Are all finite extensions of perfect fields cyclic?












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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.










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$endgroup$












  • $begingroup$
    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.
    $endgroup$
    – user623904
    Jan 4 at 7:40












  • $begingroup$
    There are noncyclic extensions of fields of characteristic zero.
    $endgroup$
    – Lord Shark the Unknown
    Jan 4 at 7:47










  • $begingroup$
    Not to mention that your link does not solve the problem, since not every perfect field of positive characteristic is finite...
    $endgroup$
    – Kenny Lau
    Jan 4 at 7:55










  • $begingroup$
    @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.
    $endgroup$
    – user623904
    Jan 5 at 5:50








  • 1




    $begingroup$
    Try $Bbb Q(sqrt 2, sqrt 3) / Bbb Q$
    $endgroup$
    – Watson
    Jan 5 at 9:47
















1












$begingroup$


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.










share|cite|improve this question











$endgroup$












  • $begingroup$
    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.
    $endgroup$
    – user623904
    Jan 4 at 7:40












  • $begingroup$
    There are noncyclic extensions of fields of characteristic zero.
    $endgroup$
    – Lord Shark the Unknown
    Jan 4 at 7:47










  • $begingroup$
    Not to mention that your link does not solve the problem, since not every perfect field of positive characteristic is finite...
    $endgroup$
    – Kenny Lau
    Jan 4 at 7:55










  • $begingroup$
    @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.
    $endgroup$
    – user623904
    Jan 5 at 5:50








  • 1




    $begingroup$
    Try $Bbb Q(sqrt 2, sqrt 3) / Bbb Q$
    $endgroup$
    – Watson
    Jan 5 at 9:47














1












1








1





$begingroup$


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.










share|cite|improve this question











$endgroup$




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.







algebraic-number-theory extension-field






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 8:17







user623904

















asked Jan 4 at 7:39









user623904user623904

234




234












  • $begingroup$
    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.
    $endgroup$
    – user623904
    Jan 4 at 7:40












  • $begingroup$
    There are noncyclic extensions of fields of characteristic zero.
    $endgroup$
    – Lord Shark the Unknown
    Jan 4 at 7:47










  • $begingroup$
    Not to mention that your link does not solve the problem, since not every perfect field of positive characteristic is finite...
    $endgroup$
    – Kenny Lau
    Jan 4 at 7:55










  • $begingroup$
    @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.
    $endgroup$
    – user623904
    Jan 5 at 5:50








  • 1




    $begingroup$
    Try $Bbb Q(sqrt 2, sqrt 3) / Bbb Q$
    $endgroup$
    – Watson
    Jan 5 at 9:47


















  • $begingroup$
    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.
    $endgroup$
    – user623904
    Jan 4 at 7:40












  • $begingroup$
    There are noncyclic extensions of fields of characteristic zero.
    $endgroup$
    – Lord Shark the Unknown
    Jan 4 at 7:47










  • $begingroup$
    Not to mention that your link does not solve the problem, since not every perfect field of positive characteristic is finite...
    $endgroup$
    – Kenny Lau
    Jan 4 at 7:55










  • $begingroup$
    @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.
    $endgroup$
    – user623904
    Jan 5 at 5:50








  • 1




    $begingroup$
    Try $Bbb Q(sqrt 2, sqrt 3) / Bbb Q$
    $endgroup$
    – Watson
    Jan 5 at 9:47
















$begingroup$
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.
$endgroup$
– user623904
Jan 4 at 7:40






$begingroup$
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.
$endgroup$
– user623904
Jan 4 at 7:40














$begingroup$
There are noncyclic extensions of fields of characteristic zero.
$endgroup$
– Lord Shark the Unknown
Jan 4 at 7:47




$begingroup$
There are noncyclic extensions of fields of characteristic zero.
$endgroup$
– Lord Shark the Unknown
Jan 4 at 7:47












$begingroup$
Not to mention that your link does not solve the problem, since not every perfect field of positive characteristic is finite...
$endgroup$
– Kenny Lau
Jan 4 at 7:55




$begingroup$
Not to mention that your link does not solve the problem, since not every perfect field of positive characteristic is finite...
$endgroup$
– Kenny Lau
Jan 4 at 7:55












$begingroup$
@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.
$endgroup$
– user623904
Jan 5 at 5:50






$begingroup$
@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.
$endgroup$
– user623904
Jan 5 at 5:50






1




1




$begingroup$
Try $Bbb Q(sqrt 2, sqrt 3) / Bbb Q$
$endgroup$
– Watson
Jan 5 at 9:47




$begingroup$
Try $Bbb Q(sqrt 2, sqrt 3) / Bbb Q$
$endgroup$
– Watson
Jan 5 at 9:47










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