From: rusin@washington.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: Another Weierstrass Normal Form Needed Date: 8 Mar 1995 22:11:24 GMT In article <3ji37p$15dk@coyote.csusm.edu>, Randall Rathbun wrote: >The Weierstrass Normal Form for the elliptic curve given below is >needed. Please show intermediate steps also ^^^^^^^^^^^^^^^^^^^^^^^ A wise choice, since my chance of performing all the substitutions without error is pretty low. > 4a / 1 \ / 1 \ >(2) -- = | x - - | | y - - | > b \ x / \ y / > > a,b are constants, x,y are variables from Q (rationals) Well, there are high-falutin' ways to do this, I guess, but a pedestrian approach will get you what you want. I'd first write this as a polynomial (x^2 y^2) - (x^2+y^2) + 1 - (4a/b) (xy) = 0. Noticing the symmetry I'd write it all in terms of u=x+y and v=xy: v^2-(u^2-2v) + 1 - (4a/b)v=0, i.e. v^2-u^2+Av+1=0 where A = 2-4a/b. This change is not quite invertible: given u and v you find x and y as the two solutions to X^2 - u X + v = 0, but of course (x,y) and (y,x) map to the same pair (u,v). But this is perfect, since the usual description of an elliptic curve y^2=f(x) is precisely a presentation as a branched 2-sheeted covering of the affine plane. Now, the pairs (u,v) which result may not look like the plane, but they're on a conic, which is almost as good. Indeed, writing it in the form (v+ A/2)^2 - u^2 = (A/2)^2 - 1 = B, say, we may use the standard rationalization of a hyperbola: write v+A/2 = w + z, u = w-z, where w and z must satisfy wz = B/4 so that v = -A/2 + w + (B/4w), u = w - (B/4w), and we may recover our original points x and y=u-x=w-(B/4w)-x as the roots of x^2 - [w - B/4w]x + [-A/2+w+B/4w] = 0, i.e. (wx)^2 - (w^2-B/4)(wx) + (w^3-(A/2)w^2+(B/4)w)=0. With the substitution t=2wx - (w^2-B/4), we're almost there: we have t^2 = w^4 -4w^3 + (2A - B/2)w^2 - Bw + (B/4)^2 If you don't mind t^2=quartic, you're all set. Otherwise I guess the standard practice is to do a linear-fractional transformation on w moving one of the roots r0 of the quartic to infinity: let t = p(w-r0)^2, so that t^2=(w-r0)(w-r1)(w-r2)(w-r3) becomes p^2=[(w-r1)/(w-r0)].[(w-r2)/(w-r0)].[(w-r3)/(w-r0)] Since (w-r)/(w-r0) = 1 - (r-r0)/(w-r0), we can let q=1/(w-r0) and si=ri-r0 and see, finally, the equation p^2 = (1-s1 q)(1-s2 q)(1-s3 q), with the right side a cubic in q. (Incidentally, the polynomial g(w)=(w-s1)(w-s2)(w-s3) is just w^3 + (f''(r0)/2) w^2 + f'(r0) w + f(r0), where f(w) = (w-r1)(w-r2)(w-r3) = (the quartic)/(w-r0). ) Of course you can go further and replace p^2=Cq^3+... with (Cp)^2=(Cq)^3+... and then replace p^2=q^3+Dq^2+... with p^2=(q+D/3)^3 + ... To recap: given (x,y) satisfying the original equation, one computes (u,v) then (w,z) and t as above and gets a pair (t,w) satisfying t^2=quartic in w. The t and w are in fact cubic and quadratic polynomials in x and y respectively. Conversely, given a solution pair (t,w) with w not zero, we get a pair (x,w), from which we can compute z, then (u,v), and finally y = u-x; these x and y satisfy the original equation if not zero. They are rational functions of t and w. Maple indicates that the quartic does not factor rationally, that is, your choice of root r0 will not be a rational function of (4a/b). Best I could do was to write the quartic as [ ww^2 - (r^2+7r+2)/2 ]^2 - (10r-6r^2)ww - (4r^3-18r^2-10r), where ww = w-1 and r = a/b. dave