Is $\sqrt{5-2\sqrt{5}}$ in $\mathbb{Q}(\sqrt{5+2\sqrt{5}},\sqrt{2})$?
I'm told that $\mathbb{Q} \subseteq \mathbb{Q}(\alpha,\sqrt{2})$ is a Galois extension, and so the minimal polynomial of $\alpha$ must split in $\mathbb{Q}(\alpha,\sqrt{2})$.
The minimal polynomial of $\alpha$ over $\mathbb{Q}$ is $x^4-10x^2+5$, whose four roots in $\mathbb{C}$ are:
$\pm \sqrt{5 \pm 2\sqrt{5}}$
And I'm trying to explicitly show that these roots are all contained in $\mathbb{Q}(\sqrt{5+2\sqrt{5}},\sqrt{2})$... I'm pretty sure that they are not in $\mathbb{Q}(\sqrt{5+2\sqrt{5}})$ but I don't know how to make $\sqrt{2}$ useful.
Thanks
Let $k=\mathbb{Q}(\sqrt{2},\sqrt{5})$
Let $\mu, \nu \in k$ and we ask when is $\sqrt{\mu}\in k(\sqrt{\nu})$ if this happens, $$\sqrt{\mu}=a+b\sqrt{\nu}$$ where $a, b\in k$ and so $$\mu=a^2+b^2\nu+2ab\sqrt{\nu}$$ and assuming $\sqrt{\mu}, \sqrt{\nu}\not\in k$ we have $a=0$ and thus $$\mu=b^2\nu$$
So we ask is $\sqrt{5-2\sqrt{5}}\in k(\sqrt{5+2\sqrt{5}})$
This is the same as
$$5-2\sqrt{5}=b^2(5+2\sqrt{5})$$ so $$b^2=\frac{5-2\sqrt{5}}{5+2\sqrt{5}}=9-4\sqrt{5}$$
Thus the problem is reduced to whether $9-4\sqrt{5}$ is a perfect square in $\mathbb{Q}(\sqrt{2},\sqrt{5})$, which in turn can be reduced to $$9-4\sqrt{5}=(n+m\sqrt{5})^2$$ and this has the solution $$9-4\sqrt{5}=(2-\sqrt{5})^2$$
So
$$5-2\sqrt{5}=(2-\sqrt{5})^2(5+2\sqrt{5})$$ and $$\sqrt{5-2\sqrt{5}}=(2-\sqrt{5})\sqrt{5+2\sqrt{5}}$$ And thus in the unlikely event that I have made no mistake the answer is yes, $\sqrt{5-2\sqrt{5}}\in k(\sqrt{5+2\sqrt{5}})$.