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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.


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 $\GalFpn$.

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This proof is correct. – hunter Jan 14 '14 at 10:25
@hunter: Thank you very much! This was the last part of the subject of finite fields covered in our course. The lecturer gave us no notes and the textbook also has none, so I had to reconstruct from memory and check it. Thanks again. – String Jan 14 '14 at 10:29
up vote 5 down vote accepted

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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That is actually very nice! How do you know that if there were $m<n$ with the said property then $\Fpn$ would have only $p^m$ elements. This has to have something to do with $\Fpn^*$ being cyclic, right? – String Jan 14 '14 at 10:54
No. Actually I wanted to emphasize that we don't need this result. It's very simple, remember how many roots a polynomial of degree $d$ has (at most). – Martin Brandenburg Jan 14 '14 at 10:57
But I really like the proof that it fixes $\Fp$ by using its homomorhpism properties on $1$'s. I think I get the idea that we have found an inverse. The inverse must be $F^{n-1}$ given by $\alpha\mapsto\alpha^{p^{n-1}}$, correct? – String Jan 14 '14 at 10:57
That's correct. Observe that this applies to any extension of $\mathbb{F}_p$, not just finite fields: Every automorphism fixed $\mathbb{F}_p$ automatically (and in characteristic $0$, we have the same for $\mathbb{Q}$). – Martin Brandenburg Jan 14 '14 at 10:59
So you are saying that if we have $\alpha^{p^m}=\alpha$ for some $m<n$ then $\Fpn^*$ and even $\Fpn$ would be only roots of $X^{p^m}-X$. These roots are to few, right? – String Jan 14 '14 at 11:02

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