## 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 $\mathsf{Aut}(\mathbb{F}_{p^n})$.