$alpha = sqrt2 + sqrt3 in V$ then $dim_Bbb Q V=4$
2
$begingroup$
Let $alpha = sqrt2 + sqrt3 in V$ where $V$ is a field and $V:=langle 1,sqrt2, sqrt3 , sqrt6 rangle_Bbb Q subset V$.
My textbook says
- $$langle 1,alpha, alpha^2 , alpha^3 rangle_Bbb Q subset V$$
- $$dim_Bbb Qlangle 1,alpha, alpha^2 , alpha^3 rangle = 4$$
Hence $dim_Bbb Q V=4$.
To prove 2. is true, it suffices to show that $ 1,alpha, alpha^2 , alpha^3 $ are linearly independent and $langle 1,alpha, alpha^2 , alpha^3 rangle_Bbb Q$ can be mapped bijectively to $V$. I'm having trouble finding such a linear map.
Any help is much appreciated!
linear-algebra abstract-algebra vector-spaces field-theory
asked Dec 4 '18 at 23:37
wtnmathwtnmath
16412
$endgroup$
|
show 1 more comment
2
$begingroup$
Let $alpha = sqrt2 + sqrt3 in V$ where $V$ is a field and $V:=langle 1,sqrt2, sqrt3 , sqrt6 rangle_Bbb Q subset V$.
My textbook says
- $$langle 1,alpha, alpha^2 , alpha^3 rangle_Bbb Q subset V$$
- $$dim_Bbb Qlangle 1,alpha, alpha^2 , alpha^3 rangle = 4$$
Hence $dim_Bbb Q V=4$.
To prove 2. is true, it suffices to show that $ 1,alpha, alpha^2 , alpha^3 $ are linearly independent and $langle 1,alpha, alpha^2 , alpha^3 rangle_Bbb Q$ can be mapped bijectively to $V$. I'm having trouble finding such a linear map.
Any help is much appreciated!
linear-algebra abstract-algebra vector-spaces field-theory
asked Dec 4 '18 at 23:37
wtnmathwtnmath
16412
$endgroup$
1
$begingroup$
If $V$ is a field and $alpha in V$, then by closure under multiplication, it is trivial that $1,alpha,alpha^2,alpha^3 in V$ and thus that $langle 1, alpha, alpha^2, alpha^3 rangle_mathbb Q subset V$. However, unless I'm misunderstanding the definitions, it's not necessarily the case that $dim_mathbb Q V = 4$, but rather $dim_mathbb Q V ge 4$. Of course it's possible that $V = mathbb R$ in which case $dim_mathbb Q V = infty$.
$endgroup$
– User8128
Dec 4 '18 at 23:49
$begingroup$
If the dimension were less than 4 then $1, alpha, alpha^2, alpha^3$ would be linearly dependent. Write down what this means and see what you can conclude.
$endgroup$
– Trevor Gunn
Dec 4 '18 at 23:53
$begingroup$
@TrevorGunn Thanks!. As User8128 mentioned, since $1, sqrt2, sqrt3, sqrt6$ are linearly independent, so the dimension has to be greater or equal to 4. But how is it exactly 4?
$endgroup$
– wtnmath
Dec 5 '18 at 0:05
1
$begingroup$
Of course from the information that you just given above it is wrong to say that $text{dim}_{mathbb{Q}} V=4$. You can have another field which is bigger than $V$ which is of course still contain $alpha$. There must be some more informations about $V$ that you didn't write down.
$endgroup$
– user9077
Dec 5 '18 at 0:17
1
$begingroup$
$[mathbb{Q}(sqrt 2,sqrt3):mathbb{Q}]=[mathbb{Q}(sqrt 2,sqrt3):mathbb{Q}(sqrt 2)]cdot[mathbb{Q}(sqrt 2):mathbb{Q}]$
$endgroup$
– random
Dec 5 '18 at 0:47
|
show 1 more comment
2
2
2
3
$begingroup$
Let $alpha = sqrt2 + sqrt3 in V$ where $V$ is a field and $V:=langle 1,sqrt2, sqrt3 , sqrt6 rangle_Bbb Q subset V$.
