[Math] My proof: Frobenius Map generates $\mbox{Gal}(\mathbb F_{p^n}/\mathbb F_p)$

abstract-algebrafinite-fieldsgalois-theoryproof-verification

I would like to ask, whether anyone can confirm or correct the following version of the proof that the Frobenius map generates the Galois Group of a finite field.

Proof

First note that $\newcommand{\Fpn}{\mathbb F_{p^n}}\newcommand{\Fp}{\mathbb F_p}\newcommand{\GalFpn}{\mbox{Gal}(\Fpn/\Fp)}\#\GalFpn=[\Fpn:\Fp]=n$. I want to show that the Frobenius map $\newcommand{\Fr}{\mbox{Fr}}\Fr:\Fpn\rightarrow\Fpn$ given by $\alpha\mapsto\alpha^p$ generates the Galois Group of the extension $\Fpn\supset\Fp$. To see this I will show the following two statements.

  1. $\Fr$ is an element of $\GalFpn$
  2. $\Fr$ has order $n$

Statement 1

Since the group $\Fp^*$ has $p-1$ elements is follows that $a\in\Fp$ satisfies $a^{p-1}=1$ so that $\Fr(a)=a^p=a$. This shows that $\Fr$ fixes $\Fp$.

Now note that $\Fr$ is injective since $\alpha^p=0$ if and only if $\alpha=0$ (there are no zero divisors in a field). Since we have a finite field $\Fpn$ an injective map is also surjective. This shows that $\Fr$ is bijective so $\Fr$ is contained in $\GalFpn$.

Statement 2

Next we must establish that $\Fr$ has order $n$. Consider the multiplicative group $\Fpn^*$. This is a cyclic group with $p^n-1$ elements, so we have an $\alpha\in\Fpn$ of order $p^n-1$. Thus
$$
\alpha^p,\alpha^{p^2},…,\alpha^{p^n}
$$
will be $n$ distinct elements which is really just to say that
$$
\Fr(\alpha),\Fr^2(\alpha),…,\Fr^n(\alpha)
$$
are $n$ distinct elements. This shows that $\Fr$ has order $n$ so that it generates $\GalFpn$.

Best Answer

The proof is correct, but you can do it faster:

Every element of $\mathbb{F}_p$ is fixed by any automorphism - it is just a sum of $1$'s. By Lagrange we have $\alpha^{p^n-1}=1$ for all $\alpha \in \mathbb{F}_{p^n}^*$, hence $\alpha^{p^n}=\alpha$ for all $\alpha \in \mathbb{F}_{p^n}$. If there is some $m<n$ with this property, then $\mathbb{F}_{p^n}$ would have only $p^m$ elements, a contradiction. Thus, $\mathrm{Frob}$ has order $p^n$. In particular, $\mathrm{Frob}$ is an isomorphism (since we have found an inverse) and it generates the Galois group.

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