Find the units in the ring of integers over $\mathbb{Q}(e^{i\pi/6})$

Question

Let $\omega=e^{i\pi/6}$, $L=\mathbb{Q}(\omega)$ and $O_L$ the ring of integers of $L$. Let $K=\mathbb{Q}(\sqrt{3})$ and $O_K$ the ring of integers in $K$. Show that the units in $O_L$ are

$O_L^{\times} = \{(1+\omega)^nu, 0 \leq n \leq 11, u \in O_K^{\times}\}$

Then find a primitive unit in $L$.

What I know so far:

• $\omega=(i+\sqrt{3})/2$, so $L=\mathbb{Q}(i,\sqrt{3})$ has two complex embeddings and no real embeddings. So by Dirichlet's unit theorem, $O_L^{\times} = \{\omega^j \alpha^n, 0 \leq j \leq 11, n \in \mathbb{Z}\}$, since the units in $L$ are just the powers of $\omega$. $\alpha$ is a primitive unit generating the (non-torsion) unit group.
• $O_K^{\times} = \{\pm(2+\sqrt{3})^n, n \in \mathbb{Z}\}$
• $N_{L/K}(1+\omega) = 2+\sqrt{3}$, which itself has norm 1 over $\mathbb{Q}$. So $1+\omega$ is a unit.
• $1+\omega=\sqrt{2+\sqrt{3}} \cdot e^{i\pi/12}$, which means that $(1+\omega)^j$ is not real for $1 \leq j \leq 11$, which means that $(1+\omega)^j$, $0 \leq j \leq 11$ are distinct in $O_K^{\times}/O_L^{\times}$.

Answer 1

$L=\Bbb{Q}(\zeta_{12}), K=\Bbb{Q}(\sqrt3)$

The kernel of $u\to |u|^2$, $O_L^\times \to O_K^\times$ are the $z$ such that $|\sigma(z)| = 1$ for all complex embedding, which implies that $z$ is a root of unity.

So it suffices to show that $O_K^\times = \pm (2+\sqrt3)^\Bbb{Z}$ and $|1+\zeta_{12}|^2=2+\sqrt3$ to obtain that $O_L^\times = \langle \zeta_{12} \rangle (1+\zeta_{12})^\Bbb{Z}$.