From: Dave Rusin Date: Thu, 9 Jul 1998 01:23:15 -0500 (CDT) To: crbd@alcor.cs.purdue.edu Subject: Re: rational parametrization of a plane curve Newsgroups: sci.math,sci.math.symbolic,sci.math.num-analysis In article <35A19D0A.C2CB9620@alcor.cs.purdue.edu> you write: >I am not an expert in algebraic geometry, but >I need to know if the plane curve below can be parametrized >by rational functions. [curve omitted] You're right, a curve needs to be of genus zero to admit a parameterization, so there isn't one for your curve. Most curves aren't parameterizable; so? On the other hand, you can improve the look of your curve a lot. (85%!) Experiments such as > factor(taylor(f,y,99)); show a clear pattern to the coefficients, so we decide to simplify, that is, we get a new curve of which the original is a 2-fold cover: > f2:=factor(simplify(f/x^6,{x^2-1=u*x})); #x drops out Putting the shoe on the other foot suggests a couple of minor simpifications (in this case a birational equivalences): > f3:=factor(subs(y=(u^2+4)/z,f2/(u^2+4)^3))*z^3; This doesn't look bad: > f4:=convert(factor(taylor(",z,4)),polynom); and a glance at the terms of heighest weight (weighing z twice as much as u) suggests this: > f5:=convert(factor(taylor(subs(z=w+u^2,f4),w,4)),polynom); 4 4 3 2 f5 := 16 u + (-3 u + 56 u + 664 u - 224 u + 16) w 2 2 3 + (110 u + 2296 u - 456) w + 1713 w that is, f5:=16*u^4+(-3*u^4+56*u^3+664*u^2-224*u+16)*w+(110*u^2+2296*u-456)*w^2+1713*w^3 This last curve is a cubic in w, with coefficients which are functions of u. Its only singular point is the point at infinity. It has genus 4. It's easily birationally equivalent to f2 (that is, each is as easily parameterized as the other) and its relationship to your f is clear: any parameterization of x leads to a parameterization of u; conversely one cannot go backwards unless one is willing to use a single square root in the parameterization. dave