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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

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1 Answer 1

up vote 1 down vote accepted

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.

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OK! So $res_{s=1}\zeta_K = res_{s=1}\zeta_{K,\nu}(s)*h_K = lim_{s\to 1}(s-1)\zeta_{K,\nu}(s)*h_K$. But how do I see then that $res_{s=1}\zeta_K = lim_{s\to 1}(s-1)\zeta_K(s)$ as in the paper, and not $= lim_{s\to 1}(s-1)\zeta_{K,\nu}(s)*h_K$? And thanks for the quick reply! –  Luca Dec 9 '13 at 21:37
    
@Luca What is your definition of the residue? In any case, it shouldn't even matter as long as you can see that the residue is linear: $\text{Res }\zeta_K = \text{Res }\sum \zeta_{K, \nu} = \sum \text{Res } \zeta_{K, \nu} = \sum \rho_\nu = h_K \rho_\nu$ (where the last equality uses the fact that the $\rho_\nu$'s are all equal)... –  Bruno Joyal Dec 9 '13 at 21:41
    
We defined the residue for $\zeta_K$ with the formula (*) times $h_K$. I'm sorry, but I still don't see how I have then $res\zeta_K = lim(s-1)\zeta_K(s)$. I only get $res\zeta_K = h_K * lim(s-1)\zeta_{K,\nu}(s)$ with what you write... how can I conclude that this equals $lim(s-1)\zeta_K(s)$? –  Luca Dec 9 '13 at 22:13
1  
@Luca That is the definition of the residue: en.wikipedia.org/wiki/Residue_(complex_analysis)#Simple_poles –  Bruno Joyal Dec 9 '13 at 22:16
    
It is? Ohhh.... than everything is clear :D Thanks a lot for your patience and explanation! –  Luca Dec 9 '13 at 22:27

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