Mixing
$mathbb{Q}(sqrt[3]{2},e^{frac{2pi i}{3}})=mathbb{Q}(sqrt[3]{2},sqrt[3]{2}e^{frac{2pi i}{3}})$
$begingroup$
I have to show that those two field extensions are the same, however I'm struggling to get the job done.
$mathbb{Q}(sqrt[3]{2},e^{frac{2pi i}{3}})=mathbb{Q}(sqrt[3]{2},sqrt[3]{2}e^{frac{2pi i}{3}})$
abstract-algebra field-theory extension-field
asked Feb 2 at 19:05
Christian SingerChristian Singer
429313
$endgroup$
add a comment |
$begingroup$
I have to show that those two field extensions are the same, however I'm struggling to get the job done.
$mathbb{Q}(sqrt[3]{2},e^{frac{2pi i}{3}})=mathbb{Q}(sqrt[3]{2},sqrt[3]{2}e^{frac{2pi i}{3}})$
abstract-algebra field-theory extension-field
asked Feb 2 at 19:05
Christian SingerChristian Singer
429313
$endgroup$
$begingroup$
Do you know that $Bbb Q(sqrt[3] 2 , e^{frac {2 pi i} {3}})$ is the smallest field containing $Bbb Q, sqrt[3] 2$ and $e^{frac {2 pi i} {3}}$?
$endgroup$
– Dbchatto67
Feb 2 at 19:10
1
$begingroup$
$K(a,b)=K(a,ab)$ always.
$endgroup$
– Lord Shark the Unknown
Feb 2 at 19:12
$begingroup$
Show that each of them is a subset of the other.
$endgroup$
– Dbchatto67
Feb 2 at 19:12
$begingroup$
Yeah, but I can't just say that since $sqrt[3]{2}e^{frac{2pi i}{3}}$ is a product of elements from the set on the left the right set is in the left one, or can I ?
$endgroup$
– Christian Singer
Feb 2 at 19:15
add a comment |
$begingroup$
I have to show that those two field extensions are the same, however I'm struggling to get the job done.
$mathbb{Q}(sqrt[3]{2},e^{frac{2pi i}{3}})=mathbb{Q}(sqrt[3]{2},sqrt[3]{2}e^{frac{2pi i}{3}})$
abstract-algebra field-theory extension-field
asked Feb 2 at 19:05
Christian SingerChristian Singer
429313
$endgroup$
I have to show that those two field extensions are the same, however I'm struggling to get the job done.
$mathbb{Q}(sqrt[3]{2},e^{frac{2pi i}{3}})=mathbb{Q}(sqrt[3]{2},sqrt[3]{2}e^{frac{2pi i}{3}})$
abstract-algebra field-theory extension-field
abstract-algebra field-theory extension-field
asked Feb 2 at 19:05
Christian SingerChristian Singer
429313
asked Feb 2 at 19:05
Christian SingerChristian Singer
429313
asked Feb 2 at 19:05
Christian SingerChristian Singer
429313
asked Feb 2 at 19:05
Christian SingerChristian Singer
429313
asked Feb 2 at 19:05
Christian SingerChristian Singer
429313
429313
$begingroup$
Do you know that $Bbb Q(sqrt[3] 2 , e^{frac {2 pi i} {3}})$ is the smallest field containing $Bbb Q, sqrt[3] 2$ and $e^{frac {2 pi i} {3}}$?
$endgroup$
– Dbchatto67
Feb 2 at 19:10
1
$begingroup$
$K(a,b)=K(a,ab)$ always.
$endgroup$
– Lord Shark the Unknown
Feb 2 at 19:12
$begingroup$
Show that each of them is a subset of the other.
$endgroup$
– Dbchatto67
Feb 2 at 19:12
$begingroup$
Yeah, but I can't just say that since $sqrt[3]{2}e^{frac{2pi i}{3}}$ is a product of elements from the set on the left the right set is in the left one, or can I ?
$endgroup$
– Christian Singer
Feb 2 at 19:15
add a comment |
$begingroup$
Do you know that $Bbb Q(sqrt[3] 2 , e^{frac {2 pi i} {3}})$ is the smallest field containing $Bbb Q, sqrt[3] 2$ and $e^{frac {2 pi i} {3}}$?
