Is $ mathrm{Aut}(mathrm{Gal}(bar{mathbb{Q}}/mathbb{Q})) $ known?












6












$begingroup$


Following my previous question about the outer automorphism group, I would like to know if the structure of the automorphism group of the absolute Galois group of the rationals is known. Specifically, is it isomorphic to the cyclic group with two elements ?






abstract-algebra group-theory galois-theory







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$endgroup$








  • 2




    $begingroup$
    There is no direct description known so far. This contains a few results: math.bu.edu/people/jsweinst/CEB/CEBTalk.pdf
    $endgroup$
    – AnalysisStudent0414
    Sep 17 '18 at 11:16
















6












$begingroup$


Following my previous question about the outer automorphism group, I would like to know if the structure of the automorphism group of the absolute Galois group of the rationals is known. Specifically, is it isomorphic to the cyclic group with two elements ?






abstract-algebra group-theory galois-theory







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    There is no direct description known so far. This contains a few results: math.bu.edu/people/jsweinst/CEB/CEBTalk.pdf
    $endgroup$
    – AnalysisStudent0414
    Sep 17 '18 at 11:16














6












6








6


2



$begingroup$


Following my previous question about the outer automorphism group, I would like to know if the structure of the automorphism group of the absolute Galois group of the rationals is known. Specifically, is it isomorphic to the cyclic group with two elements ?






abstract-algebra group-theory galois-theory







share|cite|improve this question











$endgroup$




Following my previous question about the outer automorphism group, I would like to know if the structure of the automorphism group of the absolute Galois group of the rationals is known. Specifically, is it isomorphic to the cyclic group with two elements ?






abstract-algebra group-theory galois-theory




abstract-algebra group-theory galois-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 17 '18 at 10:39









cansomeonehelpmeout

7,0873935




7,0873935










asked Sep 17 '18 at 10:13









Sylvain JulienSylvain Julien

1,140918




1,140918








  • 2




    $begingroup$
    There is no direct description known so far. This contains a few results: math.bu.edu/people/jsweinst/CEB/CEBTalk.pdf
    $endgroup$
    – AnalysisStudent0414
    Sep 17 '18 at 11:16














  • 2




    $begingroup$
    There is no direct description known so far. This contains a few results: math.bu.edu/people/jsweinst/CEB/CEBTalk.pdf
    $endgroup$
    – AnalysisStudent0414
    Sep 17 '18 at 11:16








2




2




$begingroup$
There is no direct description known so far. This contains a few results: math.bu.edu/people/jsweinst/CEB/CEBTalk.pdf
$endgroup$
– AnalysisStudent0414
Sep 17 '18 at 11:16




$begingroup$
There is no direct description known so far. This contains a few results: math.bu.edu/people/jsweinst/CEB/CEBTalk.pdf
$endgroup$
– AnalysisStudent0414
Sep 17 '18 at 11:16










1 Answer
1






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2












$begingroup$

By the Neukirch–Uchida theorem , Aut(Gal($mathbb{bar{Q}/Q}$))= Gal($mathbb{bar{Q}/Q}$).






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Isn't the theorem that $operatorname{Out}(operatorname{Gal}(overline{mathbb{Q}}/mathbb{Q})) = operatorname{Gal}(overline{mathbb{Q}}/mathbb{Q})$?
    $endgroup$
    – anomaly
    Dec 12 '18 at 13:48












  • $begingroup$
    @anomaly Every element in Gal($mathbb{bar{Q}/Q}$) acts by conjugation, which is obvious inner. Isn't it?
    $endgroup$
    – Bonbon
    Dec 12 '18 at 13:58






  • 1




    $begingroup$
    @anomaly I'm not familiar with Neukirch-Uchida, but it often happens for a non-abelian group that all its automorphisms are inner. A well known example is the symmetric group $S_n$, $nge5$, whose automorphisms are all inner except in the case $n=6$.
    $endgroup$
    – Jyrki Lahtonen
    Dec 13 '18 at 4:51






  • 1




    $begingroup$
    Thanks for the clarification @anomaly.
    $endgroup$
    – Jyrki Lahtonen
    Dec 13 '18 at 5:18






  • 1




    $begingroup$
    @anomaly In fact Out(Gal($mathbb{bar{Q}/Q}$))={1}. Gal($mathbb{bar{Q}/Q}$) is center-free, Aut(Gal($mathbb{bar{Q}/Q}$))=Inn(Gal($mathbb{bar{Q}/Q}$))$cong$Gal($mathbb{bar{Q}/Q}$)
    $endgroup$
    – Bonbon
    Dec 13 '18 at 7:02













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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

By the Neukirch–Uchida theorem , Aut(Gal($mathbb{bar{Q}/Q}$))= Gal($mathbb{bar{Q}/Q}$).






