Rational Solutions of First Order Algebraic Ordinary Differential Equations
aa r X i v : . [ c s . S C ] M a y Rational Solutions of First Order Algebraic OrdinaryDifferential Equations
Ruyong Feng and Shuang Feng
KLMM,Academy of Mathematics and Systems Science, Chinese Academy of Sciences andSchool of Mathematics, University of Chinese Academy of Sciences, 100190, [email protected], [email protected]
Abstract
Let f ( t, y, y ′ ) = P di =0 a i ( t, y ) y ′ i = 0 be a first order ordinary differential equa-tion with polynomial coefficients. Eremenko in 1999 proved that there exists aconstant C such that every rational solution of f ( t, y, y ′ ) = 0 is of degree notgreater than C . Examples show that this degree bound C depends not only onthe degrees of f in t, y, y ′ but also on the coefficients of f viewed as polynomialin t, y, y ′ . In this paper, we show that if d max i =0 { deg( a i , y ) − d − i ) } > C only depends on the degrees of f , and furthermore wepresent an explicit expression for C in terms of the degrees of f . Keywords: first order AODE, rational solution, degree bound,height
1. Introduction
The study of first order algebraic ordinary differential equations (AODEs inshort) has a long history, which can be at least tracked back to the time of Fuchsand Poincar´e. Fuchs presented a sufficient and necessary condition so calledFuchs’ criterion for a first order AODE having no movable singularities. Roughlyspeaking, an AODE is said to have movable singularities if it has a solution (witharbitrary constants) whose branch points depend on arbitrary constants. Forinstance the solution y = √ t + c of 2 yy ′ − t = − c ,where c is an arbitrary constant, so 2 yy ′ − ✩ This work was supported by NSFC under Grants No.11771433 and No.11688101, and byBeijing Natural Science Foundation (Z190004).
Preprint submitted to Journal of L A TEX Templates May 5, 2020 rst order AODEs. In particular, he presented an algebraic definition of movablesingularities. In 1999, combining Matsuda’s results and height estimates ofpoints on plane algebraic curves, Eremenko showed that rational solutions offirst order AODEs have bounded degrees. In [7], we proved that if a first orderAODE has movable singularities then it has only finitely many rational solutions.As for algebraic solutions of first order AODEs, Freitag and Moosa [8] showedthat they are of bounded heights.On the other hand, the algorithmic aspects of computing closed form solu-tions of AODEs have been extensively studied in the past decades. Several algo-rithms have been developed for computing closed form solutions (e.g. liouvilliansolutions) of linear homogeneous differential equations (see [2, 12, 19, 17, 18]etc). Yet, the situation is different in the nonlinear case. Existing algorithmswere only valid for AODEs of special types. Based on parametrization of al-gebraic curves, Aroca et al [1, 6] gave two complete methods for finding ratio-nal and algebraic solutions of first order autonomous AODEs. Their methodswere generalized by Winkler and his collegues to the class of first order non-autonomous AODEs whose rational general solutions involve arbitrary constantsrationally as well as some other certain classes of AODEs (see [21, 20, 3, 22]etc). Particularly, in [21], the authors introduced a class of first order AODEscalled maximally comparable AODEs and presented an algorithm to computea degree bound for rational solutions of this kind of equations as well as firstorder quasi-linear AODEs. Readers are referred to [22] for a survey of recentdevelopments in this direction. Theoretically, it suffices to compute a degreebound for all rational solutions of a first order AODE to find all its rationalsolutions. The following example implies that the degrees of rational solutionsmay depend not only on the degrees of the original equation but also on itsconstant coefficients.
Example 1.1.
Let n be an integer. Then y = t n is a rational solution of ty ′ − ny = 0 . The degree of t n depends on the constant coefficient n of ty ′ − ny . Let f = P di =0 a i ( t, y ) y ′ i = 0 be an irreducible first order AODE. Setm . s . index( f ) = d max i =0 { deg( a i , y ) − d − i ) } . Fuchs’ criterion (see Remark on page 14 of [14]) implies that f = 0 has movablesingularities if m . s . index( f ) >
0. On the other hand, it was proved in [5] thatif f = 0 has movable singularities then it can be transferred into an AODE g with positive m . s . index. This motivates us to focus on first order AODEs withpositive m . s . index. We prove that for an irreducible first order AODE f = 0with m . s . index( f ) > f = 0 are independentof the constant coefficients of f and furthermore we present an explicit degreebound in terms of the degrees of f . The key step to obtain this degree bound isto estimate the heights of points on plane algebraic curves. This height estimateis a special case of the result about heights on complete nonsingular varieties(see for instance Proposition 3 on page 89 of [13]). Eremenko in [5] provided2 simple proof for this special case based on the Riemann-Roch Theorem. Wefollow Eremenko’s proof but present explicit bounds for each step.The paper is organized as follows. In Section 2, we introduce some basicmaterials used in the later sections. In Sections 3, we estimate the degrees andheights for elements in a Riemann-Roth space. In Section 4, we present anexplicit bound for the heights of points on a plane algebraic curve. Finally, inSection 5, we apply the results in Section 4 to first order AODEs.Throughout this paper, Z stands for the ring of integers, k, K for alge-braically closed fields of characteristic zero, and R and R for algebraic functionfields over k and K respectively. P m ( · ) denotes the projective space of dimen-sion m over a field and V ( · ) denotes the variety in a projective space defined bya set of homogeneous polynomials.
2. Basic materials
In this section, we will introduce some basic materials used in this paper,including differential rings, algebraic function fields of one variable and heights.Readers are referred to [15, 14, 4, 13] for details.
In this subsection, we introduce some basic notations of differential algebra.
Definition 2.1.
A derivation on a ring R is a map δ : R → R satisfying thatfor all a, b ∈ R , δ ( a + b ) = δ ( a ) + δ ( b ) , δ ( ab ) = δ ( a ) b + aδ ( b ) . A ring (resp. field) equipped with a derivation is called a differential ring (resp.differential field). An ideal I ⊂ R is called a differential ideal if δ ( I ) ⊂ I . The field k ( t ) of rational functions in t can be endowed with a structure ofdifferential field whose derivation δ is the usual derivation with respect to t , i.e. δ = dd t . Set y = y and denote k ( t ) { y } = k ( t )[ y , y , . . . ]where y , y , . . . are indeterminates. One can extend the derivation δ on k ( t )to a derivation δ ′ on k ( t ) { y } by assigning y i = δ ′ i ( y ) so that k ( t ) { y } be-comes a differential ring. For the sake of notations, we use δ in place of δ ′ .Elements in k ( t ) { y } are called differential polynomials over k ( t ). Let f be adifferential polynomial not in k ( t ). Then there is a unique integer d such that f ∈ k ( t )[ y , . . . , y d ] \ k ( t )[ y , . . . , y d − ]. This integer is called the order of f .We shall use [ · ] (resp. h·i ) to stand for the differential (resp. algebraic) idealgenerated by a set of differential polynomials (resp. polynomials) respectively.Suppose that f is irreducible viewed as an algebraic polynomial. SetΣ f = { A ∈ k ( t ) { y }| ∃ m > S m A m ∈ [ f ] } S = ∂f /∂y d and d is the order of f . It was proved on page 30 of [15]that Σ f is a prime differential ideal and so k ( t ) { y } / Σ f is a differential domain.Lemma 2.2 of [7] implies that the field of fractions of k ( t ) { y } / Σ f is isomorphicto that of k ( t )[ y , . . . , y d ] / h f i . Under this isomorphism, the field of fractionsof k ( t )[ y , . . . , y d ] / h f i can be endowed with a structure of differential field. Weshall still use δ , or ′ in short, to denote the induced derivation on the field offractions of k ( t )[ y , . . . , y d ] / h f i .In this paper, the first order AODEs under consideration are differentialequations of the following form f ( y, y ′ ) = 0 (1)where f ( y, y ′ ) ∈ k ( t )[ y, y ′ ] \ k ( t ). Definition 2.2.
An element r ( t ) ∈ k ( t ) satisfying f ( r ( t ) , r ′ ( t )) = 0 is called arational solution of f ( y, y ′ ) = 0 . Remark that the derivation δ in k ( t ) can be uniquely extended to a deriva-tion in k ( t ) which we shall still denote by δ . Assume that viewed as a poly-nomial in k ( t )[ y, y ′ ], f is irreducible over k ( t ). Then the field of fractions of k ( t )[ y, y ′ ] / h f ( y, y ′ ) i is not only an algebraic function field over k ( t ) but also adifferential field. Let K be an algebraically closed field of characteristic zero and R an ex-tension field of K . We say R is an algebraic function field of one variable over K if R satisfies the following conditions: there is an element a of R which istranscendental over K , and R is algebraic of finite degree over K ( a ). Assume R is an algebraic function field of one variable over K . A valuation ring of R over K is a subring V satisfying that1. K ⊂ V = R ; and2. if a ∈ R \ V , then a − ∈ V .All non-invertible elements of V form a maximal ideal P which is called a placeof R , and V is called the corresponding valuation ring of P . Let V be a valuationring with P as place. There is an element u ∈ V , called a local uniformizer of P or V , such that P = uV and T ∞ n =1 u n V = { } . The factor ring V / P is equalto K since K is algebraically closed. For every valuation ring V with place P ,we define a map π P : R −→ K ∪ {∞} satisfying if a ∈ V then π P ( a ) = a + P ∈ V / P = K , otherwise π P ( a ) = ∞ .It is well-known that R admits infinitely many places, and there is one-to-onecorrespondence between places and valuation rings.Let P be a place of R and V the corresponding valuation ring of P . Let u be a local uniformizer of P . Then for every non-zero element a of R , there is aunique integer n such that a = u n v v ∈ V . It is easy to see that the integer n isindependent of the choice of local uniformizers. Such n is called the order of a at P and denoted by ν P ( a ). We make the convention to write ν P (0) = ∞ .Then the place P induces a map ν P from R to Z sending a to ν P ( a ). This map ν P is called the order function at P . For a, b ∈ R , we have ν P ( ab ) = ν P ( a ) + ν P ( b ) , ν P ( a + b ) ≥ min { ν P ( a ) , ν P ( b ) } where the equality in the later formula holds if ν P ( a ) = ν P ( b ). Let a ∈ R and P be a place. We say P is a zero of a if ν P ( a ) >
0, and a pole of a if ν P ( a ) < R admits only finitely many zeros and poles.A divisor in R is a formal sum D = X P n P P for all the places of R , where n P ∈ Z and n P = 0 for all but finitely many P . Itis easy to see that the set of divisors in R forms an abelian group. D is effective if n P ≥ P . The degree of D , denoted by deg( D ), is defined to be P n P and the support of D , denoted by supp( D ), is defined to be { P | n P = 0 } . Forbrief, we denote D + = X n P > n P P , D − = X n P < − n P P . Let D = P P n P P and D = P P m P P be two divisors in R , we write D ≥ D provided D − D is effective. For every non-zero element a of R , we denotediv( a ) = X P ν P ( a ) P where P ranges over all places of R . Then div( a ) is a divisor of degree 0. Fora divisor D , we denote L ( D ) = { a ∈ R | div( a ) + D ≥ } ∪ { } , which is called the Riemann-Roch space of D . It is well-known that eachRiemann-Roch space is a K -vector space of finite dimension. The Riemann-Roch Theorem implies that if D is a divisor whose degree is not less than thegenus of R then L ( D ) is of positive dimension.Let f ∈ K [ x , x ] \ K be irreducible. One sees that the field of fractionsof K [ x , x ] / h f i is an algebraical function field of one variable over K which iscalled the algebraic function field of f . For an irreducible homogeneous poly-nomial F in K [ x , x , x ], the corresponding algebraic function field is definedto be the algebraic function field of F ( x , x , F (1 , x , x ) , F ( x , , x ) and F ( x , x ,
1) are all isomorphic.5 .3. Models of algebraic function fields of one variable
Let R be an algebraic function field of one variable over K . The set of allplaces of R can be viewed as a nonsingular model of R . On the other hand, let F be an irreducible homogeneous polynomial F ∈ K [ x , x , x ] whose algebraicfunction field is R . Then the projective curve F = 0 is another model of R .There is a surjective map from a nonsingular model of R to the curve F = 0.To describe this map precisely, let ξ , ξ , ξ be three nonzero elements of R satisfying that R = K ( ξ /ξ , ξ /ξ ) and F ( ξ , ξ , ξ ) = 0 . Set ξ = ( ξ , ξ , ξ ). Let P be a place of R with u as local uniformizer. Denote by ℓ = min i { ν P ( ξ i ) } . One sees that ν P ( u − ℓ ξ i ) ≥ π P ( u − ℓ ξ i )are zero. Therefore ( π P ( u − ℓ ξ ) , π P ( u − ℓ ξ ) , π P ( u − ℓ ξ )) defines a point of P ( K ).Remark that this point does not depend on the choice of u . Definition 2.3.
