Consider the splitting field $K$ of $x^4-2x^2-2$ on $\mathbb Q$. Find $[K:\mathbb Q] $ and the structure of the Galois group $Aut(K/\mathbb Q)$.

Question

I find the degree of $K$ on $\mathbb Q$ which is 8. But i have problem to find the structure of the Galois group.

Answer 1

We can determine the Galois group with theorem:

Let be $q(x)=x^4+bx^2+c\in k[x]$ irreducible polynomial over $k$ and has two distict roots $\alpha, \beta$ and $G=Gal(k(\alpha, \beta),k)$ then

$1)$ if $\sqrt c \in k$ then $G\cong\mathbb Z_2 \times \mathbb Z_2$
$2)$ if $\sqrt c \notin k$ and $\sqrt{c(b^2-4c)} \in k $ then $G\cong\mathbb Z_4$ and $q_1(x)=x^2+bx+c$ irreducible over $k(\sqrt c)$

$3)$ if $\sqrt c \notin k$ and $\sqrt{c(b^2-4c)} \notin k $ then $G\cong D_4$ and $q_1(x)=x^2+bx+c$ irreducible over $k(\sqrt c)$

Answer 2

Since you've found the roots and have already computed the degree of the extension, here is how you can compute the Galois group.

The roots of your polynomial are $$\pm\sqrt{1\pm\sqrt3}$$ and $K = \mathbb Q(\sqrt{1+\sqrt3}, \sqrt{1-\sqrt3})=K(\sqrt{-6},\sqrt{1+\sqrt3})$.

Let $\theta$ be the non-trivial automorphism which fixes $\mathbb Q(\sqrt{1+\sqrt3})$. It is given by $$\begin{align}\theta:&\sqrt{1+\sqrt3}\mapsto\sqrt{1+\sqrt3}\\&\sqrt{1-\sqrt3}\mapsto -\sqrt{1-\sqrt3}.\end{align}$$

Let $\phi$ be the automorphism fixing $\mathbb Q(\sqrt{-6})$ given by $$\begin{align}\phi:&\sqrt{1+\sqrt3}\mapsto-\sqrt{1-\sqrt3}\\&\sqrt{1-\sqrt3}\mapsto \sqrt{1+\sqrt3}\end{align}\\$$

Then $\theta^2=\phi^4 = 1$, and $\theta\phi\theta=\phi^{-1}$. It follows that the Galois group must be dihedral.

Answer 3

Consider the polynomial $g(x)=x^2-2x-2$ (which is irreducible), so that you are looking for the splitting field of $f(x)=g(x^2)$. So, letting $\alpha,\beta$ be the roots of $g$, your field is $K=\mathbb{Q}[\pm \sqrt{\alpha}, \pm \sqrt{\beta}]$. Note that $K/L$ where $L=\mathbb{Q}[\alpha,\beta]=\mathbb{Q}[\alpha]$ is Galois with group $\mathbb{Z}_2^2$ that switches the signs before $\sqrt{\alpha},\sqrt{\beta}$. Since the group has order 8, there are only 5 such possibilities, and you can already discard the cyclic group and the quaternions because both of them have only one element of order 2. If we can show that the group is not abelian, then the group must be dihedral.

Consider the element $\lambda = \sqrt{\alpha}+\sqrt{\beta}$. Its square is $\alpha+\beta+2\sqrt{\alpha\beta}=2+2\sqrt{-2}$ so that $(\lambda^2-2)^2+8=0$. It follows that $[\mathbb{Q}[\lambda]:\mathbb{Q}]\leq 4$, and in particular $\mathbb{Q}[\lambda]\neq K$. In order to show that the Galois group is not abelian, it is enough to show that $\mathbb{Q}[\lambda]/\mathbb{Q}$ is not Galois. Applying the Galois action that fixes $\sqrt{\alpha}$ and multiply by -1 the element $\sqrt{\beta}$ we get the field $\mathbb{Q}[\sqrt{\alpha}-\sqrt{\beta}]$ which is different than $\mathbb{Q}[\lambda]$, since otherwise they would contain both $\sqrt{\alpha}$ and $\sqrt{\beta}$ and then $\mathbb{Q}[\lambda]=K$ which is not true. It follows that $\mathbb{Q}[\lambda]/\mathbb{Q}$ is not Galois, so that your Galois group is not abelian and therefore must be dihedral.

EDIT: The Galois group must be of order 8. First of all $f(x)$ is irreducible by Eisenstein's criterion. The polynomial $f(x)$ has a real root $\gamma$ because $f(0)=-2$ and $lim_{x\to \infty}f(x)=\infty$. It follows that adding this root gives us $[\mathbb{Q}[\gamma]:\mathbb{Q}]=4$ with $\mathbb{Q}[\gamma]\subseteq \mathbb{R}$. Since $g$ has a negative root (the free coefficient is negative) and the roots of $f$ are the plus minus square roots of the roots of $g$, then $f$ must also have a nonreal root, so that $[K:\mathbb{Q}]>4$ and it is divisible by 4. Using the notation from the beginning of the answer, we have that $$[K:\mathbb{Q}]=[\mathbb{Q}[\sqrt{\alpha},\sqrt{\beta}]:\mathbb{Q}[\sqrt{\alpha},\beta]][\mathbb{Q}[\sqrt{\alpha},\beta]:\mathbb{Q}[\alpha,\beta=2-\alpha]][\mathbb{Q}[\alpha]:\mathbb{Q}]\leq 2\cdot 2\cdot 2=8$$ so the extension must have degree 8.