Finding a basis for a field extension $mathbb{Q}(i,sqrt[4]{2})/mathbb{Q}(sqrt{2}).$
$begingroup$
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 '18 at 23:29
UsernameInvalidUsernameInvalid
574
$endgroup$
add a comment |
$begingroup$
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 '18 at 23:29
UsernameInvalidUsernameInvalid
574
$endgroup$
$begingroup$
What's the degree of the extension?
$endgroup$
– leibnewtz
Dec 6 '18 at 23:41
$begingroup$
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?
$endgroup$
– UsernameInvalid
Dec 6 '18 at 23:46
$begingroup$
But then the same could be said for $sqrt[4]{2}^3$ so I am unsure
$endgroup$
– UsernameInvalid
Dec 6 '18 at 23:47
add a comment |
$begingroup$
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 '18 at 23:29
UsernameInvalidUsernameInvalid
574
$endgroup$
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 '18 at 23:29
UsernameInvalidUsernameInvalid
574
asked Dec 6 '18 at 23:29
UsernameInvalidUsernameInvalid
574
asked Dec 6 '18 at 23:29
UsernameInvalidUsernameInvalid
574
asked Dec 6 '18 at 23:29
UsernameInvalidUsernameInvalid
574
asked Dec 6 '18 at 23:29
UsernameInvalidUsernameInvalid
574
574
$begingroup$
What's the degree of the extension?
$endgroup$
– leibnewtz
Dec 6 '18 at 23:41
$begingroup$
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?
$endgroup$
– UsernameInvalid
Dec 6 '18 at 23:46
$begingroup$
But then the same could be said for $sqrt[4]{2}^3$ so I am unsure
$endgroup$
– UsernameInvalid
Dec 6 '18 at 23:47
add a comment |
$begingroup$
What's the degree of the extension?
$endgroup$
– leibnewtz
Dec 6 '18 at 23:41
$begingroup$
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?
$endgroup$
– UsernameInvalid
Dec 6 '18 at 23:46
$begingroup$
But then the same could be said for $sqrt[4]{2}^3$ so I am unsure
$endgroup$
– UsernameInvalid
Dec 6 '18 at 23:47
$begingroup$
What's the degree of the extension?
$endgroup$
– leibnewtz
Dec 6 '18 at 23:41
$begingroup$
What's the degree of the extension?
$endgroup$
– leibnewtz
Dec 6 '18 at 23:41
$begingroup$
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?
$endgroup$
– UsernameInvalid
Dec 6 '18 at 23:46
$begingroup$
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?
$endgroup$
– UsernameInvalid
Dec 6 '18 at 23:46
$begingroup$
But then the same could be said for $sqrt[4]{2}^3$ so I am unsure
$endgroup$
– UsernameInvalid
Dec 6 '18 at 23:47
$begingroup$
But then the same could be said for $sqrt[4]{2}^3$ so I am unsure
$endgroup$
– UsernameInvalid
Dec 6 '18 at 23:47
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 '18 at 0:01
egregegreg
181k1485203
$endgroup$
$begingroup$
after thinking about the problem would $1,i, sqrt[4]{2}, i sqrt[4]{2}$ be a suitable basis?
$endgroup$
– UsernameInvalid
Dec 7 '18 at 0:02
$begingroup$
@UsernameInvalid Yes, correct.
$endgroup$
– egreg
Dec 7 '18 at 0:04
add a comment |
$begingroup$
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 '18 at 11:45
user499117user499117
409
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
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active
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votes
$begingroup$
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 '18 at 0:01
egregegreg
181k1485203
$endgroup$
$begingroup$
after thinking about the problem would $1,i, sqrt[4]{2}, i sqrt[4]{2}$ be a suitable basis?
$endgroup$
– UsernameInvalid
Dec 7 '18 at 0:02
$begingroup$
@UsernameInvalid Yes, correct.
$endgroup$
– egreg
Dec 7 '18 at 0:04
add a comment |
$begingroup$
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 '18 at 0:01
egregegreg
181k1485203
$endgroup$
$begingroup$
after thinking about the problem would $1,i, sqrt[4]{2}, i sqrt[4]{2}$ be a suitable basis?
$endgroup$
– UsernameInvalid
Dec 7 '18 at 0:02
$begingroup$
@UsernameInvalid Yes, correct.
$endgroup$
– egreg
Dec 7 '18 at 0:04
add a comment |
$begingroup$
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 '18 at 0:01
egregegreg
181k1485203
$endgroup$
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 '18 at 0:01
egregegreg
181k1485203
answered Dec 7 '18 at 0:01
egregegreg
181k1485203
answered Dec 7 '18 at 0:01
egregegreg
181k1485203
answered Dec 7 '18 at 0:01
egregegreg
181k1485203
181k1485203
$begingroup$
after thinking about the problem would $1,i, sqrt[4]{2}, i sqrt[4]{2}$ be a suitable basis?
$endgroup$
– UsernameInvalid
Dec 7 '18 at 0:02
$begingroup$
@UsernameInvalid Yes, correct.
$endgroup$
– egreg
Dec 7 '18 at 0:04
add a comment |
$begingroup$
after thinking about the problem would $1,i, sqrt[4]{2}, i sqrt[4]{2}$ be a suitable basis?
$endgroup$
– UsernameInvalid
Dec 7 '18 at 0:02
$begingroup$
@UsernameInvalid Yes, correct.
$endgroup$
– egreg
Dec 7 '18 at 0:04
$begingroup$
after thinking about the problem would $1,i, sqrt[4]{2}, i sqrt[4]{2}$ be a suitable basis?
$endgroup$
– UsernameInvalid
Dec 7 '18 at 0:02
$begingroup$
after thinking about the problem would $1,i, sqrt[4]{2}, i sqrt[4]{2}$ be a suitable basis?
$endgroup$
– UsernameInvalid
Dec 7 '18 at 0:02
$begingroup$
@UsernameInvalid Yes, correct.
$endgroup$
– egreg
Dec 7 '18 at 0:04
$begingroup$
@UsernameInvalid Yes, correct.
$endgroup$
– egreg
Dec 7 '18 at 0:04
add a comment |
$begingroup$
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 '18 at 11:45
user499117user499117
409
$endgroup$
add a comment |
$begingroup$
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 '18 at 11:45
user499117user499117
409
$endgroup$
add a comment |
$begingroup$
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 '18 at 11:45
user499117user499117
409
$endgroup$
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 '18 at 11:45
user499117user499117
409
answered Dec 12 '18 at 11:45
user499117user499117
409
answered Dec 12 '18 at 11:45
user499117user499117
409
answered Dec 12 '18 at 11:45
user499117user499117
409
409
add a comment |
add a comment |
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$begingroup$
What's the degree of the extension?
$endgroup$
– leibnewtz
Dec 6 '18 at 23:41
$begingroup$
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?
$endgroup$
– UsernameInvalid
Dec 6 '18 at 23:46
$begingroup$
But then the same could be said for $sqrt[4]{2}^3$ so I am unsure
$endgroup$
– UsernameInvalid
Dec 6 '18 at 23:47