ON A FAMILY OF THUE EQUATIONS OVER FUNCTION FIELDS

CLEMENS FUCHS^{∗} AND VOLKER ZIEGLER^{‡}

Abstract. Many families of parametrized Thue equations over number fields have been solved recently. In this paper we consider for the first time a family of Thue equations over a polynomial ring. In particular, we calculate all solutions of

X(X − Y)(X − (T + ξ)Y) + Y^{3} = 1 + ξT (1 − T)

over ℂ[T] for all ξ ∈ ℂ.

1. Introduction and Main Result

Let F ∈ R[X, Y] be a binary irreducible form of degree d ≥ 3 over some ring R. An equation of the form

F (X, Y) = m, m ∈ R

is called a Thue equation, due to Thue who proved the finiteness of solutions in the case of R = ℤ in his famous paper [17]. Using the ideas of Thue one may prove that a Thue equation over a ring R that is finitely generated over ℤ has only finitely many solutions in R (see [10]), e.g. when R is an order of some number field or R = ℤ[T], where T is transcendental. Nowadays it is known how to solve such a Thue equation over R algorithmically. Using Baker’s method, the above result was made effective by Győry [3]. Let us mention that for R = ℤ the most efficient algorithm is due to Bilu and Hanrot [1].

Now, let K be a function field of characteristic 0, S a finite set of places of K and R the ring of S-integers. By proving an analogue of Baker’s method of linear form in logarithms (see the ABC-Theorem e.g. [15, Theorem 7.17]) for function fields, Mason [13, 14] could prove effective bounds for the height of the solutions. Furthermore, he described how to determine effectively all solutions of a given Thue equation over R. Since R is not finitely generated over ℤ we cannot hope that such a Thue equation has finitely many solutions, e.g. the equation

X^{3} − (T + 1)X^{2}Y + T XY^{2} = 6T^{2}(T − 1)^{2}

2000 Mathematics Subject Classification. 11D59,11D25,11Y50.

Key words and phrases. Thue equation, function fields, ABC-theorem.

∗The first author was supported by the Austrian Science Foundation, grants S8307-MAT and J2407-N12.

‡The second author was supported by the Austrian Science Foundation, grant S8307-MAT.

has infinitely many solutions in ℚ[T], namely (x, y) = ((3c_{1} −2c_{2})T, 3c_{1}T −2c_{2}), where (c_{1}, c_{2}) run through the ℚ-rational points of the elliptic curve C_{1}C_{2}(3C_{1} − 2C_{2}) = 1 (see [12]). This was shown by Lettl [12], who recently considered Thue equations over function fields and gave criteria for which such a Thue equation has infinitely many solutions.

In 1990, Thomas [16] went a step further and considered a parametrized family of Thue equations with positive discriminant. In the last decade, several families of parametrized Thue equations F_{ξ}(X, Y) = m have been investigated up to degree 8 (cf. [8]). Also more general results have been proved (cf. [5, 6]) and furthermore families over imaginary quadratic number fields were considered (cf. [7]) ; a survey containing further references is given in [4]. Typically, such a family of equations has finitely many families of solutions, that means solutions depending e.g. polynomially on the parameter ξ, and finitely many “sporadic” solutions for certain values of ξ. We mention that it was show by Lettl in [11] that a family of Thue equations can certainly have infinitely many sporadic solutions by considering single Thue equations over function fields which have solutions truely lying in the function field.

The aim of this paper is to determine - for the first time - all solutions of a family of Thue equations over the function field ℂ(T). We consider the equation

X(X − Y)(X − (T + ξ)Y) + Y^{3} = 1 + ξT (1 − T),

which has the trivial solution (T, 1) for every value of ξ ∈ ℂ. The left-hand-side of the equation is a so-called splitting form, i.e. X^{3} − (T + ξ + 1)X^{2}Y + (T + ξ)XY^{2} + Y^{3} = X(X − Y)(X − (T + ξ)Y) + Y^{3} (many of the families studied previously in the integer case are of such a shape; we refer to [4]) and the right-hand-side was chosen such that the equation has at least one non-trivial solution for every value of the parameter. We have chosen the simplest equation for which a non-trivial result can be obtained, especially, in order to see which method is needed to prove the following theorem.

Theorem 1. The Thue equation

X^{3} − (T + ξ + 1)X^{2}Y + (T + ξ)XY^{2} + Y^{3} = −ξT^{2} + ξT + 1  (1)

has only finitely many solutions over ℂ[T] for all ξ ∈ ℂ. Let L_{ξ} be the set of solutions (x, y) ∈ ℂ[T] × ℂ[T] to (1) for fixed ξ ∈ ℂ, then

L_{ξ} = {ζ(T, 1) : ζ^{3} = 1},

if ξ ≠ −1, 0, 1. Furthermore,

L_{1} ={ζ(T, 1), ζ(T + 1, T) : ζ^{3} = 1},
L_{−1} ={ζ(T, 1), μ(T − i, T − 1 − i), ν(T + i, T − 1 + i) : ζ^{3} = 1, μ^{3} = −i, ν^{3} = i},
L_{0} ={ζ(T, 1), ζ(1, 0), ζ(1, 1), ζ(0, 1), ζ(1, −T − 1) : ζ^{3} = 1}.

