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 $ℝ$. Note that $K\subseteq ℝ$. 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