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Calculate $[mathbbQ (sqrt[3]3 + sqrt 2) : mathbbQ]$
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up vote
2
down vote
favorite
Good morning,
I'd like to understand the following concept and possibly being able to solve different types of exercises with it.
The exercise is the following :
Calculate $[mathbbQ (sqrt[3]3 + sqrt 2) : mathbbQ]$
The solution seems pretty standard except for the point that i would like to understand.
It says more or less that once called $alpha = sqrt[3]3 , beta = sqrt 2$ the set $ 1, alpha, alpha^2 , beta, alphabeta , alpha^2beta $ is a basis of $mathbbQ(alpha, beta)$ over $mathbbQ$, instead $A= 1, gamma dots , gamma^4 $ where $gamma = alpha + beta$ is linearly dependent over $mathbbQ$.
So it writes the coordinates of $A$ respect to the basis of $mathbbQ(alpha, beta)$ :
$1=(1,0,0,0,0,0)$
$gamma = alpha + beta = (0,1,0,1,0,0)$
$gamma^2 = alpha^2 + 2alphabeta + 2 = (2,0,1,0,2,0)$, where $2 = beta^2$
$gamma^3=(3,6,0,2,0,3)$.
After that it says that we can stop at $gamma^3$ because we can see passed in coordinates and so in $mathbbQ^6$ that the four element are linearly independent and finishes saying that $mathbbQ(gamma) = mathbbQ(alpha, beta)$ and writing down $alpha$ and $beta$ as linear combination of element of $gamma$, which I din't understood at all, because starts to put coefficients in matrices.
I'd like to understand the method and when I could apply it,and the last part on the matrices.
Thank you all, any help would be appreciated.
P.S.
If you need the text of the solution,please comment it and let me know.
linear-algebra abstract-algebra field-theory extension-field splitting-field
asked Jul 14 at 13:05
jacopoburelli
1007
add a comment |Â
up vote
2
down vote
favorite
Good morning,
I'd like to understand the following concept and possibly being able to solve different types of exercises with it.
The exercise is the following :
Calculate $[mathbbQ (sqrt[3]3 + sqrt 2) : mathbbQ]$
The solution seems pretty standard except for the point that i would like to understand.
It says more or less that once called $alpha = sqrt[3]3 , beta = sqrt 2$ the set $ 1, alpha, alpha^2 , beta, alphabeta , alpha^2beta $ is a basis of $mathbbQ(alpha, beta)$ over $mathbbQ$, instead $A= 1, gamma dots , gamma^4 $ where $gamma = alpha + beta$ is linearly dependent over $mathbbQ$.
So it writes the coordinates of $A$ respect to the basis of $mathbbQ(alpha, beta)$ :
$1=(1,0,0,0,0,0)$
$gamma = alpha + beta = (0,1,0,1,0,0)$
$gamma^2 = alpha^2 + 2alphabeta + 2 = (2,0,1,0,2,0)$, where $2 = beta^2$
$gamma^3=(3,6,0,2,0,3)$.
After that it says that we can stop at $gamma^3$ because we can see passed in coordinates and so in $mathbbQ^6$ that the four element are linearly independent and finishes saying that $mathbbQ(gamma) = mathbbQ(alpha, beta)$ and writing down $alpha$ and $beta$ as linear combination of element of $gamma$, which I din't understood at all, because starts to put coefficients in matrices.
I'd like to understand the method and when I could apply it,and the last part on the matrices.
Thank you all, any help would be appreciated.
P.S.
If you need the text of the solution,please comment it and let me know.
linear-algebra abstract-algebra field-theory extension-field splitting-field
asked Jul 14 at 13:05
jacopoburelli
1007
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Good morning,
I'd like to understand the following concept and possibly being able to solve different types of exercises with it.
The exercise is the following :
Calculate $[mathbbQ (sqrt[3]3 + sqrt 2) : mathbbQ]$
The solution seems pretty standard except for the point that i would like to understand.
