On Thue equations of splitting type over function fields. Volker Ziegler. Project Area(s): Algorithmische Diophantische Probleme

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1 FoSP Algorithmen & mathematische Modellierung FoSP Forschungsschwerpunkt Algorithmen und mathematische Modellierung On Thue equations of splitting type over function fields Volker Ziegler Project Areas): Algorithmische Diophantische Probleme nstitut für Analysis und Computational Number Theory Math A) Report , December 2007

2 ON THUE EQUATONS OF SPLTTNG TYPE OVER FUNCTON FELDS VOLKER ZEGLER Abstract n this paper we consider Thue equations of splitting type over the ring k[t ], ie they have the form XX p 1 Y ) X p Y ) Y d = ξ, with p 1,, p k[t ] and ξ k n particular we show that such Thue equations have only trivial solutions provided the degree of p is large, with respect to the degree of the other parameters p 1,, p d 2 1 ntroduction Let F Z[X, Y ] be a homogeneous, irreducible polynomial of degree d 3 Then the Diophantine equation F X, Y ) = m, m Z \ {0} is called a Thue equation in honour of Axel Thue [23] who proved the finiteness of the number of solutions Since then several Thue equations and also families of Thue equations were solved n particular families of Thue equations of the form 1) XX a 1 Y ) X a Y ) + Y d = ±1, with a 1,, a were studied by several authors, eg Heuberger [9], Lee [12], Mignotte and Tzanakis [16], Pethő [18], Pethő and Tichy [19], Thomas [22] and Wakabayshi [24] This type of Thue equations is called splitting type Obviously these Thue equations have solutions ±1, 0), ±0, 1), ±a 1, 1),, ±a, 1), which are called trivial Thomas [22] investigated Thue equations of splitting type of degree d = 3 with a 1 = p 1 n), a 2 = p 2 n), where p 1, p 2 are monic polynomials with degp 1 ) < degp 2 ) and n Z large Under some complicated degree conditions for p 1 and p 2 Thomas proved that the trivial solutions are the only solutions provided n Z is large These investigations led him to the conjecture that for a 1 = p 1 n),, a = p n), and n sufficiently large, p 1,, p monic polynomials with degp 1 ) < < degp ) Diophantine equation 1) has only trivial solutions This conjecture was finally settled by Heuberger [10] under some complicated degree conditions However, there exist counter examples to Thomas conjecture for d = 3 Ziegler [25] observed that the Thue equations and XX ny )X n 4 + 3n)Y ) + Y 3 = ±1 XX ny )X n 4 2n)Y ) + Y 3 = ± Mathematics Subject Classification 11D59,11D25,11Y50 Key words and phrases Thue equations, Thomas conjecture, Function fields The author is supported by the Austrian Science Fund FWF project numbers: S9603) 1

3 2 V ZEGLER have the non-trivial solutions ±n 9 + 3n 6 + 4n 3 + 1, n 8 + 3n 5 + 3n 2 ) and ±n 3 1, n 8 +3n 5 3n 2 ) respectively These counter examples were found by solving Thue equation 1) for d = 3 over the function field CT ), ie assume X, Y, a 1, a 2 C[T ] see [25]) Thue equations over function fields were investigated by Gill [7], Osgood [17], Schmidt [21], Mason [14] see also [15]), Lettl [13] and many others Also the case of global function fields has been considered by Gaál and Pohst [5, 6] recently Families of Thue equations were investigated by Fuchs and Ziegler [3, 4] and Fuchs and Jadrijević [2] A first attempt to prove a function field analogon of Thomas conjecture was made by Ziegler [25], who considered equation 1) in the case of d = 3 The purpose of this paper is to prove an analogon of Thomas conjecture for general d Therefore we consider the equation 2) XX p 1 Y ) X p Y ) + Y d = ξ, over k[t ], where p 1,, p k[t ], ξ k and k an algebraic closed field of characteristic 0 n particular we prove the following theorem: Theorem 1 Let 0 < degp 1 ) < < degp d 2 ) < degp ) and assume c d degp d 2 ) < degp ) with c d = 1031d) 2 2d 3)4 Then Thue equation 2) has only trivial solutions ζ1, 0), ζ0, 1), ζp 1, 1),, ζp, 1), with ζ d = ξ This is an analogon of a result of Halter-Koch, et al [8] for function fields The plan of the paper is as follows n the next section we give a short overview of the tools we need for a proof of the theorems Then we investigate the unit structure of the relevant function fields see section 3) With the knowledge of section 3 we are able to adopt a method described by Heuberger et al [11] and find a lower bound for deg Y An upper bound for deg Y is found by Mason s fundamental lemma [15, Lemma 2,Chapter 1] Compairing upper and lower bounds yields Theorem 1 Note that for the rest of the paper we will assume k is an algebraic closed field of characteristic 0 2 Preliminaries First, we state Mason s fundamental lemma [15, Lemma 2,Chapter 1], which is a special case of the ABC-theorem for function fields see eg [20, Theorem 717]) Lemma 1 Let K/kT ) be a function field of genus g, let us denote by M K the set of all valuations in K, let H K α) := ω M K min0, ωα)) denote the height of α K and let γ 1, γ 2, γ 3 K with γ 1 + γ 2 + γ 3 = 0 Let V be a finite set of valuations such that for all ω V we have ωγ 1 ) = ωγ 2 ) = ωγ 3 ), then H K γ 1 /γ 2 ) max0, 2g 2 + V ) Let K/kT ) be an extension of function fields, then we are interested in the integral closure of k[t ] in K, which is denoted by O K Obviously O K is a Dedekind ring and all primes of k[t ] are tamely ramified Assume K is Galois over kt ), then we know 3) N K/kT ) D OK /k[t ]) = a kt a) ea 1)ga = δ OK /k[t ], where D B/A and δ B/A denote the different and the discriminant of B over A respectively Moreover, T a)o L = p 1 p ga ) ea, ie e a is the ramification index of T a) in O L The equation above holds since the residue class degree is 1 in

