Let $\mathbb{K}$ the the splitting field of $x^4 -2 x^2 -2$ over $\mathbb{Q}$. Determine all the subgroups of the Galois group Let $\mathbb{K}$ the the splitting field of $x^4-2x^2-2$ over $\mathbb{Q}$. Determine all the subgroups of the Galois group and give their corresponding fixed subfields of $\mathbb{K}$ containing $\mathbb{Q}$.
$\mathbb{K}$= $\mathbb{Q}(\alpha,i\sqrt{2})$ where $\alpha$ is a root. The others roots are $-\alpha, \frac{i\sqrt{2}} {\alpha},- \frac{i\sqrt{2}} {\alpha}$.
I proved that $Gal(\mathbb{K}/\mathbb{Q})$ is (isomorphic to) $D_4$ and the generators are $\sigma,\tau$ with $\sigma^4=\tau^2=id$ and $\sigma^3\tau=\tau\sigma$, for $\sigma(\alpha)=\frac{i\sqrt{2}}{\alpha}$, $\sigma(i\sqrt{2})=-i\sqrt{2}$ and $\tau(\alpha)=\alpha,\tau(i\sqrt{2})=-i\sqrt{2}$.
I considered all the subgroups of $Gal(\mathbb{Q}(\alpha,i\sqrt{2})$/$\mathbb{Q}$) but I didn't know which are the corresponding fixed subfields. What would the fixed field of {$id,\sigma^2,\sigma\tau,\sigma^3\tau$} be?
 A: First, note that $x^4-2x^2-2=(x^2-1)^2-3$ has a positive real root, $\sqrt{1+\sqrt3}$. I'm going to call this $\alpha$, so $\alpha^2-2=2\alpha^{-2}$, and the pure imaginary root in the upper half plane (on the positive imaginary axis) is $\beta=\sqrt{1-\sqrt3}=i\sqrt2/\alpha$, and the roots are $\{\pm\alpha,\pm\beta\}$.
Let's start with a table of the actions of elements of the Galois group $G=Gal(K/\mathbb{Q})=\langle\sigma,\tau\rangle$
on some generators of our splitting field,
$K=\mathbb{Q}(\alpha,\beta)$
$=\mathbb{Q}(\alpha,\alpha\beta=i\sqrt{2})$.
$$
\begin{array}{c|cccccccc}
\phi         &\text{id} &\sigma &\sigma^2 &\sigma^3
             &\tau   &\sigma\tau &\sigma^2\tau &\sigma^3\tau \\ \hline
\phi(\alpha) &\alpha & \beta  &-\alpha &-\beta
             &\alpha & \beta  &-\alpha &-\beta \\
\phi(\beta)  & \beta &-\alpha &-\beta  & \alpha
             &-\beta & \alpha & \beta  &-\alpha \\
\phi(i\sqrt2)&\alpha\beta &-\alpha\beta &\alpha\beta &-\alpha\beta
             &-\alpha\beta &\alpha\beta &-\alpha\beta &\alpha\beta \\
\end{array}
$$
Note that odd powers of $\sigma$ have order $4$,
and the other non-identity elements have order $2$.
The subgroup $\langle\sigma^2,\sigma\tau\rangle$
you are asking about consists of the $4$ elements that fix
$\alpha\beta=i\sqrt2$ (and its opposite, $-\alpha\beta$).
These don't fix the whole pure imaginary axis, since $\beta$ is pure imaginary and each of $\sigma^2,\sigma\tau,\sigma^3\tau$ take it to one of the other roots. So the fixed field of this subgroup is
$\mathbb{Q}(\alpha\beta)$.
