Diophantine equations over global function fields I: The Thue equation

ARTICLE IN PRESS UNCORRECTED PROOF S0022-314X(05)00211-8/FLA AID:3277 Vol.•••(•••) [DTD5] P.1 (1-17) YJNTH:m1 v 1.50 Prn:17/11/2005; 9:56 yjnth3277 by:Vita p. 1 Journal of Number Theory ••• (••••)•••–••• www.elsevier.com/locate/jnt 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 14 15 15 16 16 17 17 18 18 19 19 20 20 21 21 22 22 23 23 24 24 25 25 26 26 27 27 28 28 29 29 30 30 31 31 32 32 33 33 34 34 35 35 36 36 37 37 38 38 39 39 40 40 41 41 42 42 43 43 44 44 45 45 Diophantine equations over global function fields I: The Thue equation István Gaála,1, Michael Pohstb,∗,2 aUniversity of Debrecen, Mathematical Institute, H-4010 Debrecen Pf. 12, Hungary bTechnische Universtät Berlin, Institut für Mathematik, Straße des 17. Juni 136, Berlin, Germany Received 19 January 2004; revised 5 July 2005 Communicated by David Goss Abstract We solve completely Thue equations in function fields over arbitrary finite fields. In the function field case such equations were formerly only solved over algebraically closed fields (of characteristic zero and positive characteristic). Our method can be applied to similar types of Diophantine equations, as well. 2005 Published by Elsevier Inc. MSC: 11D59; 11Y50; 11R58 Keywords: Thue equations; Global function fields 1. Introduction Classical Diophantine equations like Thue equation (cf. Thue [10]) are traditionally solved over the rings of rational integers or over the ring of integers of a number field *Corresponding author. E-mail addresses: [email protected] (I. Gaál), [email protected] (M. Pohst). 1Research supported in part by Grants T 037367 and T 042985 from the Hungarian National Foundation for Scientific Research. 2Research supported by the Deutsche Forschungsgemeinschaft. 0022-314X/$ – see front matter 2005 Published by Elsevier Inc. doi:10.1016/j.jnt.2005.10.009 ARTICLE IN PRESS UNCORRECTED PROOF S0022-314X(05)00211-8/FLA AID:3277 Vol.•••(•••) [DTD5] P.2 (1-17) YJNTH:m1 v 1.50 Prn:17/11/2005; 9:56 yjnth3277 by:Vita p. 2 2I. Gaál, M. Pohst / Journal of Number Theory ••• (••••)•••–••• 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 14 15 15 16 16 17 17 18 18 19 19 20 20 21 21 22 22 23 23 24 24 25 25 26 26 27 27 28 28 29 29 30 30 31 31 32 32 33 33 34 34 35 35 36 36 37 37 38 38 39 39 40 40 41 41 42 42 43 43 44 44 45 45 (cf. Baker [1]). Several authors considered the analogous problem over function fields: more exactly over the ring of integers of a function field over an algebraically closed field, see, e.g., [6,8], both in case the ground field is of zero or positive characteristic. These results have also common generalizations (cf. Gy˝ ory [5]). Our purpose is now to investigate this problem in function fields over arbitrary finite fields. It is well known that Diophantine equations over finite fields play an important role in cryptography, cf. Niederreiter and Xing [7]. Also, from a practical point of view this case is much more straightforward than the case of algebraically closed ground fields (which mostly occur in theory only). As it will turn out, also in this case the unit equation plays a crucial role and the solutions can be computed easily. However, for constructing the appropriate function fields, performing calculations in them, determining heights, etc. we intensively use the computer algebra package KASH [2]. 