I have come upon with the following problem:
Let $M$ be an $A$-module (assume that $A$ is commutative with identity) which is both free and finitely-generated. My question is whether there is an elementary way to prove that $M$ is isomorphic to the free finitely-generated $A$-module $A^{n}$, for some $n\in\mathbb{Z}_{\geq 0}$.
Here is a proof using tensor-products: First, since $M$ is free, we have an isomorphism $M\cong \bigoplus_{I}A$ for a set $I$. Then, since it is finitely-generated, say by $x_{1},\dots,x_{n}$, we get a surjective map $\phi:A^{n}\twoheadrightarrow M\cong \bigoplus_{I}A$ (sending each basis vector $e_{i}$ to the generator $x_{i}$). Let then $\mathfrak{m}\subseteq A$ be maximal ideal and consider the field $k=A/\mathfrak{m}$ which is an $A$-module. Then, since tensoring is right-exact, we have that the induced map $\phi\otimes\mathrm{id}_{k}:A^{n}\otimes_{A}k\to (\bigoplus_{I}A)\otimes_{A}k$ is surjective. Now, as tensor product commutes (distributes) naturally with direct sums we have a commutative square: $\require{AMScd}$ \begin{CD} A^{n}\otimes_{A}k @>{\phi\otimes\mathrm{id}_{k}}>> \left(\bigoplus_{I}A\right)\otimes_{A}k\\ @VVV @VVV\\ k^{n} @>{\psi}>> \bigoplus_{I}k \end{CD} where the vertical arrows are isomorphisms. Thus, $\psi:k^{n}\to \bigoplus_{I}k$ is a surjective map of $k$-vector spaces and so $n\geq |I|$, as desired.
I am wondering though if there is a more elementary proof, e.g. by using the basis explicitly.