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  "body": "\n## Statement\n\nLet $L$ be a field and let $G$ be a finite subgroup of the [[automorphism group|0K1SCPBNKJ]] of $L$.\nLet $K = L^G$ be the [[fixed field|0KMJ6919WS]] of $G$.\nThen:\n\n1. the [[degree|0GS74FGFEK]] $[L:K]$ equals the order $\\abs{G}$;\n2. $L/K$ is a [[Galois extension|0DS2F7GK05]];\n3. the [[Galois group|0XVR72M51M]] $\\Gal(L/K)$ equals $G$.\n\n## Remarks\n\nFor $\\alpha \\in L$, let $\\alpha_1, \\ldots, \\alpha_r$ be the distinct elements of the orbit $G\\alpha$.\nThe polynomial $\\prod_{i=1}^r (x - \\alpha_i)$ is fixed by $G$, so it has coefficients in $K$; it is separable and splits in $L$.\nHence every element of $L$ is separable over $K$ of degree at most $\\abs{G}$, and $L/K$ is normal and separable.\nBy the [[primitive element theorem|0SVX98HYV1]], every finite subextension of $L/K$ is then generated by one element, so it has degree at most $\\abs{G}$.\nHence $[L:K] \\le \\abs{G}$.\nConversely, $G \\subseteq \\Gal(L/K)$ and $\\abs{\\Gal(L/K)} \\le [L:K]$ give the reverse inequality.\nAnother proof uses [[Dedekind's lemma on independence of characters|N43G50YHG2]].\n\nExample: let $L = \\C(t)$ and let $G = \\set{\\id, \\tau}$ with $\\tau(f)(t) = f(-t)$.\nThe fixed field is $\\C(t^2)$, and indeed $[\\C(t) : \\C(t^2)] = 2$.\n\nThe hypothesis that $G$ is finite is necessary.\nThe automorphism of $\\C(t)$ with $t \\mapsto t + 1$ generates an infinite group whose fixed field is $\\C$, and $\\C(t)/\\C$ is not even algebraic.\n\n## Explanations\n\n### L1 human\n\nStart with a field and a finite group of its symmetries.\nThe elements that no symmetry moves form a smaller field.\nArtin's theorem says that the big field is a Galois extension of the small one.\nIts degree is the number of symmetries, and its Galois group consists exactly of these symmetries.\n\n### L2 human\n\nThis theorem is one half of the [[fundamental theorem of Galois theory|0T1Q8D0FW8]].\nIt shows that every subgroup $H$ of a finite Galois group $\\Gal(L/K)$ is the Galois group of $L$ over the fixed field $L^H$, so the map $H \\mapsto L^H$ is injective.\nIt also produces Galois extensions directly: the symmetric group $S_n$ acting on $k(x_1, \\ldots, x_n)$ shows that $k(x_1, \\ldots, x_n)$ is Galois over the field of symmetric rational functions with group $S_n$.\n",
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