units of real cubic fields with exactly one real embedding
Let $K\subseteq \mathbb{R}$ be a number field^{} with $[K:\mathbb{Q}]=3$ such that $K$ has exactly one real embedding. Thus, $r=1$ and $s=1$. Let $\mathcal{O}_{K}{}^{*}$ denote the group of units of the ring of integers^{} of $K$. By Dirichlet’s unit theorem, $\mathcal{O}_{K}{}^{*}\cong \mu (K)\times \mathbb{Z}$ since $r+s1=1$. The only roots of unity^{} in $K$ are $1$ and $1$ because $K\subseteq \mathbb{R}$. Thus, $\mu (K)=\{1,1\}$. Therefore, there exists $u\in \mathcal{O}_{K}{}^{*}$ with $u>1$, such that every element of $\mathcal{O}_{K}{}^{*}$ is of the form $\pm {u}^{n}$ for some $n\in \mathbb{Z}$.
Let $\rho >0$ and $$ such that the conjugates^{} of $u$ are $\rho {e}^{i\theta}$ and $\rho {e}^{i\theta}$. Since $u$ is a unit, $N(u)=\pm 1$. Thus, $\pm 1=N(u)=u(\rho {e}^{i\theta})(\rho {e}^{i\theta})=u{\rho}^{2}$. Since $u>0$ and ${\rho}^{2}>0$, it must be the case that $u{\rho}^{2}=1$. Thus, $u={\displaystyle \frac{1}{{\rho}^{2}}}$. One can then deduce that $\mathrm{disc}u=4{\mathrm{sin}}^{2}\theta {\left({\rho}^{3}+{\displaystyle \frac{1}{{\rho}^{3}}}2\mathrm{cos}\theta \right)}^{2}$. Since the maximum value of the polynomial^{} $4{\mathrm{sin}}^{2}\theta {(x2\mathrm{cos}\theta )}^{2}4{x}^{2}$ is at most $16$, one can deduce that $\mathrm{disc}u\le 4\left({u}^{3}+{\displaystyle \frac{1}{{u}^{3}}}+4\right)$. Define $d=\mathrm{disc}{\mathcal{O}}_{K}$. Then $d\le \mathrm{disc}u\le 4\left({u}^{3}+{\displaystyle \frac{1}{{u}^{3}}}+4\right)$. Thus, ${u}^{3}\ge {\displaystyle \frac{d}{4}}4{\displaystyle \frac{1}{{u}^{3}}}$. From this, one can obtain that ${u}^{3}\ge {\displaystyle \frac{d16+\sqrt{{d}^{2}32d+192}}{8}}$. (Note that a higher lower bound on ${u}^{3}$ is desirable, and the one stated here is much higher than that stated in Marcus.) Thus, ${u}^{2}\ge {\left({\displaystyle \frac{d16+\sqrt{{d}^{2}32d+192}}{8}}\right)}^{\frac{2}{3}}$. Therefore, if an element $x\in \mathcal{O}_{K}{}^{*}$ can be found such that $$, then $x=u$.
Following are some applications:

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The above is most applicable for finding the fundamental unit^{} of a ring of integers of a pure cubic field. For example, if $K=\mathbb{Q}(\sqrt[3]{2})$, then $d=108$, and the lower bound on ${u}^{2}$ is ${\left({\displaystyle \frac{23+10\sqrt{21}}{2}}\right)}^{\frac{2}{3}}$, which is larger than $9$. Note that $\left(\sqrt[3]{4}+\sqrt[3]{2}+1\right)\left(\sqrt[3]{2}1\right)=21=1$. Since $$, it follows that $\sqrt[3]{4}+\sqrt[3]{2}+1$ is the fundamental unit of ${\mathcal{O}}_{K}$.

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The above can also be used for any number field $K$ with $[K:\mathbb{Q}]=3$ such that $K$ has exactly one real embedding. Let $\sigma $ be the real embedding. Then the above produces the fundamental unit $u$ of $\sigma (K)$. Thus, ${\sigma}^{1}(u)$ is a fundamental unit of $K$.
References
 1 Marcus, Daniel A. Number Fields. New York: SpringerVerlag, 1977.
Title  units of real cubic fields with exactly one real embedding 

Canonical name  UnitsOfRealCubicFieldsWithExactlyOneRealEmbedding 
Date of creation  20130322 16:02:25 
Last modified on  20130322 16:02:25 
Owner  Wkbj79 (1863) 
Last modified by  Wkbj79 (1863) 
Numerical id  13 
Author  Wkbj79 (1863) 
Entry type  Application 
Classification  msc 11R27 
Classification  msc 11R16 
Classification  msc 11R04 
Related topic  NormAndTraceOfAlgebraicNumber 