$endgroup$
– Dbchatto67
Feb 2 at 19:10
1
$begingroup$
$K(a,b)=K(a,ab)$ always.
$endgroup$
– Lord Shark the Unknown
Feb 2 at 19:12
$begingroup$
Show that each of them is a subset of the other.
$endgroup$
– Dbchatto67
Feb 2 at 19:12
$begingroup$
Yeah, but I can't just say that since $sqrt[3]{2}e^{frac{2pi i}{3}}$ is a product of elements from the set on the left the right set is in the left one, or can I ?
$endgroup$
– Christian Singer
Feb 2 at 19:15
$begingroup$
Do you know that $Bbb Q(sqrt[3] 2 , e^{frac {2 pi i} {3}})$ is the smallest field containing $Bbb Q, sqrt[3] 2$ and $e^{frac {2 pi i} {3}}$?
$endgroup$
– Dbchatto67
Feb 2 at 19:10
$begingroup$
Do you know that $Bbb Q(sqrt[3] 2 , e^{frac {2 pi i} {3}})$ is the smallest field containing $Bbb Q, sqrt[3] 2$ and $e^{frac {2 pi i} {3}}$?
$endgroup$
– Dbchatto67
Feb 2 at 19:10
1
1
$begingroup$
$K(a,b)=K(a,ab)$ always.
$endgroup$
– Lord Shark the Unknown
Feb 2 at 19:12
$begingroup$
$K(a,b)=K(a,ab)$ always.
$endgroup$
– Lord Shark the Unknown
Feb 2 at 19:12
$begingroup$
Show that each of them is a subset of the other.
$endgroup$
– Dbchatto67
Feb 2 at 19:12
$begingroup$
Show that each of them is a subset of the other.
$endgroup$
– Dbchatto67
Feb 2 at 19:12
$begingroup$
Yeah, but I can't just say that since $sqrt[3]{2}e^{frac{2pi i}{3}}$ is a product of elements from the set on the left the right set is in the left one, or can I ?
$endgroup$
– Christian Singer
Feb 2 at 19:15
$begingroup$
Yeah, but I can't just say that since $sqrt[3]{2}e^{frac{2pi i}{3}}$ is a product of elements from the set on the left the right set is in the left one, or can I ?
$endgroup$
– Christian Singer
Feb 2 at 19:15
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Since $sqrt[3] 2 , e^{frac {2 pi i} {3}} in Bbb Q (sqrt[3] 2 , e^{frac {2 pi i} {3}})$, so $sqrt[3] 2 e^{frac {2 pi i} {3}} in Bbb Q(sqrt[3] 2 , e^{frac {2 pi i} {3}})$. So $Bbb Q(sqrt[3] 2 , sqrt[3] 2e^{frac {2 pi i} {3}}) subseteq Bbb Q(sqrt[3] 2 , e^{frac {2 pi i} {3}})$ because $Bbb Q (sqrt[3] 2 , sqrt[3] 2 e^{frac {2 pi i} {3}})$ is the smallest field containing $Bbb Q, sqrt[3] 2$ and $sqrt[3] 2 e^{frac {2 pi i} {3}}$. Can you show the other inclusion?
answered Feb 2 at 19:18
Dbchatto67Dbchatto67
2,924622
$endgroup$
$begingroup$
Ah ok, I get it. It's a field, so it's closed under multiplication aswell (Concerning your inclusion).
$endgroup$
– Christian Singer
Feb 2 at 19:24
$begingroup$
Yeah! @Christian Singer. Precisely it is!
$endgroup$
– Dbchatto67
Feb 2 at 19:33
$begingroup$
Could you say that since $mathbb{Q}(sqrt[3]{2},sqrt[3]{2}e^{frac{2pi i}{3}})$ is closed under multiplication, that it has to contain the inverse of $sqrt[3]{2}$, hence the product $(sqrt[3]{2})^{-1}cdotsqrt[3]{2}e^{frac{2pi i}{3}} = e^{frac{2pi i}{3}}$. If my train of thought is correct, this should have shown the other inclusion
$endgroup$
– Christian Singer
Feb 2 at 19:38
$begingroup$
Yes of course. That is precisely what I was trying to say.
