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Let $a$ be an ideal in $K=\mathbb{Q}(\sqrt{D})$ , $D>0$ and $Q \in \mathbb{N}$ . $u'$ denotes the complex conjugate of $u$ and S is the trace. My problem is to understand how to get the second equation after (10) .

I do not understand how the sum $\sum_{v\equiv0(a')}$ was split up .

What I know is the following:

$v\equiv0(a')$, then since a devides $\rho$ , $v\equiv\rho(a')$. Now I can calculate the bilinear form $(v,\rho')=(l+\rho',\rho')=(l,\rho')+(\rho',\rho')$,

where the second term is in $\mathbb{Z} $.

Thanks for the help.


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