The following is an exercise from Algebra by Dummit and Foote Sec 10.3:
Let $R$ be a commutative ring and let $F$ be a free $R$-module of finite rank. Prove the following isomorphism of $R$-modules: $\mathrm{Hom}_{R}(F, R) \cong F$.
What I have so far is:
Let $A=\{a_1, \dots, a_{n}\} \subseteq F$ be a basis over $R$ for $F$. That is, $F \cong \bigoplus_{i=1}^{n} Ra_{i}$ and any $0 \neq x \in F$ has a unique representation as $$
x = r_1a_1+\cdots+r_{n}a_{n}.
$$
Using the universal property given in Theorem 6 of the text:
there exists a unique $\Phi \colon F \to \mathrm{Hom}_{R}(F, R)$ such that $\Phi(a_{i})=\varphi(a_{i})$ for all $1 \le i \le n$ and $\Phi \circ \mathrm{inclusion} = \varphi$.
We now define
\begin{align*}
\Psi\colon \mathrm{Hom}_{R}(F, R) &\to F\\
\end{align*}
Problem is I'm not sure how to define $\Psi$ so that $\Psi \circ \Phi$ is the identity.
Best Answer
First, notice the analogy between the isomorphism $V\cong{V^*=Hom_K(V,K)}$ of a $K$-vector space with its dual, and the isomorphism $F\cong{Hom_R(F,R)}$ we are looking for.
As you stated, we consider $A=\{a_1,\dots,a_n\}$ an $R$-base of F. Observe that $Hom_R(F,R)$ is also a free R-module of finite rank, and in particular $rk_R(Hom_R(F,R))=rk_R(F)=n.$ Then, we consider the analogue of a dual basis, meaning $\{a^i\}_{i=0,\dots,n}$, with $a^i : F\longrightarrow{R}$ homomorphism, defined as $a^i(a_j)=\delta_{ij}$ for all $j=1,\dots,n$, where $\delta_{ij}$ is Kronecker's delta.
Finally, consider $$\phi : F\longrightarrow{Hom_R(F,R)} \space\text{ such that }\space\phi(a_i)=a^i$$ Now, we can extend $\phi$ by linearity, and you can check the injectivity. Thus, $\phi$ is an isomorphism because it's injective between free R-modules of the same rank.