Cardinality of minimal generating set of a module is constant

Question

Let $R$ be a commutative ring with unity and $M$ be a finitely presented module over $R$. Then how to show that for any minimal generating set $S$, the cardinality is same?

Edit: Thanks to Martin to bring into notice about $\mathbb{Z}$ case. I guess I should restrict the statement only to the modules over polynomial rings over a field. So the modified statement:

Let $R$ be a polynomial ring over a field and $M$ be a finitely presented module over $R$. Then how to show that for any minimal generating set $S$, the cardinality is same?

Answer 1

This is not true. For example, $\{1\}$ and $\{2,3\}$ are minimal generating sets of the $\mathbb{Z}$-module $\mathbb{Z}$.

Answer 2

The same idea for $\mathbb{Z}$ extends to a polynomial ring over a field: for the ideal $(x) \subseteq k[x]$, $\{x\}$ and $\{x^2, x + x^2\}$ are both minimal generating sets.

In general, given a generating set for an ideal $I$, say $I = (a_1, \ldots, a_n)$, one cannot conclude that $I$ can be generated by a proper subset of the $a_i$, even if $I$ is known to be principal (and even if the ring is a PID).