pure cubic field

A pure cubic field is an extension of $\mathbb{Q}$ of the form $\mathbb{Q}(\sqrt{n})$ for some $n\in\mathbb{Z}$ such that $\sqrt{n}\notin\mathbb{Q}$. If $n<0$, then $\sqrt{n}=\sqrt{-|n|}=-\sqrt{|n|}$, causing $\mathbb{Q}(\sqrt{n})=\mathbb{Q}(\sqrt{|n|})$. 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 $|n|\neq 1$, then $x^{3}-n$ is its minimal polynomial  over $\mathbb{Q}$. This polynomial  factors as $(x-\sqrt{n})(x^{2}+x\sqrt{n}+\sqrt{n^{2}})$ over $K=\mathbb{Q}(\sqrt{|n|})$. The discriminant    (http://planetmath.org/PolynomialDiscriminant) of $x^{2}+x\sqrt{n}+\sqrt{n^{2}}$ is $\left(\sqrt{n}\right)^{2}-4(1)\left(\sqrt{n^{2}}\right)=\sqrt{n^{2}}-% 4\sqrt{n^{2}}=-3\sqrt{n^{2}}$. Since the of $x^{2}+x\sqrt{n}+\sqrt{n^{2}}$ 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 PureCubicField 2013-03-22 16:02:19 2013-03-22 16:02:19 Wkbj79 (1863) Wkbj79 (1863) 14 Wkbj79 (1863) Definition msc 11R16