By the work of Lettl [12, Corollary 2] it is immediate that equation (1) has only finitely many solutions over ℂ[T] for each ξ ∈ ℂ, since X^{3} − (T + ξ + 1)X^{2}Y + (T + ξ)XY^{2} + Y^{3} is irreducible, as is shown in Section 2. So the main work is to determine those finitely many solutions for each ξ ∈ ℂ.

In order to prove this Theorem we will follow the original ideas of Mason [14, Chapter 2]. In Section 2 we present some well known results and prove some auxiliary lemmas. In Section 3 we compute the places where ramification occurs in the splitting field of equation (1) and furthermore compute the genus of this splitting field. From these data one obtains a bound for the height of solutions (x, y) to (1) from a theorem of Mason (Proposition 2). By some computations in Section 4 we sharpen these bounds. In particular we prove that deg(x), deg(y) ≤ 2. From this bound we compute in Section 6 all solutions to (1) provided ξ ≠ 0. The special case ξ = 0 is treated in Section 5.

2. Auxiliary Results

Let us first remind the ABC-Theorem for function fields (see e.g. [15, Theorem 7.17]).

Proposition 1 (ABC-Theorem). Let K be a function field with characteristic 0 and genus g_{K}. Let u, v ∈ K^{×} satisfying u + v = 1 and put A = (u)_{0}, B = (v)_{0} and C = (u)_{∞} = (v)_{∞}, where (·)_{0} denotes the zero divisor and (·)_{∞} denotes the polar divisor. Then

deg A = deg B = deg C ≤ max (0, 2g_{K} − 2 + ∑_{P ∈Supp(A+B+C)} deg_{K} P ) .

If the constant field k is algebraically closed and of characteristic 0, Mason [14, chapter 1, Lemma 2] proved the following special case.

Corollary 1. Let H(f) := − ∑_{v∈M_{K}} min(0, v(f)) denote the height of an element f ∈ 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 v ∉ V we have v(γ_{1}) = v(γ_{2}) = v(γ_{3}), then

H(γ_{1}/γ_{2}) ≤ max(0, 2g − 2 + |V|).

Here we denote the set of all valuations in K by M_{K}. Usually, M_{K} denotes the set of places of some field K. Since in the function field case valuations and places are one-to-one we use this notation.

We can deduce Corollary 1 by observing

deg_{K} P = [O_{P} /P : k] = [O_{P} /P : O_{p}/p][O_{p}/p : k] = f(P |p) · 1 = 1,

where O_{P} is a discrete valuation ring with quotient field K, maximal ideal P and p = P ∩ k[T]. From Puiseux’s Theorem (see [2] or [9]) we deduce f(P |p) = 1. The relation [O_{p}/p : k] = 1 is obvious since p is generated by T − a or T^{−1} with a ∈ k.

If F (X, Y) = m is a Thue equation over the ring of integers of some function field L, then Mason [13] could prove an effective bound for the height of solutions (x, y) to (1). By integers we mean the set of all elements in L that may only have negative valuations above ∞. Let us remind Mason’s bound on the height of the solutions.

Proposition 2. Let

F (X, Y) := (X − α_{1}Y) ⋯ (X − α_{d}Y) = m

be a Thue equation over the ring of integers of some function field L with algebraic closed constant field and characteristic 0. Then all solutions (x, y) satisfy

H(x, y) ≤ 8H + 2g_{L} + r − 1,

where g_{L} is the genus of L, r is the number of infinite valuations and H denotes the height of the polynomial (X − α_{1}) ⋯ (X − α_{d})/m.

For the rest of the paper we define

F_{ξ}(X, Y) := X^{3} − X^{2}Y (T + ξ + 1) + XY^{2}(T + ξ) + Y^{3},
f_{ξ}(X) := F_{ξ}(X, 1) = X^{3} − X^{2}(T + ξ + 1) + X(T + ξ) + 1.

The polynomial f_{ξ} is irreducible over ℂ(T). Otherwise f_{ξ} must have a root, which is a constant α, because the constant term is 1. Furthermore, we have f_{ξ}(α) = 0 and in particular the coefficient of T in f_{ξ}(α) must vanish, hence, α = 0, 1, but in both cases α is not a root of f_{ξ}. First we want to compute the Galois group and the splitting field of f_{ξ}.

Proposition 3. The Galois group of f_{ξ} is the symmetric group S_{3}. Let α_{1}, α_{2}, and α_{3} be the roots of f_{ξ} in some algebraic closure of ℂ(T). Then L = ℂ(T)(α_{1} − α_{2}) = ℂ(T)(α_{1}, α_{2}, α_{3}) is the splitting field of f_{ξ} and the polynomial

G_{ξ}(X) :=X^{6} − (2(T + ξ + 1)^{2} − 6(T + ξ)) X^{4} + (T^{2} − T + ξ(2T − 1) + ξ^{2} + 1)^{2}X^{2}

− (T + ξ)^{2}(T + ξ + 1)^{2} − 4(T + ξ + 1)^{3} + 4(T + ξ)^{3} + 18(T + ξ)(T + ξ + 1) + 27

is the minimal polynomial of α_{1} − α_{2}.

