58 2. Elliptic curves
Proof. Suppose a prime p divides abc. Then p divides a, b or c. Let us
assume that p | a (the same argument works if p divides b or c). Let pk
be the exact power of p that appears in the factorization of the rational
number x0 e1 =
au2.
Notice that k may be positive or negative,
depending on whether p divides the numerator or denominator of
au2.
Notice, however, that k must be odd, because p | a, and a is
square-free.
Suppose first that k 0, i.e.,
p|k|
is the exact power of p that
divides the denominator of x0 e1. Since ei Z, it follows that
p|k|
must divide the denominator of x0 too, and therefore
p|k|
is the
exact power that divides the denominators of x0 e2 and x0 e3 as
well. Hence,
p3|k|
is the exact power of p dividing the denominator
of y0
2
= (x0 ei), but this is impossible because y0
2
is a square and
3|k| is odd. Thus, k must be positive.
If k 0 and p divides x0 e1, then the denominator of x0 is
not divisible by p, so it makes sense to consider x0 mod p, and x0
e1 mod p. Similarly, the denominators of x0 e2 and x0 e3 are not
divisible by p and
bv2
x0 e2 e1 e2, and
cw2
x0 e3 e1 e3 mod p.
Since y0 2 = abc(uvw)2 and p divides a, then p must also divide one
of b or c. Let’s suppose it also divides b. Then 0 bv2 x0 e2
e1 e2 mod p and Δ = (e1 e2)(e2 e3)(e1 e3) 0 mod p, as
claimed.
The definition of the map δ and the previous proposition yield
the following immediate corollary:
Corollary 2.9.6. With notation as in the previous Theorem and
Proposition, define a subgroup Γ of Q×/(Q×)2 by
Γ = {n Z : 0 = n is square-free and if p | n, then p |
Δ}/(Z×)2.
Then, δ induces an injection of E(Q)/2E(Q) into
ΓΔ = {(δ1,δ2,δ3) Γ × Γ × Γ : δ1 · δ2 · δ3 = 1
Q×/(Q×)2}
Γ × Γ × Γ .
We are ready to prove the weak Mordell-Weil theorem (Thm.
2.4.5), at least in our restricted case:
Previous Page Next Page