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I am working on the analytic class number formula. There is a step that I want to understand and need some help. Here is the setup:

Let $K$ be a number field with class group $Cl_K$, $C$ be a class. I am considering the sum $\displaystyle\sum_{ \mathfrak{a} \in C}\cfrac{1}{N(\mathfrak{a})^s}$. We will write this sum in a different way, so, consider the inverse class $C^{-1}$ consisting of ideals $\mathfrak{a}^{-1}$ where $\mathfrak{a} \in C$. Choose any $\mathfrak{b} \in C^{-1}$, the inverse class. Then for any $\mathfrak{a} \in C$, $\mathfrak{a}\mathfrak{b} \in P_K$ so that $\mathfrak{a}\mathfrak{b} = <\alpha>$ for some $\alpha \in K$.

I claim that the map $\mathfrak{a} \rightarrow \mathfrak{a}\mathfrak{b} = <\alpha>$ is bijective - I mean there is a one-to-one correspondence between the ideals $\mathfrak{a}$ and $\alpha_\mathfrak{a} \in \mathfrak{b}$- so instead of counting the ideals $\mathfrak{a}$ in $C$, we can say the following: $\displaystyle\sum_{ \mathfrak{a} \in C}\cfrac{1}{N(\mathfrak{a})^s} =N(\mathfrak{b})^s\sum_{ \mathfrak{b} | <\alpha>}\cfrac{1}{|N(\alpha)^s|} $.

However, I could not prove it. On the RHS, we are summing over ideals $<\alpha>$ however, if $\alpha'$ and $\alpha$ are associates, we may count twice. Can you give me some example that for ideals $\mathfrak{a}$ and $\mathfrak{b}$, there exist $\alpha,\alpha'$ with $\mathfrak{a}\mathfrak{b} = <\alpha> = <\alpha'>.$

I also appreciate it if you clarify how the above sum is working.

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  • $\begingroup$ It's a bijection between $C$ and the set of principal ideals contained in $\frak b$ ? (or with ${\frak b}^*/U$ where $U$ is the unit group) $\endgroup$ – mercio Oct 2 '17 at 14:35
  • $\begingroup$ Yes, the first one I think. Since $\mathfrak{b}$ divides $<\alpha>$, it contains the $<\alpha>$. Where are the other comments of ours? $\endgroup$ – Ninja Oct 2 '17 at 22:46

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