The field $K = \mathbb{Q}(i)$ has ring of integers $\mathcal{O}_K = \mathbb{Z}[i]$. This is a unique factorization domain. The units are $\{\pm 1, \pm i\}$, and the discriminant is $\Delta_K = -4$.

The prime factorization in $\mathbb{Z}[i]$ reveals:

• Primes $p \equiv 1 \pmod{4}$ split: $p = \pi\bar{\pi}$ for some prime $\pi \in \mathbb{Z}[i]$
• The prime $2$ ramifies: $2 = -i(1+i)^2$
• Primes $p \equiv 3 \pmod{4}$ remain prime in $\mathbb{Z}[i]$

For $n \geq 1$, let $\zeta_n = e^{2\pi i/n}$ be a primitive $n$-th root of unity. The $n$-th cyclotomic field is $K = \mathbb{Q}(\zeta_n)$, with degree $[K : \mathbb{Q}] = \varphi(n)$ (Euler's totient function).

The ring of integers is $\mathcal{O}_K = \mathbb{Z}[\zeta_n]$, and the discriminant is $\Delta_K = (-1)^{\varphi(n)/2}\frac{n^{\varphi(n)}}{\prod_{p|n}p^{\varphi(n)/(p-1)}}$

Cyclotomic fields have rich arithmetic: they played a crucial role in the proof of Fermat's Last Theorem for regular primes.

Quadratic fields $K = \mathbb{Q}(\sqrt{d})$ split into two types:

• Real quadratic fields ($d > 0$): Have infinitely many units. For example, $\mathbb{Q}(\sqrt{2})$ has fundamental unit $1 + \sqrt{2}$ with $N(1 + \sqrt{2}) = -1$
• Imaginary quadratic fields ($d < 0$): Have finitely many units. Only nine imaginary quadratic fields have class number 1: $d \in \{-1, -2, -3, -7, -11, -19, -43, -67, -163\}$

Consider $K = \mathbb{Q}(\alpha)$ where $\alpha^3 - \alpha - 1 = 0$. This is a totally real cubic field (all three embeddings are real). Computing $\mathcal{O}_K$ requires checking when elements of $K$ satisfy monic polynomials over $\mathbb{Z}$.

For this field, $\mathcal{O}_K = \mathbb{Z}[\alpha]$, meaning the ring of integers is monogenic. However, not all number fields are monogenic; some require multiple generators.