It says more or less that once called $alpha = sqrt[3]3 , beta = sqrt 2$ the set $ 1, alpha, alpha^2 , beta, alphabeta , alpha^2beta $ is a basis of $mathbbQ(alpha, beta)$ over $mathbbQ$, instead $A= 1, gamma dots , gamma^4 $ where $gamma = alpha + beta$ is linearly dependent over $mathbbQ$.
So it writes the coordinates of $A$ respect to the basis of $mathbbQ(alpha, beta)$ :
$1=(1,0,0,0,0,0)$
$gamma = alpha + beta = (0,1,0,1,0,0)$
$gamma^2 = alpha^2 + 2alphabeta + 2 = (2,0,1,0,2,0)$, where $2 = beta^2$
$gamma^3=(3,6,0,2,0,3)$.
After that it says that we can stop at $gamma^3$ because we can see passed in coordinates and so in $mathbbQ^6$ that the four element are linearly independent and finishes saying that $mathbbQ(gamma) = mathbbQ(alpha, beta)$ and writing down $alpha$ and $beta$ as linear combination of element of $gamma$, which I din't understood at all, because starts to put coefficients in matrices.
I'd like to understand the method and when I could apply it,and the last part on the matrices.
Thank you all, any help would be appreciated.
P.S.
If you need the text of the solution,please comment it and let me know.
linear-algebra abstract-algebra field-theory extension-field splitting-field
asked Jul 14 at 13:05
jacopoburelli
1007
Good morning,
I'd like to understand the following concept and possibly being able to solve different types of exercises with it.
The exercise is the following :
Calculate $[mathbbQ (sqrt[3]3 + sqrt 2) : mathbbQ]$
The solution seems pretty standard except for the point that i would like to understand.
It says more or less that once called $alpha = sqrt[3]3 , beta = sqrt 2$ the set $ 1, alpha, alpha^2 , beta, alphabeta , alpha^2beta $ is a basis of $mathbbQ(alpha, beta)$ over $mathbbQ$, instead $A= 1, gamma dots , gamma^4 $ where $gamma = alpha + beta$ is linearly dependent over $mathbbQ$.
So it writes the coordinates of $A$ respect to the basis of $mathbbQ(alpha, beta)$ :
$1=(1,0,0,0,0,0)$
$gamma = alpha + beta = (0,1,0,1,0,0)$
$gamma^2 = alpha^2 + 2alphabeta + 2 = (2,0,1,0,2,0)$, where $2 = beta^2$
$gamma^3=(3,6,0,2,0,3)$.
After that it says that we can stop at $gamma^3$ because we can see passed in coordinates and so in $mathbbQ^6$ that the four element are linearly independent and finishes saying that $mathbbQ(gamma) = mathbbQ(alpha, beta)$ and writing down $alpha$ and $beta$ as linear combination of element of $gamma$, which I din't understood at all, because starts to put coefficients in matrices.
I'd like to understand the method and when I could apply it,and the last part on the matrices.
Thank you all, any help would be appreciated.
P.S.
If you need the text of the solution,please comment it and let me know.
linear-algebra abstract-algebra field-theory extension-field splitting-field
asked Jul 14 at 13:05
jacopoburelli
1007
asked Jul 14 at 13:05
jacopoburelli
1007
asked Jul 14 at 13:05
jacopoburelli
1007
asked Jul 14 at 13:05
jacopoburelli
1007
1007
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
We know that $[mathbbQ(alpha, beta) : mathbbQ] = 6$ and $mathbbQ(gamma) subseteq mathbbQ(alpha, beta)$. Therefore
$$6 = [mathbbQ(alpha, beta) : mathbbQ] = [mathbbQ(alpha, beta) : mathbbQ(gamma)][mathbbQ(gamma) : mathbbQ]$$
so $[mathbbQ(gamma) : mathbbQ] mid 6$.
Also you know that one basis for $mathbbQ(gamma)$ over $mathbbQ$ is of the form $$1, gamma, gamma^2, ldots, gamma^n-1$$ where $n = [mathbbQ(gamma) : mathbbQ]$.
