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Let $E/k$ be an elliptic curve over algebraically closed field of characteristic $p$ with CM, for simplicity, by the maximal order of a quadratic imaginary field $K/\mathbb{Q}$.

Suppose that $p$ is split in $K$ so that $E$ is ordinary, and write $p\cal{O}_K=\frak{p}\overline{\frak{p}}$. I would like to consider two subgroup schemes $E[\frak{p}]$ and $E[\overline{\frak{p}}]$ of $E[p]$ and ask the following two questions:

  • Is it true in general that $\text{rank}\ E[\frak{p}]$= $\text{rank}\ E[\overline{\frak{p}}]=p$ ? I think I have an argument that goes through lifting to characteristic zero and the Main Theorem of CM but probably there is a simpler way.

Secondly,

  • Is it true that $E[\frak{p}]$ and $E[\overline{\frak{p}}]$ can be seen as connected and étale parts (or viceversa) of $E[p]$? In the setting I have chosen for this question I don't see why this should be true (for example what distinguish $\frak{p}$ from $\overline{\frak{p}} $ ?) but it seems reasonable when I see $E/k$ as the reduction at the appropriate prime (above $\frak{p}$ or $\overline{\frak{p}} $) of a CM elliptic curve over a number field.

Thank you for your help.

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  • $\begingroup$ Hmm... let $\phi \in \mathrm{End}(E)$ be the Frobenius, and let $\hat{\phi} \in \mathrm{End}(E)$ be its dual. Then $\phi$ and $\hat{\phi}$ each have degree $p$, and clearly one of them lies in $\mathfrak{p}$ and the other in $\bar{\mathfrak{p}}$. This may be one way of seeing that the ranks of $E[\mathfrak{p}]$ and $E[\bar{\mathfrak{p}}]$ are $\leq p$ without lifting to characteristic $0$. $\endgroup$ May 13, 2016 at 16:07
  • $\begingroup$ As for "what distinguishes $\mathfrak{p}$ from $\bar{\mathfrak{p}}$", this probably has something to do with the choice of CM-type of a lifting of $E$ to $\tilde{E} / \mathbb{C}$; that is, a choice of embedding $\mathcal{O}_{\mathcal{K}} \hookrightarrow \mathbb{C}$. But at the moment I'm not sure of a full answer to your questions. $\endgroup$ May 13, 2016 at 16:23

1 Answer 1

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Formulating precisely this question helped me to figure it out (probably).

Fix an isomorphism $O_K\cong\text{End}(E)$ identifying complex conjugation with "taking the dual isogeny". Then we can write $\frak{p}=(p,\alpha)$, $\overline{\frak{p}}=(p,\overline{\alpha})$ (by the theory of Dedekind domains) where $\alpha$, $\overline{\alpha}$ are isogenies with $\alpha\circ \overline{\alpha}=\overline{\alpha}\circ\alpha=[p\cdot m]_E$ for some integer $m$. It follows that $E[\frak{p}]$$=E[\alpha]\times_E E[p]$ and $E[\overline{\frak{p}}]=E[\overline{\alpha}]\times_E E[p]$ finite flat group schemes.

The fact that distinguish between the two ideal is the action on the tangent space at the origin of $E$. Indeed $\underline{\omega}_{E/k}$ is one dimensional over $k$ and so the $O_K$-action is through an algebra homomorphism $\iota: O_K\to k$. Suppose that $\frak{p}=\ker\iota$ then $\overline{\alpha}$ is étale because induces an isomorphism between tangent spaces, hence $E[\overline{\alpha}]$ is étale over $\text{Spec}(k)$ and so also $E[\overline{\frak{p}}]$.

Finally, restricting the isogeny $\alpha$ to $E[p]$ gives an exact sequence of finite free group schemes $0\to E[\frak{p}]$$\to E[p]\overset{\alpha}{\to} E[\overline{\frak{p}}]\to 0$ because $p\nmid\alpha$ and $p\nmid\overline{\alpha}$. Moreover it is the connected-ètale sequence for $E[p]$ because $E[\overline{\frak{p}}]$ is ètale.

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