Given a commutative ring $R$, I want to show
If $R^n$ is generated by $\{e_1, \cdots e_m\}$ as an $R$-module, then $m \geq n$.
This follows from the standard result that if there exists a surjection from $R^m \rightarrow R^n$, then $m\geq n$. Then standard proof is tensoring with $R/\mathfrak{m}$ which gives a surjection of vector spaces $(R/\mathfrak{m})^m \rightarrow R/\mathfrak{m})^n$.
Other argument will need to use Cayley-Hamilton.
Like a vector space $R^n$ has a free basis $\{k_1, \cdots, k_n\}$. What is the main obstacle that there cannot be a simple argument like in vector spaces?
Edit: For vector space, the argument I had in mind is using basis. Given $\mathbb F^n$, we know any basis has $n$ elements, but if $n>m$, and some set $\{k_1, \cdots k_m\}$ spans $\mathbb F^n$, then this is a contradiction. Now any (free) basis of $R^n$ have $n$ elements, if there exists a small set which spans $R^n$, isn't this a contradiction?