2. Global function fields In the following we shall strongly rely on the argument used by Mason [6] for function fields over algebraically closed fields. We show how his ideas can be transferred to our situation. A general description of properties of function fields (also over finite fields) can be found in the book of Stichtenoth [9]. We introduce some notations. k=Fqdenotes a finite field with q=pdelements. The rational function field of kis k(t) as usual, and Kis a finite extension of k(t) of degree n0 and genus g0. The integral closure of k[t]in Kis denoted by oK. We assume that K is separably generated over k(t) by an element ybelonging to oKand that kis the full constant field of K. Any element f∈Khas a unique presentation f= n0  i=1 hiyi−1,h i∈k(t). Conjugates of elements (fields) are denoted by upper case indices. Let A:= ((y(j))i−1)1⩽i,j⩽n∈Kn×nhave determinant D. We note that Dis the discriminant of y. It is nonzero since Kis separably generated. We obtain the system of linear equations: f(1),...,f(n)=(h1,...,h n)A. Hence, the hiare rational functions in the f(j),(y(j))i−1. The set of all (exponential) valuations of Kis denoted by V, the subset of infinite valuations by V∞. By abuse of notation we do not distinguish between places and valuations. For example, we write degvfor the degree of the divisor belonging to the valuation v∈V. For a nonzero element f∈Kwe denote by v(f ) the value of fat v. For integral elements this is the highest power of the divisor belonging to vthat divides the divisor (f ), and this ARTICLE IN PRESS UNCORRECTED PROOF S0022-314X(05)00211-8/FLA AID:3277 Vol.•••(•••) [DTD5] P.3 (1-17) YJNTH:m1 v 1.50 Prn:17/11/2005; 9:56 yjnth3277 by:Vita p. 3 I. Gaál, M. Pohst / Journal of Number Theory ••• (••••)•••–••• 3 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 14 15 15 16 16 17 17 18 18 19 19 20 20 21 21 22 22 23 23 24 24 25 25 26 26 27 27 28 28 29 29 30 30 31 31 32 32 33 33 34 34 35 35 36 36 37 37 38 38 39 39 40 40 41 41 42 42 43 43 44 44 45 45 concept is extended to rational elements in the usual way. For the normalized valuations vN(f ) =v(f) ·degvthe product formula holds:  v∈V vN(f ) =0∀f∈K\{0}. The height of a nonzero element fof Kis defined to be H(f):=  v∈V max0,v N(f ). Because of the product formula this is tantamount to H(f)=− v∈V min0,v N(f ) which then holds for all elements of Kincluding 0. 3. Unit equations Let V0be a finite subset of V. Then the nonzero elements γ∈Ksatisfying v(γ) =0 for all v/∈V0form a multiplicative group in K. These elements are called V0-units.For V0=V∞the V0-units are just the units of the ring oK. The resolution of Thue equations (as well as several other types of classical Diophantine equations) is usually reduced to equations of the form γ1+γ2+γ3=0(1) where the γiare V0-units for a suitable set V0. The crucial inequality of Mason on the above equation becomes in our case: Lemma 3.1. Let V0be a finite subset of Vand let γi(1⩽i⩽3)be V0-units satisfying (1). Then either γ1 γ3is in Kpor its height is bounded: Hγ1 γ3⩽2g−2+ v∈V0 degv. (2) Proof. The proof is along the lines of the proof of Lemma 2 in [6] (see [6, p. 14]). Since in our case the field of constants kis not algebraically closed we encounter a few additional difficulties which we will point out in what follows. ARTICLE IN PRESS UNCORRECTED PROOF S0022-314X(05)00211-8/FLA AID:3277 Vol.