Is $mathbb{Q}(sqrt{2}, sqrt{5})=mathbb{Q}(sqrt{10})$ in field extension?
0
Is $mathbb{Q}(sqrt{2}, sqrt{5})=mathbb{Q}(sqrt{10})$ in field extension?
We know that, $mathbb{Q}(sqrt{2}, sqrt{5})=mathbb{Q}(sqrt{2}+ sqrt{5})$.
Also, $mathbb{Q}(sqrt{2}, sqrt{5})=mathbb{Q}(asqrt{2}+ bsqrt{5})$ where $a,b in mathbb{Q}$.
If the question is true, then for which values of $a$ & $b$ can $sqrt{10}$ be expressed in the form $asqrt{2}+ bsqrt{5}$ where $a,b in mathbb{Q}$
abstract-algebra
asked Sep 23 '14 at 18:04
Empty
8,07252559
add a comment |
0
Is $mathbb{Q}(sqrt{2}, sqrt{5})=mathbb{Q}(sqrt{10})$ in field extension?
We know that, $mathbb{Q}(sqrt{2}, sqrt{5})=mathbb{Q}(sqrt{2}+ sqrt{5})$.
Also, $mathbb{Q}(sqrt{2}, sqrt{5})=mathbb{Q}(asqrt{2}+ bsqrt{5})$ where $a,b in mathbb{Q}$.
If the question is true, then for which values of $a$ & $b$ can $sqrt{10}$ be expressed in the form $asqrt{2}+ bsqrt{5}$ where $a,b in mathbb{Q}$
abstract-algebra
asked Sep 23 '14 at 18:04
Empty
8,07252559
3
Look at degrees.
– Hurkyl
Sep 23 '14 at 18:06
No. $mathbb{Q}[sqrt{2}, sqrt{5}] = { a+bsqrt{2}+csqrt{5}+ dsqrt{10} : a,b,c,d, in mathbb{Q}}$.
– Crostul
Sep 23 '14 at 18:10
For no values $(a,b)$. And the question has answer No.
– Orest Bucicovschi
Sep 23 '14 at 18:16
1
You can show $sqrt{10} in mathbb{Q}(sqrt{2}, sqrt{5})$ but that $sqrt{2} notin mathbb{Q}(sqrt{10})$ and $sqrt{5} notin mathbb{Q}(sqrt{10})$. Can you see why? Also, as someone else noted, look at the degrees of the field extension.
– Islands
Sep 23 '14 at 18:18
Possible duplicate of Is $mathbf{Q}(sqrt{2},sqrt{3}) = mathbf{Q}(sqrt{6})$?
– Watson
Nov 26 at 12:44
add a comment |
0
0
0
0
Is $mathbb{Q}(sqrt{2}, sqrt{5})=mathbb{Q}(sqrt{10})$ in field extension?
We know that, $mathbb{Q}(sqrt{2}, sqrt{5})=mathbb{Q}(sqrt{2}+ sqrt{5})$.
Also, $mathbb{Q}(sqrt{2}, sqrt{5})=mathbb{Q}(asqrt{2}+ bsqrt{5})$ where $a,b in mathbb{Q}$.
If the question is true, then for which values of $a$ & $b$ can $sqrt{10}$ be expressed in the form $asqrt{2}+ bsqrt{5}$ where $a,b in mathbb{Q}$
abstract-algebra
asked Sep 23 '14 at 18:04
Empty
8,07252559
Is $mathbb{Q}(sqrt{2}, sqrt{5})=mathbb{Q}(sqrt{10})$ in field extension?
We know that, $mathbb{Q}(sqrt{2}, sqrt{5})=mathbb{Q}(sqrt{2}+ sqrt{5})$.
Also, $mathbb{Q}(sqrt{2}, sqrt{5})=mathbb{Q}(asqrt{2}+ bsqrt{5})$ where $a,b in mathbb{Q}$.
