Effective results for hyper- and superelliptic equations over number fields

Abstract

Abstract. Let f be a polynomial with coefficients in the ring O S of S-integers of a given number field K, b a non-zero S-integer, and m an integer ≥ 2. Suppose that f has no multiple zeros. We consider the equation (*) f (x) = by m in x, y ∈ O S . In the present paper we give explicit upper bounds in terms of K, S, b, f, m for the heights of the solutions of (*). Further, we give an explicit bound C in terms of K, S, b, f such that if m > C then (*) has only solutions with y = 0 or a root of unity. Our results are more detailed versions of work of Trelina, Brindza, and Shorey and Tijdeman. The results in the present paper are needed in a forthcoming paper of ours on Diophantine equations over integral domains which are finitely generated over Z.

Effective results for unit points on curves over finitely generated domains

“…, and R S is the S-regulator of K. For the S-regulator by using Lemma 3 of [11] and Lemma 2.1 of [2] (for the original result see Louboutin [21]) we can derive the bound…”

Multiplicative Dependence Among Iterated Values of Rational Functions Modulo Finitely Generated Groups

“…In [1,Equations (5.11) and (5.12)], we argue as follows. We may assume without loss of generality that…”

Galois groups of iterates of some unicritical polynomials

“…u t of height at most (2/π) 2p−2 |D p |. Considering the critical orbit Orb ϕp (1) as a subset of O K,S , we can write (2) ϕ n p (1) = d n y p n , for some d n , y n ∈ O K,S with 0 ≤ v p (d n ) ≤ p − 1 for all p / ∈ S. We now use Lemma 2.2 and our decomposition in (2) to study primitive prime divisors in Orb ϕp (1). To do this, note that ϕ p (0) = 1 − ζ p and ϕ p (1 − ζ p ) = 1 − ζ p , from which it follows that the ideal generated by (1 − ζ p ) is the only prime dividing the nontrivial elements of the orbit of 0 (it is well known that (1 − ζ p ) is the unique prime ideal above p).…”

“…However, if q = (2 − ζ 7 ), then ϕ 7 (x) ≡ (x − 1) 7 (mod q) and hence ϕ n 7 (1) ≡ −1 (mod q) for all n ≥ 2; here we use that Z[ζ 7 ]/q = F 127 . However, setting x ≡ i + k (mod 7) and y ≡ j + 2k (mod 7), one checks manually that (x, y) = {(0, 0), (1,5), (2, 3), (3, 1), (4,6), (5,4), (6,2)} are the only pairs of exponents with solutions −1 ≡ 2 i · 3 j · 18 k · y 7 n (mod 127): here 2, 3 and 18 are the images of the unit generators. In particular, there are only 49 possible tuples (i, j, k) that must be ruled out: each choice of 0 ≤ k ≤ 6 and (x, y) in the collection above uniquely determines i and j.…”

“…If k = 2, 3, 4, 6 and (2) holds, then one has seven possible pairs (i, j) coming from the restrictions on (x, y) above. For example, (i, j) = {(5, 3), (6, 1), (0, 6), (1,4), (2,2), (3, 0), (4, 5)} when k = 2. As in the previous cases k = 0 and k = 1, only one pair remains after working modulo q = (2 + ζ 7 ), and this exceptional case is ruled out modulo q = (3 + ζ 7 ).…”