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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. $$\dim_\Bbb Q\langle 1,\alpha, \alpha^2 , \alpha^3 \rangle = 4$$

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!

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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ $[\mathbb{Q}(\sqrt 2,\sqrt3):\mathbb{Q}]=[\mathbb{Q}(\sqrt 2,\sqrt3):\mathbb{Q}(\sqrt 2)]\cdot[\mathbb{Q}(\sqrt 2):\mathbb{Q}]$ $\endgroup$ – 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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