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)