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 $f$. 



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


