$\alpha = \sqrt2 + \sqrt3 \in V$ then $\dim_\Bbb Q V=4$

Question

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!

Answer 1

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?