Galois group of splitting field of $x^4-16x^2+4$ isomorphic to $\mathbb{Z}/(2)\times \mathbb{Z}/(2)$

Question

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..

Answer 1

$\dfrac{1}{\sqrt{5}-\sqrt{3}}=\dfrac{\sqrt{5}+\sqrt{3}}{2}$.

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 $\sqrt{5},\sqrt{3},\sqrt{15}$.

Answer 2

Let $K$ be the splitting field of $x^4 - 16x^2 + 4$

$x^2 - 16x + 4$ has roots $8 \pm \sqrt{60}$. Therefore, the roots of $x^4 - 16x^2 + 4$ are $\pm \sqrt{8 \pm \sqrt{60}}$. This is a degree 4 extension since we have $[K : \mathbb{Q}(\sqrt{60})] = 2$ and $[\mathbb{Q}(\sqrt{60}) : \mathbb{Q}] = 2$.

The Galois group is therefore either $\mathbb{Z} / 4 \mathbb{Z}$ or $\mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 2 \mathbb{Z}$. We look at each one : $\mathbb{Z} / 4 \mathbb{Z}$ has one subgroup of order 2, and $\mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 2 \mathbb{Z}$ has 3. $\mathbb{Q}(\sqrt{60})$ is a subfield of $K$. If we can find another subfield (not $\mathbb{Q}$ or $K$ itself), then we can eliminate the case that the Galois group is $\mathbb{Z} / 4 \mathbb{Z}$ .

Now, $\sqrt{60} = 2 \sqrt{15} $ and $15$ factors as $3 * 5$, so we take a guess that possibly $8 \pm \sqrt{60}$ might be a perfect square. We guess that $(a \sqrt{3} + b \sqrt{5})^2 = 8 \pm \sqrt{60}$, and try to solve this for $a$ and $b$. Indeed, we have that $(\sqrt{3} + \sqrt{5})^2 = 8 + \sqrt{60}$.

Therefore, one pair of roots are $\pm (\sqrt{3} + \sqrt{5})$. To find the other pair, we guess that $(a \sqrt{3} + b \sqrt{5})^2 = 8 - \sqrt{60}$. Again, solving for $a$ and $b$, we get $(\sqrt{3} - \sqrt{5})^2 = 8 - \sqrt{60}$. Therefore, $\pm (\sqrt{3} - \sqrt{5})$ is another pair of roots.

$K$ contains $\sqrt{3} + \sqrt{5}$ and $\sqrt{3} - \sqrt{5}$, so must contain their sum which is $2\sqrt{3}$. Therefore $K$ contains $\sqrt{3}$ and $\mathbb{Q}(\sqrt{3})$ is another subfield. To show that $\mathbb{Q}(\sqrt{3})$ and $\mathbb{Q}(\sqrt{60})$ are distinct, it suffices to show that the equation $(a + b \sqrt{3})^2 = 60$ has no solutions. Expanding, we get $a^2 + 3b^2 = 60, 2ab\sqrt{3} = 0$. This requires either $a$ or $b$ = 0$.

If $a = 0$, then the first equation becomes $3b^2 = 60 \rightarrow b^2 = 20$. Since $20$ is not a perfect square, this has no rational solutions.

If $b = 0$, then the first equation becomes $a^2 = 60$. Since $60$ is not a perfect square, this has no solutions either.

Therefore, $K$ has the two subfields $\mathbb{Q}(\sqrt{3})$ and $\mathbb{Q}(\sqrt{60})$.