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

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 $$O_K^{\times}/O_L^{\times}$$.

$$L=\Bbb{Q}(\zeta_{12}), K=\Bbb{Q}(\sqrt3)$$
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 $$O_L^\times = \langle \zeta_{12} \rangle (1+\zeta_{12})^\Bbb{Z}$$.
• For the part $u \mapsto |u|^2=N_{L/K}(u)$, doesn't this immediately imply the kernel have norm 1 and so are roots of unity? Apr 25, 2022 at 17:39
• Not sure of what you mean, there are two pairs of complex embeddings, for $u$ in the kernel at first we only know that $|\sigma(u)|=1$ for $\sigma$ in on of those. For the other pair of complex embeddings we need to say that the Galois group is abelian which implies that $|\sigma(u)|^2=\sigma(|u|^2)=1$. Apr 25, 2022 at 18:22
• Isn't the kernel of $u \mapsto |u|^2$ exactly $u$ such that $|u|=1$, which are precisely the roots of the unity? Also if $|\sigma(u)|=1$ for one $\sigma$, doesn't this hold for all $\sigma$? Apr 25, 2022 at 19:07
• The kernel are the $u$ such that $|u|=1$ but it is not obvious that they are the roots of unity. We need $|\sigma(u)|=1$ for the 4 complex embeddings, not only 2. Yes it holds for all $\sigma$ because the Galois group is abelian. Apr 25, 2022 at 19:52