•••(•••) [DTD5] P.4 (1-17) YJNTH:m1 v 1.50 Prn:17/11/2005; 9:56 yjnth3277 by:Vita p. 4 4I. Gaál, M. Pohst / Journal of Number Theory ••• (••••)•••–••• 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 14 15 15 16 16 17 17 18 18 19 19 20 20 21 21 22 22 23 23 24 24 25 25 26 26 27 27 28 28 29 29 30 30 31 31 32 32 33 33 34 34 35 35 36 36 37 37 38 38 39 39 40 40 41 41 42 42 43 43 44 44 45 45 We may assume that f:=γ1/γ3and therefore γ2/γ3=−f−1 do not belong to k.Let V1:=v∈V:v(f) < 0,V 2:=v∈V:v(f) > 0, V3:=v∈V:v(f) =0∧v(f +1)>0. These are disjoint subsets of the set V0. Then we have H(f)= v∈V1−v(f)degv= v∈V2 v(f)degv= v∈V3 v(f +1)degv. This is true because of 1 =(1+f)+(−f), the product formula, and the property v(f) < 0 ⇔v(1+f)<0. If zdenotes a prime element for the valuation v∈Vthen the differential 1df satisfies v(1df ) =v(df/dz) (see [9, Chapter IV]). If fis not a pth power then the divisor 1df is canonical, i.e., deg(1df ) =2g−2, since the field kof constants is complete (see [9, Chapter I.5]). This yields 2g−2= v∈V v(df)degv= v∈V0 v(df)degv= v∈V1∪V2∪V3 v(df)degv ⩾ v∈V1∪V2v(f) −1degv+ v∈V3v(1+f)−1degv =−H(f)+H(f)− v∈V1∪V2 degv+H(f)− v∈V3 degv ⩾H(f)− v∈V1∪V2∪V3 degv⩾H(f)− v∈V0 degv whence the assertion follows. 2 Taking Φ=−γ1/γ3,Ψ=−γ2/γ3Eq. (1) gives the unit equation in two variables Φ+Ψ=1(3) where Φ,Ψ are V0-units. Because of characteristic pthe number of solutions of such a unit equation can be infinite. For example, if V0is just the set of infinite valuations and η,1−ηare both units of oK then also ηκ,(1−η)κis a solution of (3) for every exponent κ=p. Hence, there exist solutions of arbitrary large heights in this situation. The subsequent lemma shows that for any finite subset V0of V, the group of V0-units of Kcontains only a finite number, say s,ofV0-units ηwhich are not pth powers and for which also 1 −ηis a V0-unit. We denote the set of these units by {η1,...,η s}. Lemma 3.2. Let V0be a finite subset of V. Assume that a V0-unit Φin Kis a solution of (3).IfΦis not a pth power of ηi(1⩽i⩽s,  ∈Z⩾0)then Φbelongs to a finite subset of Kwhich can be calculated. ARTICLE IN PRESS UNCORRECTED PROOF S0022-314X(05)00211-8/FLA AID:3277 Vol.•••(•••) [DTD5] P.5 (1-17) YJNTH:m1 v 1.50 Prn:17/11/2005; 9:56 yjnth3277 by:Vita p. 5 I. Gaál, M. Pohst / Journal of Number Theory ••• (••••)•••–••• 5 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 14 15 15 16 16 17 17 18 18 19 19 20 20 21 21 22 22 23 23 24 24 25 25 26 26 27 27 28 28 29 29 30 30 31 31 32 32 33 33 34 34 35 35 36 36 37 37 38 38 39 39 40 40 41 41 42 42 43 43 44 44 45 45 For the proof we refer to [6, Lemma 11, p. 98]. The proof starts by assuming that Φis not a pth power in Kand therefore also provides the means to calculate the ηi. 4. Application to Thue equations 4.1. Preliminaries We want to apply these results to (relative) Thue equations over K.Let F(X,Y):= n  i=0 AiXn−iYi∈oK[X, Y ] be a binary homogeneous form of degree at least 3. Without loss of generality (cf. [4, p. 20]) we can assume that Fis monic in X, i.e., A0=1. The polynomial F(X,1)∈oK[X] is required to be separable and irreducible. Then for arbitrary m∈oKthe equation F(x,y) =min x,y ∈oK is called a Thue equation (over K). Denote by αa zero of F(x,1)in K,letL=K(α) and oLthe integral closure of k[t]in L. Assume that kis the full constant field of L, too. If x,y ∈oKis a solution of the Thue equation then F(x,y) =NL/K (x −αy) =m. (4) Denote by γ(j) (j =1,...,n) the conjugates of any γ∈Lover K. Assume that (x, y) ∈o2 Kis a solution of (4). Then β=x−αy is of norm m, that is βcan be represented in the form β=x−αy =µ·η(5) where ηis a unit in Land µis an element of a finite set Sof non-associated elements of L of norm mover K. For the solution of the corresponding norm equation we use the usual methods from algebraic number theory, i.e., calculate suitable S-units [3]. Those, together with Dirichlet’s unit theorem (cf., e.g., [11]), can be easily transfered to the function field case, too. Denote by η1,...,η ra set of fundamental units in L(that can be calculated by the computer algebra system KASH [2]). Setting k∗=η0, there are integer exponents a0,a 1,...,a rsuch that β=x−αy =µ·ηa0 0·ηa1 1···ηar r.