R-Module Homomorphisms from Free Modules (10.3.13)

Dummit and Foote Abstract Algebra, section 10.3, exercise 13:

Let $R$ be a commutative ring. Prove $Hom_R(R^n,R)≅R^n$ as $R$-modules.

Proof: Define $ψ : Hom_R(R^n,R) → R^n$ by $φ \mapsto (φ(e_1),...,φ(e_n))$. We have $$ψ(φ_1+rφ_2) = ((φ_1+rφ_2)(e_1),...,(φ_1+rφ_2)(e_n)) =$$$$(φ_1(e_1),...,φ_1(e_n)) + r(φ_2(e_1),...,φ_2(e_n)) = ψ(φ_1)+rψ(φ_2)$$so $ψ$ is a homomorphism of $R$-modules. If $ψ(φ)=(0,...,0)$, then for arbitrary $r^*=(r_1,...,r_n)=r_1e_1+...+r_ne_n∈R^n$ we have $φ(r^*)=r_1φ(e_1)+...+r_nφ(e_n)=0$ so that $φ=0$ and now $ψ$ is injective. Again for arbitrary $r^*∈R^n$, define $φ∈Hom_R(R^n,R)$ by $φ(e_i)=r_i$ and extending linearly, so that $ψ(φ)=r^*$ and now $ψ$ is surjective and an isomorphism of $R$-modules.$~\square$