7
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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. $\endgroup$
    – Will Sawin
    Jul 23 at 19:15
  • $\begingroup$ That's exactly where I get stuck; I would like to know what extra information could finish off the proof. $\endgroup$
    – Rdrr
    Jul 23 at 19:50
  • 1
    $\begingroup$ 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. $\endgroup$
    – Will Sawin
    Jul 23 at 21:23
5
+300
$\begingroup$

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 $p$.

$\endgroup$
5
  • $\begingroup$ 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 $p$. $\endgroup$
    – Rdrr
    Jul 28 at 14:26
  • $\begingroup$ @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. $\endgroup$
    – Will Sawin
    Jul 28 at 17:45
  • $\begingroup$ I must be missing a basic algebraic number theory fact. How does one get a contradiction from a high power? $\endgroup$
    – Rdrr
    Jul 28 at 19:57
  • 1
    $\begingroup$ @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 $\mathfrak{p}$. $\endgroup$ Jul 29 at 3:28
  • $\begingroup$ Thank you so much Bjorn. This is a very insightful proof. $\endgroup$
    – Rdrr
    Jul 29 at 14:46
4
$\begingroup$

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 $K=Q(\alpha)$.
$\endgroup$
6
  • $\begingroup$ 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. $\endgroup$
    – Will Sawin
    Jul 24 at 14:54
  • $\begingroup$ 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. $\endgroup$ Jul 24 at 15:11
  • $\begingroup$ @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). $\endgroup$
    – Rdrr
    Jul 26 at 16:28
  • $\begingroup$ @Rdrr What do you mean by "limits" in this context? $\endgroup$
    – Will Sawin
    Jul 26 at 16:57
  • $\begingroup$ Limits in the categorical sense; the limits involved in the definitions of $\mathbb{Q}_p = \lim \mathbb{Z}/p^n$ and $T_p(E)= \lim E[p^n]$. $\endgroup$
    – Rdrr
    Jul 26 at 17:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.