For $K = \mathbb{Q}(\sqrt{d})$ with $d > 0$ square-free, the fundamental unit is obtained from the continued fraction expansion of $\sqrt{d}$.

For $d = 13$: $\sqrt{13} = [3; \overline{1, 1, 1, 1, 6}]$ with period 5.

The convergents give solutions to $x^2 - 13y^2 = \pm 1$:

• $p_4/q_4 = 18/5$: gives $18^2 - 13 \cdot 5^2 = -1$
• $p_9/q_9 = 649/180$: gives $649^2 - 13 \cdot 180^2 = 1$

Therefore $\epsilon_0 = 649 + 180\sqrt{13}$ (or $\epsilon_0 = 18 + 5\sqrt{13}$ if allowing norm $-1$).

Consider $K = \mathbb{Q}(\alpha)$ where $\alpha^3 - \alpha - 1 = 0$ (totally real cubic field).

The three real embeddings give $r = 2$ fundamental units. Using LLL algorithm on the unit lattice: $\epsilon_1 = \alpha, \quad \epsilon_2 = \alpha^2 - 1$

Verification requires checking these are independent under the logarithmic embedding and that they generate $\mathcal{O}_K^*$ modulo torsion.

For $K = \mathbb{Q}(\zeta_7)$ (cyclotomic field of conductor 7):

• Degree $[\mathbb{Q}(\zeta_7) : \mathbb{Q}] = 6$
• Signature: $r_1 = 0, r_2 = 3$, so unit rank $r = 2$

Fundamental units can be taken as: $\epsilon_1 = \frac{1 - \zeta_7^2}{1 - \zeta_7}, \quad \epsilon_2 = \frac{1 - \zeta_7^3}{1 - \zeta_7}$

Computing embeddings numerically and evaluating the logarithmic map gives regulator $R_K \approx 0.5279$.

In abelian extensions of imaginary quadratic fields, Stark units provide explicit generators for unit groups.

For $K = \mathbb{Q}(i)$ and its ray class field $K_4 = K(\sqrt{1+i})$, the Stark unit is: $\epsilon = \sqrt{1+i}$

This unit, predicted by Stark's conjectures, has special properties: its $L$-function values encode arithmetic data of the extension.

For a finite set $S$ of primes, the $S$-units are elements invertible outside $S$: $\mathcal{O}_{K,S}^* = \{\alpha \in K^* : v_\mathfrak{p}(\alpha) = 0 \text{ for all } \mathfrak{p} \notin S\}$

By Dirichlet's $S$-unit theorem: $\mathcal{O}_{K,S}^* \cong \mu_K \times \mathbb{Z}^{r + |S| - 1}$.

For $K = \mathbb{Q}$ and $S = \{2, 3\}$: $\mathbb{Z}_{(S)}^* = \{\pm 2^a 3^b : a, b \in \mathbb{Z}\} \cong \{\pm 1\} \times \mathbb{Z}^2$.

$S$-units are crucial for solving unit equations and studying integral points on varieties.

If $K_1, K_2$ are number fields with unit groups $\mathcal{O}_{K_1}^*, \mathcal{O}_{K_2}^*$, the unit group of the compositum $K_1 K_2$ relates via: $\mathcal{O}_{K_1 K_2}^* \supseteq \mathcal{O}_{K_1}^* \cdot \mathcal{O}_{K_2}^*$

The index is bounded and connected to ramification. For linearly disjoint extensions, the regulator satisfies: $R_{K_1 K_2} \geq R_{K_1}^{[K_2:\mathbb{Q}]} \cdot R_{K_2}^{[K_1:\mathbb{Q}]}$

with equality when $K_1, K_2$ are linearly disjoint over $\mathbb{Q}$.