A finite field with $p^s$ elements has only $s$ automorphisms.

Question

Let $F$ be a finite field with $p^s$ elements, where $p \in \mathbb{N}$ is the characteristic of $F$. I know that $\varphi : x \mapsto x^p$ is an automorphism and also ${\varphi}^2 = \varphi \circ \varphi$ is an automorphism; in fact ${\{{\varphi}^i\}}_{i = 1}^s$ is a finite sequence of automorphisms, being ${\varphi}^s(x) = x^{p^s} = x$ for all $x \in F$. How can I show that they are the only automporphisms of $F$? If $F_p$ is the prime field of $F$, I know that $F / F_p$ is a finite Galois extension, so the number of automorphisms of $F$ must be $$ [F : F_p] = |G(F / F_p)| $$ because each automorphism fixes the prime field $F_p$. How can I get this degree?

Answer 1

Once you know that $F$ has a primitive element $\alpha$, any automorphism $\varphi : F \to F$ is determined by what it does to $\alpha$, so if it agrees with a power of Frobenius on $\alpha$ then it must be a power of Frobenius. Let $S = \{ \alpha, \alpha^p, \dots \}$ be the orbit of $\alpha$ under the action of Frobenius, and consider

$$f(x) = \prod_{\beta \in S} (x - \beta).$$

By construction, $f(x)$, and hence all of its coefficients, is stable under the action of Frobenius, and since the fixed field of Frobenius is the prime subfield $\mathbb{F}_p$, we have $f(x) \in \mathbb{F}_p[x]$. On the other hand, if $\alpha$ is a root of any polynomial in $\mathbb{F}_p[x]$ then so are the elements of $S$. Hence $f$ must be the minimal polynomial of $\alpha$, so $\varphi(\alpha) \in S$, so $\varphi$ is a power of Frobenius on $\alpha$ and hence is a power of Frobenius.