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  "body": "\n## Statement\n\nLet $f \\in \\mathbb{Z}[x]$ be a [[monic polynomial|0BDX4MBKT4]] and let $p$ be a prime number.\nLet $\\bar{f} \\in \\mathbb{F}_p[x]$ be the polynomial whose coefficients are those of $f$ modulo $p$.\nIf $\\bar{f}$ is [[irreducible|00X2A9G9W3]] in $\\mathbb{F}_p[x]$, then $f$ is irreducible in $\\mathbb{Z}[x]$ and in $\\mathbb{Q}[x]$.\n\n## Remarks\n\nThe proof: a factorisation $f = g h$ into monic factors of positive degree gives $\\bar{f} = \\bar{g} \\bar{h}$, with $\\deg \\bar{g} = \\deg g$ and $\\deg \\bar{h} = \\deg h$.\nIrreducibility in $\\mathbb{Q}[x]$ then follows by [[Gauss's lemma|0QFV5Y1Z5X]].\nThe converse is false: $x^4 + 1$ is irreducible over $\\mathbb{Q}$, but it is reducible modulo every prime.\nExample: $x^3 + x + 1$ has no root in $\\mathbb{F}_2$, so it is irreducible over $\\mathbb{F}_2$, and so over $\\mathbb{Q}$.\n\n## Explanations\n\n### L1 human\n\nA factorisation over the integers stays a factorisation after reduction modulo $p$.\nSo if the reduced polynomial does not factor, the original one does not factor either.\nOver a small finite field, you can check irreducibility by a finite search.\n",
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