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  "body": "\n## Statement\n\nLet $R$ be a [[unique factorisation domain|06YMXJMXYS]] with [[field of fractions|0VF11ZTTMD]] $K$ and let $f \\in R[x]$ be a [[primitive polynomial|0SP14DZN0G]].\nThen $f$ is [[irreducible|00X2A9G9W3]] in $R[x]$ if and only if $f$ is irreducible in $K[x]$.\n\n## Remarks\n\nThe key step: if $f = g h$ in $K[x]$, then one can scale $g$ and $h$ by elements of $K$ so that they are in $R[x]$ and $f = g h$ still holds. This uses that the [[product of primitive polynomials is primitive|08C1NWBNHA]].\nFor $R = \\mathbb{Z}$: a primitive integer polynomial that factors over $\\mathbb{Q}$ already factors over $\\mathbb{Z}$, into factors of the same degrees.\nThe hypothesis that $f$ is primitive is necessary: $2x$ is irreducible in $\\mathbb{Q}[x]$, but $2x = 2 \\cdot x$ is a product of two nonunits in $\\mathbb{Z}[x]$.\n\n## Explanations\n\n### L1 human\n\nTo show that an integer polynomial has no factorisation over the rationals, it is enough to check factorisations over the integers.\nFractions do not help to factor a primitive polynomial.\nThis is why criteria such as Eisenstein's work with integer coefficients.\n",
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