Multiplicative dependence of cubic algebraic numbers

Here, n ∈ N and m ∈ N∪{0}, where the right–hand side is assumed to be 1 for m = 0. Assume that α is M -dependent. We call its length of multiplicative dependence (and denote it by l(α,M)) the smallest n+m for which there are x1, . . . , xn, y1, . . . , ym ∈ M satisfying (1). Denote by Zt the set of integers greater than or equal to t. Z0-dependence of algebraic numbers is important in the theory of Hurwitz zeta–function ζ(α, s) =


Introduction and results
Throughout we denote by Z, Q and C the sets of integers, rational numbers and complex numbers respectively.Let α be an algebraic number and let M ⊂ Q.We say that a complex number α is M -dependent if there are two distinct collections x 1 , . . ., x n ∈ M and y 1 , . . ., y m ∈ M such that Here, n ∈ N and m ∈ N∪{0}, where the right-hand side is assumed to be 1 for m = 0. Assume that α is M -dependent.We call its length of multiplicative dependence (and denote it by ℓ(α, M )) the smallest n+m for which there are x 1 , . . ., x n , y 1 , . . ., y m ∈ M satisfying (1).Denote by Z t the set of integers greater than or equal to t. Z 0 -dependence of algebraic numbers is important in the theory of Hurwitz zeta-function ζ(α, s) = ∞ j=0 (j + α) −s (see, e. g., [6]) as well as in the in the investigation of zero-distribution and the universality property of Lerch zeta-function (see, e.g., [4,5]).
The first named author and Dubickas [1] raised the following question.
Question.Is every algebraic number Z 0 -dependent?
It was proved in [1] that every quadratic algebraic number α is Z t -dependent and ℓ(α, Z t ) 8 for any t ∈ Z.On the other hand the equality α(α+x) = α+y shows that every quadratic algebraic integer is Z-dependent and ℓ(α, Z) 3.Moreover, it was noted in [1] that every quadratic algebraic integer is Z t -dependent and ℓ(α, Z t ) 5 for every t ∈ Z.
For quadratic algebraic numbers, which are not algebraic integers, the inequality ℓ(α, Z t ) 8 is not sharp.Indeed, it was stated in [1] that if α is a root of 2x 2 + 3 and ℓ(α, Z 1 ) = 6 then there exist two distinct collections x 1 , x 2 , x 3 ∈ Z 1 and y 1 , y 2 , y 3 ∈ Z 1 such that and at least one of x i , y i is > 1000.However, we have found that so that ℓ(α, Z 1 ) = 6.
Let α be an algebraic number whose minimal polynomial (the monic polynomial in Q[x] of smallest degree whose root is α) is Let M ⊂ Q.Then M -dependence of α implies multiplicative dependence of the set {P (−m) | m ∈ M } of values of the minimal polynomial of α.Indeed, let K := Q(α) be the number field generated by α and let Nm K/Q : K → Q be the norm in the extension K/Q (see, e.g., [3] or [7]).Assume that α is M -dependent.Since the norm map Nm K/Q is multiplicative, (1) implies In view of Nm K/Q (α + x) = (−1) d P (−x), which is valid for all x ∈ Q, we obtain and hence Therefore the set {P (−m) | m ∈ M } is multiplicatively dependent.Dubickas [2] proved that for any quadratic polynomial P (x) ∈ Q[x] and any t ∈ Z the set {P (n) | n ∈ Z t } is multiplicatively dependent.It was proved in [1] that every cubic algebraic number α is Q-dependent and ℓ(α, Q) 8. We prove the following theorem.
Theorem 1. Suppose that a, c and m are rational integers, c = 0, such that Let α be a root of the polynomial Proof.Since α is a root of P (x), it follows that On the other hand, Let r := −2m 3 + am 2 − c.Then ( 2) and ( 3) imply so that α is Z-dependent and ℓ(α, Z) 4.
We are left to prove that the polynomial P (x) is irreducible.Indeed, assume that it is reducible.Consequently, P (x) is divisible by a linear polynomial from Q[x], and therefore it has a root x 0 ∈ Q.Since P (x) is monic with integer coefficients, x 0 ∈ Z and x 0 is a divisor of P (0) = c = 0. Hence 0 < |x 0 | |c| and which is a contradiction.Therefore P (x) is irreducible.
Corollary 1.For any a, c ∈ Z, c = 0, there exist infinitely many negative integers b such that the polynomial P (x) := x 3 + ax 2 + bx + c is irreducible (over Q) and if α is a root of P (x) then it is Z-dependent and ℓ(α, Z) 4.