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  "body": "\n## Statement\n\nLet $K$ be a field, let $n \\ge 1$ be an integer that is not divisible by the [[characteristic|1ZB13N78XW]] of $K$, and let $a \\in K$.\nThen the [[Galois group|0JZY3R908H]] of $x^n - a$ over $K$ is a [[solvable group|0P3KZK0FTV]].\n\n## Remarks\n\nIf $a = 0$, the polynomial $x^n$ splits over $K$ and the group is trivial.\nIf $a \\ne 0$, let $L$ be a splitting field and $\\beta \\in L$ a root; the roots are $\\beta \\zeta^k$, where $\\zeta$ is a [[primitive $n$-th root of unity|0J817RTVG6]], so $\\zeta \\in L$.\nPut $E = K(\\zeta)$; $E/K$ is a [[cyclotomic extension|VD6VC3ZMMY]], and $\\Gal(E/K)$ embeds in $(\\Z/n\\Z)^\\times$, so it is abelian.\nThe map $\\sigma \\mapsto \\sigma(\\beta)/\\beta$ embeds $\\Gal(L/E)$ in the cyclic group of $n$-th roots of unity, so $L/E$ is a [[cyclic extension|VFE5B71E7Y]].\nThen $\\Gal(L/E)$ is a normal subgroup of $\\Gal(L/K)$ with abelian quotient $\\Gal(E/K)$, so $\\Gal(L/K)$ is solvable.\nExample: for $x^3 - 2$ over $\\Q$, the subgroup is cyclic of order $3$ and the quotient has order $2$, giving $S_3$.\n\n## Explanations\n\n### L1 human\n\nThe equation $x^n = a$ is the simplest equation that needs an $n$-th root.\nIts roots are one $n$-th root of $a$ times the $n$-th roots of unity.\nThe proposition says that its Galois group is solvable, as it should be, since its roots are written with radicals.\n\n### L2 human\n\nThis is the base case of the easy direction of [[Galois' theorem on solvability|VHCWCFA3MX]].\nThe two-step tower $K \\subseteq K(\\zeta) \\subseteq L$ is the model for every radical extension: first roots of unity, then pure roots.\nThe same argument shows that $\\Gal(L/K)$ embeds in the group of maps $x \\mapsto ux + v$ on $\\Z/n\\Z$ with $u$ a unit.\n",
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