$endgroup$
– Dbchatto67
Feb 2 at 19:39
$begingroup$
Well, I edited at the same time that you posted your comment, so am I right that this basically shows the other inclusion ?
$endgroup$
– Christian Singer
Feb 2 at 19:40
|
show 4 more comments
Your Answer
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%2f3097665%2fmathbbq-sqrt32-e-frac2-pi-i3-mathbbq-sqrt32-sqrt32%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
$begingroup$
Since $sqrt[3] 2 , e^{frac {2 pi i} {3}} in Bbb Q (sqrt[3] 2 , e^{frac {2 pi i} {3}})$, so $sqrt[3] 2 e^{frac {2 pi i} {3}} in Bbb Q(sqrt[3] 2 , e^{frac {2 pi i} {3}})$. So $Bbb Q(sqrt[3] 2 , sqrt[3] 2e^{frac {2 pi i} {3}}) subseteq Bbb Q(sqrt[3] 2 , e^{frac {2 pi i} {3}})$ because $Bbb Q (sqrt[3] 2 , sqrt[3] 2 e^{frac {2 pi i} {3}})$ is the smallest field containing $Bbb Q, sqrt[3] 2$ and $sqrt[3] 2 e^{frac {2 pi i} {3}}$. Can you show the other inclusion?
answered Feb 2 at 19:18
Dbchatto67Dbchatto67
2,924622
$endgroup$
$begingroup$
Ah ok, I get it. It's a field, so it's closed under multiplication aswell (Concerning your inclusion).
$endgroup$
– Christian Singer
Feb 2 at 19:24
$begingroup$
Yeah! @Christian Singer. Precisely it is!
$endgroup$
– Dbchatto67
Feb 2 at 19:33
$begingroup$
Could you say that since $mathbb{Q}(sqrt[3]{2},sqrt[3]{2}e^{frac{2pi i}{3}})$ is closed under multiplication, that it has to contain the inverse of $sqrt[3]{2}$, hence the product $(sqrt[3]{2})^{-1}cdotsqrt[3]{2}e^{frac{2pi i}{3}} = e^{frac{2pi i}{3}}$. If my train of thought is correct, this should have shown the other inclusion
$endgroup$
– Christian Singer
Feb 2 at 19:38
$begingroup$
Yes of course. That is precisely what I was trying to say.
$endgroup$
– Dbchatto67
Feb 2 at 19:39
$begingroup$
Well, I edited at the same time that you posted your comment, so am I right that this basically shows the other inclusion ?
$endgroup$
– Christian Singer
Feb 2 at 19:40
|
show 4 more comments
$begingroup$
Since $sqrt[3] 2 , e^{frac {2 pi i} {3}} in Bbb Q (sqrt[3] 2 , e^{frac {2 pi i} {3}})$, so $sqrt[3] 2 e^{frac {2 pi i} {3}} in Bbb Q(sqrt[3] 2 , e^{frac {2 pi i} {3}})$. So $Bbb Q(sqrt[3] 2 , sqrt[3] 2e^{frac {2 pi i} {3}}) subseteq Bbb Q(sqrt[3] 2 , e^{frac {2 pi i} {3}})$ because $Bbb Q (sqrt[3] 2 , sqrt[3] 2 e^{frac {2 pi i} {3}})$ is the smallest field containing $Bbb Q, sqrt[3] 2$ and $sqrt[3] 2 e^{frac {2 pi i} {3}}$. Can you show the other inclusion?
answered Feb 2 at 19:18
Dbchatto67Dbchatto67
2,924622
$endgroup$
$begingroup$
Ah ok, I get it. It's a field, so it's closed under multiplication aswell (Concerning your inclusion).
$endgroup$
– Christian Singer
Feb 2 at 19:24
$begingroup$
Yeah! @Christian Singer. Precisely it is!
$endgroup$
– Dbchatto67
Feb 2 at 19:33
$begingroup$
Could you say that since $mathbb{Q}(sqrt[3]{2},sqrt[3]{2}e^{frac{2pi i}{3}})$ is closed under multiplication, that it has to contain the inverse of $sqrt[3]{2}$, hence the product $(sqrt[3]{2})^{-1}cdotsqrt[3]{2}e^{frac{2pi i}{3}} = e^{frac{2pi i}{3}}$. If my train of thought is correct, this should have shown the other inclusion
$endgroup$
– Christian Singer
Feb 2 at 19:38
$begingroup$
Yes of course. That is precisely what I was trying to say.
