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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?

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    Well, how did you prove that $\sqrt{2-\sqrt{2}}\in \mathbb{Q}(\alpha)$? You may be able to turn that proof into an explicit representation. – 2017-01-16
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    @EricWofsey I showed first that $\sqrt{2}\in\mathbb{Q}(\alpha)$ and then wrote $\sqrt{2-\sqrt{2}}=\alpha^{-1}\sqrt{2}$. I don't see how this helps. – 2017-01-16
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    It helps quite a lot. If you can express $\alpha^{-1}$ and $\sqrt{2}$ as powers of $\alpha$, then you can just multiply them and then reduce higher powers of $\alpha$ using the minimal polynomial. – 2017-01-16

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First observe that $$ \sqrt{2-\sqrt{2}}=\frac{\sqrt{2}}{\alpha}=\frac{\alpha^2-2}{\alpha}=\alpha-\frac{2}{\alpha}$$

Next, the minimal polynomial for $\alpha$ is $f(x)=x^4-4x^2+2$, hence $$ -\frac{2}{\alpha}=\alpha^3-4\alpha$$ and therefore $$ \sqrt{2-\sqrt{2}}=\alpha+(\alpha^3-4\alpha)=\alpha^3-3\alpha$$

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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 $$ \beta=2\left(4\frac{\alpha^3}{8}-3\frac{\alpha}{2}\right)= \alpha^3-3\alpha $$

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$$\begin{align} \dfrac{2-\sqrt2}{2+\sqrt2}\, &=\, 3-2\sqrt 2\,=\, (\sqrt 2 - 1)^{\large 2}\\[.5em] \Rightarrow\ \sqrt{2-\sqrt2} \,&=\, \sqrt{2+\sqrt2}\,\ \Big(\sqrt 2\, -\, 1\Big)\\ &=\,\sqrt{2+\sqrt2}\left(\sqrt{2+\sqrt 2}^{\large \,2}-3\right)\\[.1em] &=\qquad\qquad\! \alpha\ (\alpha^{\large 2}\ -\ 3) \end{align}$$