Proof: Let us compute the discriminant Δ_{ξ} of f_{ξ}. We get

Δ_{ξ}^{2} = (T + ξ)^{2}(T + ξ + 1)^{2} + 4(T + ξ + 1)^{3} − 4(T + ξ)^{3} − 18(T + ξ)(T + ξ + 1) − 27.

A well known criteria for cubic polynomials says that the Galois group is not the symmetric group, if and only if the square of the discriminant is a square in the ground field. Hence, if the Galois group of f_{ξ} is not S_{3}, we would have Δ_{ξ}^{2} = (aT^{2} + bT + c)^{2} for some a, b, c ∈ ℂ. Comparing the coefficients of T yields a system of equations in a, b, c and ξ. A short computation shows that this system has no solutions, hence the Galois group is the symmetric group S_{3}.

Let L := ℂ(T)(α_{1}, α_{2}, α_{3}) be the splitting field. Obviously, L ⊃ M = ℂ(T)(α_{1} −α_{2}). We have

# {σ(α_{1} − α_{2}) : σ ∈ G(L|ℂ(T))} = 6,

otherwise α_{i} = α_{j} for i ≠ j or α_{i} = 0 for some i = 1, 2, 3. Hence, [M : ℂ(T)] = 6 and furthermore M = L. Since we know all conjugates of α_{1} − α_{2}, it is not difficult to compute the minimal polynomial G_{ξ}. ■

3. Ramification

The aim of this section is to compute which places are ramified in L/ℂ(T) and furthermore to compute the genus g_{L} of L, where L = ℂ(T)(α_{1}, α_{2}, α_{3}) is the splitting field of f_{ξ}. From this we obtain a first upper bound for the degree of solutions (x, y) to (1). In a first step we determine which places are ramified in K := ℂ(T)(α_{1}). Note that since ℂ is algebraically closed all discrete valuation rings with quotient field ℂ(T) are isomorphic to O_{a} := {f(T)/g(T) : f(T), g(T) ∈ ℂ[T], g(a) ≠ 0} for some a ∈ ℂ or O_{∞} := {f(T)/g(T) : f(T), g(T) ∈ ℂ[T], deg f ≤ deg g}. Valuations corresponding to O_{a} are denoted by v_{a} (finite valuations) and the valuation corresponding to O_{∞} is denoted by v_{∞} (infinite valuation).

Lemma 1. Let K = ℂ(T)(α_{1}). The only places that are ramified in K correspond to valuations v_{a} with a ∈ R ⊂ ℂ ∪ {∞} and

R = \{\frac{1}{2} (−1 + i^{n}\sqrt{(−1)^{n}13 + 16\sqrt{2}} − 2ξ) : 1 ≤ n ≤ 4\} .

Proof: Assume K is not ramified at valuations that lie over v_{a} and a ≠ ∞. By Puiseux’s Theorem [2, 9] there exists a formal Power series

α(T) := ∑_{n=0}^{∞} a_{n}(T − a)^{n}

such that f_{ξ}(α(T)) = 0. On the other hand we know given an equation f(X, T) = 0 with f holomorphic and \frac{∂f}{∂x} |_{T =a} ≠ 0, then there exists a holomorphic function X(T) = ∑_{n=0}^{∞} a_{n}(T − a)^{n} in an open neighborhood U ⊂ ℂ of a, such that f(X(T), T) = 0. We conclude that ramification over v_{a} may only occur if

\frac{∂f_{ξ}}{∂X} |_{T =a} = 3X^{2} − 2X(a + ξ + 1) + (a + ξ) = 0 and
f_{ξ}(X)|_{T =a} = X^{3} − X^{2}(a + ξ + 1) + X(a + ξ) + 1 = 0.

Eliminating X from these equations and solving for a yields a ∈ R. Computing the Puiseux expansion of f_{ξ} at some a ∈ R we see that one valuation is ramified with index 2 and one valuation is unramified.

We are left to prove that v_{∞} is unramified in K. Since \frac{1}{T} f_{ξ}(X) is “holomorphic at T = ∞” we may substitute T by 1/T and obtain a new function

f̃_{ξ}(X) = T X^{3} − X^{2}(T (ξ + 1) + 1) + X(ξT + 1) + T

that is holomorphic in X and T. From basic calculus we know that there exists a Laurent series α(T) = ∑_{n=−1}^{∞} a_{n}T^{−n} with f_{ξ}(α(T)) = 0 if \frac{∂f̃_{ξ}}{∂X} |_{T =0} ≠ 0. Hence we have to prove the impossibility of

\frac{∂f̃_{ξ}}{∂X} |_{T =0} = −2X + 1 = 0 and simultaneously
f̃_{ξ}(X)|_{T =0} = −X^{2} + X = 0.

