# units of real cubic fields with exactly one real embedding

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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