Let $K=\mathbb{Q}(\sqrt{5},\sqrt{-2})$. I already showed that the ring of integers in $K$ is $\mathbb{Z}(\frac{\sqrt{5}+1}{2},\sqrt{-2})$ and the discriminant is $1600$.
Now the goal is to show that the roots of unity in $\mathcal{O}_{K}$ are $\pm1$:
We know $[K:\mathbb{Q}]=4$, $K$ is Galois with Galois group $\mathbb{Z}/2\mathbb{Z} \times\mathbb{Z}/2\mathbb{Z}$ and that the only quadratic subfields are $\mathbb{Q}(\sqrt{5}),\mathbb{Q}(\sqrt{-2}),\mathbb{Q}(\sqrt{-10})$.
$K$ is not cyclotomic because there is no cyclotomic field of degree $4$ with discriminant 1600.
From this it follows that any root of unity is contained in a quadratic subfield. In the three quadratic subfields the only roots of unity are $\pm1$.
How would one show that the extension is Galois?
(I assume we show that there are exactly $4$ automorphisms. But what would those be?)
Why exactly does it follow that any root of unity is contained in a quadratic subfield?