quartic polynomial with Galois group $D_8$

The polynomial $f(x)=x^4-2x^2-2$ is Eisenstein at $2$ and thus irreducible over $\mathbb{Q}$. Solving $f(x)$ as a quadratic in $x^2$, we see that the roots of $f(x)$ are

${\alpha }_1=\sqrt{1+\sqrt{3}}$	${\alpha }_3=-\sqrt{1+\sqrt{3}}$
${\alpha }_2=\sqrt{1-\sqrt{3}}$	${\alpha }_4=-\sqrt{1-\sqrt{3}}$

Note that the discriminant of $f(x)$ is $-4608=-2^9\cdot 3^2$, and that its resolvent cubic is

$$x^3+4x^2+12x=x(x^2+4x+12)=0$$

which factors over $\mathbb{Q}$ into a linear and an irreducible quadratic. Additionally, $f(x)$ remains irreducible over $\mathbb{Q}(\sqrt{-4608})=\mathbb{Q}(\sqrt{-2})$, since none of the roots of $f(x)$ lie in this field and the discriminant of $f(x)$, regarded as a quadratic in $x^2$, does not lie in this field either, so $f(x)$ cannot factor as a product of two quadratics. So according to the article on the Galois group of a quartic polynomial, $f(x)$ should indeed have Galois group isomorphic to $D_8$. We show that this is the case by explicitly examining the structure of its splitting field.

Let $K$ be the splitting field of $f(x)$ over $\mathbb{Q}$, and let $G=\mathrm{Gal}(K/\mathbb{Q})$.

Let $K_1=\mathbb{Q}({\alpha }_1)=\mathbb{Q}({\alpha }_3)$ and $K_2=\mathbb{Q}({\alpha }_2)=\mathbb{Q}({\alpha }_4)$. Clearly $K$ contains both $K_1$ and $K_2$ and thus contains $K_1{K}_2=\mathbb{Q}({\alpha }_1,{\alpha }_2)$. But obviously $f(x)$ splits in $K_1{K}_2$, so that $K=K_1{K}_2$. We next determine the degree of $K$ over $\mathbb{Q}$.

Note that $K_1\ne K_2$ since $K_1$ is a real field while $K_2$ is not. Thus $K_1\cap K_2⊊K_1,K_2$. Clearly $[K_1:\mathbb{Q}]=[K_2:\mathbb{Q}]=4$, so $[K_1\cap K_2:\mathbb{Q}]\le 2$. But

$$\sqrt{3}={\left(\sqrt{1+\sqrt{3}}\right)}^2-1=-{\left(\sqrt{1-\sqrt{3}}\right)}^2+1$$

so $\sqrt{3}\in K_1\cap K_2$. Hence $K_1\cap K_2=\mathbb{Q}(\sqrt{3})$; call this field $F$.

Since $K_1\ne K_2$, we also have $K=K_1{K}_2\ne K_1$ and $K=K_1{K}_2\ne K_2$; thus $K$ is a quadratic extension of each and $[K:F]=4$.

Putting these results together, we see that

$$[K:\mathbb{Q}]=[K:F][F:\mathbb{Q}]=8$$

so that $G$ has order $8$.

Now, neither $K_1$ nor $K_2$ is Galois over $\mathbb{Q}$ (since the Galois closure of either one is $K$), so that the subgroup of $G$ fixing (say) $K_1$ is a nonnormal subgroup of $G$. Thus $G$ must be nonabelian, so must be isomorphic to either $D_8$ or $Q_8$ (the quaternions). But the subgroups of $G$ corresponding to $K_1$ and $K_2$ are distinct subgroups of order $2$ in $G$, and $Q_8$ has only one subgroup of order $2$. Thus $G\cong D_8$. (Alternatively, note that all subgroups of $Q_8$ are normal, so $G\cong D_8$ since it has a nonnormal subgroup).

Title	quartic polynomial with Galois group $D_8$
Canonical name	QuarticPolynomialWithGaloisGroupD8
Date of creation	2013-03-22 17:44:09
Last modified on	2013-03-22 17:44:09
Owner	rm50 (10146)
Last modified by	rm50 (10146)
Numerical id	6
Author	rm50 (10146)
Entry type	Example
Classification	msc 12D10