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:
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