S-integral points on hyperelliptic curves Homero Renato Gallegos ...

3.1.2 The Even Degree CaseWe are assuming the existence of a rational point P 0 . If P 0 is one of the two <strong>points</strong> atinfinity, let ɛ 0 = 1. Otherwise, y(P 0 ) ≠ 0 since f is irreducible. Write x 0 = γ 0 /d 2 0 withγ 0 ∈ Z and d 0 ∈ Z ≥1 and let ɛ 0 = γ 0 − αd 2 0 .In this even degree case one needs to modify the homomorphism given in theodd degree case.Since we are assuming the existence of a rational point on thecurve, we can choose a representative of every coset of J(Q)/2J(Q) having the form∑ mi=1 (P i − P 0 ) where the set {P 1 , . . . , P m } is stable under the action of the Galoisgroup Gal( ¯Q/Q), and such that all y(P i ) are non-zero (see Section 5 of [45]). Again,as in [15, Section 3], we can write x(P i ) = γ i /d 2 i where γ i is an algebraic integer andd i ∈ Z ≥1 ; and if P i , P j are conjugate, we may suppose that d i = d j and in consequenceγ i , γ j are conjugate. We associate to such a coset representative the algebraicnumber(m mod 2)ɛ = ɛ0m∏(γ i − αd 2 i ).Lemma 3.1.2. Let E be a set of ɛ associated as above to a complete set of cosetrepresentatives for J(Q)/2J(Q). Let ∆ be the discriminant of the polynomial f. Foreach ɛ ∈ E let B ɛ be the set of square-free rational integers supported only by primesdividing a∆ Norm K/Q (ɛ) ∏ p∈S p. Let K = {ɛb : ɛ ∈ E, b ∈ B ɛ}. Then K is a finitesubset of O K and if (x, y) is an S-<strong>integral</strong> point on the model (3.1.1), then x − α = κξ 2for some κ ∈ K, ξ ∈ K.Proof. In this even degree case there is also a well defined homomorphismi=1θ : J(Q)/2J(Q) → K ∗ /(Q ∗ K ∗2)given by ( m)∑∏mθ (P i − P 0 ) = a m (x(P i ) − α) (mod Q ∗ K ∗2 )i=1i=1for coset representatives ∑ (P i − P 0 ) with y(P i ) ≠ 0 (see [35] and Section 5 of [45]).Let P = (x, y) be an S-<strong>integral</strong> point on the curve. Under θ, the coset of P − P 0 is22