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Suppose I have an ordinary elliptic curve $E$ over $\overline{\mathbb{F}}_p$. Then its $p$-torsion $E[p]$ is a finite flat group scheme of order $p^2$. My understanding is that it has $p+1$ subgroups of order $p$, and there is a "special" one, the canonical subgroup, which is the connected component of the identity in $E[p]$.

I've read that if $C$ a $p$-subgroup that is non-canonical, then the canonical subgroup of $E / C$ is $E[p] / C$. However, suppose I have two distinct non-canonical $p$-subgroups $C_1, C_2$. So $C_1$ has trivial intersection with $C_2$ and hence the map $$C_1 \hookrightarrow E[p] \twoheadrightarrow E[p]/C_2 $$ has trivial kernel. So doesn't that mean it's an isomorphism? Clearly this cannot be true since $C_1$ is etale and $E[p]/C_2$ is multiplicative, but what's going on?

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    $\begingroup$ It has two subgroups of order $p$, one etale and one multiplicative. The notion of canonical subgroups is usually reserved for lifts of $E$ to characteristic 0, e.g. for elliptic curves over $\mathbb{Z}_p$. $\endgroup$
    – Ari Shnidman
    Commented Sep 11, 2019 at 13:45
  • $\begingroup$ @AriShnidman Do you have a reference for that? I guess this means that if I take $E$ and the $C_i$ over $O_{\mathbb{C}_p}$, then I get a map $C_1 \to C_2$ which is an iso. on the generic fibre but trivial on the special fibre? $\endgroup$
    – Crocodile
    Commented Sep 11, 2019 at 14:04
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    $\begingroup$ I don't know an explicit reference but it follows quickly using arguments similar to yours. Ordinary means $E[p](\overline{\mathbb{F}}_p)$ has order $p$. Since the total rank is $p^2$, there is a unique connected subgroup of order $p$. Any other subgroup must be etale, hence must be the (unique) subgroup generated by any rational point of order $p$. $\endgroup$
    – Ari Shnidman
    Commented Sep 11, 2019 at 15:21
  • $\begingroup$ Well, my viewpoint is dreadfully old-fashioned and partial, but as @AriShnidman says, for $E$ over $\Bbb F_p$ or $\Bbb Z_p$, if ordinary, $E$ has just the one canonical subgroup; if supersingular, there is none. Things get interesting if $E$ is defined over a finite extension of $\Bbb Z_p$ and supersingular: there, there may or may not be a canonical subgroup, depending partly on issues of ramification. $\endgroup$
    – Lubin
    Commented Sep 11, 2019 at 22:32
  • $\begingroup$ @Lubin I wouldn't dream of opposing your viewpoint on this, but I don't think that's really the question -- the issue is how many p-subgroups there are other than the canonical one. $\endgroup$
    – Crocodile
    Commented Sep 12, 2019 at 12:05

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As mentioned by Ari Shnidman in the comments, an ordinary elliptic curve will only have $2$ subgroups of order $p$ over $\overline{\mathbf{F}}_p$. There is also a geometric manifestation of this fact: the map $\pi: X_0(p) \rightarrow X$ has degree $p+1$, but the special fibre of $X_0(p)$ consists of two copies of $X$ meeting transversally at the supersingular points. The corresponding projection maps have degree $1$ and $p$ respectively, with the latter map being purely inseparable. That's why the number of geometric (over $\overline{\mathbf{F}}_p$) points in the preimage is just $2$, rather than $p+1$ (which is what happens over $\overline{\mathbf{Q}}_p$, for example). In fact, this was known to Kronecker; if you take the classical modular equation $\Phi(x,y)$ relating $j(\tau)$ and $j(p \tau)$, then $\Phi(x,y)$ gives a model for $X_0(p)$, but there is Kronecker's congruence

$$\Phi(x,y) \equiv (x^p - y)(y^p - x) \mod p,$$

in which the geometric claims above are manifest. (Pertinent to this story here is the geometric definition of the Hecke operator at $p$; probably Diamond and Sherman talks about all of this in detail.)

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