Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Can someone indicate how to prove the following equivalences for a CM elliptic curve $E$:

(i) $p$ is inert in End($E$) (ii) $E_p$ is supersingular (iii) The trace of the Frobenius at $p$ is $0$ [correction: $\equiv 0 \pmod{p}$]

Also, are there generalisations of this to abelian varieties?

share|improve this question
3  
Warning: the third condition is not quite correct. Rather, you want the trace of Frobenius to be divisible by p. (This comes up e.g. for reduction modulo 2 and 3 for elliptic curves $E/\mathbb{Q}$.) –  Pete L. Clark Mar 18 '10 at 15:10
    
I think i) should allow p to ramifie in End(E): the important thing is that p not be split –  Tommaso Centeleghe Feb 9 '12 at 17:34
add comment

4 Answers 4

A nice modern perspective on this is to use the Honda-Tate classification of endomorphism algebras of abelian varieties over a finite field. This allows you to recover the result in question (which, as Sam Derbyshire points out, is usually called Deuring's Criterion) as well as craft generalizations to certain higher-dimensional abelian varieties. For instance, the generalization which was of most use to me in my thesis work was that if $A_{/\mathbb{Q}}$ is an abelian surface whose endomorphism algebra is a nonsplit indefinite quaternion algebra $B_{/\mathbb{Q}}$ (or a "QM abelian surface") then the reduction modulo $p$ is always isogenous to a square of an elliptic curve.

For a bit on Honda-Tate theory, see e.g. the appendix of my PhD thesis:

http://math.uga.edu/~pete/thesis.pdf

The exposition here is pretty clunky by my present day standards, and it does not directly address the Deuring Criterion. However I need to run along to class now. I can provide more details later on today if you like; please let me know.

share|improve this answer
    
Yes, I'm interested. –  Timo Keller Mar 18 '10 at 15:59
add comment

I'm not sure what definition of supersingular you're taking; I'll assume you mean that the endomorphism ring is an order in a quaternion algebra.

Now, suppose $E$ is a supersingular elliptic curve over $\mathbb{F}_{q}$ of characteristic $p$, and $\varphi$ is the Frobenius endomorphism. You can deduce from noncommutativity of the endomorphism ring (over the algebraic closure) that $[p^n] = \varphi^m$ for some $m,n \in \mathbb{Z}$. The argument goes along the lines of: if $[p^n]$ is never equal to $\varphi^m$, this forces the endomorphism ring to commute as every endomorphism commutes with some power of the Frobenius. You can find more details in Elliptic Curves by Husemoller, including a proof of the converse.

From this, we deduce that multiplication by $p$ is purely inseparable on a supersingular elliptic curve, and hence there is no $p$-torsion.

We can also deduce the following, by degrees of endomorphisms. Let $\alpha,\beta$ be two endomorphisms, then define $\langle \alpha, \beta \rangle = \frac{1}{2} \left( \deg(\alpha + \beta ) - \deg(\alpha) - \deg(\beta) \right)$, this defines an inner product on the endomorphism ring. Using the above property that $[p^n] = \varphi^m$, some basic algebra shows that $p \mid t$, where $t$ is the trace of the Frobenius, which is equal to $\langle \varphi, 1 \rangle = \#E(\mathbb{F}_q) - (q+1)$. It is however not necessarily the case that $t=0$; this happens for $q=p$, $p \geq 5$ by Hasse's inequality, but there are examples in which $t \neq 0$. Going the other way, we see that $\#E(\mathbb{F}_q) \equiv 1 \pmod{p}$, so by elementary group theory $E(\mathbb{F}_q)$ has no $p$-torsion, which is equivalent to supersingularity as shown above.

For the question about reduction modulo $p$, the full criterion of Deuring is as follows:

Theorem: Let $L$ be a number field, and $E$ an elliptic curve with complex multiplication by an order in the imaginary quadratic field $K$.
Let $p$ be a (rational) prime, and $P$ a prime above $p$ at which $E$ has good reduction.
Then $E$ has supersingular reduction at $P$ iff there is a unique prime of $K$ above $p$. Otherwise, write $c$ for the conductor of the endomorphism ring of $E$ in $K$, and let $c = c_0 p^k$ with $p \nmid c_0$. Then the ring of endomorphisms of the reduction mod $p$ is $\mathbb{Z} + c_0 \mathcal{O}_K$, the order of $K$ with conductor $c_0$.

I have taken this from Lang's book Elliptic Functions, which contains a proof (page 182 in my edition).

share|improve this answer
add comment

Let $k$ be a finite field of char. $p$ and $E$ an elliptic curve over $k$. Denote by $R_E$ the endomorphism ring of $E$ over $k$. This ring has a distinguished element given by the purely inseparable $k$-isogeny $\pi_E:E\rightarrow E$ given by the Frobenius endomorphism relative to $k$. Its degree is $|k|$. We know that $\pi_E$ satisfies the Weil polynomial $f(x)=x^2-a_Ex+|k|$, where $a_E$ is the error term $|k|+1-|E(k)|$ (it is not hard to deduce this from the fact that $|E(k)|$ is equal to the degree of the separable $k$-isogeny $1-\pi_E$).

Assume now that $F$ is an imaginary quadratic field that embeds inside $R_E\otimes\mathbf{Q}$. If $E$ is supersingular, then there is a finite extension $k'$ of $k$ such that $R_E':={\rm End}_{k'}(E\otimes_k k')$ becomes a maximal order of "the" quaternion over $\mathbf{Q}$ ramified precisely at $p$ and infinity (in fact $k'$ of degree $2$ suffices "most times"). Considering the embedding $F\rightarrow R_E\otimes\mathbf{Q}\subset R_E'\otimes\mathbf{Q}$ you see that $F$ cannot be split at $p$ (and at infinity) for otherwise it would not embed in such a quaternion algebra (Vigneras' book has all of this basics).

If $E$ is ordinary then $R_E\otimes\mathbf{Q}$ is isomorphic to the quadratic field obtained by joining to $\mathbf{Q}$ a root of $f(x)$, therefore so is $F$ and since $p\nmid a_E$ we see easily that the discriminant of $f(x)$ is not divisible by $p$ and it is a square mod $p$, thus $p$ splits in $F$.

Back to your original question, the reduction mod $p$ of a CM ellipitic curve induce a nice, injective reduction map at the endomorphisms level. Therefore you should get what you wanted!

(Waterhouse thesis "Abelian varieties over finite fields" is a great source to learn these things. The Bourbaki talk by Tate on abelian varieties over finite fields is a must to learn these things)

share|improve this answer
add comment

I came up with the following partial proof:

(i) => (ii) $End(E) \otimes \mathbf{Q} \hookrightarrow End(E_p) \otimes \mathbf{Q}$ is not surjective since $Frob_p$ is not in the image. For, otherwise, we would have $N(Frob_p) = p$, so $p$ would be split in $End(E)$.

(ii) <=> $\hat{Frob_p}$ purely inseparable <=> $Tr(Frob_p) = Frob_p + \hat{Frob_p}$ purely inseparable <=> $Tr(Frob_p) \equiv 0 \pmod{p}$.

Can someone complete this?

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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