From: rusin@math.niu.edu (Dave Rusin) Newsgroups: sci.math.research Subject: Re: Another question about elliptic curves Date: 19 Nov 1998 22:30:02 -0600 Hauke Reddmann wrote: > We all :-) know how to "double" a rational point > on a EC (draw the tangent at the point and compute > the intersection), but under which condition can > a point be "halved" You need only write treat "doubling" symbolically, computing the coordinates of 2P from P=(x0,y0); set the result equal to your particular (x,y) and solve for x0,y0. But in fact when the curve has 2-torsion of rank 2, that is, for curves > y^2=(x-x1)*(x-x2)*(x-x3) then it is necessary and sufficient that each of the three factors on the right be squares. > Example: > x1=(4/3)*(s^2+1) x2=(4/3)*(1-2*s^2) x3=(4/3)*(s^2-2) > y^2=(x-x1)*(x-x2)*(x-x3) > x4=(4/3)*(s^2+3s+1) y4=8s(s+1) > The point (x1,0) is the double of (x4,y4). > But (x2,0) can't be "halved" Careful: Do you mean this to be a single EC over a function field K(s), or a family of curves over a field K? In the former sense, you're right: x2-x3 = 4(1-s^2) isn't a square. But in the latter sense, you may be wrong for some s, such as s=(3/5)i. Certainly there are no rational s making your (x2,0) a double. If halved, it would give the EC a torsion subgroup isomorphic to Z/4 x Z/4, violating Mazur's theorem. Indeed, that theorem permits no supergroup of the Z/2 x Z/4 you already have as torsion subgroup, so none of your points can be divided by 3 (say) either. >BTW, can anyone find me another rational point on >this special EC, aside from those mentioned already? Again you need to specify the ground field. By specializing s we find that most small rational values of s give a curve of rank 0, so the torsion group you've found is all there is. We surmise the curve (probably) has rank 0 over the function field K(s). On the other hand, there are some s for which the rank is positive, e.g. if s=7 we have a point (-100/3, 960) of infinite order. (As usual, the _real_ question is, how on earth did you come to this surface?) dave ============================================================================== From: "Noam D. Elkies" Newsgroups: sci.math.research Subject: Tangent Re: Another question about elliptic curves Date: 20 Nov 1998 02:00:02 -0600 In article <7320ij$6q0$1@gannett.math.niu.edu>, Dave Rusin wrote: >Certainly there are no rational s making your (x2,0) a double. >If halved, it would give the EC a torsion subgroup isomorphic >to Z/4 x Z/4, violating Mazur's theorem. [...] Whoa -- that's a huge hammer you've just hit this fly with. Mazur's deep theorem is not needed here, only a very simple argument: the real locus of any elliptic curve is either a circle or the union of two circles, so clearly it's impossible for the curve to have full n-torsion even over R once n>2. In fact thanks to the Weil pairing we know that if an elliptic curve has full n-torsion over any field F then F must contain the n-th roots of unity. >Indeed, that theorem permits no supergroup of the Z/2 x Z/4 >you already have as torsion subgroup, That's not quite true: Z/2 x Z/8 is allowed, and does in fact occur on occasion (the corresponding modular curve is rational). --Noam D. Elkies (delete the Wiles-Taylor theorem from e-address to reply)