reduction of CM elliptic curves - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T18:37:35Z http://mathoverflow.net/feeds/question/18591 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/18591/reduction-of-cm-elliptic-curves reduction of CM elliptic curves norondion 2010-03-18T14:37:13Z 2012-02-09T17:28:53Z <p>Can someone indicate how to prove the following equivalences for a CM elliptic curve $E$:</p> <p>(i) $p$ is inert in End($E$) (ii) $E_p$ is supersingular (iii) The trace of the Frobenius at $p$ is $0$ [correction: $\equiv 0 \pmod{p}$]</p> <p>Also, are there generalisations of this to abelian varieties?</p> http://mathoverflow.net/questions/18591/reduction-of-cm-elliptic-curves/18598#18598 Answer by Sam Derbyshire for reduction of CM elliptic curves Sam Derbyshire 2010-03-18T15:07:01Z 2010-03-18T15:17:55Z <p>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.</p> <p>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 <em>Elliptic Curves</em> by Husemoller, including a proof of the converse.</p> <p>From this, we deduce that multiplication by $p$ is purely inseparable on a supersingular elliptic curve, and hence there is no $p$-torsion.</p> <p>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 $q=p$, $p \geqslant 5$ 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.</p> <p>For the question about reduction modulo $p$, the full criterion of Deuring is as follows:</p> <p><strong>Theorem</strong>: Let $L$ be a number field, and $E$ an elliptic curve with complex multiplication by an order in the imaginary quadratic field $K$.<br> Let $p$ be a (rational) prime, and $P$ a prime above $p$ at which $E$ has good reduction.<br> 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 $c_0$. </p> <p>I have taken this from Lang's book <em>Elliptic Functions</em>, which contains a proof (page 182 in my edition). </p> http://mathoverflow.net/questions/18591/reduction-of-cm-elliptic-curves/18601#18601 Answer by Pete L. Clark for reduction of CM elliptic curves Pete L. Clark 2010-03-18T15:20:55Z 2010-03-18T15:20:55Z <p>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 <strong>Deuring's Criterion</strong>) 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.</p> <p>For a bit on Honda-Tate theory, see e.g. the appendix of my PhD thesis:</p> <p><a href="http://math.uga.edu/~pete/thesis.pdf" rel="nofollow">http://math.uga.edu/~pete/thesis.pdf</a></p> <p>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.</p> http://mathoverflow.net/questions/18591/reduction-of-cm-elliptic-curves/18613#18613 Answer by norondion for reduction of CM elliptic curves norondion 2010-03-18T16:06:04Z 2010-03-18T16:06:04Z <p>I came up with the following partial proof:</p> <p>(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 $End(E)$.</p> <p>(ii) &lt;=> $\hat{Frob_p}$ purely inseparable &lt;=> $Tr(Frob_p) = Frob_p + \hat{Frob_p}$ purely inseparable &lt;=> $Tr(Frob_p) \equiv 0 \pmod{p}$.</p> <p>Can someone complete this?</p> http://mathoverflow.net/questions/18591/reduction-of-cm-elliptic-curves/88016#88016 Answer by Tommaso Centeleghe for reduction of CM elliptic curves Tommaso Centeleghe 2012-02-09T17:28:53Z 2012-02-09T17:28:53Z <p>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 $1-\pi_E$).</p> <p>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).</p> <p>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 $F$.</p> <p>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!</p> <p>(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)</p>