Sub-fields of $\mathbb{Q}(a)$ ,$a=\sqrt{4+\sqrt{15}}$ using Galois theory.

Question

Let $K\subseteq\mathbb{C}$ be a splitting field of $f(x)=x^4-8x^2+1$, $a=\sqrt{4+\sqrt{15}}$ , $b=\sqrt{4-\sqrt{15}}$ and $G=Gal(K,\mathbb{Q})$.Find every subfield of K.

What I've done so far :
I found out that $f(x)$ is irreducible using Gauss lemma .Then I showed that $\mathbb{Q}(a)=K$ since b is also a solution of $f(x)$ and also $ab=1 \Rightarrow b=a^{-1} \Rightarrow b \in \mathbb{Q}(a)$ and thus $\mathbb{Q}(a)$ contains all four solutions of $f(x)$ ,meaning it is a splitting of $f$ over $\mathbb{Q}$.Of course $[K:\mathbb{Q}]=4$ and using the fundmental Galois theorem we conclude that $[K:\mathbb{Q}]=|G|=4$.

From here I'm getting a bit unsure on how to continue.

I found that $G=\{s_1,s_2,s_3,s_4|s_1(a)=a,s_2(a)=-a,s_3(a)=b,s_4(a)=-b\}$ (I'm not sure about these).I calculated that every s has an order of 2 and thus $G\simeq\mathbb{Z}_2 \times \mathbb{Z}_2$.Of course every $<s_i>,i\neq1$ is a subfield of G.From the Galois correspondence we know that $K$ has 3 subfields :$Fix<s_2>,Fix<s_3>,Fix<s_4>$ .I want to express them as $\mathbb{Q}(-)$ but I'm a little bit lost .I expressed $Fix<s_2>=\mathbb{Q}(4+\sqrt{15})$ since $s_2(a^2)=4+\sqrt{15}$.I have no idea what to do for the 2 remaining $Fix-$ and I think I might have done something wrong.

Any kind of help would be really appreciated!

Answer 1

Everything goes well, so it strongly suggests that $\sqrt{4+\sqrt{15}}$ can be broken into a sum of roots, and indeed

$$\sqrt{4\pm\sqrt{15}}=\frac{\sqrt{10}\pm\sqrt{6}}{2}$$

Thus the field contains $\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}=\sqrt{10}$ and also $\sqrt{6}$.

Once we find those intermediate extensions, we may conclude the Galois group is isomorphic to $(\mathbb Z/2\mathbb Z)^2$ instead of $\mathbb Z/4\mathbb Z$ without earlier work.

Here is one way to show any element $\sigma$ in the Galois group satisfying $\sigma^2=\text{id}$. Let $a=\sqrt{4+\sqrt{15}}$, then we have $\sigma(a) = \pm a, \pm\frac{1}{a}$. If $\sigma(a)=-a$, apply $\sigma$ on both sides, we get $\sigma^2(a) = \sigma(-a)=\sigma(-1)\sigma(a)=-\sigma(a)=a$, hence $\sigma^2=\text{id}$. The other cases can be solved in similar fashion. Enjoy.