Explicit Chevalley-Weil Theorem for Affine Plane Curves

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Explicit Chevalley-Weil Theorem for Affine Plane Curves

Explicit Chevalley-Weil Theorem forAffine Plane CurvesKonstantinos Draziotis ∗ and Dimitrios Poulakis (Thessaloniki)AbstractLet φ : ˜C → C be an unramified morphism of plane affine curvesdefined over a number field K. In this paper we obtain a quantitative versionof the classical Chevalley-Weil theorem for curves giving an effectiveupper bound for the norm of the relative discriminant of the field K(Q)over K for any integral point P ∈ C(K) and Q ∈ φ −1 (P ).1 Introduction and Statement of ResultsIn this paper we revisit the classical theorem of Chevalley-Weil for unramifiedmaps [2], [18], [7, Theorem 8.1, page 45], [4, page 292], [14, pages 50 and109]. This theorem has quite interesting applications to the study of integralpoints of algebraic curves [1], [14, Section 8.4], [6, Chapter VI] and [5, §1].Partial quantitative versions on it have been used for the effective analysis ofintegral points on some families of algebraic curves [10, 12]. Note also thatan interesting alternative application of the unramified maps in the study ofDiophantine equations is given in [3].Let K be an algebraic number field and O K be the ring of integers. Inthis paper we deal with the following version derived from [14, page 109]: Ifφ : ¯C → C is an unramified morphism of affine plane curves defined over K,then there exists a finite extension L/K such that φ −1 (C(O K )) ⊆ ¯C(L). Moreprecisely, if Q ∈ φ −1 (P ), where P ∈ C(O K ), and K(Q) is the field generatedover K by the coordinates of Q, then we calculate, following a new approach,an effective upper bound for the norm of the relative discriminant of K(Q) overK, depending only on C, ¯C, and φ.We consider the set of standard absolute values on Q containing the ordinaryabsolute value | · | and for every prime p the p-adic absolute value | · | p with|p| p = p −1 . By an absolute value of K we will always understand an absolutevalue that is an extension to K of one of the above absolute values of Q. Wedenote by M(K) a set of symbols v such that with every v ∈ M(K) there isprecisely one associated absolute value |·| v on K. For v ∈ M K , we denote by K vand Q v the completion of the indicated field with respect to the absolute value| · | v . The local degree at v is given by d v = [K v : Q v ]. Let x = (x 0 : . . . : x n )be a point of the projective space P n (K) over K. We define the field height∗ The research of the first author was financial supported by Hellenic State ScholarshipsFoundation-I.K.Y1


H K (x) of x byH K (x) =∏max{|x 0 | v , . . . , |x n | v } dv .v∈M(K)Let d be the degree of K. Then we define the absolute height H(x) by H(x) =H K (x) 1/d . Furthermore, for x ∈ K we put H K (x) = H K ((1 : x)) and H(x) =H((1 : x)). If G ∈ K[X 1 , . . . , X m ], then we define the field height H K (G) andthe absolute height H(G) of G as the field height and the absolute height ofthe point whose coordinates are the coefficients of G (in any order). Givenv ∈ M(K), we denote by |G| v the maximum of |c| v over all the coefficients cof G. For an account of the properties of heights see [17, chapter VIII] or [7,chapter 3].Let K be an algebraic closure of K. Recall that a morphism φ : ¯C → C ofalgebraic curves is said to be unramified, if it is finite, C is nonsingular and forevery P ∈ C(K) the number of elements of φ −1 (P ) is equal to the degree ofφ [15, page 117]. If M is a finite extension of K, then we denote by N M andD M/K the norm and the relative discriminant of M over K, respectively. Forany subfield L of K, we denote by L[C] and L(C) the ring of regular functionsand the function field of an affine curve C over L, respectively.Let F (X, Y ) and ¯F (X, Y ) be absolutely irreducible polynomials of K[X, Y ]with degrees deg F = N ≥ 2 and deg ¯F = ¯N ≥ 2, respectively. We denote by Cand ¯C the affine curves defined by the equations F (X, Y ) = 0 and ¯F (X, Y ) = 0,respectively. Let φ : ¯C → C be a morphism of affine curves defined over Kof degree m > 1 and φ 1 (X, Y ), φ 2 (X, Y ) be polynomials of K[X, Y ] such thatφ(P ) = (φ 1 (P ), φ 2 (P )) for every P ∈ C(K). We set M = max{deg φ 1 , deg φ 2 }and we denote by Φ a point of the projective space having as coordinates 1 andthe coefficients of φ i (X, Y ) (i = 1, 2).Theorem 1.1 Assume that C is nonsingular and the morphism φ : ¯C → Cunramified. Then, for any P ∈ C(O K ) and Q ∈ φ −1 (P ), we haveN K (D K(Q)/K ) < Ω (H(F ) 5 ¯NN 3 H(Φ) ¯NH( ¯F ) M ) ωdm6 ¯N 6 N 22 ,where Ω is an effectively computable constant in terms of N, ¯N, M, m, d and ωis an effectively computable numerical constant.Remarks. 1) By [15, Corollary 3, page 120], the curve ¯C is nonsingular.2) Since m > 1, the quantity M is > 1.3) Since ¯F (X, Y ) divides F (φ 1 (X, Y ), φ 2 (X, Y )), the quantities H( ¯F ) and ¯Nare bounded by constants depending only on F and φ. Thus, our bound can bemade independent of ¯F .4) Theorem 1.1 can be easily generalised for S-integral points. For simplicitywe have chosen to present it only for integral points.The proof of Theorem 1.1 relies on the construction of two primitive elementsu 1 and u 2 for the field extension K( ¯C)/K(C) and polynomials P i (X, Y, U) (i =1, 2) of K[X, Y, U] which represent the irreducible polynomials of u i over K(C)having the following property: The algebraic set defined by F (X, Y ) and thediscriminants D i (X, Y ) of P i (X, Y, U) (i = 1, 2) (considered as polynomial withcoefficients in K[X, Y ]) is empty. Then we obtain a Bézout identity involvingD i (X, Y ) (i = 1, 2) and F (X, Y ) and we deduce that when ℘ is a prime ideal2


