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: