Finding a basis for a field extension $mathbb{Q}(i,sqrt[4]{2})/mathbb{Q}(sqrt{2}).$
I am looking to find a basis for the field extension $$mathbb{Q}(i,sqrt[4]{2})/mathbb{Q}(sqrt{2}).$$
Clearly as far as I can tell the elements $i$ and $sqrt[4]{2}$ would be in the basis, as neither of these elements is in the field $mathbb{Q}(sqrt{2})$. However, when it comes to finding other elements in the basis, all that I could really think of at first was $isqrt[4]{2}$. Then I supposed that we would also need $sqrt[4]{2}^3$ and $isqrt[4]{2}^3$ in the basis too, and later wondered whether $i sqrt{2}$ would also be required. I think not because it would be covered under the element $i$ as $sqrt{2}$ is contained in the field but I am not certain.
As it took me a while to find each of these 5 elements, I am wondering whether this is actually all the elements that would be required to form a basis, whether I am missing any elements or have too many. Any help would be appreciated thanks :)
galois-theory extension-field
asked Dec 6 at 23:29
UsernameInvalid
524
add a comment |
I am looking to find a basis for the field extension $$mathbb{Q}(i,sqrt[4]{2})/mathbb{Q}(sqrt{2}).$$
Clearly as far as I can tell the elements $i$ and $sqrt[4]{2}$ would be in the basis, as neither of these elements is in the field $mathbb{Q}(sqrt{2})$. However, when it comes to finding other elements in the basis, all that I could really think of at first was $isqrt[4]{2}$. Then I supposed that we would also need $sqrt[4]{2}^3$ and $isqrt[4]{2}^3$ in the basis too, and later wondered whether $i sqrt{2}$ would also be required. I think not because it would be covered under the element $i$ as $sqrt{2}$ is contained in the field but I am not certain.
As it took me a while to find each of these 5 elements, I am wondering whether this is actually all the elements that would be required to form a basis, whether I am missing any elements or have too many. Any help would be appreciated thanks :)
galois-theory extension-field
asked Dec 6 at 23:29
UsernameInvalid
524
What's the degree of the extension?
– leibnewtz
Dec 6 at 23:41
I think it should be degree 4 by the Tower Law, and now upon further examination would that mean that as $isqrt[4]{2}^3 = isqrt[4]{2} cdot sqrt{2}$ that this term is unnecessary?
– UsernameInvalid
Dec 6 at 23:46
But then the same could be said for $sqrt[4]{2}^3$ so I am unsure
– UsernameInvalid
Dec 6 at 23:47
add a comment |
I am looking to find a basis for the field extension $$mathbb{Q}(i,sqrt[4]{2})/mathbb{Q}(sqrt{2}).$$
Clearly as far as I can tell the elements $i$ and $sqrt[4]{2}$ would be in the basis, as neither of these elements is in the field $mathbb{Q}(sqrt{2})$. However, when it comes to finding other elements in the basis, all that I could really think of at first was $isqrt[4]{2}$. Then I supposed that we would also need $sqrt[4]{2}^3$ and $isqrt[4]{2}^3$ in the basis too, and later wondered whether $i sqrt{2}$ would also be required. I think not because it would be covered under the element $i$ as $sqrt{2}$ is contained in the field but I am not certain.
As it took me a while to find each of these 5 elements, I am wondering whether this is actually all the elements that would be required to form a basis, whether I am missing any elements or have too many. Any help would be appreciated thanks :)
galois-theory extension-field
asked Dec 6 at 23:29
UsernameInvalid
524
I am looking to find a basis for the field extension $$mathbb{Q}(i,sqrt[4]{2})/mathbb{Q}(sqrt{2}).$$
Clearly as far as I can tell the elements $i$ and $sqrt[4]{2}$ would be in the basis, as neither of these elements is in the field $mathbb{Q}(sqrt{2})$. However, when it comes to finding other elements in the basis, all that I could really think of at first was $isqrt[4]{2}$. Then I supposed that we would also need $sqrt[4]{2}^3$ and $isqrt[4]{2}^3$ in the basis too, and later wondered whether $i sqrt{2}$ would also be required. I think not because it would be covered under the element $i$ as $sqrt{2}$ is contained in the field but I am not certain.
As it took me a while to find each of these 5 elements, I am wondering whether this is actually all the elements that would be required to form a basis, whether I am missing any elements or have too many. Any help would be appreciated thanks :)
galois-theory extension-field
galois-theory extension-field
asked Dec 6 at 23:29
UsernameInvalid
524
asked Dec 6 at 23:29
UsernameInvalid
524
asked Dec 6 at 23:29
UsernameInvalid
524
asked Dec 6 at 23:29
UsernameInvalid
524
asked Dec 6 at 23:29
UsernameInvalid
524
524
What's the degree of the extension?