We define the height of A(X) over K(C) byH C (A) = ∏max{|a 0 | V , . . ., |a k | V }.V ∈ΣIf a 0 = 1, then [7, Proposition 3.1, p. 62] givesH C (A) = exp{deg(sup i (a i ) ∞ )}.For the proof of the Proposition 2.1 we shall need the following lemmas.Lemma 2.1 Suppose that C is smooth. Let u be an integral element over K[C]of degree µ > 1 and G(X, U) be a polynomial of K[X, U] such that G(x, u) = 0.Put γ = deg X G. Then the irreducible polynomial of u over K(C) has the formP (U) = U µ + p 1 (x, y)U µ−1 + · · · + p µ (x, y),where p i (x, y) ∈ K[C] (i = 1, . . . , µ) withdeg(p i (x, y)) ∞ ≤ 2γn.Proof. Since C is smooth, it follows that the ring K[C] is integrally closed.Thus the irreducible polynomial of u over K(C) has the formP (U) = U µ + p 1 (x, y)U µ−1 + · · · + p µ (x, y),where p i (x, y) ∈ K[C] (i = 1, . . . , µ). WriteG(X, U) = g 0 (X)U ν + · · · + g ν (X)where g j (X) ∈ K[X] (j = 0, . . . , ν) and g 0 (X) ≠ 0. Without loss of generalitywe may consider that the polynomials g j (X) are relatively prime. SinceG(x, u) = 0, it follows that P (x, y, U) divides G(x, U). Thus, [7, Proposition2.4, p. 57] yieldsH C (P (x, y, U)) ≤ H C (G(x, U)),whence it followsdeg(p i (x, y)) ∞ ≤ deg(sup 1≤j≤ν (g j (x)/g 0 (x)) ∞ ).Let g 0 (X) = b 0 (X − b 1 ) c1· · ·(X − b s ) cs and V i,j (j = 1, . . ., r(i)) be the ringsof Σ lying above the discrete valuation ring of K(X) defined by X − b i and e i,jbe the ramification index of V i,j . The divisor of poles of g i /g 0 satisfiesdeg((g i (x)/g 0 (x)) ∞ ) ≤ deg( ∑V ∈Σ ∞(deg g i )ord V (1/x)V + ∑ i,jc i e i,j V i,j ).Thusdeg(p i (x, y)) ∞ ≤ 2γn.Lemma 2.2 Let a(x, y) ∈ K[C] with deg(a(x, y)) ∞ ≤ 2γn, where γ is a positiveinteger. Then there is A(X, Y ) ∈ K[X, Y ] with deg A < 5γN 3 such thata(x, y) = A(x, y).4


Proof. Consider the divisorD = 2γn ∑V ∈Σ ∞Vand denote by l(D) the dimension of the Riemann-Roch space L(D). By [13,Theorems A2 and B2], it follows that there are polynomials b 1 (X, Y ), . . .,b l(D) (X, Y ), q(X) with coefficients in K anddeg q ≤ m(n − 1), deg Y b i ≤ n − 1, deg X b i ≤ 6γn 2 + 2mn,such that the functions φ 1 , . . ., φ l(D) defined by the fractions b 1 (X, Y )/q(X),. . ., b l(D) (X, Y )/q(X), respectively, on C, form a basis of L(D). The leadingcoefficient of q(X) is 1 and if q(X) ≠ 1, then its roots are among the roots ofthe discriminant of F (X, Y ) (considered as polynomial in Y ).Since a(x, y) ∈ K[C] there is A(X, Y ) ∈ K[X, Y ] with a(x, y) = A(x, y).Further, a(x, y) is an element of L(D) defined over K. Thusa(x, y) = f(x, y)/q(x) (i = 1, . . . , µ),where f(X, Y ) is a linear combination of b 1 (X, Y ), . . . , b l(D) (X, Y ) with coefficientsin K. It follows that there are polynomials B i (X, Y ) (i = 1, . . ., µ) withcoefficients in K such thatf(X, Y ) = q(X)A(X, Y ) + B i (X, Y )F (X, Y ) (i = 1, . . ., l(D)).Let q h (X, Z) and F h (X, Y, Z) be the homogenizations of q(X) and F (X, Y ),respectively. Since F (X, Y ) is absolutely irreducible, {F h , q h } is a regular sequence.So, by [16, Proposition 2], it follows that we can takedeg A i , deg B i < 5γN 3 .Lemma 2.3 Let a j,s ∈K (j = 1, . . ., q, s = 1, . . ., p). Suppose that the homogeneouslinear systema j,1 X 1 + · · · + a j,p X p = 0 (j = 1, . . ., q)has rank r. If the system has a solution x 1 , . . ., x p ∈K with x t ≠ 0, then there isa solution y 1 , . . ., y p ∈K with y t ≠ 0, such that for every absolute value |·| v of Kwe have|y i | v ≤ max{1, |r!| v }(maxi,j {|a j,i| v }) r (i = 1, . . . , p).If a j,s ∈O K (j = 1, . . ., q, s = 1, . . ., p), then there is a solution y 1 , . . ., y p ∈O Kwith y t ≠ 0 such that for every absolute value of K the above inequality holds.Proof. We may suppose, without loss of generality, that the matrix M =(a i,j ) 1≤i,j≤r has rank r. Thus, the above system is equivalent to the system∆X i = ∆ i,r+1 X r+1 + · · · + ∆ i,p X p(i = 1, . . ., r),where ∆ = det M and ∆ i,j (i = 1, . . ., r, j = r + 1, . . ., p) are minors of order rof the matrix (a i,j ). Hence, ∆ i,j ∈K and|∆ i,j | v ≤ max{1, |r!| v }(maxi,j {|a j,i| v }) r (i = 1, . . ., r, j = r + 1, . . ., p).5


Furthermore, if a j,s ∈O K (j = 1, . . ., q, s = 1, . . ., p), then ∆ i,j ∈O K (i =1, . . ., r, j = r + 1, . . ., p).If 1 ≤ t ≤ r, since there is a solution x 1 , . . ., x p with x t ≠ 0, we obtain thatthere is s∈{r + 1, . . ., p} such that ∆ t,s ≠ 0. Thus, a solution of the system withthe required properties is given byX s = ∆, X j = 0 (j = r + 1, . . ., p, j ≠ s), X i = ∆ i,s (i = 1, . . ., r).If r + 1 ≤ t ≤ p, then the required solution is given byX t = ∆, X j = 0 (j = r + 1, . . ., p, j ≠ t), X i = ∆ i,t (i = 1, . . ., r).Proof of the Proposition 2.1. By Lemmata 2.1 and 2.2, there is a polynomialP (X, Y, U) = U µ + p 1 (X, Y )U µ−1 + · · · + p µ (X, Y ),where p 1 (X, Y ), . . ., p µ (X, Y )∈K[X, Y ] such that P (x, y, U) is the irreduciblepolynomial of u over K(C) anddeg p i < 5γN 3 .Put θ = 5γN 3 . Consider θN distinct points (x i , y i ) (i = 1, . . ., θN) on Cwith x i ∈Z and |x i | ≤ θN. Since P (x i , y i , U) divides G(x i , U), [4, PropositionB.7.3, p. 228] givesThusH(1, p 1 (x i , y i ), . . ., p µ (x i , y i )) ≤ e ν H(g 0 (x i ), . . ., g ν (x i )).H(1, p 1 (x i , y i ), . . ., p µ (x i , y i )) ≤ e ν H(G)|x i | γ (γ + 1) (i = 1, . . ., θN).Writep r (X, Y ) = ∑k+l


and using [11, Lemma 7] we obtainFurther, we haveThereforeWriteH(A) ≤ Λ 2 (N, γ)H(F ) 25γ2 N 7 .H(B) < Λ 3 (N, γ, ν)H(G) 5γN 4 .H(R) < Λ 4 (N, γ, ν, µ)(H(G)H(F ) 5γN 3 ) 25µγ2 N 8 .Q r (X, Y ) = ∑(ρ r,k,l /ζ)X k Y l .k+l(deg φ)(deg F ), Bézout’s theorem implies that F (X, Y ) divides φ(X, Y ) andtherefore p r (x, y) = q r (x, y). Thus, we can take p i (X, Y ) = q i (X, Y ) and if Qis a point of the projective space having as coordinates 1 and the coefficients ofq r (X, Y ), then H(Q) = H(R). The lemma follows.3 Polynomials with no Common ZeroLet φ : ¯C → C be an unramified morphism of affine plane (irreducible) curvesdefined over K of degree m > 1. Since the morphism φ is finite, the homomorphismφ ∗ : K[C] → K[ ¯C], defined by φ ∗ (f) = φ ◦ f for every f ∈ K[C], isinjective and is extended to a field homomorphism φ ∗ : K(C) → K( ¯C). We alsoidentify K(C) with its image into K( ¯C). Let x, y be the coordinate functionson C and ¯x, ȳ be the coordinate functions on ¯C. Suppose that the curve C isdefined by the polynomial F (X, Y ) ∈ K[X, Y ]. Let u ρ = ȳ + ρ¯x, where ρ ∈ K,andP ρ (X, Y, U) = U µ(ρ) + p ρ,1 (X, Y )U µ(ρ)−1 + · · · + p ρ,µ(ρ) (X, Y )is a polynomial of K[X, Y, U] such that P ρ (x, y, U) is the irreducible polynomialof u ρ over K(ρ)(C). We have µ(ρ) ≤ m. We denote by Π the maximum ofthe degrees of the polynomials p ρ,j (j = 1, . . . , µ(ρ)) and by R the degree ofP ρ (x, y, U), considered as polynomial in ρ.Lemma 3.1 Let D ρ (X, Y ) be the discriminant of P ρ (X, Y, U), considered aspolynomial with coefficients in K[X, Y ]. Let r ∈ K such that D r (X, Y ) is nota constant. Then there is a set A ⊂ K with |A| ≤ (2m − 1)R((2m − 1)NΠ + 1)such that for every t ∈ K \ A, with t ≠ r, D t (X, Y ) is not a constant and thepolynomials D r (X, Y ), D t (X, Y ), and F (X, Y ) have no common zero.Proof. Let V ρ be the set defined by the equationsD ρ (X, Y ) = F (X, Y ) = 0.Since the polynomial P ρ (x, y, U) is irreducible, it follows that F (X, Y ) does notdivide D ρ (X, Y ) and so the set V ρ is finite. We have deg D ρ (X, Y ) ≤ (2m − 1)Πand so Bézout’s theorem implies that |V ρ | ≤ (2m − 1)NΠ.Let r ∈ K. Suppose that V r is not empty and let (α r,j , β r,j ) (j = 1, . . ., s(r))be all its points (hence D r (X, Y ) is not a constant). Then for every j =7


