Show that every subfield of F is the set of fixed points under some power of $\phi$, $\phi(a) = a^p$

Question

Given: $p$ is prime. and $F$ is a finite field of characteristic $p$.

I've already shown that the map defined by $\phi(a) = a^p \quad \forall a \in F$ is an automorphism. (Also called: Frobenius automorphism)

My question: How to show that every subfield of F is the set of fixed points under some power of $\phi$?

Answer 1

$F$ contains $\Bbb{F}_p$ so it is a $\Bbb{F}_p$-vector space, of finite dimension $n$, so it has $p^n$ elements. Thus $F^\times$ is a group with $p^n-1$ elements and any $a \in F$ is a root of $x(x^{p^n-1}-1) = x^{p^n}-x \in \Bbb{F}_p[x]$. But this polynomial has at most $p^n$ roots in the field $F$, thus $F$ is exactly the splitting field of $x^{p^n}-x \in \Bbb{F}_p[x]$ the latter being the unique field with $p^n$ elements.

Which means $F = \{ a \in \overline{\Bbb{F}}_p, \phi^n(a) = a\}$.

Also note if $F^\times$ is not cyclic, let $e < p^n-1$ its exponent, any $a \in F$ would be a root of $x^{e+1}-x$ having at most $e+1$ roots contradicting that $F$ has $p^n$ elements. Whence $F^\times$ is cyclic.

Answer 2

Let $k$ be a subfield. What is its characteristic ? What is therefore its cardinal ? The cardinal of $k^\times$ ?

What does Lagrange's theorem then tell you about its elements ? And then about the elements of $k=k^\times\cup\{0\}$ ? How does that relate to powers of $\phi$ ?