = \left\{f \in \mathrm{Hom}_R(R^n,B)\ |\ I \subset \ker f \right\}$.

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By the universal property of quotients of modules, I agree on the reverse inclusion of your last sentence. But I didn't know the direct inclusion held (and that's the one we care about, anyway), i.e.: Every morphism $\frac{R^n}{I} \to B$ comes from a map $R^n \to B$ that annihilates $I$...? – lentic catachresis Nov 5 '10 at 1:56
Of course. If you have $\varphi: \frac{R^n}{I} \to B$, define $f: R^n \to B$ as $f(x)=\varphi([x])$. It is well defined, etc. How silly of me, I didn't have this simple fact clear in my mind, in fact I had even thought it would be false. Somehow in my mind I had thought that the map would be ill-defined. Ah well. Thank you for your quick, concise and clear answer. – lentic catachresis Nov 5 '10 at 2:08
@user6495: Indeed, and I like that point of view better! It's cleaner to apply exactness arguments than to mess with elements :) Thanks. – lentic catachresis Apr 20 '12 at 10:51