4 THUE EQUATONS OF SPLTTNG TYPE 3 this case see also [25]) Beside the valuations which are obtained from the primes T a) with a k there also exist the infinite valuations which are obtained by the unique maximal ideal of the discrete valuation ring O := {ft )/gt ) : f, g C[T ], degf) degg)} kt ) The following result is useful to determine ramifications and valuations: Proposition 1 Puiseux) Let the function field K/kT ) be defined by the polynomial P X, T ) = X d + P T )X + + P 0 T ) with coefficients P 0,, P kt ) Then for each a k there exist formal Puiseux series y i,j = c h,i ζ hj i T a) h/ea,i h=m i 1 j e a,i, 1 i r a ), where c h,i k and ζ i k is an e a,i -th root of unity such that P X, T ) = r a e a,i i=1 j=1 X y i,j ) Moreover, let P 1,, P ra be the primes of K lying above the prime T a) Then e a,i = ep i T a)) for i = 1,, r a for some appropriate order of the indices Note that a similar statement holds also for infinite valuations Furthermore, the m i are the valuations of α with respect to the primes above T a), where α is a root of P X, T ) An essential tool in our proof is Mason s fundamental lemma see Lemma 1) But for an application of this lemma, we need a tool to compute the genus of function fields The Riemann-Hurwitz formula see eg [20, Theorem 716]) yields such a tool Proposition 2 Riemann-Hurwitz) Let L/K be a geometric extension of function fields with constant field k Let g K and g L be the genera of K and L, respectively Then 4) 2g L 2 = [L : K]2g K 2) + w M L e w 1), where M L is the set of valuations of L and e w denotes the ramification index of w M L in the extension L/K A geometric extension L/K is a finite algebraic extension of function fields such that L k = k, where k is the constant field of K and L Note, if k is algebraic closed, then every finite algebraic extension is geometric 3 Unit structure Let F X) = XX p 1 ) X p ) + 1 k[t, X], α a root of F X), K = kt, α) and L K the splitting field of F over kt ) Moreover, let us assume K and L are imbedded by a fixed morphism into K := r=1kt 1/r )), which is the algebraic closure of kt ) We fix this embedding for the rest of the paper Furthermore, let ν be the valuation such that νf) = degf) for any f K and ν0) = n the sequel we will use both the deg-notation and the ν-notation

5 4 V ZEGLER n order to distinguish between polynomials k[t ] and algebraic functions K we will use the the deg-notation only for polynomials Let d i = να i ), for 0 i d 1, with d 0 d, where α 0,, α L K are the conjugates of α Then we have: Lemma 2 Let d i = degp i ) and d 0 = i=1 d i Then we have να i ) = d i = d i for 0 i d 1 Proof Suppose x = να i ) and d k > x > d k+1 for some k = 0, 1,, d 2 or d > x and k = d 1 Then we have 0 = ν1) = να i α i p 1 ) α i p )) = k + 1)x j=k+1 d j < 0 a contradiction Similarly we get a contradiction if x > d 0 Hence, each d j is equal to some d i Therefore we have to prove that d j d i if i j Obviously d 0 = d 0 respectively d = d since otherwise 0 = ν1) = να 0 α ) < 0 respectively d = degp p ) = να α ) να ) = d > d Let j be the largest index such that d j = d j 1 = = d j k+1 for some k > 1 Then we obtain the following equations: d = deg j p j = ν j α j = να ) = d, d d d 2 = deg p j1 p j2 = ν α j1 α j2 = να α d 2 ) j 1<j 2 j 1<j 2 = d + d d 2 d d j+1 = deg p j1 p jd j 1, j 1< <j d j 1 =ν α j1 α jd j 1 = να α j+1 ) j 1< <j d j 1 = d + + d j+1

