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I am not well trained in number theory. If there is any mistake, please straightly edit this question or devote me if the question is too trivial.

According to https://en.wikipedia.org/wiki/Perfect_field, a field $k$ is perfect iff $char(k)=0$ or $char(k)=p>0$ with Frobenius endomorphism being an isomorphism of $k$.

I think at least the Pete L. Clark's answer in Are all extensions of finite fields cyclic? having gently solved the $char(k)=p>0$ case of the problem, but the same method can not be applied to characteristic zero cases.

user623904
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  • Addendum: This is a fact mentioned in Serre's $\it{Local,, Field}$, giving the motivation for a claim. Let $k_{n}$ be a cyclic extension of $k$ of degree $n$ and $F^{n}$ be the generator of $G_{k_{n}/k}$. This claim is, if $k$ is perfect and $k^{sep}$ is the union of all cyclic extensions $k^{n}$ for all $n$ with $F^{mn}|{k{n}}=F^{n}$, then $k$ is a quasi-finite field. – user623904 Jan 04 '19 at 07:40
  • There are noncyclic extensions of fields of characteristic zero. – Angina Seng Jan 04 '19 at 07:47
  • Not to mention that your link does not solve the problem, since not every perfect field of positive characteristic is finite... – Kenny Lau Jan 04 '19 at 07:55
  • @Lord Shark the Unknown I will grateful if you can provide any explicit example for a finite extension of a characteristic zero field with Galois group which is not cyclic, and I think a counterexample is also an answer for this problem. I failed to find an example at present. – user623904 Jan 05 '19 at 05:50
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    Try $\Bbb Q(\sqrt 2, \sqrt 3) / \Bbb Q$ – Watson Jan 05 '19 at 09:47
  • @Watson As far as I’m concerned, you gave a counterexample for this claim, which means the answer to this question should be NO! Thanks a lot. As for the calculations, I think it is the same manner with the calculations here – user623904 Jan 05 '19 at 10:24

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