$\varinjlim\operatorname{Hom}_R(N,M_i) = \operatorname{Hom}_R(N, \varinjlim M_i)$

Question

Show that $\varinjlim \operatorname{Hom}_R(N,M_i) = \operatorname{Hom}_R(N, \varinjlim M_i)$ is true when $N$ is finitely generated and $R$ is noetherian. Do you think the noetherian condition is helpful?

Thanks for help.

Answer 1

It can be seen easily that there is a natural map $$\varinjlim \operatorname{Hom}(N, M_i) \to \operatorname{Hom}(N,\varinjlim M_i),$$ and this is equivalence when $N\cong R^n, n\in \mathbb{N}.$

As $N$ is f.g. and $R$ is noetherian, one can write $N$ as a quotient of a finite free $R$ module by a f.g. submodule. So there exists an exact sequence: $$R^m\to R^n \to N\to 0.$$ Now composition of this sequence with two functors $\varinjlim \operatorname{Hom}(-, M_i)$ and $\operatorname{Hom}(-,\varinjlim M_i)$ gives two exact sequences. The above natural map between this two sequences is equivalence is zeroth, second and third place and thus should be equivalence also for the first place which gives the desired equivalence for $N$.