6 THUE EQUATONS OF SPLTTNG TYPE 5 This yields d i = d i for all i > j But we have d d j = deg p j1 p jd j j 1< <j d j =ν α j1 α jd j j 1< <j d j να α j ) = d d j+1 + d j d d j+1 d j and therefore d j = d j 1 = = d j k+1 = d j This implies d d j k+1 = deg p j1 p jd j+k 1 j 1< <j d j+k 1 =ν α j1 α jd j+k 1 j 1< <j d j+k 1 =να α j k+1 ) = d d j+1 kd j which is a contradiction to the assumption d 0 < d 1 < < d unless k = 1 The lemma above implies F X) is irreducible Otherwise F X) = P 1 X)P 2 X) such that P 1 α 0 ) 0 and {α i : i N {1, 2,, d 1}} is a complete set of conjugate units such that να i ) < 0 for each i N which is a contradiction Furthermore, we conclude that K is unramified at infinity since all the infinite valuations of α are distinct by the lemma above Similarly we deduce that L is not ramified at infinity, since for each automorphism σ GalL/kT )) not the identity we can find two indices j and k such that να j α k ) νσα j α k )) Therefore we have: Lemma 3 The polynomial F X) is irreducible and the fields K and L are unramified over kt ) at infinity Let us denote by i the valuation that is induced by the imbedding σ i : K K such that α α i and let α i := νσ i α) Then we consider the free Abelian group D generated by 0,, Because of the previous lemma we know that all the valuations i are distinct The group D is called the group of polar divisors We also define the polar height H ) K a a ) = min{0, a i }) Similar we can define for every a k the group of local divisors D a, which is freely generated by the valuations above the finite valuation induced by the prime T a) Let ω 1,, ω ga be these valuations, then H a) K a 1ω a ga ω ga ) = min{0, a i }) is called the local height For every α K we can define the principal divisor α) = ω M K ωα)ω of α and similar we can define the principal) polar and local divisors of α Note M K is the set of all valuations of K Proposition 3 Let ɛ k[t, α] O K with ɛ k, then we have H K ɛ) d 0 = d i i=1

7 6 V ZEGLER Proof Let us write ɛ = h 0 + h 1 α + + h α, with h 0,, h k[t ] Moreover, we define m 0 = min{ degh 0 ), degh 1 ) d 0,, degh ) d 1)d 0 }, m 1 = min{ degh 0 ), degh 1 ) d 1,, degh ) d 1)d 1 }, m = min{ degh 0 ), degh 1 ) d,, degh ) d 1)d } We see that the polar divisor of ɛ has j -coefficient m j if the corresponding minimum occurs only once Note that the minimal index j with m i = degh j ) jd i is decreasing with i Now, let us assume m l = deg h j jd l is the singular minimum with l maximal Then we have degh j ) jd l degh j1 ) j 1 d l degh j1 ) j 1 d l+1 degh j2 ) j 2 d l+1 degh jd l 1 ) j d l 1 d = degh jd l ) j d l d with j = j 0 j 1 < < j d l 1 < j d l and each j i minimal with 1 i d l 1 Moreover j d l is maximal such that degh jd l ) j d l d = m Since j d l d 1 we conclude j l and by the inequalities above we obtain degh j ) j d i j d i 1 )d i degh jd l ) i=l i= j d l j) Therefore m l = degh j ) jd l i= j d l d i Furthermore, there are at least d j d l singular minima m i and at least d j d l 1 such minima with i > 0 ndeed, if m i = degh ji ) + j i d i and m i+1 = degh ji+1 ) + j i+1 d i+1 with j i and j i+1 minimal and m i not a singular minimum, then j i < j i+1 Therefore we have singular minima m ik other than m l with 1 i 1 < < i d 2 jd l We claim m ik d ik, with k = 1,, d 2 j d l Each minima m i is of the form degh j ) jd i f j 0 we are done Otherwise degh 0 ) < degh j ) jd i d i for some j with degh j ) Note that the case h j = 0 for each j > 0 leads to ɛ = h 0 k[t ] and this leads to ɛ k Altogether we deduce m ik d k and d 2 j d l H K ɛ) m l k=1 m ik k= j d l d k d 2 j d l k=1 d i m ik = d k By an analog of Dirichlet s unit theorem for function fields we know that there are at most d 1 multiplicatively independent units that generate the unit group of k[t, α] This fundamental system of units spans a lattice, ie consider the map log : k[t, α] Z with logɛ) = ɛ 1,, ɛ ), then the image of k[t, α] is a lattice Λ Z Moreover, logɛ) is the vector of components of the polar divisor of ɛ except the component of 0 t is obvious that a set of units is multiplicatively k=1

8 THUE EQUATONS OF SPLTTNG TYPE 7 independent if and only if the corresponding vectors in the lattice are independent Let us consider the set E = {ɛ 1,, ɛ } of d 1 independent units Then the absolute value of the determinant of the matrix whose rows are logɛ 1 ),, logɛ ) is usually called the regulator of E Note that the regulator is the same as the lattice constant of the lattice spanned by the vectors logɛ 1 ),, logɛ ) Let us fix the following set of units E = {α, α p 1,, α p d 2 } Then we have Lemma 4 The set E is a system of multiplicatively independent units and the corresponding regulator R E is 2d 1 + d 2 + d )3d 2 + d d ) d 1)d d 2 + d )d Proof Note, if l k then α p l k = { dl if k < l, d k if k > l Since the sum of all valuations of an element is zero and α p l has non-zero valuations only at infinity we deduce α p k k = ld l + k=l+1 d k Therefore d 1 d 2 d 3 d d 1 + k=2 d k d 2 d 3 d d 2 2d 2 + k=3 d k d 3 d d 3 d 3 3d 3 + k=4 d k d d d 2 d d 2 d d 2 d is the matrix which consists of the vectors logα), logα p 1 ),, logα p d 2 ) After a 180 rotation and Gaussian elimination we obtain the upper triangular matrix d 0 d 1)d d 2 + d 0 0 2d 1 + k=2 d k This yields the lemma Lemma 5 Let Σ = k=1 d k = d 0 Then we have R E 2 2 ) d 2 Σ) < 2Σ) d 1 Proof By Lemma 4 we have R E =2d 1 + d 2 + d )3d 2 + d d ) d 1)d d 2 + d )d =Σ + d 1 )Σ + 2d 2 d 1 ) Σ + d 2)d d 2 d 1 d d 3 ) Σ d 1 d d 2 ) ) Σ + d1 ) + + Σ d 1 d d 2 ) d 1 ) d1 + 3d 2 + 5d d 5)d d 2 + d 1)d = d 1

9 8 V ZEGLER The inequality is due to the arithmetic-geometric mean We have to maximize 5) d 1 + 3d d 5)d d 2 + d 1)d under the condition 6) 0 d 1 d 2 d Σ d 1 d 2 d d 2 Since this describes a linear program the maximum lies in a corner of the polytop defined by 6) Let us consider the corner d 1 = d 2 = = d k = 0 and d k+1 = = d = Σ/d 1 k) for some k = 0,, d 1 For this corner assume k < d 1) function 5) turns into Σ d 1 k d 2) + j=k+1 Σ 2j 1) = Σd 1)2 k 2 ) Σd 2) d k 1 d 1 k = d 1 + k d 2 ) Σ2d 1) 2 d 2) d 1 k For k = d 1 the quantity 5) is zero and yields no maximum Let U be the group of units of k[t, α] and Ũ the group of units generated by E, then we know that := [U : Ũ] = R E R, where R is the regulator of U Therefore we have to find a lower bound for R in order to find an upper bound for Before we determine such an upper bound we prove the following lemma 2 Lemma 6 There is a symmetric convex region S with λs) Σ around 0 such that logu) S = {0} and λ denotes the d 1-dimensional Lebesgue measure Proof By Proposition 3 for any unit ɛ U \ k we have H K ɛ) Σ Let logɛ) = e 1,, e ), then { max max{0, ej }, } min{0, e j } Σ We consider the set S = {x = x 0,, x ) R : xi Σ} Then S consists of 2 simplices of volume Σ and fullfills logu) S = {0} Proposition 4 We have d 1)4 Proof We apply Minkowski s convex body theorem see [1][Theorem ]) to the region S of Lemma 6 This yields R Σ/2) On the other hand we have by Lemma 5 the inequality R E < 2Σ), hence = R E R d 1)4 4 Application of Mason s fundamental Lemma Let X, Y ) be a solution to 2) n the classical approach to Thue equations the terms X α i Y and α k α l )X α i Y ) are denoted by β i and γ k,l,i respectively Then Siegel s identity can be written as γ k,l,i + γ l,i,k + γ i,k,l = 0

