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.

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