From: Adam Meredith Logan Newsgroups: sci.math.research Subject: Question about the Tate-Shafarevich Group Date: Tue, 31 Jan 95 21:21:31 EST Has anyone ever produced an elliptic curve (over the rationals, let's say, though I'd be interested in examples over number fields in general) which has nontrivial p-component of Sha for some prime greater than 3? Examples where this has been proved rigorously would be of greatest interest, but I'd very much like to see cases where this is predicted by the Birch- Swinnerton-Dyer conjecture or the Gross-Zagier formula as well. (Of course, the Gross-Zagier formula is a theorem, but deductions from it are based on the assumption of finite Sha, which, unless something remarkable has been done recently, is still unprovable for a general elliptic curve.) Thanks, Adam ============================================================================== From: elkies@ramanujan.harvard.edu (Noam Elkies) Newsgroups: sci.math.research Subject: Re: Question about the Tate-Shafarevich Group Date: 1 Feb 1995 17:23:14 GMT Summary: examples follow Keywords: elliptic curve, Sha In article <9502010221.AA25532@tucson.Princeton.EDU> Adam Meredith Logan writes: >Has anyone ever produced an elliptic curve (over the rationals, let's say, >though I'd be interested in examples over number fields in general) which >has nontrivial p-component of Sha for some prime greater than 3? >Examples where this has been proved rigorously would be of greatest interest, An e-mail message from John Cremona dated April 15, 1992 gives the following: SHA=25: 275B3 570L3 570L4 870I3 870I4 SHA=49: 546F2 858K2 The numbers refer to his table in _Algorithms for Modular Elliptic Curves_. For instance 275B3 is the curve y^2 + a1 x y + a3 y = x^3 + a2 x^2 + a4 x + a6 where [a1,a2,a3,a4,a6]=[0,1,1,-195508,-33338481], a twist of one of the curves of conductor 11; and 546F2 has [a1,a2,a3,a4,a6]=[1,0,0,-3674496,-2711401518]. These curves were originally found by using the conjectural analytic formula for Sha of Birch and Swinnerton-Dyer, but Dick Gross soon observed that the fact that SHA has 5- or 7-torsion in those cases follows from Cassels' theorem that the BSD Conjecture is compatible with isogenies: each of these curves has a 5- or 7-isogenous curve the product of whose c_p's is 5^2 or 7^2 times smaller. >but I'd very much like to see cases where this is predicted by the Birch- >Swinnerton-Dyer conjecture or the Gross-Zagier formula as well. The curves in Cremona's list (even with SHA=4,9,16 which I didn't reproduce above) all have rank zero. I have found some examples with rank 1 where the Gross-Zagier formula suggests |SHA|=25: the "congruent number" curves y+2 = x^3 - p^2 x where p = 12269, 24133, 26423, 28183, 31799, 32119, 39511, 40487. Note that here there is no 5-isogeny to explain these Sha's (though the curves are all 5-isogenous to themselves over Q(i) so it may be possible to produce explicit homogeneous spaces). To see how I went about computing these, see my paper on "Heegner point computations" in the proceedings of the first international symposium on algorithmic number theory (ANTS-I), edited by Adleman and Huang (Lecture Notes in Computer Science #877). --Noam D. Elkies (elkies@ramanujan.harvard.edu) Dept. of Mathematics, Harvard University ============================================================================== From: amlogan@phoenix.Princeton.EDU (Adam Meredith Logan) Date: Tue, 7 Feb 1995 01:36:45 EST To: rusin@math.niu.edu (Dave Rusin) Subject: Re: Question about Tate-Shafarevich group [ME-TOO] Following is the only reply I have received. } Subject: Re: Question about the Tate-Shafarevich Group } }I don't have a copy to hand, but I think that there is an }article by Nelson Stephens, possibly with Bryan Birch, in }"Applications of Modular Forms", conference proceedings }edited by Robert Rankin, published by Ellis & Horwood, }(with John Wiley) [I always think of it as "that ghastly }purple book"], in which he computes the size of Sha for }various twists of modular curves (maybe even complex }multiplication) and gets some powers of 5 and 7 in the }order. } Adam ============================================================================== Newsgroups: sci.math.research From: brock@ccr-p.ida.org (Bradley Brock) Subject: Nontrivial Sha Date: Tue, 21 Feb 1995 21:18:07 GMT Someone recently asked about Shas with torsion outside of the primes 2 and 3. Noam Elkies responded with some elliptic curves found by Cremona with |Sha|=25 and 49 from his exhaustive search of curves with small conductor. To this I can add that Brumer and McGuinness found a curve with |Sha|=289. See Bull. AMS (1990). -- (c) Copyright 1995 Bradley Brock, IDA/CCR-P, Thanet Road, Princeton, NJ 08540 brock@ccr-p.ida.org,brock@alumni.caltech.edu,609-279-6350(w),609-924-3061(fax) "Football exemplifies the worst features of American life: it's violence punctuated by committee meetings."--George Will ============================================================================== Newsgroups: sci.math.research From: elkies@ramanujan.math.harvard.edu (Noam Elkies) Subject: Re: Nontrivial Sha Date: Wed, 22 Feb 1995 03:36:29 GMT In article <3idlaf$pl0@runner.ccr-p.ida.org> brock@ccr-p.ida.org (Bradley Brock) writes: >Someone recently asked about Shas with torsion outside of the primes 2 and 3. >Noam Elkies responded with some elliptic curves found by Cremona with >|Sha|=25 and 49 from his exhaustive search of curves with small conductor. >To this I can add that Brumer and McGuinness found a curve with |Sha|=289. >See Bull. AMS (1990). How hard it is to find large Sha depends on where you look. B & McG did an exhaustive search for curve with prime conductor, and used the conjecture of Birch and Swinnerton-Dyer to surmise |Sha|; their curve with |Sha|=289 had rank zero (as do all the curve with large |Sha| from the Cremona tables). If you're content with rank 0 and constant j-invariant, probably the easiest thing is to look for curves y^2=Dx^3-x with |D| congruent to 1,2,3 mod 8; the work of Waldspurger, made explicit in this case by Tunnell, makes it easy to compute conjectural |Sha| values en masse, and the record |Sha| should grow roughly as the square root of |D|. Searching all |D|<10^6 congruent to 1 mod 8 for odd |Sha| values, I find the following records (where conjectural |Sha| is S^2): |D| = 32673, S = 17 |D| = 92873, S = 23 |D| = 132513, S = 25 |D| = 235673, S = 27 |D| = 269337, S = 29 |D| = 321513, S = 31 |D| = 376113, S = 33 |D| = 499073, S = 35 |D| = 629457, S = 47 The whole computation takes a fraction of a minute on a workstation. (That all of these have last digit 3 or 7, i.e. are quadratic nonresidues of 5, could probably be surmised from the local factor at 5 of the L-series; likewise quadratic residues of 13 and 17 predominate. There seems to be no preference modulo the primes 7 and 11, at which the local factor is the same unless |D| is a multiple, and even then doesn't change much.) >(c) Copyright 1995 Bradley Brock, IDA/CCR-P, Thanet Road, Princeton, NJ 08540 I think the above quote constitutes fair use... --Noam D. Elkies (elkies@math.harvard.edu) Dept. of Mathematics, Harvard University ============================================================================== Newsgroups: sci.math.research From: elkies@ramanujan.math.harvard.edu (Noam Elkies) Subject: Re: Nontrivial Sha Date: Wed, 22 Feb 1995 04:19:04 GMT [In a few more minutes I extended the search to |D|<10^7, still congruent to 1 mod 8. The further records are: |D| = 1451433: S = 51 |D| = 1470057: S = 55 |D| = 1518513: S = 57 |D| = 1523913: S = 59 |D| = 2529537: S = 63 |D| = 3360153: S = 81 |D| = 5381457: S = 95 |D| = 9943233: S = 101 Note that this means |Sha|=3^8 for |D|=3360153. Also noteworthy is S=49 for |D|=1452057, i.e. |Sha|=7^4. The Lenstra heuristics suggest that these Shas are still likely to be (Z/SZ)^2, though. Actually deciding between (say) Sha=(Z/49Z)^2 and Sha=(Z/7Z)^4 looks like a much harder problem. Carrying out such computations for |D| congruent to 2 or 3 mod 8, or for non-CM modular elliptic curves such as X0(11), is left as an exercise.] --Noam