10 THUE EQUATONS OF SPLTTNG TYPE 9 We apply Mason s fundamental lemma Lemma 1) in order to find an upper bound for H L γ k,l,i γ l,i,k ) Since L/kT ) is Galois and k is of characteristic zero we have by equation 3) δ OL /k[t ] = a kt a) ea 1)ga Let δ k[t,α]/k[t ] be the discriminant of k[t, α] over k[t ] Since δ k[t,α]/k[t ] = i<jα i α j ) 2 k[t ] we have degδ k[t,α]/k[t ] ) = 2 i<j max{d i, d j } = 2 j=1 jd j Moreover, algebraic number theory shows that δ k[t,α]/k[t ] and δ OL /k[t ] differ only by a square factor say R 2 Let degr) = r By Hurwitz formula we obtain 2g L 2 = deg L2g kt ) 2) + a e a 1)g a = 2 deg L + a e a 1)g a By Mason s fundamental lemma we get ) γk,l,i H L 2 deg L + γ l,i,k a e a 1)g a + V + V = deg L + a e a 1)g a + {p O L : p δ} deg L + a e a 1)g a + a ) = deg L r a g a + r deg L All the sums are taken over all a such that T a) δ OL /k[t ] Moreover, V denotes the set of all valuations ω of L such that ωα i α j ) 0 for some i, j and V denotes the set of all infinite valuations of L This enables us to compute a bound for H K β): Lemma 7 H K β) d 1) r 1 + a 1 ) + d 1 d d + 4d 1) d d 2 27 Proof Let b 0 b 1 b be the infinite valuations of β = X αy over K Since β is a unit we have j=0 b j = 0 Moreover, we have νx α i Y ) 0 unless deg X deg Y + d i, hence H K β) = b 0 ndeed there is at most one positive valuation of β, which is by definition b 0 Furthermore, we have ) ) βi α k α l ) H L H ) βi α k α l ) L β k α l α i ) β k α l α i ) =H L βi β k ) H ) L αk α l α l α i )

11 10 V ZEGLER Let H be a system of representatives of S d / GalL/kT )) Note that the symmetric group S d acts on L by permuting the conjugates of α We obtain ) max H γk,l,i L G H L ρ β ) 1 H L ρ α ) 2 α 3 i,j,k γ l,i,k S d β 2 α 3 α 1 ρ H ρ H 7) = G max 0, ν σ β )) 1 max 0, ν σ α )) ) 2 α 3 S d β 2 α 3 α 1 σ S d σ S d Let us consider the first sum of 7): G 0, ν σ β )) 1 = deg L d 2)! i b j ) S d β 2 d! i<jb σ S d max = deg L dd 1) [d 1)b 0 + d 3)b d + 1)b ] = deg L dd 1) [ 2d 2)b 2b 1 ] deg L dd 1) db 0 = deg L d 1 H Kβ) The last inequality is true, since 2d 2)b 2b 1 with 0 b 1 b b 0 = j>0 b j takes its minimum for b 1 = = b = b0 Now we investigate the second sum of 7) Let us remark that max 0, α ) k α l = α l α i 0 if l > k or i > k, d k d i if k > i > l, d k d l if k > l > i This yields G S d 0, ν σ α )) 2 α 3 α 3 α 1 deg L dd 1)d 2) σ S d max = i=1 = deg L d! i 1 2d 3)! jd i d j ) i=2 j=1 i 1 ii 1)d i 2 jd j i=2 j=1 deg L = ii 1) 2id i 1))d i dd 1)d 2) i=1 deg L = d i)d 3i + 1)d d i dd 1)d 2) i=1 deg L deg L d + d dd 1)d 2) d d 2 deg L d d + 4 deg L d d 2 27 [ d+1 3 ] d i)d 3i + 1) Combining these estimates with 7) we get the desired result i=2

