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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 y2 0 = 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: