Can u help me show that if $M$ is a finitely generated free module and $A$ a commutative ring then $M\cong\operatorname{Hom}_A(M,A)$?
Using the fact A has invariant dimension property and prove their bases have the same cardinality I tried to create generators for $Hom_A(M,A)$ that I name $e_{vi}(x)$ where its $1_A$ if $v_i=x$ and $0$ otherwise (where all the $v_i$ are the generators of M), but I'm not sure they generate the whole set of homomorphisms. Well any help is appreciated.
Let $\{e_1,\ldots,e_n\}$ generate $M$. By linearity, any $\phi\in\textrm{Hom}_A(M,A)$ is defined by where it sends $\phi(e_i)$ for each $i$. This allows us to construct $$\begin{align} M & \longrightarrow \textrm{Hom}_A(M,A) \\ \sum\lambda_ie_i & \longmapsto \{\phi : e_i \mapsto \lambda_i\}\end{align} $$ with inverse $$\begin{align} \textrm{Hom}_A(M,A)& \longrightarrow M \\ \phi & \longmapsto \sum\phi(e_i)e_i\end{align} $$