# Find the canonical subgroup of a CM curve with ordinary reduction!

Let $p$ be a prime and $K$ a quadratic imaginary field in which $p$ splits. Let $\mathcal{O}$ be an order in $K$ and $A$ an elliptic curve with CM by $\mathcal{O}$. Then $A$ can be defined over the ring class field $H$ of the order $\mathcal{O}$. Assume there exists a prime $\mathfrak{p}$ of $H$ above $p$ at which $A$ has good reduction. Then this reduction is in fact ordinary, and thus $A$ has a distinguished subgroup $\Lambda$ of order $p$, the so-called canonical subgroup (see Chapter 3 of Katz in LNM350).

Can one give a different description of the isogeny $A\rightarrow A/\Lambda$? And what is the relation between ${\rm End}(A)$ and ${\rm End}(A/\Lambda)$?

1. I am thinking in an answer generalizing the following: suppose $A$ defined over $K$ of class number one, and $p=PQ$ with $P\neq Q$. Then $\Lambda=A[P]$, so $A\rightarrow A/\Lambda$ is induced by the complex multiplication of $P$ on $A$, and ${\rm End}(A/\Lambda)={\rm End}(A)$.
2. I pressume the answer in general depends on whether the conductor of $\mathcal{O}$ is divisible by $p$ or not.
Here's what worries me a little bit about this question. You say "$A$ can be defined over $H$", which is true, but of course there will probably be infinitely many non-isomorphic elliptic curves over $H$, all of which become isomorphic over the complexes. Which one are you choosing? Should I be worried about this? – Kevin Buzzard Apr 29 '11 at 15:29
Maybe I shouldn't be worried by this. My guess is that the answer will be that if $P$ is the prime of $K$ under your fixed prime of $H$, then the canonical subgroup will be $A[P]$. I think I can even sketch a proof of this when the conductor is coprime to $p$: canonical subgroups don't change under isogenies of degree prime to $p$ (because such isogenies induce isomorphisms on $p$-divisible groups and can. subgroups only depend on the underlying $p$-div group) so WLOG the order is maximal, and you want an isogeny lifting absolute Frob. Now Prop 10.4 of Silverman 2 gives you the isogeny... – Kevin Buzzard Apr 29 '11 at 15:41
I am especially interested in the case in which the conductor of the order is divisible by an arbitrary power of $p$... – monodromy Apr 30 '11 at 3:20