[Dummit and Foote, Exercises 10.3, problem 13] Suppose $R$ is a commutative ring with $1$. Suppose $F$ is a free $R$-module of finite rank. Then show that $\operatorname{Hom}_R(F,R) \cong F$.
Suppose the basis of $F$ is $\{a_1, a_2,\dots,a_n\}$. I want to show that $\operatorname{Hom}_R(F,R)$ is a free $R$-module of rank $n$. Suppose we define $\{e_i\mid i=1,2,\dots,n\}\subseteq \operatorname{Hom}_R(F,R)$ such that $e_i(a_j)=1$ when $i=j$, and $e_i(a_j)=0$ when $i\neq j$. I am guessing that any homomorphism from $F$ to $R$ can be written as an element of the linear span of $\{e_i\mid i=1,2,...n\}$, but I am stuck here.
Any help would be appreciated. Thank you.
This is the correct idea. As a hint to help you finish, notice that any $v\in F$ can be written as $v=\sum r_ja_j$, for some elements $r_j\in R$. For each $1\le i \le n$, if we apply $e_i$ to $v$, we see $e_i(v) = r_i$, so we can write $$ v = \sum e_j(v)a_j. $$ Now let $\varphi\colon F\to R$ be an arbitrary homomorphism, and consider $\varphi(v)$...
As a side note, the basis for $\operatorname{Hom}(F,R)$ you have described is commonly called the dual basis for $\operatorname{Hom}(F,R)$ given the basis $\{a_1,\dots,a_n\}$ for $F$, and people will also sometimes use the notation $a_i^*$ for the elements $e_i$ you defined.