D. Elkies (elkies@math.harvard.edu) Dept. of Mathematics, Harvard University ============================================================================== Newsgroups: sci.math.research From: brock@ccr-p.ida.org (Bradley Brock) Subject: Re: Nontrivial Sha Date: Fri, 24 Feb 1995 22:14:20 GMT In article <3iebft$1fu@decaxp.harvard.edu> you write: >In article <3idlaf$pl0@runner.ccr-p.ida.org> >brock@ccr-p.ida.org (Bradley Brock) writes: > >If you're content with rank 0 and constant j-invariant, probably the >easiest thing is to look for curves y^2=Dx^3-x with |D| congruent to That should be y^2=D^2x^3-x or Dy^2=x^3-x. >1,2,3 mod 8; the work of Waldspurger, made explicit in this case by >Tunnell, makes it easy to compute conjectural |Sha| values en masse, >and the record |Sha| should grow roughly as the square root of |D|. >Searching all |D|<10^6 congruent to 1 mod 8 for odd |Sha| values, >I find the following records (where conjectural |Sha| is S^2): I noticed that 1 mod 8 Shas tended to be even and 3 mod 8 Shas tended to be odd. Any idea why? Consequently I find records at smaller values for 3 mod 8. D S 2707 13 8147 15 14083 19 15227 -21 >|D| = 32673, S = 17 >|D| = 92873, S = 23 >|D| = 132513, S = 25 >|D| = 235673, S = 27 >|D| = 269337, S = 29 >|D| = 321513, S = 31 >|D| = 376113, S = 33 >|D| = 499073, S = 35 >|D| = 629457, S = 47 > >The whole computation takes a fraction of a minute on a workstation. How did you program it? I tried finding solutions to D=(2x+1)^2+8y^2+2z^2 by running through y's and using factorization in Q(sqrt(-2)). Is there a more efficient way? Note: If A is the sum over all such solutions with 2x+1>0 of (-1)^y and B is the number of divisors of D, then S=A/B if A\ne0 and D is odd and squarefree. In this case it is easy to see that D must be 1 or 3 mod 8. -- (c) Copyright 1995 Bradley Brock, IDA/CCR-P, Thanet Road, Princeton, NJ 08540 brock@ccr-p.ida.org,brock@alumni.caltech.edu,609-279-6350(w),609-924-3061(fax) "Football exemplifies the worst features of American life: it's violence punctuated by committee meetings."--George Will ============================================================================== Newsgroups: sci.math.research From: elkies@abel.math.harvard.edu (Noam Elkies) Subject: Re: Nontrivial Sha Date: Sat, 25 Feb 1995 19:05:17 GMT In article <3illns$g7@wahoo.ccr-p.ida.org>, Bradley Brock wrote: >[I wrote:] >>If you're content with rank 0 and constant j-invariant, probably the >>easiest thing is to look for curves y^2=Dx^3-x with |D| congruent to >That should be y^2=D^2x^3-x or Dy^2=x^3-x. Right, the D belongs with the y. (Though in the Antwerp and Cremona tables both the x^3 and y^2 coefficients are 1 so the curve would appear as y^2=x^3-D^2x.) >>the work of Waldspurger, made explicit in this case by Tunnell, I neglected to give the reference: Tunnell, J.B.: A Classical Diophantine problem and Modular Forms of Weight 3/2 Invent. Math. 72 (1983), 323-334. >I noticed that 1 mod 8 Shas tended to be even and 3 mod 8 Shas tended >to be odd. Any idea why? Consequently I find records at smaller values >for 3 mod 8. The 2-torsion of the Selmer group (which is the same as the 2-torsion in Sha in the rank zero case, once you account for the 2-torsion points on the curve) can be obtained by a 2-descent; its dimension as a (Z/2)-vector space is the sum of the dimensions of the row and column nullspaces of a matrix made from Legendre symbols of the primes in 2*D. In particular the parity of the dimension is predictable, and turns out to be even iff |D| is 1,2,3 mod 8 -- same as the condition for the sign of the L-function of the curve to be +1. If Selmer group is to be accounted for by the 2-torsion point, with rank 0 and odd Sha, then the 2-descent matrix must be almost nonsingular -- i.e. its only row and column null vectors should come from the 2-torsion of the curve. Now I don't remember the details of the entries of the matrix (though it can be worked out in under an hour), but the class of |D| mod 8 determines where and whether there was a row or a column with Legendre symbols