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

1 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 $O_L^\times = \langle \zeta_{12} \rangle (1+\zeta_{12})^\Bbb{Z}$.

  • $\begingroup$ 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? $\endgroup$
    – Steven Mai
    Apr 25, 2022 at 17:39
  • $\begingroup$ 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 $|\sigma(u)|^2=\sigma(|u|^2)=1$. $\endgroup$
    – reuns
    Apr 25, 2022 at 18:22
  • $\begingroup$ 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 $\sigma$? $\endgroup$
    – Steven Mai
    Apr 25, 2022 at 19:07
  • $\begingroup$ 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. $\endgroup$
    – reuns
    Apr 25, 2022 at 19:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .