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 $r+s-1=1$ 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 $0<\theta<\pi$ 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, $\displaystyle u=\frac{1}{\rho^2}$ One can then deduce that $\displaystyle \operatorname{disc}u=-4\sin^2\theta\left(\rho^3+\frac{1}{\rho^3}- 2\cos\theta\right)^2$ Since the maximum value of the polynomial $4\sin^2\theta(x-2\cos\theta)^2-4x^2$ is at most $16$ one can deduce that $\displaystyle |\operatorname{disc}u| \le 4\left(u^3+\frac{1}{u^3}+4\right)$ Define $d=|\operatorname{disc}\mathcal{O}_K|$ Then $\displaystyle d\le|\operatorname{disc}u| \le 4\left(u^3+\frac{1}{u^3}+4\right)$ Thus, $\displaystyle u^3 \ge \frac{d}{4}-4-\frac{1}{u^3}$ From this, one can obtain that $\displaystyle u^3 \ge \frac{d-16+\sqrt{d^2-32d+192}}{8}$ (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, $\displaystyle u^2 \ge \left( \frac{d-16+\sqrt{d^2-32d+192}}{8} \right)^{\frac{2}{3}}$ Therefore, if an element $x \in {\mathcal{O}_K}^*$ can be found such that $\displaystyle 1<x<\left( \frac{d-16+\sqrt{d^2-32d+192}}{8} \right)^{\frac{2}{3}}$ then $x=u$

Following are some applications:

• 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 $\displaystyle \left( \frac{23+10\sqrt{21}}{2} \right)^{\frac{2}{3}}$ which is larger than $9$ Note that $\displaystyle \left( \sqrt[3]{4}+\sqrt[3]{2}+1 \right) \left( \sqrt[3]{2}-1 \right)=2-1=1$ Since $1<\sqrt[3]{4}+\sqrt[3]{2}+1<9$ it follows that $\sqrt[3]{4}+\sqrt[3]{2}+1$ is the fundamental unit of $\mathcal{O}_K$
• 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 $K$

Bibliography

1

Marcus, Daniel A. Number Fields. New York: Springer-Verlag, 1977.