of 2. Otherwise, the columns are indexed by all the odd prime factors of D, and the rows by the prime factors congruent to 1 mod 4. Thus odd Sha can only happen if D has few or no prime factors congruent to 3 mod 4, and how little is "few" depends on the class of |D| mod 8. So it would make sense for one of these classes to have odd Sha occur more commonly, at least for "small" |D|. >>The whole computation takes a fraction of a minute on a workstation. >How did you program it? I tried finding solutions to >D=(2x+1)^2+8y^2+2z^2 by running through y's and using >factorization in Q(sqrt(-2)). Is there a more efficient way? Gulp. This takes about F*sqrt(D) steps where F is the cost of factoring. It's much simpler to just multiply the three q-expansions -- certainly for D Subject: Re: Nontrivial Sha Date: Mon, 27 Feb 1995 22:52:56 GMT Since the paper of Brumer and myself didn't contain the equation of the curve with Tate-Shafarevich group of size 289, here are some more details. (The full account of our calculations is still "in progress".) The output below was generated from what Brumer and I calculated for all our curves, just giving a pretty-print form. Cremona's program (recently advertised in this forum) would probably let you get all this information easily... -------------------------------------------------------------- The elliptic curve [1, 1, 0, -103324, -12826653] has equation: $y^2 + xy = x^3 + x^2 -103324 x -12826653$ The invariants are: b2 = 5, b4 = -206648, b6 = -51306612, b8 = -10739982241, c4 = 4959577, c6 = 11045031427. Discriminant is 4959593 Rank parity is setzer Number of components is 2 The two-division points are -186.001796, -186.000000, 370.751796 Real period is 0.266286 The analytic rank is: 0 The L-series value (or derivative) is: 19.239175 Dividing this by the period gives the predicted regulator: 288.999919 The point rank is: 0 The regulator is: 1.000000 The analytic and point ranks agree. The quotient of the L-value by the period and regulator is: 288.999919 Thus the predicted size of the Tate-Shafarevich group is: 289 No points were found. --------------------------------------------------------------- Some remarks: 1) By "Rank parity is setzer" I mean simply that the curve is a Setzer-Neumann curve: it has a 2-torsion point, the conductor has the form $u^2+64" where u = 17 * 131 here. These have rank 0. We had found it convenient to separate (apart from the few very small curves with non-trivial torsion) the curves of prime conductor into 3 classes: odd, even and setzer. 2) The L-series value was calculated with 2000 terms. 3) For other conventions, see our paper, exact reference is @Article{Br:Mc, Author = "Armand Brumer and Ois\'{\i}n McGuinness", Title = "The behavior of the {M}ordell-{W}eil group of elliptic curves", Journal = "Bulletin of the American Mathematical Society", Volume = "23", Series ="New Series", Year = "1990", Pages = "375--382"} (MR 91b:11076) ------------------------------------ Oisin McGuinness email: oisin@sbcm.com ============================================================================== Newsgroups: sci.math.research From: brock@ccr-p.ida.org (Bradley Brock) Subject: Re: Nontrivial Sha Date: Sat, 25 Feb 1995 17:57:48 GMT I wrote: >In article <3iebft$1fu@decaxp.harvard.edu> Noam Elkies writes: >> >>If you're content with rank 0 and constant j-invariant, probably the >>easiest thing is to look for curves y^2=Dx^3-x with |D| congruent to > >That should be y^2=D^2x^3-x or Dy^2=x^3-x. > >>1,2,3 mod 8; the work of Waldspurger, made explicit in this case by >>Tunnell, makes it easy to compute conjectural |Sha| values en masse, >>and the record |Sha| should grow roughly as the square root of |D|. >>Searching all |D|<10^6 congruent to 1 mod 8 for odd |Sha| values, >>I find the following records (where conjectural |Sha| is S^2): > >I noticed that 1 mod 8 Shas tended to be even and 3 mod 8 Shas tended >to be odd. Any idea why? Consequently I find records at smaller values >for 3 mod 8. Oops, this is fairly obvious because Sha is even if D is prime and 1 