= Solution
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 $g=\pm1$. 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.
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