REVIEW SHEET FOR LECTURE 11/3 AND 11/7
MING-LUN HSIEH
1. Diophantine problems for quadratic equations Let f (x, y)∈ Q[x, y]. For k = Q or Qp, we let
Xf(k) :={
(x, y)∈ k2 | f(x, y) = 0} .
The basic question is to know if Xf(Q)̸= ∅. When f(x, y) is a quadratic form, i.e. f(x, y) = x2− ay2− b, Hasse-Minkowski’s theorem tells us that
Xf(Q)̸= ∅ ⇐⇒ Xf(Qp)̸= ∅ for all p
⇐⇒ f′ = x2− ay2 represents b in Qp for all p
⇐⇒ (b, −d(f′))p = hp(f′) = (1,−a)p for all p
⇐⇒ (a, b)p = 1 for all odd p|ab and p = ∞, (why don’t we need to consider p = 2?).
Therefore, to know if Xf(Q) is empty not not, it boils down to a finite amount of computation of Hilbert symbols at odd primes, which can be computed effectively via Gauss quadratic reciprocity law. We can also ask the following natural questions:
(1) How to obtain a point in Xf(Q) if it is not empty?
(2) How many solutions in Xf(Q)?
When f (x, y) = x2 − ay2− b, the answers to the above questions is fairly easy. The first one follows from the proof of the proof of Hasse-Minkowski’s theorem for n = 3. As for the second one, we can show if Xf(Q)̸= ∅, then #Xf(Q) is infinite, and we can write down all solutions.
Example 1.1. Let f (x, y) = x2− 5y2− 19. Write down all solutions in Xf(Q).
2. Cubic equations: Elliptic curves 2.1. Now let us consider cubic equations, namely
f (x, y) = ax3+ bx2y + cxy2+ dy3+ ex2+ f xy + gy2+ hx + iy + j ∈ Q[x, y].
Let k be a field, so k may be one of Fp, Qp, Q, R, C. Let
f (x) = x3+ ax2+ bx + c with a, b, c∈ k.
In what follows, we shall only consider a simpler but difficult enough equation f (x, y) = y2− f(x).
Let F(X, Y, Z) be the homogeneous polynomial attached to f defined by F(X, Y, Z) := Z3 · f(X
Z,Y
Z) = ZY2− (X3+ aX2Z + bXZ2+ cZ3).
Date: December 1, 2011.
1
2 M.L. HSIEH
We let
Xf(k) ={
(x, y)∈ k2 | y2 = f (x)} , Cf(k) ={
[X, Y, Z]∈ P2(k)| F(X, Y, Z) = 0} . We have an obvious embedding:
Xf(k) ,→ Cf(k) (x, y)7→ [x, y, 1].
It is clear thatCf(k) contains the point [0, 1, 0], which is the only point with zero Z-coordinate.
Therefore, we can decompose Cf(k) into a disjoint union:
Cf(k) = Xf(k)⊔
{[0, 1, 0]} .
We can ask the following natural questions regarding the size of Cf(Q) (or Xf(Q)):
(1) When is #Cf(Q) > 1?
(2) How to determine effectively if #Cf(Q) =∞?
Before we can study the above questions seriously, we need to understand some basic structural theory for Cf(k).
2.2. Singular points on Cf(k).
Definition 2.1. We say a point P = [a, b, c]∈ Cf(k) is a singular point if F = FX = FY = FZ = 0 at [a, b, c] (FX := ∂F
∂X(X, Y, Z)).
By definition, the point [0, 1, 0] is not a singular point. If (a, b)∈ Xf(k) is a singular point, then
f (a, b) = fx(a, b) = fy(a, b) = 0
⇐⇒ f′(a) = f (a) = b = 0
⇐⇒ the equation f(x) = 0 has a multiple root.
Therefore, we can conclude that there is at most one singular point in Cf(k). Let α1, α2, α3
are three roots of f (x) = 0 and let ∆f be the discriminant of f (x) given by
∆f := (α1− α2)2(α2− α3)2(α3− α1)2. The above discussion shows that
Lemma 2.2. The cubic curve Cf(k) has a singular point if and only if ∆f = 0.