My textbook says
- $$langle 1,alpha, alpha^2 , alpha^3 rangle_Bbb Q subset V$$
- $$dim_Bbb Qlangle 1,alpha, alpha^2 , alpha^3 rangle = 4$$
Hence $dim_Bbb Q V=4$.
To prove 2. is true, it suffices to show that $ 1,alpha, alpha^2 , alpha^3 $ are linearly independent and $langle 1,alpha, alpha^2 , alpha^3 rangle_Bbb Q$ can be mapped bijectively to $V$. I'm having trouble finding such a linear map.
Any help is much appreciated!
linear-algebra abstract-algebra vector-spaces field-theory
asked Dec 4 '18 at 23:37
wtnmathwtnmath
16412
$endgroup$
Let $alpha = sqrt2 + sqrt3 in V$ where $V$ is a field and $V:=langle 1,sqrt2, sqrt3 , sqrt6 rangle_Bbb Q subset V$.
My textbook says
- $$langle 1,alpha, alpha^2 , alpha^3 rangle_Bbb Q subset V$$
- $$dim_Bbb Qlangle 1,alpha, alpha^2 , alpha^3 rangle = 4$$
Hence $dim_Bbb Q V=4$.
To prove 2. is true, it suffices to show that $ 1,alpha, alpha^2 , alpha^3 $ are linearly independent and $langle 1,alpha, alpha^2 , alpha^3 rangle_Bbb Q$ can be mapped bijectively to $V$. I'm having trouble finding such a linear map.
Any help is much appreciated!
linear-algebra abstract-algebra vector-spaces field-theory
linear-algebra abstract-algebra vector-spaces field-theory
asked Dec 4 '18 at 23:37
wtnmathwtnmath
16412
asked Dec 4 '18 at 23:37
wtnmathwtnmath
16412
edited Dec 5 '18 at 2:37
wtnmath
asked Dec 4 '18 at 23:37
wtnmathwtnmath
16412
asked Dec 4 '18 at 23:37
wtnmathwtnmath
16412
asked Dec 4 '18 at 23:37
wtnmathwtnmath
16412
16412
1
$begingroup$
If $V$ is a field and $alpha in V$, then by closure under multiplication, it is trivial that $1,alpha,alpha^2,alpha^3 in V$ and thus that $langle 1, alpha, alpha^2, alpha^3 rangle_mathbb Q subset V$. However, unless I'm misunderstanding the definitions, it's not necessarily the case that $dim_mathbb Q V = 4$, but rather $dim_mathbb Q V ge 4$. Of course it's possible that $V = mathbb R$ in which case $dim_mathbb Q V = infty$.
$endgroup$
– User8128
Dec 4 '18 at 23:49
$begingroup$
If the dimension were less than 4 then $1, alpha, alpha^2, alpha^3$ would be linearly dependent. Write down what this means and see what you can conclude.
$endgroup$
– Trevor Gunn
Dec 4 '18 at 23:53
$begingroup$
@TrevorGunn Thanks!. As User8128 mentioned, since $1, sqrt2, sqrt3, sqrt6$ are linearly independent, so the dimension has to be greater or equal to 4. But how is it exactly 4?
$endgroup$
– wtnmath
Dec 5 '18 at 0:05
1
$begingroup$
Of course from the information that you just given above it is wrong to say that $text{dim}_{mathbb{Q}} V=4$. You can have another field which is bigger than $V$ which is of course still contain $alpha$. There must be some more informations about $V$ that you didn't write down.
$endgroup$
– user9077
Dec 5 '18 at 0:17
1
$begingroup$
$[mathbb{Q}(sqrt 2,sqrt3):mathbb{Q}]=[mathbb{Q}(sqrt 2,sqrt3):mathbb{Q}(sqrt 2)]cdot[mathbb{Q}(sqrt 2):mathbb{Q}]$
$endgroup$
– random
Dec 5 '18 at 0:47
|
show 1 more comment
1
$begingroup$
If $V$ is a field and $alpha in V$, then by closure under multiplication, it is trivial that $1,alpha,alpha^2,alpha^3 in V$ and thus that $langle 1, alpha, alpha^2, alpha^3 rangle_mathbb Q subset V$. However, unless I'm misunderstanding the definitions, it's not necessarily the case that $dim_mathbb Q V = 4$, but rather $dim_mathbb Q V ge 4$. Of course it's possible that $V = mathbb R$ in which case $dim_mathbb Q V = infty$.
