example of free module
ExampleOfFreeModule
2003-07-29 13:11:13
2003-07-30 19:13:30
Example
free module
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\PMlinkescapetext{Clearly} from the definition, $\Z^n$ is \PMlinkid{free}{FreeModule} as a $\Z$-module for any positive integer $n$.
A more interesting example is the following:
\begin{Theo}
The set of rational numbers $\Q$ do \emph{not} form a \PMlinkid{free}{FreeModule} $\Z$-module.
\end{Theo}
\begin{proof}
First note that any two elements in $\Q$
are $\Z$-linearly dependent. If $x=\frac{p_1}{q_1}$ and
$y=\frac{p_2}{q_2}$, then $q_1p_2x-q_2p_1y=0$. Since \PMlinkname{basis}{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 $\Z$-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 $n,p\in\Z$ ,$q\in\Z^+$, giving us
a contradiction.
\end{proof}