We call ( π P ( u − ℓ ξ ) , π P ( u − ℓ ξ ) , π P ( u − ℓ ξ )) the center of P with respect to ξ . Denote by C ( ξ ) the set of centers with respect to ξ . We claim that C ( ξ ) is the plane projective curve in P ( K ) defined by F and the map sending P to the center of P with respect to ξ is the requiredmap. It is easy to verify that F ( c ) = 0 for all c ∈ C ( ξ ). Conversely, let( c , c , c ) be a point of F = 0. Without loss of generality, we may assumethat c = 0. Then F (1 , c /c , c /c ) = 0. Remark tha R = K ( ξ /ξ , ξ /ξ ).As F (1 , ξ /ξ , ξ /ξ ) = 0, due to Corollary 2 on page 8 of [4], there is a place P containing ξ /ξ − c /c and ξ /ξ − c /c . For this place, one has that ν P ( ξ ) ≥ ν P ( ξ ) , ν P ( ξ ) ≥ ν P ( ξ ) and furthermore π P ( ξ i /ξ ) = c i /c . Write ℓ = ν P ( ξ ). Then the center of P with respect to ξ is( π P ( u − ℓ ξ ) , π P ( u − ℓ ξ ) , π P ( u − ℓ ξ )) = π P ( u − ℓ ξ )(1 , c /c , c /c ) . This implies that ( c , c , c ) ∈ C ( ξ ). Definition 2.4.
We call C ( ξ ) or F = 0 a plane projective model of R . The plane projective models of R usually have singularities. Let C be anirreducible projective curve in P ( K ) defined by a homogeneous polynomial F .A point c of C is said to be of multiplicity r , if all derivatives of F up to andincluding the ( r − c but not all the r -th derivatives vanish at c . Suppose that c is a point of C with multiplicity r . If r = 1, then c is calleda simple point of C , otherwise a singular point of C . A point of multiplicity r is ordinary if the r tangents to C at this point are distinct, otherwise it is non-ordinary. Due to Propositon on page of [9], R has always a plane projectivemodel with only ordinary singularities.Let Φ = ( φ , φ , φ ) be an invertible transformation, where φ , φ , φ arehomogeneous polynomials in K [ x , x , x ] of the same degree and they have nocommon factors. We further assume that φ i ( ξ ) = 0 for all i = 0 , ,
2. Then R = K (cid:18) φ ( ξ ) φ ( ξ ) , φ ( ξ ) φ ( ξ ) (cid:19) . roposition 2.5. Let Φ , ξ be as above and P a place of R . Assume that c isthe center of P with respect to ξ . If Φ( c ) = (0 , , , then Φ( c ) is the center of P with respect to Φ( ξ ) .Proof. Let u be a local uniformizer of P and ℓ = min i =0 { ν P ( ξ i ) } . One has that( π P ( u − ℓ ξ ) , π P ( u − ℓ ξ ) , π P ( u − ℓ ξ )) = λ c for some nonzero λ ∈ K . Denote m = tdeg( φ i ). Then π P (Φ( u − ℓ ξ )) = Φ( π P ( u − ℓ ξ )) = Φ( λ c ) = λ m Φ( c ) = (0 , , . This implies that min i =0 { ν P ( φ i ( u − ℓ ξ )) } = 0 . In other words, min i =0 { ν P ( φ i ( ξ )) } = mℓ. Then the center of P with respect to Φ( ξ ) is π P ( u − mℓ Φ( ξ )) = π P (Φ( u − ℓ ξ )) = Φ( λ c ) = λ m Φ( c ) . Hence Φ( c ) is the center of P with respect to Φ( ξ ). All algebraic function fields under consideration in this subsection are finiteextensions of k ( t ). They are algebraic function fields of one variable over k andthe places and order functions in them are defined as the same as in the previoussubsection. Definition 2.6.
All points are considered as points in some suitable projectivespaces over k ( t ) . Given a = ( a , . . . , a m ) ∈ P m ( k ( t )) , let R be a finite extension of k ( t ) containing all a i . We define the height of a , denoted by T ( a ) , to be P p max mi =0 {− ν p ( a i ) } [ R : k ( t )] where p ranges over all places of R . For A = ( a i,j ) ∈ GL ( k ( t )) , we define T ( A ) = T (( a , , a , , a , , . . . , a , )) . For a ∈ k ( t ) , we define the height of a to be T ((1 , a )) , denoted by T ( a ) . Let F be a polynomial in k ( t )[ x , . . . , x m ] . Suppose that F contains atleast two terms. We define the height of F , denoted by T ( F ) , to be T ( c ) where c is the point in a suitable projective space formed by the coefficientsof F . For convention, when F only contains one term, we defined T ( F ) to be zero. Let V be a hypersurface in P m ( k ( t )) defined by F ∈ k ( t )[ x , . . . , x m ] . Wedefine the height of V , denoted by T ( V ) , to be T ( F ) . Remark 2.7.
Assume that a = ( a , . . . , a m ) , b = ( b , . . . , b m ) ∈ P m ( k ( t )) . One sees that T ( a ) is independent of the choice of homogeneous coordi-nates and the choice of R . Without loss of generality, we suppose a = 1 ,then T ( a ) = P p max { , − ν p ( a ) , . . . , ν p ( a m ) } [ R : k ( t )] ≥ . Assume R is a finite extension of k ( t ) containing all a i and b i . Then onesees that if max i {− ν p ( a i ) } ≥ max i {− ν p ( b i ) } for all places p of R then T ( a ) ≥ T ( b ) . Suppose that a , a , . . . , a m ∈ k [ t ] and gcd( a , . . . , a m ) = 1 . Then T ( a ) = max { deg( a ) , . . . , deg( a m ) } . To see this, let R = k ( t ) . Then T ( a ) = X p max {− ν p ( a ) , . . . , − ν p ( a m ) } where p ranges over all the places of R . Note that every place of R has alocal uniformizer of the form /t or t − c for some c ∈ k . Suppose the place p has t − c as a local uniformizer. Then ν p ( a i ) > if and only if ( t − c ) | a i .Since gcd( a , . . . , a m ) = 1 , there is some i such that ν p ( a i ) = 0 . Thisimplies that for places p with t − c, c ∈ k as local uniformizers, max {− ν p ( a ) , . . . , − ν p ( a m ) } = 0 . For the place with /t as local uniformizer, one has that ν p ( a i ) = − deg( a i ) .So for this place, max {− ν p ( a ) , . . . , − ν p ( a m ) } = max { deg( a ) , . . . , deg( a m ) } . Consequently, T ( a ) = max { deg( a ) , . . . , deg( a m ) } . Let a ∈ k ( t ) and R = k ( t, a ) . Let g ( t, x ) be a nonzero irreducible poly-nomial over k such that g ( t, a ) = 0 . It is clear that T ( a ) = 0 if a ∈ k .Assume that a / ∈ k and p , . . . , p s are all distinct poles of a in R , then T ( a ) = − P si =1 ν p i ( a )[ R : k ( t )] = [ R : k ( a )][ R : k ( t )] = deg( g, t )deg( g, x ) . In particular, if a ∈ k ( t ) then T ( a ) = deg( a ) which is defined to be themaximun of the degrees of the denominator and numerator of a . From the above remark, it is easy to see that for a ∈ k ( t ) \ { } and i ∈ Z T ( a i ) = | i | T ( a ) . roposition 2.8. Let a, b ∈ k ( t ) , c , . . . , c ∈ k with c c − c c = 0 . Then T (cid:16) c a + c c a + c (cid:17) = T ( a ) if c a + c = 0 ; T ( ab ) ≤ T ( a ) + T ( b ) ; T ( a + λb ) ≤ T ( a ) + T ( b ) for all λ ∈ k .Proof. . If a ∈ k then the assertion is obvious. Suppose that a / ∈ k . Let R = k ( t, a ). Then R = k ( t, ( c a + c ) / ( c a + c )). The assertion follows fromRemark 2.7 and the fact that (cid:20) R : k (cid:18) c a + c c a + c (cid:19)(cid:21) = [ R : k ( a )] . . Let R = k ( t, a, b ). For each place p of R , − ν p ( ab ) = − ν p ( a ) − ν p ( b ) andthus max { , − ν p ( ab ) } ≤ max { , − ν p ( a ) } + max { , − ν p ( b ) } . The assertion then follows from Remark 2.7.3 . Use an argument similar to that in 2. and the fact that − ν p ( a + λb ) ≤ max {− ν p ( a ) , − ν p ( b ) } . The following examples show that the equalities may hold in 2 and 3 ofProposition 2.8.
Example 2.9.
Let a = t , b = t + 1 and λ ∈ k \ { } . Then T ( ab ) = 5 = T ( a ) + T ( b ) , T (1 /a + λ/b ) = 5 = T (1 /a ) + T (1 /b. ) Moreover both of them are greater than the maximun of T ( a ) , T ( b ) . In the following, y stands for the vector with indeterminates y , . . . , y s and y d denotes Q si =1 y d i i for d = ( d , . . . , d s ) ∈ Z s . Proposition 2.10.
Let f and g be polynomials in k ( t )[ y ] , then T ( f g ) ≤ T ( f ) + T ( g );2. If both f and g have 1 as a coefficient, then T ( f + g ) ≤ T ( f ) + T ( g ) . Proof.
Write f = P d a d y d and g = P d b d y d with a d , b d ∈ k ( t ). Let R be afinite extension of k ( t ) containing all a d , b d , and p a place of R .1. Each coefficient of f g is of the form P si =1 a d i b d i , where s ≥
1. Since − ν p s X i =1 a d i b d i ! ≤ s max i =1 {− ν p ( a d i ) − ν p ( b d i ) } , we have max c ∈ C {− ν p ( c ) } ≤ max d {− ν p ( a d ) } + max d {− ν p ( b d ) } C is the set of all coefficients of f g . It follows that T ( f g ) ≤ T ( f ) + T ( g ) .
2. The assertion follows from the fact that − ν p ( a d + b d ) ≤ max d {− ν p ( a d ) , − ν p ( b d ) } ≤ max d { , − ν p ( a d ) , − ν p ( b d ) }≤ max d { , − ν p ( a d ) } + max d { , − ν p ( b d ) } . Proposition 2.11.
Let f = x n + a n − x n − + · · · + a where n > and a i ∈ k ( t ) .Suppose that α is a zero of f in k ( t ) and R is a finite extension of k ( t ) containing α and all a i . Then for each place p of R , max { , − ν p ( α ) } ≤ max { , − ν p ( a ) , . . . , − ν p ( a n − ) } . Proof.
The assertion is clear if α ∈ k or p is not a pole of α . Assume α ∈ k ( t ) \ k and p is a pole of α . Then ν p ( α n ) = ν p n − X i =0 a i α i ! ≥ n − min i =0 { iν p ( α ) + ν p ( a i ) } = i ν p ( α ) + ν p ( a i )for some i with 0 ≤ i ≤ n −
1. This together with the fact that ν p ( α ) < ν p ( α ) ≥ ν p ( a i ) n − i ≥ ν p ( a i ) ≥ n − min i =0 { ν p ( a i ) } , i.e. − ν p ( α ) ≤ n − max i =0 {− ν p ( a i ) } . Consequently, one has thatmax { , − ν p ( α ) } ≤ max { , − ν p ( a ) , . . . , − ν p ( a n − ) } . Corollary 2.12.
Let f be a polynomial in k ( t )[ x ] and α a zero of f in k ( t ) .Then T ( α ) ≤ T ( f ) . The equality in Corollary 2.12 may hold as shown in the following example.
Example 2.13.
Let f = x − ( t + 1) x + t = ( x − t )( x − , then T ( t ) = T ( f ) = 3 . Lemma 2.14.
Assume f, g ∈ k ( t )[ x ] and g is a factor of f . Then T ( g ) ≤ T ( f ) . roof. Without loss of generality, we may assume both f and g are monic. Write f = x n + n − X i =0 a i x i , a i ∈ k ( t ) . We first show that if f = gh and deg g = 1 then T ( h ) ≤ T ( f ). Suppose that h = x n − + n − X j =0 b j x j , g = x + α where α, b j ∈ k ( t ). Let R = k ( t, α, b , . . . , b n − ). For each place p of R , denote N p = max { , − ν p ( a ) , . . . , − ν p ( a n − ) } . Then max { , − ν p ( a i ) } ≤ N p for all i = 0 , . . . , n − p of R . From f = gh , one has that b n − + α = a n − , b α = a , b i α + b i − = a i , i = 1 , . . . , n − . We claim that max { , − ν p ( b j ) } ≤ N p for all j = 0 , . . . , n − p of R . For j = n −
2, one has that ν p ( b n − ) = ν p ( a n − − α ) ≥ min { ν p ( a n − ) , ν p ( α ) } . This implies that max { , − ν p ( b n − ) } ≤ max { , − ν p ( a n − ) , − ν p ( α ) } . By Propo-sition 2.11, we have that max { , − ν p ( α ) } ≤ N p . Hence max { , − ν p ( b n − ) } ≤ N p for all places p of R . Now assume that there is a place q of R and j with0 ≤ j < n − { , − ν q ( b j ) } > N q but max { , − ν q ( b j +1 ) } ≤ N q .Then one has that ν q ( b j ) < ν q ( b j ) < ν q ( a i ) for all i = 0 , . . . , n −
1. Onthe other hand, since max { , − ν q ( b j +1 ) } ≤ N q , − ν q ( b j ) > N q ≥ − ν q ( b j +1 )i.e. ν q ( b j +1 ) < ν q ( b j ) <
0. From αb j +1 = a j +1 − b j , one has that ν q ( αb j +1 ) = ν q ( α ) + ν q ( b j +1 ) = min { ν q ( a j +1 ) , ν q ( b j ) } = ν q ( b j ) . The last equality holds because ν q ( b j ) < ν q ( a i ) for all i . Thus ν q ( α ) < ν q ( αb j ) = ν q ( α ) + ν q ( b j ) < ν q ( a i )for all i = 0 , . . . , n −
1. As b j − = a j − αb j , one has that ν q ( b j − ) = ν q ( α ) + ν q ( b j ) < ν q ( b j − ) < ν q ( a i ) for all i = 0 , . . . , n −
1. In otherwords, max { , − ν q ( b j − ) } > N q . Applying a similar argument to the equalities b j = a j +1 − αb j +1 for j = j − , . . . , ν q ( b j ) < { , − ν q ( b j ) } > N q for all j = j − , . . . ,
0. However, one has that αb = a .This implies that ν q ( a ) < ν q ( b ) < N q ≥ − ν q ( a ) > − ν q ( b ) > N q ≥ max { , − ν q ( b ) } , a contradiction. This proves the claim.This claim and Remark 2.7 imply that T ( h ) ≤ T ( f ).Now we prove the assertion by induction on deg( f ). The base case deg( f ) =1 is obvious. Suppose that the assertion holds for deg( f ) ≤ n . Consider the11ase deg( f ) = n + 1. If f = g then there is nothing to prove. Suppose that f = g . Then there is β ∈ k ( t ) such that ( x + β ) g divides f . Let f = ( x + β ) ˜ f forsome ˜ f ∈ k ( t )[ x ]. Then T ( ˜ f ) ≤ T ( f ) and g divides ˜ f . By induction hypothesis,one has that T ( g ) ≤ T ( ˜ f ) ≤ T ( f ). Corollary 2.15.
