# pure cubic field

A pure cubic field is an extension of $\mathbb{Q}$ of the form $\mathbb{Q}(\sqrt[3]{n})$ for some $n\in \mathbb{Z}$ such that $\sqrt[3]{n}\notin \mathbb{Q}$. If $$, then $\sqrt[3]{n}=\sqrt[3]{-|n|}=-\sqrt[3]{|n|}$, causing $\mathbb{Q}(\sqrt[3]{n})=\mathbb{Q}(\sqrt[3]{|n|})$. Thus, without loss of generality, it may be assumed that $n>1$.

Note that no pure cubic field is Galois (http://planetmath.org/GaloisExtension) over $\mathbb{Q}$. For if $n\in \mathbb{Z}$ is cubefree^{} with $|n|\ne 1$, then ${x}^{3}-n$ is its minimal polynomial^{} over $\mathbb{Q}$. This polynomial^{} factors as $(x-\sqrt[3]{n})({x}^{2}+x\sqrt[3]{n}+\sqrt[3]{{n}^{2}})$ over $K=\mathbb{Q}(\sqrt[3]{|n|})$. The discriminant^{} (http://planetmath.org/PolynomialDiscriminant) of ${x}^{2}+x\sqrt[3]{n}+\sqrt[3]{{n}^{2}}$ is ${\left(\sqrt[3]{n}\right)}^{2}-4(1)\left(\sqrt[3]{{n}^{2}}\right)=\sqrt[3]{{n}^{2}}-4\sqrt[3]{{n}^{2}}=-3\sqrt[3]{{n}^{2}}$. Since the of ${x}^{2}+x\sqrt[3]{n}+\sqrt[3]{{n}^{2}}$ is negative, it does not factor in $\mathbb{R}$. Note that $K\subseteq \mathbb{R}$. Thus, ${x}^{3}-n$ has a root (http://planetmath.org/Root) in $K$ but does not split completely in $K$.

Note also that pure cubic fields are real cubic fields with exactly one real embedding. Thus, a possible method of determining all of the units of pure cubic fields is outlined in the entry regarding units of real cubic fields with exactly one real embedding.

Title | pure cubic field |
---|---|

Canonical name | PureCubicField |

Date of creation | 2013-03-22 16:02:19 |

Last modified on | 2013-03-22 16:02:19 |

Owner | Wkbj79 (1863) |

Last modified by | Wkbj79 (1863) |

Numerical id | 14 |

Author | Wkbj79 (1863) |

Entry type | Definition |

Classification | msc 11R16 |