– leibnewtz
Dec 6 at 23:41
I think it should be degree 4 by the Tower Law, and now upon further examination would that mean that as $isqrt[4]{2}^3 = isqrt[4]{2} cdot sqrt{2}$ that this term is unnecessary?
– UsernameInvalid
Dec 6 at 23:46
But then the same could be said for $sqrt[4]{2}^3$ so I am unsure
– UsernameInvalid
Dec 6 at 23:47
add a comment |
What's the degree of the extension?
– leibnewtz
Dec 6 at 23:41
I think it should be degree 4 by the Tower Law, and now upon further examination would that mean that as $isqrt[4]{2}^3 = isqrt[4]{2} cdot sqrt{2}$ that this term is unnecessary?
– UsernameInvalid
Dec 6 at 23:46
But then the same could be said for $sqrt[4]{2}^3$ so I am unsure
– UsernameInvalid
Dec 6 at 23:47
What's the degree of the extension?
– leibnewtz
Dec 6 at 23:41
What's the degree of the extension?
– leibnewtz
Dec 6 at 23:41
I think it should be degree 4 by the Tower Law, and now upon further examination would that mean that as $isqrt[4]{2}^3 = isqrt[4]{2} cdot sqrt{2}$ that this term is unnecessary?
– UsernameInvalid
Dec 6 at 23:46
I think it should be degree 4 by the Tower Law, and now upon further examination would that mean that as $isqrt[4]{2}^3 = isqrt[4]{2} cdot sqrt{2}$ that this term is unnecessary?
– UsernameInvalid
Dec 6 at 23:46
But then the same could be said for $sqrt[4]{2}^3$ so I am unsure
– UsernameInvalid
Dec 6 at 23:47
But then the same could be said for $sqrt[4]{2}^3$ so I am unsure
– UsernameInvalid
Dec 6 at 23:47
add a comment |
2 Answers
2
active
oldest
votes
Note that $inotinmathbb{Q}(sqrt[4]{2})$, so $mathbb{Q}(i,sqrt[4]{2})$ has degree $8$ over $mathbb{Q}$. Hence it has degree $4$ over $mathbb{Q}(sqrt{2})$.
A basis for $mathbb{Q}(sqrt[4]{2})$ over $mathbb{Q}(sqrt{2})$ is ${1,sqrt[4]{2})$.
Can you finish? You still have to add $i$.
answered Dec 7 at 0:01
egreg
177k1484198
after thinking about the problem would $1,i, sqrt[4]{2}, i sqrt[4]{2}$ be a suitable basis?
– UsernameInvalid
Dec 7 at 0:02
@UsernameInvalid Yes, correct.
– egreg
Dec 7 at 0:04
add a comment |
I think (2^¼)^2=√2 this imply that √2∈Q(2^¼) that is Q(√2) Contained in Q(2^¼) therefore degree of i over Q(√2) is 2 that is degree of Q(2^¼, i) over Q(√2) is 2 therefore basis of Q(2^¼, i) over Q(√2) contains only two element namely {1,i}.
answered Dec 12 at 11:45
user499117
409
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%2f3029210%2ffinding-a-basis-for-a-field-extension-mathbbqi-sqrt42-mathbbq-sqr%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Note that $inotinmathbb{Q}(sqrt[4]{2})$, so $mathbb{Q}(i,sqrt[4]{2})$ has degree $8$ over $mathbb{Q}$. Hence it has degree $4$ over $mathbb{Q}(sqrt{2})$.
A basis for $mathbb{Q}(sqrt[4]{2})$ over $mathbb{Q}(sqrt{2})$ is ${1,sqrt[4]{2})$.
Can you finish? You still have to add $i$.
answered Dec 7 at 0:01
egreg
177k1484198
after thinking about the problem would $1,i, sqrt[4]{2}, i sqrt[4]{2}$ be a suitable basis?
– UsernameInvalid
Dec 7 at 0:02
@UsernameInvalid Yes, correct.
– egreg
Dec 7 at 0:04
add a comment |
Note that $inotinmathbb{Q}(sqrt[4]{2})$, so $mathbb{Q}(i,sqrt[4]{2})$ has degree $8$ over $mathbb{Q}$. Hence it has degree $4$ over $mathbb{Q}(sqrt{2})$.
A basis for $mathbb{Q}(sqrt[4]{2})$ over $mathbb{Q}(sqrt{2})$ is ${1,sqrt[4]{2})$.
Can you finish? You still have to add $i$.
answered Dec 7 at 0:01
egreg
177k1484198
after thinking about the problem would $1,i, sqrt[4]{2}, i sqrt[4]{2}$ be a suitable basis?
