Set of all R-homomorphism is finitely generated

Question

I have the following proposition:

Proposition: If $R$ is a principal ideal domain (PID) and $M$, $N$ two finitely generated (fg) $R-$modules, then $\text{Hom}_R(M,N)$ is finitely generated.

My idea: since $M$, $N$ are fg $R-$modules over a PID, then each one is $R-$isomorphic to a direct sum of cyclic modules:

$$M \cong R^{r} \oplus R/\left<a_1\right> \oplus R/\left<a_2\right> \oplus \cdots \oplus R/\left<a_m\right>$$ $$N \cong R^{s} \oplus R/\left<b_1\right> \oplus R/\left<b_2\right> \oplus \cdots \oplus R/\left<b_n\right>$$

where $a_i,b_j$ are the invarian factors in $R$ and $r,s$ are the rank of $M$ and $N$ respectively. To simplify define $R_{a_0} := R^r$, $R_{a_i} := R/\left<a_i\right>$, $R_{b_0} := R^s$ and $R_{b_j} := R/\left<b_j\right>$. So, $M\cong\sum_{i=0}^m\oplus R_{a_i}$ and $N\cong\sum_{j=0}^{n}\oplus R_{b_j}$. Therefore, we have the following $R-$isomorphism:

$$\text{Hom}_R(M,N) \cong \text{Hom}_R\left(\sum_{i=0}^m\oplus R_{a_i},\sum_{j=0}^n\oplus R_{b_j}\right) \cong \text{Hom}_R\left(\bigoplus_{i=0}^m R_{a_i},\bigoplus_{j=0}^n R_{b_j}\right) \cong \bigoplus_{i,j}\text{Hom}_R(R_{a_i},R_{b_j}) \cong \sum_{i,j}\oplus\text{Hom}_R(R_{a_i},R_{b_j})$$

Here I have to calculate $(r+m)\times(n+s)$ Hom-groups between cosets of $R$ and copies of $R$ to finally get that $\text{Hom}_R(M,N)$ is $R-$isomorphic to a direct sum of cyclic modules and therefore is fg. But, I think there is a more simple way to do this. I'm right? Any hit? Thanks.

Answer 1

A PID is Noetherian. Let us do the proof more in general for a commutative Noetherian ring $R$:

Let $M,N$ be finitely generated modules. Pick $n,m\in\mathbb{N}$ and $K\leq R^m$, $L\leq R^n$ such that $$M\cong R^m/K, N\cong R^n/L.$$ Let us write these identities as exact sequences: $$0\rightarrow K\rightarrow R^m \rightarrow M\rightarrow 0, \,\,\,\,(1)$$ $$0\rightarrow L\rightarrow R^n \rightarrow N\rightarrow 0. \,\,\,\,\,\,\,\, (2)$$

Let us apply the functor Hom$(\cdot, N)$ to (1). Since it is a left-exact contravariant functor and $R$ is commutative, we get the exact sequence of $R$-modules $$0\rightarrow\text{Hom}(M,N)\rightarrow^f\text{Hom}(R^m,N)\rightarrow\text{Hom}(K,N).$$ From this we see that $f$ is injective, so Hom$(M,N)$ is isomorphic to a submodule of Hom$(R^m,N)$; since $R$ is Noetherian, if Hom$(R^m,N)$ were finitely generated then so would be Hom$(M,N)$. In order to prove this, apply now the functor Hom$(R^m,\cdot)$ to (2). Since it is a left covariant functor, we get the exact sequence $$\text{Hom}(R^m,L)\rightarrow\text{Hom}(R^m,R^n)\rightarrow^g\text{Hom}(R^m,N)\rightarrow 0.$$ From this we see that $g$ is surjective, so Hom$(R^m,N)$ is a quotient of Hom$(R^m,R^n)$. If this last $R$-module were finitely generated, then so would be Hom$(R^m,N)$ and our proof would be complete. But Hom$(R^m,R^n)$ is isomorphic to the $R$-module of $n\times m$ matrices over $R$, which is finitely generated (it is actually isomorphic to $R^{nm}$ as $R$-modules). Therefore Hom$(M,N)$ is finitely generated.

Observe that we have only actually used the sequences $$R^m \rightarrow M\rightarrow 0,$$ $$R^n \rightarrow N\rightarrow 0. $$