Past exam of the mathematics course of the University of Cambridge / 2017 / ib / Paper 2 / 2E / b

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.

For the same pair, $\boxed{S=\mathbb{Z}[X]\subset \mathbb{Q}[X]}$. Since $\mathbb{Z}$ is a unique factorization domain, Gauss lemma for polynomials proves that $S$ is a unique factorization domain.

To prove that $S$ is not a principal ideal domain, consider the ideal $I=(2,X)$. A generator $g$ would divide 2 and $X$. Divisibility of 2 forces $g$ to be a nonzero constant integer; divisibility of $X$ then forces that integer to divide its coefficient 1, so $g=\pm 1$. This would make $I=S$. But evaluation at $X=0$ followed by reduction modulo 2 maps every element of $I$ to zero and maps 1 to one. Hence $1\notin I$, a contradiction. This is the integer polynomial ring is not a principal ideal domain example, with both clauses satisfied by a single inclusion.