2
$\begingroup$

enter image description heres://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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.