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> **[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 $\operatorname{Hom}\_R(F,R) \cong F$.
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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.