12 THUE EQUATONS OF SPLTTNG TYPE 11 Before we establish an upper bound for deg Y let us note that we may exclude the case X = 0 n this case equation 2) turns into Y d = ξ, which yields a trivial solution Therefore we may assume deg X 0 Let us note νx α j Y ) < 0 unless deg X = deg Y + d j f β is not a constant there exists an index j such that νx α j Y ) > 0 Let us fix this index j and let b 0 b the infinite valuations of β Then we have b 1 = = b j = deg X, b k = deg Y d k for k > j and b 0 = j deg X + d 1 j) deg Y + d k Hence H K β) = b 0 = d 1) deg Y + jd j + Together with Lemma 7 we obtain deg Y r 1 + a j<k j<k d k d 1) deg Y + d k d j 1 d 1 + d + 4d d 2 d 27 j=1 Counting the number of ramifications yields: Proposition 5 We have deg Y j 1 ) d 1 j=1 First, we exclude the case Y k d j 1 + d d 5 A lower bound for deg Y Lemma 8 f Y k then X, Y ) is a trivial solution + 4d d 2 27 Proof Let us write Y = ζ k and assume X ζp j for all 1 j d 1 and X 0 Then the right hand side of 2) has degree at least j=1 d j = d 0 > 0 which is a contradiction Therefore X = ζp j for some j or X = 0 This yields Y d = ζ d = ξ, hence X, Y ) is a trivial solution As mentioned above there is some index j such that νβ j ) = νx α j Y ) > 0 Let us fix this index j Then we have for k j )) β j νβ k ) =νx Y α k ) = ν Y α j α k ) 1 + = Y α k α j ) να j α k ) deg Y β j Y α k α j) Note that ν unit and therefore ) > 0, hence ν1 + β k=1 j Y α k α j) ) = 0 On the other hand β k is a β i = α k ) B0/ α k p 1 ) B1/ α k p d 2 ) B d 2/ where B 0, B 1,, B d 2 Z and 4 d 1) This yields 8) να k ) B 0 + να k p 1 ) B να k p d 2 ) B d 2 = να j α k ) deg Y

13 12 V ZEGLER for k j Solving system 8) with k j for B i we obtain 9) where B k = v k u k deg Y R u k = detνα i ),, να i p k 1 ), 1, να i p k+1 ), να i p d 2 )) i j, v k = detνα i ),, να i p k 1 ), να i α j ), να i p k+1 ), να i p d 2 )) i j, R = det να i ), να i p 1 ),, να i p d 2 )) i j Lemma 9 We have with R = 1) d+j R E R k = j d 1; v k = R k = d 1; 0 otherwise; Proof Let k = j d 1 Then the matrices which determine v k and R are identical Moreover, the matrix corresponding to R can be transformed to the matrix corresponding to the regulator by summing up all lines in the last line, multiplying the last line by 1 and exchanging d 1 k lines Now let us assume j k and k < d 1 The j-th respectively k-th column of the matrix corresponding to v k are να j ), να j p 1 ),, να j p )) T and να j α 0 ),, να j α )) T, where we omit the j + 1-th entry Since these two columns are equal we deduce v k = 0 n the case of k = we sum up all columns of the corresponding matrix to v k in the k-th column We multiply this column by 1 and obtain the matrix which corresponds to R Lemma 10 We have and where and u d 2 R = u R = d d 2 d d )d d 2 +d ) if j < d 2, d d 2 +)d d )d d 2 +d ) if j = d 2, 1 d if j = d 1, U d d d 2)d d 3 +d d 2 +d ))d d 2 +d ) if j < d 3, U d d d 2)d d 3 +d d 2 +d ))d d 2 +d ) if j = d 3,, d d 2 d d )d d 2 +d ) if j = d 2, 1 d if j = d 1, U d = d 2)d d 3 d d 2 + d 2 d 2 + 2d d 3 d dd d 2 d d 2 U d = d d 2 d d 3 )d d 2 d ) + d 1)d d 2 + d )d d 3 + d 1)d ) Moreover, u k = R d if j = d 1