$endgroup$
– User8128
Dec 4 '18 at 23:49
$begingroup$
If the dimension were less than 4 then $1, alpha, alpha^2, alpha^3$ would be linearly dependent. Write down what this means and see what you can conclude.
$endgroup$
– Trevor Gunn
Dec 4 '18 at 23:53
$begingroup$
@TrevorGunn Thanks!. As User8128 mentioned, since $1, sqrt2, sqrt3, sqrt6$ are linearly independent, so the dimension has to be greater or equal to 4. But how is it exactly 4?
$endgroup$
– wtnmath
Dec 5 '18 at 0:05
1
$begingroup$
Of course from the information that you just given above it is wrong to say that $text{dim}_{mathbb{Q}} V=4$. You can have another field which is bigger than $V$ which is of course still contain $alpha$. There must be some more informations about $V$ that you didn't write down.
$endgroup$
– user9077
Dec 5 '18 at 0:17
1
$begingroup$
$[mathbb{Q}(sqrt 2,sqrt3):mathbb{Q}]=[mathbb{Q}(sqrt 2,sqrt3):mathbb{Q}(sqrt 2)]cdot[mathbb{Q}(sqrt 2):mathbb{Q}]$
$endgroup$
– random
Dec 5 '18 at 0:47
1
1
$begingroup$
If $V$ is a field and $alpha in V$, then by closure under multiplication, it is trivial that $1,alpha,alpha^2,alpha^3 in V$ and thus that $langle 1, alpha, alpha^2, alpha^3 rangle_mathbb Q subset V$. However, unless I'm misunderstanding the definitions, it's not necessarily the case that $dim_mathbb Q V = 4$, but rather $dim_mathbb Q V ge 4$. Of course it's possible that $V = mathbb R$ in which case $dim_mathbb Q V = infty$.
$endgroup$
– User8128
Dec 4 '18 at 23:49
$begingroup$
If $V$ is a field and $alpha in V$, then by closure under multiplication, it is trivial that $1,alpha,alpha^2,alpha^3 in V$ and thus that $langle 1, alpha, alpha^2, alpha^3 rangle_mathbb Q subset V$. However, unless I'm misunderstanding the definitions, it's not necessarily the case that $dim_mathbb Q V = 4$, but rather $dim_mathbb Q V ge 4$. Of course it's possible that $V = mathbb R$ in which case $dim_mathbb Q V = infty$.
$endgroup$
– User8128
Dec 4 '18 at 23:49
$begingroup$
If the dimension were less than 4 then $1, alpha, alpha^2, alpha^3$ would be linearly dependent. Write down what this means and see what you can conclude.
$endgroup$
– Trevor Gunn
Dec 4 '18 at 23:53
$begingroup$
If the dimension were less than 4 then $1, alpha, alpha^2, alpha^3$ would be linearly dependent. Write down what this means and see what you can conclude.
$endgroup$
– Trevor Gunn
Dec 4 '18 at 23:53
$begingroup$
@TrevorGunn Thanks!. As User8128 mentioned, since $1, sqrt2, sqrt3, sqrt6$ are linearly independent, so the dimension has to be greater or equal to 4. But how is it exactly 4?
$endgroup$
– wtnmath
Dec 5 '18 at 0:05
$begingroup$
@TrevorGunn Thanks!. As User8128 mentioned, since $1, sqrt2, sqrt3, sqrt6$ are linearly independent, so the dimension has to be greater or equal to 4. But how is it exactly 4?
$endgroup$
– wtnmath
Dec 5 '18 at 0:05
1
1
$begingroup$
Of course from the information that you just given above it is wrong to say that $text{dim}_{mathbb{Q}} V=4$. You can have another field which is bigger than $V$ which is of course still contain $alpha$. There must be some more informations about $V$ that you didn't write down.
$endgroup$
– user9077
Dec 5 '18 at 0:17
$begingroup$
Of course from the information that you just given above it is wrong to say that $text{dim}_{mathbb{Q}} V=4$. You can have another field which is bigger than $V$ which is of course still contain $alpha$. There must be some more informations about $V$ that you didn't write down.
