Page 151 -

128 6. Proof of Mordell’s Finite Generation Theorem

§3. Finiteness of (E E E(Q) :2E E E(Q)) for E E E = E E E[a a a,b b b]

Using the factorization of multiplication by 2 on the elliptic curve E = E[a, b] con-
sidered in 4(5.3), we will show by relatively elementary means that (E(Q) :2E(Q))
is finite for an elliptic curve of this form over the rational numbers. Note E[a, b]is
3
2
2
defined by y = x + ax + bx for a, b ∈ k.
∗ 2
∗
Recall from 4(5.6) the function α : E[a, b] → k /(k ) defined by
α(0) = 1,
∗ 2
α((0, 0)) = b mod(k ) ,
∗ 2
α((x, y)) = x mod(k ) for x = 0,
and having the property, see 4(5.7), that the sequence
ϕ 2 α ∗ 2
∗
E[a, b] → E[−2a, a − 4b] → k /(k )
is exact. In the case k = Q, the field of rational numbers, the quotient group
∗ 2
Q /(Q ) is additively an F 2 -vector space with a basis consisting of −1 and all
∗
the positive prime numbers p.
(3.1) Proposition. Let E = E[a, b] be an elliptic curve over the rational numbers
Q. The homomorphism
∗ 2
∗
α : E[a, b] → Q /(Q )
has image im(α) contained in the F 2 -vector subspace generated by −1 and all
primes p dividing b. If r distinct primes divide b, then the cardinality of im(α) is
less than 2 r+1 .
2
3
Proof. Let x = m/e and y = n/e be the coordinates of (x, y) on E[a, b] with
rational coefficients where the representations are reduced to lowest terms. Note we
can always choose denominators in this form. The equation of the curve gives
2 2
2
3
n = m + am e = bme 4
2
4
2
= m(m + ame + be ).
2
2
4
If m and m + ame + be are relatively prime, then it follows that m is a square up
to sign and α(x, y) =±1.
2
2
4
More generally, let d be the greatest common divisor of m and m +ame +be .
4
Then d divides be , and since m and e are relatively prime, the integer d also divides
2
∗ 2
∗
b. Moreover, m = M d up to sign, and α(x, y) = d mod Q /(Q ) . This proves the
proposition.
4 2
2
(3.2) Theorem. Let a and b be two integers with = 2 b (a − 4b) unequal to
zero, and let r be the number of distinct prime divisors of b and s the number of
2
a − 4b. Then for E = E[a, b] we have
r+s+2
(E(Q) :2E(Q)) ≤ 2 .