1, . . ., s(r) there are m distinct points (a r,j,i , b r,j,i ) (i = 1, . . ., m) of ¯C withφ(a r,j,i , b r,j,i ) = (α r,j , β r,j ) and ζ r,j ∈Z such that a r,j,k ζ r,j + b r,j,k ≠ a r,j,l ζ r,j +b r,j,l , for k ≠ l. Hence P ζr,j (α r,j , β r,j , U) has exactly m distinct roots and soD ζr,j (α r,j , β r,j ) ≠ 0. It follows that the degree in ρ of D ρ (α r,j , β r,j ) is ≥ 1 andso we obtain that D ρ (x, y) is not independent from ρ.If D ρ (x, y) is independent from x and y, then there is e ∈ K such thatD e (x, y) = 0 and hence P e (x, y, U) divides its derivative with respect to Uwhich is a contradiction. So, D ρ (x, y) is not independent from x and y.The degree in ρ of D ρ (x, y) is ≤ (2m − 1)R. So, for every j = 1, . . ., s(r)there are at most (2m − 1)R integers ρ with D ρ (α r,j , β r,j ) = 0. Also, there areat most (2m − 1)R integers ρ such that D ρ (x, y) is independent from x and y.Thus, there is a set A ⊂ K with |A| ≤ (2m − 1)R((2m − 1)NΠ + 1) suchthat for every t ∈ K \ A with t ≠ r we have D t (α r,j , β r,j ) ≠ 0 (j = 1, . . ., s(r)))and D t (X, Y ) is not a constant. So, the polynomials D r (X, Y ), D t (X, Y ), andF (X, Y ) have no common zero.Lemma 3.2 Let F j (j = 1, 2, 3) be polynomials in O K [X 1 , X 2 ], with deg F j =d j and d 1 ≥ d 2 ≥ d 3 ≥ 1, having no common zero in K 2 . We denote by Φ apoint in the projective space having as coordinates the coefficients of F j (in anyorder) and we set δ = max{d 1 d 3 , d 1 + d 2 + d 3 − 2}. Then there are polynomialsA j (j = 1, 2, 3) in O K [X 1 , X 2 ] and c∈O K \ {0} such thatA 1 F 1 + A 2 F 2 + A 3 F 3 = c.Furthermore, deg A j F j ≤ δ and for every archimedean absolute value |·| v of Kwe have|A j | v , |c| v ≤ ((δ + 2)(δ + 1)/2)! |Φ| (δ+2)(δ+1)/2v .Proof. By [16, Corollary 3], there are polynomials A j of K[X 1 , X 2 ] (j = 1, 2, 3),with deg A j = e j , such thatand deg A j F j ≤ δ. We putandHence∑F j =A j =s 1+s 2≤δ(A 1 F 1 + A 2 F 2 + A 3 F 3 = 1∑i 1+i 2≤d jf j,i1,i 2X i11 Xi2 2 (j = 1, 2, 3)∑i 1+i 2≤e ja j,i1,i 2X i11 Xi2 2 (j = 1, 2, 3).3∑j=1∑k j+p j=s jf j,k1,k 2a j,p1,p 2)X s11 Xs2 2 = 1.Thus, the numbers a j,p1,p 2and 1 is a solution of the homogeneous linear system3∑j=1∑f j,k1,k 2X j,p1,p 2= 0 (s 1 + s 2 = 1, . . ., δ),k j+p j=s j8


3∑f j,0,0 X j,0,0 − Z = 0.j=1The number of the equations of the above system is at most equal to (δ +2)(δ + 1)/2. Lemma 2.3 implies that there is a non trivial solution b j,p1,p 2andc in O K with c ≠ 0 such that for every archimedean absolute |·| v of K we have|b j,p1,p 2| v , |c| v ≤ ((δ + 2)(δ + 1)/2)! |Φ| (δ+2)(δ+1)/2v ,where Φ is a point in the projective space having as coordinates the coefficientsof F j (in any order). Furthermore, the polynomialsB j =∑i 1+i 2≤e jb j,i1,i 2X i11 Xi2 2 (j = 1, 2, 3)satisfy the equalityB 1 F 1 + B 2 F 2 + B 3 F 3 = c.Lemma 3.3 Let F j (j = 1, 2) be polynomials in O K [X 1 , X 2 ], with deg F j ≥ 1,having no common zero in K 2 . Suppose that F 1 is absolutely irreducible and doesnot divide F 2 . We denote by Φ the point in the projective space having as coordinatesthe coefficients of F 1 and F 2 (in any order) and put δ = deg F 1 deg F 2 .Then there are A j ∈ O K [X 1 , X 2 ] (j = 1, 2), with deg A j F j ≤ δ, and c∈O K \ {0}such thatA 1 F 1 + A 2 F 2 = c.Furthermore, for every archimedean absolute value |·| v of K we have|A j | v , |c| v ≤ ((δ + 2)(δ + 1)/2)! |Φ| (δ+2)(δ+1)/2v .Proof. Let F i,h be the homogenization of F i . Then, F 1,h is absolutely irreducibleand does not divide F 2,h . Then the class of F 2,h is not zero in the integral domainK[X, Y, Z]/(F 1,h ) and hence {F 1,h , F 2,h } is a regular sequence. By [16, Corollary5], there are polynomials A j ∈ K[X 1 , X 2 ] (j = 1, 2), with deg A j F j ≤ δ, suchthatA 1 F 1 + A 2 F 2 = 1.Next, working as in the previous lemma, the result follows.4 Some Auxiliary LemmataIn this section we give some lemmata useful for the proof of Theorem 1.1.Lemma 4.1 Let a ∈ K. Then there is an integer δ > 0 with δ ≤ H K (a) suchthat δa ∈ O K .Proof. Put e = [Q(a) : Q]. Then a is a root of a polynomial P (X) = c 0 X e +· · · + c e with integer pairwise prime coefficients and c 0 > 0. So, c 0 a is analgebraic integer. By [7, page 54], we haveH K (a) = c 0e∏max{1, |a i |},i=19


where a 1 , . . . , a d are the distinct conjugates of a. Thus c 0 ≤ H K (a).Next lemma deals with the resultant of two polynomials. For a formal definitionof the resultant see [8, Chapter V, §10]Lemma 4.2 Let P and S be polynomials of O K [X 1 , . . ., X n ] \ K and R be itsresultant with respect to X n . Put p i = deg XiP and s i = deg XiS (i = 1, . . ., n).Suppose that R ≠ 0. Then deg XiR ≤ s n p i + p n s i (i = 1, . . ., n − 1) andn−1∏n−1∏H(R) ≤ (p n + s n )!( (p i + 1)) sn ( (s i + 1)) pn H(P ) sn H(S) pn .i=1If R, P, S are points in the projective space having as coordinates 1 and thecoefficients of R, P , S respectively, then the above inequality holds with thequantities R, P, S in the places of R, P , S respectively.Proof. Writei=1P = Π 0 (X 1 , . . ., X n−1 )X pnn + · · · + Π pn (X 1 , . . ., X n−1 ),S = Σ 0 (X 1 , . . ., X n−1 )X snn + · · · + Σ sn (X 1 , . . ., X n−1 ),where Π i (X 1 , . . ., X n−1 ) (i = 0, . . ., p n ) and Σ j (X 1 , . . ., X n−1 ) (j = 0, . . ., s n )are polynomials with coefficients in K. The resultant R is a homogeneouspolynomial with integer coefficients of degree s n in Π i (X 1 , . . ., X n−1 ) and ofdegree p n in Σ j (X 1 , . . ., X n−1 ). Let |·| v be an absolute value of K. If |·| v is notarchimedean, then|R| v ≤ |P | snv |S| pnv .Now, suppose that |·| v is archimedean. If M(X 1 , . . ., X n−1 ) is a monomial ofdegree s n in Π i (X 1 , . . ., X n−1 ) and of degree p n in Σ j (X 1 , . . ., X n−1 ), thenThusn−1∏n−1∏|M| v ≤ ( (p i + 1)|P | v ) sn ( (s i + 1)|S| v ) pn .i=1i=1i=1n−1∏n−1∏|R| v ≤ (p n + s n )!( (p i + 1)|P | v ) sn ( (s i + 1)|S| v ) pn .Hence, we have the inequalitiesandn−1∏n−1∏H(R) ≤ (p n + s n )!( (p i + 1)) sn ( (s i + 1)) pn H(P ) sn H(S) pn ,i=1n−1∏n−1∏H(R) ≤ (p n + s n )!( (p i + 1)) sn ( (s i + 1)) pn H(P) sn H(S) pn .i=1Furthermore, from the definition of the resultant, we obtain that deg XiR ≤s n p i + p n s i (i = 1, . . ., n − 1).i=1i=1i=110


Lemma 4.3 Let L be a finite extension of K of degree µ and R be the set ofprime ideals of O K which are ramified in L. Then the discriminant D L/K ofthe extension L/K satisfiesN K (D L/K ) < ∏ ℘∈RN K (℘) µ−1 exp(2µ 2 d).Proof. For every prime p and every prime ideal ℘ of O K dividing p we denote bye ℘ and f ℘ the ramification index and the residuel class degree of ℘ respectively.Furthermore, if P is a prime ideal of O L dividing ℘, then we denote by e(P/℘)and by f(P/℘) the ramification index and the residuel class degree of P over℘ respectively. Finally, let ord P be the discrete valuation of L associated to Pand ord p be the discrete valuation of Q associated to p.By [9, Proposition 6.3], we getD L/K = ∏ ℘∈R℘ a℘ ,wherea ℘ ≤ ∑ P |℘(e(P/℘) − 1 + ord P (e(P/℘)))f(P/℘).If p > µ, then ord P (e(P/℘)) = 0 and hence a p ≤ µ − 1. Suppose that p ≤ µ.We have∑∑ord P (e(P/℘))f(P/℘) = e ℘ e(P/℘)f(P/℘)ord p (e(P/℘))℘|pP |℘P |℘= (e ℘ / log p) ∑ P |℘e(P/℘)f(P/℘) log e(P/℘)≤ (e ℘ / log p)µ log µ.Thus∏ ∑∑N K (℘)ord P (e(P/℘))f(P/℘) (µ log µ/ log p)P |℘ ≤ p℘|p f℘e℘ = exp(dµ log µ).Hence, we obtainN K (D L/K ) ≤ ∏ ℘∈RN K (℘) µ−1 exp(π(µ)dµ log µ),where π(µ) is the number of primes ≤ µ. Since π(µ) < 2µ/ log µ, the resultfollows.Lemma 4.4 Let g j (X, Y ) (j = 1, . . ., k) be polynomials of K[X, Y ] of degree≤ M and p, q, r, s, t, u ∈ K. We setG j (U, V ) = g j (pU + qV + r, sU + tV + u), (j = 1, . . ., k)and we denote by Γ and γ the two points in the projective space having ascoordinates the coefficients of G j (X, Y ) and g j (X, Y ) respectively. ThenH(Γ) ≤ 2 M(1+2e) (M + 1)H(1, p, q, s, t) M H(1, r, u) M H(γ),where e = 1 if (r, u) ≠ (0, 0) and e = 0 otherwise.11


Proof. Put b i (X, Y ) = g i (X + r, Y + u). By [4, Proposition B.7.4(e), page 234],for every absolute value |·| v of K, we haveWrite|b i | v ≤ max{1, |2| v } 2M max{1, |r| v , |u| v } M |g i | v .b i (X, Y ) =∑i+j≤Ma i,j X i Y j .Since G i (U, V ) = b i (pU + qV, sU + tV ) we have∑G i (U, V ) = a i,j (pU + qV ) i (sU + tV ) j=i+j≤M∑i∑a i,ji+j≤M π=0 ρ=0j∑( ) (i jπ ρ)p π q i−π s ρ t j−ρ U π+ρ V i+j−π−ρ .The coefficient of U S V T isc S,T =∑i∑a i,ji+j=S+T π=0 ρ=0j∑( iπ) ( jρ)p π q i−π s ρ t j−ρ .If | · | v is an archimedean absolute value of K, then∑|c S,T | v ≤i∑(|a i,j | v max{1, |p| v , |q| v , |s| v , |t| v } i+j iπi+j=S+Tπ=0≤ 2 S+T |b i | v (S + T + 1) max{1, |p| v , |q| v , |s| v , |t| v } S+T .) j∑ρ=0( jρ)If | · | v is a non archimedean absolute value of K, then |c S,T | v ≤ |b i | v . Thus, forevery absolute value | · | v of K we haveThe result follows.|G i | v ≤ max{1, |2| v } M(1+2e(r)) max{1, |M + 1| v }×max{1, |p| v , |q| v , |s| v , |t| v } M max{1, |r| v , |u| v } M |g i | v .5 Proof of Theorem 1.1Since the morphism φ is finite, the homomorphism φ ∗ : K[C] → K[ ¯C], definedby φ ∗ (f) = φ ◦ f for every f ∈ K[C], is injective and so we may identify K[C]with its image into K[ ¯C]. Thus, if x, y are the coordinate functions on C and ¯x,ȳ the coordinate functions on ¯C, then x = φ 1 (¯x, ȳ) and y = φ 2 (¯x, ȳ). Further,the morphism φ ∗ is extended to a field homomorphism φ ∗ : K(C) → K( ¯C).We also identify K(C) with its image into K( ¯C). Thus, for every ρ ∈ K thefunction u ρ = ȳ + ρ¯x is an integral element over K[C].Consider the polynomialsE(W, X, ¯X, U) = X − φ 1 ( ¯X, U − W ¯X),¯F1 (W, ¯X, U) = ¯F ( ¯X, U − W ¯X).We haveE(ρ, x, ¯x, u ρ ) = ¯F 1 (ρ, ¯x, u ρ ) = 0.12


We denote by G(W, X, U) the resultant of E(W, X, ¯X, U) and ¯F 1 (W, ¯X, U)with respect to ¯X. Then G(ρ, x, u ρ ) = 0. If G(W, X, U) is equal to zero,then ¯F 1 (W, ¯X, U) divides E(W, X, ¯X, U) which is not the case. Therefore,G(W, X, U) ≠ 0.By Lemma 4.2, deg X G ≤ ¯N, deg U G ≤ 2M ¯N, and deg W G ≤ 2M ¯N. Forρ ∈ K, we put G ρ (X, U) = G(ρ, X, U), E ρ (X, ¯X, U) = E(ρ, X, ¯X, U) and¯F ρ ( ¯X, U) = ¯F 1 (ρ, ¯X, U). Then Lemma 4.2 impliesH(G ρ ) ≤ (M + ¯N)!(1 + ¯N) M (2(1 + M)) ¯NH(Eρ ) ¯NH( ¯F ρ ) M .On the other hand, Lemma 4.4 yieldsH( ¯F ρ ) ≤ 2 ¯N( ¯N + 1)H(ρ) ¯N H( ¯F ), H(E ρ ) ≤ 2 M (M + 1)H(ρ) M H(Φ),where Φ is a point of the projective space having as coordinates the coefficientsof φ i (X, Y ) (i = 1, 2). ThereforeH(G ρ ) < (M + ¯N + 1) 3M+3 ¯N (2H(ρ)) 2 ¯NM H(Φ) ¯NH( ¯F ) M .Let I(T ) = T s + b 1 T s−1 + · · · + b s , where b j ∈K[C] (j = 1, . . ., s), theirreducible polynomial of u ρ over K(C). For every σ∈Gal(K/K(ρ)) we setI σ (T ) = T s + σ(b 1 )T s−1 + · · · + σ(b s ). Since the function u ρ is defined overK(ρ), we have I σ (u ρ ) = 0 for every σ∈Gal(K/K(ρ)). Thus, σ(b j ) = b j , forevery σ∈Gal(K/K(ρ)) and hence b j ∈K(ρ)[C]. Therefore, u ρ is integral overK(ρ)[C] and [K(C)(u ρ ) : K(C)] = [K(ρ)(C)(u ρ ) : K(ρ)(C)]. By Proposition2.1 and the bound for the quantity H(G ρ ) we obtain that there is a polynomialP ρ (X, Y, U) = U µ(ρ) + p ρ,1 (X, Y )U µ(ρ)−1 + · · · + p ρ,µ(ρ) (X, Y )of K(ρ)[X, Y, U] with µ(ρ) = [K(ρ)(C)(u ρ ) : K(ρ)(C)] ≤ m, andH(P ρ ) < Λ 5 (N, ¯N, M, m, ρ) (H(F ) 5 ¯NN 3 H(Φ) ¯NH( ¯F ) M ) 25m ¯N 2 N 8such that P ρ (x, y, U) is the irreducible polynomial of u ρ over K(ρ)(C). Moreover,we have deg p ρ,i < 5 ¯NN 3 (i = 1, . . . , µ).Let P = (a, b) be a point in C(O K ) and Q ∈ φ −1 (P ). We consider theK(C)-embedings σ 1 , . . ., σ m of K( ¯C) into an algebraic closure of K(C). Wedenote by Γ the set of integers ρ with σ i (u ρ ) ≠ σ j (u ρ ) for i ≠ j. Note thatat most m(m − 1)/2 integers do not lie in Γ. It follows that for every ρ∈Γ wehave K( ¯C) = K(C)(u ρ ) and hence for every ρ∈Γ we have m = µ(ρ). Further,we similarly deduce that there are at most m(m − 1)/2 integers ρ ∈ Γ suchthat K(u ρ (Q)) ≠ K(Q). Hence, there is r∈Γ with |r| ≤ m 2 /2, such thatK(u r (Q)) = K(Q). Since P ρ (x, y, U) divides G ρ (x, U) and deg W G ≤ 2M ¯N,we deduce that the degree of P ρ (x, y, U), considered as polynomial in ρ, is≤ 2M ¯N. Thus, the degree in ρ of D ρ (x, y) is ≤ 2(2m − 1)M ¯N.Suppose that the polynomials D ρ (X, Y ) and F (X, Y ) have common zeros.Thus, Lemma 3.1 implies that there is t ∈ K with |t| < 21m 2 MN 4 ¯N 2 such thatD t (X, Y ) is not a constant, the polynomials F (X, Y ), D r (X, Y ), D t (X, Y ) haveno common zero and we have K(u r (Q)) = K(Q) = K(u t (Q)).Let D ρ,1 be a point of the projective space having as coordinates 1 and thecoefficients of D ρ . By Lemma 4.2, we haveH(D ρ,1 ) < m 3m−1 (9 ¯NN 2 ) 4m−2 H(P ρ ) 2m−1 .13