$endgroup$
– Dbchatto67
Feb 2 at 19:39
$begingroup$
Well, I edited at the same time that you posted your comment, so am I right that this basically shows the other inclusion ?
$endgroup$
– Christian Singer
Feb 2 at 19:40
|
show 4 more comments
$begingroup$
Since $sqrt[3] 2 , e^{frac {2 pi i} {3}} in Bbb Q (sqrt[3] 2 , e^{frac {2 pi i} {3}})$, so $sqrt[3] 2 e^{frac {2 pi i} {3}} in Bbb Q(sqrt[3] 2 , e^{frac {2 pi i} {3}})$. So $Bbb Q(sqrt[3] 2 , sqrt[3] 2e^{frac {2 pi i} {3}}) subseteq Bbb Q(sqrt[3] 2 , e^{frac {2 pi i} {3}})$ because $Bbb Q (sqrt[3] 2 , sqrt[3] 2 e^{frac {2 pi i} {3}})$ is the smallest field containing $Bbb Q, sqrt[3] 2$ and $sqrt[3] 2 e^{frac {2 pi i} {3}}$. Can you show the other inclusion?
answered Feb 2 at 19:18
Dbchatto67Dbchatto67
2,924622
$endgroup$
Since $sqrt[3] 2 , e^{frac {2 pi i} {3}} in Bbb Q (sqrt[3] 2 , e^{frac {2 pi i} {3}})$, so $sqrt[3] 2 e^{frac {2 pi i} {3}} in Bbb Q(sqrt[3] 2 , e^{frac {2 pi i} {3}})$. So $Bbb Q(sqrt[3] 2 , sqrt[3] 2e^{frac {2 pi i} {3}}) subseteq Bbb Q(sqrt[3] 2 , e^{frac {2 pi i} {3}})$ because $Bbb Q (sqrt[3] 2 , sqrt[3] 2 e^{frac {2 pi i} {3}})$ is the smallest field containing $Bbb Q, sqrt[3] 2$ and $sqrt[3] 2 e^{frac {2 pi i} {3}}$. Can you show the other inclusion?
answered Feb 2 at 19:18
Dbchatto67Dbchatto67
2,924622
edited Feb 2 at 19:24
answered Feb 2 at 19:18
Dbchatto67Dbchatto67
2,924622
answered Feb 2 at 19:18
Dbchatto67Dbchatto67
2,924622
answered Feb 2 at 19:18
Dbchatto67Dbchatto67
2,924622
2,924622
$begingroup$
Ah ok, I get it. It's a field, so it's closed under multiplication aswell (Concerning your inclusion).
$endgroup$
– Christian Singer
Feb 2 at 19:24
$begingroup$
Yeah! @Christian Singer. Precisely it is!
$endgroup$
– Dbchatto67
Feb 2 at 19:33
$begingroup$
Could you say that since $mathbb{Q}(sqrt[3]{2},sqrt[3]{2}e^{frac{2pi i}{3}})$ is closed under multiplication, that it has to contain the inverse of $sqrt[3]{2}$, hence the product $(sqrt[3]{2})^{-1}cdotsqrt[3]{2}e^{frac{2pi i}{3}} = e^{frac{2pi i}{3}}$. If my train of thought is correct, this should have shown the other inclusion
$endgroup$
– Christian Singer
Feb 2 at 19:38
$begingroup$
Yes of course. That is precisely what I was trying to say.
$endgroup$
– Dbchatto67
Feb 2 at 19:39
$begingroup$
Well, I edited at the same time that you posted your comment, so am I right that this basically shows the other inclusion ?
$endgroup$
– Christian Singer
Feb 2 at 19:40
|
show 4 more comments
$begingroup$
Ah ok, I get it. It's a field, so it's closed under multiplication aswell (Concerning your inclusion).
$endgroup$
– Christian Singer
Feb 2 at 19:24
$begingroup$
Yeah! @Christian Singer. Precisely it is!