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Isn't the theorem that $operatorname{Out}(operatorname{Gal}(overline{mathbb{Q}}/mathbb{Q})) = operatorname{Gal}(overline{mathbb{Q}}/mathbb{Q})$?
    $endgroup$
    – anomaly
    Dec 12 '18 at 13:48












  • $begingroup$
    @anomaly Every element in Gal($mathbb{bar{Q}/Q}$) acts by conjugation, which is obvious inner. Isn't it?
    $endgroup$
    – Bonbon
    Dec 12 '18 at 13:58






  • 1




    $begingroup$
    @anomaly I'm not familiar with Neukirch-Uchida, but it often happens for a non-abelian group that all its automorphisms are inner. A well known example is the symmetric group $S_n$, $nge5$, whose automorphisms are all inner except in the case $n=6$.
    $endgroup$
    – Jyrki Lahtonen
    Dec 13 '18 at 4:51






  • 1




    $begingroup$
    Thanks for the clarification @anomaly.
    $endgroup$
    – Jyrki Lahtonen
    Dec 13 '18 at 5:18






  • 1




    $begingroup$
    @anomaly In fact Out(Gal($mathbb{bar{Q}/Q}$))={1}. Gal($mathbb{bar{Q}/Q}$) is center-free, Aut(Gal($mathbb{bar{Q}/Q}$))=Inn(Gal($mathbb{bar{Q}/Q}$))$cong$Gal($mathbb{bar{Q}/Q}$)
    $endgroup$
    – Bonbon
    Dec 13 '18 at 7:02


















2












$begingroup$

By the Neukirch–Uchida theorem , Aut(Gal($mathbb{bar{Q}/Q}$))= Gal($mathbb{bar{Q}/Q}$).






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Isn't the theorem that $operatorname{Out}(operatorname{Gal}(overline{mathbb{Q}}/mathbb{Q})) = operatorname{Gal}(overline{mathbb{Q}}/mathbb{Q})$?
    $endgroup$
    – anomaly
    Dec 12 '18 at 13:48












  • $begingroup$
    @anomaly Every element in Gal($mathbb{bar{Q}/Q}$) acts by conjugation, which is obvious inner. Isn't it?
    $endgroup$
    – Bonbon
    Dec 12 '18 at 13:58






  • 1




    $begingroup$
    @anomaly I'm not familiar with Neukirch-Uchida, but it often happens for a non-abelian group that all its automorphisms are inner. A well known example is the symmetric group $S_n$, $nge5$, whose automorphisms are all inner except in the case $n=6$.
    $endgroup$
    – Jyrki Lahtonen
    Dec 13 '18 at 4:51






  • 1




    $begingroup$
    Thanks for the clarification @anomaly.
    $endgroup$
    – Jyrki Lahtonen
    Dec 13 '18 at 5:18






  • 1




    $begingroup$
    @anomaly In fact Out(Gal($mathbb{bar{Q}/Q}$))={1}. Gal($mathbb{bar{Q}/Q}$) is center-free, Aut(Gal($mathbb{bar{Q}/Q}$))=Inn(Gal($mathbb{bar{Q}/Q}$))$cong$Gal($mathbb{bar{Q}/Q}$)
    $endgroup$
    – Bonbon
    Dec 13 '18 at 7:02
















2












2








2





$begingroup$

By the Neukirch–Uchida theorem , Aut(Gal($mathbb{bar{Q}/Q}$))= Gal($mathbb{bar{Q}/Q}$).






share|cite|improve this answer









$endgroup$



By the Neukirch–Uchida theorem , Aut(Gal($mathbb{bar{Q}/Q}$))= Gal($mathbb{bar{Q}/Q}$).







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 12 '18 at 12:47









BonbonBonbon

47118




47118








  • 1




    $begingroup$
    Isn't the theorem that $operatorname{Out}(operatorname{Gal}(overline{mathbb{Q}}/mathbb{Q})) = operatorname{Gal}(overline{mathbb{Q}}/mathbb{Q})$?
    $endgroup$
    – anomaly
    Dec 12 '18 at 13:48












  • $begingroup$
    @anomaly Every element in Gal($mathbb{bar{Q}/Q}$) acts by conjugation, which is obvious inner. Isn't it?
    $endgroup$
    – Bonbon
    Dec 12 '18 at 13:58






  • 1




    $begingroup$
    @anomaly I'm not familiar with Neukirch-Uchida, but it often happens for a non-abelian group that all its automorphisms are inner. A well known example is the symmetric group $S_n$, $nge5$, whose automorphisms are all inner except in the case $n=6$.
    $endgroup$
    – Jyrki Lahtonen
    Dec 13 '18 at 4:51






  • 1




    $begingroup$
    Thanks for the clarification @anomaly.
    $endgroup$
    – Jyrki Lahtonen
    Dec 13 '18 at 5:18