Hence the valuation v_{∞} is not ramified. ■

Now we consider the function field L. Recall that we defined L = ℂ(T)(α_{1}, α_{2}, α_{3}) to be the splitting field of f_{ξ}(X).

Proposition 4. The only places that are ramified in L correspond to valuations v_{a} with a ∈ R ⊂ ℂ ∪ {∞} and

R = \{\frac{1}{2} (−1 + i^{n}\sqrt{(−1)^{n}13 + 16\sqrt{2}} − 2ξ) : 1 ≤ n ≤ 4\} .

There are exactly three valuations lying above v_{a} with ramification index 2 for each a ∈ R. Furthermore g_{L} = 1, where g_{L} denotes the genus of the function field L.

Proof: Since L/K is a Galois extension the statement concerning the ramification index is obvious from Lemma 1. With similar arguments as in the proof of Lemma 1 we see that there does not occur any further ramification. In order to compute the genus we recall the Hurwitz-Formula [15, Theorem 7.16, page 90]. Let L/K be a finite, geometric extension of function fields of characteristic 0 and let g_{K} and g_{L} be the genus of K and L, respectively, then

2g_{L} − 2 = [L : K](2g_{K} − 2) + ∑_{w∈M_{L}} e_{w},  (2)

where M_{L} is the set of valuations of L and e_{w} denotes the ramification index of w in the extension L/K. If we put K = ℂ(T), then g_{L} = 1 is computed from (2). ■

We are now able to prove a first bound for the degrees of solutions X and Y to (1).

Corollary 2. Let (x, y) be a solution to (1) with ξ ≠ 0, then max(deg x, deg y) ≤ 17 and if ξ = 0 we have max(deg x, deg y) ≤ 9.

Proof: We have to make our computations in L, since in this function field the Thue equation (1) splits and we want to use Proposition 2. From Proposition 4 we know r = 6 and g_{L} = 1. Suppose first ξ ≠ 0. One computes

H = max (H (\frac{1}{−ξT^{2} + ξT − 1}) , H (\frac{T + ξ + 1}{−ξT^{2} + ξT − 1}) , H (\frac{T + ξ}{−ξT^{2} + ξT − 1})) = 12.

Proposition 2 yields now max(H(x), H(y)) ≤ 103. Notice that we assume x, y ∈ ℂ[T], hence deg x = 6H(x) and deg y = 6H(y), so we have proved the corollary in the case ξ ≠ 0. If ξ = 0 we similarly compute H = 6, hence max(H(x), H(y)) ≤ 55, respectively, max(deg x, deg y) ≤ 9. ■

4. Reduction of Height

From Corollary 2 one could try to compare coefficients and so deduce all solutions. However, the number of equations and unknowns is too large to do this efficiently and therefore, we first calculate a better lower bound for deg x and deg y. In order to achieve this we use the ABC-Theorem (Proposition 1) and the method of the proof of Mason for the finiteness of a single Thue equation over function fields instead of applying Mason’s theorem (Proposition 2) at once. First we will fix some notations usually used in the number field case.

Let us consider the Thue equation

F (X, Y) = m  (3)

over the ring of integers 𝔬 of some function field K. Let L be the splitting field of F (X, 1), thus we have

F (X, Y) = (X − α_{1}Y) ⋯ (X − α_{d}Y).

Let (x, y) ∈ 𝔬^{2} be some solution to equation (3). We put for pairwise distinct indices i, j, l ∈ {1, …, d}

β_{i}(x, y) = x − α_{i}y,
γ_{i,j,l}(x, y) = β_{i}(x, y)(α_{j} − α_{l}) = (x − α_{i}y)(α_{j} − α_{l}).

It is easy to check that

γ_{i,j,l}(x, y) + γ_{j,l,i}(x, y) + γ_{l,i,j}(x, y) = 0.  (4)

This identity is usually called Siegel’s identity. Furthermore, the β’s are S-integers in the function field L. Moreover, we have from equation (3)

β_{1} ⋯ β_{d} = m.  (5)

4.1. Preliminaries. Before we go on we pause for a moment to introduce another notation. Let M/ℂ(T) be a Galois extension of degree d, α ∈ M and for each a ∈ ℂ∪{∞} let us fix a d-tuple (w_{1}, …, w_{d}) of valuations in M with w_{i}|v_{a} for i ∈ {1, …, d}, where the valuations are used with multiplicity, i.e. a valuation with ramification index e is written down e times. Let us define

(·)_{a} : M → ℤ^{d}, f ↦ (f)_{a} := (w_{1}(f), …, w_{d}(f)).

