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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 $f$.

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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.

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  • $\begingroup$ Here is the relevant passage from Meyer's book drive.google.com/file/d/1M9tALE9eAJH7c7NTfz5vKZ_aPDhdrcGS/…, I would appreciate if you took a look, for it seems to me that your conclusion contradicts that of Meyer. $\endgroup$ – Shimrod Aug 8 '19 at 10:41
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    $\begingroup$ @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 $20$). $\endgroup$ – GH from MO Aug 8 '19 at 11:45
  • $\begingroup$ I presume that by $\mathcal O$ you mean the maximal order? $\endgroup$ – Shimrod Aug 8 '19 at 12:27
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    $\begingroup$ @Shimrod: Sure. I updated my response accordingly. $\endgroup$ – GH from MO Aug 8 '19 at 13:17
  • $\begingroup$ Thanks. Do you know about some (more modern) reference concerning characters for orders in imaginary quadratic fields? $\endgroup$ – Shimrod Aug 8 '19 at 13:32

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