Let's take a simple example of a $Z$-module: $Z_n$, the cyclic group of order $n$ = $\{0, 1, ..., n-1\}$.

This is generated by the number $1$, since by taking all linear combinations of 1 -- i.e. all multiples of 1 -- we get everything in $Z_n$. \{1\} spans the whole module. But \{1\} is not linearly independent, because $n * 1 = 0$.

Put differently, it is not the case that every member of $Z_n$ is a unique linear combination of the "vectors" in $\{1\}$ -- the mapping which sends $k$ in $Z$ to $k * 1$ in $Z_n$ is not one-to-one.

So $Z_n$ is a module that has no basis -- it is not "free." Whereas a basic theorem tells us that every vector space is free, has a basis.

This is generated by the number $1$, since by taking all linear combinations of 1 -- i.e. all multiples of 1 -- we get everything in $Z_n$. \{1\} spans the whole module. But \{1\} is not linearly independent, because $n * 1 = 0$.

Put differently, it is not the case that every member of $Z_n$ is a unique linear combination of the "vectors" in $\{1\}$ -- the mapping which sends $k$ in $Z$ to $k * 1$ in $Z_n$ is not one-to-one.

So $Z_n$ is a module that has no basis -- it is not "free." Whereas a basic theorem tells us that every vector space is free, has a basis.