(6) ARTICLE IN PRESS UNCORRECTED PROOF S0022-314X(05)00211-8/FLA AID:3277 Vol.•••(•••) [DTD5] P.6 (1-17) YJNTH:m1 v 1.50 Prn:17/11/2005; 9:56 yjnth3277 by:Vita p. 6 6I. Gaál, M. Pohst / Journal of Number Theory ••• (••••)•••–••• 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 14 15 15 16 16 17 17 18 18 19 19 20 20 21 21 22 22 23 23 24 24 25 25 26 26 27 27 28 28 29 29 30 30 31 31 32 32 33 33 34 34 35 35 36 36 37 37 38 38 39 39 40 40 41 41 42 42 43 43 44 44 45 45 For fixed distinct i, j, k (with 1 ⩽i, j, k ⩽n)setLij k =L(α(i),α(j),α(k))with genus g. Denote by V0a finite set of valuations of Lij k containing the infinite valuations and such that vα(i) −α(j)=0,v α(j) −α(k)=0,v α(k) −α(i)=0ifv/∈V0 and vµ(i)=0,v µ(j)=0,v µ(k)=0ifv/∈V0. Siegel’s identity (holding trivially for any solution, see [4, Chapter 3]) gives α(i) −α(j)β(k) +α(j) −α(k)β(i) +α(k) −α(i)β(j) =0.(7) By the fundamental Lemma 3.1 τij k =(α(j) −α(k))β(i) (α(i) −α(j))β(k) is either of bounded height or is contained in Lp ij k . In the following the height function is applied always in Lij k . 4.2. Effective upper bounds for the solutions of Thue equations In case Eq. (4) has only finitely many solutions we derive an upper bound for the heights of the solutions. If the equation has only finitely many solutions, then there must be i, j, k such that τij k is not a pth power in Lij k . We keep the above notation and set A=max(H(α(i),H(α(j)), H (α(k))). Theorem 4.1. If τij k is not a pth power, then Eq. (4) has only finitely many solutions and for all solutions (x, y) we have maxH(x),H(y)⩽11A+1 nH(µ)+4g−4+2 v∈V0 degv. Proof. Applying Lemma 3.1 we get H(τij k )⩽2g−2+ v∈V0 degv=c1. This implies Hβ(i) β(k) =Hx−α(i)y x−α(k)y⩽H(τij k )+Hα(i) −α(j) α(j) −α(k) ⩽c1+4A=c2. ARTICLE IN PRESS UNCORRECTED PROOF S0022-314X(05)00211-8/FLA AID:3277 Vol.•••(•••) [DTD5] P.7 (1-17) YJNTH:m1 v 1.50 Prn:17/11/2005; 9:56 yjnth3277 by:Vita p. 7 I. Gaál, M. Pohst / Journal of Number Theory ••• (••••)•••–••• 7 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 14 15 15 16 16 17 17 18 18 19 19 20 20 21 21 22 22 23 23 24 24 25 25 26 26 27 27 28 28 29 29 30 30 31 31 32 32 33 33 34 34 35 35 36 36 37 37 38 38 39 39 40 40 41 41 42 42 43 43 44 44 45 45 Using an argument of Mason [6, Chapter II.1] for y=0wehave x y=α(k)β(i)/β(k) −α(i) β(i)/β(k) −1 whence Hx y⩽2A+2c2. By yn=µ n h=1x y−α(h) we derive nH(y) ⩽H(µ)+nHx y+A whence the assertion follows for y. The bound for xcan be obtained similarly. 2 4.3. An algorithm for calculating the solutions of Thue equations We now turn to finding the solutions of Eq. (4). Case I. Consider first the case when τij k is of bounded height. Similarly as in the proof of Theorem 4.1 we obtain Hx−α(i)y x−α(k)y⩽H(τij k )+Hα(i) −α(j) α(j) −α(k) ⩽c1+Hα(i) −α(j) α(j) −α(k) =c 2.(8) By (6) we have x−α(i)y x−α(k)y=µ(i) µ(k) η(i) 1 η(k) 1a1 ···η(i) r η(k) rar whence using (8) we obtain Hη(i) 1 η(k) 1a1 ···η(i) r η(k) rar⩽c 2+Hµ(k) µ(i) =c3. This means for any infinite valuation vof Lij k we have a1·vη(i) 1 η(k) 1+···+ar·vη(i) r η(k) r⩽c3. ARTICLE IN PRESS UNCORRECTED PROOF S0022-314X(05)00211-8/FLA AID:3277 Vol.•••(•••) [DTD5] P.8 (1-17) YJNTH:m1 v 1.50 Prn:17/11/2005; 9:56 yjnth3277 by:Vita p. 8 8I. Gaál, M. Pohst / Journal of Number Theory ••• (••••)•••–••• 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 14 15 15 16 16 17 17 18 18 19 19 20 20 21 21 22 22 23 23 24 24 25 25 26 26 27 27 28 28 29 29 30 30 31 31 32 32 33 33 34 34 35 35 36 36 37 37 38 38 39 39 40 40 41 41 42 42 43 43 44 44 45 45 Note that by interchanging iand kwe get the same expression on the left-hand side with opposite sign: for this reason the inequalities are also valid with absolute values:  a1·vη(i) 1 η(k) 1+···+ar·vη(i) r η(k) r ⩽c3.(9) Note that the units in the above formula have zero values at finite valuations. The inequalities of type (9) (obtained for different choices of i, k) can be used to determine all possible values of the exponents a1,...,a r. For any possible exponent vector a1,...,a rwe can determine η=ηa1 1···ηar rin (5). Then the system of equations x−α(1)y=µ(1)·η(1),x−α(2)y=µ(2)·η(2) can be used to determine the corresponding x,y. Case II. If in (7) we have τij k =(α(j) −α(k))β(i) (α(i) −α(j))β(k) ∈Lp ij k , then using (5) we obtain (α(j) −α(k))µ(i) (α(i) −α(j))µ(k) ·η(i) η(k) ∈Lp ij k . Here the last term is a unit in Lij k hence for any finite valuation vof Lij k v(α(j) −α(k))µ(i) (α(i) −α(j))µ(k)  must be divisible by p. This usually does not hold and there is no Case II solution. Otherwise, τij k is a pth power, say τij k =ψp ij k , we replace τij k by ψp ij k and repeat the argument. Remark. In the above calculations several elements (e.g., α(i) −α(j)) are contained in subfields of type Lij =K(α(i),α(j))of Lij k . Since for elements in Lij the values at any valuation of Lij k can be easily calculated from the values of the corresponding valuations of Lij , hence in fact almost all calculations can be performed in the subfields Lij which are much easier to deal with, especially for large degrees n. 5. Examples Example 1. In the first example we do not need to apply the fundamental lemma. ARTICLE IN PRESS UNCORRECTED PROOF S0022-314X(05)00211-8/FLA AID:3277 Vol.•••(•••) [DTD5] P.9 (1-17) YJNTH:m1 v 1.50 Prn:17/11/2005; 9:56 yjnth3277 by:Vita p. 9 I. Gaál, M. Pohst / Journal of Number Theory ••• (••••)•••–••• 9 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 14 15 15 16 16 17 17 18 18 19 19 20 20 21 21 22 22 23 23 24 24 25 25 26 26 27 27 28 28 29 29 30 30 31 31 32 32 33 33 34 34 35 35 36 36 37 37 38 38 39 39 40 40 41 41 42 42 43 43 44 44 45 45 Let k=F5,letK=k(t),letαbe a root of y3+t7+t=0 and let L=K(α). Consider the Thue equation NL/k(t)(x −αy) =1inx,y ∈k[t].(10) Denote by α(1),α(2),α(3)the conjugates of α. Using symmetric polynomials we have α(3)=−α(1)−α(2) and substituting it into α(1)α(3)+α(2)α(3)+α(1)α(2)=0 we obtain α(2)2+α(1)α(2)+α(1)2=0 whence α(2)=4α(1)±α(1)√−3 2=34α(1)±α(1)√2.(11) Observe that √2 is contained in F25, a quadratic extension of K, hence in this case M=Lα(1),α(2),α(3)=F25(t)(α). Denote by wa generating element of the multiplicative group F∗ 25 of F25 with 2=w6,3=w18,4=w12. By (11) we have α(2)=w16α(1),α (3)=w8α(1). Siegel’s identity gets the form w8x−α(1)y+x−α(2)y+w16x−α(3)y=0.(12) In our case Mhas one infinite valuation. In the above equation all terms are units having zero values at all finite valuations. By the product formula their value at the infinite valuation is also 0, hence they are contained in the constant field F25. Equation (12) leads to the unit equation w4x−α(1)y x−α(3)y+w20 x−α(2)y x−α(3)y=1 ARTICLE IN PRESS UNCORRECTED PROOF S0022-314X(05)00211-8/FLA AID:3277 Vol.•••(•••) [DTD5] P.16 (1-17) YJNTH:m1 v 1.50 Prn:17/11/2005; 9:56 yjnth3277 by:Vita p. 16 16 I. Gaál, M. Pohst / Journal of Number Theory ••• (••••)•••–••• 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 14 15 15 16 16 17 17 18 18 19 19 20 20 21 21 22 22 23 23 24 24 25 25 26 26 27 27 28 28 29 29 30 30 31 31 32 32 33 33 34 34 35 35 36 36 37 37 38 38 39 39 40 40 41 41 42 42 43 43 44 44 45 45 γ1=α(2)−α(3)x−α(1)y, γ2=α(3)−α(1)x−α(2)y, γ3=α(1)−α(2)x−α(3)y, then we have γ1+γ2+γ3=0.