We may suppose that one of the coefficients of F (X, Y ) is equal to 1. By Lemma4.1, there are positive integers a ρ and b withH K (a ρ ) ≤ H K (P ρ ) 25m ¯N 2 N 6 and H K (b) ≤ H K (F ) 2N 2such that the polynomials a ρ P ρ (X, Y, U) and bF (X, Y ) have all its coefficientsin O K . Then the polynomial a 2m−2ρ D ρ (X, Y ) is in O K [X, Y ]. Since F (X, Y ),D r (X, Y ), and D t (X, Y ) have no common zero, Lemma 3.2 implies that thereare polynomials A j (j = 1, 2, 3) of O K [X, Y ] and c∈O K , c ≠ 0, such thatA 1 a 2m−1r D r + A 2 a 2m−1t D t + A 3 bF = c.Further, for every archimedean absolute value |·| v of K we have|c| v ≤ ((δ + 1)(δ + 2)/2)!|Ψ| (δ+1)(δ+2)/2v ,where δ = (2m − 1)5N 4 ¯N and Ψ is a point of the projective space with coordinates1 and the coefficients of a 2m−1r D r (X, Y ), a 2m−1t D t (X, Y ) and bF (X, Y ).Thus|N K (c)| ≤(100m 2 ¯N 2 N 8 (H(a r )H(a t )) 2m−1 H(b)H(D r,1 )H(D t,1 )H(F )) 100dm2 ¯N 2 N 8 .Using the bounds for all the quantities involved in this estimate, we obtain|N K (c)| < Λ 6 (N, ¯N, M, m, d) (H(F ) 5 ¯NN 3 H(Φ) ¯N H( ¯F ) M ) λdm5 ¯N 6 N 22 ,where λ is a computable numerical constant. Since D r (X, Y ) and D t (X, Y )have no common zero on C, it follows that either D r (a, b) ≠ 0 or D t (a, b) ≠ 0.On the other hand, we haveA 1 (a, b)a 2m−1r D r (a, b) + A 2 (a, b)a 2m−1t D t (a, b) = c.Let ℘ be a prime ideal that does not divide c, O K,℘ the local ring at ℘ and˜℘ = ℘O K,℘ . We put L = K(Q) and ν = [L : K]. We have L = K(u r (Q)) =K(u t (Q)). We denote by D ℘ the discriminant of the integral closure of O K,℘in L over O K,℘ . Since ℘ does not divide c, it follows that ˜℘ does not divideat least one of the elements a 2m−1rD r (a, b) and a 2m−1t D t (a, b) (in O K,℘ ). WeD r (a, b). Then ˜℘ does not divide the al-suppose that ˜℘ does not divide a 2m−1rgebraic integers a r and a 2m−2r D r (a, b). Thus, the coefficients of a r P r (X, Y, U)lie in O K and the coefficient of the highest power of U, a r , is a unit in O K,℘ .Therefore, u r (Q) is an integral element over O K,℘ and hence D ℘ divides thediscriminant D(1, u r (Q), . . ., u r (Q) ν−1 ) of 1, u r (Q), . . ., u r (Q) ν−1 in O K,℘ . Further,D(1, u r (Q), . . ., u r (Q) ν−1 ) divides D r (a, b). Thus, D ℘ divides D r (a, b) inO K,℘ . Since ˜℘ does not divide D r (a, b) (in O K,℘ ), it follows that does not divideD ℘ and hence ℘ is not ramified in L. Using Lemma 4.3, we deduceN K (D L/K ) < ∏ ℘|cN K (℘) m−1 exp(2m 2 d)< Λ 7 (N, ¯N, M, m, d) (H(F ) 5 ¯NN 3 H(Φ) ¯NH( ¯F ) M ) λdm6 ¯N 6 N 22 .Suppose now that D ρ (X, Y ) and F (X, Y ) have no common zero. If D r (X, Y )is not a constant, then we work as previously (using Lemma 3.3 instead Lemma14


3.2) and we obtain a sharper upper bound for N K (D L/K ). Finally, let D r (X, Y ) =D r ∈ K − {0}. If η = a 2m−1r D r , then it follows that L is not ramified at everyprime ideal ℘ of O K which is not divide ηa r and so we deduce a sharper upperbound for N K (D L/K ).Acknowledgements. We thank Professor Marc Hindry for interesting discussionsand the referee for helpful comments.References[1] C. Chabauty, Démonstration de quelques lemmes de rehaussement, C. R.Acad. Sci. Paris 217, (1943) 413-415.[2] C. Chevalley, Un théorème d’ arithmétique sur les courbes algébriques, C.R. Acad. Sci. Paris 195 (1932), 570-572.[3] K. R. Coobes and D. R. Grant, On the heterogenous Spaces, J. LondonMath. Soc. 40 (1989), 385-397.[4] M. Hindry - J. Silverman, Diophantine Geometry, New-York Inc.: Springer-Verlag 2000.[5] D. Kubert - S. Lang, Units in the Modular Function Field. I, Math. Ann.218 (1975), 67-96.[6] S. Lang, Elliptic Curves. Diophantine Analysis, Springer Verlag 1978.[7] S. Lang, Diophantine Geometry, Springer Verlag 1983.[8] S. Lang, Algebra (2nd edition), Addison-Wesley, 1984.[9] W. Narkiewicz, Elementary and Analytic Theory of Algebraic Numbers,Springer and PWN 1990.[10] D. Poulakis, Estimation effective de points entiers d’ une famille de courbesalgébriques, Ann. Fac. Sci. Toulouse, Vol. V, no 4 (1996), 691-725.[11] D. Poulakis, Polynomial bounds for the solutions of a class of Diophantineequations, J. Number Theory, 66, 2 (1997), 271-281.[12] D. Poulakis, Bounds for the size of integral solutions to Y m = f(X), Proc.Edinburg Math. Soc., 42 (1999), 127-141.[13] W. M. Schmidt, Construction and estimation of bases in function fields, J.Number Theory 39, 2, (1991), 181-224.[14] J. P. Serre, Lectures on the Mordell-Weil Theorem, Vieweg 1989.[15] I. Shafarevich, Basic Algebraic Geometry, Berlin-Heidelberg-New York:Springer Verlag 1977.[16] B. Shiffman, Degree bounds for the division problem in polynomial ideals,Michigan Math. J. 36 (1989), 163-171.[17] J. H. Silverman, The arithmetic of elliptic curves, Springer Verlag 1986.15


[18] A. Weil, Arithmétique et géométrie sur les variétés algébriques, Act. Sci.et Ind. No 206, Paris: Hermann 1935.Aristotle University of Thessaloniki,Department of Mathematics,54124 Thessaloniki, GreeceD. Poulakis e.mail: poulakis@math.auth.grK. Draziotis e.mail: drazioti@math.auth.gr16

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