Computation of the degree of rational surface parametrizations

Transcription

1 Journal of Pure and Applied Algebra 193 (004) Computation of the degree of rational surface parametrizations Sonia Perez-Daz, J. Rafael Sendra Departamento de Matematicas, Facultad de Ciencias, Universidad de Alcala, Apartado de Correos 0, E-8871 Madrid, Spain Received 15 September 003; received in revised form 30 January 004 Communicated by C.A. Weibel Abstract A rational ane parametrization of an algebraic surface establishes a rational correspondence of the ane plane with the surface. We consider the problem of computing the degree of such a rational map. In general, determining the degree of a rational map can be achieved by means of elimination theoretic methods. For curves, it is shown that the degree can be computed by gcd computations. In this paper, we show that the degree of a rational map induced by a surface parametrization can be computed by means of gcd and univariate resultant computations. The basic idea is to express the elements of a generic bre as the nitely many intersection points of certain curves directly constructed from the parametrization, and dened over the algebraic closure of a eld of rational functions. c 003 Elsevier B.V. All rights reserved. MSC: 14Q10; 68W30; 14E05 0. Introduction Let P( t ) be a rational ane parametrization of a unirational surface V over an algebraically closed eld K of characteristic zero. Associated with the parametrization P(t ), we have the rational map P : K V ; t P(t ), where P (K ) V is dense. This work is partially supported by BMF C0-01 (Curvas y Supercies: Fundamentos, Algoritmos y Aplicaciones), Accion Integrada Hispano-Austriaca HU (Computer Aided Geometric Design by Symbolic-Numerical Methods), and GAIA II (IST ). Corresponding author. addresses: sonia.perez@uah.es (S. Perez-Daz), rafael.sendra@uah.es (J.R. Sendra) /$ - see front matter c 003 Elsevier B.V. All rights reserved. doi: /j.jpaa

2 100 S. Perez-Daz, J.R. Sendra / Journal of Pure and Applied Algebra 193 (004) P induces, over the elds of rational functions, the monomorphism P : K(V ) K(t ); R(x; y; z) R(P(t )). Then, the degree of the rational map P is dened as the degree of the nite eld extension P (K(V )) K( t ); that is deg( P )=[K(t ): P (K(V ))]: In fact, deg( P ) is the cardinality of a generic bre of P. The aim of this paper is to compute deg( P ). The notion of degree appears in some applied and computational problems, for instance in plotting since the degree measures how often the parametrization traces the image. Also, when computing the implicit equation of a surface by means of resultants, the degree appears as the power of the dening polynomial []. In addition, the birationality of P is characterized by deg( P )=1, and therefore it provides a method for testing the rationality of V or the properness of P(t ). Moreover, when the degree is 1, the expression of the ber of a generic element provides the inverse mapping of P (see [5]). In general, the problem of determining the degree of a rational map can be approached by means of elimination theoretic methods. For the case of curves, in [8] a method for computing the degree by means of gcds is presented. Moreover, it is also shown how deg( P ) is related to the degree of the rational parametrization; that is to the maximum degree of the rational components in the parametrization. This fact provides a nice characterization for the birationality in terms of the degrees of the rational components of P(t) and the partial degrees of the implicit equation of the curve (see [9]). This behavior does not hold for the case of surfaces (see [6]). Furthermore, in [7] the problem of simplifying the degree of a proper surface parametrization is analyzed. In addition a properness criteria for the surface case can be found in [5]. In this paper we prove that the degree of the induced rational map of a rational surface parametrization can be expressed as the degree of a polynomial that is directly computed from the parametrization by means of gcds and univariate resultants. Therefore, we extend the results presented in [8] to the case of unirational surfaces. Geometrically, the idea is to prove that the elements of a generic ber of P are essentially the nitely many intersection points of three plane algebraic curves dened over the algebraic closure of K(h 1 ;h ). These associated plane curves are directly generated from the parametrization. In fact the elements of the bre are those intersection points, of these three associated plane curves, not lying on the curve dened by the least common multiple of the denominators of P( t ). Then, preparing appropriately the input parametrization, by means of a linear change, one ensures that no pair of intersection points of the associated curves is on the same vertical line. From this fact, one deduces that the cardinality of the ber, i.e. the degree of P, is the degree of a polynomial which roots correspond to the rst coordinate of the ber element, and that is computed as the content of a univariate resultant. These ideas are also applied to the explicit computation of the elements of the bre. In addition, from these results we derive two algorithms for computing the degree of P, that have been implemented in Maple and which running times are very satisfactory. The structure of the paper is the following. In Section 1, we give the terminology, we present the general assumptions showing that they can be assumed without loss of generality, and we state some preliminary results. Section focuses on the problem of computing the degree of the induced map. For this purpose several subsections are

3 S. Perez-Daz, J.R. Sendra / Journal of Pure and Applied Algebra 193 (004) considered. First the computation of the elements in the ber is analyzed. Secondly we show how to compute the cardinality of the ber when working within a suitable non-empty Zariski open subset of K. Finally, in the last subsection, we see how these results can be generalized to compute the degree. In Section 3 we present the algorithms, and in Section 4 a brief experimental analysis of the algorithms is given. We also introduce an appendix with the data information used in Section Preliminaries and terminology In this section, we introduce the notation and terminology that will be use throughout this paper, as well as some preliminaries results Notation Let K be an algebraically closed eld of characteristic zero, let t =(t 1 ;t ), and let ( p1 (t ) P(t )= q 1 (t ) ; p (t ) q (t ) ; p ) 3(t ) K(t ) 3 q 3 (t ) be a rational parametrization of a surface V, where gcd(p i ;q i )=1, i {1; ; 3}. Associated with the parametrization P( t ), we have the rational map P : K V t P(t ); where P (K ) V is dense. We denote by deg( P ) the degree of P. Moreover, we introduce the following polynomials: G i (t; h)=p i (t )q i ( h) p i ( h)q i (t ) K[ h][t]; i {1; ; 3}; where h=(h 1 ;h ), and G 4 (t )=lcm(q 1 (t );q (t );q 3 (t )) K[t]: Note that the polynomials G 1 ;G ;G 3 are in K[ h][t], while G 4 belongs to K[t]. The reason for this construction is that G 1 ;G ;G 3 will control the intersection points that give information on the degree, while G 4 will take care of the zeros of the denominators of the parametrization. Also, for i =1; ; 3 we denote by i (t 1 ; h) = lc(g i (t; h);t ) the leading coecient of G i (t; h) w.r.t t. In addition, for every K, and i =1; ; 3, we denote by Gi (t ) the polynomial G i (t; ), and by Vi the algebraic set dened over K by Gi (t ). Also, we denote by V 4 the algebraic set dened by G 4 (t ). Note that V 4 is empty if and only if P(t ) is a polynomial parametrization, and otherwise it is a plane curve. In order to analyze deg( P ) we will study the cardinality of a generic bre. For this purpose, for every K such that P() is dened, we denote by F P () the bre of via P ; i.e F P ()={t K P(t )=P()}: Throughout the paper, we will work with a non-empty open subset of K associated to the parametrization P( t ), where certain properties are satised. This open subset will be denoted by P, and it is dened as follows. First, we consider a non-empty open

