Show that $mathbb{Q}(alpha+ialpha)/mathbb{Q}$ is not Galois
Question: Let $alpha$ be a real root of $x^4-5 in mathbb{Q}[X].$
Show that $mathbb{Q}(alpha+ialpha)/mathbb{Q}$ is not Galois extension.
To show that an extension is not Galois, we just need to show that $mathbb{Q}(alpha+ialpha)$ is not a splitting field of any polynomial over $mathbb{Q}.$
So I guess we start with supposing it is the splitting field of some polynomial.
But I do not know what contradiction to be derived here.
Any hint is appreciated.
abstract-algebra field-theory galois-theory
asked Nov 19 '18 at 6:18
Idonknow
2,284749112
add a comment |
Question: Let $alpha$ be a real root of $x^4-5 in mathbb{Q}[X].$
Show that $mathbb{Q}(alpha+ialpha)/mathbb{Q}$ is not Galois extension.
To show that an extension is not Galois, we just need to show that $mathbb{Q}(alpha+ialpha)$ is not a splitting field of any polynomial over $mathbb{Q}.$
So I guess we start with supposing it is the splitting field of some polynomial.
But I do not know what contradiction to be derived here.
Any hint is appreciated.
abstract-algebra field-theory galois-theory
asked Nov 19 '18 at 6:18
Idonknow
2,284749112
add a comment |
Question: Let $alpha$ be a real root of $x^4-5 in mathbb{Q}[X].$
Show that $mathbb{Q}(alpha+ialpha)/mathbb{Q}$ is not Galois extension.
To show that an extension is not Galois, we just need to show that $mathbb{Q}(alpha+ialpha)$ is not a splitting field of any polynomial over $mathbb{Q}.$
So I guess we start with supposing it is the splitting field of some polynomial.
But I do not know what contradiction to be derived here.
Any hint is appreciated.
abstract-algebra field-theory galois-theory
asked Nov 19 '18 at 6:18
Idonknow
2,284749112
Question: Let $alpha$ be a real root of $x^4-5 in mathbb{Q}[X].$
Show that $mathbb{Q}(alpha+ialpha)/mathbb{Q}$ is not Galois extension.
To show that an extension is not Galois, we just need to show that $mathbb{Q}(alpha+ialpha)$ is not a splitting field of any polynomial over $mathbb{Q}.$
So I guess we start with supposing it is the splitting field of some polynomial.
But I do not know what contradiction to be derived here.
Any hint is appreciated.
abstract-algebra field-theory galois-theory
abstract-algebra field-theory galois-theory
asked Nov 19 '18 at 6:18
Idonknow
2,284749112
asked Nov 19 '18 at 6:18
Idonknow
2,284749112
asked Nov 19 '18 at 6:18
Idonknow
2,284749112
asked Nov 19 '18 at 6:18
Idonknow
2,284749112
asked Nov 19 '18 at 6:18
Idonknow
2,284749112
2,284749112
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
The splitting field of $K$ of $X^4-5$ is $Bbb Q(i,sqrt[4]5)$. Its Galois
group is dihedral order $8$, generated by $sigma$ and $tau$, where $sigma(i)=i$
and $sigma(sqrt[4]5)=isqrt[4]5$, and where $tau$ is complex conjugation.
Set $alpha=sqrt[4]5$. Then $sigma(tau((1+i)alpha))=sigma((1-i)alpha)
=(1+i)alpha$. Therefore $beta=(1+i)alpha$ is fixed by $sigmatau$. The
powers of $tau$ take $beta$ to four distinct values, so the stabiliser
of $beta$ under the action of $G$ is $H={text{id},sigmatau}$.
Therefore $K$ is the fixed field of $H$. But $H$ is not a normal
subgroup of $G$. Therefore $K$ is not Galois over $Bbb Q$.
answered Nov 19 '18 at 6:34
Lord Shark the Unknown
101k958132
add a comment |
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
});
}
});
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%2f3004586%2fshow-that-mathbbq-alphai-alpha-mathbbq-is-not-galois%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
The splitting field of $K$ of $X^4-5$ is $Bbb Q(i,sqrt[4]5)$. Its Galois
group is dihedral order $8$, generated by $sigma$ and $tau$, where $sigma(i)=i$
and $sigma(sqrt[4]5)=isqrt[4]5$, and where $tau$ is complex conjugation.
