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 $n<0$ , 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 over $\mathbb{Q}$ . For if $n \in \mathbb{Z}$ is cubefree with $|n| \neq 1$ , 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 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 discriminant 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 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.