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Suppose $K$ is a field. Then we call $f: K\to K$ a (field) automorphism if $f$ is a one-to-one, onto and unital (i.e. $f(1)=1$) homomorphism of rings.

The following results are well-known.

  • There are no non-trivial automorphisms of $\mathbb{Q}$ (Proof).
  • If $\mathbb{F}_{p}$ is the field of $p$ elements ($p$ is prime), then there are no nontrivial automorphisms of $\mathbb{F}_{p}$. (Proof: $f(1)=1$ implies $f(n)=f(1+1+\cdots +1)=f(1)+f(1)+\cdots + f(1) = f(n)$ for each $n\in\mathbb{F}_p$).

It turns out that the the set of automorphisms of a field $K$ forms a group, known as the automorphism group of $K$, denoted by $\textrm{Aut}(K)$. So in all the above cases, the automorphism group is trivial.

My question is the following:

What is the automorphism group of the field $\mathbb{F}_{p^2}$ for a prime $p$ ? Here $\mathbb{F}_{p^2}$ denotes the (unique) field of $p^2$ elements.

I conjecture that the answer is the group $\mathbb{Z}/p\mathbb{Z}$. In the case $p=2$, we have $\mathbb{F}=\{0, 1, a, b\}$. Then, I believe the map $f:\mathbb{F}_4\to\mathbb{F}_4$ defined by $f(a)=b$ and $f(b)=a$ is the only non-trivial automorphism besides the identity. Hence, we have $\textrm{Aut}(F_4)=\{\textrm{identity}, f\}\cong\mathbb{Z}/2\mathbb{Z}$.

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Prism
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    The automorphism group is always cyclic, generated by the Frobenius automorphism $x\mapsto x^p$. The order of the group is $r$ on the field $F_{p^r}$. In particular, you should do by hand the case of $F_9$ before making any conjectures! :-) – Mariano Suárez-Álvarez Dec 01 '13 at 09:36
  • @MarianoSuárez-Alvarez: Thanks a lot! I think you have nailed it. Feel free to post the answer :) – Prism Dec 01 '13 at 09:41
  • To the downvoter: Couldn't you wait before I hit 2000 rep? :) On a more serious note, does my question really qualify as "This question does not show any useful research effort; its unclear or not useful."? Really now? I even link a MSE link to another question... – Prism Dec 01 '13 at 09:46
  • @Prism: Don't be so sensitive! I've given you a vote to put you back to 0. Perhaps the downvoter thought that you were taking too long to state well known things, like the set of automorphisms of a field forms a group. – Derek Holt Dec 01 '13 at 12:19
  • @DerekHolt: Thanks. I have now removed the off-topic comment of mine. – Prism Dec 01 '13 at 20:18
  • If $p>2$, then $\Bbb{F}_{p^2}=\Bbb{F}_p[\sqrt a]$ for some non-square element $a$ of $\Bbb{F}_p$, and for the purposes of constructing field automorphisms the situation is analogous to the extension $\Bbb{Q}[\sqrt a]/\Bbb{Q}$. In the case $p=2$ we dont get the quadratic extension by adjoining a square root, but you already covered that case :-) – Jyrki Lahtonen Dec 02 '13 at 04:56
  • @JyrkiLahtonen: Oh very nice perspective! Yeah, so the dimension of the automorphism group for $\mathbb{F}{p^2}$ would be kind of Galois group of the extension $\Bbb{F}_p[\sqrt a]/\Bbb{F}{p}$, which is $\mathbb{Z}/2\mathbb{Z}$. Hope I got it right. Thank you :-) – Prism Dec 02 '13 at 05:17

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