14 THUE EQUATONS OF SPLTTNG TYPE 13 Proof We consider only the case k = d 1 and j < d 2 since the other cases run analogously Let us write Σ 0 = Σ = d d, Σ i = id i + d i d and Σ i = i + 1)d i + d i d for i = 1, d 1 We note that R = 1) d+j Σ 1 Σ d 2 d n order to prove the lemma we have to compute the determinant of Σ 0 d 1 d j 1 d j d j+1 d d 3 1 d 1 Σ 1 d j 1 d j d j+1 d d 3 1 d j 1 d j 1 Σ j 1 d j d j+1 d d 3 1 d j+1 d j+1 d j+1 d j+1 Σ j+1 d d 3 1 d d 3 d d 3 d d 3 d d 3 d d 3 Σ d 3 1 d d 2 d d 2 d d 2 d d 2 d d 2 d d 2 1 d d d d d d 1 which is 1) u This is done by Gaussian elimination Let us assume j 0 First, we subtract from every row except the first row) the first row and obtain the matrix, Σ 0 d 1 d j 1 d j d j+1 1 Σ + d 1 Σ Σ + d j 1 d j 1 d 1 Σ j Σ + d j+1 d j+1 d 1 d j+1 d j 1 d j+1 d j Σ j+1 0 Σ + d d d 1 d d j 1 d d j d d j+1 0 Next, we sum up all columns and write this sum instead of the first column Then the first column is of the form 1 d d 2 d, 0,, 0, d 1)d d 2 + d, d d 2 + d 1)d ) T This yields u = 1) j Σ 1 Σ d+j 1 det M, where d j+1 d j Σ j d j+2 d j d j+1 d j 0 0 M = d d 3 d j d d 3 d j+1 Σ d 3 0 d d 2 d j d d 2 d j+1 d d 2 d d 3 Σ d 2 d d j d d j+1 d d d 3 d d 2 + d 1)d We obtain M from the previous matrix if we place the first column behind the last column and delete the first j rows and columns Next, we subtract form the last row the second to last row and then from the second to last row the third to last

15 14 V ZEGLER row and so on By transposing this new matrix we obtain d j+1 d j d j+2 d j+1 d d 2 d d 3 d d d 2 Σ j+1 Σ j+1 + d j+2 d d 2 d d 3 d d d 2 0 Σ j+2 d d 2 d d 3 d d d Σ d 3 + d d 2 d d d Σ d 2 d 2)d d d 2 ) We multiply the last row by 1 and add the last row to the second to last Then we add the second to last row to the third to last row and so on This yields Σ j+1 d j jd j+1 d j+2 ) jd d 2 d ) Σ j+1 j + 1)d j+1 d j+2 ) j + 1)d d 2 d ) 0 Σ j+2 j + 2)d d 2 d ) 0 0 d 3)d d 2 d ) 0 0 d 2)d d 2 d ) We multiply the second row by j and subtract j + 1 times the first Then we divide the first line by j Therefore the new matrix has the same determinant but the first two rows are of the form Σj+1 dj j d j+1 d j+2 d d 2 d Σ j 0 0 Now we eliminate the other rows and obtain Σj+1 dj j d j+1 d j+2 d d 3 d d 2 d d 2 d Σ j Σ j Σ d 3 0, which yields the lemma in this case n the case j = 0 we have to compute the determinant of the matrix d 1 Σ 1 d 2 d d 3 1 d 2 d 2 Σ 2 d d 3 1 d d 3 d d 3 d d 3 Σ d 3 1 d d 2 d d 2 d d 2 d d 2 1 d d d d 1

16 THUE EQUATONS OF SPLTTNG TYPE 15 Subtracting the first column from all other columns except the last column yields the matrix d 1 Σ 1 d 2 d 1 d d 3 d 1 1 d 2 0 Σ 2 d d 3 d 2 1 d d Σ d 3 1 d d d whose determinant is Σ 1 Σ d 3 d d d 2 ) The last statement of the lemma can be easily deduced Multiply the j + 1-th column by d and add all other columns to this column This yields the matrix corresponding to R As indicated by Heuberger, et al see [11]) we want to find a linear combination of the equations 9) such that we get a lower bound for deg Y Due to Lemmas 9 and 10 we have for j < d 3 B d 2 B = dd d 2 d d 3 ) deg Y Σ d 3 Σ d 2, for j = d 3 1 d 1)B B d 2 = dd 2)d d 3 d d 2 + d 2 d 2 d 3)d d 3d + d 2)d d 2 d ) deg Y d Σ d 3 Σ d 2 and for j = d 2 1 B d 1)B d 2 = dd d 2 deg Y Σ d 2 d For j = d 1 we find 10) B 0 = = B = d + deg Y d We note that the right hand sides of the equations above are > 0 Since the numerator on the left side is an integer the right hand sides are at least 1/ Hence we get lower bounds for deg Y : Proposition 6 Let j < d 1 and cd d 2 d for some constant c Then we have cd 2d if j < d 3, c deg Y 2 d d+cd 2)) if j = d 3, cd d if j = d 2

