# P-torsion of elliptic curves

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?

• 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 $\mathbb{Z}_p$. – Ari Shnidman Sep 11 '19 at 13:45
• @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? – Crocodile Sep 11 '19 at 14:04
• 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 $p$. – Ari Shnidman Sep 11 '19 at 15:21
• 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. – Lubin Sep 11 '19 at 22:32
• @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. – Crocodile Sep 12 '19 at 12:05

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.)