# Supersingular elliptic curves and their “functorial” structure over F_p^2

In a letter to Tate from 1987, Serre describes a beautiful Theorem relating mod p modular forms to quaternions ("Two letters on quaternions and modular forms (mod p)", Israel J. Math. 95 (1996), 281--299). At the beginning of Remark (4) on page 284, Serre says that every supersingular elliptic curve E over \bar F_p has a "canonical and functorial" structure E_0 over the finite field F with p^2 elements. He further says (if I understand it correctly) that such F-structure is determined by the condition that the relative Frobenius endomorphism of E_0 is equal to -p in the endomorphism ring of E_0 over F.

My question is: does anyone have a detailed reference for the proof of this fact, please? Also, can one find an F-structure of a given supersingular elliptic curve E where the relative Frobenius acts as multiplication by p (instead of -p)? Thanks

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(EDIT: I've rewritten my argument in terms of the inverse functor, i.e., base extension, since it is clearer and more natural this way.)

Much of what is below is simply a reorganization of what Robin Chapman wrote.

Theorem: For each prime $p$, the base extension functor from the category $\mathcal{C}_{-p}$ of elliptic curves over $\mathbf{F}_{p^2}$ on which the $p^2$-Frobenius endomorphism acts as $-p$ to the category of supersingular elliptic curves over $\overline{\mathbf{F}}_p$ is an equivalence of categories.

Proof: To show that the functor is an equivalence of categories, it suffices to show that the functor is full, faithful, and essentially surjective. It is faithful (trivially), and full (because homomorphisms between base extensions of elliptic curves in $\mathcal{C}_{-p}$ automatically respect the Frobenius on each side). Essential surjectivity follows from Lemma 3.2.1 in

which is proved by constructing a model for one curve and getting models for the others via separable isogenies. $\square$

The same holds for the category $\mathcal{C}_p$ defined analogously, but with Frobenius acting as $+p$. Here are two approaches for proving essential surjectivity for $\mathcal{C}_p$:

1) If $G:=\operatorname{Gal}(\overline{\mathbf{F}}_p/\mathbf{F}_{p^2})$ and $E$ is an elliptic curve over $\mathbf{F}_{p^2}$, and $\overline{E}$ is its base extension to $\overline{\mathbf{F}}_{p^2}$, then the image of the nontrivial element under $H^1(G,\{\pm 1\}) \to H^1(G,\operatorname{Aut} \overline{E})$ gives the quadratic twist of $E$ (even when $p$ is $2$ or $3$, and even when $j$ is $0$ or $1728$). Applying this to each $E$ with Frobenius $-p$ gives the corresponding elliptic curve with Frobenius $+p$.

2) Use Honda-Tate theory (actually, it goes back to Deuring in this case) to find one supersingular elliptic curve over $\mathbf{F}_{p^2}$ with Frobenius $+p$, and then repeat the proof of Lemma 3.2.1 to construct the models of all other supersingular elliptic curves via separable isogenies.

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Thank you very much for providing the answer! – Tommaso Centeleghe Mar 22 '10 at 15:07

I'm not sure about what "functorial" would entail here, but at least when $p\ge5$ from a naive point of view things are quite simple. Once one knows that the $j$-invariant of a supersingular elliptic curve $E$ lies in $k=\mathbb{F}_{p^2}$ then there is a curve $E'$ defined over $k$ with the same $j$-invariant as $E$. Up to $k$-isomorphism there are two candidates for $E'$ but they are quadratic twists: one has $(p+1)^2$ points over $k$ and the other has $(p-1)^2$ points. Equivalently the $k$-Frobenius acts on one as $-p$ and the other as $+p$.

Let's pick an isomorphism $\alpha:E\to E'$ for each supersingular curve where $E'$ is defined over $k$ with Frobenius $-p$. Given an isogeny $\phi:E_1\to E_2$ then there is a corresponding isogeny $\phi':E_1'\to E_2'$ making the obvious square commute. This isogeny $\phi'$ is defined over $k$, because it commutes with the $k$-Frobenius which on both sides equal $-p$.

We could proceed in exactly the same way taking each $E'$ to have Frobenus $+p$ and come to the same conclusion. In some sense though, choosing $-p$ is more natural. If $E$ has $j$-invariant in $\mathbb{F}_p$ then $E'$ will be defined over $\mathbb{F}_p$ if we take the $-p$ option but not the $+p$ option.

Of course in characteristic $2$ and $3$ things are different, but in each case there is only one supersingular $j$-invariant.

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Thank you for your answer. Even though I do not see a complete proof of the existence of this k-structure I was asking. Given the fact that there exists a model E' of E over k, how do you show that you can pick E' so that the relative Frobenius is -p? – Tommaso Centeleghe Mar 22 '10 at 11:37
The proof is in Lang's Elliptic Functions. Briefly speaking, given a supersingular $E$ defined over $\mathbb{F}_{p^2}$ the Frobenius endomorphism and $p$ are both totally inseparable and so differ by a factor of an automorphism. If $p\ge 5$ the only automorphism is $\pm1$ (according to Lang who doesn't quite convince we :-( ). So the Frobenius is $+p$ or $-p$ and that of the quadratic twist is the other. – Robin Chapman Mar 22 '10 at 12:12
I now think Lang is wrong, since $E$ could have an automorphism of order $3$, $4$ or $6$. But then $E$ has $j$-invariant $0$ or $1728$ and so has a model over $\mathbb{F}_p$. Over $\mathbb{F}_{p^2}$ this model will have $(p+1)^2$ points. – Robin Chapman Mar 22 '10 at 12:19
I agree with you, Lang contains a mistake. Thanks for having thought about the question. – Tommaso Centeleghe Mar 22 '10 at 15:05
I thought I would give the precise reference being discussed above: the error pointed out by Robin is in the corollary following Theorem 6 on page 176 of the second edition of Lang, Elliptic functions, and the error in the proof is where he says "it follows from Appendix 1 that the only automorphisms of A are...". – Bjorn Poonen Mar 22 '10 at 18:18

I think Robin Chapman is right about the Corollary in the Lang's book. For example as we know for Deuring (see Waterhouse "Abelian varieties over finite fields" Chapter 4 Theorem 4.1) there are elliptic curves over F_{p^2} (if p not = 1 mod 3) such that Frob_{p^2}+-p Frob_{p^2}+p^2=0, i. e. Frob_{p^2} does not equal p or -p.

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Yes, and when $p \equiv 3$ (mod 4) there are $a\in\mathbb{F}_{p^2}$ such that the Frobenius of the elliptic curve $y^2=x^3+ax$ has $F^2=-p^2$ (and so the curve has $p^2+1$ points over $\mathbb{F}_{p^2}$). – Robin Chapman Mar 24 '10 at 14:54