Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hi all,

I have been looking at complex multiplication of elliptic curves for a course project in cryptography and the following question came up: Let $\mathcal{O}_K$ be the maximal order in $K$ ($K$ is an imaginary quadratic field), let $h_K (X)$ be the Hilbert class polynomial of $K$. Suppose that $\mathcal{O}$ is another order (say $\mathcal{O} =\mathbb{Z}[ \frac{1 + \sqrt{D}}{2}]$ and $\mathcal{O} = \mathbb{Z}[\sqrt{D}]$ for concreteness). Let $h_\mathcal{O} (X)$ be the hilbert class polynomial of the order $\mathcal{O}$. Is there any relation between $h_k(X)$ and $h_\mathcal{O} (X)$? For example can one obtain $h_\mathcal{O}(X)$ from $h_K(X)$ and vice verse?

share|improve this question
add comment

1 Answer

up vote 3 down vote accepted

As far as I know, the best relation between the two is the following: the field generated by the hilbert class polynomial $h_\mathcal{O} (X)$ contains the field generated by $h_K(X)$. This is implied by Proposition 25 of http://www.math.uga.edu/~pete/torspaper.pdf This implies among other things that $\deg(h_K(X)) | \deg(h_\mathcal{O}(X))$ (although this could be determined by simpler means).

Now as to your question about whether one can be generated from the other? No, unless you're in a very limited set of circumstances like $\deg(h_\mathcal{O}(X)) =1$ or such a thing. In fact it's a celebrated theorem of Heilbronn that $\deg(h_\mathcal{O}(X)) \to \infty$ as $|D| \to \infty$ where $D$ is the discriminant of $\mathcal{O}$.

share|improve this answer
add comment

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.