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) 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
Canonical name	ExampleOfFreeModule
Date of creation	2013-03-22 13:48:41
Last modified on	2013-03-22 13:48:41
Owner	mathcam (2727)
Last modified by	mathcam (2727)
Numerical id	5
Author	mathcam (2727)
Entry type	Example
Classification	msc 13C10