Splitting field and galois group of $x^4-2x^2-1$ over $\Bbb{Q}$ I have that the roots of $m(x)=x^4-2x^2-1$ are plus minus $a=\sqrt{1+\sqrt{2}}$ and $b=\sqrt{1-\sqrt{2}}$ which is complex because $1-\sqrt{2}$ is negative. Taking out $i$ and letting $b=ic$ we have that $c^{-1}=a$. I think the splitting field is $\Bbb{Q}(a,b)=\Bbb{Q}(a,ic)$ but seeing as $c^{-1}=a$ does this mean we can simplify the splitting field to $\Bbb{Q}(a,i)$. In that case the minimal poly for $a$ is $m(x)$ which has degree 4 and the minimal poly for $i$ over $\Bbb{Q}(a)$ is $x^2+1$ which has degree 2. Therefore the order of the Galois group which equals the order of the splitting field is $2\cdot 4=8$. Listing out the automorphisms I think the Galois group is isomorphic to the dihedral group of the square, $D_4$. However, I am having trouble identifying the fixed group of an order 4 subgroup of $D_4$. How should I approach this?
 A: You are correct in saying that the splitting field can be written $\Bbb{Q}(a,i)$.
However there are three subgroups of order four.
1: The cyclic subgroup
The automorphism group contains a cycle of the roots, namely the map that takes $a$ to $\frac{i}{a}$ and $i$ to $-i$, which corresponds to the cycle $$\pmatrix{a & \frac{i}{a} & -a & \frac{-i}{a}}.$$
Now the subfield fixed by the group generated by this automorphism has degree 2 over $\Bbb{Q}$ by the main theorem of Galois theory. Note in addition that $ia^2+\frac{i}{a^2}$ is fixed by this map since $ia^2\mapsto -i(\frac{i^2}{a^2})=\frac{i}{a^2}$, and $\frac{i}{a^2}\mapsto -i(\frac{a^2}{i^2})=ia^2$.
However $$ia^2+\frac{i}{a^2}=i(1+\sqrt{2})+\frac{i}{1+\sqrt{2}}=i\left(1+\sqrt{2}+\frac{1-\sqrt{2}}{1-2}\right)=2i\sqrt{2},$$ which has degree two over $\Bbb{Q}$. Thus the fixed field corresponding to this subgroup is $$\Bbb{Q}\left(ia^2+\frac{i}{a^2}\right)=\Bbb{Q}[2i\sqrt{2}]=\Bbb{Q}[\sqrt{-2}].$$
2: A Klein 4 subgroup
Another subgroup of order four is generated by the automorphism that sends $a$ to $-a$ and $i$ to $i$ and the automorphism that sends $a$ to $a$ and $i$ to $-i$. Note that both of these fix $a^2$, which has order 2 over $\Bbb{Q}$, so the fixed field corresponding to this subgroup is $$\Bbb{Q}[a^2]=\Bbb{Q}[1+\sqrt{2}]=\Bbb{Q}[\sqrt{2}].$$
3: Another Klein 4 subgroup
Finally, there must be a third subgroup of order four, which is generated by the automorphism that sends $a$ to $\frac{i}{a}$ and $i$ to $i$ and the automorphism that sends $a$ to $-a$ and $i$ to $i$. Note that these both fix $i$, so the fixed field is $\Bbb{Q}[i]$.
Note that this makes a lot of sense, subgroups of order 4 correspond to fixed fields of degree 2 over the base field, or quadratic extensions. Therefore, we expect $\Bbb{Q}[i]$ and $\Bbb{Q}[\sqrt{2}]$, since both of these are clearly contained in the splitting field. Then we notice that since $\Bbb{Q}[\sqrt{2}]$ is a real field, and since $\Bbb{Q}[i]$ contains no real irrational elements, $\Bbb{Q}[i\sqrt{2}]=\Bbb{Q}[\sqrt{-2}]$ must be the last field (if you count subgroups of $D_4$, you can check there are 3 of order 4). Then we just have to match these fields up to groups.