Assume f ( x, y ) , g ( x, y ) ∈ k ( t )[ x, y ] and g ( x, y ) is a factor of f ( x, y ) . Then T ( g ( x, y )) ≤ T ( f ( x, y )) . Proof.
Suppose that f ( x, y ) = P i,j c i,j x i y j where c i,j ∈ k ( t ). Let d be aninteger greater than tdeg( f ( x, y )). One has that f ( x, x d ) = X i,j c i,j x i + jd . Note that for 0 ≤ i, j, l, m < d , i + jd = l + md if and only if ( i, j ) = ( l, m ).This implies that the set of the coefficients of f ( x, y ) coincides with that of f ( x, x d ). Hence T ( f ( x, y )) = T ( f ( x, x d )). Similarly, T ( g ( x, y )) = T ( g ( x, x d )).It is clear that g ( x, x d ) is still a factor of f ( x, x d ). By Lemma 2.14, T ( g ( x, x d )) ≤ T ( f ( x, x d )). Thus T ( g ( x, y )) ≤ T ( f ( x, y )). Proposition 2.16.
Assume f, g ∈ k ( t )[ y , z ] , then T (res z ( f, g )) ≤ deg( g, z ) T ( f ) + deg( f, z ) T ( g ) where res z ( f, g ) is the resultant of f and g with respect to z .Proof. The assertion is clear that if res z ( f, g ) = 0. Consider the case res z ( f, g ) =0. Assume deg( f, z ) = n, deg( g, z ) = m . Write f = n X i =0 a i ( y ) z i , g = m X i =0 b i ( y ) z i where a i ( y ) , b i ( y ) ∈ k ( t )[ y ]. Denote further by C , C the sets of the coefficientsin y , z of f, g respectively. Thenres z ( f, g ) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) a n a n − · · · a . . . . . . . . . a n a n − · · · a b m b m − · · · b . . . . . . . . . b m b m − · · · b (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . By the definitions of determinant, we can writeres z ( f, g ) = X d ℓ d X j =1 β d ,j m d ,j n d ,j y d β d ,j , ℓ d ∈ Z , ℓ d ≥ m d ,j is a monomial in C with total degree m and n d ,j is a monomial in C with total degree n . Let R = k ( t, C , C ). For eachplace p of R , we have − ν p X j β d ,j m d ,j n d ,j ≤ max j {− ν p ( m d ,j n d ,j ) }≤ m max c ∈ C {− ν p ( c ) } + n max c ∈ C {− ν p ( c ) } . Therefore by Remark 2.7, T (res z ( f, g )) ≤ mT ( f ) + nT ( g ) .
3. Degrees and Heights on Riemann-Roth spaces
Throughout this section, R denotes an algebraic function field of one variableover k ( t ). Let ξ = ( ξ , ξ , ξ ) ∈ R be such that C ( ξ ) a plane projective modelof R , i.e. R = k ( t )( ξ /ξ , ξ /ξ ). Each h ∈ R can be presented by G ( ξ ) /H ( ξ )where G, H are two homogeneous polynomials in k ( t )[ x , x , x ] of the samedegree and having no common factors, and H ( ξ ) = 0. We call ( G, H ) a repre-sentation of h . This section shall focus on determining the degrees and heightsof representations of elements in Riemann-Roch spaces. There are several al-gorithms for computing the bases of Riemann-Roth spaces (see for example[10, 11]). However no existing algorithm provided explicit bounds for the de-grees and heights of G and H where ( G, H ) represents an element in these bases.These bounds play an essential role in estimating the heights of points on a planealgebraic curve. In this section, we shall follow the algorithm developed in [11]to obtain these bounds. For this purpose, we need to resolve singularities of agiven plane algebraic curve to obtain the one with only ordinary singularities.This can be done by a sequence of quadratic transformations.In this section, unless otherwise stated, F always stands for an irreduciblehomogeneous polynomial in k ( t )[ x , x , x ] of degree n > R , i.e. there is ξ = ( ξ , ξ , ξ ) ∈ R such that R = k ( t )( ξ /ξ , ξ /ξ ) and F ( ξ ) = 0. Let D be a divisor in R . Due to Proposition on page of [9], there is abirational transformation B such that the transformation of F = 0 under B is aplane projective curve with only ordinary singularties, moreover B can be chosento be the composition of a sequence of suitable quadratic transformations. Inthis subsection, we shall investigate the degree and height of the transformationof F = 0 under a quadratic transformation.13 efinition 3.1. L stands for a projective change of coordinates on P ( k ( t )) that is defined as L ( c ) = c M L where M L ∈ GL ( k ( t )) , and Q denotes thestandard quadratic transformation that is defined as Q ( c ) = ( c c , c c , c c ) where c = ( c , c , c ) . The height of L , denoted by T ( L ) , is defined as T ( M L ) . Notation 3.2. F L stands for F (( x , x , x ) M L ) . F Q stands for the irreducible polynomial ˜ F satisfying F ( x x , x x , x x ) = x d x d x d ˜ F where d i ≥ .One sees that V ( F L ) (resp. V ( F Q ) ) is the variaty L − ( V ( F )) (resp. theprojective closure of Q − ( V ( F ) \ V ( x x x ) ). Remark 3.3. Q is bijective on P ( k ( t )) \ V ( x x x ) and Q − = Q . Let us first bound the heights of the common points of two algebraic curvesin P ( k ( t )). Proposition 3.4.
Let
F, G be two homogenenous polynomials in k ( t )[ x , x , x ] of degree n, m respectively. Suppose that F and G have no common factor, and c ∈ P ( k ( t )) is a common point of F = 0 and G = 0 . Then T ( a ) ≤ mT ( F ) + nT ( G )) . Furthermore, if G = c x + c x + c x with c i ∈ k then T ( a ) ≤ T ( F ) .Proof. Let H i ( x j , x l ) = res x i ( F, G ) where { i, j, l } = { , , } . Proposition 2.16implies that T ( H i ) ≤ mT ( F ) + nT ( G ) . Without loss of generality, suppose a = (1 , a , a ). Since a is a common point of F = 0 and G = 0, H (1 , a ) = H (1 , a ) = 0 . It follows from Corollary 2.12 that T ( a i ) ≤ mT ( F ) + nT ( G ) for all i = 1 ,
2. Let R = k ( t, a , a ) and p a place of R . Thenmax { , − ν p ( a ) , − ν p ( a ) } ≤ max { , − ν p ( a ) } + max { , − ν p ( a ) } . Whence T ( a ) ≤ T ( a ) + T ( a ) ≤ mT ( F ) + nT ( G )) . It remains to pove the second assertion. Since a = 1 = 0, not all c , c arezero. Without loss of generality, assume that c = 0. Substituting a = − ( c + a c ) /c into F = 0 yields that F (1 , − ( c + a c ) /c , a ) = 0 . T ( a ) ≤ T ( F ). On the other hand, one sees that ν p ( a ) = ν p ( − ( c + a c ) /c ) ≥ min { ν p ( c ) , ν p ( a ) } . So max { , − ν p ( a ) , − ν p ( a ) } ≤ max { , − ν p ( a ) } . which results in T ( a ) ≤ T ( a ) ≤ T ( F ). Corollary 3.5. If a ∈ P ( k ( t )) is a singular point of F = 0 , then T ( a ) ≤ n − T ( F ) . Lemma 3.6.
Suppose that L is a projective change of coordinates. Then T ( F L ) ≤ T ( F ) + deg( F ) T ( L ) ; for each c ∈ P ( k ( t )) , T ( L ( c )) ≤ T ( c ) + T ( L ) .Proof. Suppose that M L = ( a i,j ) and F = n X i =0 n − i X j =0 c i,j x i x j x n − i − j where n = deg( F ) and c i,j ∈ k ( t ).1. One has that F L = n X i =0 n − i X j =0 c i,j X l =1 a l, x l − ! i X l =1 a l, x l − ! j X l =1 a l, x l − ! n − i − j . Let ρ be a coefficient of F L viewed as polynomial in x , x , x . Then ρ is a k -linear combination of monomials c i,j a e , , . . . a e , , with P e i,j = n . Let R bea finite extension of k ( t ) containing all c i,j and a i,j . Suppose that p is a placeof R . Then one has that ν p ( ρ ) ≥ min i,j,i ′ ,j ′ ν p ( c i,j ) + X i ′ ,j ′ e i ′ ,j ′ ν p ( a i ′ ,j ′ ) ≥ min i,j { ν p ( c i,j ) } + min i,j X i,j e i,j ν p ( a i,j ) , i.e. − ν p ( ρ ) ≤ max i,j {− ν p ( c i,j ) } + max i,j − X i,j e i,j ν p ( a i,j ) ≤ max i,j {− ν p ( c i,j ) } + n max i,j {− ν p ( a i,j ) } . Therefore T ( F L ) ≤ T ( F ) + nT ( L ) due to Remark 2.7.15. Suppose that c = ( c , c , c ) and L ( c ) = ( b , b , b ). Then b i = P j =1 a j,i c j − .Let R be a finite extension of k ( t ) containing all c i and a i,j , and p a place of R .Then ν p ( b i ) = ν p X j =1 a j,i c j − ≥ min j { ν p ( a j,i ) + ν p ( c j − ) } i.e. − ν p ( b i ) ≤ max j {− ν p ( a j,i ) − ν p ( c j − ) } ≤ max j {− ν p ( a j,i ) } + max j {− ν p ( c j − ) } . So T ( L ( c )) ≤ T ( c ) + T ( L ). Corollary 3.7.
Suppose that c = ( c , c , ∈ P ( k ( t )) . Let L be a projectivechange of coordinates with M L = a a a a a a c c (2) where a i ∈ k . Then L ((0 , , c ; T ( F L ) , T ( F L − ) ≤ T ( F ) + deg( F ) T ( c ) ; for each b ∈ P ( k ( t )) , T ( L ( b )) , T ( L − ( b )) ≤ T ( b ) + T ( c ) .Proof. The first assertion is obvious. The second and third assertions followsfrom Lemma 3.6 and the fact that T ( M L ) = T ( c ) and T ( M L − ) ≤ T ( c ). Definition 3.8. The points (1 , , , (0 , , , (0 , , ∈ P ( k ( t )) are calledfundamental points. Assume (0 , , is a singular point of F = 0 with multiplicity r . F = 0 issaid to be in excellent position if it satisfies the following two conditions: (a) the line x = 0 intersects F = 0 in n distinct non-fundamental points; (b) the lines x = 0 , x = 0 intersect F = 0 in n − r distinct points otherthan fundamental points. Lemma 3.9.
Suppose that c = ( c , c , c ) is a singular point of F = 0 ofmultiplicity r . There is a projective change of coordinates L with M L havingthe form (2) such that L ((0 , , c and F L = 0 is in excellent position.Proof. Denote y = ( y , . . . , y ). Let L ′ y be the projective change of coordinateswith M L ′ y of the form y y y y y y c c c . One sees that there are polynomials f , . . . , f s ∈ k ( t )[ y , . . . , y ] such that L ′ b with b ∈ k ( t ) satisfies the required conditions if and only if b ∈ S = n b ∈ k ( t ) | ∀ i = 1 , . . . , s, f i ( b ) = 0 o . L is a projective change of coordinates such that L ( c ) = (0 , , L = L ′ b for some b ∈ k ( t ) . Due to Lemma 1 on page of [9], thereare projective changes of coordinates satisfying the above conditions. In otherwords, S = ∅ . Therefore S ∩ k = ∅ . For every b ∈ S ∩ k , L ′ b is as required. Lemma 3.10. T ( F Q ) = T ( F ) ; For each a = ( a , a , a ) ∈ P ( k ( t )) , T ( Q ( a )) ≤ T ( a ) .Proof. . Assume that F = P ni =0 P n − ij =0 c i,j x i x j x n − i − j where c i,j ∈ k ( t ). Then F ( x x , x x , x x ) = n X i =0 n − i X j =0 c i,j ( x x ) i ( x x ) j ( x x ) n − i − j = n X i =0 n − i X j =0 c i,j x n − i x n − j x i + j . From this, one sees that the set of coefficients of F is equal to that of F Q . Hence T ( F ) = T ( F Q ).2 . One has that Q ( a ) = ( a a , a a , a a ). Let R be a finite exntesion of k ( t ) containing all a i and p a place of R . Note that ν p ( a i a j ) = ν p ( a i ) + ν p ( a j ) ≥ { ν p ( a ) , ν p ( a ) , ν p ( a ) } , i.e. − ν p ( a i a j ) ≤ {− ν p ( a ) , − ν p ( a ) , − ν p ( a ) } . So T ( Q ( a )) ≤ T ( a ). Definition 3.11. We call a projective change of coordinates in Lemma 3.9a projective change of coordinates centered at c . Let L c be a projective change of coordinates centered at c and Q the standquadratic transformation. We call Q c = L c ◦ Q , the composition of Q and L c , a quadratic transformation centered at c . Notation 3.12.
Let Q c be a quadratic transformation centered at c . We shalldenote F Q c = ( F L c ) Q . Corollary 3.13.
Let c be a singular point of F = 0 and Q c a quadratic trans-formation centered at c . Then T ( F Q c ) ≤ T ( F ) + deg( F ) T ( c ) ; for a ∈ P ( k ( t )) , T ( Q − c ( a )) ≤ T ( c ) + T ( a )) ;Proof. Q − = Q and Lemmas 3.6 and 3.10. Proposition 3.14.
Let C ( ξ ) be a plane projective model of R defined by F and P a place of R . Let a be the center of P with respect to ξ . Assume that Q c is aquadratic transformation centered at c for some singular point c of F = 0 and a ′ is the center of P with respect to Q − c ( ξ ) . Then T ( a ′ ) ≤ max { T ( c ) + T ( a )) , T ( F ) + deg( F ) T ( c ) } . roof. We first claim that if a = c then Q − c ( a ) = (0 , , Q − c ( a ) = Q − L − c ( a ) = (0 , , L − c ( a ) is a fundamental point of F L c = 0. Since neither (1 , ,
0) nor (0 , ,
0) is a point of F L c = 0. One has that L − c ( a ) = (0 , , a = c . This proves our claim.Suppose that a = c . Then Q − c ( a ) = (0 , ,
0) and thus a ′ = Q − c ( a ) byProposition 2.5. Corollary 3.13 then implies that T ( a ′ ) ≤ T ( a ) + T ( c )) . Nowsuppose that a = c . Denote ξ ′ = L − c ( ξ ) = ( ξ ′ , ξ ′ , ξ ′ ). By Proposition 2.5again, (0 , ,
1) is the center of P with respect to ξ ′ . Suppose that u is a localuniformizer of P and ℓ i = ν P ( ξ ′ i ). From the definition of center, one sees that ℓ i > ℓ for all i = 0 ,
1. Write ξ ′ i = u ℓ i ( c i + uη i ) where c i ∈ k ( t ) \ { } , η i ∈ R with ν P ( η i ) ≥
0. One then has that Q − ( ξ ′ ) = ( ξ ′ ξ ′ , ξ ′ ξ ′ , ξ ′ ξ ′ )= (cid:0) u ℓ + ℓ ( c c + u ˜ η ) , u ℓ + ℓ ( c c + u ˜ η ) , u ℓ + ℓ ( c c + u ˜ η ) (cid:1) where ˜ η i ∈ R with ν P ( ˜ η i ) ≥
0. Set µ = min { ν P ( ξ ′ ξ ′ ) , ν P ( ξ ′ ξ ′ ) , ν P ( ξ ′ ξ ′ ) } . Since both ℓ and ℓ are greater than ℓ , µ = ℓ + ℓ or ℓ + ℓ . In the casethat µ = ℓ + ℓ = ℓ + ℓ , one has that a ′ = ( c c , c c ,
0) = c ( c , c , a ′ ∈ V ( F Q c ) ∩ V ( x ). By Propositon 3.4, T ( a ′ ) ≤ T ( F Q c ) ≤ T ( F )+deg( F ) T ( c ).In other cases, one sees that a ′ is a fundamental point and so T ( a ′ ) = 0. Hence,in each case, one has that T ( a ′ ) ≤ max { T ( c ) + T ( a )) , T ( F ) + deg( F ) T ( c ) } . Let D = P mi =1 n i P i be a disivor in R where n i = 0. Let C ( ξ ) definedby F be a plane projective model of R . Suppose that h ∈ L ( D ) and h = G ( ξ ) /H ( ξ ) where G, H are two homogeneous polynomials of the same degree in k ( t )[ x , x , x ]. In this subsection, we shall estimate deg( G ) and T ( G ) , T ( H ) interms of deg( F ) and T ( F ). For this, we introduce the following notations anddefinitions. Notation 3.15.
Let C ( ξ ) be a plane projective model of R and D a divisor. S ξ ( D ) := { the centers of places in supp( D ) with respect to ξ } . T ξ ( D ) := max { T ( c ) | c ∈ S ξ ( D ) } . Definition 3.16.
Let
G, H be two homogeneous polynomials in k ( t )[ x , x , x ] of the same degree. Write ξ = ( ξ , ξ , ξ ) . (1) . Define ord P ( G ( ξ )) = ν P ( G ( ξ )) − deg( G ) min i =0 { ν P ( ξ i ) } . . Define div ξ ( G ) = X ord P ( G ( ξ )) P where the sum ranges over all places of R . Furthermore, define div ξ ( G/H ) = div ξ ( G ) − div ξ ( H ) . It is easy to see that ord P ( G ( ξ )) ≥ P ( G ( ξ )) > G ( c ) = 0 where c is the center of P with respect to ξ . Futhermore ord P ( G ( λ ξ )) =ord P ( G ( ξ )) for all nonzero λ ∈ R . Remark 3.17.
On page 182 of [9], ord P ( G ) is defined to be the order at P ofthe image of G in the valutaion ring of P . Remark that ord P ( G ( ξ )) = ν P ( G ( ξ ) /ξ di ) where ξ i satisfies that ν P ( ξ i ) = min i =0 { ν P ( ξ i ) } . Under the map sending x j to ξ j /ξ i for all j = 0 , , , G is sent to G ( ξ ) /ξ di which lies in the valuation ringof P . Therefore ord P given in Definition 3.16 concides with the one given in[9] and div ξ ( G ) is nothing else but the intersection cycle of G and H (see page119 of [9]). The lemma below follows easily from the definition.
Lemma 3.18.
Suppose that
G, H ∈ k ( t )[ x , x , x ] are two homogeneous poly-nomials of the same degree. Then
1. div ξ (cid:0) GH (cid:1) = div ξ ( G ) − div ξ ( H ) = div (cid:16) G ( ξ ) H ( ξ ) (cid:17) ; and
2. deg(div ξ ( G )) = deg( G ) deg( F ) . Lemma 3.19.
Suppose that c = ( c , c , is an ordinary singular point of F = 0 of multiplicity r , n = deg( F ) and S is a finite set of points of F = 0 . Let L λ = x − c x − λ ( x − c x ) . Then for all but a finite number of λ , L λ = 0 intersects F = 0 in n − r distinct points other than the points in { c } ∪ S . Write F = F r ( x − c x , x − c x ) x n − r + · · · + F n ( x − c x , x − c x ) where F i ( y , y ) is a homogeneous polynomial in y , y of degree i . Assumethat F r (1 , F r (0 , F n (1 , F n (0 , = 0 . Let G λ = α ( x − c x ) x − ( x − c x ) x − λ ( x − c x )( x − c x ) where α ∈ k ( t ) \ { } satisfies that F r ( α,
1) = 0 . Then for all but a finitenumber of λ , G λ = 0 intersects F = 0 in n − r − distinct points otherthan the points in { c } ∪ S . roof. . Under the projective change of coordinates L with L ( x ) = x + c x , L ( x ) = x + c x and L ( x ) = x , we may assume that c = (0 , , L λ = x − λx where λ ∈ k ( t ). Substituting x = λx into F yields that x r (cid:0) F r ( λ, x n − r + F r +1 ( λ, x x n − r − + · · · + F n ( λ, x n − r (cid:1) . Set H λ ( t ) = F λ ( λ, t n − r + · · · + F n ( λ, γ of H λ ( t ) = 0, one seesthat ( λ, , γ ) is a common point of L λ = 0 and F = 0 other than c . Moreover if H λ ( t ) = 0 has n − r distinct roots then L λ = 0 intersects F = 0 in n − r distinctpoints other than c . So it suffices to prove that for all but a finite number of λ , H λ ( t ) = 0 has n − r distinct roots. Note tht substituting x = λx into ∂F/∂x yields that x r n X i = r ( n − i ) F i ( λ, x n − i − x i − r . From this, one sees that if γ is a common root of H λ ( t ) = 0 and ∂H λ ( t ) /∂t = 0then ( λ, , γ ) is a common point of F = 0 and ∂F/∂x = 0. Since F = 0 and ∂F/∂x = 0 have only finitely many common points, there are only finitelymany λ such that H λ ( t ) = 0 has multiple roots. In other words, there is only afinite number of λ such that L λ = 0 intersects F = 0 in less than n − r distinctpoints other than c . It remains to show that there are only finitely many λ suchthat ( S \ { c } ) ∩ V ( L λ ) = ∅ . Assume that ( a , a , a ) ∈ S \ { c } which lies in theline L λ = 0. If a − c a = 0 then λ = ( a − c a ) / ( a − c a ) . If a − c a = 0 then a − c a = 0 and thus ( a , a , a ) = a ( c , c , a , a , a ) = (0 , ,
0) or c , which is impossible.2 . Similarly, we may assume that c = (0 , , G λ = αx x − x x − λx x ,F = F r ( x , x ) x n − r + · · · + F n ( x , x ) . Applying the standard quadratic transformation Q to G λ and F , one obtainsthat G Q λ = αx − x − λx ,F ( x x , x x , x x ) = x r (cid:0) F r ( x , x )( x x ) n − r + · · · + F n ( x , x ) x n − r (cid:1) . Since c is an ordinary singular point and F r (1 , F r (0 , = 0, by on page of [9](1 , α,
0) is a simple point of F Q = 0. Moveover, as F n (1 , F n (0 , = 0, neither(1 , ,
0) nor (0 , ,
0) is a point of F Q = 0 and so deg( F Q ) = 2 n − r . Thus F Q = F r ( x , x )( x x ) n − r + · · · + F n ( x , x ) x n − r . For every common point ( γ , γ , γ ) of G Q λ = 0 and F Q = 0 with λγ = 0,( γ γ , γ γ , γ γ ) = (0 , ,
0) and then it is a common point of G λ = 0 and20 = 0 other than c . Therefore it suffices to show that for all but a finitenumber of λ , G Q λ = 0 and F Q = 0 have 2 n − r − γ , γ , γ ) with γ = 0. Let L be the projective change of coordinates such that L ( x ) = ( x + x ) /α, L ( x ) = x , L ( x ) = x . Then ( G Q λ ) L = x − λx . Notethat L − ((1 , α, , α,
0) which is a simple point of ( F Q ) L = 0. Thus( F Q ) L = ˜ F ( x , x ) x n − r − + ˜ F ( x , x ) x n − r − + · · · + ˜ F n − r ( x , x ) . By (1), for all but a finite number of λ , ( G Q λ ) L = 0 intersects ( F Q ) L = 0 in2 n − r − γ ′ , γ ′ , γ ′ ) with γ ′ = 0. Remark that if ( γ ′ , γ ′ , γ ′ )is a common point of ( G Q λ ) L = 0 and ( F Q ) L = 0 with γ ′ = 0 then ( γ ′ + β/αγ ′ , γ ′ , γ ′ ) is a common point of G Q λ = 0 and F Q = 0 with γ ′ = 0. Theseimply that for all but a finite number of λ , G Q λ = 0 intersects F Q = 0 in 2 n − r − γ , γ , γ ) with γ = 0.Finally, we need to prove that there are only finitely many λ such that S \ { c } ∩ V ( G λ ) = ∅ . Assume that a = ( a , a , a ) ∈ S \ { c } which lies in G λ = 0. We claim that ( a − c a )( a − c a ) = 0. Suppose on the contrarythat ( a − c a )( a − c a ) = 0. Then by G λ ( a ) = 0, one sees that either a = 0or both a − c a and a − c a are zero. This implies that a must be one ofthree points (1 , , , (0 , , , a ( c , c , λ is uniquely determined by a . Proposition 3.20.
Suppose that F = 0 has only ordinary singularities and D is an effective divisor in R . Let D ′ be a divisor in R . Assume further that D = P ri =1 P i where all P i have the same centerwhich is a point of F = 0 with multiplicity r . Then there is a linearhomogeneous polynomial G in k ( t )[ x , x , x ] such that div ξ ( G ) = D + A where A is a very simple and effective divisor of degree n − r , supp( A ) ∩ (supp( D ′ ) ∪ { P , . . . , P r } ) = ∅ , and T ( G ) ≤ T ξ ( D ) , T ξ ( A ) ≤ T ( F ) + nT ξ ( D )) . Assume that D = P where the center of P is a singular point of F = 0 .Then there are two homogeneous polynomials G, H ∈ k ( t )[ x , x , x ] ofdegree two such that div ξ ( G/H ) = D + A where A is a very simple divisor, supp( A ) ∩ (supp( D ′ ) ∪ { P } ) = ∅ and T ( G ) , T ( H ) ≤ T ( F ) + nT ξ ( D ) , T ξ ( A ) ≤ (2 n + 4) T ( F ) + 2 n T ξ ( D ) . Proof. . Suppose c = ( c , c , c ) is the center of P i with respect to ξ . Withoutloss of generality, we assume that c = 0 and c = ( c , c , L λ = x − c x − λ ( x − c x ). Due to Lemma 3.19, for all but a finite number of λ , L λ = 021ntersects F = 0 in n − r distincet points other than the points in { c } ∪ S ξ ( D ′ ).Let λ ′ ∈ k be such that L λ ′ = 0 satisfies the above condition. Thendiv ξ ( L λ ′ ) = r X i =1 P i + A where A is an effective divisor of degree n − r and supp( A ) ∩ (supp( D ′ ) ∪{ P , . . . , P r } ) = ∅ . It is clear that A is very simple since L λ ′ = 0 intersects F in n − r distinct points other than c . Finally, one easily sees that T ( L λ ′ ) ≤ T ( c ) = T ξ ( D ). As the points in T ξ ( A ) are the intersection points of F = 0 and L λ ′ = 0, T ξ ( A ) ≤ T ( F ) + nT ξ ( D )) by Proposition 3.4.2 . Suppose that c = ( c , c , c ) is the center of P with respect to ξ , and c is of multiplicity r >
0. Since c is an ordinary singular point, there are exactly r places of R with c as the center with respect to ξ . Denote these r places by P = P , . . . , P r . Without loss of generality, we may assume that c = 0 and c = ( c , c , F = F r ( x − c x , x − c x ) x n − r + · · · + F n ( x − c x , x − c x )where F i ( y , y ) is a homogeneous polynomial of degree i . Choose a projectivechange of coordinates L with M L = diag( B, , B ∈ GL ( k ) such that F L = ˜ F r ( x − ˜ c x , x − ˜ c x ) x n − r + · · · + ˜ F n ( x − ˜ c x , x − ˜ c x )satisfies that ˜ F r (1 ,
0) ˜ F r (0 ,
1) ˜ F n (1 ,
0) ˜ F n (0 , = 0, where ˜ F i = F i (( y , y ) B ) and(˜ c , ˜ c ) = ( c , c ) B − . By Lemma 3.6, T ( F L ) ≤ T ( F ). Denote˜ ξ = ( ˜ ξ , ˜ ξ , ˜ ξ ) = ξ M − L . Then ˜ c = (˜ c , ˜ c ,
1) = c M − L is the center of P with respect to ˜ ξ . For i = 0 , ξ i / ˜ ξ = ˜ c i + α i u d + u d +1 η i where u is a local uniformizer of P , d ≥ α i ∈ k ( t ) not all zero, and ν P ( η i ) ≥ F L ( ˜ ξ / ˜ ξ , ˜ ξ / ˜ ξ ,
1) = u dr ˜ F r ( α , α ) + u dr +1 β where ν P ( β ) ≥
0. This implies that ˜ F r ( α , α ) = 0. Since ˜ F r (0 ,
1) ˜ F r (1 , = 0, α α = 0. Set ¯ α = α /α and˜ G λ = ¯ α ( x − ˜ c x ) x − ( x − ˜ c x ) x − λ ( x − ˜ c x )( x − ˜ c x ) . Due to Lemma 3.19, for all but a finite number of λ , ˜ G λ intersects F L = 0 in2 n − r − { ˜ c } ∪ S ˜ ξ ( D ′ ). Let A λ be the verysimple divisor consisting of the 2 n − r − ξ are the intersection points of ˜ G λ = 0 and F L = 0 other than ˜ c respectively.22hen supp( A λ ) ∩ (supp( D ′ ) ∪ { P , . . . , P r } ) = ∅ . We claim that for the above˜ G λ , div ˜ ξ ( ˜ G λ ) = P + r X i =1 P i + A λ Note that ˜ G λ ( ˜ ξ )˜ ξ = ¯ α (cid:0) α u d + u d +1 η (cid:1) − ( α u d + u d +1 η ) − λ ( α u d + u d +1 η )( α u d + u d +1 η ) = u d +1 γ where ν P ( γ ) ≥
0. This implies that ord P ( ˜ G λ ( ˜ ξ )) ≥ d + 1 ≥
2. Hencediv ˜ ξ ( ˜ G λ ) ≥ P + r X i =1 P i + A λ . On the other hand, since deg(div ˜ ξ ( ˜ G λ )) = 2 n , one has thatdiv ξ ′ ( ˜ G λ ) = P + r X i =1 P i + A λ . This proves our claim. Now set G λ = ˜ G L − λ . As ˜ ξ = ξ , one sees thatmin j { ν P i ( ˜ ξ j ) } = ν P i ( ˜ ξ ) = ν P i ( ξ ) = min j { ν P i ( ξ j ) } . This implies thatord P i ( G λ ( ξ )) = ν P i ( G λ ( ξ )) − i { ν P i ( ξ i ) } = ν P i (cid:16) ˜ G L − λ ( ξ ) (cid:17) − i { ν P i ( ˜ ξ i ) } = ν P i (cid:16) ˜ G λ ( ˜ ξ ) (cid:17) − i { ν P i ( ˜ ξ i ) } = ord P i ( ˜ G λ ( ˜ ξ )) . Therefore div ξ ( G λ ) = P + P ri =1 P i + A . Note that we can choose λ ∈ k . Forsuch λ , one has that T ( G λ ) ≤ T ( ˜ G λ ) ≤ T ( c ) + T (( α , α , ≤ T ( c ) + T ( ˜ F r ) . Since deg( ˜ F r ) = r ≥
2, one sees that T ( ˜ F r ) ≤ T ( F ) + ( n − T ( c ). This impliesthat T ( G λ ) ≤ T ( F ) + nT ( c ) and T ( A λ ) ≤ T ( F ) + nT ( G λ )) ≤ (2 n + 4) T ( F ) + 2 n T ( c ) . Now applying 1 . to the case that D = P ri =1 P i and D ′ = A λ , one gets a linearhomogeneous polynomial L such that div ξ ( L ) = P ri =1 P i + C , where C is avery simple divisor satisfying that supp( C ) ∩ { supp( A λ ) ∪ { P , . . . , P r }} = ∅ .23oreover T ( L ) ≤ T ξ ( P ) = T ( c ). Let L be a linear homogeneous polynomialin k [ x , x , x ] such that L = 0 intersects F = 0 in n distinct points otherthan the points in T ξ ( A λ + C ). For such L , one has that div ξ ( L ) is a verysimple divisor satisfying that supp(div ξ ( L )) ∩ supp( A + C + D ′ ) = ∅ . Set H = L L and A = A − C − div ξ ( L ). Then T ( H ) ≤ T ( c ) and we obtaintwo polynomials G, H as required. Note that T ξ ( C ) ≤ T ( F ) + nT ξ ( D )) and T ξ (div ξ ( L )) ≤ T ( F ). Hence T ( G λ ) , T ( H ) ≤ T ( F ) + nT ( c ) and T ξ ( A ) ≤ (2 n + 4) T ( F ) + 2 n T ( c ) . Definition 3.21.
Suppose that F has only ordinary singularities, say q , . . . , q ℓ ,and r i is the multiplicity of q i . Suppose further that for each i = 1 , . . . , ℓ , Q i, , . . . , Q i,r i are all places of R with q i as center with respect to ξ . Set E ξ = ℓ X i =1 ( r i − r i X j =1 Q i,j . A homogeneous polynomial G such that div ξ ( G ) ≥ E ξ is called an adjoint of F . We have the following two corollaries of Proposition 3.20.
Corollary 3.22.
Suppose that D is a simple and effective divisor in R . Let D ′ be a divisor in R . Then there is a homogeneous polynomial G of degree notgreater than deg( D ) + ( n − / such that div ξ ( G ) = D + E ξ + A where A is a very simple and effective divisor of degree not greater than deg( D )( n −
1) + n ( n − / such that supp( A ) ∩ (supp( D + E ξ ) ∪ supp( D ′ )) = ∅ and T ξ ( A ) ≤ T ( F ) + nT ξ ( D + E ξ )) . Moreover T ( G ) ≤ (cid:0) deg( D ) + ( n − / (cid:1) T ξ ( D + E ξ ) . Proof.
Denote µ = deg( D )+ P ℓi =1 ( r i −
1) where r i is given as in Definition 3.21.Note that P ℓi =1 ( r i − ≤ ( n − / µ ≤ deg( D ) + ( n − /
2. Write D + E ξ = deg( D ) X i =1 P i + µ X s =deg( D )+1 D s where the centers of P i is a simple point of F = 0, D s = P r i j =1 Q i,j for some1 ≤ i ≤ ℓ . Applying succesively Proposition 3.20 to P i and D s , one obtains µ linear homogeneous polynomials L , . . . , L µ such that div ξ ( L i ) = P i + A i if i ≤ m , or div ξ ( L i ) = D i + A i if i > m , where A i is a very simple and effectivedivisor such thatsupp( A i ) ∩ (supp( D ′ ) ∪ supp( D + E ξ + A + · · · + A i − )) = ∅ . G = P µi =1 L i and A = P µi =1 A i . Then one has thatdiv ξ ( G ) = D + E ξ + A. Moreover by Proposition 3.20, T ( L i ) ≤ T ξ ( D + E ξ ) for all i = 1 , . . . , µ andthen Proposition 2.10 implies that T ( G ) ≤ µT ξ ( D + E ξ ). It is obvious that T ξ ( A ) is not greater than 2( T ( F ) + nT ξ ( D + E ξ )) because so is T ξ ( A i ) for all i = 1 , . . . , µ . Corollary 3.23.
Suppose that
D, D ′ are two divisors in R . Then there are twohomogeneous polynomials G, H of the same degree ≤ D + + D − ) such that ˆ D = div ξ ( G/H ) + D is very simple and supp(div ξ ( G/H ) + D ) ∩ supp( D ′ ) = ∅ .Moreover deg( ˆ D + ) , deg( ˆ D − ) ≤ n (deg( D + + D − )) and T ξ (div ξ ( G/H ) + D ) ≤ (2 n + 4) T ( F ) + 2 n T ξ ( D ) T ( G ) , T ( H ) ≤ deg( D + + D − )( T ( F ) + nT ξ ( D )) , where n = deg( F ) .Proof. We first show the case that − D is effective. Denote µ = deg( − D ) andwrite − D = s X i =1 P i + µ X i = s +1 Q i where the center of P i (resp. Q j ) with respect to ξ is a simple (resp. singular)point of F = 0. Applying Proposition 3.20 to P si =1 P i yields a homogenenouspolynomial G of degree s such that div ξ ( G ) = P si =1 P i + A where A is avery simple and effective divisor such that supp( A ) ∩ (supp( D ) ∪ supp( D ′ )) = ∅ . Moreover deg( A ) = ns − deg( P si =1 P i ). Construct s linear homogeneouspolynomials L , . . . , L s in k [ x , x , x ] such that div ξ ( L · · · L s ) is very simpleand supp(div ξ ( L · · · L s )) ∩ (supp(div ξ ( G ) ∪ supp( D ′ )) = ∅ . It is easy to seethat deg(div ξ ( L · · · L s )) = ns . Set H = L · · · L s . By Proposition 3.20 again,one obtains µ − s pairs ( G , H ) , . . . , ( G µ − s , H µ − s ) of homogeneous polynomialsof degree two such that div ξ ( G i /H i ) = Q i + A i where A i is a very simple divisorsuch that supp( A i ) ∩ (supp( D ′ ) ∪ supp( D + A + · · · + A i − )) = ∅ and deg( A + i ) = 2 n − deg( Q i ), deg( A − i ) = 2 n . Set ˜ G = G G · · · G µ − s and˜ H = H H · · · H µ − s . Thenˆ D = div ξ ( ˜ G/ ˜ H ) + D = A + A + · · · + A µ − s which is very simple. It is clear that deg( ˜ G ) = deg( ˜ H ) ≤ − D ), and byProposition 3.20 T ( ˜ G ) ≤ T ( G ) + ( µ − s ) T ( G i ) ≤ µ ( T ( F ) + nT ξ ( D )) . T ( ˜ H ) ≤ µ ( T ( F ) + nT ξ ( D )). Furthermore, one has that T ξ (cid:16) div ξ ( ˜ G/ ˜ H ) + D (cid:17) ≤ (2 n + 4) T ( F ) + 2 n T ξ ( D )and deg( ˆ D + ) = (2 µ − s ) n − deg( − D ) ≤ nµ, deg( ˆ D − ) = (2 µ − s ) n ≤ nµ. For the general case, write D = D + − D − . The previous discussion implies thatwe can obtain ˜ G i , ˜ H i such that div ξ ( ˜ G / ˜ H ) − D + and div ξ ( ˜ G / ˜ H ) − D − arevery simple. Moreoversupp (cid:16) div ξ ( ˜ G / ˜ H ) − D + (cid:17) \ supp (cid:16) div ξ ( ˜ G / ˜ H ) − D − (cid:17) = ∅ , supp (cid:16) div ξ ( ˜ G / ˜ H ) − D − + div ξ ( ˜ G / ˜ H ) − D + (cid:17) \ (supp( D ) ∪ supp( D ′ )) = ∅ . Set G = ˜ G ˜ H and H = ˜ G ˜ H . Then div ξ ( G/H ) + D satisfies the requiredcondition. Furthermore deg( ˆ D + ) , deg( ˆ D − ) ≤ n (deg( D + + D − )) and T ( G ) , T ( H ) ≤ deg( D + + D − )( T ( F ) + nT ξ ( D )) ,T ξ (div ξ ( G/H ) + D ) ≤ (2 n + 4) T ( F ) + 2 n T ξ ( D ) . Now we are ready to pove the main results of this section. Let us start withtwo lemmas.
Lemma 3.24.
Suppose that F = 0 has only ordinary singularities and D is adivisor in R . Let H be a homogeneous polynomial in k ( t )[ x , x , x ] such that div ξ ( H ) = D + + E ξ + A , where A is an effective divisor. Then L ( D ) = (cid:26) G ( ξ ) H ( ξ ) (cid:12)(cid:12)(cid:12)(cid:12) G are homogeneous polynomials of deg( H ) with div ξ ( G ) ≥ D − + E ξ + A (cid:27) . (3) Proof.
Note that div( G ( ξ ) /H ( ξ )) = div ξ ( G ) − div ξ ( H ). It is obvious that theright hand side of (3) is a subspace of L ( D ). Suppose that h ∈ L ( D ) \ { } , i.e. D ′ = div( h ) + D is effective. Thendiv ξ ( H ) + div( h ) = D − + D ′ + E ξ + A. By the Residuce Theorem (see page of [9]), there is a homogeneous polynomial G of degree deg( H ) such thatdiv ξ ( G ) = D − + D ′ + E ξ + A ≥ D − + E + A. One sees thatdiv( hH ( ξ ) /G ( ξ )) = div( h ) + div ξ ( H/G ) = div( h ) + div ξ ( H ) − div ξ ( G ) = 0 . Thus hH ( ξ ) /G ( ξ ) ∈ k ( t ), i.e. h belongs to the right hand side of (3).26 emma 3.25. Assume that M = ( a i,j ) is an l × m matrix with a i,j ∈ k ( t ) .Assume further that for each place p of k ( t, a , , . . . , a l,m ) , − ν p ( a i,j ) ≤ m p where m p ≥ and for all but finite number of p , m p = 0 . Then there is a basis B ofthe solution space of M Y = 0 satisfying that T ( b ) ≤ min { l, m } P p m p [ k ( t, a , , . . . , a l,m ) : k ( t )] for all b ∈ B .Proof. Assume that r = rank( M ). Then r ≤ min { l, m } . Without loss ofgenerality, we may assume the first r -rows of M are linearly independent anddenote by ˜ M the matrix formed by them. Then the solution space of ˜ M Y = 0 isthe same as that of
M Y = 0. Hence it suffices to consider the system ˜
M Y = 0.We may further assume that the matrix ˜ M formed by the first r -columns of˜ M is invertible. For every i = 1 , . . . , r and j = r + 1 , . . . , m , set d i,j to be thedeterminant of the matrix obtained from ˜ M by replacing the i -th column of˜ M by the j -th column of ˜ M . For each j = r + 1 , · · · , m , denote c j = ( d ,j , . . . , d r,j , , . . . , , det( M ) | {z } j , , . . . , t where ( · ) t denotes the transpose of a vector. Then by Cramer’s rule, the c j are solutions of ˜ M Y = 0 and thus they form a basis of the solution space of˜
M Y = 0. Note that d i,j as well as det( ˜ M ) is an integer combination of themonomials in the entries of ˜ M of total degree r . So for all i = 1 , . . . , r and j = r + 1 , . . . , m , − ν p ( ˜ M ) , − ν p ( d i,j ) ≤ rm p ≤ min { l, m } m p where p is a place of k ( t, a , , . . . , a l,m ). This together Remark 2.7 implies thelemma. Theorem 3.26.
Suppose that F = 0 has only ordinary singularities. Let D bea divisor in R . Denote µ = deg( D + + D − ) and N = max { T ξ ( D ) , T ( F ) } . Thenthere is a k ( t ) -basis B of L ( D ) such that every element of B can be representedby G ( ξ ) /H ( ξ ) where G, H are two homogeneous polynomials of the same degreenot greater than ≤ n + 1) µ + ( n − / and T ( G ) , T ( H ) ≤ n ( n + 1) (2 µ + ( n − / N. Proof.
By Corollary 3.23, there are two homogeneous polynomials G , H of thesame degree ≤ µ such that ˆ D = div ξ ( G /H ) + D is very simple. Moreover T ( G ) , T ( H ) ≤ µ ( n + 1) N,T ξ ( ˆ D ) ≤ (2 n + 4) T ( F ) + 2 n T ξ ( D ) ≤ (2 n + 2 n + 4) N, deg( ˆ D + ) , deg( ˆ D − ) ≤ nµ. G of degree notgreater than 2 nµ + ( n − / ξ ( G ) = ˆ D + + E ξ + A , where A isa very simple and effective divisor and supp( A ) ∩ supp( ˆ D − ) = ∅ . Moreover T ( G ) ≤ (2 nµ + ( n − / T ξ ( ˆ D + + E ξ ) ≤ (2 nµ + ( n − / n + 2 n + 4) N and T ξ ( A ) ≤ T ( F ) + nT ξ ( ˆ D + + E ξ )) ≤ n + 2 n + 4 n + 1) N. Denote d = deg( G ). By Lemma 3.24, to compute L ( D ), it suffices to computeall homogeneous polynomials H of degree d satisfying thatdiv P ( H ) ≥ ˆ D − + E ξ + A. Assume that H = d X i =0 d − i X j =0 c i,j x i x j x d − i − j where c i,j are indeterminates. There are ( d +1)( d +2) / P ∈ supp( ˆ D − + A ), div ξ ( H ) ≥ P if and only if the center of P withrespect to ξ is a zero of H . This imposes deg( ˆ D − + A ) linear constraints on H . At the same time, div ξ ( H ) ≥ ( r i − P r i j =1 Q i,j if and only if the centerof Q i, with respect to ξ is a common zero of ∂ j + j + j ( H ) ∂x j x j x j for all nonnegative integers j , j , j satisfying that j + j + j = r i −
2, where Q i,j is as in Definition 3.21. This imposes r i ( r i − / H .So there are totally deg( ˆ D − + A ) + deg( E ξ ) / H . Theproblem of finding H is reduced to that of solving the system M Y = 0, where Y is a vector with indeterminates entries and M is a (deg( ˆ D − + A ) + deg( E ξ ) / × ( d + 1)( d + 2) / c P = ( c , P , c , P , c , P ) the center of P insupp( ˆ D − + E ξ + A ). Then the entries in the same row of M are monomials oftotal degree ≤ d in c , P , c , P , c , P for some P in supp( ˆ D − + E ξ + A ). Withoutloss of generality, we may assume that one of c , P , c , P , c , P is 1. Let R be afinite extension of k ( t ) containing all c i, P . For each place p of R , set m p = d X P ∈ supp( ˆ D − + E ξ + A ) max {− ν p ( c , P ) , − ν p ( c , P ) , − ν p ( c , P ) } . Since max {− ν p ( c , P ) , − ν p ( c , P ) , − ν p ( c , P ) } ≥ P , m p ≥ − ν p ( a i,j ) ≤ d max P ∈ supp( ˆ D − + E ξ + A ) max {− ν p ( c , P ) , − ν p ( c , P ) , − ν p ( c , P ) } ≤ m p where M = ( a i,j ). Note thatdeg( ˆ D − + E ξ + A ) ≤ deg(div ξ ( H )) = nd. M yields that T ( H ) ≤ deg( ˆ D − + E ξ + A ) X p m p [ R : k ( t )] ≤ nd X P X p max {− ν p ( c , P ) , − ν p ( c , P ) , − ν p ( c , P ) } [ R : k ( t )] ≤ nd X P T ( c P ) ≤ nd deg( ˆ D − + E ξ + A ) max P T ( c P ) ≤ n d T ξ ( ˆ D − + E ξ + A ) ≤ n d (2 n + 2 n + 4 n + 1) N ≤ n (2 µ + ( n − / (2 n + 2 n + 4 n + 1) N. The last inequality holds because d ≤ nµ + ( n − / ≤ n (2 µ + ( n − / . Set G = H G and H = G H . Thendeg( G ) = deg( H ) ≤ n + 1) µ + ( n − / ,T ( G ) , T ( H ) ≤ T ( H ) + T ( G ) < T ( H ) + µ ( n + 1) N ≤ n (2 µ + ( n − / (2 n + 2 n + 4 n + 2) N ≤ n ( n + 1) (2 µ + ( n − / N. Next, we consider the case that F = 0 may have non-ordinary singularpoints. Let C ( ξ ) be a plane projective model of R . Suppose that C ( ξ ) is definedby F and for i = 1 , . . . , s , F i is the quadratic transformation of F i − underthe quadratic transformation Q c i − , where c i − is a singular point of F i − = 0.Denote n = deg( F ) and set ξ = ξ , ξ i +1 = Q − c i ( ξ i ) and N i = 2 i ( i − n i max { nT ( F ) , T ξ ( D ) } . (4) Proposition 3.27.
Let D be a divisor in R and the notations F i , ξ i , N i asabove. One has that
1. tdeg( F i ) ≤ n i − i +1 + 2 ; T ξ i ( D ) , T ( F i ) ≤ N i .Proof.
1. Set n i = deg( F i ). Since every c i is a singular point, one has that n i ≤ n i − −
2. This implies n i ≤ i n − i +1 + 2.2. Denote by S i the maximum of the heights of singular points of F i = 0. Wefirst prove by induction on i that T ( F i ) , S i ≤ N i for all i = 0 , . . . , s . Note that n i +1 < i n for all i = 1 , . . . , s . Since N = 8 nT ( F ), it is clear that T ( F ) < N and S < nT ( F ) < N by Corollary 3.5. Now assume that T ( F i ) , S i ≤ N i i = ℓ ≥
0. Consider the case i = ℓ + 1. By Corollary 3.13 and inductionhypothesis, one has that T ( F ℓ +1 ) ≤ T ( F ℓ ) + n ℓ S ℓ ≤ (1 + n ℓ ) N ℓ < ℓ nN ℓ = N ℓ +1 . Note that n = n > F = 0 has singularities. One sees that4 S ℓ ≤ N ℓ < ℓ nN ℓ = N ℓ +1 if ℓ > S < nT ( F ) < n T ( F ) ≤ N . Consequently, 4 S j < N j +1 for all j ≥
0. On the other hand, one has alreadyseen that T ( F ℓ ) + n ℓ S ℓ < N ℓ +1 . By Corollary 3.13 again, S ℓ +1 ≤ max { S ℓ , T ( F ℓ ) + n ℓ S ℓ } < N ℓ +1 . For the divisor D , it is obvious that T ξ ( D ) ≤ N . Suppose that T ξ i ( D ) ≤ N i for i = ℓ ≥
0. By Corollary 3.13 and the induction hypothesis, T ξ ℓ +1 ( D ) ≤ max { S ℓ + N ℓ ) , T ( F ℓ ) + n ℓ T ξ ℓ ( D ) } ≤ N ℓ +1 . Notation 3.28.
Let F be the defining polynomial of C ( ξ ) . Denote by s ( ξ ) thenumber of quadratic transformations such that C ( ˜ ξ ) has only ordinary singu-larities, where ˜ ξ is the image of ξ under these quadratic transformations. ByTheorem 2 in Chapter 7 of [9], s ( ξ ) can be chosen to be an integer not greaterthan m + ( n − n − − X r c ( r c − ≤ ( n − n − where n = deg( F ) , m is the number of non-ordinary singularities of F = 0 , c ranges over all singularities of F = 0 and r c is the multiplicity of c . Theorem 3.29.
Let D be a divisor in R . Denote n = deg( F ) , s = s ( ξ ) , µ = deg( D + + D − ) . Then there is a k ( t ) -basis B of L ( D ) such that every element of B can berepresented by G ( ξ ) /H ( ξ ) where G, H are two homogeneous polynomials of thesame degree not greater s +1 ( n + 1)( µ + 2 s − n ) and T ( G ) , T ( H ) ≤ s + s +5 n s +5 ( n + 1) ( µ + 2 s − n ) max { nT ( F ) , T ξ ( D ) } . Proof. If F = 0 has only ordinary singularities, i.e. s = 0, then the assertion isclear by Theorem 3.26. Suppose F = 0 has non-ordinary singularities, i.e. s ≥ ξ be the image of ξ under s quadratic transformations Q − c , . . . , Q − c s − suchthat C ( ˜ ξ ) has only ordinary singularities. Let ˜ F be the defining polynomial of C ( ˜ ξ ) and set κ = 2 s ( s − / n s max { nT ( F ) , T ξ ( D ) } . n = deg( ˜ F ) ≤ s ( n −
2) + 2 and T ˜ ξ ( D ) , T ( ˜ F ) ≤ κ. By Theorem 3.26, there is a k ( t )-basis B of L ( D ) satisfying that each element in B can be represented by ˜ G ( ˜ ξ ) / ˜ H ( ˜ ξ ) where ˜ G, ˜ H are homogeneous polynomialsof degree not greater than 2(˜ n + 1) µ + (˜ n − / T ( ˜ G ) , T ( ˜ H ) ≤ n (˜ n + 1) (2 µ + (˜ n − / κ. It remains to represent elements of B in terms of ξ . We use the same notations asin the proof of Propositoin 3.27. Let ξ = ξ and ξ i = Q − c i − ( ξ i − ). Denote by n i the degree of the defining polynomial of C ( ξ i ) and S i the maximum of the heightsof singular points of C ( ξ i ). Let G s = ˜ G and G i − = G i ( Q − c i − (( x , x , x ))) forall i = 1 , . . . , s. One sees that deg( G i ) = 2 s − i deg( ˜ G ). By Lemmas 3.6 and 3.10, T ( G i − ) ≤ T ( ¯ G i ) + deg( ¯ G i ) T ( c i − ) = T ( G i ) + 2 deg( G i ) T ( c i − )where ¯ G i = G i ( Q − (( x , x , x ))). From the proof of Proposition 3.27, we have T ( G ) ≤ T ( ˜ G ) + s − X i =0 G i +1 ) T ( c i ) ≤ T ( ˜ G ) + ( s − X i =0 s − i ) deg( ˜ G ) N s − ≤ T ( ˜ G ) + 2 s +1 deg( ˜ G ) N s − , where N s − is given as in (4). Note that ˜ n ≤ n s . One has thatdeg( G ) ≤ s deg( ˜ G ) ≤ s (2(˜ n + 1) µ + (˜ n − / ≤ s (˜ n + 1)(2 µ + (˜ n − / ≤ s +1 ( n + 1)( µ + n s − ); T ( G ) ≤ n (˜ n + 1) (cid:18) µ + ˜ n − (cid:19) κ + 2 s +1 (cid:18) n + 1) µ + (˜ n − (cid:19) N s − ≤ n (˜ n + 1) (cid:18) µ + ˜ n − (cid:19) κ + 2 s +1 (˜ n + 1) (cid:18) µ + ˜ n − (cid:19) κ ≤ (cid:18) n (cid:18) µ + ˜ n − (cid:19) + 2 s +1 (cid:19) (˜ n + 1) (cid:18) µ + ˜ n − (cid:19) κ ≤ (cid:18) n s (cid:18) µ + n s − (cid:19) + 2 s +1 (cid:19) (˜ n + 1) (cid:18) µ + ˜ n − (cid:19) κ ≤ n s ( µ + n s − )(˜ n + 1) (cid:18) µ + ˜ n − (cid:19) κ ≤ s +5 n ( n + 1) ( µ + n s − ) κ ≤ s + s +5 n s +5 ( n + 1) ( µ + 2 s − n ) max { nT ( F ) , T ξ ( D ) } . Similarly, we obtain bounds for deg( H ) and T ( H ).31 . Heights on plane algebraic curves Let f ∈ k [ x , x ] be an irreducible polynomial over k and a, b ∈ k ( t ) \ { } satisfy f ( a, b ) = 0, i.e. ( a, b ) is a rational parametrization of f = 0. The resulton parametrization (see [16] for instance) implies thatdeg( a ) = m deg( f, x ) , deg( b ) = m deg( f, x ) . In other words, T ( a ) deg( f, x ) = T ( b ) deg( f, x ). A similar relation holds forpoints in algebraic curves defined over k ( t ), i.e there is a constant C only de-pending on f such that if ( a, b ) is a point of f ( x , x ) = 0 with coordinates in k ( t ) then deg( f, x ) T ( a ) − C ≤ deg( f, x ) T ( b ) ≤ deg( f, x ) T ( a ) + C. This is a special case of a general result for points in complete nonsingularvarieties over a field with valuations. In the case of algebraic curves definedover k ( t ), Eremenko in 1999 presented another proof which actually providesa procedure to find C explicitly. In this section, we shall present an explicitformula for C following Eremenko’s proof. Throughout this subsection, f is an irreducible polynomial in k ( t )[ x , x ]and R is the algebraic function field over k ( t ) associated to f . Let us start witha refinement of Lemma 1 of [5]. Lemma 4.1.
Assume that f ∈ k ( t )[ x , x ] is irreducible over k ( t ) and α, β ∈R \ k ( t ) satisfying f ( α, β ) = 0 . If div( α ) − ≤ div( β ) − , then for every place P of R with ν P ( β ) ≥ , we have that T ( π P ( α )) ≤ T ( π P ( β )) + T ( f ) . Proof.
Since div( α ) − ≤ div( β ) − , by Proposition 2 of [5], f can be written to beof the form f = x n + a n − ( x ) x n − + · · · + a ( x ) x + a ( x ) , where a i ∈ k ( t )[ x ] with deg( a i ) ≤ n − i . Write a i = P n − ij =0 a i,j x j with a i,j ∈ k ( t ).Let R be a finite extension of k ( t ) containing all a i,j and π P ( α ) , π P ( β ). Supposethat p is a place of R . Then ν p ( π P ( α n )) = ν p − n − X i =0 n − i X j =0 a i,j π P ( β ) j π P ( α ) i ≥ min ≤ i ≤ n − , ≤ j ≤ n − i { ν p ( a i,j ) + jν p ( π P ( β )) + iν p ( π P ( α )) } = ν p ( a i ′ ,j ′ ) + j ′ ν p ( π P ( β )) + i ′ ν p ( π P ( α ))32or some 0 ≤ i ′ ≤ n − , ≤ j ′ ≤ n − i ′ . Equivalently, ν p ( π P ( α )) ≥ n − i ′ ν p ( a i ′ ,j ′ ) + j ′ n − i ′ ν p ( π P ( β )) . Thereforemax { , − ν p ( π P ( α )) } ≤ max (cid:26) , − ν p ( a i ′ ,j ′ ) n − i ′ − j ′ ν p ( π P ( β )) n − i ′ (cid:27) ≤ max { , − ν p ( a i ′ ,j ′ ) } + max { , − ν p ( π P ( β )) }≤ max i,j { , − ν p ( a i,j ) } + max { , − ν p ( π P ( β )) } . This implies that T ( π P ( α )) ≤ T ( π P ( β )) + T ( f ). Lemma 4.2.
Let S be a finite set of places in R and α ∈ R . Then there are a , a ∈ k with a = 0 such that supp div (cid:18) αa α + a (cid:19) − ! ∩ S = ∅ . Proof.
Set M = { π P ( α ) | ∀ P ∈ S with ν P ( α ) ≥ } . Then M is a finite set in k ( t ). Let a , a ∈ k satisfy that a = 0 and a c + a = 0for all c ∈ M . For P ∈ S with ν P ( α ) ≥
0, one has that π P ( a α + a ) = a π P ( α ) + a = 0 , i .e. ν P ( a α + a ) = 0 . This implies that ν P ( α/ ( a α + a )) = ν P ( α ) ≥
0. On the other hand, for P ∈ S with ν P ( α ) <
0, one has that ν P ( α/ ( a α + a )) = ν P ( α ) − ν P ( a α + a ) = ν P ( α ) − ν P ( α ) = 0 . In both cases, P is not a pole of α/ ( a α + a ). Thus a , a satisfy the require-ment.The main result of this section is the following theorem which is a specialcase of Lemma 2 of [5]. The original proof of Lemma 2 of [5] contains a smallgap. We shall fill in this gap in the proof. Theorem 4.3.
Let f be an irreducible polynomial in k ( t )[ x , x ] of degree n with respect to x and of degree n with respect to x . Suppose that n = tdeg( f ) and N ≥ . Then for every c , c ∈ k ( t ) satisfying f ( c , c ) = 0 , one has that (cid:18) − nN + n (cid:19) n T ( c ) − C ≤ n T ( c ) ≤ (cid:16) nN (cid:17) n T ( c ) + C where C = 2 s / s/ (2 N n + n + 2 s − ) n s +9 ( n + 1) T ( f ) /N (5) and s is the number of quadratic transformations which are applied to resolvethe singularities of f = 0 . roof. If one of c i is in k then the height of the other one is not greater than T ( f ). The inequalities then obviously hold. In the following, we assume thatneither c nor c is in k .Let R be the algebraic function field associated to f and α, β ∈ R \ k ( t )satisfy that f ( α, β ) = 0. Choose a , a ∈ k such that a = 0 andsupp(div( α/ ( a α + a )) − ) ∩ div( β ) − = ∅ . Such a , a exist due to Lemma 4.2. Set ¯ α = α/ ( a α + a ). Consider the divisor D = ( N + n ) n div( β ) − − N n div(¯ α ) − . Note that deg(div(¯ α ) − ) = n , deg(div( β ) − ) = n and n n ≥ n + n − ≥ n − . So deg( D ) = nn n ≥ n ( n − . This implies that deg( D ) is greater than the genus of f = 0 and thus L ( D ) = { } . Denote ξ = ( α, β,
1) and by F ( x , x , x ) the homogenization of f . Weclaim that T ξ ( D ) ≤ T ( f ). Note that T ξ ( D ) = max { T ξ (div(¯ α ) − ) , T ξ (div( β ) − ) } . For each a ∈ S ξ (div( β ) − ), a is of the form ( b , b ,
0) where b , b satisfies that F ( b , b ,
0) = 0. So T ( a ) ≤ T ( F ) = T ( f ) and then T ξ (div( β ) − ) ≤ T ( f ). If a = 0 then each point in S ξ (div(¯ α ) − ) is of the form ( b , b ,
0) too and so T ξ (div(¯ α ) − ) ≤ T ( f ). Otherwise, for each P ∈ div(¯ α ) − , one has that ν P ( α ) = ν P ( a ¯ α/ (1 − a ¯ α )) = ν P (¯ α ) − ν P (1 − a ¯ α ) = 0 . Moreover π P ( α ) = − a /a . This implies that each point of S ξ (div(¯ α ) − ) is of theform ( − a /a , b,
1) whose height is not greater than T ( f ). Thus T ξ (div(¯ α ) − ) ≤ T ( f ). Our claim is proved. Note thatdeg( D + + D − ) = 2 N n n + nn n ≤ (2 N + n ) n . Suppose that γ ∈ L ( D ) \ { } . Due to Theorem 3.29, γ = G ( ξ ) /H ( ξ ) where G, H are two homogeneous polynomials of degree not greater than2 s +1 ( n + 1) (cid:0) deg( D + + D − ) + 2 s − n (cid:1) ≤ s +1 n ( n + 1)(2 N n + n + 2 s − )and T ( G ) , T ( H ) ≤ s / s/ n s +6 ( n + 1) (cid:0) deg( D + + D − ) + 2 s − n (cid:1) T ( f ) ≤ s / s/ n s +9 ( n + 1) (cid:0) N n + n + 2 s − (cid:1) T ( f ) . Set ˜ C = 2 s / s/ n s +9 ( n + 1) (2 N n + n + 2 s − ) T ( f ) . Without loss of generality, we assume that G ( x , x ,
1) and H ( x , x ,
1) haveno common factor. Otherwise, by Corollary 2.15, we may replace G and H by34 /W and H/W where W is the greatest common factor of G and H . Moreover,multiplying by suitable elements in k ( t ) if necessary, we can assume that both G ( x , x ,
1) and H ( x , x ,
1) have 1 as a coefficient. Let P be a place of R containing α − c and β − c . Then ν P ( α ) = 0 and ν P ( β ) = 0. As γ ∈ L ( D ), ν P ( γ ) ≥
0. If ν P ( γ ) >
0, then ν P ( G ( α, β, > G ( c , c ,
1) = 0.Consequently, ( c , c ) is a common point of G ( x , x ,
1) = 0 and f ( x , x ) = 0.Proposition 3.4 implies that T ( c i ) ≤ deg( G ) T ( f ) + nT ( G ). It is easy to verifythat in this case T ( c ) , T ( c ) satisfy the required inequalities. Therefore we onlyneed to prove the case ν P ( γ ) = 0.Set h ( x , y ) = res x ( f ( x , x ) , H ( x , x , y − G ( x , x , h ( x , y ) = res x ( h ( x , y ) , x − x ( N + n ) n )where res x ( f, g ) denotes the resultant of f and g with respect to x . Note that h ( x , y ) = 0, because G ( x , x ,
1) and H ( x , x ,
1) have no common factor. As D is not effective, γ / ∈ k ( t ). Furthermore, as h ( x , γ ) = 0, deg( h , x ) >
0. It iseasy to see that h = 0 and h ( β ( N + n ) n , γ ) = 0. Let ˜ h be an irreducible factorof h in k ( t )[ x , y ] such that ˜ h ( β ( N + n ) n , γ ) = 0. Propositions 2.16 and 2.10imply that T (˜ h ) ≤ T ( h ) ≤ ( N + n ) n T ( h ) ≤ ( N + n ) n (deg( H ) T ( f ) + n ( T ( G ) + T ( H ))) ≤ ( N + n ) n (2 s +1 n ( n + 1)(2 N n + n + 2 s − ) T ( f )+ 2 n s / s/ n s +9 ( n + 1) (cid:0) N n + n + 2 s − (cid:1) T ( f )) ≤ ( N + n ) n s / s/ n s +9 ( n + 1) (cid:0) N n + n + 2 s − (cid:1) T ( f ) ≤ s / s/ n s +9 ( n + 1) (cid:0) N n + n + 2 s − (cid:1) T ( f ) = ˜ C. Remark thatdiv( β ( N + n ) n ) − = ( N + n ) n div( β ) − ≥ D + ≥ div( γ ) − . Note that π P ( β ) = c . By Lemma 4.1, T ( π P ( γ )) ≤ T (cid:16) π P ( β ) ( N + n ) n (cid:17) + T (˜ h ) (6) ≤ ( N + n ) n T ( c ) + ˜ C. Similarly, let r ( x , y ) = res x ( f ( x , x ) , G ( x , x , y − H ( x , x , r ( x , y ) = res x (cid:16) r ( x , y ) , ( a x + a ) Nn x − x Nn (cid:17) . Then r = 0 and r (¯ α Nn , γ − ) = 0. Let ˜ r be an irreducible factor of r in k ( t )[ x , y ] such that ˜ r (¯ α Nn , γ − ) = 0. Applying Propositions 2.16 and 2.10again yields that T (˜ r ) ≤ T ( r ) ≤ N n T ( r ) ≤ N n (deg( G ) T ( f ) + n ( T ( H ) + T ( G ))) .
35n argument similar to the above implies that T (˜ r ) ≤ ˜ C . Since supp(div(¯ α ) − ) ∩ supp(div( β ) − ) = ∅ , one has that N n δ (¯ α ) − = D − and thusdiv(¯ α Nn ) − = N n div(¯ α ) − = D − ≤ div( γ ) + = div( γ − ) − . Furthermore since a c + a = 0, π P (¯ α ) = c / ( a c + a ). By Lemma 4.1, N n T (cid:18) c a c + a (cid:19) = T ( π P (¯ α ) Nn ) ≤ T ( π P ( γ − )) + T (˜ r )= T ( π P ( γ )) + T (˜ r ) ≤ T ( π P ( γ )) + ˜ C. which together with (6) gives N n T (cid:18) c a c + a (cid:19) ≤ ( N + n ) n T ( c ) + 2 ˜ C. Proposition 2.8 implies that (cid:18) − nN + n (cid:19) n T (cid:18) c a c + a (cid:19) − CN + n ≤ (cid:18) − nN + n (cid:19) n T ( c ) − CN + n ≤ n T ( c ) . To prove the inequality in the opppsite direction, consider˜ D = ( N + n ) n div(¯ α ) − − N n div( β ) − . Remark that deg( D + + D − ) = deg( ˜ D + + ˜ D − ) and T ξ ( D ) = T ξ ( ˜ D ). We havethe same bounds for elements in L ( ˜ D ). A similar argument then implies that n T ( c ) ≤ N + nN n T ( c ) + 2 ˜ CN ≤ (cid:16) nN (cid:17) n T ( c ) + 2 ˜ CN .
Set C = 2 ˜ C/N . Then one gets the required inequalities.
5. Main results
In this section, we always ssume that f ( y, y ′ ) = P di =0 a i ( y ) y ′ is irreducibleover k ( t ) and ℓ = m . s . index( f ) = d max i =0 { deg( a i ) − d − i ) } > . Pick c ∈ k such that a ( c ) = 0. Set y = ( cz + 1) /z . Then y ′ = − z ′ /z . Set b i ( z ) = a i (( cz + 1) /z ) z ℓ +2 d − i ( − i where i = 0 , . . . , d . Then an easy calculation yields that g ( z, z ′ ) = d X i =0 b i ( z ) z ′ i = z d + ℓ f (cid:18) cz + 1 z , − z ′ z (cid:19) . a ( c ) = 0, deg( a ( cz + 1) /z ) z deg( a ) ) = deg( a ) . This implies thatdeg( b ) = ℓ + 2 d > d. Then tdeg( g ) = 2 d + ℓ because2 d + ℓ ≤ tdeg( g ) = max { deg( b i ) + i } ≤ max { d + ℓ − i } = 2 d + ℓ. We claim that g ( z, z ′ ) is irreducible over k ( t ). First of all, assume that ℓ =deg( a i ) − d − i ) for some 0 ≤ i ≤ d . Then we have that b i (0) = ( − i · the leading coefficient of a i ( y ) = 0 . If gcd( b , . . . , b d ) = 1 then the b i ( z ) have common zeroes and none of commonzeroes is zero. It is easy to see that ( cη +1) /η is a common zero of all a i ( y ) if η isa common zero of all b i . This contradicts with the fact that gcd( a , . . . , a d ) = 1.Secondly, if g ( z, z ′ ) has a factor with positive degree in z ′ then f ( y, y ′ ) willhave a factor with positive degree in y ′ , a contradiction. This proves our claim.Remark that r ( t ) is a nontrivial rational solution of g ( z, z ′ ) = 0 if and only if( cr ( t ) + 1) /r ( t ) is a nontrivial rational solution of f ( y, y ′ ) = 0. The main resultof this paper is the following theorem. Theorem 5.1.
Assume that f ( y, y ′ ) = 0 is a first order AODE with positive m . s . index and assume further that f ( y, y ′ ) is irreducible over k ( t ) . Then if r ( t ) is a rational solution of f ( y, y ′ ) = 0 then deg( r ( t )) ≤ (54 n + 9 n + 2 n ) n n +12 n +43 n +34 T ( f ) . where n = tdeg( f ) .Proof. We shall use the notations as above. Due to the above discussion, weonly need to consider the differential equation g ( z, z ′ ) = 0. Denote n = tdeg( f )and d = deg( f, y ′ ). One sees that T ( g ) ≤ T ( f ) , d = deg( g, z ′ ) anddeg( g, z ) = 2 d + ℓ = tdeg( g ) ≤ n. Suppose that g = h h . . . h m where h i is irreducible over k ( t ). Since g is irreducible over k ( t ), one has thatall h i are conjugate to each other and thendeg( h i , z ) = deg( g, z ) /m, deg( h i , z ′ ) = deg( g, z ′ ) /m = d/m. By Corollary 2.15, T ( h i ) ≤ T ( g ) ≤ T ( f ). Assume that r ( t ) is a rational solutionof g ( z, z ′ ) = 0 then r ( t ) is a rational solution of all h i = 0. In particular, h ( r ( t ) , r ′ ( t )) = 0. Denote ˜ n = tdeg( h ) and ˜ d = deg( h , z ′ ). Then˜ n = tdeg( g ) /m = (2 d + ℓ ) /m ≤ n/m ˜ d = deg( g, z ′ ) /m = d/m. N = ˜ n . By Theorem 4.3 and Remark 2.7, one has that NN + ˜ n deg( g, z )deg( g, z ′ ) deg( r ( t )) − C ˜ d ≤ deg( r ′ ( t ))where C = (2˜ n + ˜ n + 2 s − ) ˜ n s +7 (˜ n + 1) s + s +10 T ( f ) . Note that s is the number of quadratic transformations applied to transfer h = 0 to an algebraic curve with only ordinary singularities. Due to Theorem2 in Chapter 7 of [9], s can be chosen to be an integer not greater than(˜ n − n − / ≤ (3 n − n − / ≤ n / . Remark that deg( r ′ ( t )) ≤ r ( t )). Thus (cid:18) NN + ˜ n deg( g, z )deg( g, z ′ ) − (cid:19) deg( r ( t )) ≤ C ˜ d . (7)As m divides both deg( g, z ) and deg( g, z ′ ), m divides ℓ . Set ℓ = m ¯ ℓ . Then NN + ˜ n deg( g, z )deg( g, z ′ ) − n (2 d + ℓ )(˜ n + ˜ n ) d − nℓ − d (˜ n + 1) d ≥ ¯ ℓ deg( g, z ) − d (˜ n + 1) d ≥ n + 1) d . This together with (7) implies thatdeg( r ( t )) ≤ m (˜ n + 1) C ≤ n (2˜ n + ˜ n + 2 s − ) ˜ n s +7 (˜ n + 1) s + s +10 T ( f ) ≤ n (54 n + 9 n + 2 s − ) (4 n ) s +11 s + s +12 T ( f ) ≤ (54 n + 9 n + 2 s − ) n s +12 s + s +34 T ( f ) ≤ (54 n + 9 n + 2 n ) n n +12 n +43 n +34 T ( f ) . The second inequality holds because m (˜ n + 1) ≤ n + m ≤ n . Remark 5.2.
Theorem 5.1 implies that an autonomous first order AODE f = 0 with positive m . s . index has no nontrival rational solutions, because T ( f ) = 0 .In fact, suppose that f = 0 has a nontrival rational solution. Then it will haveinfinitely many rational solutions. By Corollary 4.6 of [7], f = 0 has no movablesingularities. However, as f = 0 has positive m . s . index , Fuchs’ criterion impliesthat f = 0 has movable singularities, a contradiction. In [21], the authors developed two algorithms to compute rational solutionsof maximally comparable first-order AODEs and first order quasi-linear AODEs38espectively. Let us first recall the definition of maximally comparable first orderAODEs. Suppose that f = P i,j a i,j y i y ′ j is a differential polynomial over k ( t ).Denote S ( f ) = { ( i, j ) ∈ N | a i,j = 0 } . If there is ( i , j ) ∈ S ( f ) satisfying that i + j ≥ i + j and i + 2 j > i + 2 j forevery ( i, j ) ∈ S ( f ), then we say that f is maximally comparable. The followingexamples shows that their algorithms can not deal with all first order AODEswith positive m . s . index. Example 5.3.
Let f = yy ′ m + y m +1 + t where m ≥ . Then S ( f ) = { (1 , m ) , (2 m +1 , , (0 , } . Since m +1+0 ≥ m but m + 1 + 2 · · m , f is not maximally comparable. On the otherhand, we have that m . s . index( f ) = 1 > . ReferencesReferences [1] J. M. Aroca, J. Cano, R. Feng, and X. S. Gao. Algebraic general solutions ofalgebraic ordinary differential equations. In
ISSAC’05 , pages 29–36. ACM,New York, 2005.[2] Moulay A. Barkatou. On rational solutions of systems of linear differen-tial equations. volume 28, pages 547–567. 1999. Differential algebra anddifferential equations.[3] L. X. Chˆau Ngˆo and Franz Winkler. Rational general solutions of first ordernon-autonomous parametrizable ODEs.
J. Symbolic Comput. , 45(12):1426–1441, 2010.[4] Claude Chevalley.
Introduction to the theory of algebraic functions of onevariable . Mathematical Surveys, No. VI. American Mathematical Society,Providence, R.I., 1963.[5] A. Eremenko. Rational solutions of first-order differential equations.
Ann.Acad. Sci. Fenn. Math. , 23(1):181–190, 1998.[6] Ruyong Feng and Xiao-Shan Gao. A polynomial time algorithm for findingrational general solutions of first order autonomous ODEs.
J. SymbolicComput. , 41(7):739–762, 2006.[7] Shuang Feng and Ruyong Feng. Descent of ordinary differential equationswith rational general solutions.
J. Syst. Sci. Complex. , 2020.[8] James Freitag and Rahim Moosa. Finiteness theorems on hypersurfaces inpartial differential-algebraic geometry.
Adv. Math. , 314:726–755, 2017.399] William Fulton.
Algebraic curves . Advanced Book Classics. Addison-Wesley Publishing Company, Advanced Book Program, Redwood City, CA,1989. An introduction to algebraic geometry, Notes written with the col-laboration of Richard Weiss, Reprint of 1969 original.[10] F. Hess. Computing Riemann-Roch spaces in algebraic function fields andrelated topics.
J. Symbolic Comput. , 33(4):425–445, 2002.[11] Ming-Deh Huang and Doug Ierardi. Efficient algorithms for the Riemann-Roch problem and for addition in the Jacobian of a curve.
J. SymbolicComput. , 18(6):519–539, 1994.[12] Jerald J. Kovacic. An algorithm for solving second order linear homoge-neous differential equations.
J. Symbolic Comput. , 2(1):3–43, 1986.[13] Serge Lang.
Fundamentals of Diophantine geometry . Springer-Verlag, NewYork, 1983.[14] Michihiko Matsuda.
First-order algebraic differential equations , volume804 of
Lecture Notes in Mathematics . Springer, Berlin, 1980. A differentialalgebraic approach.[15] Joseph Fels Ritt.
Differential algebra . Dover Publications, Inc., New York,1966.[16] J. Rafael Sendra and Franz Winkler. Tracing index of rational curveparametrizations.
Comput. Aided Geom. Design , 18(8):771–795, 2001.[17] Michael F. Singer. Liouvillian solutions of n th order homogeneous lineardifferential equations. Amer. J. Math. , 103(4):661–682, 1981.[18] Marius van der Put and Michael F. Singer.
Galois theory of linear dif-ferential equations , volume 328 of
Grundlehren der Mathematischen Wis-senschaften [Fundamental Principles of Mathematical Sciences] . Springer-Verlag, Berlin, 2003.[19] Mark van Hoeij, Jean-Fran¸cois Ragot, Felix Ulmer, and Jacques-ArthurWeil. Liouvillian solutions of linear differential equations of order threeand higher. volume 28, pages 589–609. 1999. Differential algebra anddifferential equations.[20] N. Thieu Vo, Georg Grasegger, and Franz Winkler. Deciding the existenceof rational general solutions for first-order algebraic ODEs.
J. SymbolicComput. , 87:127–139, 2018.[21] Thieu N. Vo, Georg Grasegger, and Franz Winkler. Computation of allrational solutions of first-order algebraic ODEs.
Adv. in Appl. Math. , 98:1–24, 2018.[22] Franz Winkler. The algebro-geometric method for solving algebraic differ-ential equations—a survey.