most recent 30 from http://mathoverflow.net 2013-05-21T14:19:19Z http://mathoverflow.net/feeds/question/23295 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/23295/relation-between-the-hilbert-class-polynomial-of-mathcalo-k-and-an-order AP 2010-05-02T23:32:45Z 2010-05-03T12:25:00Z

<p>Hi all, </p> <p>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?</p>

http://mathoverflow.net/questions/23295/relation-between-the-hilbert-class-polynomial-of-mathcalo-k-and-an-order/23302#23302 stankewicz 2010-05-03T00:18:31Z 2010-05-03T12:25:00Z

<p>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 <a href="http://www.math.uga.edu/~pete/torspaper.pdf" rel="nofollow">http://www.math.uga.edu/~pete/torspaper.pdf</a> This implies among other things that $\deg(h_K(X)) | \deg(h_\mathcal{O}(X))$ (although this could be determined by simpler means).</p> <p>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 $\mathcal{O}$.</p>