If the question is true, then for which values of $a$ & $b$ can $sqrt{10}$ be expressed in the form $asqrt{2}+ bsqrt{5}$ where $a,b in mathbb{Q}$
abstract-algebra
abstract-algebra
asked Sep 23 '14 at 18:04
Empty
8,07252559
asked Sep 23 '14 at 18:04
Empty
8,07252559
edited Nov 26 at 15:51
asked Sep 23 '14 at 18:04
Empty
8,07252559
asked Sep 23 '14 at 18:04
Empty
8,07252559
asked Sep 23 '14 at 18:04
Empty
8,07252559
8,07252559
3
Look at degrees.
– Hurkyl
Sep 23 '14 at 18:06
No. $mathbb{Q}[sqrt{2}, sqrt{5}] = { a+bsqrt{2}+csqrt{5}+ dsqrt{10} : a,b,c,d, in mathbb{Q}}$.
– Crostul
Sep 23 '14 at 18:10
For no values $(a,b)$. And the question has answer No.
– Orest Bucicovschi
Sep 23 '14 at 18:16
1
You can show $sqrt{10} in mathbb{Q}(sqrt{2}, sqrt{5})$ but that $sqrt{2} notin mathbb{Q}(sqrt{10})$ and $sqrt{5} notin mathbb{Q}(sqrt{10})$. Can you see why? Also, as someone else noted, look at the degrees of the field extension.
– Islands
Sep 23 '14 at 18:18
Possible duplicate of Is $mathbf{Q}(sqrt{2},sqrt{3}) = mathbf{Q}(sqrt{6})$?
– Watson
Nov 26 at 12:44
add a comment |
3
Look at degrees.
– Hurkyl
Sep 23 '14 at 18:06
No. $mathbb{Q}[sqrt{2}, sqrt{5}] = { a+bsqrt{2}+csqrt{5}+ dsqrt{10} : a,b,c,d, in mathbb{Q}}$.
– Crostul
Sep 23 '14 at 18:10
For no values $(a,b)$. And the question has answer No.
– Orest Bucicovschi
Sep 23 '14 at 18:16
1
You can show $sqrt{10} in mathbb{Q}(sqrt{2}, sqrt{5})$ but that $sqrt{2} notin mathbb{Q}(sqrt{10})$ and $sqrt{5} notin mathbb{Q}(sqrt{10})$. Can you see why? Also, as someone else noted, look at the degrees of the field extension.
– Islands
Sep 23 '14 at 18:18
Possible duplicate of Is $mathbf{Q}(sqrt{2},sqrt{3}) = mathbf{Q}(sqrt{6})$?
– Watson
Nov 26 at 12:44
3
3
Look at degrees.
– Hurkyl
Sep 23 '14 at 18:06
Look at degrees.
– Hurkyl
Sep 23 '14 at 18:06
No. $mathbb{Q}[sqrt{2}, sqrt{5}] = { a+bsqrt{2}+csqrt{5}+ dsqrt{10} : a,b,c,d, in mathbb{Q}}$.
– Crostul
Sep 23 '14 at 18:10
No. $mathbb{Q}[sqrt{2}, sqrt{5}] = { a+bsqrt{2}+csqrt{5}+ dsqrt{10} : a,b,c,d, in mathbb{Q}}$.
– Crostul
Sep 23 '14 at 18:10
For no values $(a,b)$. And the question has answer No.
– Orest Bucicovschi
Sep 23 '14 at 18:16
For no values $(a,b)$. And the question has answer No.
– Orest Bucicovschi
Sep 23 '14 at 18:16
1
1
You can show $sqrt{10} in mathbb{Q}(sqrt{2}, sqrt{5})$ but that $sqrt{2} notin mathbb{Q}(sqrt{10})$ and $sqrt{5} notin mathbb{Q}(sqrt{10})$. Can you see why? Also, as someone else noted, look at the degrees of the field extension.
