How to determine $[\mathbb{Q}(\sqrt{2}+\sqrt{-2}) : \mathbb{Q}(\sqrt2)]$ i know that $\mathbb{Q}(\sqrt{2}+\sqrt{-2}) = \mathbb{Q}[\sqrt{2}+\sqrt{-2}] = \{a+b(\sqrt{2}+\sqrt{-2}) \space \text{with} \space a,b\in\mathbb{Q}\}$. And i think the answer is 2 because we got a lineair combination of $a+b\sqrt{2}$ and $b\sqrt{-2}$ but i don't know how to prove this correctly.... can i get hints/tips/tricks?
Thanks!
 A: Hint: What is the minimal polynomial of $\sqrt{2}+\sqrt{-2}$ over $\mathbb{Q}(\sqrt{2})$? In particular, think about the fact that $( \sqrt{2}+\sqrt{-2})\notin \mathbb{Q}(\sqrt{2})$ (Why?)(It might be nice to use the fact that you have a quadratic to prove irreducibility)
A: $\alpha =\sqrt{2}+\sqrt{-2}\Rightarrow \alpha -\sqrt{2}=\sqrt{-2}\Rightarrow (\alpha -\sqrt{2})^2=-2\Rightarrow (\alpha -\sqrt{2})^2+2=0$
$\alpha$ satisfies a second degree polynomial in $\mathbb{Q}(\sqrt{2})$
So.....
A: Note that $\mathbb{Q}(\sqrt{2})(\sqrt{2} + \sqrt{-2}) = \mathbb{Q}(\sqrt{2})(\sqrt{-1})$, since $\sqrt{-1} = ((\sqrt{2} + \sqrt{-2}) - \sqrt{2})/\sqrt{2}$.  So, the degree is at most 2, and it is at least 2 because $\sqrt{-1} \notin \mathbb{Q}(\sqrt{2})$.
A: $$ \frac{1}{\sqrt 2+\sqrt{-2}}=\frac{\sqrt 2-\sqrt{-2}}{4} $$
Since $4\in \Bbb Q(\sqrt 2+\sqrt{-2})$, $\frac{4}{\sqrt 2+\sqrt{-2}}=\sqrt{2}-\sqrt{-2}\in \Bbb Q(\sqrt 2+\sqrt{-2})\Rightarrow \sqrt{2},\sqrt{-2}\in\Bbb Q(\sqrt 2+\sqrt{-2})$.
$ \frac{\sqrt{-2}}{\sqrt 2}=i\in\Bbb Q$.  It's plain to see that $\Bbb Q(\sqrt 2+\sqrt{-2})\subseteq\Bbb Q(\sqrt 2,i)$, so we have $\Bbb Q(\sqrt 2+\sqrt{-2})\cong\Bbb Q(\sqrt 2, i)$.  $[\Bbb Q(\sqrt 2,i):\Bbb Q(\sqrt 2)]=2.$
