Z[x] is not a principal ideal domain

Counterexample

The ring $\mathbb{Z}[x]$ is an integral domain, but it is not a principal ideal domain. The ideal $(2, x)$, the set of the polynomials with an even constant term, is not principal. Suppose $(2, x) = (f)$. Then $f \mid 2$, so $f$ is a constant $c$ with $c \in \{\pm 1, \pm 2\}$. If $c = \pm 1$, then $1 \in (2, x)$, but $1$ has an odd constant term. If $c = \pm 2$, then $x$ is a multiple of $2$ in $\mathbb{Z}[x]$, which is false.

Remarks

The quotient $\mathbb{Z}[x]/(2, x)$ is $\mathbb{F}_2$, so $(2, x)$ is a maximal ideal. The ring $\mathbb{Z}[x]$ is still a unique factorization domain, by the theorem on polynomial rings over a UFD. Over a field $K$, the ring $K[x]$ is a principal ideal domain.

Explanations

L1 human

In $\mathbb{Z}[x]$, the ideal generated by $2$ and $x$ needs two generators. No single polynomial divides both $2$ and $x$ except the units $\pm 1$, and $\pm 1$ generates too much. The cause is that $\mathbb{Z}$ is not a field.