– Islands
Sep 23 '14 at 18:18
You can show $sqrt{10} in mathbb{Q}(sqrt{2}, sqrt{5})$ but that $sqrt{2} notin mathbb{Q}(sqrt{10})$ and $sqrt{5} notin mathbb{Q}(sqrt{10})$. Can you see why? Also, as someone else noted, look at the degrees of the field extension.
– Islands
Sep 23 '14 at 18:18
Possible duplicate of Is $mathbf{Q}(sqrt{2},sqrt{3}) = mathbf{Q}(sqrt{6})$?
– Watson
Nov 26 at 12:44
Possible duplicate of Is $mathbf{Q}(sqrt{2},sqrt{3}) = mathbf{Q}(sqrt{6})$?
– Watson
Nov 26 at 12:44
add a comment |
1 Answer
1
active
oldest
votes
1
As comment suggested, look at degrees over the rationals:
$$[Bbb Q(sqrt2,sqrt5):Bbb Q]=[Bbb Q(sqrt2)(sqrt5):Bbb Q(sqrt2)]cdot [Bbb Q(sqrt2):Bbb Q]=2cdot2=4$$
since $;x^2-5inBbb Q(sqrt2)[x];$ is irreducible here (proof?)
On the other side
$$[Bbb Qsqrt{10}:Bbb Q]=2$$
since x$;x^2-10inBbb Q[x];$ is irreducible here.
answered Sep 23 '14 at 19:11
Timbuc
30.9k22045
OK. I understand it. Due to the degree, Q(square root(2),cube root(7))=Q(square root(2)+cube root(7)) as well as =Q(square root(2).cube root(7)).As the degrees are equal. Am I right?
– Empty
Sep 23 '14 at 21:43
Don't understand your question, @SayantanPanja
– Timbuc
Sep 23 '14 at 22:31
I want to say that, [(Q(square root(2),cube root(7)):Q]=6=[(Q(square root(2).cube root(7)):Q]=[(Q(square root(2)+cube root(7)):Q]. So can we write that, Q(square root(2),cube root(7))=Q(square root(2).cube root(7))=Q(square root(2)+cube root(7))?
– Empty
Sep 24 '14 at 3:45
1
Well, yes we can...but only if you prove something else. For example, $;[Bbb Q(sqrt2):Bbb Q]=[Bbb Q(sqrt5):Bbb Q]=2;$ , yet $;Bbb Q(sqrt2)neqBbb Q (sqrt5);$ . Equality follows if for example one of the fields contains the other one. In your example, since clearly $;Bbb Q(sqrt2+sqrt7)subsetBbb Q(sqrt2,sqrt7);$ , if the degrees over $;Bbb Q;$ are equal then the fields are the same.
– Timbuc
Sep 24 '14 at 5:26
Sir, you have some mistake...My question is not root(7),it is cube root(7)
– Empty
Sep 24 '14 at 5:58
|
show 3 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%2f943250%2fis-mathbbq-sqrt2-sqrt5-mathbbq-sqrt10-in-field-extension%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
1
As comment suggested, look at degrees over the rationals:
$$[Bbb Q(sqrt2,sqrt5):Bbb Q]=[Bbb Q(sqrt2)(sqrt5):Bbb Q(sqrt2)]cdot [Bbb Q(sqrt2):Bbb Q]=2cdot2=4$$
since $;x^2-5inBbb Q(sqrt2)[x];$ is irreducible here (proof?)
On the other side
$$[Bbb Qsqrt{10}:Bbb Q]=2$$
since x$;x^2-10inBbb Q[x];$ is irreducible here.
answered Sep 23 '14 at 19:11
Timbuc
30.9k22045
OK. I understand it. Due to the degree, Q(square root(2),cube root(7))=Q(square root(2)+cube root(7)) as well as =Q(square root(2).cube root(7)).As the degrees are equal. Am I right?