You showed that $1, gamma, gamma^2, gamma^3$ is linearly independent over $mathbbQ$ so $[mathbbQ(gamma) : mathbbQ] ge 4$.
Therefore $[mathbbQ(gamma) : mathbbQ] = 6$.
This now implies that $[mathbbQ(alpha, beta) : mathbbQ(gamma)] = 1$ so $mathbbQ(alpha, beta) = mathbbQ(gamma)$. Hence, $alpha$ and $beta$ can be expressed in the basis $1, gamma, ldots, gamma^5$.
answered Jul 14 at 15:39
mechanodroid
22.3k52041
Almost understood, just one question, i’d like to understand how could i determine that $gamma$‘s power linear indipendant with the method of the matrices in the solution
– jacopoburelli
Jul 14 at 17:56
1
@jacopoburelli If you have a vector space $V$ over a field $F$ with a basis $e_1, ldots, e_n$ then you can represent any vector $v in V$ as a unique linear combination $v = sum_i=1^n alpha_ie_i$. For a set of vectors $v_1, ldots, v_k subseteq V$, express $v_j = sum_i=1^nalpha_ije_i$ and write all the coefficients in a matrix $$pmatrixv_1 & v_2 & cdots & v_k = pmatrixalpha_11 & alpha_12 & cdots & alpha_1k \ alpha_21 & alpha_22 & cdots & alpha_2k \ vdots & vdots & ddots & vdots \ alpha_k1 & alpha_k2 & cdots & alpha_kk$$
– mechanodroid
Jul 14 at 23:03
1
Then $v_1, ldots, v_k$ is linearly independent if and only if the columns of this matrix are linearly independent, which means that its rank is $k$.
– mechanodroid
Jul 14 at 23:03
add a comment |Â
up vote
1
down vote
I think the argument is just that
$$
mathbbQ(gamma) subseteq mathbbQ(alpha, beta),
[mathbbQ(gamma): mathbbQ]ge 4
[mathbbQ(alpha, beta): mathbbQ]=6,
implies
[mathbbQ(gamma): mathbbQ]=6
$$
answered Jul 14 at 13:31
lhf
156k9160367
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
We know that $[mathbbQ(alpha, beta) : mathbbQ] = 6$ and $mathbbQ(gamma) subseteq mathbbQ(alpha, beta)$. Therefore
$$6 = [mathbbQ(alpha, beta) : mathbbQ] = [mathbbQ(alpha, beta) : mathbbQ(gamma)][mathbbQ(gamma) : mathbbQ]$$
so $[mathbbQ(gamma) : mathbbQ] mid 6$.
Also you know that one basis for $mathbbQ(gamma)$ over $mathbbQ$ is of the form $$1, gamma, gamma^2, ldots, gamma^n-1$$ where $n = [mathbbQ(gamma) : mathbbQ]$.
You showed that $1, gamma, gamma^2, gamma^3$ is linearly independent over $mathbbQ$ so $[mathbbQ(gamma) : mathbbQ] ge 4$.
Therefore $[mathbbQ(gamma) : mathbbQ] = 6$.
This now implies that $[mathbbQ(alpha, beta) : mathbbQ(gamma)] = 1$ so $mathbbQ(alpha, beta) = mathbbQ(gamma)$. Hence, $alpha$ and $beta$ can be expressed in the basis $1, gamma, ldots, gamma^5$.
answered Jul 14 at 15:39
mechanodroid
22.3k52041
Almost understood, just one question, i’d like to understand how could i determine that $gamma$‘s power linear indipendant with the method of the matrices in the solution
– jacopoburelli
Jul 14 at 17:56
1
@jacopoburelli If you have a vector space $V$ over a field $F$ with a basis $e_1, ldots, e_n$ then you can represent any vector $v in V$ as a unique linear combination $v = sum_i=1^n alpha_ie_i$. For a set of vectors $v_1, ldots, v_k subseteq V$, express $v_j = sum_i=1^nalpha_ije_i$ and write all the coefficients in a matrix $$pmatrixv_1 & v_2 & cdots & v_k = pmatrixalpha_11 & alpha_12 & cdots & alpha_1k \ alpha_21 & alpha_22 & cdots & alpha_2k \ vdots & vdots & ddots & vdots \ alpha_k1 & alpha_k2 & cdots & alpha_kk$$
– mechanodroid
Jul 14 at 23:03
1
Then $v_1, ldots, v_k$ is linearly independent if and only if the columns of this matrix are linearly independent, which means that its rank is $k$.
