Class number formula

Remarks

The proof counts ideals of norm at most $x$ in each ideal class. This count reduces to counting lattice points in a region of $\R^{r_1} \times \C^{r_2}$ cut out using the units. So its leading term is a volume; the volume brings in $R_K$ and $\sqrt{\abs{d_K}}$, and the classes bring in $h_K$. Example: for $K = \Q$ one has $r_1 = 1$, $r_2 = 0$, $h = R = 1$, $w = 2$, $d = 1$, and the formula gives residue $1$, as for the Riemann zeta function. For an imaginary quadratic field of discriminant $d < 0$, the formula becomes $h_K = \frac{w_K \sqrt{\abs{d}}}{2\pi} L(1, \chi_d)$, where $\chi_d$ is the Kronecker character of $K$ and $L$ its Dirichlet L-function. For $K = \Q(i)$: $w = 4$, $d = -4$ and $L(1, \chi_{-4}) = 1 - \frac13 + \frac15 - \dots = \frac{\pi}{4}$, so $h_K = 1$.

Going deeper

For a student

For an abelian field $K$, the function $\zeta_K(s)$ is the product of $\zeta(s)$ and Dirichlet $L$-functions, so the residue becomes a product of values $L(1, \chi)$. This gives Dirichlet's class number formulas for quadratic fields, and the fact that $L(1, \chi) \ne 0$ for real characters, which is the key step in Dirichlet's theorem on arithmetic progressions. The units enter through the regulator, which is defined by means of Dirichlet's unit theorem.