Isomorphism of two $\operatorname{Hom}$ modules

Question

Let $R$ be a ring (associative, commutative, with unity) and $I\subset R$ is an ideal. Let $M$ be an $R/I$-module and $N$ an $R$-module. Is it true that $$\operatorname{Hom}_R(M,N)\cong \operatorname{Hom}_{R/I}(M,\operatorname{Hom}(R/I,N))\ ?$$

I've tried to use some isomorphisms like $\operatorname{Hom}(R,N)\cong N$ but this doesn't help.

Thanks.

Answer 1

By viewing $M$ as an $R$-module, you are in fact considering the $R$-module $M\otimes_{R/I}(R/I)$. Since the functor $?\otimes_{R/I}(R/I)$ is left adjoint to the functor $\operatorname{Hom}_{R}(R/I, ?)$, you get the isomorphism $$ \operatorname{Hom}_{R}(M\otimes_{R/I}(R/I), N) \cong \operatorname{Hom}_{R/I}(M, \operatorname{Hom}_{R}(R/I, N)).$$

Answer 2

It is very easy even from an more elementary point of view:

We have $\operatorname{Hom}_R(R/I,N) = \{n \in N \mid In=0 \} =: N[I]$ by the fundamental homomorphism theorem.

Now consider $f \in \operatorname{Hom}_R(M,N)$. We have $If(M) = f(IM) = f(0)=0$, because $M$ is a $R/I$-module. This shows $f(M) \subset N[I]$, hence $f: M \to N$ is actually a map $M \to N[I]$, thus is an element of $\operatorname{Hom}_{R/I}(M,N[I]) = \operatorname{Hom}_{R/I}(M,\operatorname{Hom}_R(R/I,N))$.