Possible difference between $\mathbb{Z}$-modules and vector spaces Suppose $G$ is a free abelian groups, i.e. a free $\mathbb{Z}$-module; we have a set $S \subset G $ such that $S$ spans $G$.
Can we conclude that the rank of $G$ as a  $\mathbb{Z}$-module is $ \leq |S| $ as in the vector-space case ? Why ?

Suppose $S$ spans $G$ as $\mathbb{Z}$-module. $G$ is contained in a free $\mathbb{Q}$-module $V$; $S$ spans $V$ over $\mathbb{Q}$, so $\operatorname{rank}(V) \leq |S| $ .
Suppose $B$ is a $\mathbb{Z}$-basis for $G$; then $B$ is linearly independent over $\mathbb{Z} \Rightarrow $ $B$ is obviously linearly independent over $\mathbb{Q} $ $\Rightarrow |B|  \leq \operatorname{rank}(V) \leq |S| $
Observe that $|B|$ is well-defined because $\mathbb{Z}$ is a commutative ring.
