Fourth day of class - 9/3
Thm: If C is a curve of genus zero, and C is a rational point that is not singular, then there
are infinitely many rational points on C, and they can be parametrized.
Thm: If C is a curve of genus one, and C has a rational point, then C can be put in the form of
an elliptic curve y 2 = x3 + ax + b. There can be either finitely many or infinitely many rational
points.
Thm (Faltings, 1985): If C is a curve of genus > 1, then C only has finitely many rational
points on it.
Examples:
1. Any conic has genus zero.
2. Any non-singular cubic curve has genus one. (If a cubic curve is singular it has genus zero.)
3. Are there infinitely many rational numbers a, b ∈ Q for which
256(a2 − 3)3
27(b + 1)(b + 9)3
=
?
a2 − 4
b3
This is an odd question to ask, it is connected with the study of elliptic curves. We are asking
about the curve
256(a2 − 3)3 b3 = 27(a2 − 4)(b + 1)(b + 9)3 .
It turns out that this curve has genus zero. (It has many singular points.) We can parametrize
2 −1)
4
2 +3
it with a = −t +6t
and b = 27(t
. This shows that any rational number of the form
4t
t6 −9t4
(t2 − 3)3 (t6 − 9t4 + 3t2 − 3)3
(t2 − 9)(t2 − 1)3 t4
is of the form 256(a2 − 3)3 /(a2 − 4) and also of the form 27(b + 1)(b + 9)3 /b3 .
4. What are the rational solutions to
1
= b2 .
a
This equation defines a genus 1 curve, and there is only one rational point (namely a = b = 1).
a+
5. The curve y 2 = 2x5 − 2 has genus 2. This curve has only three rational points: (1, 0) and
(3, ±22). It takes quite a bit of work to prove this.
Plan: Start talking about adding points on elliptic curves. We won’t be able to properly do it
now, because there are a couple of things we’ll need to talk about (group theory and projective
space), but we’ll start going over those as needed.
Elliptic curves:
Let E : y 2 = x3 + Ax + B, where A, B ∈ Z.
1
2
If we take the intersection of a line with E, y = mx + b, we get a cubic equation. If we choose
the line to go through a rational point P on E, even one with rational slope, the line will
usually intersect the curve in P , and in two other points that are both irrational.
Instead, what we do is we take two rational points P = (c, d) and Q = (e, f ) on E and let
L be the line that goes through both P and Q. This line y = mx + b gives a cubic equation
(mx + b)2 = x3 + Ax + B. Two roots of this cubic equation are e and f . This means that
when we factor out (x − e) and (x − f ) from the cubic, we’re left with a linear factor, which
gives a third rational root. Let R = (g, h) be the third intersection point on the curve. Define
P + Q = (g, −h). This is called the “group law”.
Example: Let y 2 = x3 − 16x + 16. Let P = (1, 1) and Q = (0, 4). The line through P and Q
has slope (1 − 4)/(1 − 0) = −3. It has the equation y − 1 = −3(x − 1) or y = −3x + 4. When
we plug this in to the equation we get (−3x + 4)2 = x3 − 16x + 16. How do we solve this??
Well, we know that x = 0 and x = 1 are already solutions. When we expand the left hand side
we get 9x2 + ... = x3 − 16x + 16, so our cubic equation is x3 − 9x2 + .... That’s all we need to
know!
This factors as (x − 0)(x − 1)(x − g), where g is the x-coordinate of R, the third intersection
point. When we multiply out this cubic we get x3 − (0 + 1 + g)x2 + ... and this must equal
x3 − 9x2 + ... and so 0 + 1 + g = 9. Thus, g = 8. We plug 8 into y = −3x + 4 and get
(−3)(8) + 4 = −24 + 4 = −20. So R = (8, −20). Thus, P + Q = (8, 20).
Note: We are guaranteed that if P and Q are rational points, the third intersection will also
have a rational x-coordinate. Since it is on a line with a rational slope and a rational y-intercept,
the y-coordinate will be rational as well.
Properties of the group law:
P + Q = Q + P . (Easy to see. The line through P and Q is the same as the line through Q
and P .)
(P + Q) + R = P + (Q + R). (Hard to see. Maybe the easiest way to prove this is with a
computer algebra system.)
If E : y 2 = x3 + Ax + B is an elliptic curve, we let E(Q) be the set of all rational points on E.
We really want E(Q) to be a group. (We’ll need to tweak the definition a bit to make it one.)
Note: At this point, I talked a little bit about projective space, the point at infinity, and how
these things make E(Q) into a group. I didn’t start the group theory at all.