Mixing
Degree of splitting fields
$begingroup$
I'm learning about splitting fields but I'm not sure if I am right. Hopefully I can get some insights on whether I have been learning correctly.
The question asks to find the degree of the splitting fields over $mathbb Q$.
In $x^4+x^3+x+1$, we can factorize it to become $(x+1)^2(x^2-x+1)=(x+1)^2(x-frac{1+sqrt 3i}2)(x-frac {1-sqrt 3i}{2})$. Am I right to say that we want to find $[mathbb Q(sqrt 3i):mathbb Q]$? Hence the answer is just $2$.
In $x^4+x^2+x+1$, we can factorize it to become $(x-frac 14sqrt5+frac 14-frac14isqrt2sqrt{5+sqrt5})(x+frac 14sqrt5+frac 14-frac14isqrt2sqrt{5-sqrt5})\(x+frac 14sqrt5+frac 14+frac14isqrt2sqrt{5-sqrt5})(x-frac 14sqrt5+frac 14+frac14isqrt2sqrt{5+sqrt5})$.
Hence, we want to find $mathbb Q(sqrt5,isqrt2sqrt{5+sqrt5},isqrt2sqrt{5-sqrt5})$. But how do we go about doing this?
abstract-algebra field-theory galois-theory extension-field splitting-field
asked Jan 9 at 10:10
IcycarusIcycarus
4751313
$endgroup$
add a comment |
$begingroup$
I'm learning about splitting fields but I'm not sure if I am right. Hopefully I can get some insights on whether I have been learning correctly.
The question asks to find the degree of the splitting fields over $mathbb Q$.
In $x^4+x^3+x+1$, we can factorize it to become $(x+1)^2(x^2-x+1)=(x+1)^2(x-frac{1+sqrt 3i}2)(x-frac {1-sqrt 3i}{2})$. Am I right to say that we want to find $[mathbb Q(sqrt 3i):mathbb Q]$? Hence the answer is just $2$.
In $x^4+x^2+x+1$, we can factorize it to become $(x-frac 14sqrt5+frac 14-frac14isqrt2sqrt{5+sqrt5})(x+frac 14sqrt5+frac 14-frac14isqrt2sqrt{5-sqrt5})\(x+frac 14sqrt5+frac 14+frac14isqrt2sqrt{5-sqrt5})(x-frac 14sqrt5+frac 14+frac14isqrt2sqrt{5+sqrt5})$.
Hence, we want to find $mathbb Q(sqrt5,isqrt2sqrt{5+sqrt5},isqrt2sqrt{5-sqrt5})$. But how do we go about doing this?
abstract-algebra field-theory galois-theory extension-field splitting-field
asked Jan 9 at 10:10
IcycarusIcycarus
4751313
$endgroup$
$begingroup$
Yes, the degree of the spliting field of $(X-1)^2(X^2-X+1)$ is of course equal to $2$. Why do you want to involve $mathbb Q(sqrt5,isqrt2sqrt{5+sqrt5},isqrt2sqrt{5-sqrt5})$?
$endgroup$
– Dietrich Burde
Jan 9 at 12:11
add a comment |
$begingroup$
I'm learning about splitting fields but I'm not sure if I am right. Hopefully I can get some insights on whether I have been learning correctly.
The question asks to find the degree of the splitting fields over $mathbb Q$.
In $x^4+x^3+x+1$, we can factorize it to become $(x+1)^2(x^2-x+1)=(x+1)^2(x-frac{1+sqrt 3i}2)(x-frac {1-sqrt 3i}{2})$. Am I right to say that we want to find $[mathbb Q(sqrt 3i):mathbb Q]$? Hence the answer is just $2$.
In $x^4+x^2+x+1$, we can factorize it to become $(x-frac 14sqrt5+frac 14-frac14isqrt2sqrt{5+sqrt5})(x+frac 14sqrt5+frac 14-frac14isqrt2sqrt{5-sqrt5})\(x+frac 14sqrt5+frac 14+frac14isqrt2sqrt{5-sqrt5})(x-frac 14sqrt5+frac 14+frac14isqrt2sqrt{5+sqrt5})$.
Hence, we want to find $mathbb Q(sqrt5,isqrt2sqrt{5+sqrt5},isqrt2sqrt{5-sqrt5})$. But how do we go about doing this?
abstract-algebra field-theory galois-theory extension-field splitting-field
asked Jan 9 at 10:10
IcycarusIcycarus
4751313
$endgroup$
I'm learning about splitting fields but I'm not sure if I am right. Hopefully I can get some insights on whether I have been learning correctly.
The question asks to find the degree of the splitting fields over $mathbb Q$.
