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.

Let $\mathcal{O}_l=\mathbb{Z}[\frac{1+\sqrt{-l}}{2}]$ with $l$ a prime number congruent to 3 mod 4. Let $\mathfrak{a}$ be a non-principal fractional ideal of $\mathcal{O}_l$.

My questions are: Why N$(x)$/N$(\mathfrak{a})$ is less than $\frac{1+l}{4}$, being $x$ a generator of $\mathfrak{a}$? Why the fundamental parallelogram of $\mathfrak{a}$ have area equal to $\frac{\sqrt{-l}}{2}$?

I've been told that it is crucial that the discriminant is prime, so I thought about Minkowsky's theorems but I do not get to do it.

I would be thankful if you could help me.

share|improve this question
    
What is a generator of a non-principal ideal? The only ideal (non-fractional) with the area that you mention is $\mathcal{O}_l$ (in general the area is $N(\mathfrak{a})\sqrt{l}/2$). –  minu May 28 '12 at 17:44
    
By generator I mean the elements which span the lattice when we see $\mathfrak{a}$ as a lattice in $\mathbb{C}$. Thank you for the comment about the area. Could you tell me where can I find that kind of theorem? –  MPI May 28 '12 at 17:55
    
All this problem came to me while studying an article written by Noam Elkies about supersingular primes for elliptic curves. In this article, and it is him who says that the area is going to be $\frac{\sqrt{l}}{2}$. But the only thing we know in the beginning is that $\mathfrak{a}$ is a non-principal fractional ideal, he does not say anywhere that it is going to be exactly $\mathcal{O}_l$... –  MPI May 28 '12 at 17:58

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.