# 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?

• Are you defining $\alpha$ to be the positive real root, $(1+\sqrt3)^{1/2}$? – bgins Mar 25 '16 at 19:39

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 $\mathbb{Q}(\alpha\beta)$.