For every σ ∈ G(M|ℂ(T)) there exists obviously a permutation τ(σ) ∈ S_{d} such that

(σα)_{a} = (w_{τ(1)}(α), …, w_{τ(d)}(α))

and furthermore T = {τ(σ) : σ ∈ G(M|ℂ(T))} is a transitive subgroup of S_{d}. Let w|v_{a} have ramification index e and let f ∈ ℂ(T) with v_{a}(f) = m_{a}, then we have (f)_{a} = (em_{a}, …, em_{a}). Collecting these facts together we obtain:

Lemma 2. Let α ∈ M be an integer in M and suppose there exists a conjugate β = σα, with some σ ∈ G(M/ℂ(T)) \ {id} such that there is a valuation w ∤ v_{∞} with w(α) = w(β) ≠ 0 and suppose that for the corresponding permutation τ we have w_{τ(1)} ≠ w_{1}. Then the norm N(α) = ∏_{i=1}^{d} σ_{i}α has a non constant quadratic factor.

4.2. Calculation of the heights. Let us return to our problem. We assume from now on ξ ≠ 0. In a first step we want to get a bound for H (\frac{γ_{1,2,3}}{γ_{2,3,1}} ). Since

β_{1}β_{2}β_{3} = −ξT^{2} + ξT + 1 = −ξ (T − \frac{\sqrt{ξ} + \sqrt{4 + ξ}}{2\sqrt{ξ}} ) (T − \frac{\sqrt{ξ} − \sqrt{4 + ξ}}{2\sqrt{ξ}} )

the β’s may only have non zero valuations at w if w|v_{a} or w|v_{∞} with

a ∈ B := \{\frac{\sqrt{ξ} + \sqrt{4 + ξ}}{2\sqrt{ξ}} , \frac{\sqrt{ξ} − \sqrt{4 + ξ}}{2\sqrt{ξ}} \} .

Since the α_{i,j} := α_{i} − α_{j} with distinct i, j ∈ {1, 2, 3} are roots of G_{ξ} the α_{i,j} may only have non zero valuations at w if w|v_{a} or w|v_{∞}, where a is a complex root of the constant term of G_{ξ}, i.e. a ∈ R. We conclude that the γ′s may only have non zero valuations at w lying above v_{a} with a ∈ R∪B ∪{∞}. By the considerations above and Proposition 4 there are exactly 30 such valuations. Let us consider Siegel’s identity

γ_{1,2,3} + γ_{2,3,1} + γ_{3,2,1} = 0

and we obtain by Corollary 1 that H (\frac{γ_{1,2,3}}{γ_{2,3,1}} ) ≤ 30.

Next, we want to compute an upper bound for H(β_{1}/β_{2}). Therefore, we conclude from Lemma 2, that there is no finite valuation w, such that w(α_{2,3}) = w(α_{3,1}) ≠ 0, otherwise the constant term of G_{ξ} would have a non constant quadratic factor. From the above considerations we further deduce that H(α_{2,3}) = H(α_{3,1}) = \frac{1}{6}H_{G} = 4, where H_{G} denotes the height of the constant term of G_{ξ}. Furthermore, we know that the β’s may only have positive finite valuations w|v_{a} if a ∈ B. Let us denote by

H_{a}(α) := − ∑_{w|v_{a}} min(0, w(α)), a ∈ ℂ ∪ {∞}

the local height. Obviously, we have

H(α) = ∑_{a∈ℂ∪{∞}}H_{a}(α).  (6)

From equation (6) one computes

\begin{matrix}H (\frac{γ_{1,2,3}}{γ_{2,3,1}} ) = & ∑_{a∈R} H_{a} (\frac{α_{2,3}}{α_{3,1}} ) + ∑_{b∈B} H_{b} (\frac{β_{1}}{β_{2}} ) + H_{∞} (\frac{γ_{1,2,3}}{γ_{2,3,1}} ) \\ = & 4 + ∑_{b∈B} H_{b} (\frac{β_{1}}{β_{2}} ) + H_{∞} (\frac{β_{1}}{β_{2}} · \frac{α_{2,3}}{α_{3,1}} ) \\ ≥ & 4 + ∑_{b∈B} H_{b} (\frac{β_{1}}{β_{2}} ) + H_{∞} (\frac{β_{1}}{β_{2}} ) − H_{∞}(α_{3,1}) \\ = & H (\frac{β_{1}}{β_{2}} ) . \end{matrix}  (7)

The second equation in (7) holds, since there is no finite valuation w such that w(α_{2,3}) = w(α_{3,1}) ≠ 0 as remarked above. Inequality (7) yields now H(β_{1}/β_{2}) ≤ 30.

Let us consider the valuation structure of β_{1}. Since β_{1} ∈ K we have (after a suitable permutation of the 6-tuple of valuations defining (·)_{a})

