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

Next assume that B > 2e max((s 1 + s 2 − 2)π/ √ 2, A 1 , . . . , A s1 +s 2 −2)A 0 . Then, usingagain Lemma 3.5.1 and (3.6.8) we get the following bound.∥ logν 3 ε 3 ∥∥∥υ∥ > −2d υ dHc 4 (s 1 , d 1 )c 4 (s 2 , d 2 )c 7 (n, d)R 1 R 2 ×ν 1 ε 1log(B/(2 √ 2dH)).(3.6.11)Now assume υ is finite. Let p be a corresponding prime ideal to υ. Then∥ logν 3 ε 3 ∥∥∥υ∥ = − ord p (α b 00ν 1 ε µ−b 11 · · · µ −b s 1 −1s 1 −1 ρ f 11 · · · ρf s 2 −1s 2 −1 − 1) log N(p). (3.6.12)1Again, we distinguish two cases. First assume thatB < 4c 9 (n, d)H h(µ 1 ) · · · h(µ s1 −1) h(ρ 1 ) · · · h(ρ s2 −1).In this case we will directly compute an upper bound for h(ν 3 ε 3 /(ν 1 ε 1 )). From (3.6.4)and Lemma 3.3.1 we deduce that( )ν3 ε 3h ≤ log 2 + 2H + B(n − 1) max{c 2 (s 1 , d 1 )R 1 , c 2 (s 2 , d 2 )R 2 }.ν 1 ε 1Then, from our assumption for B and (3.6.2) we get( )ν3 ε 3h ≤ log 2 + 2H + 4(n − 1)Hc 1 (s 1 , d 1 )c 1 (s 2 , d 2 )c 9 (n, d)×ν 1 ε 1R 1 R 2 max{c 2 (s 1 , d 1 )R 1 , c 2 (s 2 , d 2 )R 2 }.(3.6.13)We next assumeB ≥ 4c 9 (n, d)H h(µ 1 ) · · · h(µ s1 −1) h(ρ 1 ) · · · h(ρ s2 −1).Since 2H ≥ 1 we have h ′ 0 = max{h(α 0), 1/(16e 2 d 2 )} ≤ 2H. Moreover, using Lemma3.2.4, for i = 1, . . . , s 1 − 1 and j = 1, . . . , s 2 − 1 we haveh ′ i = max{h(µ i ), (16e 2 d 2 ) −1 } = h(µ i ),andh ′ j+s 1 −1 = max{h(ρ j ), (16e 2 d 2 ) −1 } = h(ρ j ).36