– Empty
Sep 23 '14 at 21:43
Don't understand your question, @SayantanPanja
– Timbuc
Sep 23 '14 at 22:31
I want to say that, [(Q(square root(2),cube root(7)):Q]=6=[(Q(square root(2).cube root(7)):Q]=[(Q(square root(2)+cube root(7)):Q]. So can we write that, Q(square root(2),cube root(7))=Q(square root(2).cube root(7))=Q(square root(2)+cube root(7))?
– Empty
Sep 24 '14 at 3:45
1
Well, yes we can...but only if you prove something else. For example, $;[Bbb Q(sqrt2):Bbb Q]=[Bbb Q(sqrt5):Bbb Q]=2;$ , yet $;Bbb Q(sqrt2)neqBbb Q (sqrt5);$ . Equality follows if for example one of the fields contains the other one. In your example, since clearly $;Bbb Q(sqrt2+sqrt7)subsetBbb Q(sqrt2,sqrt7);$ , if the degrees over $;Bbb Q;$ are equal then the fields are the same.
– Timbuc
Sep 24 '14 at 5:26
Sir, you have some mistake...My question is not root(7),it is cube root(7)
– Empty
Sep 24 '14 at 5:58
|
show 3 more comments
1
As comment suggested, look at degrees over the rationals:
$$[Bbb Q(sqrt2,sqrt5):Bbb Q]=[Bbb Q(sqrt2)(sqrt5):Bbb Q(sqrt2)]cdot [Bbb Q(sqrt2):Bbb Q]=2cdot2=4$$
since $;x^2-5inBbb Q(sqrt2)[x];$ is irreducible here (proof?)
On the other side
$$[Bbb Qsqrt{10}:Bbb Q]=2$$
since x$;x^2-10inBbb Q[x];$ is irreducible here.
answered Sep 23 '14 at 19:11
Timbuc
30.9k22045
OK. I understand it. Due to the degree, Q(square root(2),cube root(7))=Q(square root(2)+cube root(7)) as well as =Q(square root(2).cube root(7)).As the degrees are equal. Am I right?
– Empty
Sep 23 '14 at 21:43
Don't understand your question, @SayantanPanja
– Timbuc
Sep 23 '14 at 22:31
I want to say that, [(Q(square root(2),cube root(7)):Q]=6=[(Q(square root(2).cube root(7)):Q]=[(Q(square root(2)+cube root(7)):Q]. So can we write that, Q(square root(2),cube root(7))=Q(square root(2).cube root(7))=Q(square root(2)+cube root(7))?
– Empty
Sep 24 '14 at 3:45
1
Well, yes we can...but only if you prove something else. For example, $;[Bbb Q(sqrt2):Bbb Q]=[Bbb Q(sqrt5):Bbb Q]=2;$ , yet $;Bbb Q(sqrt2)neqBbb Q (sqrt5);$ . Equality follows if for example one of the fields contains the other one. In your example, since clearly $;Bbb Q(sqrt2+sqrt7)subsetBbb Q(sqrt2,sqrt7);$ , if the degrees over $;Bbb Q;$ are equal then the fields are the same.
– Timbuc
Sep 24 '14 at 5:26
Sir, you have some mistake...My question is not root(7),it is cube root(7)
– Empty
Sep 24 '14 at 5:58
|
show 3 more comments
1
1
1
As comment suggested, look at degrees over the rationals:
$$[Bbb Q(sqrt2,sqrt5):Bbb Q]=[Bbb Q(sqrt2)(sqrt5):Bbb Q(sqrt2)]cdot [Bbb Q(sqrt2):Bbb Q]=2cdot2=4$$
since $;x^2-5inBbb Q(sqrt2)[x];$ is irreducible here (proof?)
