Past exam of the mathematics course of the University of Cambridge / 2018 / ib / Paper 2 / 11G / b

Any unit in a ring of $R$ must also be a unit of $\mathbb{Q}[X]$, hence a nonzero constant $q\in \mathbb{Q}$. Membership of both $q$ and $q^{-1}$ in $R$ requires $q,q^{-1}\in \mathbb{Z}$, so

Let $f\in R$ be irreducible. If $\mathrm{deg}f=0$, then $f\in \mathbb{Z}$. A nonzero integer that is neither a unit nor, up to sign, a prime number factors nontrivially in $R$, so necessarily $f=\pm p$ for a prime $p$.

Now suppose $\mathrm{deg}f\ge 1$ and write $m=f(0)\in \mathbb{Z}$. If $m=0$, thenand both factors lie in $R$ and are nonunits. If $|m|>1$, choose a prime $p\mid m$; then $f=p(f/p)$ is again a factorization into two nonunits of $R$, since $(f/p)(0)=m/p\in \mathbb{Z}$. Both cases contradict irreducibility. Hence

Suppose $X=f_1\cdots f_s$ were a product of irreducible elements of $R$. Since degree of a polynomial is additive under products, exactly one factor has positive degree and all the others are constant. By the preceding classification, the positive-degree factor has constant term $\pm 1$, while every constant irreducible is $\pm p$ for a prime $p$. Their product therefore has a nonzero constant term, contradicting $X(0)=0$.

Thus $X$ is not a product of irreducibles, so this integral domain is not an atomic domain and in particular is not a unique factorization domain.