– mechanodroid
Jul 14 at 23:03
add a comment |Â
up vote
3
down vote
accepted
We know that $[mathbbQ(alpha, beta) : mathbbQ] = 6$ and $mathbbQ(gamma) subseteq mathbbQ(alpha, beta)$. Therefore
$$6 = [mathbbQ(alpha, beta) : mathbbQ] = [mathbbQ(alpha, beta) : mathbbQ(gamma)][mathbbQ(gamma) : mathbbQ]$$
so $[mathbbQ(gamma) : mathbbQ] mid 6$.
Also you know that one basis for $mathbbQ(gamma)$ over $mathbbQ$ is of the form $$1, gamma, gamma^2, ldots, gamma^n-1$$ where $n = [mathbbQ(gamma) : mathbbQ]$.
You showed that $1, gamma, gamma^2, gamma^3$ is linearly independent over $mathbbQ$ so $[mathbbQ(gamma) : mathbbQ] ge 4$.
Therefore $[mathbbQ(gamma) : mathbbQ] = 6$.
This now implies that $[mathbbQ(alpha, beta) : mathbbQ(gamma)] = 1$ so $mathbbQ(alpha, beta) = mathbbQ(gamma)$. Hence, $alpha$ and $beta$ can be expressed in the basis $1, gamma, ldots, gamma^5$.
answered Jul 14 at 15:39
mechanodroid
22.3k52041
Almost understood, just one question, i’d like to understand how could i determine that $gamma$‘s power linear indipendant with the method of the matrices in the solution
– jacopoburelli
Jul 14 at 17:56
1
@jacopoburelli If you have a vector space $V$ over a field $F$ with a basis $e_1, ldots, e_n$ then you can represent any vector $v in V$ as a unique linear combination $v = sum_i=1^n alpha_ie_i$. For a set of vectors $v_1, ldots, v_k subseteq V$, express $v_j = sum_i=1^nalpha_ije_i$ and write all the coefficients in a matrix $$pmatrixv_1 & v_2 & cdots & v_k = pmatrixalpha_11 & alpha_12 & cdots & alpha_1k \ alpha_21 & alpha_22 & cdots & alpha_2k \ vdots & vdots & ddots & vdots \ alpha_k1 & alpha_k2 & cdots & alpha_kk$$
– mechanodroid
Jul 14 at 23:03
1
Then $v_1, ldots, v_k$ is linearly independent if and only if the columns of this matrix are linearly independent, which means that its rank is $k$.
– mechanodroid
Jul 14 at 23:03
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
We know that $[mathbbQ(alpha, beta) : mathbbQ] = 6$ and $mathbbQ(gamma) subseteq mathbbQ(alpha, beta)$. Therefore
$$6 = [mathbbQ(alpha, beta) : mathbbQ] = [mathbbQ(alpha, beta) : mathbbQ(gamma)][mathbbQ(gamma) : mathbbQ]$$
so $[mathbbQ(gamma) : mathbbQ] mid 6$.
Also you know that one basis for $mathbbQ(gamma)$ over $mathbbQ$ is of the form $$1, gamma, gamma^2, ldots, gamma^n-1$$ where $n = [mathbbQ(gamma) : mathbbQ]$.
You showed that $1, gamma, gamma^2, gamma^3$ is linearly independent over $mathbbQ$ so $[mathbbQ(gamma) : mathbbQ] ge 4$.
Therefore $[mathbbQ(gamma) : mathbbQ] = 6$.
