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80 3. Elliptic Curves and Their Isomorphisms

8. For each of the five curves E i in (6.3) determine for which q = 2, 4, 8, 16, or 256 the
group E(F q ) is noncyclic. Describe F 8 as a cubic extension of F 2 .
9. Find all elliptic curves over the field F 5 of five elements, determine the structure of their
group of points over F 5 ,and findtheir j values.

§7. Singular Cubic Curves

Singular cubic curves will arise naturally in Chapter 5 when an elliptic curve is re-
duced modulo a prime. For this reason and as a rounding off of our discussion of
cubics in the plane, we study cubics over a field k wih irreducible equation
2 3 2
F(x, y) = y + a 1 xy + a 3 y − x − a 2 x − a 4 x − a 6 = 0
having a singular point which is rational.
The singular point can be transformed to the affine origin (x, y) = (0, 0) by an
admissible change of variables as discussed in (2.4). Observe that (0, 0) is on the
curve, i.e., F(0, 0) = 0 if and only if a 6 = 0. To determine whether (0, 0) is a
singular point, we substitute into the partial derivatives

2
F y = 2y + a 1 x + a 3 and F x = a 1 y − 3x − 2a 2 x − a 4 .
(7.1) Remark. The point (0, 0) is a singular point on the Weierstrass cubic with
equation F(x, y) = 0ifand onlyif

a 3 = a 4 = a 6 = 0.

This follows by just substituting F y (0, 0) = a 3 = 0and F x (0, 0) =−a 4 = 0. With
reference to the formulas (3.1) we see that (0, 0) a singular point on the cubic curve
implies that

b 4 = b 6 = b 8 = 0and = 0.

The Weierstrass form becomes
2
3
2
y = a 1 xy − a 2 x = x ,
2
and the discriminant of the quadratic form on the left is the coefficient b 2 = a +4a 2 .
1
As in the Introduction, we go to the (t, s)-plane with t =−x/y and s =−1/y.
3
2
The singular Weierstrass equation becomes s = t + a 1 ts + a 2 t s, and thus s is a
rational function of t, namely
t 3
s = 2 .
1 − a 1 t − a 2 t
This means that a singular cubic C is a rational curve in the geometric sense, and the
2
3
set of nonsingular points C ns consists of all (t, s) with s = t /(1 − a 1 t − a 2 t ) and