17 16 V ZEGLER 6 Proof of the Theorem 1 We have to consider two cases: j < d 1 and j = d 1 The first case is solved by comparing the bounds for deg Y given by Propositions 5 and 6 Let us assume cd d 2 d for some constant c Then we get by Proposition 5 and on the other hand we have by Proposition 6 This yields deg Y d 1)d + d 2)d 1) + 8/27)d 2c c d2d 3) deg Y But this inequality holds only for c c d with c d2d 3) d 2)d 1) + 8/27 d 1 + 2c 9dd 1)2d 3) c d = 18 3d22d 3) d 3)d) + 27d2d2 5d + 3) + 2 ) dd 1)2d 3), where d 1)4 and d 3 Therefore Theorem 1 is proved for j < d 1 Now let us assume j = d 1 By 10) we know that B 0 = = B d 2 =: B, ie β = ξα B/ α p 1 ) B/ α p d 2 ) B/ = α p ) B/ with ξ k, which yields νβ) = B να p ) = B d 1)d < 0, hence B > 0 Let us consider the case d = 3 By Lemma 7 and r + a 1 = j>0 jd j we have H K β) = H K α p 2 ) B/) = 2Bd 2 2 jd j 1 < 6d 2, j>0 B ie < 3 On the other hand we know = 1 for d = 3 see [25]) Hence, we have B = 1, 2 The case B = 1 yields a trivial solution and in the case B = 2 we obtain β = ξα 2 2αp 2 + p 2 2) which is not a solution Therefore we may assume d 4 We compute α k p ) B/ = p B/ B α kp B/ 1 + BB ) 2 2 αkp 2 B/ 2 +,

18 THUE EQUATONS OF SPLTTNG TYPE 17 where the remaining terms have lower valuations provided k < d 1 β = α p ) B/ yields a solution, then by Siegel s identity we have 0 =α 0 α 1 )α 2 p ) B/ + α 1 α 2 )α 0 p ) B/ + α 2 α 0 )α 1 p ) B/ =α 0 α 1 ) p B/ B α 2p B/ α 1 α 2 ) p B/ B α 0p B/ α 2 α 0 ) p B/ B α 1p B/ 1 + ) BB ) 2 2 α2p 2 B/ 2 + BB ) 2 2 α0p 2 B/ 2 + BB ) 2 2 α1p 2 B/ 2 + ) ) Assume BB )pb 2 = α0 2 2 α2 2 α 1 α2 2 + α 1 α0 2 α 2 α0 2 + α 2 α1 2 α 0 α1 2 ) + where the remaining terms have lower valuations Therefore some cancellation occurs in the main term Since να 2 ) < να 1 ) < να 0 ) we deduce BB ) = 0, ie B = 0 or B = n both cases we obtain only trivial solutions Therefore Theorem 1 is proved Remark 1 Note that in the case of d = 3 we have = 1 Therefore we find c 3 = = This improves the bound c 3 = 34 found by Ziegler [25, Theorem 1] Acknowledgement The author wants to thank Clemens Fuchs, who draw the attention of the author to this problem References [1] J W S Cassels An introduction to the geometry of numbers Springer-Verlag, Berlin, 1971 Second printing, corrected, Die Grundlehren der mathematischen Wissenschaften, Band 99 [2] C Fuchs and B Jadrijević On a parametric family of Thue inequalities over function fields to appear in Math Proc Cambridge Philos Soc [3] C Fuchs and V Ziegler On a family of Thue equations over function fields Monatsh Math, 1471):11 23, 2006 [4] C Fuchs and V Ziegler Thomas family of Thue equations over function fields Q J Math, 571):81 91, 2006 [5] Gaál and M Pohst Diophantine equations over global function fields The Thue equation J Number Theory, 1191):49 65, 2006 [6] Gaál and M Pohst Diophantine equations over global function fields R-integral solutions of Thue equations Experiment Math, 151):1 6, 2006 [7] B P Gill An analogue for algebraic functions of the Thue-Siegel theorem Ann of Math 2), 312): , 1930 [8] F Halter-Koch, G Lettl, A Pethő, and R F Tichy Thue equations associated with Ankeny- Brauer-Chowla number fields J London Math Soc 2), 601):1 20, 1999 [9] C Heuberger On a family of quintic Thue equations J Symbolic Comput, 262): , 1998 [10] C Heuberger On a conjecture of E Thomas concerning parametrized Thue equations Acta Arith, 98: , 2001 [11] C Heuberger, A Togbé, and V Ziegler Automatic solution of families of Thue equations and an example of degree 8 J Symbolic Computation, 38: , 2004 [12] E Lee Studies on Diophantine Equations PhD thesis, Cambridge University, 1992 [13] G Lettl Thue equations over algebraic function fields Acta Arith, 1172): , 2005

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