mod 8 and odd if D is prime and 3 mod 8. It is a matter of noting that (x,y,2z) is a solution to D=(2x+1)^2+8y^2+2z^2 iff (x,z,2y) is, which can occur iff D is 1 mod 8. In fact we have Sha is: odd for D=p_1, 2*q_1, p_1*p_2, 2*p_1*p_2, or 2*q_1*q_2 where p_i\equiv 3 mod 8 and q_i\equiv 5 mod 8 are primes, and even for D=r_1, 2*r_1, r_1*r_2, 2*r_1*r_2, q_1*q_2, s_1*s_2, or 2*s_1*s_2 where r_i\equiv 1 mod 8 and s_i\equiv 7 mod 8 are primes. > D S >14083 19 >15227 -21 34883 23 37963 27 39443 -31 And a few D\equiv 2 mod 8 23962 17 36178 18 >>|D| = 32673, S = 17 >>|D| = 92873, S = 23 >>|D| = 132513, S = 25 >>|D| = 235673, S = 27 >>|D| = 269337, S = 29 >>|D| = 321513, S = 31 -- Bradley Brock, IDA/CCR-P, Thanet Road, Princeton, NJ 08540 brock@ccr-p.ida.org,brock@alumni.caltech.edu,609-279-6350(w),609-924-3061(fax) "Football exemplifies the worst features of American life: it's violence punctuated by committee meetings."--George Will ============================================================================== Newsgroups: sci.math.research From: elkies@abel.math.harvard.edu (Noam Elkies) Subject: Re: Nontrivial Sha Date: Sat, 25 Feb 1995 22:51:41 GMT In article <3inr2s$11f@growler.ccr-p.ida.org>, Bradley Brock wrote: >I noticed that 1 mod 8 Shas tended to be even and 3 mod 8 Shas tended >to be odd. Any idea why? Consequently I find records at smaller >values for 3 mod 8. [later,] >Oops, this is fairly obvious because Sha is even if >D is prime and 1 mod 8 and odd if D is prime and 3 mod 8. >It is a matter of noting that (x,y,2z) is a solution >to D=(2x+1)^2+8y^2+2z^2 iff (x,z,2y) is, which can occur iff D is 1 mod 8. Whoa. You seem to be taking the Waldspurger-Tunnell formula for the normalized value S of L(E,1) as the *definition* of the (square root of the order of) the Tate-Shafarevich group! I'm not even sure that it has by now been proved that if that S is odd then Sha has trivial 2-torsion. Sure, if you take for granted the formula for S and this part of the conjecture of Birch and Swinnerton-Dyer then it is "fairly obvious" that the parity of |Sha| in these special cases can be obtained by counting representations of D by ternary quadratic forms. But that's a very big iceberg you're supporting yourself on, considering that "Sha is odd" simply means (in the rank-zero case) "the fact that Dy^2=x^3-x has no nontrivial solutions follows from the first 2-descent", i.e. from the methods of Fermat! That this is confirmed by Waldspurger-Tunnell assuming the 2-part of Birch-Swinnerton-Dyer is best regarded as only a nice sanity check on our arithmetic. >In fact we have Sha is: >odd for D=p_1, 2*q_1, p_1*p_2, 2*p_1*p_2, or 2*q_1*q_2 >where p_i\equiv 3 mod 8 and q_i\equiv 5 mod 8 are primes, and >even for D=r_1, 2*r_1, r_1*r_2, 2*r_1*r_2, q_1*q_2, s_1*s_2, or 2*s_1*s_2 >where r_i\equiv 1 mod 8 and s_i\equiv 7 mod 8 are primes. NB Assuming that the 2-descent and/or ternary form computations here are correct, this would be true only in the rank zero case; a curve in one of the families you list under "even" may actually have rank 2 and odd (usually trivial) Sha. [It will never have rank 1 unless the parity conjecture fails.] >And a few D\equiv 2 mod 8 >23962 17 >36178 18 18? I thought we were only looking at odd values. Anyway I also get |S|=17 for 23962, and furthermore (this time preserving the sign as you have done) D = 35018: S = 19 D = 53642: S = 23 D = 75178: S = -25 D = 82378: S = 29 D = 200378: S = -37 D = 250202: S = -39 D = 464218: S = -41 D = 491258: S = 43 D = 500218: S = 45 D = 686618: S = -47 D = 732538: S = -49 D = 880442: S = 51 D = 1129402: S = -59 D = 1167722: S = 63 D = 1376618: S = 65 D = 2159018: S = -67 D = 2490842: S = 75 D = 2507482: S = 85 D = 3412282: S = -91 D = 4042762: S = -95 D = 4741162: S = -99 D = 4880602: S = 101 D = 5093338: S = 109 D = 6989722: S = 117 D = 7004378: S =-125 D = 8285978: S =-127 > Bradley Brock, IDA/CCR-P, Thanet Road, Princeton, NJ 08540 What, no copyright notice anymore? :-) --Noam D. Elkies (elkies@ramanujan.math.harvard.edu) Dept. of Mathematics, Harvard University