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 $\mathbb{Z}[\alpha]$?
Best Answer
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.