Why is the basis the same for $\mathbb{Q}(\sqrt{2},\sqrt{3})$ and $\mathbb{Q}(\sqrt{3},\sqrt{6})$?

Question

Why is the basis the same for $\mathbb{Q}(\sqrt{2},\sqrt{3})$ and $\mathbb{Q}(\sqrt{3},\sqrt{6})$?

Here is the basis I am talking about: $\left \{1,\sqrt{2},\sqrt{3},\sqrt{6}\right \}$.

Would the set $\left \{1,\sqrt{3},\sqrt{6},\sqrt{18}\right \}$ be another valid basis for $\mathbb{Q}(\sqrt{3},\sqrt{6})$?

Answer 1

That would also be valid because you've just scaled $\sqrt{2}$. The two fields have the same basis because of proofs such as $\sqrt{2}=\frac{1}{3}\sqrt{3}\sqrt{6}$.

Answer 2

$\frac{\sqrt6}{\sqrt3}=\sqrt2$ and $\sqrt2\cdot \sqrt3 =\sqrt6$. Hence the fields are equal.