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Let $\phi:(R,m,k) \rightarrow (S,n,l)$ be a morphism of local Noetherian rings. Let $M$ be a finite $R$-module and $N$ a finite $S$-module that is flat over $R$.

Question: Why is it true that $Hom_R(k,M) \otimes_R N \cong Hom_S(k \otimes_R S, M \otimes_R N)$?

Remark 1: I am aware of the result that $Hom_A(M,N) \otimes B \cong Hom_B(M \otimes B, N \otimes B)$, whenever $M,N$ are $A$-modules, $M$ has a finite presentation and $B$ is a flat $A$-algebra. However, the proof of this result does not seem to apply to the question at hand.

Remark 2: By identifying $Hom_R(k,M)$ with the elements of $M$ that are annihilated by $m$, and similarly identifying $Hom_S(k \otimes_R S, M \otimes_R N)$ with the elements of $M \otimes N$ that are annihilated by $mS$ i could show that $Hom_R(k,M) \otimes_R N \subset Hom_S(k \otimes_R S, M \otimes_R N)$.

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Ok, i figured it out:

Theorem 1 (Rotman, Introduction to Homological Algebra, Thm 4.86): Let $L$ be flat $R$-module, $E$ finitely-presented $R$-module and $M$ any $R$-module. Then $L \otimes_R Hom_R(E,M) \cong Hom_R(E,M \otimes L)$.

Theorem 2 (Rotman, Introduction to Homological Algebra, Thm 4.85): Let $M,E$ be $R$-modules, $E$, finitely presented, $S$ an $R$-algebra. Then $Hom_R(E,M) \cong Hom_S(E \otimes_R S, M)$.

Apply Theorem 1 with $E=k$, $L=N$ to get $N \otimes_R Hom_R(k,M) \cong Hom_R(k,M \otimes N)$. Now apply Theorem 2 to $Hom_R(k,M \otimes N)$ to get $Hom_R(k,M \otimes N) \cong Hom_S(k \otimes S,M \otimes N)$.

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