How can you find a basis of $\Bbb{Q}(\sqrt{3},\sqrt{5})$ over $\Bbb{Q}(\sqrt{15})$?
I thought every element of $\Bbb{Q}(\sqrt{3},\sqrt{5})$ is of the form $a+b\sqrt{3}+c\sqrt{5}+d\sqrt{15}$ with $a$, $b$, $c$ and $d$ rational numbers, but I don't see why we have {$1,\sqrt{3}$} as basis for $\Bbb{Q}(\sqrt{3},\sqrt{5})$ over $\Bbb{Q}(\sqrt{15})$.
I also need a basis of $\Bbb{Q}(\sqrt{2+\sqrt{2}})$ over $\Bbb{Q}$. I think that, because $\sqrt{2+\sqrt{2}}$ is an algebraic element of $\Bbb{Q}$, we have that $\Bbb{Q}(\sqrt{2+\sqrt{2}})=\Bbb{Q}[\sqrt{2+\sqrt{2}}]$ (here, I use () for field and [] for ring) and this is equal to the set of all polynomials over $\Bbb{Q}$ were you replace $X$ with $\sqrt{2+\sqrt{2}}$, but then I'm stuck. Can someone explain me please how you can find in gerenal a basis of a field $K$ over $F$ (where $F \subset K$)?
Thanks!