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  "body": "\n## Statement\n\nLet $K$ be a [[field|0QA8CHH8MG]], let $a \\in K$ and let $n \\ge 2$ be an integer.\nThe polynomial $x^n - a$ is [[irreducible|00X2A9G9W3]] in $K[x]$ if and only if both of the following hold:\n\n1. for every prime $p$ dividing $n$, $a$ is not of the form $b^p$ with $b \\in K$;\n2. if $4$ divides $n$, then $a$ is not of the form $-4 c^4$ with $c \\in K$.\n\n## Remarks\n\nThe conditions are necessary: if $a = b^p$ with $p \\mid n$, then $y^p - b^p$ has the factor $y - b$, and $y = x^{n/p}$ gives a factor of $x^n - a$.\nIf $a = -4c^4$ and $4 \\mid n$, the identity $y^4 + 4c^4 = (y^2 + 2cy + 2c^2)(y^2 - 2cy + 2c^2)$ with $y = x^{n/4}$ gives a factorisation.\nThe proof of sufficiency first treats $n$ prime with a norm argument, then prime powers, and combines them with the multiplicativity of degrees in towers of extensions.\nExample: $x^4 + 4 = (x^2 + 2x + 2)(x^2 - 2x + 2)$ over $\\Q$, although $-4$ is not a square in $\\Q$; this is the case $c = 1$ of clause 2.\nExample: $x^n - 2$ is irreducible over $\\Q$ for every $n \\ge 1$.\nFor $n$ odd, clause 2 is empty and the test reduces to $p$-th powers for the primes $p \\mid n$.\n\n## Explanations\n\n### L1 human\n\nThe polynomial $x^n - a$ factors in two obvious ways: when $a$ is a perfect $p$-th power for a prime $p$ dividing $n$, and in one special case built from $x^4 + 4$.\nCapelli's theorem says that there are no other ways.\n\n### L2 human\n\nThe theorem is the basic tool to study binomial equations $x^n = a$ and the radical extensions $K(\\sqrt[n]{a})$.\nIt gives the degree $[K(\\sqrt[n]{a}) : K] = n$ when the two clauses hold.\nThis is used in the theory of [[radical extensions|V74WMGQQ9D]] and of Kummer extensions.\n",
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