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  "body": "\n## Counterexample\n\nThe ring $\\mathbb{Z}[x]$ is an integral domain, but it is not a [[principal ideal domain|0RZYDVZ475]].\nThe ideal $(2, x)$, the set of the polynomials with an even constant term, is not [[principal|00A2EDFT9Q]].\nSuppose $(2, x) = (f)$. Then $f \\mid 2$, so $f$ is a constant $c$ with $c \\in \\{\\pm 1, \\pm 2\\}$.\nIf $c = \\pm 1$, then $1 \\in (2, x)$, but $1$ has an odd constant term.\nIf $c = \\pm 2$, then $x$ is a multiple of $2$ in $\\mathbb{Z}[x]$, which is false.\n\n## Remarks\n\nThe quotient $\\mathbb{Z}[x]/(2, x)$ is $\\mathbb{F}_2$, so $(2, x)$ is a maximal ideal.\nThe ring $\\mathbb{Z}[x]$ is still a [[unique factorization domain|06YMXJMXYS]], by [[the theorem on polynomial rings over a UFD|0D6KH8Y59N]].\nOver a field $K$, the ring $K[x]$ is a principal ideal domain.\n\n## Explanations\n\n### L1 human\n\nIn $\\mathbb{Z}[x]$, the ideal generated by $2$ and $x$ needs two generators.\nNo single polynomial divides both $2$ and $x$ except the units $\\pm 1$, and $\\pm 1$ generates too much.\nThe cause is that $\\mathbb{Z}$ is not a field.\n",
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  "name": "Z[x] is not a principal ideal domain",
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