## If R is a commutative unital ring and F a free unital left R-module of finite rank then Hom(F,R) and F are isomorphic as R-modules

Let $R$ be a commutative ring with 1 and let $F$ be a free left unital $R$-module of finite rank. Prove that $\mathsf{Hom}_R(F,R) \cong_R F$, where $\cong_R$ means “isomorphic as $R$-modules”.

Suppose $F$ is free on the set $\{a_i\}_{i=1}^n$; then every element of $F$ can be written uniquely as $\sum r_i \cdot a_i$ for some $r_i \in R$. Now define $\Phi : \mathsf{Hom}_R(F,R) \rightarrow F$ by $\Phi(\varphi) = \sum \varphi(a_i) \cdot a_i$. We claim that $\Phi$ is an $R$-module isomorphism.

Note that if $\varphi, \psi \in \mathsf{Hom}_R(F,R)$ and $r \in R$, then $\Phi(\varphi + r \cdot \psi) = \sum (\varphi + r \cdot \psi)(a_i) \cdot a_i$ $= (\sum \varphi(a_i) \cdot a_i) + r \cdot (\sum \psi(a_i) \cdot a_i)$ $= \Phi(\varphi) + r \cdot \Phi(\psi)$. Thus $\Phi$ is an $R$-module homomorphism.

Now suppose $\varphi \in \mathsf{ker}\ \Phi$. Then $0 = \Phi(\varphi) = \sum \varphi(a_i) \cdot a_i$. Since $F$ is free on the $a_i$, we have $\varphi(a_i) = 0$ for all $a_i$, and thus $\varphi = 0$. So $\mathsf{ker}\ \Phi = 0$, and thus $\Phi$ is injective.

Finally, let $\sum t_i \cdot a_i \in F$. Define $\theta : F \rightarrow R$ by $\theta(\sum r_i \cdot a_i) = \sum r_it_i$. Evidently, $\theta$ is an $R$-module homomorphism. Moreover, $\theta(a_i) = t_i$, so that $\Phi(\theta) = \sum t_i \cdot a_i$. Hence $\Phi$ is surjective.

Thus $\Phi$ is an $R$-module isomorphism, so that we have $\mathsf{Hom}_R(F,R) \cong_R F$.