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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 ${\rm End}(A/\Lambda)$?

Comments:

  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.

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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
    
...lifting some power of absolute Frob and it should come out from that relatively easily, I think. Another approach, probably better, would be to work on the formal group and use some Lubin-Tate theory. It wouldn't surprise me if this were all standard. One doesn't need all this Katz stuff because one is in the ordinary situation. Katz' insight was that the story also worked in the supersingular reduction case, as long as the reduction was "only just" supersingular. –  Kevin Buzzard Apr 29 '11 at 15:44
    
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
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