4 10 S. Perez-Daz, J.R. Sendra / Journal of Pure and Applied Algebra 193 (004) subset of V such that for all the bre F P () is zero dimensional. Note that always exists (see, e.g [10, p. 76]). Now, included in the constructible set P 1 () of K, we take a non-empty open subset that we denote by. In this situation, P is dened as P =(K \ V 4 ) : Observe that for every P the cardinality of F P() is equal to deg( P ). Finally, if A is a subset of any ane algebraic set we will denote by A its Zariski closure. 1.. General assumptions Since P(t ) is a surface parametrization, at least two of the component gradients are not parallel. Let us assume w.l.o.g that { (p 1 =q 1 ); (p =q )} are linearly independent as vectors in K(t 1 ;t ) 3. This in particular implies that p 1 =q 1, p =q are both not constant, and ensures that V1 and V are plane curves, while V3 might be either a plane curve or K, if the third parametrization component is constant. Also, we assume that none of the projective curves dened by each of the numerators and denominators of the parametrization components passes through the point at innity (0:1:0), where the homogeneous variables are (t 1 ;t ;w). Note that the degree of a rational map is multiplicative under composition, therefore the composition of P and linear transformation will preserve the degree of P. Thus, we also can assume w.l.o.g. the previous condition. Also, note that under these conditions, if Gi H (t 1 ;t ;w;h 1 ;h ) denotes the homogenization of G i (t 1 ;t ;h 1 ;h ) as a polynomial in K[ h][t], it holds that Gi H (0; 1; 0; h) 0 for i =1;. This in particular implies that, for i =1;, the leading coecients i (t 1 ; h) only depends on h. It also holds that 3 (t 1 ; h) only depends on h if V3 is a curve, i.e. if the third parametrization component is not constant; if it is not a curve then G 3 (t 1 ;t ;h 1 ;h ) is identically zero and thus 3 (t 1 ; h) = 0. Therefore, in the following we simplify the notation and we write i ( h), by taking into account that 3 could be identically zero. Finally note, that in these conditions we have that gcd ( i ( h); j ( h))=1; for i; j {1; ; 3} with i j: K( h)[t 1] 1.3. Preliminaries results In the following we show that for every in a non-empty open subset of K, there exists a one to one relationship between the intersection points of the three varieties V1 ;V ;V3, and the roots of a univariate polynomial computed by a resultant. We start with the following lemmas. Lemma 1. For every P, the varieties V1, V, V 3 do not have common components. Proof. Let us assume that there exists f P such that V1, V, V 3 have common components, and let M K[t] be the dening polynomial of all the common components; note that V1, V are plane curves and V3 is either a plane curve or the whole

5 S. Perez-Daz, J.R. Sendra / Journal of Pure and Applied Algebra 193 (004) plane. Then, there exist N i K[t] such that q i (t )p i () p i (t )q i ()=M(t )N i (t ); for i =1; ; 3: Observe that gcd(q i (t );M(t )) = 1 for i =1;, since otherwise it would imply that gcd(q i ;p i ) 1. It also holds that gcd(q 3 (t );M(t )) 1; indeed: if V3 is a curve, i.e. if the third parametrization component is not constant, the same reasoning works; if it is not a curve then q 3 is a non-zero constant and the results trivially holds. We consider the set = { K =M( )=0; q i ( ) 0 for i =1; ; 3}: is an open subset of the curve dened by M(t ), and since it holds that gcd(q i (t ); M(t )) = 1 one has that is not empty. Moreover one has that F P () which is impossible since Card( )= and F P () is zero dimensional because P. On the other hand, since the leading coecients i only depend on h the next lemma follows trivially. Lemma. Let P be the non-empty open subset of K dened as P = P { K i () 0 for i =1; }: For all P the polynomials G1 ( t ), and G ( t )+ZG3 ( t ) K[Z; t] do not have factors depending only on the variable t 1. Lemma 3. Let L be a subeld of K, and let C 1, C, C 3 be plane algebraic curves over F, with no common components, dened by the polynomials F 1, F, F 3 L[t 1 ;t ], respectively. Let F 1 ;F ;F 3 be such that each two of their leading coecients, w.r.t. one the variables, have trivial gcd. Let F 1 do not have a factor in L[t 1 ]. The t 1 -coordinates of the intersection points of C 1, C, C 3 are the roots of the content w.r.t Z of the resultant w.r.t. t of the polynomials F 1 ;F + ZF 3. Proof. See Proposition 1, in [5]. Lemma 3 can be easily extended to more than three curves. Applying these results one has the following theorem. Theorem 1. For all P it holds that: 1. The t 1 -coordinates of the intersection points of V1, V, V 3 are the roots of the polynomial S1 (t 1) = Content Z (Res t (G1 ( t ); G ( t )+ZG3 ( t ))):. The t 1 -coordinates of the intersection points of V1, V, V 3, V 4 are the roots of the polynomial T (t 1 ) = Content {Z;W } (Res t (G1 ( t ); G ( t )+ZG3 ( t )+WG 4 (t ))): 3. The t 1 -coordinates of the points in (V1 V V 3 ) \ (V1 V V 3 V 4) are the roots of the polynomial S (t 1 )= S 1 (t 1) T (t 1 ) :

6 104 S. Perez-Daz, J.R. Sendra / Journal of Pure and Applied Algebra 193 (004) A similar reasoning can be done taking a generic element h of K. For this purpose, h we denote by F the algebraic closure of the eld K( h), and by V i the algebraic set dened over F by the polynomials G i (t; h) F[t], i =1; ; 3. By Lemmas 1 and, we h deduce that V h 1, V h, V 3 do not have common components, and that the polynomials G 1 (t; h), and G Z = G (t; h)+zg 3 (t; h) do not have factors in F[t 1 ]. Therefore, one gets the next result. h Theorem. (1) The t 1 -coordinates of the intersection points of V h 1, V h, V 3 are the roots of the polynomial S 1 (t 1 ; h) = Content Z (Res t (G 1 (t; h); G (t; h)+zg 3 (t; h))): h () The t 1 -coordinates of the intersection points of V h 1, V h, V 3, V 4 are the roots of the polynomial T (t 1 ; h) = Content {Z;W } (Res t (G 1 (t; h); G (t; h)+zg 3 (t; h)+wg 4 (t ))): h (3) The t 1 -coordinates of the points in (V h 1 V h V h 3 ) \ (V h 1 V h V 3 V 4 ) are the roots of the polynomial S(t 1 ; h)= S 1(t 1 ; h) T (t 1 ; h) : These theorems motivate the following denition. Denition 1. The polynomial S (t 1 ) in Theorem 1 is called the t 1 -coordinate polynomial associated to the pair (P(t ); ), and the polynomial S(t 1 ; h) in Theorem is called the t 1 -coordinate polynomial associated to P(t ). The polynomial S(t 1 ; h) has been introduced in Theorem as a factor of the polynomial S 1 (t 1 ; h), and it has been expressed as a quotient. Nevertheless, at the end of the next section, we will see that one may replace this quotient computation by crossing out the factors of S 1 (t 1 ; h) ink[t 1 ].. Degree of the induced rational map In this section we show how to compute the degree of the rational map induced by the surface parametrization. This problem may be approached by means of elimination techniques as Grobner basis. However, we see how this can be done by means of univariate resultants and gcds. For this purpose, the section is structured as follows. First, we study the problem of actually determining the elements in the bre. Secondly, we adapt these results to compute the cardinality of the bre without computing explicitly its elements. Finally, we show that the degree of the t 1 -coordinate polynomial associated to the pair (P(t ); ), is preserved under almost all specializations of the variables h 1, h. Then, the degree of the induced rational map is proved to be equal to the degree in t 1 of the t 1 -coordinate polynomial associated to P(t )..1. Computation of the bre We rst observe that for every P (see Lemma ) the bre F P () can be expressed as F P ()={t K G1(t )=G(t )=G3(t )=0 and G 4 (t ) 0};

7 S. Perez-Daz, J.R. Sendra / Journal of Pure and Applied Algebra 193 (004) that is, F P ()=(V1 V V3 ) \ (V1 V V3 V 4 ): Therefore, from Theorem 1, one deduces that Theorem 3. The t 1 -coordinates of the elements in F P () vare the roots of the t 1 -coordinate polynomial associated to the pair (P(t ); ). A similar reasoning to the one we have done in the previous section, for introducing the t 1 -coordinate polynomial associated to the pair (P(t ); ), might be also considered in order to compute the t -coordinates of the elements in the bre. Afterwards combining those t 1 and t -coordinates one can determine the elements in F P (). Nevertheless, in order to avoid the above combinatorial checking one can proceed in the dierent way. Let P and let S (t 1 )bethet 1 -coordinate polynomial associated to (P(t ); ). Then, for every root, a, ofs (t 1 ), i.e. for the t 1 -coordinates of the elements in the bre, we consider the polynomial Ma gcd K[t] (t )= (G 1 (a; t );G (a; t );G3 (a; t )) gcd (G K[t] 1 (a; t );G (a; t );G3 (a; t );G 4 (a; t )) K[t ]: In these conditions, it is clear that the following theorem holds: Theorem 4. If P then F P ()={(a; b) K S (a)=0;m a (b)=0}: Example 1. Let V the surface parametrized by P(t 1 ;t ), dened by ( t P(t 1 ;t )= 1 1+t t + t ;t t 1 4 3t1 +t1t +1 t + t 4 + t ;t + t1 4 t1 1 ) +t1t + t 4 : We consider P, for instance =(3; ). We apply Theorem 4 to compute the elements of F P (). For this purpose, rst of all, for i =1; ; 3; we determine the polynomials Gi (t ) dening the algebraic curves Vi : G1(t )= 3t1 1+5t +4t ; G (t )=t t 1 +t 1t 140 t + t 4 + t ; G3(t )=t + t1 4 t1 +t1t + t 4 153; as well as the polynomial G 4 (t )=lcm(q 1 (t ); q (t ); q 3 (t ))=t +t t 1 ; that denes the curve V 4. Now, we compute the polynomial S1 (t 1) dening the t 1 -coordinates of the intersection points of V1, V, V 3, and the polynomial T (t 1 ) dening the t 1 -coordinates of the intersection points of V1, V, V 3, V 4 (see Theorem 1). We get S1 (t 1 ) = Content Z (Res t (G1(t ); G(t )+ZG3(t )))=(t 1 3)(t 1 + 3)(4t1 7); T (t 1 ) = Content {Z;W } (Res t (G 1(t ); G (t )+ZG 3(t )+WG 4 (t )))=1:

8 106 S. Perez-Daz, J.R. Sendra / Journal of Pure and Applied Algebra 193 (004) Therefore, the t 1 -coordinate polynomial associated to the pair (P(t ); ) is given by S (t 1 )= S 1 (t 1) T (t 1 ) =(t 1 3)(t 1 + 3)(4t 1 7): The roots of the polynomial S, i.e. the t 1 -coordinates of the elements in the bre, are a 1 = 3, a =3, a 3 = 3 3, a4 = 3 3. Now, for every root ai we compute the polynomial M a i (t )= gcd K[t] (G 1 (a i;t );G (a i;t );G 3 (a i;t )) gcd K[t] (G 1 (a i;t );G (a i;t );G 3 (a i;t );G 4 (a i ;t )) K[t ]; which roots provide the t -coordinates of the elements in the bre. Hence, one gets that { F P ()= (3; ); ( 3; ); which implies that deg( P )=4:.. Cardinality of the bre ( 3 ) ( 5 3; ; 3 )} 5 3; ; In this subsection we deal with the problem of computing the cardinality of the bre, without determining explicitly its elements. For this purpose, we characterize the cardinality by means of the degree of the t 1 -coordinate polynomial associated to the pair (P(t ); ), where is in a non-empty open subset of K. From this result we will derive a probabilistic method to compute deg( P ). Lemma 4. Let q K[t 1 ;t ], q not identically zero, and q ={ a P q(a)=0} K. Then it holds that 1. q = (V \ P( q)) is a non-empty open subset of V.. There exists a non-empty open subset q of P 1 ( q ), such that for every q, it holds that F P () q =. Proof. (1) Clearly q is open. Furthermore, if q =, one has that V decomposes as a union of two closed sets, namely V =(V \ ) P( q): Moreover, since V, one has that V \ V. Then, P( q). Therefore, since V is irreducible one has that P( q)=v, which is impossible because dim(p( q)) 6 1 (see [4, Theorem 11.1; p. 139]). () Since P 1 ( q ) is a constructible set (see [4]), there exists a non-empty open subset of K in P 1 ( q ). We denote it by q. Now, we take q, and we prove that F P () q =. First of all note that since q P 1 ( q ), one has that P() q ; and hence P() P( q): Now, let us assume that there exists F P () q.

9 S. Perez-Daz, J.R. Sendra / Journal of Pure and Applied Algebra 193 (004) Then, P() =P( ); with contradiction. q: Therefore, P() P( q) P( q), which is a In the following, we will work with a non-empty open subset of K associated to the parametrization P( t ), where certain properties are satised. This open subset will be denote by P 1, and it is dened as follows. First, we consider the Jacobian matrix: J P (t )= ( p1 q 1 ( p q ( p3 q 3 ) (t ) ) (t ) ) (t ) and we denote by M(t ) the principal minor of J P. Since we have assumed that { (p 1 =q 1 ); (p =q )} are linearly independent as vectors in K(t ) 3, one has that M is not identically zero. Thus, applying Lemma 4 to M = { a P M(a) =0}; there exists a non-empty open subset M P 1 ( M ), where M = (V \ P( M )); such that for every M, it holds that F P () M =. In these conditions, 1 P is dened as 1 P = P M : Observe that 1 P is a non-empty open subset of K. Lemma 5. For every P 1, there exists a non-empty open subset 1 K such that for every Z 0 1 it holds that every point T F P() is a simple point of transversal intersection of the two plane curves with equation G1, and G + Z 0G3. Proof. Let 1 P, and let us consider T F P(). Observe that in particular P, and Card(F P ()) = deg( P ). Now, we note that the following two statements hold: 1. Since P(T )=P(), it holds that the kth row of J P (T) is(1=q k (T)q k ()) G k (T). Indeed: The kth row of J P (T ), is given by ( pk q k ) ( (@pk =@t 1 )(T )q k (T) (@q k =@t 1 )(T)p k (T) (T )= qk (T) ; ) (@p k =@t )(T )q k (T) (@q k =@t )(T)p k (T) qk (T) = 1 (T k (T) p k(t) q k (T 1 q k (T) k k (T) p ) q k (T)

10 108 S. Perez-Daz, J.R. Sendra / Journal of Pure and Applied Algebra 193 (004) Therefore, since (p k =q k )(T )=(p k =q k )( ), one deduces that ( ) pk 1 (T )= q k q k (T )q k () (T )q k k (T)p k k (T)q k ) k (T)p k @t 1 = q k (T )q k () G k (T):. Since M, and T F P (), by Lemma 4 one has that T M ; that is M(T) 0. Thus, rank(j P (T ))=. Taking into account statements (1) and (), one deduces that for i =1; the gradient of Gi is not zero at T. Therefore, the gradient of G + Z 0G3 at T is not zero for every Z 0 in the non-empty open subset of K dened as A 1 = {Z K G(T )+Z G3(T ) 0}: Thus for every Z 0 A 1, T is a simple point of the two plane curves with equation G1 = 0 and G + Z 0G3 =0. Now let us prove that there exists an open subset of K containing the Z 0 values for which, G1, and G Z 0 intersect transversally at the points T in the bre. First of all, let us see that V1, V, V 3 do not have a common tangent at T. Note that, because of the previous reasoning, T is a simple intersection point of the three curves. Now, let us assume that the three curves do not intersect transversally at T. This implies that for every i; j {1; ; 3}, with i j, Gi (T ) and Gj (T) are parallel. Then taking into account statement (1), one has that rank(j P (T)), which is impossible because of statement (). Now, we consider the subset of K dened as A = {Z K G(T )+Z G3(T ) is not parallel to G1(T)}: We prove that K \ A is a closed subset of K. One has to nd those values of Z for which there exists Z K such that Z G1(T ) Z G3(T )= G(T ): This last equality generates a linear system in the variables { Z ;Z} over K(t 1 ;t ). But since rank(j P ) =, one has that this linear system cannot have innitely many solutions. Therefore A is non-empty and open subset of K, and for every Z 0 A, the curves G1, and G Z 0 intersect transversally at T. Thus, A 1 ensures that the points in the bre are simple intersection points, and A guarantees that these intersections are transversal. Therefore, 1 = A 1 A K. Lemma 6. For every P, there exists a non-empty open subset K such that for every Z 0 it holds that (0:1:0) is neither on the projective curve dened by G1 nor on the projective curve dened by G + Z 0G3. Proof. We have assumed that for i =1; ; it holds that G H i (0; 1; 0) 0. Thus, in particular, for every P, we have that i () 0, and then we deduce that (0:1:0)

11 S. Perez-Daz, J.R. Sendra / Journal of Pure and Applied Algebra 193 (004) H; H; G1 G. Therefore, one may take 0}. H; = {Z K GH; (0; 1; 0) + ZG3 (0; 1; 0) Theorem 5. For every 1 P, it holds that Card(F P()) = deg t1 (S (t 1 )): Proof. First of all note that since F P ()=(V 1 V V 3 ) \ (V 1 V V 3 V 4 ); one has that Card(F P ()) = Card((V 1 V V 3 ) \ (V 1 V V 3 V 4 )): This cardinal is invariant under any linear change of variables done on P(t ) preserving the general assumptions introduced in Section 1. Thus, in particular, for every 1 P we may consider a linear change of variables (see [3]) such that the point at innity (0:1:0) is not on any line connecting two points on F P (). Taking into account Theorem 1 and the previous lemmas, if one shows that for every 1 P, the t 1-coordinate polynomial, S (t 1 ), associated to the pair (P(t ); ) is squarefree, one would deduce that Card(F P ()) = deg t1 (S (t 1 )): Therefore, it only remains to prove that S (t 1 ) is squarefree. Indeed, let us assume that S is not squarefree. This implies that S (t 1 ) can be written as S (t 1 )=(t 1 a) r S (t 1 ) with r 1; where a denotes the t 1 -coordinate of a point T F P (). Furthermore, since S divides R(t 1 ;Z) = Res t (G1 ;G + ZG 3 ) one has that R(t 1 ;Z)=(t 1 a) r S (t 1 ) R(t 1 ;Z) with R(t 1 ;Z) K[t 1 ;Z]. Now, let 1 and be the non-empty open subsets considered in Lemma 5 and Lemma 6, respectively. We also consider the non-empty open subset of K dened by 3 = {Z K R(t 1 ;Z) 0}: Lemma 1 guarantees that R is not identically zero because the varieties V1, V, V 3 do not have common components. Let Z By well-known properties of the resultants concerning to the multiplicity of intersection (see [1]), it holds that those factors of R 0 (t 1 ) = Res t (G 1;G + Z 0 G 3); dening the t 1 -coordinates of the points T of F P () are simple. Now, let Z0 the natural evaluation homomorphism of K[t 1 ;Z] into K[t 1 ]; that is, Z0 : K[t 1 ;Z] K[t 1 ] f(t 1 ;Z) f(t 1 ;Z 0 ): denote

12 110 S. Perez-Daz, J.R. Sendra / Journal of Pure and Applied Algebra 193 (004) Then, taking into account the behavior of the resultant under an homomorphism (see e.g Lemma 4.3.1, p. 96 in [11]), one deduces that Z0 (R(t 1 ;Z))=(t 1 a) r S (t 1 ) R(t 1 ;Z 0 )= k 1Res t (G 1;G + Z 0 G 3)= k 1R 0 (t 1 ) with 1 K \{0}, and k Z. Therefore, since the factors of R 0 dening the t 1 -coordinates of the points T of F P () are simple, one concludes that r =1. Theorem 5 provides a method to compute the degree. In the following we illustrate this approach by some examples. Example. Let V the surface parametrized by ( t P(t 1 ;t )= t1 + t t ; ( t 1 + t t ) 3 t ; t 1 + t t 4 Let us compute deg( P ). For this purpose, one possibility is to consider 1 P, and to compute the elements of the F P () by applying Theorem 4 (see Example 1). The second possibility is to apply Theorem 5. For this purpose, we consider 1 P, for instance =(3; ), and we determine the polynomials G i (t ) for i =1; ; 3, and G 4 (t ), G 1(t )= 11t +4t 1 +4t ; ) : G (t )=4t 6 1 1t 4 1t +1t 4 1t +1t 1t 4 4t 1t 3 +1t 1t 4t 6 +1t 5 1t 4 +4t 3 343t ; G 3(t )=16t 1 +16t 11t 4 ; G 4 (t ) = lcm(q 1 (t );q (t );q 3 (t )) = t 4 ( t 1 + t t ): Now, we compute the polynomials S 1 (t 1) and T (t 1 ) (see Theorem 1). We get S 1 (t 1 ) = Content Z (Res t (G 1(t );G (t )+ZG 3(t ))) = t 4 1(t 1 3)(t 1 + 3)(t 1 13); T (t 1 ) = Content {Z;W } (Res t (G 1(t );G (t )+ZG 3(t )+WG 4 (t ))) = t 4 1: Therefore, the t 1 -coordinate polynomial associated to the pair (P(t ); ) is given by S (t 1 )= S 1 (t 1) T (t 1 ) =(t 1 3)(t 1 + 3)(t1 13) and thus, by Theorem 5 one deduces that deg( P ) = deg t1 (S (t 1 ))=4. Example 3. Let V the surface parametrized by ( t 4 P(t 1 ;t )= (t 1 1)3 (t1 1); t 4 ; t4 + t 1 1 ) t 8 :

13 S. Perez-Daz, J.R. Sendra / Journal of Pure and Applied Algebra 193 (004) Let us apply Theorem 5 in order to determine deg( P ) by avoiding the requirement of computing the explicit values of the bre. For this purpose, we consider 1 P, for instance =(; 1), and we determine the polynomials G i (t ) for i =1; ; 3, and G 4 (t ): G 1(t )=3t 4 +1 t 1; G (t )=t 6 1 3t t 1 1 7t 4 ; G 3(t )=t 4 + t 1 1 4t 8 ; G 4 (t )=(t 1 1)t 4 : Now, we compute the polynomials S 1 (t 1) and T (t 1 ) (see Theorem 1). We get S 1 (t 1 )=(t 1 1) 4 (t 1 +1) 4 (t 1 ) 4 (t 1 +) 4 ; T (t 1 )=(t 1 1) 4 (t 1 +1) 4 : Therefore, the t 1 -coordinate polynomial associated to the pair (P(t ); ) is given by S (t 1 )= S 1 (t 1) T (t 1 ) =(t 1 ) 4 (t 1 +) 4 and thus, by Theorem 5, one deduces that deg( P ) = deg t1 (S (t 1 ))=8:.3. Preservation of the degree In the previous subsections we have seen that taking a generic, i.e. an element in the open subset P 1, one may compute the degree. In fact, taking into account how the set P 1 is constructed one might choose deterministically. In this subsection, we show how this diculty can be avoided. For this purpose, we rst recall some results on gcds and resultants. More precisely, for A =(a; b) K, we consider the natural evaluation homomorphism: A : K[h 1 ;h ;y 1 ;:::;y n ]; K[y 1 ;:::;y n ] f(h 1 ;h ;y 1 ;:::;y n ) f(a; b; y 1 ;:::;y n ): Then the following two lemmas on gcds and resultants hold (see Lemma 3 in [9], and Lemma 4.3.1, p. 96 in [11], respectively). Lemma 7. Let f; g K[h 1 ;h ][t 1 ], f = f gcd(f; g), g =g gcd(f; g). Let A K be such that not both leading coecients of f and g w.r.t. t 1 vanish at A. Then 1. deg t1 (gcd( A (f); A (g))) deg t1 ( A (gcd(f; g))) = deg t1 (gcd(f; g)).. If Res t1 ( f; g) does not vanish at A, then A (gcd(f; g)) = gcd( A (f); A (g)): Lemma 8. Let f; g K[h 1 ;h ][t 1 ], let A K be such that deg t1 ( A (f)) = deg t1 (f), and deg t1 ( A (g)) = deg t1 (g) k. Then, A (Res t1 (f; g)) = A (lc(f; t 1 )) k Res t1 ( A (f); A (g)); where lc(f; t 1 ) denotes the leading coecient of f w.r.t t 1.

14 11 S. Perez-Daz, J.R. Sendra / Journal of Pure and Applied Algebra 193 (004) Using Lemma 8 one can easily generalize Lemma 7 for more than two polynomials as follows: Lemma 9. Let f i K[h 1 ;h ][t 1 ], f i = f i gcd(f 1 ;:::;f m ), i =1;:::;m. Let A K be such that the leading coecient of f 1 w.r.t. t 1 does not vanish at A. 1. deg t1 (gcd( A (f 1 );:::; A (f m ))) deg t1 ( A (gcd(f 1 ;:::;f m ))) = deg t1 (gcd(f 1 ;:::; f m )):. If Res t1 ( f 1 ; f + m i=3 X i f i )(A) 0,where X j, j=1;:::;m ; are new variables, then A (gcd(f 1 ;:::;f m )) = gcd( A (f 1 );:::; A (f m )): Proof. The proof of (1) is direct. Let us see (). Since A is a homomorphism, one has that A (f i )= A ( f i ) A (gcd(f 1 ;:::;f m )): Thus, gcd( A (f 1 );:::; A (f m )) = A (gcd(f 1 ;:::;f m )) gcd( A ( f 1 );:::; A ( f m )): By hypotheses, Res t1 ( f 1 ; f + m i=3 X i f i )(A) 0. Since the leading coecient of f 1 w.r.t. t 1 does not vanish at A one has that the leading coecient of f 1 w.r.t. t 1 does not vanish either at A. Thus, applying Lemma 8, one deduces that Res t1 ( A ( f 1 ); A ( f )+ m i=3 X i A ( f i )) is not identically zero. Therefore one concludes that gcd ( A ( f 1 );:::; A ( f m ))=1; and then gcd( A (f 1 );:::; A (f m ))= A (gcd(f 1 ;:::;f m )): In the following, we introduce a new non-empty open subset of K associated to the parametrization P(t ), that will be denoted by P. First, since (see Theorem ) S 1 (t 1 ; h) = Content Z (Res t (G 1 (t; h);g (t; h)+zg 3 (t; h))); we may write and Res t (G 1 (t; h);g (t; h)+zg 3 (t; h)) = a 0 (t 1 ; h)+a 1 (t 1 ; h)z + + a m (t 1 ; h)z m a i (t 1 ; h)=s 1 (t 1 ; h)a i (t 1 ; h); for i =0;:::;m; with gcd( a 0 ;:::; a m )=1; where a i K[t 1 ; h]. Moreover, applying Lemma 1 and Lemma 9 (1) we have that the Res t (G 1 ;G (t; h)+zg 3 (t; h)) is not identically zero, and therefore one of the coecients a i is not identically zero. Let us assume w.l.o.g that a 0 (t 1 ; h) is not identically zero. Now we consider the subset of K dened as 1 = lc(a 0 ;t 1 )( ) 0 K Res t1 (a 0 (t 1 ; h); a 1 (t 1 ; h)+ m X i 1 a i (t 1 ; h))( ) 0 where lc(a 0 ;t 1 ) denotes the leading coecient of a 0 (t 1 ; h) w.r.t. t 1. Observe that 1 is a non-empty open subset of K, since lc(a 0 ;t 1 ), Res t1 (a 0 ; a 1 + m i= X i 1 a i ), are non-identically zero polynomials. Reasoning similarly for the polynomial T(t 1 ; h) i= ;

15 S. Perez-Daz, J.R. Sendra / Journal of Pure and Applied Algebra 193 (004) (see Theorem ), we get another non-empty open subset, K. In this conditions, P is dened as P = 1 P 1 : Observe that P is a non-empty open subset of K. Theorem 6. For every value P it holds that deg t1 (S(t 1 ; h)) = deg t1 (S (t 1 )): Proof. Let G Z (t; h) =G (t; h) +ZG 3 (t; h) and G Z = G Z(t; ). First we see that for P, deg t 1 (S 1 (t 1 ; h)) = deg t1 (S 1 (t 1)). For every P, one has that 1() 0 and then by Lemma 8, we deduce that (Res t (G 1 (t; h);g Z (t; h))) = ( 1 ( h)) k Res t (G1(t );GZ(t )); where k is a constant depending on. Thus, if we express the resultant as Res t (G 1 ( t ); G Z ( t )) = b 0 (t 1)+b 1 (t 1)Z + + b n(t 1 )Z n ; then we have that a i (t 1 ; )= 1 () k bi (t 1 ) (I): On the other hand, for every P one has that lc(a 0;t 1 )( ) 0. Thus, taking into account that S 1 (t 1 ; h) = gcd K( h)[t 1] (a 0(t 1 ; h);:::;a n (t 1 ; h)), and Lemma 9 (1), we deduce that for every P, deg t1 (S 1 (t 1 ; h)) = deg t1 ( (S 1 (t 1 ; h)) (II): Moreover since for every P, one has that Res t 1 (a 0 ; a 1 + m i= X i 1 a i )( ) 0. Thus Lemma 9 () implies that (S 1 (t 1 ; h))=gcd K[t1] ( (a 0 );:::; (a n )); and therefore by (I), one deduces that (S 1 (t 1 ; h)) = gcd( 1 () k b0(t 1 );:::; 1 () k bn(t 1 )) = 1 () k S1 (t 1 ): K[t 1] In particular, this implies that deg t1 (S 1 (t 1)) = deg t1 ( (S 1 (t 1 ; h)); and from (II) one concludes that deg t1 (S 1 (t 1 )) = deg t1 (S 1 (t 1 ; h)): A similar reasoning shows that for every P, deg t1 (T (t 1 )) = deg t1 (T (t 1 ; h)): Then taking into account that S = S 1 =T, S = S1 =T, that T divides S 1, and that T divides S1, the statement holds. The following corollary follows directly from Theorem 6.

16 114 S. Perez-Daz, J.R. Sendra / Journal of Pure and Applied Algebra 193 (004) Corollary. deg( P ) = deg t1 (S(t 1 ; h)). Moreover, for every P deg( P ) = deg t1 (S (t 1 )). it also holds that Corollary to Theorem 6 shows how to compute the degree of the rational map by means of univariate resultants and gcds. For this purpose, one computes deg t1 (S(t 1 ; h)), and for computing the t 1 -coordinate polynomial, S(t 1 ; h), associated to P(t ), one needs to compute the polynomials S 1 and the polynomial T (see Theorem ). In the following we see that the computation of the polynomial T and the quotient of S 1 and T can be avoided by crossing out the constant roots of S 1 (t 1 ; h); i.e. the roots over K. For h this purpose, we denote by F the algebraic closure of the eld K( h), and by V i the algebraic set dened over F by the polynomials G i (t; h) (see Section 1). First, we state some technical results. Lemma 10. (V h 1 V h V h 3 ) \ (V h 1 V h V h 3 V 4 ) (V h 1 V h V h 3 ) (F \ K). Proof. Let us assume that there exists A =(a; b) (V h 1 V h V h 3 ) \ (V h 1 V h V h 3 V 4 ) but A (F \ K). That is, at least one component of A is constant. Let us say that a K, similarly for b. Then, P(A) =P( h). Thus, b cannot be constant since P( h) is a surface parametrization. Moreover, the above equality also implies that almost all points on the surface dened by P( h) are contained in the curve dened by P(a; h ) which is impossible. Lemma 11. V h 1 V h V h 3 V 4 K. h Proof. Let (a; b) V h 1 V h V 3 V 4. Then, G 4 (a; b)=0. Thus, there exists a denominator h q i (t ) that vanishes on (a; b); let us say w.l.o.g that it is q 1. Therefore, since (a; b) V 1 one deduces that p 1 (a; b)=q 1 (a; b)=0: Now, since gcd(p 1 ;q 1 ) = 1, it holds that the resultant of p 1, and q 1 w.r.t. t is not identically zero. Furthermore, this resultant is in K[t 1 ], and its roots are the t 1 -coordinates of the intersection points of the curves dened by p 1 and q 1 over K. Hence, since K is algebraically closed one has that a is in K. A similar reasoning shows that b K. The following result follows from Lemmas 10 and 11. Theorem 7. S(t 1 ; h)=pp h (S 1(t 1 ; h)), where pp h denotes the primitive part w.r.t. h. Remark. (1) Since the t 1 -coordinate polynomial, S(t 1 ; h), associated to P(t ) is the result of crossing out the factors over K[t 1 ]ofs 1 (t 1 ; h), one may consider an alternative approach to Theorem 7 that avoids the primitive part computation. Namely, one may specialize the variables h in S 1 (t 1 ; h) at two appropriate dierent values and then compute the gcd of the resulting univariate polynomials to determine the factors depending only on t 1. () A similar reasoning might be also considered in order to compute the t -coordinates of the elements in the generic bre. To be more precisely, for every

17 S. Perez-Daz, J.R. Sendra / Journal of Pure and Applied Algebra 193 (004) root A( h) F of S(t 1 ; h), i.e. for the t 1 -coordinates of the elements in F P ( h), we consider the polynomial M A (t ; h)= gcd F[t] (G 1(A; t ; h);g (A; t ; h);g 3 (A; t ; h)) gcd F[t] (G 1(A; t ; h);g (A; t ; h);g 3 (A; t ; h);g 4 (A; t )) : Lemmas 10 and 11 imply that one may compute M A (t ; h), by crossing out the constant roots in gcd(g 1 (A; t ; h);g (A; t ; h);g 3 (A; t ; h)): F[t ] Example 4. Let V the surface parametrized by ( t P(t 1 ;t )= 1 t 1 ; t 1 1 (t 1 t 1+t 1)t1 t 1 (t1 t 1); (t1 t 1) Let us compute deg( P ). For this purpose, we determine the polynomials G i (t; h) for i =1; ; 3. We get G 1 (t; h)=h 1 t 1 t h 1 t 1 h h 1 + t 1, G (t; h)=h 3 1 h h 1 t 3 1 t + t 1; and G 3 (t; h)=t 4 1 t h4 1 h4 t4 1 t h 1 h + t4 1 t t 1 h4 1 h4 +t 1 h 1 h t 1 + t3 1 h4 1 h4 t3 1 h 1 h + t 3 1 h4 1 h t4 1 t4 +h4 1 t 1 t h h4 1 h + h 1 t4 1 t4 h 1 t 1 t + h 1 h3 1 t4 1 t4 +h3 1 t 1 t h3 1 : Now, we compute the polynomial S 1 (t 1 ; h) (see Theorem ), S 1 (t 1 ; h) = Content Z (Res t (G 1 (t; h);g (t; h)+zg 3 (t; h))) = t 8 1(t 1 h 1 ) : Applying Theorem 7 one obtains that the t 1 -coordinate polynomial associated to P(t ) is given by S(t 1 ; h)=(t 1 h 1 ) : Thus, by Corollary to Theorem 6 one deduces that deg( P ) = deg t1 (S(t 1 ; h)) =. Furthermore, in this case, one gets that F P ( h) = {(h 1 ;h ); ( h 1 ;h )}: ) : 3. Algorithm and examples The results obtained in Section can be applied to derive two algorithms to compute deg( P ) by avoiding the requirement of computing explicitly the elements in the bre. One may either compute the degree deterministically using that deg( P )= deg t1 (S(t 1 ; h))) (Algorithm-1) or probabilistically using that deg( P ) = deg t1 (S (t 1 )) for P (Algorithm-). The result of Algorithm- is correct with probability almost one. In the following, we outline these two approaches. Also, we nish this section illustrating the algorithms with two examples. Algorithm-1. Given a surface rational parametrization P(t )=(p 1 =q 1 ;p =q ;p 3 =q 3 ), in reduced form, the algorithm computes deg( P ): (1) Check whether the general assumptions introduced in Section 1 are satised, if not, apply a suitable linear change of variables to P(t ). () Compute the polynomials G i (t; h)=p i ( h)q i (t ) p i (t )q i ( h) for i =1; ; 3.