(β_{1})_{b} = (1, 1, 0, 0, 0, 0)  (β_{1})_{∞} = (−b_{1}, −b_{1}, −b_{2}, −b_{2}, −b_{3}, −b_{3})
(β_{2})_{b} = (0, 0, 1, 1, 0, 0)  (β_{2})_{∞} = (−b_{3}, −b_{3}, −b_{1}, −b_{1}, −b_{2}, −b_{2})
(β_{3})_{b} = (0, 0, 0, 0, 1, 1)  (β_{3})_{∞} = (−b_{2}, −b_{2}, −b_{3}, −b_{3}, −b_{1}, −b_{1})
(β_{1}/β_{2})_{b} = (1, 1, −1 − 1, 0, 0) (β_{1}/β_{2})_{∞} = (b_{3} − b_{1}, b_{3} − b_{1}, b_{1} − b_{2}, b_{1} − b_{2}, b_{2} − b_{3}, b_{2} − b_{3})

where b ∈ B and b_{1} ≥ b_{2} ≥ b_{3}. Since the sum over all valuations must be zero we have b_{3} ≤ 0, b_{1} ≥ 1 and b_{1} + b_{2} + b_{3} = 1. From H(β_{1}/β_{2}) = 2 + 2(b_{1} − b_{3}) ≤ 30 we deduce 14 ≥ b_{1} − b_{3}. This yields

H(β_{1}) = 2 · max(b_{1}, b_{1} + b_{2}) ≤ 2 · max(14 + b_{3}, 14 − b_{1} + 1) ≤ 28.  (8)

Next, we want to prove that H(α_{1}) = 2. Since the constant term of f_{ξ} is 1, it is clear that α_{1} is a unit in K, respectively in L, hence α_{1} has only non zero valuations at w|v_{∞}. We may assume

(α_{1})_{∞} = (a_{1}, a_{1}, a_{2}, a_{2}, a_{3}, a_{3}),

with a_{1} ≥ a_{2} ≥ a_{3} and since H(α_{1}) = H(α_{1}^{−1}) we may further assume a_{1}, a_{2} ≥ 0. This yields

6 = H(T + ξ + 1) = H(α_{1} + α_{2} + α_{3}) = 6 · min(a_{1}, a_{2}, a_{3}) = 3H(α_{1}),

hence H(α_{1}) = 2. Another unit is given by α_{1} − 1, since α_{1}(α_{1} − 1)(α_{1} − (T + ξ)) = 1. Because of α_{1}, α_{1} − 1 ∈ K, hence 2|H(α_{1} − 1), H(α_{1}) + H(1) ≥ H(α_{1} − 1) and α_{1} − 1 is not a constant, we conclude H(α_{1} − 1) = 2. This yields H_{K}(α_{1}) = H_{K}(α_{1} − 1) = 1, where H_{K} denotes the height associated to K. Since α_{1} ≠ c(α_{1} − 1) for any c ∈ ℂ, we have proved that α_{1} and α_{1} − 1 generate the unit group of K factored by ℂ^{×}.

Let β̃ ∈ K such that H(β̃) = 4 and after a suitable permutation of valuations

(β̃)_{b} = (1, 1, 0, 0, 0, 0), (b ∈ B),
(β̃)_{∞} = (0, 0, −2, −2, 0, 0),
(α_{1})_{∞} = (1, 1, −1, −1, 0, 0).

Then all β_{1}’s that might yield solutions to (1) are of the form β̃α_{1}^{a_{1}}(α_{1} − 1)^{a_{2}} with |a_{1} + a_{2}|, |a_{1}|, |a_{2}| ≤ 16. We want to construct β̃. Therefore let us set β̃_{1} := β̃, β̃_{2} and β̃_{3} the conjugates of β̃. We have

β̃_{i} = h_{0} + h_{1}α_{i} + h_{2}α_{i}^{2} (1 ≤ i ≤ 3),  (9)

with h_{0}, h_{1}, h_{2} ∈ ℂ(T). Solving this linear system by Cramer’s rule one obtains

\begin{aligned}h_{0}δ_{ξ}^{2} & = δ_{ξ} (β̃_{1}α_{2}α_{3}(α_{3} − α_{2}) + β̃_{2}α_{3}α_{1}(α_{1} − α_{3}) + β̃_{3}α_{1}α_{2}(α_{2} − α_{1})) , \\ h_{1}δ_{ξ}^{2} & = δ_{ξ} (β̃_{1}(α_{2} + α_{3})(α_{2} − α_{3}) + β̃_{2}(α_{3} + α_{1})(α_{3} − α_{1}) + β̃_{3}(α_{1} + α_{2})(α_{1} − α_{2})) , \\ h_{2}δ_{ξ}^{2} & = δ_{ξ} (β̃_{1}(α_{3} − α_{2}) + β̃_{2}(α_{1} − α_{3}) + β̃_{3}(α_{2} − α_{1})) , \end{aligned} ; (10)

where

δ_{ξ} = det(α_{j}^{i−1})_{1≤i,j≤3} = (α_{1} − α_{2})(α_{2} − α_{3})(α_{3} − α_{1})

is the discriminant of f_{ξ}. On the right side of (10) only integers in L occur, hence h_{i}δ_{ξ}^{2} ∈ ℂ[T] with i = 1, 2, 3. Furthermore, some analysis on the infinite valuations yields deg h_{i}δ_{ξ}^{2} ≤ 4. This is done by computing each infinite valuation on the right side of (10) using the facts

v(αβ) = v(α) + v(β),
v(α + β) = v(α),  if v(α) < v(β),
v(α + β) ≥ v(α),  if v(α) = v(β).

