I try to solve following question.
Show that $R=\mathbf{Q}[x,x^{-1}]$ is not a PID. Construct a free module over $R$ having a non free submodule.
One may give some examples for free modules but not PIDs which has a non free submodule for different $R$s. For such $R$, i don't know the answer.
Oops. Let me revise my answer. Perhaps I won't make such a silly mistake this time. :)
The polynomial ring $\mathbf{Q}[x]$ is a subring of $R$ (and is a known PID), and all ideals of $R$ are extended from $\mathbf{Q}[x]$. What this means is that an ideal $I$ of $R$ is of the form $JR$ where $J$ is an ideal of $\mathbf{Q}[x]$. In view of this fact, $R$ is a PID since $\mathbf{Q}[x]$ is.
To prove the above claim, let $I$ be an ideal of $R$. If we put $J = I \cap \mathbf{Q}[x]$, then note $J$ is an ideal of $\mathbf{Q}[x]$ and hence $J = (f)$ for some $f\in \mathbf{Q}[x]$. Then it's not difficult to show $I = JR$. But $JR$ is the ideal in $R$ generated by $f$.
In fancier language, $R$ can be viewed as the localization of $\mathbf{Q}[x]$ at powers of $x$ (we are inverting $x$), and the ideal structures of a ring and its localizations are tightly linked.