# Is $\mathcal{O}_K^{\times}$ a cyclic group just like $\mathbb{Z}^{\times}$?

The $$\text{units}$$ in a ring of integers are those elements whose multiplicative inverse exists. That is, $$u$$ is unit if $$u^{-1}$$ also exists in the ring such that $$uu^{(-1)}=u^{(-1)}u=\text{multiplicative identity}$$.

For example, consider the ring of integers $$\mathbb{Z}$$ of the rational field $$\mathbb{Q}$$, then $$\mathbb{Z}^{\times}=$$ units in $$\mathbb{Z}=\{1,-1 \}.$$ This is a cyclic group. In fact, this is trivial and $$\mathbb{Z}$$ is infinite cyclic group.

Now consider the ring of integers $$\mathcal{O}_K$$ in a finite extension $$K \supset \mathbb{Q}$$ or the ring of integers $$\mathcal{O}_K$$ of $$p$$-adic field $$K \supset \mathbb{Q}_p$$.

Now denote the units of $$\mathcal{O}_K$$ by $$\mathcal{O}_K^{\times}$$.

Is $$\mathcal{O}_K^{\times}$$ a cyclic group just like $$\mathbb{Z}^{\times}$$ ?

If $$K$$ is a global number field, i.e. some finite extension of $$\mathbf Q$$, then $$\mathcal O_K^{\times}$$ is cyclic precisely when $$K = \mathbf Q$$ or $$K$$ is an imaginary quadratic number field. The unit group is the group of roots of unity lying in $$K$$. If $$K$$ is a local number field, i.e. an extension of $$\mathbf Q_p$$ for some prime $$p$$, then the unit group $$\mathcal O_K^{\times}$$ is never cyclic. An easy way to see this is that this unit group has both an element of order $$2$$ and an element of infinite order, which can't happen in any cyclic group.

• Thanks, but I didn't understand the last sentence. Which two particular elements are they ?
– MAS
Sep 22, 2020 at 14:01
• @Masmath If $p$ is odd, then take $-1$ and $2$. If $p = 2$, then take $-1$ and $3$. Sep 22, 2020 at 14:01
• Thank you very much
– MAS
Sep 22, 2020 at 14:03

As mentioned by Marktmeister in the comments, Dirichlet's unit theorem tells us that $$\mathcal{O}_K^{\times}$$ is finitely generated with rank $$r_1 + r_2 - 1$$ where $$r_1$$ is the number of real embeddings $$K \to \mathbb{R}$$ and $$r_2$$ is the number of conjugate pairs of complex embeddings $$K \to \mathbb{C}$$ ("complex" here means that their image is not contained in $$\mathbb{R}$$). Since $$-1 \in \mathcal{O}_K^{\times}$$ is always torsion it follows that the unit group is cyclic iff it's finite (since then it's a finite subgroup of $$K$$, hence cyclic), and Dirichlet's theorem tells us this happens iff $$r_1 + r_2 = 1$$.

• If $$r_1 = 1, r_2 = 0$$ then the degree of the extension is $$n = r_1 + 2r_2 = 1$$ so $$K = \mathbb{Q}$$.
• If $$r_1 = 0, r_2 = 1$$ then the degree of the extension is $$n = r_1 + 2r_2 = 2$$ so $$K = \mathbb{Q}(\sqrt{-d})$$ is imaginary quadratic. This recovers Ege's claim.

$$K = \mathbb{Q}(\sqrt{2})$$ is a minimal example where the rank is positive. Here $$r_1 = 2, r_2 = 0$$ so the unit group has rank $$1$$. A fundamental unit (a generator of the torsion-free part) is given by $$1 + \sqrt{2}$$, but $$-1$$ is also a unit (the only nontrivial root of unity), so the group of units is abstractly isomorphic to $$\mathbb{Z} \times \mathbb{Z}/2$$.

• Thank you for your nice elaboration.
– MAS
Sep 23, 2020 at 14:00