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118 5. Reduction mod p and Torsion Points

Exercises

2 3
1. For the elliptic curve E : y − y = x − 7, see 1(2.5), show that E is nonsingular mod
2. Show that E(Q) has no point of order 2, and deduce that E(Q) tors is cyclic of order 3.
2
3
2
2. Study E : y + xy + y = x − x both modulo 2 and modulo 3. Determine E(Q) tors .
2
2
3
3. Study E : y +xy = x −x −2x −1 both modulo 2 and modulo 3. Determine E(Q) tors .
3
2
2
4. In 1(2. Ex.5) we noted that (1, 0) has order 7 on E : y + xy + y = x − x − 3x + 3.
Show that E is nonsingular modulo 3 and determine E(Q) tors .
5. Let M n (R) denote the algebra of n by n matrices over the ring R. Reduction modulo q
of integral matrices defines a ring homomorphism r q : M n (Z) → M n (Z/qZ),and show
that it restricts to r q :GL n (Z) → GL n (Z/qZ) on the invertible matrices. For X ∈ M n (Z)
such that r p (X) = 0 show that
a n a a+b+1
(I n + p X) ≡ I n + np X mod p ,
where b = ord p (n) and p > 2or a ≥ 1. Prove that for a finite subgroup G of GL n (Z)
that G ∩ ker(r p ) = 1 for p > 2and G ∩ ker(r 4 ) = 1at p = 2. Compare with (5.1).
3
2
6. Show that the curve y + y = x − 7x + 6 has no torsion in its group of rational points.
§6. Computability of Torsion Points on Elliptic Curves from
Integrality and Divisibility Properties of Coordinates
The first theorem says that when looking for torsion points we have only to look at
integral points.
(6.1) Theorem. Let E be an elliptic curve defined over Q with an equation in nor-
mal form with integer coefficients. If (x, y) ∈ E(Q) tors , then the coordinates x and y
are integers.

Proof. If y = 0, then x is a solution of the cubic equation
3 2
0 = x + a 2 x + a 4 x + a 6
with integer coefficients. Since x is rational, it is also an integer, for x = m/n re-
duced to lowest terms satisfies
2
3
3
2
0 = m + a 2 m n + a 4 mn + a 6 n ,
and any prime dividing n must divide m. Thus x = m is an integer.

If y = 0, then the point with homogeneous coordinates has the form (w : x :

1) = (1: x : y), where w = 1/y and x = x/y. By Theorem (4.1), we have that

(w : x :1) ∈ r −1 (0) for p odd and (w : x :1) ∈ E (2) (Q) at 2. In other words
p
we have ord p (w) ≤ 0 for p odd and ord 2 (w) ≤−1 at 2. This condition becomes
for y the relation ord p (y) ≥ 0 for all odd p and ord 2 (y) ≥−1at2.Thus y has
the form y = h/2 for an integer h. Again write x = m/n,and x satisfies a cubic
3
2
equation with coefficient of x one, coefficient of x an integer, coefficient of x an
integer over 2, and constant coefficent an integer over 4. A modification, using 2, of