Let $E= \mathbb{Q}(\sqrt{2} + \sqrt{7})$
For every $H < \mathrm{Gal}(E/\mathbb{Q})$ identify $\mathcal{F}(H)$
My attempt:
Notice that we need to adjoin four elements to reach the splitting field, namely $\pm \sqrt{2}, \pm \sqrt{7}$ so then the automorphisms can only permute these four elements. Naturally we would have the identity in this group
$\begin{align*} \sigma_0: \sqrt{2} \to \sqrt{2} & \hspace{10pt} \sqrt{7} \to \sqrt{7} \\ \sigma_1: \sqrt{2} \to \sqrt{2} & \hspace{10pt} \sqrt{7} \to -\sqrt{7} \\ \sigma_2: \sqrt{2} \to -\sqrt{2} & \hspace{10pt} \sqrt{7} \to \sqrt{7} \\ \sigma_3: \sqrt{2} \to -\sqrt{2} & \hspace{10pt} \sqrt{7} \to -\sqrt{7} \end{align*}$
Since $E/\mathbb{Q}$ is a Galois extension,
$$\therefore \mathrm{Gal}(E/\mathbb{Q}) = \mathrm{Aut}(E/\mathbb{Q}) = \{ \sigma_0, \sigma_1, \sigma_2,\sigma_3\}$$
Now we have that $\forall i = 0,...,3$,
$\begin{align*} \sigma_i(\sigma_i(\sqrt{2})) = \sqrt{2} = \sigma_0(\sqrt{2}),\hspace{10pt} \mbox{ and } & \hspace{10pt} \sigma_i(\sigma_i(\sqrt{7})) = \sqrt{7} = \sigma_0(\sqrt{7}) \end{align*}$
Hence, each element is of order two (it's own inverse) and
$$\therefore \mathrm{Gal}(E/\mathbb{Q}) \cong \mathbb{Z}_2\times \mathbb{Z}_2$$
Now that we have that $\mathrm{Gal}(E/\mathbb{Q}) \cong \mathbb{Z}_2\times \mathbb{Z}_2$, we must describe $\mathcal{F}(H)$ in terms of the subgroups of the Klein group. All possible proper subgroups H of $\mathrm{Gal}(E/\mathbb{Q})$ are given by
$$ H_0 = \{\sigma_0\}, H_1 = \{\sigma_0,\sigma_1\}, H_2 = \{\sigma_0,\sigma_2\}, H_3 = \{\sigma_0,\sigma_3\}$$
Now we must describe for each $H_i$ $$\mathcal{F}(H) = \{ x \in E: \sigma(x) = x, \forall \sigma \in H \} $$
Now for $H_0$ $$\mathcal{F}(H_0) = \mathbb{Q}(\sqrt{2},\sqrt{7})$$
Similarly, $$\mathcal{F}(H_1) = \mathbb{Q}(\sqrt{2})$$ $$\mathcal{F}(H_2) = \mathbb{Q}(\sqrt{7})$$ However, my problem is this one:
$$\mathcal{F}(H_3) = ??$$
I am correct, or this are not the fixed fields ? Can anyone help me computing the one for $H_3$ ?
I guess it must be $\mathbb{Q}(\sqrt{2},\sqrt{7})$ since $\sigma_3 = -\sigma_0$, but I dont know how to prove it.