– UsernameInvalid
Dec 7 at 0:02
@UsernameInvalid Yes, correct.
– egreg
Dec 7 at 0:04
add a comment |
Note that $inotinmathbb{Q}(sqrt[4]{2})$, so $mathbb{Q}(i,sqrt[4]{2})$ has degree $8$ over $mathbb{Q}$. Hence it has degree $4$ over $mathbb{Q}(sqrt{2})$.
A basis for $mathbb{Q}(sqrt[4]{2})$ over $mathbb{Q}(sqrt{2})$ is ${1,sqrt[4]{2})$.
Can you finish? You still have to add $i$.
answered Dec 7 at 0:01
egreg
177k1484198
Note that $inotinmathbb{Q}(sqrt[4]{2})$, so $mathbb{Q}(i,sqrt[4]{2})$ has degree $8$ over $mathbb{Q}$. Hence it has degree $4$ over $mathbb{Q}(sqrt{2})$.
A basis for $mathbb{Q}(sqrt[4]{2})$ over $mathbb{Q}(sqrt{2})$ is ${1,sqrt[4]{2})$.
Can you finish? You still have to add $i$.
answered Dec 7 at 0:01
egreg
177k1484198
answered Dec 7 at 0:01
egreg
177k1484198
answered Dec 7 at 0:01
egreg
177k1484198
answered Dec 7 at 0:01
egreg
177k1484198
177k1484198
after thinking about the problem would $1,i, sqrt[4]{2}, i sqrt[4]{2}$ be a suitable basis?
– UsernameInvalid
Dec 7 at 0:02
@UsernameInvalid Yes, correct.
– egreg
Dec 7 at 0:04
add a comment |
after thinking about the problem would $1,i, sqrt[4]{2}, i sqrt[4]{2}$ be a suitable basis?
– UsernameInvalid
Dec 7 at 0:02
@UsernameInvalid Yes, correct.
– egreg
Dec 7 at 0:04
after thinking about the problem would $1,i, sqrt[4]{2}, i sqrt[4]{2}$ be a suitable basis?
– UsernameInvalid
Dec 7 at 0:02
after thinking about the problem would $1,i, sqrt[4]{2}, i sqrt[4]{2}$ be a suitable basis?
– UsernameInvalid
Dec 7 at 0:02
@UsernameInvalid Yes, correct.
– egreg
Dec 7 at 0:04
@UsernameInvalid Yes, correct.
– egreg
Dec 7 at 0:04
add a comment |
I think (2^¼)^2=√2 this imply that √2∈Q(2^¼) that is Q(√2) Contained in Q(2^¼) therefore degree of i over Q(√2) is 2 that is degree of Q(2^¼, i) over Q(√2) is 2 therefore basis of Q(2^¼, i) over Q(√2) contains only two element namely {1,i}.
answered Dec 12 at 11:45
user499117
409
add a comment |
I think (2^¼)^2=√2 this imply that √2∈Q(2^¼) that is Q(√2) Contained in Q(2^¼) therefore degree of i over Q(√2) is 2 that is degree of Q(2^¼, i) over Q(√2) is 2 therefore basis of Q(2^¼, i) over Q(√2) contains only two element namely {1,i}.
answered Dec 12 at 11:45
user499117
409
add a comment |
I think (2^¼)^2=√2 this imply that √2∈Q(2^¼) that is Q(√2) Contained in Q(2^¼) therefore degree of i over Q(√2) is 2 that is degree of Q(2^¼, i) over Q(√2) is 2 therefore basis of Q(2^¼, i) over Q(√2) contains only two element namely {1,i}.
answered Dec 12 at 11:45
user499117
409
I think (2^¼)^2=√2 this imply that √2∈Q(2^¼) that is Q(√2) Contained in Q(2^¼) therefore degree of i over Q(√2) is 2 that is degree of Q(2^¼, i) over Q(√2) is 2 therefore basis of Q(2^¼, i) over Q(√2) contains only two element namely {1,i}.
answered Dec 12 at 11:45
user499117
409
answered Dec 12 at 11:45
user499117
409
answered Dec 12 at 11:45
user499117
409
answered Dec 12 at 11:45
user499117
409
409
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%2f3029210%2ffinding-a-basis-for-a-field-extension-mathbbqi-sqrt42-mathbbq-sqr%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
What's the degree of the extension?
– leibnewtz
Dec 6 at 23:41
I think it should be degree 4 by the Tower Law, and now upon further examination would that mean that as $isqrt[4]{2}^3 = isqrt[4]{2} cdot sqrt{2}$ that this term is unnecessary?
– UsernameInvalid
Dec 6 at 23:46
But then the same could be said for $sqrt[4]{2}^3$ so I am unsure
– UsernameInvalid
Dec 6 at 23:47