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Assume that $M$ and $N$ are two finitely generated $R$-modules. Then $\operatorname{Hom}_{R}(M,N)$ is a finitely generated $\mathbb Z$-module and/or $R$-module (in this case, assume that $R$ is commutative)?

Note that $R$ is not Noetherian ring. Is there any counterexample (for any cases)?

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$M = N = R = \mathbb{Q}[x_1, x_2, x_3, \ldots]$ has $\textrm{Hom}_R(M, N) = R$, which is certainly not a f.g. $\mathbb{Z}$-module. –  Zhen Lin Nov 10 '12 at 22:57
How can i prove $Hom_{R}(M,N)=R$? –  Deliasaghi Nov 10 '12 at 23:10
@Ali, you would have better chance if you replace $\mathbb Z$ by $R$ and add finite presentation hypothesis. –  user18119 Nov 10 '12 at 23:14
@QiL, Please explain.. –  Deliasaghi Nov 10 '12 at 23:28
@Ali, $\text{Hom}_R(R,N)$ is isomorphic to $N$ (prove this!). Then, if you take $N=R$ you get that $\text{Hom}_R(R,R)$ is isomorphic to $R$, so no chance (in general) to be finitely generated ovez $\mathbb{Z}$. (Take, for example, $R=\mathbb{Q}$.) –  user26857 Nov 10 '12 at 23:51

1 Answer 1

up vote 6 down vote accepted

If $R$ is not noetherian it is not clear (to me) what kind of finiteness conditions on modules would imply that $\operatorname{Hom}_R(M,N)$ is finitely generated. Even for finitely presented modules this property fails. An example is the following: $R=K[X_1,\dots,X_n,\dots]/(X_1,\dots,X_n,\dots)^2$, $M=R/I$, where $I=(x_1)$, and $N=R$.

However, there are some trivial cases when $\operatorname{Hom}_R(M,N)$ is finitely generated, e.g. $M$ finitely generated and projective and $N$ finitely generated.

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If $M$ is finitely-generated and $N$ is noetherian, then $\textrm{Hom}_R (M, N)$ is also finitely-generated (indeed, noetherian). –  Zhen Lin Apr 18 '13 at 20:41
How can we see that Hom(M,N) is not finitely generated in this example? –  Joel R. Sep 28 '13 at 23:30
$\operatorname{Hom}_R(R/I,R)\simeq (0:_RI)=(x_1,\dots,x_n,\dots)$. –  user26857 Jul 26 at 19:19

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