$endgroup$
– user9077
Dec 5 '18 at 0:17
1
1
$begingroup$
$[mathbb{Q}(sqrt 2,sqrt3):mathbb{Q}]=[mathbb{Q}(sqrt 2,sqrt3):mathbb{Q}(sqrt 2)]cdot[mathbb{Q}(sqrt 2):mathbb{Q}]$
$endgroup$
– random
Dec 5 '18 at 0:47
$begingroup$
$[mathbb{Q}(sqrt 2,sqrt3):mathbb{Q}]=[mathbb{Q}(sqrt 2,sqrt3):mathbb{Q}(sqrt 2)]cdot[mathbb{Q}(sqrt 2):mathbb{Q}]$
$endgroup$
– random
Dec 5 '18 at 0:47
|
show 1 more comment
1 Answer
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Think about what radicals $sqrt{n}$ you can create using polynomials in $alpha$. Two of them are staring you in the face. Is there another one?
answered Dec 4 '18 at 23:49
zoidbergzoidberg
1,065113
$endgroup$
add a comment |
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Think about what radicals $sqrt{n}$ you can create using polynomials in $alpha$. Two of them are staring you in the face. Is there another one?
answered Dec 4 '18 at 23:49
zoidbergzoidberg
1,065113
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add a comment |
0
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Think about what radicals $sqrt{n}$ you can create using polynomials in $alpha$. Two of them are staring you in the face. Is there another one?
answered Dec 4 '18 at 23:49
zoidbergzoidberg
1,065113
$endgroup$
add a comment |
0
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$begingroup$
Think about what radicals $sqrt{n}$ you can create using polynomials in $alpha$. Two of them are staring you in the face. Is there another one?
answered Dec 4 '18 at 23:49
zoidbergzoidberg
1,065113
$endgroup$
Think about what radicals $sqrt{n}$ you can create using polynomials in $alpha$. Two of them are staring you in the face. Is there another one?
answered Dec 4 '18 at 23:49
zoidbergzoidberg
1,065113
answered Dec 4 '18 at 23:49
zoidbergzoidberg
1,065113
answered Dec 4 '18 at 23:49
zoidbergzoidberg
1,065113
answered Dec 4 '18 at 23:49
zoidbergzoidberg
1,065113
1,065113
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1
$begingroup$
If $V$ is a field and $alpha in V$, then by closure under multiplication, it is trivial that $1,alpha,alpha^2,alpha^3 in V$ and thus that $langle 1, alpha, alpha^2, alpha^3 rangle_mathbb Q subset V$. However, unless I'm misunderstanding the definitions, it's not necessarily the case that $dim_mathbb Q V = 4$, but rather $dim_mathbb Q V ge 4$. Of course it's possible that $V = mathbb R$ in which case $dim_mathbb Q V = infty$.
$endgroup$
– User8128
Dec 4 '18 at 23:49
$begingroup$
If the dimension were less than 4 then $1, alpha, alpha^2, alpha^3$ would be linearly dependent. Write down what this means and see what you can conclude.
$endgroup$
– Trevor Gunn
Dec 4 '18 at 23:53
$begingroup$
@TrevorGunn Thanks!. As User8128 mentioned, since $1, sqrt2, sqrt3, sqrt6$ are linearly independent, so the dimension has to be greater or equal to 4. But how is it exactly 4?
$endgroup$
– wtnmath
Dec 5 '18 at 0:05
1
$begingroup$
Of course from the information that you just given above it is wrong to say that $text{dim}_{mathbb{Q}} V=4$. You can have another field which is bigger than $V$ which is of course still contain $alpha$. There must be some more informations about $V$ that you didn't write down.
$endgroup$
– user9077
Dec 5 '18 at 0:17
1
$begingroup$
$[mathbb{Q}(sqrt 2,sqrt3):mathbb{Q}]=[mathbb{Q}(sqrt 2,sqrt3):mathbb{Q}(sqrt 2)]cdot[mathbb{Q}(sqrt 2):mathbb{Q}]$
$endgroup$
– random
Dec 5 '18 at 0:47