Hilbert class field

Remarks

By class field theory, $H/K$ is a finite abelian extension, unramified at every place, and the Artin map gives an isomorphism from the ideal class group $\mathrm{Cl}(K)$ onto $\Gal(H/K)$. So $[H : K] = h_K$, and a prime ideal of $\mathcal{O}_K$ splits completely in $H$ if and only if it is principal. Every ideal of $\mathcal{O}_K$ becomes principal in $\mathcal{O}_H$ (the principal ideal theorem).

Example: $K = \Q(\sqrt{-5})$ and $L = K(i) = \Q(\sqrt{-5}, i)$. The three quadratic subfields of $L$ are $\Q(i)$, $\Q(\sqrt{5})$ and $\Q(\sqrt{-5})$, with discriminants $-4$, $5$ and $-20$, so $d_L = (-4) \cdot 5 \cdot (-20) = 400 = d_K^2$. Hence the relative discriminant of $L/K$ has norm $1$ and no finite prime ramifies; $K$ has no real embedding, so $L/K$ is unramified at every place. Since $h_K = 2 = [L : K]$, the field $L$ is the Hilbert class field of $K$.

Non-example: $L' = \Q(\sqrt{-5}, \sqrt{2})$ is abelian of degree $2$ over $K$ but not unramified. Its quadratic subfields have discriminants $8$, $-20$ and $-40$, so $d_{L'} = 6400 = 16 \, d_K^2$, and some prime of $K$ above $2$ ramifies in $L'$.