In $x^4+x^3+x+1$, we can factorize it to become $(x+1)^2(x^2-x+1)=(x+1)^2(x-frac{1+sqrt 3i}2)(x-frac {1-sqrt 3i}{2})$. Am I right to say that we want to find $[mathbb Q(sqrt 3i):mathbb Q]$? Hence the answer is just $2$.
In $x^4+x^2+x+1$, we can factorize it to become $(x-frac 14sqrt5+frac 14-frac14isqrt2sqrt{5+sqrt5})(x+frac 14sqrt5+frac 14-frac14isqrt2sqrt{5-sqrt5})\(x+frac 14sqrt5+frac 14+frac14isqrt2sqrt{5-sqrt5})(x-frac 14sqrt5+frac 14+frac14isqrt2sqrt{5+sqrt5})$.
Hence, we want to find $mathbb Q(sqrt5,isqrt2sqrt{5+sqrt5},isqrt2sqrt{5-sqrt5})$. But how do we go about doing this?
abstract-algebra field-theory galois-theory extension-field splitting-field
abstract-algebra field-theory galois-theory extension-field splitting-field
asked Jan 9 at 10:10
IcycarusIcycarus
4751313
asked Jan 9 at 10:10
IcycarusIcycarus
4751313
asked Jan 9 at 10:10
IcycarusIcycarus
4751313
asked Jan 9 at 10:10
IcycarusIcycarus
4751313
asked Jan 9 at 10:10
IcycarusIcycarus
4751313
4751313
$begingroup$
Yes, the degree of the spliting field of $(X-1)^2(X^2-X+1)$ is of course equal to $2$. Why do you want to involve $mathbb Q(sqrt5,isqrt2sqrt{5+sqrt5},isqrt2sqrt{5-sqrt5})$?
$endgroup$
– Dietrich Burde
Jan 9 at 12:11
add a comment |
$begingroup$
Yes, the degree of the spliting field of $(X-1)^2(X^2-X+1)$ is of course equal to $2$. Why do you want to involve $mathbb Q(sqrt5,isqrt2sqrt{5+sqrt5},isqrt2sqrt{5-sqrt5})$?
$endgroup$
– Dietrich Burde
Jan 9 at 12:11
$begingroup$
Yes, the degree of the spliting field of $(X-1)^2(X^2-X+1)$ is of course equal to $2$. Why do you want to involve $mathbb Q(sqrt5,isqrt2sqrt{5+sqrt5},isqrt2sqrt{5-sqrt5})$?
$endgroup$
– Dietrich Burde
Jan 9 at 12:11
$begingroup$
Yes, the degree of the spliting field of $(X-1)^2(X^2-X+1)$ is of course equal to $2$. Why do you want to involve $mathbb Q(sqrt5,isqrt2sqrt{5+sqrt5},isqrt2sqrt{5-sqrt5})$?
$endgroup$
– Dietrich Burde
Jan 9 at 12:11
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your field is $mathbb{Q}(sqrt{5})(isqrt{10-2sqrt{5}})$, because the product of $isqrt{2}sqrt{5 pm sqrt{5}}$ is $-2(5-sqrt{5})in mathbb{Q}(sqrt{5})$.
So you have a biquadratic extension and the degree is $4$.
answered Jan 9 at 11:36
MindlackMindlack
3,49217
$endgroup$
$begingroup$
How did we get the degree to be 4? Should it be 8?
$endgroup$
– Icycarus
Jan 9 at 12:03
$begingroup$
I edited, now there shouldn’t be any reason why the degree is $8$.
$endgroup$
– Mindlack
Jan 9 at 12:06
$begingroup$
ahh! okay I understand now! Thank you much! Can I ask if my first question is correct as well?
$endgroup$
– Icycarus
Jan 9 at 12:10
$begingroup$
I think your own answer to the first point is correct.
$endgroup$
– Mindlack
Jan 9 at 12:11
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%2f3067281%2fdegree-of-splitting-fields%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
$begingroup$
Your field is $mathbb{Q}(sqrt{5})(isqrt{10-2sqrt{5}})$, because the product of $isqrt{2}sqrt{5 pm sqrt{5}}$ is $-2(5-sqrt{5})in mathbb{Q}(sqrt{5})$.
So you have a biquadratic extension and the degree is $4$.
answered Jan 9 at 11:36
MindlackMindlack
3,49217
$endgroup$
$begingroup$
How did we get the degree to be 4? Should it be 8?
$endgroup$
– Icycarus
Jan 9 at 12:03
$begingroup$
I edited, now there shouldn’t be any reason why the degree is $8$.
$endgroup$
– Mindlack
Jan 9 at 12:06
$begingroup$
ahh! okay I understand now! Thank you much! Can I ask if my first question is correct as well?
$endgroup$
– Icycarus
Jan 9 at 12:10
$begingroup$
I think your own answer to the first point is correct.