Set $alpha=sqrt[4]5$. Then $sigma(tau((1+i)alpha))=sigma((1-i)alpha)
=(1+i)alpha$. Therefore $beta=(1+i)alpha$ is fixed by $sigmatau$. The
powers of $tau$ take $beta$ to four distinct values, so the stabiliser
of $beta$ under the action of $G$ is $H={text{id},sigmatau}$.
Therefore $K$ is the fixed field of $H$. But $H$ is not a normal
subgroup of $G$. Therefore $K$ is not Galois over $Bbb Q$.
answered Nov 19 '18 at 6:34
Lord Shark the Unknown
101k958132
add a comment |
The splitting field of $K$ of $X^4-5$ is $Bbb Q(i,sqrt[4]5)$. Its Galois
group is dihedral order $8$, generated by $sigma$ and $tau$, where $sigma(i)=i$
and $sigma(sqrt[4]5)=isqrt[4]5$, and where $tau$ is complex conjugation.
Set $alpha=sqrt[4]5$. Then $sigma(tau((1+i)alpha))=sigma((1-i)alpha)
=(1+i)alpha$. Therefore $beta=(1+i)alpha$ is fixed by $sigmatau$. The
powers of $tau$ take $beta$ to four distinct values, so the stabiliser
of $beta$ under the action of $G$ is $H={text{id},sigmatau}$.
Therefore $K$ is the fixed field of $H$. But $H$ is not a normal
subgroup of $G$. Therefore $K$ is not Galois over $Bbb Q$.
answered Nov 19 '18 at 6:34
Lord Shark the Unknown
101k958132
add a comment |
The splitting field of $K$ of $X^4-5$ is $Bbb Q(i,sqrt[4]5)$. Its Galois
group is dihedral order $8$, generated by $sigma$ and $tau$, where $sigma(i)=i$
and $sigma(sqrt[4]5)=isqrt[4]5$, and where $tau$ is complex conjugation.
Set $alpha=sqrt[4]5$. Then $sigma(tau((1+i)alpha))=sigma((1-i)alpha)
=(1+i)alpha$. Therefore $beta=(1+i)alpha$ is fixed by $sigmatau$. The
powers of $tau$ take $beta$ to four distinct values, so the stabiliser
of $beta$ under the action of $G$ is $H={text{id},sigmatau}$.
Therefore $K$ is the fixed field of $H$. But $H$ is not a normal
subgroup of $G$. Therefore $K$ is not Galois over $Bbb Q$.
answered Nov 19 '18 at 6:34
Lord Shark the Unknown
101k958132
The splitting field of $K$ of $X^4-5$ is $Bbb Q(i,sqrt[4]5)$. Its Galois
group is dihedral order $8$, generated by $sigma$ and $tau$, where $sigma(i)=i$
and $sigma(sqrt[4]5)=isqrt[4]5$, and where $tau$ is complex conjugation.
Set $alpha=sqrt[4]5$. Then $sigma(tau((1+i)alpha))=sigma((1-i)alpha)
=(1+i)alpha$. Therefore $beta=(1+i)alpha$ is fixed by $sigmatau$. The
powers of $tau$ take $beta$ to four distinct values, so the stabiliser
of $beta$ under the action of $G$ is $H={text{id},sigmatau}$.
Therefore $K$ is the fixed field of $H$. But $H$ is not a normal
subgroup of $G$. Therefore $K$ is not Galois over $Bbb Q$.
answered Nov 19 '18 at 6:34
Lord Shark the Unknown
101k958132
answered Nov 19 '18 at 6:34
Lord Shark the Unknown
101k958132
answered Nov 19 '18 at 6:34
Lord Shark the Unknown
101k958132
answered Nov 19 '18 at 6:34
Lord Shark the Unknown
101k958132
101k958132
add a comment |
add a comment |
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
To learn more, see our tips on writing great answers.
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%2f3004586%2fshow-that-mathbbq-alphai-alpha-mathbbq-is-not-galois%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
