I am trying to find $Gal(L:\mathbb Q)$ where $L$ is the splitting field of $f(X)= X^4-6X^2-2$ over $\mathbb Q$.
My working so far:
I have worked out that the four roots are $\pm \sqrt{3+\sqrt{11}} $ and $\pm i\sqrt{-3+\sqrt{11}}$. And I'm quite stuck on how to find the splitting field, intuition tells me that it could be $\mathbb Q((\sqrt{3+\sqrt{11}}),i)$, let $\alpha=\sqrt{3+\sqrt{11}} $ so $\mathbb Q(\alpha):\mathbb Q$ but I have doubts and the radicals have really confused me.
$\alpha=\sqrt{3+\sqrt{11}}$ and $\beta = \pm i\sqrt{-3+\sqrt{11}}$
What I think are the automorphisms:
id: $\alpha \mapsto \alpha, \beta \mapsto \beta$
$\theta_1: \alpha \mapsto \alpha, \beta \mapsto -\beta$
$\theta_2: \alpha \mapsto \beta, \beta \mapsto \alpha$
$\theta_3 : \alpha \mapsto -\beta, \beta \mapsto -\alpha$
$\theta_3\theta_2: \alpha \mapsto -\alpha, \beta \mapsto \beta$
$\theta_3^2: \alpha \mapsto -\alpha, \beta \mapsto -\beta$
$\theta_1\theta_2: \alpha \mapsto -\beta, \beta \mapsto \alpha$
$\theta_3^2\theta_2: \alpha \mapsto -\beta, \beta \mapsto -\alpha$
I'm very stumped, can anyone help?
The splitting field is $\Bbb Q(\sqrt{3+\sqrt{11}},i\sqrt{-3+\sqrt{11}})$. Complex conjugation induces an automorphism of $L$ fixing $\sqrt{3+\sqrt{11}}$ and sending $i\sqrt{-3+\sqrt{11}}$ to its negative. This should help explicate the action of the Galois group on the four zeros of $f$.