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Let $K$ be a fixed field in $\mathbb C$ (complex numbers) of an automorphism of $\mathbb C$. Prove that every finite extension of $K$ in $\mathbb C$ is cyclic.

Thank you for your help!

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Do u mean field fixed by an automorphism of C? –  Dinesh May 3 '11 at 6:39
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Let $\sigma$ be the automorphism. Given an extension $L$ of $K$, first show that a power of $\sigma$ acts as identity on $L$. Now you have the following problem: let $L$ be a field, $G$ a finite subgroup of $\mathrm{Aut}(L)$. If $K$ is the subfield of $L$ fixed by $G$, then $L/K$ is Galois with Galois group $G$. –  Jiangwei Xue May 3 '11 at 6:42
    
That's exactly what I mean. –  Danny May 3 '11 at 6:42
    
Thank you Jiangwei! I will try that. –  Danny May 3 '11 at 6:43
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I might be confused about something but surely K is either $\mathbb R$ or $\mathbb C$? (So the group is either trivial or $C_2$) –  quanta May 3 '11 at 7:33

1 Answer 1

As Jyrki Lahtonen kindly pointed out in a comment below, there was a gap in the previous version of this answer. After having read Jyrki's comment, I also noticed that an answer, which (unlike mine) was correct, had been given by Jiangwei Xue in comment of May $3$, $2011$. I'll write Jiangwei Xue's solution below, and turn this answer into a community wiki.

Recall the following.

Let $G$ be a finite group of automorphisms of a field $L$, and let $K$ be the fixed field. Then $L/K$ is Galois with Galois group $G$.

If $G$ is abelian, any sub-extension $S/K$ of $L/K$ is Galois with Galois group a quotient of $G$. Thus $S/K$ is abelian.

If $G$ is cyclic, then so is any sub-extension $S/K$ of $L/K$.

Now here is Jiangwei Xue's argument.

Let $\sigma$ be an automorphism of a field $M$, let $K$ be the fixed field, and let $L/K$ be a finite degree sub-extension of $M/K$. Then the restriction of $\sigma$ to $M$ generates a finite group $G$ of automorphisms of $M$ whose fixed field is $K$, and the above observations show that $L/K$ is cyclic.

Aside. If $G$ be a finite group of automorphisms of an algebraically closed field $L$, then $G$ has order $1$ or $2$. Here are three references for this:

$\bullet$ These notes of our friend Pete L. Clark on Field Theory (see Theorem $124$ p. $75$, called "Grand Artin-Schreier Theorem"),

$\bullet$ The sub-entry Definitions of the entry Real closed field in Wikipedia,

$\bullet$ This nLab entry.

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All of this is correct, but the group generated by a single automorphism of $\mathbf{C}$ is infinite more often than not, right? –  Jyrki Lahtonen Jan 14 '12 at 14:28
    
Dear @Jyrki: Thank you very much! We, MSE users, are very lucky to have people like you around! I hope the new version is correct. (I don't know why, I figured that the automorphism was supposed to be of finite order.) –  Pierre-Yves Gaillard Jan 14 '12 at 15:40
    
You are too kind (you had my +1) already. My infinite Galois theory is too rusty for me to reliably answer a question like. You deserve the credit, so "Credit Waived", was IMHO somewhat overkill, but kudos to you! –  Jyrki Lahtonen Jan 15 '12 at 9:58

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