$endgroup$
– Dbchatto67
Feb 2 at 19:33
$begingroup$
Could you say that since $mathbb{Q}(sqrt[3]{2},sqrt[3]{2}e^{frac{2pi i}{3}})$ is closed under multiplication, that it has to contain the inverse of $sqrt[3]{2}$, hence the product $(sqrt[3]{2})^{-1}cdotsqrt[3]{2}e^{frac{2pi i}{3}} = e^{frac{2pi i}{3}}$. If my train of thought is correct, this should have shown the other inclusion
$endgroup$
– Christian Singer
Feb 2 at 19:38
$begingroup$
Yes of course. That is precisely what I was trying to say.
$endgroup$
– Dbchatto67
Feb 2 at 19:39
$begingroup$
Well, I edited at the same time that you posted your comment, so am I right that this basically shows the other inclusion ?
$endgroup$
– Christian Singer
Feb 2 at 19:40
$begingroup$
Ah ok, I get it. It's a field, so it's closed under multiplication aswell (Concerning your inclusion).
$endgroup$
– Christian Singer
Feb 2 at 19:24
$begingroup$
Ah ok, I get it. It's a field, so it's closed under multiplication aswell (Concerning your inclusion).
$endgroup$
– Christian Singer
Feb 2 at 19:24
$begingroup$
Yeah! @Christian Singer. Precisely it is!
$endgroup$
– Dbchatto67
Feb 2 at 19:33
$begingroup$
Yeah! @Christian Singer. Precisely it is!
$endgroup$
– Dbchatto67
Feb 2 at 19:33
$begingroup$
Could you say that since $mathbb{Q}(sqrt[3]{2},sqrt[3]{2}e^{frac{2pi i}{3}})$ is closed under multiplication, that it has to contain the inverse of $sqrt[3]{2}$, hence the product $(sqrt[3]{2})^{-1}cdotsqrt[3]{2}e^{frac{2pi i}{3}} = e^{frac{2pi i}{3}}$. If my train of thought is correct, this should have shown the other inclusion
$endgroup$
– Christian Singer
Feb 2 at 19:38
$begingroup$
Could you say that since $mathbb{Q}(sqrt[3]{2},sqrt[3]{2}e^{frac{2pi i}{3}})$ is closed under multiplication, that it has to contain the inverse of $sqrt[3]{2}$, hence the product $(sqrt[3]{2})^{-1}cdotsqrt[3]{2}e^{frac{2pi i}{3}} = e^{frac{2pi i}{3}}$. If my train of thought is correct, this should have shown the other inclusion
$endgroup$
– Christian Singer
Feb 2 at 19:38
$begingroup$
Yes of course. That is precisely what I was trying to say.
$endgroup$
– Dbchatto67
Feb 2 at 19:39
$begingroup$
Yes of course. That is precisely what I was trying to say.
$endgroup$
– Dbchatto67
Feb 2 at 19:39
$begingroup$
Well, I edited at the same time that you posted your comment, so am I right that this basically shows the other inclusion ?
$endgroup$
– Christian Singer
Feb 2 at 19:40
$begingroup$
Well, I edited at the same time that you posted your comment, so am I right that this basically shows the other inclusion ?
$endgroup$
– Christian Singer
Feb 2 at 19:40
|
show 4 more comments
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%2f3097665%2fmathbbq-sqrt32-e-frac2-pi-i3-mathbbq-sqrt32-sqrt32%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
$begingroup$
Do you know that $Bbb Q(sqrt[3] 2 , e^{frac {2 pi i} {3}})$ is the smallest field containing $Bbb Q, sqrt[3] 2$ and $e^{frac {2 pi i} {3}}$?
$endgroup$
– Dbchatto67
Feb 2 at 19:10
1
$begingroup$
$K(a,b)=K(a,ab)$ always.
$endgroup$
– Lord Shark the Unknown
Feb 2 at 19:12
$begingroup$
Show that each of them is a subset of the other.
$endgroup$
– Dbchatto67
Feb 2 at 19:12
$begingroup$
Yeah, but I can't just say that since $sqrt[3]{2}e^{frac{2pi i}{3}}$ is a product of elements from the set on the left the right set is in the left one, or can I ?
$endgroup$
– Christian Singer
Feb 2 at 19:15