Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 123 / 2

For $\mathrm{Re}s>1$, the Dedekind zeta function of $K$ isUnique factorization of ideals gives its Euler productover the nonzero prime ideals of ${\mathcal{O}}_K$.

The analytic class number formula states that ${\zeta }_K(s)$ has a simple pole at $s=1$ withwhere $(r_1,r_2)$ is the signature, $h_K$ the class number, $R_K$ the regulator, $w_K$ the number of roots of unity in $K$, and $d_K$ the field discriminant.

A Dirichlet character modulo $m$ is a group homomorphismextended periodically to $\mathbb{Z}$ by setting $\chi (n)=0$ when $(n,m)>1$. Its Dirichlet L-function is

Let $K/\mathbb{Q}$ be quadratic with discriminant $d_K$, and letbe its primitive quadratic Dirichlet character. For an unramified prime $p$, the Euler factor of ${\zeta }_K$ isSince ${\chi }_{d_K}(p)$ is respectively $1$ or $-1$, this equals the product of the Euler factors of $\zeta (s)$ and $L(s,{\chi }_{d_K})$. At a ramified prime ${\chi }_{d_K}(p)=0$, and the same equality holds. Thus

Suppose $d_K<-4$. Then $K$ has signature $(0,1)$, regulator $1$, and exactly $w_K=2$ roots of unity. The analytic class number formula becomesBecause ${\zeta }_K(s)=\zeta (s)L(s,{\chi }_{d_K})$ and ${\mathrm{res}}_{s=1}\zeta (s)=1$,Putting $m=|d_K|$ into the supplied odd-character identity yields the quadratic class number formula

For $K=\mathbb{Q}(\sqrt{-11})$, the fundamental discriminant is $-11$ and ${\chi }_{-11}(2)=-1$ because $-11\equiv 5(\mathrm{mod}8)$. The quadratic residues modulo $11$ areThereforeThe quadratic class number formula gives