I understand that I can split it thus far: $x^4-2x^2-1=x^4-2x^2+1-2=(x^2-1)^2-(\sqrt{2})^2=(x^2-1-\sqrt{2})(x^2-1+\sqrt{2})$.
So $L=\mathbb{Q}(\sqrt{1+\sqrt{2}},\sqrt{\sqrt{2}-1}i)$.
But then how to compute $G(L,\mathbb{Q})$, I understand that this is the group of automorphisms on $L$ above $\mathbb{Q}$, such that every element in $L$ is invariant under those automorphisms.
But I need to split into four options, with the signs of the roots, shouldn't this group stays as the example for the dihedral group of $D_4$?
Appreciate your help.
There is a well known classification of Galois groups of polynomials of degree $4$ without computing it's roots. We only need to know a cubic resolvent of the polynomial and a square root of the discriminant of the polynomial. Since the polynomial is biquadratic the cubic resolvent is on the form $ax^3+bx^2+cx$, so it is reducible over rationals. Let's denote $\delta=\sqrt{\alpha}$, where $\alpha=-1024$ which is the discriminant of the polynomial. Since $\delta=32i\not \in \mathbb{R}$ and also the polynomial is irreducible in $\mathbb{Q}(\delta)$ we have that the Galois group must be $D_4$.
If you don't know about this classification note that the splitting field of the polynomial is $\mathbb{Q}(a,b)$ where $a=\sqrt{1+\sqrt{2}}$ and $b=\sqrt{1-\sqrt{2}}$. Note that $b=ia^{-1}$ so we just have that $\mathbb{Q}_f=\mathbb{Q}(a,i)$. Since $f$ is irreducible we know that the Galois group have order $8$. Now note that there are exactly three quadratic subfields $\mathbb{Q}(\sqrt 2),\mathbb{Q}(i\sqrt 2),\mathbb{Q}(i)$, which corresponds with three subgroups of order $4$ by the fundamental theorem of Galois theory. The only group of order $8$ with exactly $3$ subgroups of order $4$ is $D_4$