# 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 $G=\operatorname{Gal}(K/\mathbb{Q})$.

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 $K_{1}\neq K_{2}$ since $K_{1}$ is a real field while $K_{2}$ is not. Thus $K_{1}\cap K_{2}\subsetneq K_{1},K_{2}$. Clearly $[K_{1}:\mathbb{Q}]=[K_{2}:\mathbb{Q}]=4$, so $[K_{1}\cap K_{2}:\mathbb{Q}]\leq 2$. 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 $K_{1}\neq K_{2}$, we also have $K=K_{1}K_{2}\neq K_{1}$ and $K=K_{1}K_{2}\neq K_{2}$; 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 $D_{8}$ QuarticPolynomialWithGaloisGroupD8 2013-03-22 17:44:09 2013-03-22 17:44:09 rm50 (10146) rm50 (10146) 6 rm50 (10146) Example msc 12D10