# 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 $\mathbb{Q}$ do not form a http://planetmath.org/node/FreeModulefree $\mathbb{Z}$-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) elements must be linearly independent, this shows that any basis must consist of 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 the set 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 us a contradiction   . ∎

Title example of free module ExampleOfFreeModule 2013-03-22 13:48:41 2013-03-22 13:48:41 mathcam (2727) mathcam (2727) 5 mathcam (2727) Example msc 13C10