Further analysis yields

h_{2}δ_{ξ}^{2}α_{1}^{a_{1}}(α_{1} − 1)^{a_{2}} =δ_{ξ} (x((α_{3} − α_{2}) + (α_{1} − α_{3}) + (α_{2} − α_{1}))
−y(α_{1}(α_{3} − α_{2}) + α_{2}(α_{1} − α_{3}) + α_{3}(α_{2} − α_{1}))) = 0,

hence h_{2} = 0. Now, we have

δ_{ξ}^{2}x − δ_{ξ}^{2}α_{1}y = δ_{ξ}^{2}β_{1} = (∑_{i=0}^{4} ∑_{j=0}^{1} c_{i,j}T^{i}α_{1}^{j}) α_{1}^{a_{1}}(α_{1} − 1)^{a_{2}} = H_{0} + H_{1}α_{1} + H_{2}α_{1}^{2}, (11)

where c_{i,j} ∈ ℂ are not specified yet and H_{i} ∈ ℂ[T], i = 1, 2, 3. Obviously, H_{2} = 0 and δ_{ξ}^{2}|H_{0}, H_{1}. By comparing coefficients we obtain for each admissible pair of exponents (a_{1}, a_{2}) a linear system with unknowns c_{i,j}. By solving each system we obtain that only if (a_{1}, a_{2}) ∈ E with

\begin{matrix}E = & \{(−2, 1), (−1, −2), (−1, −1), (−1, 0), (−1, 1), (−1, 2), \\ (0, −1), (0, 0), (0, 1), (1, −1), (1, 0), (2, −1), (3, −2), (3, −1)\}, \end{matrix}

then the corresponding system has a non-trivial solution. Since a trivial solution yields β_{1} = 0, which is a contradiction, the last result may be reformulated in the following way:

Proposition 5. We have H(β_{1}) ≤ 10, i.e. H_{K}(β_{1}) ≤ 5.

From this bound we find now a bound for the degree of solutions (x, y) to (1).

Corollary 3. Let (x, y) be a solution to (1) then max{deg(x), deg(y)} ≤ 2.

Proof: Since x, y ∈ ℂ[T] we have

(x)_{∞} = (𝔵, 𝔵, 𝔵, 𝔵, 𝔵, 𝔵),
(y)_{∞} = (𝔶, 𝔶, 𝔶, 𝔶, 𝔶, 𝔶),
(yα_{1})_{∞} = (𝔶 + 1, 𝔶 + 1, 𝔶 − 1, 𝔶 − 1, 𝔶, 𝔶),

with 𝔵 = − deg x, 𝔶 = − deg y ≤ 0. In the computation of (β_{1})_{∞} we distinguish 5 cases.

(β_{1})_{∞} ≥ \{ \begin{matrix}(𝔵, 𝔵, 𝔵, 𝔵, 𝔵, 𝔵) & if 𝔵 < 𝔶 − 1, \\ (𝔵, 𝔵, ∞, ∞, 𝔵, 𝔵) & if 𝔵 = 𝔶 − 1, \\ (𝔵, 𝔵, 𝔶 − 1, 𝔶 − 1, ∞, ∞) & if 𝔵 = 𝔶, \\ (∞, ∞, 𝔶 − 1, 𝔶 − 1, 𝔶, 𝔶) & if 𝔵 = 𝔶 + 1, \\ (𝔶 + 1, 𝔶 + 1, 𝔶 − 1, 𝔶 − 1, 𝔶, 𝔶) & if 𝔵 > 𝔶 + 1, \end{matrix}

where ≥ is considered componentwise. From the fact that β_{1} is an integer and hence 10 ≥ H(β_{1}) = H_{∞}(β_{1}) and taking into account that 𝔵, 𝔶 ∈ ℤ, we get the following bounds;

𝔵 ≥ −1, 𝔶 ≥ 1, if 𝔵 < 𝔶 − 1,
𝔵 ≥ −2, 𝔶 ≥ −1, if 𝔵 = 𝔶 − 1,
𝔵 ≥ −2, 𝔶 ≥ −2, if 𝔵 = 𝔶,
𝔵 ≥ −1, 𝔶 ≥ −2, if 𝔵 = 𝔶 + 1,
𝔵 ≥ 1, 𝔶 ≥ −1, if 𝔵 > 𝔶 + 1.
■

5. The special case ξ = 0

By similar considerations as in Section 4 we want to solve the case ξ = 0. Using Mason’s version of the ABC-Theorem (Corollary 1) we obtain H(γ_{1}/γ_{2}) ≤ 18, since the β’s are units and have non-zero valuations only above v_{∞}. The same computations as in Section 4 yield H(β_{1}) ≤ 16. We further obtain

x + α_{1}y = β_{1} = α_{1}^{a_{1}}(α_{1} − 1)^{a_{2}} = H_{0} + α_{1}H_{1} + α_{1}^{2}H_{2},

with |a_{1} + a_{2}|, |a_{1}|, |a_{2}| ≤ 8. Obviously, H_{2} = 0, but this only holds for exponents (a_{1}, a_{2}) ∈ E′ with

E′ = {(−1, −1), (0, 0), (0, 1), (1, 0), (3, −1)}.