2.3. Group structure on Cf(k).
Definition 2.3 (Collinear points). Three points P, Q, R∈ Cf(k) are called collinear if there exits a line
Lλ,µ,ν(k) :={
[X, Y, Z]∈ P2(k)| λX + µY + νZ = 0}
([λ, µ, ν]∈ P2(k)) such that
Lλ,µ,ν(k)∩ Cf(k) ={P, Q, R} .
In the class, we explained the following proposition by direct computation:
REVIEW SHEET FOR LECTURE 11/3 AND 11/7 3
Proposition 2.4. Let P and Q be two non-singular points in Cf(k). There exists a unique point R∈ Cf(k) such that P, Q, R are collinear.
Proof. Let P = (x1, y1) and Q = (x2, y2) are in Xf(k) ⊂ Cf(k). Suppose that y1 ̸= −y2. Let
λ =y1− y2
x1− x2
if x1 ̸= x2, and λ =f′(x1)
2y1 if x1 = x2, y1 = y2 (so P = Q).
and let ν = y1− λx1. Then we have
Lλ,−1,ν(k)∩ Cf(k) ={P, Q, R} , where R = (x3, y3) with
x3 =λ2− a − x1− x2, y3 = λx3+ ν.
If x1 = x2 and y1 =−y2, then R = [0, 1, 0] and we have
L1,0,−x1(k)∩ Cf(k) ={P, Q, R} .
We leave the other cases to you as exercises.
Definition 2.5 (Group law). Suppose that Cf(k) is non-singular. Namely, ∆f ̸= 0. Then we define the group law on Cf(k) as follows:
(1) The identity element is O := [0, 1, 0]∈ Cf(k).
(2) For P ∈ Cf(k), define−P to be the unique element such that {P, O, −P } are collinear.
(3) For P, Q ∈ Cf(k), let P + Q be the unique point such that {P, Q, −(P + Q)} are collinear.
By definition, if P, Q, R are collinear, then P + Q + R = O.
Theorem 2.6. Suppose ∆f ̸= 0. Then the above definition gives an abelian group structure on Cf(k) with the identity element O.
Definition 2.7 (Elliptic curves). The pair (Cf(k), O) is called an elliptic curve defined by y2 = f (x). From now on, we shall use E(k) to denote the abelian group Cf(k) with non-zero discriminant ∆f ̸= 0.
Example 2.8. Let E : y2 = x3 + 17. Let P = (−1, 4) and Q = (2, 5). Compute P + Q, 2P and 2Q.
2.4. Torsion points in E(k). Let E : y2 = f (x) be an elliptic curve over k. For each positive integer m > 1, we let E[m](k) be the m-torsion subgroup in E(k) defined by
E[m](k) :={P ∈ E(k) | m · P = O = [0, 1, 0]} .
Proposition 2.9. If E is an elliptic curve over the complex number C, then we have E[2](C)≃ Z/2Z ⊕ Z/2Z, E[3](C) ≃ Z/3Z ⊕ Z/3Z.
Using the analytic method, in general one can show that E[m](C) ≃ Z/mZ ⊕ Z/mZ.
4 M.L. HSIEH
3. Exercises
Exercise 3.1. Show the discriminant ∆f of f (x) = x3+ ax2+ bx + c is given by the formula
∆f =−4a3c + a2b2+ 18abc− 4b3− 27c2.
Exercise 3.2. Let E : y2 = x3 + 17 be an elliptic curve over Q. Let P = (−2, 3) and Q = (2, 5). Compute the following points:
2P, P − Q, 3P − Q.
These points all have integral x, y coordinates.
Exercise 3.3. Let E : y2 = x3− 2x. Let i =√
−1 ∈ C. Define a map u : E(C) → E(C) by u(x, y) := (−x, iy) and u(O) := O.
Show that the map u is a group homomorphism. In other words, u(P + Q) = u(P ) + u(Q).
Exercise 3.4. Consider the point P = (3, 8) on the elliptic curve E : y2 = x3 − 43x + 166.
Compute P , 2P , 3P 4P and 8P . Show that 7P = O.
Exercise 3.5. Let E : y2 = x3+ 1. Find all 3-torsion points of E(C).
Department of Mathematics, National Taiwan University, Taipei, Taiwan E-mail address: mlhsieh@math.ntu.edu.tw
