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

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!

• 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 $\dim_\mathbb Q V = \infty$. – User8128 Dec 4 '18 at 23:49
• 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. – Trevor Gunn Dec 4 '18 at 23:53
• @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? – wtnmath Dec 5 '18 at 0:05
• 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. – user9077 Dec 5 '18 at 0:17
• $[\mathbb{Q}(\sqrt 2,\sqrt3):\mathbb{Q}]=[\mathbb{Q}(\sqrt 2,\sqrt3):\mathbb{Q}(\sqrt 2)]\cdot[\mathbb{Q}(\sqrt 2):\mathbb{Q}]$ – random Dec 5 '18 at 0:47

## 1 Answer

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?