= 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>.
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