Isomorphism between tensor of module's dual and Hom(M,-)

Question

For an $A$-module $M$, how can I show that the homomorphism

$M^* \bigotimes N \rightarrow \operatorname{Hom}(M,N)$

is an isomorphism when $M=A^k$ or $N=A^k$?

Where $M^*$ denotes the dual of $M$ and $k \geq 0$.

Answer 1

There is a canonical bilinear map $f\colon M^*\times N\to\operatorname{Hom}(M,N)$, defined by $$ f(\xi,y)\colon x\mapsto \xi(x)y $$ This bilinear map defines a homomorphism $\varphi\colon M^*\otimes N\to\operatorname{Hom}(M,N)$ which is generally not an isomorphism. For instance, if $A=\mathbb{Z}$, $M=N=\mathbb{Z}/2\mathbb{Z}$, the domain is the zero module and the codomain is (isomorphic to) $\mathbb{Z}/2\mathbb{Z}$.

However, if $M=A^k$, the map $\varphi$ is indeed an isomorphism, because of the commutative diagram $$\require{AMScd} \begin{CD} (A^k)^*\otimes N @>>> \operatorname{Hom}(A^k,N) \\ @VVV @VVV \\ N^k @>>> N^k \end{CD} $$ where the bottom arrow is the identity and the two vertical isomorphisms are readily written out.

For $N=A^k$, consider that $M^*\otimes A^k\cong (M^*)^k$ and also $\operatorname{Hom}(M,A^k)\cong\operatorname{Hom}(M,A)^k$ (again, easy isomorphisms to write).