# Set of all R-homomorphism is finitely generated

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:

$$M \cong R^{r} \oplus R/\left \oplus R/\left \oplus \cdots \oplus R/\left$$ $$N \cong R^{s} \oplus R/\left \oplus R/\left \oplus \cdots \oplus R/\left$$

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 $$R_{a_0} := R^r$$, $$R_{a_i} := R/\left$$, $$R_{b_0} := R^s$$ and $$R_{b_j} := R/\left$$. 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.

• It is correct for me – Federico Fallucca Feb 19 '19 at 15:15

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 $$R^m \rightarrow M\rightarrow 0,$$ $$R^n \rightarrow N\rightarrow 0.$$