Let $K = \mathbb{Q}(\sqrt{-5})$ with $\mathcal{O}_K = \mathbb{Z}\left[\frac{1+\sqrt{-5}}{2}\right]$ and discriminant $\Delta_K = -20$.

The Minkowski bound is: $M_K = \frac{2!}{2^2} \cdot \frac{4}{\pi} \cdot \sqrt{20} = \frac{4\sqrt{20}}{\pi} \approx 5.7$

So every ideal class has a representative with norm $\leq 5$. Check primes up to 5:

• $(2) = \mathfrak{p}_2^2$ where $\mathfrak{p}_2 = (2, 1+\sqrt{-5})$: non-principal (if principal, $2 = N(\alpha)$ impossible)
• $(3) = \mathfrak{p}_3\mathfrak{p}_3'$: factors as $(3, 1+\sqrt{-5})(3, 1-\sqrt{-5})$
• $(5) = (\sqrt{-5})(\sqrt{-5})$: principal

We find $\mathfrak{p}_2^2 = (2)$ is principal, so $[\mathfrak{p}_2]$ has order 2. Therefore $\text{Cl}(K) = \mathbb{Z}/2\mathbb{Z}$ and $h_K = 2$.

For $K = \mathbb{Q}(\sqrt{-23})$, we have $\Delta_K = -23$ and Minkowski bound $M_K \approx 4.8$.

Factor small primes:

• $(2) = \mathfrak{p}_2^2$: ramifies
• $(3) = \mathfrak{p}_3\mathfrak{p}_3'$: splits

Computing: $\mathfrak{p}_2$ and $\mathfrak{p}_3$ are non-principal. Check relations: $\mathfrak{p}_3^3 = (x + y\sqrt{-23})$ for some $x, y \in \mathbb{Z}$.

Working through the arithmetic shows $[\mathfrak{p}_3]$ has order 3, giving $\text{Cl}(K) = \mathbb{Z}/3\mathbb{Z}$ and $h_K = 3$.

For quadratic fields $K = \mathbb{Q}(\sqrt{d})$, Gauss developed genus theory to analyze class group structure.

The class group contains a subgroup $\text{Cl}^2(K)$ of squares of ideal classes. The quotient $\text{Cl}(K)/\text{Cl}^2(K)$ is an elementary abelian 2-group whose structure is determined by the factorization of $d$.

For $d = -p_1 \cdots p_r$ with distinct odd primes $p_i$, the genus group has order $2^{r-1}$, explaining part of the class group structure.

The Hilbert class field $H_K$ is the maximal unramified abelian extension of $K$. Its key properties:

• $\text{Gal}(H_K/K) \cong \text{Cl}(K)$
• All prime ideals of $K$ split completely or remain prime in $H_K$
• $\mathcal{O}_{H_K}$ has class number 1 (principal ideal domain)

For $K = \mathbb{Q}(i)$, since $h_K = 1$, we have $H_K = K$ itself.

For $K = \mathbb{Q}(\sqrt{-5})$ with $h_K = 2$, the Hilbert class field is $H_K = K(\sqrt{-1})$ with $[H_K : K] = 2$.

For $K_0 = \mathbb{Q}(\sqrt{-3 \cdot 5 \cdot 7 \cdot 11 \cdot 13})$, the tower of Hilbert class fields: $K_0 \subset H_{K_0} \subset H_{H_{K_0}} \subset \cdots$

can be infinite! Golod-Shafarevich constructed examples where this tower never terminates, showing class numbers can grow arbitrarily fast in such towers.