# 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 |