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  "body": "\n## Proof\n\n1. Let $p$ be prime, let $n \\ge 1$, and let $F = \\F_{p^n}$ be a [[finite field|08RTDHX1SK]] of order $p^n$. Let $P$ be its [[prime subfield|0VPFBAY7MW]]. By the [[prime power theorem|0H5WWC8RZX]], $[F : P] = n$.\n2. For $m \\ge 1$, let $S_m = \\set{a \\in F : a^{p^m} = a}$. These are the roots in $F$ of $x^{p^m} - x$, so $\\abs{S_m} \\le p^m$ by the [[bound on the number of roots|0KJVQV3E0F]].\n\nWe first show that every [[subfield|07QY5RPFBR]] has this form.\n\n3. Let $K$ be a subfield of $F$. It contains $1$, so it contains $P$, and $P$ is also the prime subfield of $K$. By the prime power theorem, $\\abs{K} = p^m$ with $m = [K : P]$.\n4. By the [[tower law|0A7KJ68XHT]], $n = [F : P] = [F : K] \\cdot [K : P] = [F : K] \\cdot m$. So $m$ divides $n$.\n5. By [[the equation $x^q = x$|0QWR8WTZEW]] applied to $K$, every $a \\in K$ satisfies $a^{p^m} = a$. So $K \\subseteq S_m$.\n6. Then $p^m = \\abs{K} \\le \\abs{S_m} \\le p^m$ by step 2, so $K = S_m$.\n7. So a subfield of order $p^m$ exists only if $m \\mid n$, and it must equal $S_m$. In particular there is at most one subfield of each order, and there are no other subfields.\n\nNow let $m$ divide $n$. We show that $S_m$ is a subfield of order $p^m$.\n\n8. The [[Frobenius endomorphism|0GHDDVAH8P]] $a \\mapsto a^p$ is a ring homomorphism, by the [[freshman's dream|0396VGFA78]]. Its $m$-th iterate $\\psi(a) = a^{p^m}$ is a ring homomorphism too.\n9. So $S_m = \\set{a : \\psi(a) = a}$ contains $0$ and $1$ and is closed under subtraction and multiplication. If $a \\in S_m$ is nonzero, then $\\psi(a^{-1}) = \\psi(a)^{-1} = a^{-1}$. So $S_m$ is a subfield.\n10. Write $n = mk$. Then $p^n - 1 = (p^m - 1) N$ with $N = \\sum_{j=0}^{k-1} p^{mj}$, since $p^n = (p^m)^k$.\n11. Let $y = x^{p^m - 1}$. Then $x^{p^n - 1} - 1 = y^N - 1 = (y - 1)(y^{N-1} + \\dots + y + 1)$. Multiplying by $x$ gives $x^{p^n} - x = (x^{p^m} - x) \\, h$ for some $h \\in F[x]$ of degree $p^n - p^m$.\n12. Every $a \\in F$ satisfies $a^{p^n} = a$, by the equation $x^q = x$. So each $a \\in F$ is a root of $x^{p^m} - x$ or a root of $h$.\n13. The nonzero polynomial $h$ has at most $p^n - p^m$ roots in $F$. So at least $p^m$ elements of $F$ are roots of $x^{p^m} - x$, that is, $\\abs{S_m} \\ge p^m$.\n14. With step 2, $\\abs{S_m} = p^m$. So for each $m \\mid n$, the field $F$ has exactly one subfield of order $p^m$, namely $S_m$, and by step 7 it has no other subfields.\n\n## Idea\n\nEvery element of a field of order $p^m$ satisfies $a^{p^m} = a$, and this equation has at most $p^m$ solutions. So a subfield of order $p^m$ is forced to be the solution set $S_m$. When $m$ divides $n$, the polynomial $x^{p^m} - x$ divides $x^{p^n} - x$, and this gives $S_m$ enough elements.\n",
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