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

Question

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 $\mathbb{Z}^{\times}$ ?

Answer 1

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.

Answer 2

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 $\mathbb{Z} \times \mathbb{Z}/2$.