Possible difference between $\mathbb{Z}$-modules and vector spaces

Question

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 ?

Answer 1

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.

Answer 2

There are also many differences between vector spaces and $\mathbb{Z}$-modules. Every vector space has a basis, but not every $\mathbb{Z}$-module. For example, any finite abelian group is not a free $\mathbb{Z}$-module, and the $\mathbb{Z}$-module $\mathbb{Q}$ is not free. Furthermore a free $\mathbb{Z}$-module may have a linear independent set which cannot be extended to a basis. It also may have a subset $S$ which spans it, but does not contain a basis.