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.