  • 1




    $begingroup$
    @anomaly In fact Out(Gal($mathbb{bar{Q}/Q}$))={1}. Gal($mathbb{bar{Q}/Q}$) is center-free, Aut(Gal($mathbb{bar{Q}/Q}$))=Inn(Gal($mathbb{bar{Q}/Q}$))$cong$Gal($mathbb{bar{Q}/Q}$)
    $endgroup$
    – Bonbon
    Dec 13 '18 at 7:02
















  • 1




    $begingroup$
    Isn't the theorem that $operatorname{Out}(operatorname{Gal}(overline{mathbb{Q}}/mathbb{Q})) = operatorname{Gal}(overline{mathbb{Q}}/mathbb{Q})$?
    $endgroup$
    – anomaly
    Dec 12 '18 at 13:48












  • $begingroup$
    @anomaly Every element in Gal($mathbb{bar{Q}/Q}$) acts by conjugation, which is obvious inner. Isn't it?
    $endgroup$
    – Bonbon
    Dec 12 '18 at 13:58






  • 1




    $begingroup$
    @anomaly I'm not familiar with Neukirch-Uchida, but it often happens for a non-abelian group that all its automorphisms are inner. A well known example is the symmetric group $S_n$, $nge5$, whose automorphisms are all inner except in the case $n=6$.
    $endgroup$
    – Jyrki Lahtonen
    Dec 13 '18 at 4:51






  • 1




    $begingroup$
    Thanks for the clarification @anomaly.
    $endgroup$
    – Jyrki Lahtonen
    Dec 13 '18 at 5:18






  • 1




    $begingroup$
    @anomaly In fact Out(Gal($mathbb{bar{Q}/Q}$))={1}. Gal($mathbb{bar{Q}/Q}$) is center-free, Aut(Gal($mathbb{bar{Q}/Q}$))=Inn(Gal($mathbb{bar{Q}/Q}$))$cong$Gal($mathbb{bar{Q}/Q}$)
    $endgroup$
    – Bonbon
    Dec 13 '18 at 7:02










1




1




$begingroup$
Isn't the theorem that $operatorname{Out}(operatorname{Gal}(overline{mathbb{Q}}/mathbb{Q})) = operatorname{Gal}(overline{mathbb{Q}}/mathbb{Q})$?
$endgroup$
– anomaly
Dec 12 '18 at 13:48






$begingroup$
Isn't the theorem that $operatorname{Out}(operatorname{Gal}(overline{mathbb{Q}}/mathbb{Q})) = operatorname{Gal}(overline{mathbb{Q}}/mathbb{Q})$?
$endgroup$
– anomaly
Dec 12 '18 at 13:48














$begingroup$
@anomaly Every element in Gal($mathbb{bar{Q}/Q}$) acts by conjugation, which is obvious inner. Isn't it?
$endgroup$
– Bonbon
Dec 12 '18 at 13:58




$begingroup$
@anomaly Every element in Gal($mathbb{bar{Q}/Q}$) acts by conjugation, which is obvious inner. Isn't it?
$endgroup$
– Bonbon
Dec 12 '18 at 13:58




1




1




$begingroup$
@anomaly I'm not familiar with Neukirch-Uchida, but it often happens for a non-abelian group that all its automorphisms are inner. A well known example is the symmetric group $S_n$, $nge5$, whose automorphisms are all inner except in the case $n=6$.
$endgroup$
– Jyrki Lahtonen
Dec 13 '18 at 4:51




$begingroup$
@anomaly I'm not familiar with Neukirch-Uchida, but it often happens for a non-abelian group that all its automorphisms are inner. A well known example is the symmetric group $S_n$, $nge5$, whose automorphisms are all inner except in the case $n=6$.
$endgroup$
– Jyrki Lahtonen
Dec 13 '18 at 4:51




1




1




$begingroup$
Thanks for the clarification @anomaly.
$endgroup$
– Jyrki Lahtonen
Dec 13 '18 at 5:18




$begingroup$
Thanks for the clarification @anomaly.
$endgroup$
– Jyrki Lahtonen
Dec 13 '18 at 5:18




1




1




$begingroup$
@anomaly In fact Out(Gal($mathbb{bar{Q}/Q}$))={1}. Gal($mathbb{bar{Q}/Q}$) is center-free, Aut(Gal($mathbb{bar{Q}/Q}$))=Inn(Gal($mathbb{bar{Q}/Q}$))$cong$Gal($mathbb{bar{Q}/Q}$)
$endgroup$
– Bonbon
Dec 13 '18 at 7:02






$begingroup$
@anomaly In fact Out(Gal($mathbb{bar{Q}/Q}$))={1}. Gal($mathbb{bar{Q}/Q}$) is center-free, Aut(Gal($mathbb{bar{Q}/Q}$))=Inn(Gal($mathbb{bar{Q}/Q}$))$cong$Gal($mathbb{bar{Q}/Q}$)
$endgroup$
– Bonbon
Dec 13 '18 at 7:02




















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