$endgroup$
– Mindlack
Jan 9 at 12:11
add a comment |
$begingroup$
Your field is $mathbb{Q}(sqrt{5})(isqrt{10-2sqrt{5}})$, because the product of $isqrt{2}sqrt{5 pm sqrt{5}}$ is $-2(5-sqrt{5})in mathbb{Q}(sqrt{5})$.
So you have a biquadratic extension and the degree is $4$.
answered Jan 9 at 11:36
MindlackMindlack
3,49217
$endgroup$
$begingroup$
How did we get the degree to be 4? Should it be 8?
$endgroup$
– Icycarus
Jan 9 at 12:03
$begingroup$
I edited, now there shouldn’t be any reason why the degree is $8$.
$endgroup$
– Mindlack
Jan 9 at 12:06
$begingroup$
ahh! okay I understand now! Thank you much! Can I ask if my first question is correct as well?
$endgroup$
– Icycarus
Jan 9 at 12:10
$begingroup$
I think your own answer to the first point is correct.
$endgroup$
– Mindlack
Jan 9 at 12:11
add a comment |
$begingroup$
Your field is $mathbb{Q}(sqrt{5})(isqrt{10-2sqrt{5}})$, because the product of $isqrt{2}sqrt{5 pm sqrt{5}}$ is $-2(5-sqrt{5})in mathbb{Q}(sqrt{5})$.
So you have a biquadratic extension and the degree is $4$.
answered Jan 9 at 11:36
MindlackMindlack
3,49217
$endgroup$
Your field is $mathbb{Q}(sqrt{5})(isqrt{10-2sqrt{5}})$, because the product of $isqrt{2}sqrt{5 pm sqrt{5}}$ is $-2(5-sqrt{5})in mathbb{Q}(sqrt{5})$.
So you have a biquadratic extension and the degree is $4$.
answered Jan 9 at 11:36
MindlackMindlack
3,49217
edited Jan 9 at 12:05
answered Jan 9 at 11:36
MindlackMindlack
3,49217
answered Jan 9 at 11:36
MindlackMindlack
3,49217
answered Jan 9 at 11:36
MindlackMindlack
3,49217
3,49217
$begingroup$
How did we get the degree to be 4? Should it be 8?
$endgroup$
– Icycarus
Jan 9 at 12:03
$begingroup$
I edited, now there shouldn’t be any reason why the degree is $8$.
$endgroup$
– Mindlack
Jan 9 at 12:06
$begingroup$
ahh! okay I understand now! Thank you much! Can I ask if my first question is correct as well?
$endgroup$
– Icycarus
Jan 9 at 12:10
$begingroup$
I think your own answer to the first point is correct.
$endgroup$
– Mindlack
Jan 9 at 12:11
add a comment |
$begingroup$
How did we get the degree to be 4? Should it be 8?
$endgroup$
– Icycarus
Jan 9 at 12:03
$begingroup$
I edited, now there shouldn’t be any reason why the degree is $8$.
$endgroup$
– Mindlack
Jan 9 at 12:06
$begingroup$
ahh! okay I understand now! Thank you much! Can I ask if my first question is correct as well?
$endgroup$
– Icycarus
Jan 9 at 12:10
$begingroup$
I think your own answer to the first point is correct.
$endgroup$
– Mindlack
Jan 9 at 12:11
$begingroup$
How did we get the degree to be 4? Should it be 8?
$endgroup$
– Icycarus
Jan 9 at 12:03
$begingroup$
How did we get the degree to be 4? Should it be 8?
$endgroup$
– Icycarus
Jan 9 at 12:03
$begingroup$
I edited, now there shouldn’t be any reason why the degree is $8$.
$endgroup$
– Mindlack
Jan 9 at 12:06
$begingroup$
I edited, now there shouldn’t be any reason why the degree is $8$.
$endgroup$
– Mindlack
Jan 9 at 12:06
$begingroup$
ahh! okay I understand now! Thank you much! Can I ask if my first question is correct as well?
$endgroup$
– Icycarus
Jan 9 at 12:10
$begingroup$
ahh! okay I understand now! Thank you much! Can I ask if my first question is correct as well?
$endgroup$
– Icycarus
Jan 9 at 12:10
$begingroup$
I think your own answer to the first point is correct.
$endgroup$
– Mindlack
Jan 9 at 12:11
$begingroup$
I think your own answer to the first point is correct.
$endgroup$
– Mindlack
Jan 9 at 12:11
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.
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%2f3067281%2fdegree-of-splitting-fields%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
$begingroup$
Yes, the degree of the spliting field of $(X-1)^2(X^2-X+1)$ is of course equal to $2$. Why do you want to involve $mathbb Q(sqrt5,isqrt2sqrt{5+sqrt5},isqrt2sqrt{5-sqrt5})$?
$endgroup$
– Dietrich Burde
Jan 9 at 12:11