552 XIII Connections with Number Theory
with coefficientsaij∈Ksuch that
F(a 11 X′+···,a 21 X′+···,a 31 X′+···)=G(X′,Y′,Z′).It is clear thatFandGnecessarily have the same degree, and that projective equiva-
lence is in fact an equivalence relation.
Consider now the affine cubic curveCdefined by the polynomial
f(X,Y)=a 30 X^3 +a 21 X^2 Y+a 12 XY^2 +a 03 Y^3 +a 20 X^2 +a 11 XY
+a 02 Y^2 +a 10 X+a 01 Y+a 00.We assume thatChas a non-singular K-point which is a flex. Without loss of general-
ity, suppose that this is the origin. Thena 00 =0,a 10 anda 01 are not both zero, and
a 20 X^2 +a 11 XY+a 02 Y^2 =(a 10 X+a 01 Y)(a′ 10 X+a′ 01 Y)for somea 10 ′ ,a 01 ′ ∈K. By an invertible linear change of variables we may suppose
thata 10 =0,a 01 =1. Thenfhas the form
f(X,Y)=Y+a 1 XY+a 3 Y^2 −a 0 X^3 −a 2 X^2 Y−a 4 XY^2 −a 6 Y^3.Ifa 0 =0, thenfis divisible byYand the corresponding projective curve is reducible.
Thus we now assumea 0 =0. In fact we may assumea 0 =1, by replacingfby a con-
stant multiple and then scalingY. The projective completionC ̄ofCis now defined
by the homogeneous polynomial
YZ^2 +a 1 XY Z+a 3 Y^2 Z−X^3 −a 2 X^2 Y−a 4 XY^2 −a 6 Y^3.If we interchangeYandZ, the flex becomes the unique point at infinity of the affine
cubic curve defined by the polynomial
Y^2 +a 1 XY+a 3 Y−(X^3 +a 2 X^2 +a 4 X+a 6 ).This can be further simplified by making mild restrictions on the fieldK.IfKhas
characteristic=2, i.e. if 1+ 1 =0, then by replacingYby(Y−a 1 X−a 3 )/2we
obtain the cubic curve defined by the polynomial
Y^2 −( 4 X^3 +b 2 X^2 + 2 b 4 X+b 6 ).IfKalso has characteristic=3, i.e. if 1+ 1 + 1 = 0, then by replacingX by
(X− 3 b 2 )/ 62 andYby 2Y/ 63 , we obtain the cubic curve defined by the polynomial
Y^2 −(X^3 +aX+b). Thus we have proved:
Proposition 7If a projective cubic curve over the field K is irreducible and has a
non-singular K -point which is a flex, then it is projectively equivalent to the projective
completionW =W(a 1 ,...,a 6 )of an affine curve of the form
Y^2 +a 1 XY+a 3 Y−(X^3 +a 2 X^2 +a 4 X+a 6 ).