# reduction of CM elliptic curves

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?

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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

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.

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Yes, I'm interested. –  Timo Keller Mar 18 '10 at 15:59

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).

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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)

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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?

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