If $$\rho_\nu := res_{s=1} \zeta_K,\nu(s)=lim_{s\to 1}(s-1)\zeta_{K,\nu}(s)=\frac{2^r(2\pi)^s}{\omega_K |disc(\mathfrak(O)_K)|^{\frac{1}{2}}} \textrm{ (*)}$$
how can I follow that
$$\rho := res_{s=1} \zeta_K(s) = \rho_{\nu} * h_K = lim_{s\to 1}(s-1)\zeta_K(s) \textrm{ ?}$$
First the definitions:
$h_K = $ the number of elements in $mathscr{C}_K$
$\zeta_K(s) = \sum\limits_{\nu\in\mathscr{C}_K} \zeta_{K,\nu}(s)$, whith $\mathscr{C}_K$ the ideal class group of the number field $K$.
$\zeta_{K,\nu}(s)=\sum\limits_{I\in\nu}N(I)^{-s}$, $N$ the absolute norm.
I know the property (*) and I saw the second property in this paper: http://modular.math.washington.edu/129-05/final_papers/Gary_Sivek.pdf in Theorem 2 on page 3.
If I consider $res_{s=1}\zeta_K(s)=\rho_\nu * h_K = lim_{s\to 1}(s-1)\zeta_{K,\nu}(s) * h_K$.
But I don't see why $\zeta_{K,\nu}(s)*h_K$ should be $\zeta_K(s)$.
I would be happy if someone could help me with it.
All the best, Luca
It's not true that $h_K \zeta_{K, \nu}(s) = \zeta_K(s)$.
It is true, however, that their residues are equal. This is true because $\zeta_K(s) = \sum_\nu \zeta_{K, \nu}(s)$, each $\zeta_{K, \nu}(s)$ has residue $\rho_\nu$ (which is independent of $\nu$ by the formula ($*$)), and there are $h_K$ terms in the sum.