I am trying to prove the following statement:
If $R$ is a commutative ring with unity and $M = \langle a_1,\ldots, a_n \rangle$ is a free $R$-module, then $M$ has rank at most $n$ .
My thought was to maybe reduce this down to the case of linear algebra by taking a maximal ideal, but I am a little stuck at the end.
We first prove that the rank is finite. Indeed, since $a_1,\ldots, a_n$ span $M$ and $B$ is a basis, then for each $k = 1, \ldots, n$ we see that $a_k = \sum_{\text{finite}}r_ie_i$ where $e_i$ are basis elements and $r_i \in R$. Thus, $B$ must be finite.
Now, suppose that $|B| = m$. This means that $M \cong R^m$ and there is an isomorphism $\phi: M \to R^m$. We wish to prove that $m \leq n$. Let $I$ be a maximal ideal in $R$. Now
\begin{equation} R^m/IR^m \cong (R/I)^m \end{equation}
(where the right hand side is the $m$-fold cartesian product of $R/I$. Since $\phi$ is an isomorphism and $a_1,a_2,\ldots a_n$ span $M$ it follows that $\phi(a_1),\phi(a_2),\ldots,\phi(a_n)$ span $R^m$. Hence, $\phi(a_1),\phi(a_2),\ldots,\phi(a_n)$ span $R^m/IR^m$ and consequently $(R/I)^m$. Since $I$ is a maximal ideal $R/I$ is a field...
It is here that I am confused. I tried to do a similar argument as in exercise 2 of Dummit and Foote in section 10.3. In this exercise you are asked to show that $R^n \cong R^m$ iff $m = n$. The hint that they give is to take a maximal ideal and then use the result from linear algebra that two finite dimensional vector spaces are isomorphic iff they have the same dimension.
I was thinking a similar strategy would work here. Namely, since the length of any spanning set of a finite dimensional vector space is longer than any basis. But as it stands it looks like $(R/I)^m$ is a $R$-module which means that I cannot invoke any results from linear algebra.
My question: Do you think that there anything else that I can do to make an argument like this work? For reference, I have knowledge of modules up to Dummit and Foote section 10.3. So, I would really appreciate answers that only use the material developed up to that section. This was not an exercise in the book, but it was that I was thinking about.
Thank you