This now implies that $[mathbbQ(alpha, beta) : mathbbQ(gamma)] = 1$ so $mathbbQ(alpha, beta) = mathbbQ(gamma)$. Hence, $alpha$ and $beta$ can be expressed in the basis $1, gamma, ldots, gamma^5$.
answered Jul 14 at 15:39
mechanodroid
22.3k52041
We know that $[mathbbQ(alpha, beta) : mathbbQ] = 6$ and $mathbbQ(gamma) subseteq mathbbQ(alpha, beta)$. Therefore
$$6 = [mathbbQ(alpha, beta) : mathbbQ] = [mathbbQ(alpha, beta) : mathbbQ(gamma)][mathbbQ(gamma) : mathbbQ]$$
so $[mathbbQ(gamma) : mathbbQ] mid 6$.
Also you know that one basis for $mathbbQ(gamma)$ over $mathbbQ$ is of the form $$1, gamma, gamma^2, ldots, gamma^n-1$$ where $n = [mathbbQ(gamma) : mathbbQ]$.
You showed that $1, gamma, gamma^2, gamma^3$ is linearly independent over $mathbbQ$ so $[mathbbQ(gamma) : mathbbQ] ge 4$.
Therefore $[mathbbQ(gamma) : mathbbQ] = 6$.
This now implies that $[mathbbQ(alpha, beta) : mathbbQ(gamma)] = 1$ so $mathbbQ(alpha, beta) = mathbbQ(gamma)$. Hence, $alpha$ and $beta$ can be expressed in the basis $1, gamma, ldots, gamma^5$.
answered Jul 14 at 15:39
mechanodroid
22.3k52041
answered Jul 14 at 15:39
mechanodroid
22.3k52041
answered Jul 14 at 15:39
mechanodroid
22.3k52041
answered Jul 14 at 15:39
mechanodroid
22.3k52041
22.3k52041
Almost understood, just one question, i’d like to understand how could i determine that $gamma$‘s power linear indipendant with the method of the matrices in the solution
– jacopoburelli
Jul 14 at 17:56
1
@jacopoburelli If you have a vector space $V$ over a field $F$ with a basis $e_1, ldots, e_n$ then you can represent any vector $v in V$ as a unique linear combination $v = sum_i=1^n alpha_ie_i$. For a set of vectors $v_1, ldots, v_k subseteq V$, express $v_j = sum_i=1^nalpha_ije_i$ and write all the coefficients in a matrix $$pmatrixv_1 & v_2 & cdots & v_k = pmatrixalpha_11 & alpha_12 & cdots & alpha_1k \ alpha_21 & alpha_22 & cdots & alpha_2k \ vdots & vdots & ddots & vdots \ alpha_k1 & alpha_k2 & cdots & alpha_kk$$
– mechanodroid
Jul 14 at 23:03
1
Then $v_1, ldots, v_k$ is linearly independent if and only if the columns of this matrix are linearly independent, which means that its rank is $k$.
– mechanodroid
Jul 14 at 23:03
add a comment |Â
Almost understood, just one question, i’d like to understand how could i determine that $gamma$‘s power linear indipendant with the method of the matrices in the solution
– jacopoburelli
Jul 14 at 17:56
1
@jacopoburelli If you have a vector space $V$ over a field $F$ with a basis $e_1, ldots, e_n$ then you can represent any vector $v in V$ as a unique linear combination $v = sum_i=1^n alpha_ie_i$. For a set of vectors $v_1, ldots, v_k subseteq V$, express $v_j = sum_i=1^nalpha_ije_i$ and write all the coefficients in a matrix $$pmatrixv_1 & v_2 & cdots & v_k = pmatrixalpha_11 & alpha_12 & cdots & alpha_1k \ alpha_21 & alpha_22 & cdots & alpha_2k \ vdots & vdots & ddots & vdots \ alpha_k1 & alpha_k2 & cdots & alpha_kk$$
– mechanodroid
Jul 14 at 23:03
1
Then $v_1, ldots, v_k$ is linearly independent if and only if the columns of this matrix are linearly independent, which means that its rank is $k$.
