From s.donnelly@iu-bremen.de Mon May 31 14:41:32 2004 Return-Path: Received: from abel.math.harvard.edu (abel [140.247.28.153]) by modular.fas.harvard.edu (8.12.11/8.12.11/Debian-3) with ESMTP id i4VIg2Kl030137; Mon, 31 May 2004 14:42:02 -0400 Received: from merkur.iu-bremen.de (merkur.iu-bremen.de [212.201.44.27]) by abel.math.harvard.edu (8.11.7-20030923/8.11.7) with ESMTP id i4VIfZp03426 for ; Mon, 31 May 2004 14:41:35 -0400 (EDT) Received: from localhost (demetrius.iu-bremen.de [212.201.44.32]) by merkur.iu-bremen.de (Postfix) with ESMTP id DB10E9FA5E for ; Mon, 31 May 2004 20:41:33 +0200 (CEST) Received: from merkur.iu-bremen.de ([212.201.44.27]) by localhost (demetrius [212.201.44.32]) (amavisd-new, port 10024) with ESMTP id 10781-07 for ; Mon, 31 May 2004 20:41:32 +0200 (CEST) Received: from Donnelly (Donnelly.iuhb02.iu-bremen.de [212.201.48.249]) by merkur.iu-bremen.de (Postfix) with SMTP id D1FD99FA4E for ; Mon, 31 May 2004 20:41:32 +0200 (CEST) From: "Stephen Donnelly" To: Subject: mod-p representations Date: Mon, 31 May 2004 20:41:32 +0200 Message-ID: MIME-Version: 1.0 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: 8bit X-Priority: 3 (Normal) X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook IMO, Build 9.0.2416 (9.0.2911.0) Importance: Normal X-MimeOLE: Produced By Microsoft MimeOLE V5.50.4939.300 X-Virus-Scanned: by amavisd-new 20030616p5 at demetrius.iu-bremen.de Status: R X-Bogosity: No, tests=bogofilter, spamicity=0.000000, version=0.91.0 X-UID: X-Status: N X-KMail-EncryptionState: X-KMail-SignatureState: X-KMail-MDN-Sent: Hi, Michael showed me part of your email about mod-p Galois representations. A while ago I looked at what is actually required for Kolyvagin's method to apply at a prime p. It seems to apply when the curve admits no p-isogenies (details below). Since this is fairly easy to determine, there might be no need to use Serre's criteria for deciding when the representation is GL(p). Incidentally, your email gives a criterion that the rep'n is GL(p) for fixed p, which raises the question how one can, at one stroke, deal with all primes larger than some determined number. Serre posed this as a question in his paper (noting that it would follow from his proof if explicit Cebotarev was already known at the time). On the practical side, for a given elliptic curve it seems one can do this fairly easily. Serre also asks a uniform version of the question, in particular for elliptic curves over Q, whether the rep'n is GL(p) for all primes p>=19. Do we know the answer today? Back to Kolyvagin's method... According to Gross' account (which ignors the CM case) there are two places where the assumption on p-torsion is used: (1) In §4, in order for the construction to give cohomology classes in H^1(K,E[p]), one needs that the restriction map H^1(K,E[p]) -> H^1(H_n,E[p])^K is an isomorphism. Here H_n is the ring class field of conductor n. (2) In §9, in order to get one's hands on the Selmer group, one wants H^1( K(E[p])/K , E[p] )=0. This was already dealt with; in earlier email, someone explained that this fails only when there is a p-isogeny (in which case it's easier to do p-descent). I'll show that (1) is OK when E(K)[p]=0 for any p>3, and also for p=3 assuming that K is not Q(mu_3). It suffices to show H^1(H_n/K , E[p](H_n) ) = 0 and H^2(H_n/K , E[p](H_n) ) = 0. As in the earlier email, one can use the trick involving a nonidentity scalar. Proof: Gal(H_n/K) is abelian, so we may write it as a direct sum P+P', where P has p-power order and P' is prime to p. CLAIM: The subgroup of E[p](H_n) invariant under P' is trivial. To see this, let G be the quotient of Gal(H_n/K) giving the action on E[p](H_n). If #G is prime to p, we are done, since in that case P' maps onto G. Otherwise, #E[p](H_n)=p^2 and G contains the subgroup ( 1 * ) ( 0 1 ) in some basis. Since G is abelian, it is contained in the normaliser of this subgroup, which is the set of matrices of the form ( a b ) ( 0 a ) Moreover, G must contain an element with a\=1; indeed det(G) is the cyclotomic character and deg(K/Q)=2, so all the squares in F_p must be values of det(G), which gives us an a\=1 assuming p>3. By lifting our element a to P', we get an element which acts on E[p] as a nontrivial scalar, proving the claim for p>3. When p=3, the case #E[p](H_n)=p^2 is impossible: by assumption K is not Q(mu_3), which means that det(G) takes all nonzero values in F_p. But that is not possible if G has the above form. The subgroup P' < Gal(H_n/K) gives us the exact sequence 0 -> H^1( P , E[p](H_n)^P' ) -> H^1( H_n/K , E[p](H_n) ) -> H^1( P' , E[p](H_n) ) . The first H^1 vanishes because E[p](H_n)^P' = 0, and the last H^1 vanishes because the order of P' is prime to p. Therefore the middle H^1 vanishes, as required. Furthermore, since the last H^1 vanishes, we have the same sequence with H^2 in place of H^1. Again, all three terms vanish, for the same reasons as before. Remark: I guess we have a lot of freedom to choose K. So that if E(Q)[p]=0, we can choose K such that E(K)[p]=0, and also K \= Q(mu_3). steve