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
edited Sep 17 '18 at 10:39
cansomeonehelpmeout
7,0873935
asked Sep 17 '18 at 10:13
Sylvain JulienSylvain Julien
1,140918
$endgroup$
add a comment |
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
edited Sep 17 '18 at 10:39
cansomeonehelpmeout
7,0873935
asked Sep 17 '18 at 10:13
Sylvain JulienSylvain Julien
1,140918
$endgroup$
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
add a comment |
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
edited Sep 17 '18 at 10:39
cansomeonehelpmeout
7,0873935
asked Sep 17 '18 at 10:13
Sylvain JulienSylvain Julien
1,140918
$endgroup$
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
edited Sep 17 '18 at 10:39
cansomeonehelpmeout
7,0873935
asked Sep 17 '18 at 10:13
Sylvain JulienSylvain Julien
1,140918
edited Sep 17 '18 at 10:39
cansomeonehelpmeout
7,0873935
asked Sep 17 '18 at 10:13
Sylvain JulienSylvain Julien
1,140918
edited Sep 17 '18 at 10:39
cansomeonehelpmeout
7,0873935
edited Sep 17 '18 at 10:39
cansomeonehelpmeout
7,0873935
edited Sep 17 '18 at 10:39
cansomeonehelpmeout
7,0873935
7,0873935
asked Sep 17 '18 at 10:13
Sylvain JulienSylvain Julien
1,140918
asked Sep 17 '18 at 10:13
Sylvain JulienSylvain Julien
1,140918
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
2
$begingroup$
By the Neukirch–Uchida theorem , Aut(Gal($mathbb{bar{Q}/Q}$))= Gal($mathbb{bar{Q}/Q}$).
answered Dec 12 '18 at 12:47
BonbonBonbon
47118
$endgroup$
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
|
show 5 more comments
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2920068%2fis-mathrmaut-mathrmgal-bar-mathbbq-mathbbq-known%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
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($mathbb{bar{Q}/Q}$).
answered Dec 12 '18 at 12:47
BonbonBonbon
47118
$endgroup$
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
|
show 5 more comments
2
$begingroup$
By the Neukirch–Uchida theorem , Aut(Gal($mathbb{bar{Q}/Q}$))= Gal($mathbb{bar{Q}/Q}$).
answered Dec 12 '18 at 12:47
BonbonBonbon
47118
$endgroup$
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
|
show 5 more comments
2
2
2
$begingroup$
By the Neukirch–Uchida theorem , Aut(Gal($mathbb{bar{Q}/Q}$))= Gal($mathbb{bar{Q}/Q}$).
answered Dec 12 '18 at 12:47
BonbonBonbon
47118
$endgroup$
By the Neukirch–Uchida theorem , Aut(Gal($mathbb{bar{Q}/Q}$))= Gal($mathbb{bar{Q}/Q}$).
answered Dec 12 '18 at 12:47
BonbonBonbon
47118
answered Dec 12 '18 at 12:47
BonbonBonbon
47118
answered Dec 12 '18 at 12:47
BonbonBonbon
47118
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
|
show 5 more comments
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
|
show 5 more comments
draft saved
draft discarded
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2920068%2fis-mathrmaut-mathrmgal-bar-mathbbq-mathbbq-known%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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