<s>[INST] Galois group of splitting field of $x^4-16x^2+4$ isomorphic to $\mathbb{Z}/(2)\times \mathbb{Z}/(2)$ I've already shown irreducible. I know I need to get intermediate fields since they correspond to subgroups (which I can use for isomorphisms). But how do I get the intermediate fields here?
edit: I'm thinking perhaps there is some way to show the order of the extension is $4$, and the only groups of order $4$ are the Klein-$4$ group and cyclic group and eliminating from there.
edit 2:
Treating $x^4-16x^2+4$ as a quadratic in $x^2$, applying the quadratic formula gives $$\frac{16\pm \sqrt{16^2-4\cdot 4}}{2}=\frac{16\pm 4\sqrt{15}}{2}=8\pm 2\sqrt{15}=(\sqrt{3}\pm\sqrt{5})^2.$$ and this is squared so $\pm(\sqrt{3}\pm\sqrt{5})$ gives the roots and there are four of them. Now what's next..
 [/INST] $\dfrac{1}{\sqrt{5}-\sqrt{3}}=\dfrac{\sqrt{5}+\sqrt{3}}{2}$.
Therefore, any root generates all three other roots. Therefore, the splitting field is of a degree-$4$ extension of $\mathbb{Q}$.
Now, since you want three distinct intermediate subfields of degree $2$,
$(\sqrt{5}+\sqrt{3})+(\sqrt{5}-\sqrt{3})=2\sqrt{5}$
$(\sqrt{5}+\sqrt{3})-(\sqrt{5}-\sqrt{3})=2\sqrt{3}$
$(\sqrt{5}+\sqrt{3})^2=8+ 2\sqrt{15}$
so you can use $\sqrt{5},\sqrt{3},\sqrt{15}$.
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