# Constructing an example s.t. $\operatorname{Hom}_R(M,N)$ is not finitely generated [duplicate]

Let $R$ be a commutative ring and $M$ and $N$ be two finitely generated $R$-modules. I wanna construct an example s.t. $\operatorname{Hom}_R(M,N)$ is not finitely generated.

It's well-known that if $R$ be a Noetherian ring and $M$ and $N$ be two finitely generated $R$-modules then $\operatorname{Hom}_R(M,N)$ is finitely generated. So in this example $R$ must be a Non-Noetherian ring.

• – Jack Schmidt Apr 18 '13 at 20:35

We know that $$\operatorname{Hom}_R(R/I,R/J)\simeq (J:I)/J.$$ If $J=(0)$ we get $$\operatorname{Hom}_R(R/I,R)\simeq (0:I).$$ Now take $R=K[X_1,\dots,X_n,\dots]/(\{X_iX_j:1\le i\leq j\})$ and $I=(x_1,\dots,x_n,\dots)$, where $x_i$ is the residue class of $X_i$. Then $(0:I)=I$ and this ideal is not finitely generated.