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Consider this question:

If K is a finite field and F is an algebraic closure of K, then $\operatorname{Aut}_K F $ is abelian. Every element of $\operatorname{Aut}_K F$ (except 1) has infinite order.

So, I chose $\sigma_1$ and $\sigma_2 \in \operatorname{Aut}_K F$ and f from F : $\sigma_{1} \sigma_2 (f) $ but I have no intuition on why it should be commutative. Also, I am really struggling to prove that Why every element has infinite order.

Kindly give some hints. I self study mathematics and don't have any help as I live in a very poor country.

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    Proof sketch for commutativity: Let $x \in F$ be arbitrary, and let $L = K(x)$. Then $(\sigma_1 \sigma_2)|_L = (\sigma_2 \sigma_1)|_L$ (figure out why!), so $\sigma_1(\sigma_2(x)) = \sigma_2(\sigma_1(x))$. Since $x$ was arbitrary, $\sigma_1$ and $\sigma_2$ commute. – diracdeltafunk Jan 15 '21 at 19:49
  • @RedditUser123 Actually I got some assignments and mid term exam questions of some institute and was solving them. I am currently working on problems . I have some pending target although I am self studying and am looking foreward to completing that and will surely read and accept answer. –  Jan 15 '21 at 20:09
  • For the second part see this. – Jyrki Lahtonen Jan 16 '21 at 07:21
  • @diracdeltafunk I am not able to figure it out. Can you please explain? –  Oct 17 '21 at 12:27

2 Answers2

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The key fact here is that any finite field $K$ has a unique degree $n$ extension, for each $n\ge1$. Denote such a field as $K_n$. We have that $F=\bigcup_{n\ge1} K_n$. [Try to show this]

Given this, for any element $\sigma\in Aut_KF$ and any $n\ge 1$, the automorphism restricts to $\sigma|_{K_n}\in Aut_K(K_n)$. [$\sigma(K_n)$ is a degree $n$ extension of $K$, so by uniqueness $\sigma(K_n)=K_n$.] Moreover, $\sigma$ must restrict to a group homomorphism $K_n^\times\to K_n^\times$, which is given by $x\mapsto x^q$ for some $q$, since $K_n^\times$ is cyclic.

$Aut_KF$ is abelian: For any two elements of $Aut_KF$, it suffices to check their restrictions to $K_n$ commute, for each field $n\ge1$, since $\sigma|_{K_n}\in Aut_K(K_n)$. Now such automorphisms have to be of the form $x\mapsto x^q$, so commutativity is immediate. Alternatively put, $Aut_K(K_n)$ can be seen as a subgroup of the abelian group $Aut(K_n^\times)$, and hence is also abelian.

Every nontrivial element of $Aut_KF$ has infinite order: Suppose $\sigma\ne1\in Aut_KF$ is such that $\sigma^N=1$. Consider $F^\sigma:=\{x\in F:\sigma(x)=x\}$.

In particular for each $n\ge1$, we have $(\sigma|_{K_{nN}})^N=1$, so for the subfield $K_{nN}^\sigma=K_{nN}\cap F^\sigma:=\{x\in K_{nN}:\sigma(x)=x\}$, each element $x\in K_{nN}$ is a root of the $K_{nN}^\sigma$-polynomial $\prod_{i=0}^{N-1}(T-\sigma^ix)$, so the index is $[K_{nN}:K_{nN}^\sigma]|N$, or $n|[K_{nN}^\sigma:K]$. In other words, $F^\sigma\supseteq K_{nN}^\sigma\supseteq K_n$. Since $n$ was arbitrary, this implies $F^\sigma=F$, that is, $\sigma=1$.

user600016
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Kenta S
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  • will have a look at this question soon –  Jan 16 '22 at 10:28
  • I have a few doubts. Why must each element of $K_{nN}$ be a root of the degree $N$ polynomial? What is the 'T" in that polynomial as well? Why is it necessary that $[K_{nN}:K_{nN}^{\sigma}]$ divides N? – user600016 Jan 22 '22 at 18:10
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I will try to elaborate on Kenta's solution for second part in a slightly different way.

Assuming same notations as Kenta.

Suppose $\sigma \neq 1 \in Aut_KF$ has finite order $N$. So $\sigma^N=1$. Consider $G:=\{1,\sigma,\sigma^2,\ldots,\sigma^{N-1}\}$.

We have the following towers: $F \supseteq F^G \supseteq K_{nN}^G \supseteq K$ and $F \supseteq K_{nN} \supseteq K$.

$[K_{nN}:K_{nN}^G]=|G|=N$, $[K_{nN}:K]=nN$ $\implies [K_{nN}^G:K]=n$.

But $[K_{n}:K]=n$ and any two finite fields of same order are isomorphic to each other and hence $K_{nN}^G \cong K_n$ and hence $K_n$ can be thought of as a subfield of $F^G$ $\forall n\geq1$(i.e. the two towers mentioned above are really the same).

Hence, $F=\bigcup_{n\geq1}K_n \subseteq F^G \implies F=F^G \implies \sigma=1$. Contradicting our assumption that there exists a finite order non-trivial element in $Aut_KF$.

user600016
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