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  "body": "\n## Statement\n\nLet $p$ be a prime number and let $n \\ge 1$.\nFor each divisor $m$ of $n$, the field $\\mathbb{F}_{p^n}$ has exactly one [[subfield|07QY5RPFBR]] of order $p^m$, namely $\\{ a : a^{p^m} = a \\}$.\nIt has no other subfields.\n\n## Remarks\n\nIf $K$ is a subfield of $\\mathbb{F}_{p^n}$ of order $p^m$, then $n = m \\cdot [\\mathbb{F}_{p^n} : K]$ by the [[tower law|0A7KJ68XHT]], so $m$ divides $n$.\nFor example, the subfields of $\\mathbb{F}_{64} = \\mathbb{F}_{2^6}$ are $\\mathbb{F}_2$, $\\mathbb{F}_4$, $\\mathbb{F}_8$ and $\\mathbb{F}_{64}$.\nThe subfield of order $p^m$ is the fixed field of $\\varphi^m$, where $\\varphi$ is the [[Frobenius automorphism|0VP4TBVJ19]].\n\n## Explanations\n\n### L1 human\n\nThe subfields of a finite field match the divisors of the exponent.\nThe field with $2^6$ elements contains the fields with $2^1$, $2^2$ and $2^3$ elements, but not one with $2^4$ elements.\n",
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