How to express $\sqrt{2-\sqrt{2}}$ in terms of the basis for $\mathbb{Q}(\sqrt{2+\sqrt{2}})$ I've already shown that $\sqrt{2-\sqrt{2}}\in\mathbb{Q}(\sqrt{2+\sqrt{2}})$ but I want to write $\sqrt{2-\sqrt{2}}=a+b\alpha+c\alpha^2+d\alpha^3$ where $\alpha=\sqrt{2+\sqrt{2}}$. By squaring $\sqrt{2-\sqrt{2}}$ I get a system of four equations with four variables, but everything I try (Sage, for instance) takes too long to solve it. Is there a simpler way to write it in terms of the basis?

There's a trigonometry trick for this particular case.
Let $\alpha=\sqrt{2+\sqrt{2}}$; then
$$
\alpha=2\sqrt{\frac{1+\sqrt{2}/2}{2}}=2\cos\frac{\pi}{8}
$$
whereas
$$
\beta=\sqrt{2-\sqrt{2}}=2\sqrt{\frac{1-\sqrt{2}/2}{2}}=2\sin\frac{\pi}{8}=
2\cos\left(\frac{\pi}{2}-\frac{\pi}{8}\right)=
2\cos\frac{3\pi}{8}
$$
Now
$$
\cos3t=4\cos^3t-3\cos t
$$
so
$$
\beta=2\left(4\frac{\alpha^3}{8}-3\frac{\alpha}{2}\right)=
\alpha^3-3\alpha
$$