On the other side
$$[Bbb Qsqrt{10}:Bbb Q]=2$$
since x$;x^2-10inBbb Q[x];$ is irreducible here.
answered Sep 23 '14 at 19:11
Timbuc
30.9k22045
As comment suggested, look at degrees over the rationals:
$$[Bbb Q(sqrt2,sqrt5):Bbb Q]=[Bbb Q(sqrt2)(sqrt5):Bbb Q(sqrt2)]cdot [Bbb Q(sqrt2):Bbb Q]=2cdot2=4$$
since $;x^2-5inBbb Q(sqrt2)[x];$ is irreducible here (proof?)
On the other side
$$[Bbb Qsqrt{10}:Bbb Q]=2$$
since x$;x^2-10inBbb Q[x];$ is irreducible here.
answered Sep 23 '14 at 19:11
Timbuc
30.9k22045
answered Sep 23 '14 at 19:11
Timbuc
30.9k22045
answered Sep 23 '14 at 19:11
Timbuc
30.9k22045
answered Sep 23 '14 at 19:11
Timbuc
30.9k22045
30.9k22045
OK. I understand it. Due to the degree, Q(square root(2),cube root(7))=Q(square root(2)+cube root(7)) as well as =Q(square root(2).cube root(7)).As the degrees are equal. Am I right?
– Empty
Sep 23 '14 at 21:43
Don't understand your question, @SayantanPanja
– Timbuc
Sep 23 '14 at 22:31
I want to say that, [(Q(square root(2),cube root(7)):Q]=6=[(Q(square root(2).cube root(7)):Q]=[(Q(square root(2)+cube root(7)):Q]. So can we write that, Q(square root(2),cube root(7))=Q(square root(2).cube root(7))=Q(square root(2)+cube root(7))?
– Empty
Sep 24 '14 at 3:45
1
Well, yes we can...but only if you prove something else. For example, $;[Bbb Q(sqrt2):Bbb Q]=[Bbb Q(sqrt5):Bbb Q]=2;$ , yet $;Bbb Q(sqrt2)neqBbb Q (sqrt5);$ . Equality follows if for example one of the fields contains the other one. In your example, since clearly $;Bbb Q(sqrt2+sqrt7)subsetBbb Q(sqrt2,sqrt7);$ , if the degrees over $;Bbb Q;$ are equal then the fields are the same.
– Timbuc
Sep 24 '14 at 5:26
Sir, you have some mistake...My question is not root(7),it is cube root(7)
– Empty
Sep 24 '14 at 5:58
|
show 3 more comments
OK. I understand it. Due to the degree, Q(square root(2),cube root(7))=Q(square root(2)+cube root(7)) as well as =Q(square root(2).cube root(7)).As the degrees are equal. Am I right?
– Empty
Sep 23 '14 at 21:43
Don't understand your question, @SayantanPanja
– Timbuc
Sep 23 '14 at 22:31
I want to say that, [(Q(square root(2),cube root(7)):Q]=6=[(Q(square root(2).cube root(7)):Q]=[(Q(square root(2)+cube root(7)):Q]. So can we write that, Q(square root(2),cube root(7))=Q(square root(2).cube root(7))=Q(square root(2)+cube root(7))?
– Empty
Sep 24 '14 at 3:45
1
Well, yes we can...but only if you prove something else. For example, $;[Bbb Q(sqrt2):Bbb Q]=[Bbb Q(sqrt5):Bbb Q]=2;$ , yet $;Bbb Q(sqrt2)neqBbb Q (sqrt5);$ . Equality follows if for example one of the fields contains the other one. In your example, since clearly $;Bbb Q(sqrt2+sqrt7)subsetBbb Q(sqrt2,sqrt7);$ , if the degrees over $;Bbb Q;$ are equal then the fields are the same.
– Timbuc
Sep 24 '14 at 5:26
Sir, you have some mistake...My question is not root(7),it is cube root(7)
– Empty
Sep 24 '14 at 5:58
OK. I understand it. Due to the degree, Q(square root(2),cube root(7))=Q(square root(2)+cube root(7)) as well as =Q(square root(2).cube root(7)).As the degrees are equal. Am I right?