The exponents (a_{1}, a_{2}) ∈ E′ determine β_{1} up to a constant factor, which is easily computed in each case, hence we have found all solutions in the case of ξ = 0.

6. Proof of the main Theorem

Let (x, y) be a solution to (1). By Corollary 3 we may assume

x =x_{0} + x_{1}T + x_{2}T^{2},
y =y_{0} + y_{1}T + y_{2}T^{2}.

Substituting this in (1) yields by comparing the coefficients of T^{7} the equation x_{2}y_{2}(x_{2}− y_{2}) = 0, hence x_{2} = 0, y_{2} = 0 or x_{2} = y_{2}. Let us assume max(deg x, deg y) = 2 then we have y_{2} ≠ 0, x_{2} ≠ 0 and x_{2} = y_{2} ≠ 0, respectively. Comparing the coefficients of lower powers of T we get the following table:

\begin{matrix}T^{7} & T^{6} & T^{5} & T^{4} & T^{3} \\ x_{2} = 0 & x_{1} = y_{2} & x_{0} = y_{2}(1 + ξ) − y_{1} & y_{0} = y_{1}(ξ + 1) − (ξ + 1)^{2}y_{2} & y_{2} = 0 \\ y_{2} = 0 & x_{2} = y_{1} & x_{1} = x_{2} + y_{0} & x_{0} = ξy_{0} & x_{2} = 0 \\ y_{2} = x_{2} & x_{1} = y_{2} + y_{1} & x_{0} = y_{0} + y_{1} − ξx_{2} & y_{0} = ξy_{1} − ξ^{2}x_{2} − 2x_{2} & y_{1} = 2ξx_{2} + x_{2}/2 \end{matrix}

The first two cases yield contradictions. Comparing further coefficients in the case y_{2} = x_{2} we obtain 9x_{2}^{3}/2 = −ξ and −ξ − 2ξ^{2} = ξ, hence ξ = 0, −1. Substituting in the constant term yields 1 = 0, −55/36, hence in both cases a contradiction.

So we may assume x and y are linear, i.e.

x = x_{0} + x_{1}T, y = y_{0} + y_{1}T.

Comparing the coefficient of T^{4} now yields x_{1}y_{1}(x_{1} − y_{1}) = 0, this is x_{1} = 0, y_{1} = 0 or x_{1} = y_{1}. By assuming max(deg x, deg y) = 1 we have y_{1} ≠ 0, x_{1} ≠ 0 and x_{1} = y_{1} ≠ 0, respectively. Solving the equations that occur by comparing coefficients of lower powers of T, we obtain following table by successive elimination of unknowns:

T^{4}  T^{3}  T^{2}  T^{1}  T^{0} = 1
x_{1} = 0 x_{0} = −y_{1}  y_{0} = \frac{ξy_{1}^{3}+y_{1}^{3}−ξ}{y_{1}^{2}} ξ = \frac{2y_{1}^{3}}{1−y_{1}^{3}}  y_{1}^{3} = −1
y_{1} = 0 x_{1} = y_{0}  x_{0} = \frac{ξy_{0}^{3}−ξ}{y_{0}^{2}}  y_{0}^{3} = 1
ξ = 1  y_{1}^{3} = 1
x_{1} = y_{1} x_{0} = y_{0} + y_{1} y_{0} = \frac{ξy_{1}^{3}−ξ}{y_{1}^{2}}  ξ = ^{\frac{−y_{1}^{3}+y_{1}^{3}\sqrt{−3+4y_{1}^{3}}}{2(y_{1}^{3}−1)}} y_{1}^{3} ≠ 1 y_{1}^{3} = ±i
ξ = ^{\frac{−y_{1}^{3}−y_{1}^{3}\sqrt{−3+4y_{1}^{3}}}{2(y_{1}^{3}−1)}} y_{1}^{3} ≠ 1 y_{1}^{3} = 1

We have either a contradiction or a solution, which is listed in Theorem 1.

So we are left to the case that the solution (x, y) is constant, i.e. x = x_{0} and y = y_{0} with x_{0}, y_{0} ∈ ℂ. Substituting this in F_{ξ}(x, y) yields

F_{ξ}(x, y) = T (x_{0}y_{0}^{2} − x_{0}^{2}y_{0}) + x_{0}^{3} − x_{0}^{2}y_{0} − ξx_{0}^{2}y_{0} + ξx_{0}y_{0}^{2} + y_{0}^{3} = −ξT^{2} + ξT + 1,

again a contradiction. ■

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Mathematisch Instituut, Universiteit Leiden, Niels Bohrweg 1, Postbus 9512, 2300 RA Leiden, The Netherlands

E-mail address: fuchs@math.leidenuniv.nl

Institut für Mathematik A, Technische Universität Graz, Steyrergasse 30, A-8010 Graz, Austria

E-mail address: ziegler@finanz.math.tugraz.at