(26) The field Mhas genus 13. It has eight infinite valuations, all of degrees 1. The x−α(i)y are units, having nonzero values only at the infinite valuations. The quotients α(1)−α(2) α(1)−α(3),α(2)−α(3) α(2)−α(1),α(3)−α(1) α(3)−α(2) have nonzero values all together at four finite valuations, two of them being of degree 4, the other two of degree 6. Denote by V0the set of the eight infinite and these four finite valuations. Then by the fundamental Lemma 3.1 γi/γjis either of bounded height, or is contained in M5. Case I. Assume Hγ1 γ3⩽2·13 −2+(8+12 +8)=52. This implies Hx−α(1)y x−α(3)y⩽52 +Hα(1)−α(2) α(2)−α(3)=65.(27) As we mentioned above, Khas unit rank 3. We denote by ε1,ε2,ε3the fundamental units. Considering the values of vε(i) h ε(k) h (at infinite valuations) for h=1,2,3, by (27) and (9) we become x−αy =µ·εa1 1·εa2 2·εa3 3 with a root of unity µin kwhere among others the exponents satisfy |50a1+6a2+54a3|⩽65, |51a1+5a2+54a3|⩽65, |49a1+3a2+54a3|⩽65. ARTICLE IN PRESS UNCORRECTED PROOF S0022-314X(05)00211-8/FLA AID:3277 Vol.•••(•••) [DTD5] P.17 (1-17) YJNTH:m1 v 1.50 Prn:17/11/2005; 9:56 yjnth3277 by:Vita p. 17 I. Gaál, M. Pohst / Journal of Number Theory ••• (••••)•••–••• 17 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 14 15 15 16 16 17 17 18 18 19 19 20 20 21 21 22 22 23 23 24 24 25 25 26 26 27 27 28 28 29 29 30 30 31 31 32 32 33 33 34 34 35 35 36 36 37 37 38 38 39 39 40 40 41 41 42 42 43 43 44 44 45 45 There are about 8000 solutions (a1,a 2,a 3)of the above system of linear inequalities. Testing all possible exponent vectors we found that Eq. (24) has only the trivial solutions (x, y) =(1,0), (2,0), (3,0), (4,0)(these yield in fact the multiplies of x−αy for x=1,y=0 with roots of unity in k). Case II. To exclude γ1 γ3∈M5we consider γ1 γ3=α(2)−α(3) α(1)−α(2)·x−α(1)y x−α(3)y. The second term on the right-hand side is a unit, hence vα(2)−α(3) α(1)−α(2) should be divisible by 5 at all finite valuations v. This is not satisfied, however. (Similarly for γ1/γ2and γ2/γ3.) Computational experiences. All computations used in the examples were performed by using the computer algebra system KASH [2], running on 1 GHz PC-s. The calculations took just some seconds with the exception of the test of about 8000 possible exponent vectors in Example 4 which took about 90 minutes. Acknowledgment The authors are thankful to the referee for his/her valuable remarks that lead to an improvement of the paper. References [1] A. Baker, Transcendental Number Theory, Cambridge Univ. Press, Cambridge, 1990. [2] M. Daberkow, C. Fieker, J. Klüners, M. Pohst, K. Roegner, K. Wildanger, KANT V4, J. Symbolic Comput. 24 (1997) 267–283. [3] C. Fieker, Über relative Normgleichungen in algebraischen Zahlkörpern, PhD thesis, Berlin, 1997. [4] I. Gaál, Diophantine Equations and Power Integral Bases, Birkhäuser Boston, Boston, 2002. [5] K. Gy˝ ory, Bounds for the solutions of norm form, discriminant form and index form equations in finitely generated integral domains, Acta Math. Hungar. 42 (1983) 45–80. [6] R.C. Mason, Diophantine Equations Over Function Fields, Cambridge Univ. Press, Cambridge, 1984. [7] H. Niederreiter, C. Xing, Rational points on curves over finite fields, in: London Math. Soc. Lecture Note Ser., vol. 285, Cambridge Univ. Press, Cambridge, 2001. [8] W.M. Schmidt, Thue’s equation over function fields, J. Austral. Math. Soc. Ser. A 25 (1978) 385–422. [9] H. Stichtenoth, Algebraic Function Fields and Codes, Springer, Berlin, 1993. [10] A. Thue, Über Annäherungswerte algebraischer Zahlen, J. Reine Angew. Math. 135 (1909) 284–305. [11] E. Weiss, Algebraic Number Theory, New York, 1963.