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§8. The Elliptic Curve Group Law 21

π
B = −A
O

A

T(O)

a group law. The zero point corresponded to a flex point of which the tangent line
has a triple intersection point.
There is another classical picture of an elliptic curve as the smooth intersection
of two quadric hypersurfaces in projective three space. The geometric construction
of the group law, which was explained to me by Gizatulin, is outlined here.

(8.1) Zero Point and Negative of an Element. The intersection curve
of the two

quadric hypersurfaces H and H can be shown to have a hyperflex point, denoted

0 ∈
.If T (0) denotes the tangent line to
at 0, then every plane π containing T (0)
intersects
in just two other points, i.e., π ∩
={0, A, B}. Or in terms of cycles
we have π ∩
= 2.0 + A + B. With this choice of zero we will make a group law
with B =−A.

(8.2) Sum of Three Points Equal to Zero. Given 0 ∈
= H ∩ H . We define the

group law by starting with P, Q ∈
, forming the plane π(0, P, Q) containing the
three points 0, P, Q. Then there are four points of intersection π(0, P, Q) ∩
=
{0, P, Q, R}. Since 0 is a hyperflex point, we have for cycles 0 + P + Q + R = 4.0
so that P + Q + R = 3.0. If π(P) is the plane through P containig T (0),thenthe
group law is given by the following cycle intersections:

π(0, P, Q) ∩
= 0 + P + Q + (−(P + Q))

and
π(P) ∩
= 2.0 + P + (−P).

These constructions should be compared with the intersection geometry of a plane
cubic curve C where

L(P, Q) ∩ C = P + Q + (−(P + Q)) and L(P) = P + (−P) + 0