# $L$-series and the $\zeta$-function of ideal classes modulo $f$

Let $$K$$ be an imaginary quadratic field and let $$\mathcal O_f$$ be an order in $$K$$ with conductor $$f$$. Let $$\chi$$ be a proper character of $$\operatorname{Cl}(\mathcal O_f)$$ (non-principal if $$f=1$$). Define $$L_f(s,\chi)=\sum_{(\mathfrak a,f)=1} \chi(\mathfrak a)N(\mathfrak a)^{-s}.$$ Here the sum is over all $$\mathcal O_f$$-ideals prime to $$f$$.

Suppose that there are only two units in $$O_f$$. For $$C\in \operatorname{Cl}(\mathcal O_f)$$ and $$\mathfrak b\in C^{-1},\mathfrak b \subset \mathcal O_f$$, let $$\zeta(s,C)=\frac{N(\mathfrak b)^s}{2}\sum_{0\neq\gamma\in \mathfrak b}N(\gamma)^{-s}.$$

How to prove that $$L_f(s,\chi)=\sum_{C\in \operatorname{Cl}(\mathcal O_f)}\chi(C)\zeta(s,C)\quad?$$ This is proved in Curt Meyer's Die Berechnung der Klassenzahl Abelscher Körper über Quadratischen Zahlkörpern, p. 24-25, but I dont understand the argument there.

The point here is that we may dispose of the condition requiring the ideals to be prime to the conductor $$f$$.

Decompose the original sum according to the ideal class of $$\mathfrak{a}$$. Fixing the ideal class of $$\mathfrak{a}$$, we can write $$\mathfrak{a}=\mathfrak{b}^{-1}\gamma$$ where $$\mathfrak{b}$$ is a fixed representative of the inverse class, and $$\gamma\in\mathfrak{b}$$. The map $$\gamma\mapsto\mathfrak{a}$$ is $$2$$-to-$$1$$, because there are only two units. The result follows if in the definition of $$\zeta(s,C)$$ we restrict $$\gamma$$ by the condition $$(\gamma,f)=1$$. However, this restriction can be dropped, using that $$f$$ is the conductor of $$\chi$$. Indeed, if we pick a divisor $$d\mid f$$, and examine the joint contribution of $$\gamma$$'s with $$(\gamma,f)=d$$ in the various series $$\zeta(s,C)$$, we end up with an expression of the shape $$\sum_{C'\in \operatorname{Cl}(\mathcal O_{f/d})}\biggl(\sum_{\substack{C\in \operatorname{Cl}(\mathcal O_{f})\\C\subset C'}}\chi(C)\biggr)\dots.$$ As $$\chi$$ has conductor $$f$$, the inner sum is zero unless $$d=1$$.
The above argument uses that $$(\gamma,f)$$ is always a rational integer. Indeed, denoting by $$\mathcal{O}$$ the maximal order, we can write any $$\gamma\in\mathcal{O}_f$$ as $$x+fy$$ with $$x\in\mathbb{Z}$$ and $$y\in\mathcal{O}$$, hence $$(\gamma,f)=(x,f)$$ is a rational integer.
• @Shimrod: OK, I see now. Meyer assumes $\chi$ to be primitive, i.e., $f$ is its conductor. The key group theoretic input is the last display on Page 24: if we take $f'|f$ with $f'<f$, and we sum $\chi$ over the ray classes mod $f$ that lie in a fixed ray class mod $f'$, then we get zero. Here is an analogy that should explain it: if we take a primitive Dirichlet character mod $100$, and we sum it over the numbers $3,23,43,63,83$, we get zero (otherwise our Dirichlet character would have conductor dividing $20$). – GH from MO Aug 8 '19 at 11:45
• I presume that by $\mathcal O$ you mean the maximal order? – Shimrod Aug 8 '19 at 12:27