From: stoll@math.uni-duesseldorf.de (Michael Stoll) Subject: Re: 3 descent on elliptic curve Date: 3 Apr 01 12:01:31 GMT Newsgroups: sci.math.numberthy Summary: How to carry out 3-descent on elliptic curves with 3-torsion Troy Kessler asks > Does anyone know how to do a 3 descent on a elliptic curve with Z3 torsion? > Is it as simple as silvermans algorithm to do a 2 descent on elliptic curve > with 2 or 4 torsion group? (given in rational points on elliptic curves) I > think one should be able to change his algorithm to work for y^2=x^3+16m^2 > with T={oo,(0,+-4m)}. I dont know what would take the place of his mod > squares homomorphism. m=19 with rank 2 would be a good example to > illustrate the algorithm Yes. -- Have a look at http://www.math.uni-duesseldorf.de/~stoll/papers/genus2/p-descent.dvi In this paper, Ed Schaefer and I describe how you can do a p-descent (p any odd prime [at least in principle]) on an elliptic curve, including p-isogeny descent and full p-descent. For p = 3, this is actually practicable (and for larger p, too, when you can use a p-isogeny). Regards, Michael Stoll ============================================================================== From: kesslert@surfree.com (Troy Kessler) Subject: More Questions on 3descent Thanks For Help Date: 12 Apr 01 18:49:01 GMT Newsgroups: sci.math.numberthy Thank you everyone for your help on 3-descent. I think I am beginning to understand 3-descent. The way I understand it one needs to find all curves of the form ax^3+by^3+cz^3=0 for a elliptic curve X^3+Y^3=abc. Then one needs to see if these abc curves are soluble much like the 2-descent in silvermans book. This was done by selmer in his papers. Do the abc curves form a group like the case of the 2-descent? Silverman constructs the alpha and alpha bar groups and uses this to calculate the rank in Rational Points on Elliptic Curves. The alpha set is a group with multiplication mod squares as the operation. For X^3+Y^3=489489, I found the following soluble abc curves (a and b are given and c=489489/(ab)) a b 1 13 1 21 1 77 1 273 1 489 3 11 3 143 7 33 7 163 11 39 13 21 13 77 33 91 Do these form a group? I think you would call it the selmer group??? I think you find the first part of the rank in this case by solving (3^g- 1)/2=13 and you would get g=3. Then you add 3 to the result of the second descent to get a rank of 5. How does one calculate the second descent? The paper by selmer is long and complicated.