# Supersingular curves over $\mathbb{F}_q$ and the splitting of $p$

I'm looking at chapter 4 of Waterhouse's "Abelian varieties over finite fields"; and Theorems 4.1 and 4.2 seem to use the following fact:

Suppose that $$E/\mathbb{F}_q$$ is an elliptic curve over a finite field with $$q=p^n$$ elements and let $$\pi_E$$ denote the $$q$$th-power Frobenius map acting on $$E$$. Suppose that $$\pi_E$$ does not act like multiplication-by-$$N$$ for any integer $$N$$ so that $$\pi_E = [\alpha]_E$$ where $$\alpha$$ is a root of the polynomial $$x^2 - \mathrm{tr}(\pi_E)x + q$$ Then $$E$$ is ordinary if and only if $$p$$ is splits in $$\mathbb{Q}(\alpha)$$.

Waterhouse uses some very general theory (namely a theorem of Honda/Tate and the general theory of abelian varieties) and I find his reasoning/terminology vague. Does anybody know of a more down to earth proof for this result?

Lang has a proof utilitizing the fact that $$End(E)\to End(T_p(E))$$ is injective; but i'm trying to find proofs that don't use local methods

EDIT: Update on attempt: Here is a proof attempt; perhaps someone can help me finish it off.

We prove $$E$$ is supersingular if and only if $$p$$ is non-split in $$\mathbb{Q}(\alpha)$$.

First suppose that $$p$$ is non-split. If $$p$$ is inert, then since $$\alpha$$ has norm $$q = p^n$$, it follows that $$p$$ divides $$(\alpha)$$ and so $$E[p]\subseteq E[\pi_E] = \{O \}$$; thus $$E$$ is supersingular. If $$p$$ ramifies then $$p$$ divides the discriminant of $$\mathbb{Q}(\alpha)$$ and so $$p$$ divides $$\mathrm{tr}(\pi_E)^2 - 4q$$, which implies that $$p$$ divides $$\mathrm{tr}(\pi_E)$$ and so $$E$$ is supersingular;

Conversely, suppose that $$E$$ is supersingular. Suppose that $$p$$ splits as $$(p) = \mathfrak{p}\overline{\mathfrak{p}}$$. Then $$(\alpha) = \mathfrak{p}^a\overline{\mathfrak{p}}^b$$ where $$a+b=n$$. However, since $$E$$ is supersingular $$\alpha^N \in \mathbb{Z}$$ for some positive integer $$N$$. Which implies that $$(\alpha)^N = \mathfrak{p}^{Na}\overline{\mathfrak{p}}^{Nb}$$ and so $$Na=Nb$$ since $$(\alpha^N) = \overline{(\alpha^N)}$$. Therefore $$a=b$$ and so $$(\alpha) = \mathfrak{p}^a \overline{\mathfrak{p}}^a = (p^a)$$ where $$a=n/2$$. This implies that $$\alpha = \zeta p^{a}$$ for some unit $$\zeta$$. This is where I am stuck.

Any help would be appreciated.

• I don't think it's possible to finish the argument from where you are because you need to rule out polynomials like $x^2 - \sqrt{q} x + q$ for $p$ congruent to $1$ mod $3$ and $q$ a square, but roots $\alpha$ of these polynomials satisfy the condition $\alpha^N \in \mathbb Z$. So you need more information about elliptic curves. Jul 23 at 19:15
• That's exactly where I get stuck; I would like to know what extra information could finish off the proof.
– Rdrr
Jul 23 at 19:50
• The way I see how to do it is to observe that the endomorphism algebra is a quaternion algebra, and (by the $\ell$-adic Tate module) split at each prime $\ell$ not $p$. If $\mathbb Q(\alpha)$ is split at $p$ then it is split everywhere, hence a matrix algebra, thus contains nilpotents, which is absurd. Is this the argument of Lang you mention? I don't see a better way. Jul 23 at 21:23

Here is a solution that avoids explicit use of $$\mathbf{Q}_p$$ and in particular does not require knowing that $$(\operatorname{End} E) \otimes \mathbf{Q}_p$$ is a division ring. The key is to use inseparable degree of endomorphisms.

Identify $$\alpha$$ with $$\pi_E$$. Let $$a = \operatorname{tr} \alpha \in \mathbf{Z}$$. Let $$\mathcal{O}$$ be the ring of integers of $$\mathbf{Q}(\alpha)$$. Let $$\mathcal{O}' := (\operatorname{End} E) \cap \mathbf{Q}(\alpha)$$. For $$\beta \in \mathcal{O}'$$, the inseparable degree $$\deg_{\text{i}} \beta$$ is multiplicative in $$\beta$$. The $$\beta \in \mathcal{O}'$$ with $$\deg_{\text{i}} \beta \ge p^m$$ are the $$\beta$$ that factor through the $$p^m$$-power Frobenius morphism $$F_m \colon E \to E^{(p^m)}$$; they form an additive subgroup. Thus $$\beta \mapsto \log_p \deg_{\text{i}} \beta$$ defines a valuation $$v \colon \mathcal{O}' \to \mathbf{Z} \cup \{\infty\}$$. Extend $$v$$ to the fraction field $$\mathbf{Q}(\alpha)$$.

