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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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    http://math.stackexchange.com/questions/8996/hom-of-finitely-generated-modules-over-a-noetherian-ring could help.2012-11-11

1 Answers 1

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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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    $\operatorname{Hom}_R(R/I,R)\simeq (0:_RI)=(x_1,\dots,x_n,\dots)$.2015-07-26