18 116 S. Perez-Daz, J.R. Sendra / Journal of Pure and Applied Algebra 193 (004) (3) Compute the polynomial S 1 (t 1 ; h) = Content Z (Res t (G 1 (t; h);g (t; h) +ZG 3 (t; h))) (see Theorem ). (4) Compute the polynomial S(t 1 ; h)=pp h (S 1(t 1 ; h)) (see Theorem 7). (5) Return deg t1 (S(t 1 ; h)). Algorithm-. Given a surface rational parametrization P(t )=(p 1 =q 1 ;p =q ;p 3 =q 3 ), in reduced form, the algorithm computes, with probability almost one, deg( P ): (1) Check whether the general assumptions introduced in Section 1 are satised, if not, apply a suitable linear change of variables to P(t ). () Take K, and compute G i (t )=p i ()q i (t ) p i (t )q i (), for i =1; ; 3, as well as G 4 (t ) = lcm(q 1 (t );q (t );q 3 (t )). (3) Compute the polynomials S 1 (t 1) and T (t 1 ) (see Theorem 1). (4) Compute S (t 1 )=S 1 (t 1)=T (t 1 ). (5) Return deg t1 (S (t 1 )). Finally we illustrate the algorithms by two examples. In addition we also provide the elements of F P ( h), where h is a generic element of K. Example 5. Let V the surface parametrized by ( t 3 P(t 1 ;t )= + t1 +3 t1 +3 ; (t1 +3)3 t 3 +t 1 +6 t 3 + t 1 +3; (t 3 + t 1 +3) Let us compute deg( P ). For this purpose, we apply Algorithm-1 and Algorithm-. In Algorithm-1 we determine the polynomials G 1 (t; h)=t 3 h 1 +3t 3 h 3 t 1 3h 3 ; ) : G (t; h)=t 6 1h 3 + t 6 1h 1 +3t t 4 1h 3 +9t 4 1h 1 +7h 3 t 1 +54t 1 +7h 3 54h 1 h 6 1t 3 +7t 4 1 h 6 1t 1 3h 6 1 9h 4 1t 3 9h 4 1t 1 7h 4 1 7t 3 h 1 7t 3 ; G 3 (t; h)= h 3 t 3 t 1 7t 3 18t 1 1t 3 t 1 6t h 1 h 1t 6 4h 1t 3 t 1 t 4 1h 3 t 4 1h 1 +t 3 h 3 h 1 + h 4 1t 3 +h 4 1t 1 +1h 3 h 1 + t 3 h 6 +t 1h 6 h 3 t 6 +6h 6 +4t 1h 3 h 1 +7h 3 6t 3 h 1 +6h 3 t 1 +6h 4 1 6t 6 : We compute the polynomial S 1 (t 1 ; h) (see Theorem ): S 1 (t 1 ; h)=(t 1 +3) 3 (t 1 h 1 ) 3 (t 1 + h 1 ) 3 : Therefore, the t 1 -coordinate polynomial associated to P(t ) is given by S(t 1 ; h)=(t 1 h 1 ) 3 (t 1 + h 1 ) 3, and then deg( P ) = deg t1 (S(t 1 ; h)) = 6. In addition, we may compute the elements of the F P (h), where h is a generic element of K. The roots of the polynomial S(t 1 ; h) are A 1 = h 1 and A = h 1. Now, for every root A i we compute the

19 S. Perez-Daz, J.R. Sendra / Journal of Pure and Applied Algebra 193 (004) polynomials M Aj (t ; h) = gcd(g 1 (A j ;t );G (A j ;t );G 3 (A j ;t )); j =1; ; K[t ] which non-constant roots provide the t -coordinates of the elements in the bre. Hence, one gets that F P ( h)={(h 1 ;h ); (h 1 ; ( 1=+1=i 3)h ); (h 1 ; ( 1= 1=i 3)h ); ( h 1 ;h ); ( h 1 ; ( 1=+1=i 3)h ); ( h 1 ; ( 1= 1=i 3)h )}: Now we apply Algorithm-. First we take =(; 1) K, and we determine the polynomials Gi (t ) for i =1; ; 3, and G 4 (t ), G1(t )=7t 3 t1 3; G (t )=8t t t t 3 ; G 3(t )= 6t 3 +38t t 6 30t 3 t 1 15t 4 1; G 4 (t )=(t 3 + t1 +3) (t1 +3): We compute the polynomials S1 (t 1) and T (t 1 ) (see Theorem 1): S1 (t 1 )=(t 1 ) 3 (t 1 +) 3 (t1 +3) 3 ; T (t 1 )=(t1 +3) 3 : Thus, the t 1 -coordinate polynomial associated to the pair (P(t ); ) is given by S (t 1 )= S 1 (t 1) T (t 1 ) =(t 1 ) 3 (t 1 +) 3 : Therefore, deg( P ) = deg t1 (S (t 1 ))=6. Example 6. Let V the surface parametrized by P(t 1 ;t ), dened by ( t1 +1+4t +6t4 +4t6 + t8 4+t ;t 1 ++3t + t 4 ; t t +6t4 +4t6 + t8 t 1 ++t Let us compute deg( P ). For this purpose, we apply Algorithm-1 and Algorithm-. In Algorithm-1 we determine the polynomials G 1 (t; h)=4t 1 + t 1 h 15h +15t +4t 4 +6th 4 +16t 6 +4th 6 +4t 8 + th 8 4h 1 h 1 t 4h 4 6h 4 t 16h 6 4h 6 t 4h 8 h 8 t ; G (t; h)=t 1 +3t + t 4 h 1 3h h 4 ; ) : G 3 (t; h)=t 8 h 1 + t 1 +1t 4 +7t 1h 4 7h + t 8 h 1 + t 8 h h 8 t 1 h 8 t +3h 1 t 6h 4 t h 8 +6t 4 h 1 3t 1 h 6t 1 h 4 +6t 4 h 4h 6 t 1 4h 6 t 8h 6 +8t 6 +4t 6 h 1 +4t 6 h :

20 118 S. Perez-Daz, J.R. Sendra / Journal of Pure and Applied Algebra 193 (004) We compute the polynomial S 1 (t 1 ; h) (see Theorem ): S 1 (t 1 ; h)=(t 1 h 1 ) : Therefore, the t 1 -coordinate polynomial associated to the parametrization P(t ) is given by S(t 1 ; h) =(t 1 h 1 ), and then deg( P ) = deg t1 (S(t 1 ; h)) =. In addition, we may compute the elements of the F P (h), where h is a generic element of K. The roots of the polynomial S(t 1 ; h) are A 1 = h 1 and A = h 1. Now, for every root A i we compute the polynomials M Aj (t ; h) = gcd(g 1 (A j ;t );G (A j ;t );G 3 (A j ;t )); j =1; ; K[t ] which non-constant roots provide the t -coordinates of the elements in the bre. Hence, one gets that F P ( h)={(h 1 ;h ); ( h 1 ;h )}: Now we apply Algorithm-. First we take =(3; ) K, and we determine the polynomials G i (t ) for i =1; ; 3, and G 4 (t ), G 1(t )=t t +1t 4 +8t 6 +t 8 ; G (t )=t t + t 4 ; G 3(t )= 619t t +54t 4 +36t 6 +9t 8 ; G 4 (t )=(4+t )(t 1 ++t ): We compute polynomials S 1 (t 1) and T (t 1 ) (see Theorem 1): S 1 (t 1 )=(t 1 3) ; T (t 1 )=1: Thus, the t 1 -coordinate polynomial associated to the pair (P(t ); ) is given by S (t 1 )= S 1 (t 1) T (t 1 ) =(t 1 3) : Therefore, deg( P ) = deg t1 (S (t 1 ))=. 4. Practical implementation Algorithm-1 and Algorithm- have been implemented in Maple, and the running times are very satisfactory. In addition, we have implemented the corresponding algorithms (deterministic and probabilistic) based on Grobner bases for the computation of the degree. Here, with probabilistic, we mean that the cardinality of the bre (i.e. the degree of the map) is computed by choosing a random point on the surface. In the following we briey outline the experimental computing times for some examples. Actual computing times are measures on a PC PENTIUM III PROCESSOR 18 MB of RAM, and times are given in seconds of CPU.

21 S. Perez-Daz, J.R. Sendra / Journal of Pure and Applied Algebra 193 (004) In the following table we illustrate the performance of our implementation, and the one based on Grobner bases, showing times in seconds for some parametrizations. In the table, we also show the degree of each parametrization, and the degree of the induced rational map. Also, in the table, we denote by Algorithm-3 and Algorithm-4 the implementation of the deterministic and probabilistic approach based in Grobner bases, respectively. In the appendix, we give the parametrizations considered in this analysis. We also remark that all degrees computed by Algorithm- were correct. Input deg(p(t )) deg( P ) Time of Time of Time of Time of Alg-1 Alg- Alg-3 Alg-4 I II III IV V VI VII VIII IX X Appendix A. Parametrizations in Section 4 In the following we give the input data of the examples analyzed in Section 4. ( t P 1 = (t 1 1)3 t1 1; t ; t + t 1 1 ) t 4 ; ( t t1 P = (t 1 ) t1 1 ; (t 1 1)3 t t1 (t 1 ) ; t t1 4 +t t1 + t 1 +1 ) t 3t4 1 (t 1 ; ) ( 54+67t P 3 = t1 134t t t +10t t1 8 t (9t t1 58t 1 3t 90t1 ) ; 40+1t t 1 4t t 1 +1t 1t t1 ; 98(t t1 t 1 + t t1 )t ) 39+95t t 1 190t t t ; ( t 4 P 4 = (t 1 1)3 t1 1; t 4 ; t4 + t 1 1 ) t 8 ; P 5 =((t1(t ) 3 )=(t1 4 +3t1 +t 1 t 8t 1 + t 8t + 19); (t1 +t 1 t 8t 1 +t 8t +19+t 4 1) ; (t t 1 +t 1 t 8t 1 + t 8t + 19)=(t 1 +t 1 t 8t 1 + t 8t +19+t 4 1));

22 10 S. Perez-Daz, J.R. Sendra / Journal of Pure and Applied Algebra 193 (004) P 6 = ((113 16t1 +9t 1 6t 14t 1 t )=( 130t1 110t t 150t 1 t ); ( 7 101t t 1 +5t 1 t +t1 3 +6t + 44t1 1t1t 533t1t 30tt t 1 t 46t1t 3 80t)=( 63+40t 4 +80t 1 t +40t1); ( t t 1 +94t 1 t + 186t t +4t t1t + 635t1t +37tt t 1 t + 56t1t 3 +93t)=( 3+4t t 1 )); P 7 = ((56 + 9t 1 t + 388t1t + 871t1t + 516t 1 t t 1 t + 710t1t t t t+19t +1314t t 3 1t +1460t 4 1t t 3 1t t 1t t 1 t t1t t1t t1t t1+436t t t 6 1 t)=( t +90t 1 t +45t t1); (17 3t 1 t 37t1t 465t1t 48t 1 t 3 186t 1 t 434t 1t t t1 11t 1 16t 4 6t 14t 4 1)=(49 3 5t1 10t 1 ); (37 60t 1 t 44t1t 169t1t 7t 1 t 1t 3 1 t 194t 1t 3 148t 1 15t1 91t1 4 30t 18t 4 170t1)=( t1 +t 1 )); P 8 =(( t t1 4 94t 1 6t1 56t 4 56t t1 94t 3 1 4t 5 1t 3 336tt 1 564tt 1 3 4t 1 t 3 376t1t 3 94t 1 t t1t tt t1t t1t t1t t1t t 1 t t1t tt t1t t1t t1t t1t t1t 7 +58tt1+51t t 8 +58t1)=( t1 4 36t1t 3 54tt 1 36t 1 t 3 9t); 4 ( 119+4t t1 4 63t 1 +5t1 7t 4 7t1 4 + t1 5 88t1t 3 43tt 1 +6tt t 1 t 3 +4t1t 3 + t 1 t 4 78t1t 7 548tt t1t t1t t1t t1t 6 78t 1 t 7 91t1 8 91t)=(39 8 4t 1 ); ( 3+3t t1 4 31t 1 49t1 +58t 4 +58t1 4 +8t t1t tt 1 +48tt t 1 t 3 +3t1t 3 +8t 1 t 4 31t1t 7 109tt t1t t1t t1t t1t 6 31t 1 t 7 39t1 8 39t)( t 1 )); P 9 = ((57 t 1 t 14t 3 1t 447t 5 1t 340t 7 1t 85t 9 1t 36t 3 1t 3 74t 1 45t t t t t 1 1 )=(78+94t 1 +47t 4 1); (85 71t 1 t + 398t t t t 1 34t 3 1t 17t 5 1t + 56t t 8 1)=(60t 1 t t 1

23 S. Perez-Daz, J.R. Sendra / Journal of Pure and Applied Algebra 193 (004) t 4 1); ( t 1 t + 366t t t t 1 +98t 3 1t +49t 5 1t +0t t 8 1)=(65(1 + t 1) )); P 10 = (((t t t 4 + t 1 t 8 +9t 1 +6tt t 16 +1t 1 ) 4 )=(t t 1 t 8 +9t 1 +6tt t t 8 +1t t); 4 ((t t t 4 + t 1 t 8 +9t 1 +6tt t 16 +1t 1 )(t 4 +3) )=(t t 1 t 8 +9t 1 +6t 4 t 1 + t t 8 +1t t 4 ); ((t t t 4 + t 1 t 8 +9t 1 +6tt t 16 +1t 1 ) )=(t t 1 t 8 +9t 1 +6tt t t 8 +1t t)): 4 References [1] E. Brieskorn, H. Knoerrer, Plane Algebraic Curves, Birkhauser Verlag, Basel, [] E.W. Chionh, R.N. Goldman, Using multivariate resultants to nd the implicit equation of a rational surface, Visual Comput. 8 (199) [3] D. Cox, J. Little, D. O Shea, Ideals, Varieties, and Algorithms, nd Edition, Springer, New York, [4] J. Harris, Algebraic Geometry. A rst Course, Springer, Berlin, [5] S. Perez-Daz, J. Schicho, J.R. Sendra, Properness and inversion of rational parametrizations of surfaces, Appl. Algebra Eng., Comm. Comput. 13 (00) [6] J. Schicho, A degree bound for the parametrization of a rational surface, J. Pure App. Algebra 145 (1999) [7] J. Schicho, Simplication of surface parametrizations, in: T. Mora (Ed.), Proceedings of ISSAC 00, ACM Press, New York, 00, pp [8] J.R. Sendra, F. Winkler, Computation of the degree of a rational map, in: B. Mourrain (Ed.), Proceedings of ISSAC 001 (London, Canada), ACM Press, New York, NY, USA, 001, pp [9] J.R. Sendra, F. Winkler, Tracing index of rational curve parametrizations, Comput. Aided Geom. Des. 18 (8) (001) [10] I.R. Shafarevich, Basic Algebraic Geometry Schemes; 1 Varieties in Projective Space, Vol. 1, Springer, Berlin, New York, [11] F. Winkler, Polynomials Algorithms in Computer Algebra, Springer, Wien, New York, 1996.