How to prove that $\sqrt{2-\sqrt{2}} \in \mathbb{Q}(\sqrt{2+\sqrt{2}})$ I am trying to prove a statement about the decomposition field of a polynomial that has both $\sqrt{2-\sqrt{2}}$ and $\sqrt{2+\sqrt{2}}$ as roots. I cannot find a way to prove that $\sqrt{2-\sqrt{2}} \in \mathbb{Q}(\sqrt{2+\sqrt{2}})$. I have tried writing it in the basis $1,\sqrt{2+\sqrt{2}}(\sqrt{2+\sqrt{2}})^2,(\sqrt{2+\sqrt{2}})^3$ but nothing works and without this I cannot prove that the decomposition field of $t^4-4t^2+2$ is $\mathbb{Q}(\sqrt{2+\sqrt{2}})$ over $\mathbb{Q}$

Hint: what is $\sqrt{2 - \sqrt 2} \cdot \sqrt{2 + \sqrt 2}$ ?

 It's $\sqrt 2$ ! Therefore $\sqrt{2 - \sqrt 2} = \frac{\sqrt 2}{\sqrt{2 + \sqrt 2}} = \frac{\left(\sqrt{2 + \sqrt 2}\right)^2 - 2}{\sqrt{2 + \sqrt 2}}$

