The order of the Frobenius map on a finite field

Let \varphi denote the Frobenius map x \mapsto x^p on \mathbb{F}_{p^n}. Prove that \varphi is an automorphism and compute its order in \mathsf{Aut}(\mathbb{F}_{p^n}).

Recall that \varphi is a homomorphism. Moreover, if \alpha \in \mathsf{ker}\ \varphi, then \alpha^p = 0. Since fields contain no nontrivial zero divisors, we have \alpha = 0 (using induction if you want). So the kernel of \varphi is trivial, and thus \varphi is injective. Since \mathbb{F}_{p^n} is finite, \varphi is surjective, and so is a field isomorphism.

Next, we claim that \varphi^t(\alpha) = \alpha^{p^t} for all \alpha and all t \geq 1, and will show this by induction. The base case certainly holds, and if \varphi^t(\alpha) = \alpha^{p^t}, then \varphi^{t+1}(\alpha) = \varphi(\varphi^t(\alpha)) = \varphi(\alpha^{p^t}) = (\alpha^{p^t})^p = \alpha^{p^{t+1}} as desired.

Now \varphi^n(\alpha) = \alpha^{p^n} = \alpha, since the elements of \mathbb{F}_{p^n} are precisely the roots of x^{p^n}-x. So we have \varphi^n = 1.

If \varphi^t = 1, then we have \alpha^{p^t} - \alpha = 0 for all \alpha, so that each \alpha is a root of x^{p^t}-x. So x^{p^n-1}-1 divides x^{p^t-1}-1, and so p^n-1 divides p^t-1 (by this previous exercise) and then n divides t (by this previous exercise). In particular, n \leq t.

So n is the order of \varphi in \mathsf{Aut}(\mathbb{F}_{p^n}).

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