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  "body": "\n## Statement\n\nLet $q$ be a prime power, let $n \\ge 1$, and let $\\F_q \\subseteq \\F_{q^n}$ be [[finite fields|08RTDHX1SK]] of orders $q$ and $q^n$.\nThen $\\F_{q^n}$ is a [[Galois extension|0DS2F7GK05]] of $\\F_q$.\nIts [[Galois group|0XVR72M51M]] is [[cyclic|0B85Q41RP6]] of order $n$, generated by the map $\\varphi(x) = x^q$.\n\n## Remarks\n\nThe map $\\varphi$ is a power of the [[Frobenius automorphism|0VP4TBVJ19]], so it is a field automorphism of $\\F_{q^n}$.\nIt fixes $\\F_q$, because [[every element of $\\F_q$ satisfies $a^q = a$|0QWR8WTZEW]].\nIf $\\varphi^k = \\id$ with $k \\ge 1$, then every element of $\\F_{q^n}$ is a root of $x^{q^k} - x$, so $q^n \\le q^k$ and $k \\ge n$; and $\\varphi^n = \\id$.\nSo $\\varphi$ has order $n$, and since [[the Galois group has order at most the degree|0QM98DNWVW]] $n$, the group is $\\langle \\varphi \\rangle$ and the extension is Galois.\nFor example, $\\Gal(\\F_4/\\F_2) = \\set{\\id, \\varphi}$ with $\\varphi(x) = x^2$, and $\\varphi$ swaps the two roots $\\omega$ and $\\omega + 1$ of $x^2 + x + 1$.\n\n## Explanations\n\n### L1 human\n\nA finite field has very few symmetries over a smaller finite field.\nAll of them are powers of one map, raising to the $q$-th power.\nSo the symmetry group is as simple as possible: a cycle of length $n$.\n\n### L2 human\n\nWith the [[fundamental theorem of Galois theory|0T1Q8D0FW8]], the subgroups of a cyclic group of order $n$ match the fields between $\\F_q$ and $\\F_{q^n}$.\nThis gives again that these fields are the $\\F_{q^m}$ with $m$ dividing $n$, as in [[subfields of a finite field|08YERZDM9X]].\nThe absolute Galois group of $\\F_q$ is the profinite completion $\\widehat{\\Z}$ of $\\Z$, topologically generated by $\\varphi$.\nThe Frobenius element of a prime in algebraic number theory comes from this theorem applied to residue fields.\n",
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