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  "body": "\n## Statement\n\nLet $f = a_n x^n + a_{n-1} x^{n-1} + \\dots + a_0 \\in \\mathbb{Z}[x]$ with $n \\ge 1$, and let $p$ be a prime number with these three properties.\n\n1. $p$ divides $a_i$ for each $i$ with $0 \\le i \u003c n$.\n2. $p$ does not divide $a_n$.\n3. $p^2$ does not divide $a_0$.\n\nThen $f$ is [[irreducible|00X2A9G9W3]] in $\\mathbb{Q}[x]$.\n\n## Remarks\n\nIf $f$ is also [[primitive|0SP14DZN0G]], then $f$ is irreducible in $\\mathbb{Z}[x]$, by [[Gauss's lemma|0QFV5Y1Z5X]].\nFor example, $x^n - 2$ is irreducible over $\\mathbb{Q}$ for each $n \\ge 1$, with $p = 2$.\nA change of variable can help: $f(x + 1)$ is irreducible if and only if $f(x)$ is. This proves that $x^{p-1} + \\dots + x + 1$ is irreducible for $p$ prime.\nThe criterion holds more generally for a prime ideal $P$ of an integral domain $R$, with $P^2$ in place of $p^2$.\n\n## History\n\nTheodor Schönemann published a version of the criterion in 1846.\nGotthold Eisenstein published a different version in 1850.\nBoth papers appeared in Crelle's Journal.\n\n## Explanations\n\n### L1 human\n\nThe criterion is a quick test of irreducibility over the rationals.\nFind one prime that divides all the coefficients except the leading one, and whose square does not divide the constant term.\nFor $x^5 + 6x + 3$, the prime $3$ works, so the polynomial is irreducible.\n",
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