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

Then( )ν1 ε 1h ≤ max{c 10 , c 12 + c 11 log(max{h(ε 1 ), h(ε 2 ), 1}), c 13 ,ν 3 ε 3c 15 + c 14 log(max{h(ε 1 ), h(ε 2 ), 1})}Proof. Let {µ 1 , . . . , µ s1 −1} and {ρ 1 , . . . , ρ s2 −1} be respectively systems of fundamentalS i -units for K 1 and K 2 as in Lemma 3.3.1; in particular we knows∏1 −1j=1h(µ j ) ≤ c 1 (s 1 , d 1 )R 1 ,s∏2 −1j=1h(ρ j ) ≤ c 1 (s 2 , d 2 )R 2 . (3.6.2)We can writeε 1 = ζ 1 µ b 11 · · · µ b s 1 −1s 1 −1 , ε 2 = ζ 2 ρ f 11 · · · ρf s 2 −1s 2 −1 ,where ζ 1 and ζ 2 are roots of unity and b 1 , . . . , b s1 −1, and f 1 , . . . , f s2 −1 are rationalintegers. SetB 1 = max{|b 1 |, . . . , |b s1 −1|},B 2 = max{|f 1 |, . . . , |f s2 −1|}We deduce from Lemma 3.3.3 thatB 1 ≤ c 5 (s 1 , d 1 ) h(ε 1 ), B 2 ≤ c 5 (s 2 , d 2 ) h(ε 2 ),and henceB := max{c 5 (s 1 , d 1 ), c 5 (s 2 , d 2 ), 1)} max{h(ε 1 ), h(ε 2 ), 1} ≥ max{B 1 , B 2 }. (3.6.3)Set α 0 = −ζ 2 ν 2 /(ζ 1 ν 1 ) and b 0 = 1. By (3.6.1) we haveν 3 ε 3= 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. (3.6.4)1Now choose the place υ of L such that ‖ε 3 /ε 1 ‖ υ is minimal. From Lemma 3.2.2 wededuce thath(ε 3 /ε 1 ) ≤ − s d log (‖ε 3/ε 1 ‖ υ ) . (3.6.5)34