# How to split this sum into a double sum.

s://i.stack.imgur.com/Uwv3v.jpg

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.