example of free module

from the definition, ${\mathbb{Z}}^n$ is http://planetmath.org/node/FreeModulefree as a $\mathbb{Z}$-module for any positive integer $n$.

A more interesting example is the following:

Theorem 1.

The set of rational numbers $\mathrm{Q}$ do not form a http://planetmath.org/node/FreeModulefree $\mathrm{Z}$-module.

Proof.

First note that any two elements in $\mathbb{Q}$are $\mathbb{Z}$-linearly dependent. If $x=\frac{p_1}{q_1}$ and$y=\frac{p_2}{q_2}$, then $q_1{p}_{2}x-q_2{p}_{1}y=0$. Since basis (http://planetmath.org/Basis) elementsmust be linearly independent, this shows that any basis must consistof only one element, say $\frac{p}{q}$, with $p$ and $q$ relatively prime, and without loss of generality, $q>0$. The $\mathbb{Z}$-span of $\{\frac{p}{q}\}$ is theset of rational numbers of the form $\frac{np}{q}$. I claim that$\frac{1}{q+1}$ is not in the set. If it were, then we would have$\frac{np}{q}=\frac{1}{q+1}$ for some $n$, but this implies that$np=\frac{q}{q+1}$ which has no solutions for $n,p\in \mathbb{Z}$ ,$q\in {\mathbb{Z}}^{+}$, giving usa contradiction.∎