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  "body": "\n## Statement\n\nLet $p$ be a [[prime number|0SN0MAEKYX]] and let $f \\in \\Q[x]$ be an [[irreducible|00X2A9G9W3]] polynomial of degree $p$.\nAssume that $f$ has exactly two nonreal roots in $\\C$.\nThen the [[Galois group|0JZY3R908H]] of $f$ over $\\Q$, acting on the $p$ roots of $f$, is the full [[symmetric group|0RJMNEN3BJ]] $S_p$.\n\n## Remarks\n\nLet $G$ be the Galois group, seen as a group of permutations of the $p$ roots.\nSince $f$ is irreducible, adjoining one root gives a field of degree $p$, so $p$ divides $\\abs{G}$.\nBy [[Cauchy's theorem|0JM7PA288B]], $G$ contains an element of order $p$, which is a $p$-cycle.\n[[Complex conjugation|0BJV24C4SY]] restricts to an element of $G$ that swaps the two nonreal roots and fixes the others, so $G$ contains a [[transposition|02DTRYKGE8]].\nFor $p$ prime, a $p$-cycle and any transposition generate $S_p$.\nExample: $x^5 - 4x + 2$ has Galois group $S_5$; see the [[quintic equation|S1WS18QEGJ]].\nThe prime degree matters: in degree $4$, the group generated by $(1\\,2\\,3\\,4)$ and $(1\\,3)$ is a dihedral group of order $8$, not $S_4$.\n\n## Explanations\n\n### L1 human\n\nThe theorem gives an easy test for when the symmetries of the roots of a polynomial include every permutation.\nCount the real roots: if the degree is prime and exactly two roots are not real, the group is as big as possible.\nThis is the usual way to produce explicit equations that cannot be solved by radicals.\n\n### L2 human\n\nCombined with [[Galois' theorem on solvability|VHCWCFA3MX]] and the fact that [[$S_p$ is not solvable for $p \\ge 5$|0J53NE674D]], the theorem proves the second part of the [[Abel–Ruffini theorem|163A2WQQJX]].\nThe number of real roots can be found with calculus or with [[Sturm's theorem|R44CV0KD9T]].\nThe proof uses only that $\\Q \\subseteq \\R$; it works for any subfield of $\\R$ in place of $\\Q$.\n",
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