# Finitely generated modules [duplicate]

Suppose that $R$ is a commutative ring and $M$ and $N$ are finitely generated $R-$modules. What we can say about $Hom_R (M,N)$? is it a finitely generated $R$-module?

• – wckronholm Dec 5 '14 at 5:53
• So If $R$ is a commutative noetherian ring, then $Hom (M,N)$ is finitely generated. but if $R$ is not noetherian, is $Hom (M,N)$ finitely generated? – A.B. Dec 5 '14 at 5:57
• It is already not necessarily the case that the dual of a finitely generated module is finitely generated: see math.stackexchange.com/questions/392620/… . – Qiaochu Yuan Dec 5 '14 at 6:41

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.