From: kramsay@aol.commangled (Keith Ramsay)
Subject: Re: Elliptic curves with singularities...
Date: 19 Jan 2000 05:20:55 GMT
Newsgroups: sci.math
Summary: [missing]

In article <8630fe$nsc$1@panix2.panix.com>,
lrudolph@panix.com (Lee Rudolph) writes:
|The cusp curve (with the cusp removed, but its points at infinity 
|all present and accounted for) can be turned into the additive group 
|of the ground field, and the node curve (with similar provisos)
|can be turned into the multiplicative group. 
|
|I think.

Sure, although I think you meant to include the point at infinity.

   y^2*z - x^3                (1)

Consider the mapping t->(x,y,z) given by t->(t, 1, t^3). It gives a
parametrization of the curve (1) everywhere except the cusp; t=0 maps
to the point at infinity with homogeneous coordinates (0,1,0). A line
in the projective plane is represented by a plane in three dimensional
space through the origin, ax+by+cz=0, and the points of intersection
of such a plane where c<>0 correspond to the roots of the cubic
ct^3+at+b=0. The three points are regarded as adding up to zero in the
group law, and lo and behold they correspond to a triple of t values
which also adds to zero, because the sum of the three roots of the
cubic is minus the coefficient of t^2 (0) divided by the coefficient
of t^3. When c=0, we get a line in the projective plane passing through
the singular point. The line intersects the singular point twice in a
sense, and there's just one other root. That's a bit like "infinity+
anything=infinity". The line y=0 (the x-axis) intersects the curve with
multiplicity three at the cusp.

So the group law then (picking any point away from the cusp as the
origin) works out to be just the same as addition.

In the other case

   y^2*z - x^3 + ax^2*z       (2)

things are more complicated because of the way factors of sqrt(-a)
occur, so to keep it simple let me set a=-1. Then

   y^2*z = x^3 + x^2*z        (2')

can be parametrized by t->(4t(t-1), 4t(t+1), (t-1)^3), where the node
at (0,0,1) is approached by letting t go either to zero or to infinity.
A line is given by Ax+By+Cz=0 again, and the three associated values
of t satisfy a cubic of the form

   C(t-1)^3 + 4At(t-1) + 4Bt(t+1) = 0

i.e. a cubic where the coefficients of t^3 and 1 add to zero. This
means that the roots of the cubic multiply to 1. So in that case the
group law is the same as multiplication on (complex, say) nonzero
numbers.

Keith Ramsay