---
tag: 06E3AG9TM2
kind: counterexample
name: Z[x] is not a principal ideal domain
msc: [13F20]
uses: [0RZYDVZ475, 0Q9EZS90GY, 00A2EDFT9Q]
sources:
  - url: https://en.wikipedia.org/wiki/Principal_ideal_domain
    locator: Non-examples
---

## Counterexample

The ring $\mathbb{Z}[x]$ is an integral domain, but it is not a [[principal ideal domain|0RZYDVZ475]].
The ideal $(2, x)$, the set of the polynomials with an even constant term, is not [[principal|00A2EDFT9Q]].
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|06YMXJMXYS]], by [[the theorem on polynomial rings over a UFD|0D6KH8Y59N]].
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.