Deduce fundamental unit $\mathbb{Z}[\alpha]$ from fundamental unit of $\mathbb{Z}[\alpha,\beta]$ Let $R=\mathbb{Z}[\sqrt[3]{19}] =\mathbb{Z}[\alpha]\subset K=\mathbb{Q}(\sqrt[3]{19})$. We know that $\mathcal{O}_K=R[\beta]=\mathbb{Z}[\alpha,\beta]$ where $\beta=\dfrac{\alpha^2-\alpha+1}{3}$, c.f Example 3.7, page 34. 
Furthermore, we have $1-\alpha-\beta$ is a fundamental unit of $\mathcal{O}_K$ but not in $R$, c.f Example 7.3, page 75. My question is

How we can deduce fundamental unit of $\mathbb{Z}[\alpha], \mathbb{Z}[\beta]$ from fundamental unit of $\mathbb{Z}[\alpha,\beta]$?

In 5.16, page 64, the author use Regulator to implies that $R^*$ is of finite index in $\mathcal{O}_K$ but i don't really understand it. How can we get the bound for $n$ such that $(1-\alpha-\beta)^n$ to be a fundamental unit of $\mathbb{Z}[\alpha]$?
 A: Have a look at the corresponding problem for quadratic fields. The fundamental unit of the field with discriminant $5$ is $\varepsilon = \frac{1+\sqrt{5}}2$; for finding the fundamental unit of the order ${\mathbb Z}[\sqrt{5}]$ you have to get rid of the $2$ in the denominator, that is, you are looking for a power of $\varepsilon$ with $\varepsilon^n \equiv 1 \bmod 2$. By Fermat's Little Theorem you have $\varepsilon^{\Phi(2)} \equiv 1 \bmod 2$; in the present case, $2$ is inert, hence $\Phi(m) = N(2)-1 = 3$.   
In ${\mathbb Z}[\sqrt{19}]$, the ideal $(2)$ splits into two prime ideals $P$ and $Q$ with norms $2$ and $4$, respectively. Since units are congruent to $1  \bmod P$, you only have to make sure that $(1-\alpha-\beta)^n \equiv 1 \bmod Q$, and again $\Phi(Q) = N(Q)-1 = 3$, so $n$ divides $3$. Observe that in general situations, the theorem of Euler-Fermat only gives an upper bound for the exponent.
A: 
About $R^\times$ being of finite index in $O_K^\times$, my argument was wrong but it is still true that it is obvious.

If $R$ is of finite index $m$ in $O_K$ then $R$ contains $mO_K$, since $R$ is a unital subring it contains $\Bbb{Z}+mO_K$ (in your case $m = 3$)
$u\in O_K^\times$ iff its minimal polynomial $f\in \Bbb{Z}[x]$ is monic and satisfies $f(0) = \pm 1$, with $g(x)=-\frac{f(x)-f(0)}{f(0) x}\in \Bbb{Z}[x]$ we get $u^{-1}=g(u)\in \Bbb{Z}[u]$ so that $R \cap O_K^\times=R^\times$.
For each $a \in O_K/(m)$ pick some $u_a \in O_K^\times, u_a\equiv a \bmod m$ if it exists. 
Let $T = \{ u \in O_K^\times, u\equiv 1 \bmod m\}$. Then 
$$O_K^\times= \bigcup_{u_a} u_a T\subset \bigcup_{u_a} u_a (R \cap O_K^\times) =\bigcup_{u_a} u_a R^\times$$ thus $R^\times$ is at most of index $|O_K/(m)|=m^3$ in $O_K^\times$.

$H=\{a \in O_K/(m)^\times, \exists u \in (\Bbb{Z}+mO_K)^\times, u \equiv a \bmod m\}$ is a subgroup of $G=\{a \in O_K/(m)^\times, \exists u \in O_K^\times, u \equiv a \bmod m\}$ which is a subgroup of $O_K/(m)^\times$ 

The previous argument shows $(\Bbb{Z}+mO_K)^\times$ is of index $\frac{|G|}{|H|}$ in $O_K^\times$. 

By Dirichlet theorem $O_K^\times = \pm u^\Bbb{Z}$ so that $R^\times =\pm u^{d\Bbb{Z}}$ where $d$ is the least integer such that $u^d \in R$ and your text tells us $u = 1-\alpha-\beta$.

The previous argument shows that $d$ divides $ |O_K/(m)^\times|$ (here $3 O_K=PQ^2$ thus $|O_K/(3)^\times| =|O_K/P^\times||O_K/(Q^2)^\times|=(3-1)(9-3)$) thus it suffices to check one by one.
