$alpha = sqrt2 + sqrt3 in V$ 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




  1. $$langle 1,alpha, alpha^2 , alpha^3 rangle_Bbb Q subset V$$

  2. $$dim_Bbb Qlangle 1,alpha, alpha^2 , alpha^3 rangle = 4$$


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







share|cite|improve this question











$endgroup$








  • 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
















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




  1. $$langle 1,alpha, alpha^2 , alpha^3 rangle_Bbb Q subset V$$

  2. $$dim_Bbb Qlangle 1,alpha, alpha^2 , alpha^3 rangle = 4$$


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







share|cite|improve this question











$endgroup$








  • 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














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




  1. $$langle 1,alpha, alpha^2 , alpha^3 rangle_Bbb Q subset V$$

  2. $$dim_Bbb Qlangle 1,alpha, alpha^2 , alpha^3 rangle = 4$$


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







share|cite|improve this question











$endgroup$




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




  1. $$langle 1,alpha, alpha^2 , alpha^3 rangle_Bbb Q subset V$$

  2. $$dim_Bbb Qlangle 1,alpha, alpha^2 , alpha^3 rangle = 4$$


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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 2:37







wtnmath

















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














  • 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










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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?






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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?






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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?






      share|cite|improve this answer









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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?






        share|cite|improve this answer









        $endgroup$



        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?







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 4 '18 at 23:49









        zoidbergzoidberg

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