# quartic polynomial with Galois group ${D}_{8}$

The polynomial^{} $f(x)={x}^{4}-2{x}^{2}-2$ 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

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

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 ${K}_{1}\cap {K}_{2}\u228a{K}_{1},{K}_{2}$. 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 |