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:
