Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How do can I determine all classes of ideals of $\mathbb{Z}[\sqrt{-104}]$? Or $\mathbb{Z}[\sqrt{-132}]$? (so a list of representatives and showing they are not equivalent, and and that we get all of them) If someone can show me these examples it would be very helpful.

Is there a general method or approach to determining all of the ideals classes of $\mathcal{O}_K$ where $K=\mathbb{Q}[\sqrt{-N}]$.

(when i say classes i mean under equi relation $I\sim J$ if $\exists$ nonzero $x,y$ s.t. $xI=yJ$.)

share|improve this question
    
Note that for $N=-104$, since $104 = 4\times 26$, $\mathcal{O}_K\neq \mathbb{Z}[\sqrt{-104}]$; the field is equal to $\mathbb{Q}(\sqrt{-26})$, and the ring of integers is $\mathbb{Z}[\sqrt{-26}]$. Similarly, $132 = 4\times 33$, so the ring of integers is $\mathbb{Z}[\sqrt{-33}]$. –  Arturo Magidin Nov 25 '11 at 20:36
2  
Can you please indicate what your background is and what books you have read or are reading that are related to the question you are asking? –  KCd Nov 26 '11 at 19:42
    
@KCd: I am learning this material. I know some ring theory and galois theory. i know that the group of classes is finite and abelian and also know minkowski's theorem. i think embedding this in a vector space and using properties of ideals there is a way to solve this but I need help. how would you do it? –  Math Student Nov 26 '11 at 21:20
    
Under certain conditions, as described in the answer by WIll Jagy, the group will be isomorphic with the group of binary quadratic forms with discriminant the same as the number field. In particular, your examples satisfy the conditions, so you are actually looking at the forms. –  awllower May 23 '13 at 2:23
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.