Is $\mathrm{End}(A) \cong A \otimes_k A^*$ for module $A$ over $k \in \mathsf{Comm}$ Let $k$ be a commutative ring, and $A$ be a $k$-module. Does $\mathrm{End}(A) \cong A \otimes_k A^*$ hold in general? It holds for finite dimensional vector spaces, but I'm not sure how to construct an element of $A \otimes_k A^*$ that corresponds to $\mathrm{id}_A$ in general, without picking a basis.
 A: Notice that $\operatorname{End}(A)$ is never zero, unless $A$ is zero: the identity map is then a not-zero element. So one way to break the isomorphism is to pick a $k$ and $A$ so that the right hand side is zero. One way to do that is to have $A^*$ zero.
For example: let $k=\mathbb Z$ and let $A=\mathbb Z/2\mathbb Z$. As there is no linear map $\mathbb Z/2\mathbb Z\to\mathbb Z$, we have $A^*=0$.
The isomorphism does work when $A$ is a finitely generated projective, though.
Later. This is treated, for example, in F. W. Anderson and K. R. Fuller's Rings and categories of modules, where it is Proposition 20.10. The general statement is:

Let $R$, $S$ be two rings and let ${}_SP$, ${}_SU_T$ ${}_TN$ be modules. Then there is a $\mathbb Z$-linear map $$\phi_{P,T,N}:\hom_S(P,U)\otimes_TN\to\hom_S(P,U\otimes_TN)$$ which is natural in $P$, $U$ and $N$ (and in $R$ and $S$, in the appropriate sense!) which is an isomorphism if $P$ is a finitely generated projective. The map $\phi$ is uniquely determined by the condition that $$\phi_{P,T,N}(f\otimes n)(p)=f(p)\otimes n$$ for all $f\in\hom_S(P,U)$, $n\in N$ and $p\in P$.

You obtain the map in the question as the inverse of what you get from this statement by taking $S=T=k$, $P=N=A$ and $U=k$. The proof of the statement is easy. One first checks that for each choice of $P$, $U$, $N$ there is indeed a map $_{P,T,N}$ satisfying the condition mentioned in the last sentence, and that it is in fact natural; this is straightforward but a bit boring. If now $P$ is finitely generated and projective, there exists a ${}_SP'$ such that $P\oplus P'$ is finitely generated and free, isomorphic to $S^n$ for some $n\geq0$. Since the map $\phi$ is a natural transformation between additive functors of $P$, so show that $\phi_{P,T,N}$ is an isomorphism it is enough to show that $\phi_{P\oplus P',T,N}$, which is conjugate to $\phi_{S^n,T,N}$. Again, additivity implies that to show that $\phi_{S^n,T,N}$ is an isomorphism it is enough to show that the map $\phi_{S,T,N}$ is an isomorphism. And the latter is trivial.
