 Solution

Take $\boxed{R=\mathbb{Q}[X]}$ and, for the next clause, its subring $S=\mathbb{Z}[X]$. Division with remainder makes $R$ a Euclidean domain, with the degree as Euclidean size for nonzero polynomials. For completeness, a nonzero ideal of $R$ contains a polynomial $g$ of least degree; division of any member by $g$ leaves a smaller-degree remainder in the ideal, so that remainder must vanish. The ideal is therefore $(g)$; the zero ideal is already principal. Thus $R$ is a principal ideal domain.