– Empty
Sep 23 '14 at 21:43
OK. I understand it. Due to the degree, Q(square root(2),cube root(7))=Q(square root(2)+cube root(7)) as well as =Q(square root(2).cube root(7)).As the degrees are equal. Am I right?
– Empty
Sep 23 '14 at 21:43
Don't understand your question, @SayantanPanja
– Timbuc
Sep 23 '14 at 22:31
Don't understand your question, @SayantanPanja
– Timbuc
Sep 23 '14 at 22:31
I want to say that, [(Q(square root(2),cube root(7)):Q]=6=[(Q(square root(2).cube root(7)):Q]=[(Q(square root(2)+cube root(7)):Q]. So can we write that, Q(square root(2),cube root(7))=Q(square root(2).cube root(7))=Q(square root(2)+cube root(7))?
– Empty
Sep 24 '14 at 3:45
I want to say that, [(Q(square root(2),cube root(7)):Q]=6=[(Q(square root(2).cube root(7)):Q]=[(Q(square root(2)+cube root(7)):Q]. So can we write that, Q(square root(2),cube root(7))=Q(square root(2).cube root(7))=Q(square root(2)+cube root(7))?
– Empty
Sep 24 '14 at 3:45
1
1
Well, yes we can...but only if you prove something else. For example, $;[Bbb Q(sqrt2):Bbb Q]=[Bbb Q(sqrt5):Bbb Q]=2;$ , yet $;Bbb Q(sqrt2)neqBbb Q (sqrt5);$ . Equality follows if for example one of the fields contains the other one. In your example, since clearly $;Bbb Q(sqrt2+sqrt7)subsetBbb Q(sqrt2,sqrt7);$ , if the degrees over $;Bbb Q;$ are equal then the fields are the same.
– Timbuc
Sep 24 '14 at 5:26
Well, yes we can...but only if you prove something else. For example, $;[Bbb Q(sqrt2):Bbb Q]=[Bbb Q(sqrt5):Bbb Q]=2;$ , yet $;Bbb Q(sqrt2)neqBbb Q (sqrt5);$ . Equality follows if for example one of the fields contains the other one. In your example, since clearly $;Bbb Q(sqrt2+sqrt7)subsetBbb Q(sqrt2,sqrt7);$ , if the degrees over $;Bbb Q;$ are equal then the fields are the same.
– Timbuc
Sep 24 '14 at 5:26
Sir, you have some mistake...My question is not root(7),it is cube root(7)
– Empty
Sep 24 '14 at 5:58
Sir, you have some mistake...My question is not root(7),it is cube root(7)
– Empty
Sep 24 '14 at 5:58
|
show 3 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.
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.
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%2f943250%2fis-mathbbq-sqrt2-sqrt5-mathbbq-sqrt10-in-field-extension%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
3
Look at degrees.
– Hurkyl
Sep 23 '14 at 18:06
No. $mathbb{Q}[sqrt{2}, sqrt{5}] = { a+bsqrt{2}+csqrt{5}+ dsqrt{10} : a,b,c,d, in mathbb{Q}}$.
– Crostul
Sep 23 '14 at 18:10
For no values $(a,b)$. And the question has answer No.
– Orest Bucicovschi
Sep 23 '14 at 18:16
1
You can show $sqrt{10} in mathbb{Q}(sqrt{2}, sqrt{5})$ but that $sqrt{2} notin mathbb{Q}(sqrt{10})$ and $sqrt{5} notin mathbb{Q}(sqrt{10})$. Can you see why? Also, as someone else noted, look at the degrees of the field extension.
– Islands
Sep 23 '14 at 18:18
Possible duplicate of Is $mathbf{Q}(sqrt{2},sqrt{3}) = mathbf{Q}(sqrt{6})$?
– Watson
Nov 26 at 12:44