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  "body": "\n## Definition\n\nLet $K$ be a [[field|0QA8CHH8MG]] and let $f \\in K[x]$ be a nonconstant polynomial.\nThe polynomial $f$ is *solvable by radicals* over $K$ if there is a [[radical extension|V74WMGQQ9D]] $R/K$ that contains a [[splitting field|02NZZ1J0FH]] of $f$ over $K$.\nThe equation $f(x) = 0$ is then also called solvable by radicals.\n\n## Remarks\n\nEquivalently, every root of $f$ lies in some radical extension of $K$, since the compositum of finitely many radical extensions inside a common field is radical.\nExample: $f = x^3 - 2$ over $\\Q$ splits in $R = \\Q(\\omega, \\sqrt[3]{2})$, with $\\omega = e^{2\\pi i/3}$.\nThis $R$ is radical, since $\\omega^3 = 1 \\in \\Q$ and $(\\sqrt[3]{2})^3 = 2 \\in \\Q(\\omega)$, so $f$ is solvable by radicals.\nNon-example: $f = x^5 - 4x + 2$ over $\\Q$ is not solvable by radicals, because its Galois group is $S_5$, which is not [[solvable|0P3KZK0FTV]]; see the [[quintic equation|S1WS18QEGJ]].\nIn characteristic $0$, [[Galois' theorem on solvability|VHCWCFA3MX]] says that $f$ is solvable by radicals if and only if its Galois group is solvable.\n",
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