Case 1: $$E$$ is ordinary. Then $$p \nmid a$$, so $$x^2-ax+q$$ mod $$p$$ has two distinct factors, so $$p$$ splits in $$\mathbf{Q}(\alpha)$$.

Case 2: $$E$$ is supersingular. Then $$p \mid a$$, so the equality $$\alpha^2 - a \alpha + q = 0$$ yields $$\alpha^2 \in p \mathcal{O}$$.

First consider the case $$\mathcal{O}'=\mathcal{O}$$. If $$\beta \in \mathcal{O}$$ satisfies $$v(\beta) \ge 2n$$, then $$\beta$$ factors through $$F_{2n} = \alpha^2 \in p \mathcal{O}$$, so $$\beta \in p \mathcal{O}$$. This implication for all $$\beta$$ shows that $$v$$ is the unique place above $$p$$.

In general, write $$(\mathcal{O}:\mathcal{O}')=p^e d$$ with $$p \nmid d$$. If $$\beta \in \mathcal{O}$$ satisfies $$v(\beta) \ge (e+1)2n$$, then the endomorphism $$(p^e d) \beta$$ is divisible (in $$\mathcal{O}'$$ or in $$\mathcal{O}$$) by $$\alpha^{2(e+1)}$$ and hence by $$p^{e+1}$$, but $$p \nmid d$$, so $$\beta \in p\mathcal{O}$$. Again, this shows that $$v$$ is the unique place above $$p$$.

• This is a wonderful proof. If you wouldn't mind I'd like some clarification on why $v(\beta)\geq 2n$ implies that $v$ is the only place above $p$.
– Rdrr
Jul 28 at 14:26
• @Rdrr If $v$ is not the only place above $p$, choose some element which is $0$ modulo $v$ but not zero modulo the other place above $p$, and then take a high power, for a contradiction. Jul 28 at 17:45
• I must be missing a basic algebraic number theory fact. How does one get a contradiction from a high power?
– Rdrr
Jul 28 at 19:57
• @Rdrr: Here is another way to say it. Let $\mathfrak{p}$ be the prime corresponding to $v$. The implication $v(\beta) \ge 2n \implies \beta \in (p)$ is equivalent to $\mathfrak{p}^{2n} \subset (p)$, which is the same as saying that $(p)$ divides $\mathfrak{p}^{2n}$. In this case, the factorization of the $\mathcal{O}$-ideal $(p)$ cannot involve any primes other than $\mathfrak{p}$. Jul 29 at 3:28
• Thank you so much Bjorn. This is a very insightful proof.
– Rdrr
Jul 29 at 14:46

Here is a proof that does not use the Honda-Tate theory.

Notice that the quadratic field $$K=Q(\alpha)$$ may be viewed as the subfield of the endomorphism algebra $$End^0(E)=End(E)\otimes Q$$ with the same identity element. Here $$End(E)$$ is the algebra of ALL endomorphisms of $$E$$ over an algebraic closure of $$F_q$$.

1. Suppose that $$E$$ is ordinary and $$T_p(E)$$ is its physical $$p$$-adic Tate module, which is a free $$Z_p$$-module of rank 1. Let us consider the corresponding $$Q_p$$-vector space $$V_p(E)=T_p(E)\otimes_{Z_p}Q_p$$, which is a one-dimensional vector space over the field $$Q_p$$ of $$p$$-adic numbers.

By functoriality, there is the natural $$Q_p$$-algebra homomomorphism

$$K_p=K\otimes_Q Q_p \to End_{Q_p}(V_p(E))=Q_p,$$ which sends $$1$$ to $$1$$ and therefore is not zero. Since $$K_p$$ has $$Q_p$$-dimension $$2>1$$, this homomorphism is not injective and therefore $$K_p$$ is NOT a field, i.e., $$p$$ splits in $$K=Q(\alpha)$$.

1. Suppose that $$E$$ is supersingular. Then $$End(E)\otimes Q_p=End^0(E)\otimes_Q Q_p$$ is a division algebra over $$Q_p$$ of dimension 4 that contains $$K_p=K\otimes_Q Q_p$$ as a $$Q_p$$-subalgebra. Since $$End^0(E)\otimes_Q Q_p$$ has no zero divisors, $$K_p$$ also has no zero divisors, i.e., $$p$$ does NOT split in $$K=Q(\alpha)$$.
• How do you check that the endomorphism algebra in the supersingular case is a division algebra over $\mathbb Q_p$? I sketched one way in the comments, but it's not (to me) obvious. Jul 24 at 14:54
• Your suggestion works just fine. I may only add that since $E$ is a simple abelian variety, its endomorphism algebra has no zero divisors and therefore cannot be isomorphic to the matrix algebra of size 2 over the rationals. Jul 24 at 15:11
• @WillSawin This is essentially the proof in Lang's Elliptic functions. I was hoping for a proof without limits, but one may not exists. This seems to be a good way to link the characteristic zero info (the splitting of $p$) with the mod p information (the p-torsion).
– Rdrr
Jul 26 at 16:28
• @Rdrr What do you mean by "limits" in this context? Jul 26 at 16:57
• Limits in the categorical sense; the limits involved in the definitions of $\mathbb{Q}_p = \lim \mathbb{Z}/p^n$ and $T_p(E)= \lim E[p^n]$.
– Rdrr
Jul 26 at 17:01