– mechanodroid
Jul 14 at 23:03
Almost understood, just one question, i’d like to understand how could i determine that $gamma$‘s power linear indipendant with the method of the matrices in the solution
– jacopoburelli
Jul 14 at 17:56
Almost understood, just one question, i’d like to understand how could i determine that $gamma$‘s power linear indipendant with the method of the matrices in the solution
– jacopoburelli
Jul 14 at 17:56
1
1
@jacopoburelli If you have a vector space $V$ over a field $F$ with a basis $e_1, ldots, e_n$ then you can represent any vector $v in V$ as a unique linear combination $v = sum_i=1^n alpha_ie_i$. For a set of vectors $v_1, ldots, v_k subseteq V$, express $v_j = sum_i=1^nalpha_ije_i$ and write all the coefficients in a matrix $$pmatrixv_1 & v_2 & cdots & v_k = pmatrixalpha_11 & alpha_12 & cdots & alpha_1k \ alpha_21 & alpha_22 & cdots & alpha_2k \ vdots & vdots & ddots & vdots \ alpha_k1 & alpha_k2 & cdots & alpha_kk$$
– mechanodroid
Jul 14 at 23:03
@jacopoburelli If you have a vector space $V$ over a field $F$ with a basis $e_1, ldots, e_n$ then you can represent any vector $v in V$ as a unique linear combination $v = sum_i=1^n alpha_ie_i$. For a set of vectors $v_1, ldots, v_k subseteq V$, express $v_j = sum_i=1^nalpha_ije_i$ and write all the coefficients in a matrix $$pmatrixv_1 & v_2 & cdots & v_k = pmatrixalpha_11 & alpha_12 & cdots & alpha_1k \ alpha_21 & alpha_22 & cdots & alpha_2k \ vdots & vdots & ddots & vdots \ alpha_k1 & alpha_k2 & cdots & alpha_kk$$
– mechanodroid
Jul 14 at 23:03
1
1
Then $v_1, ldots, v_k$ is linearly independent if and only if the columns of this matrix are linearly independent, which means that its rank is $k$.
– mechanodroid
Jul 14 at 23:03
Then $v_1, ldots, v_k$ is linearly independent if and only if the columns of this matrix are linearly independent, which means that its rank is $k$.
– mechanodroid
Jul 14 at 23:03
add a comment |Â
up vote
1
down vote
I think the argument is just that
$$
mathbbQ(gamma) subseteq mathbbQ(alpha, beta),
[mathbbQ(gamma): mathbbQ]ge 4
[mathbbQ(alpha, beta): mathbbQ]=6,
implies
[mathbbQ(gamma): mathbbQ]=6
$$
answered Jul 14 at 13:31
lhf
156k9160367
add a comment |Â
up vote
1
down vote
I think the argument is just that
$$
mathbbQ(gamma) subseteq mathbbQ(alpha, beta),
[mathbbQ(gamma): mathbbQ]ge 4
[mathbbQ(alpha, beta): mathbbQ]=6,
implies
[mathbbQ(gamma): mathbbQ]=6
$$
answered Jul 14 at 13:31
lhf
156k9160367
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I think the argument is just that
$$
mathbbQ(gamma) subseteq mathbbQ(alpha, beta),
[mathbbQ(gamma): mathbbQ]ge 4
[mathbbQ(alpha, beta): mathbbQ]=6,
implies
[mathbbQ(gamma): mathbbQ]=6
$$
answered Jul 14 at 13:31
lhf
156k9160367
I think the argument is just that
$$
mathbbQ(gamma) subseteq mathbbQ(alpha, beta),
[mathbbQ(gamma): mathbbQ]ge 4
[mathbbQ(alpha, beta): mathbbQ]=6,
implies
[mathbbQ(gamma): mathbbQ]=6
$$
answered Jul 14 at 13:31
lhf
156k9160367
answered Jul 14 at 13:31
lhf
156k9160367
answered Jul 14 at 13:31
lhf
156k9160367
answered Jul 14 at 13:31
lhf
156k9160367
156k9160367
add a comment |Â
add a comment |Â
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