Computing Birational Polynomial Surface Parametrizations without Base Points

In this paper, we present an algorithm for reparametrizing birational surface parametrizations into birational polynomial surface parametrizations without base points, if they exist. For this purpose, we impose a transversality condition to the base points of the input parametrization.


Introduction
Algebraic surfaces are mainly studied from three different, but related, points of view, namely: pure theoretical, algorithmic and because of their applications.In this paper, we deal with some computational problems of algebraic surfaces taking into account the potential applicability.
In many different applications, as for instance in geometric design (see e.g.[8]) parametric representations of surfaces are more suitable than implicit representations.Among the different types of parametric representations, one may distinguish radical parametrizations (see [24]) and rational parametrizations (see e.g.[18]), the first being tuples of fractions of nested radical of bivariate polynomials, and the second being tuples of fractions of bivariate polynomials; in both cases the tuples are with generic Jacobian of rank 2. Other parametric representations by means of series can be introduced, but this is not within the scope of this paper.One may observe that the set of rational parametrizations is a subclass of the class of radical parametrizations.Indeed, in [27], one can find an algorithm to decide whether a radical parametrization can be transformed by means of a change of the parameters into a rational parametrization; in this case, we say that a reparametrization has been performed.Now, we consider a third type of parametric representation of the surface, namely, the polynomial parametrization.That is, tuple of bivariate polynomials with generic Jacobian of rank 2. Clearly the class of polynomial parametrizations is a subclass of the class of the rational parametrizations, and the natural question of deciding whether a given rational parametrization can be reparametrized into a polynomial parametrization appears.This is, indeed, the problem we deal with in the paper.Unfortunately the inclusion of each of these classes into the next one is strict, and hence the corresponding reparametrizations are not always feasible.In some practical applications, the alternative is to use piecewise parametrizations with the desired property (see e.g.[2], [30]).
Before commenting the details of our approach to the problem, let us look at some reasons why polynomial parametrizations may be more interesting than rational ones.In general, rational parametrizations are dominant over the surface (i.e. the Zariski closure of its image is the surface), but not necessarily surjective.This may introduce difficulties when applying the parametric representation to a problem, since the answer might be within the non-covered area of the surface.For the curve case, polynomial parametrizations are always surjective (see [21]).For the surface case, the result is not so direct but there are some interesting results for polynomial parametrizations to be surjective (see [16]) as well as subfamilies of polynomial parametrizations that are surjective (see [26]).Another issue that could be mention is the numerical instability when the values, substituted in the parameters of the parametrizations, get close to the poles of the rational functions; note that, in this case, the denominators define algebraic curves which points are all poles of the parametrization.One may also think on the advantages of providing a polynomial parametrization instead of a rational parametrization, when facing surface integrals.Let us mention a last example of motivation: the algebra-geometric technique for solving non autonomous ordinary differential equations (see [6], [10]).In these cases, the differential equation is seen algebraically and hence representing a surface.Then, under the assumption that this surface is rational (resp.radical) the general rational (resp.radical) solution, if it exists, of the differential equation is determined from a rational parametrization of the surface.This process may be simplified if the associated algebraic parametrization admits a polynomial parametrization.
Next, let us introduce, and briefly comment, the notion of base points of a rational parametrization.A base point of a given rational parametrization is a common solution of all numerators and denominators of the parametrization (see e.g.[5], [15]).The presence of this type of points is a serious obstacle when approaching many theoretical, algorithmic or applied questions related to the surface represented by the parametrizations; examples of this phenomenon can be found in, e.g., [3], [4], [25], [29].In ad-dition, it happens that rational surface may admit, both, birational parametrizations with empty base locus and with non-empty base locus.Moreover, the behavior of the base locus is not controlled, at least to our knowledge, by the existing parametrization algorithms or when the resulting parametrization appears as the consequence of the intersection of higher dimension varieties, or as the consequence of cissoid, conchoid, offsetting, or any other geometric design process applied to a surface parametrization (see e.g.[1], [22], [23], [31]).
In this paper, we solve the problem, by means of reparametrizations, of computing a birational polynomial parametrization without base points of a rational surface, if it exists.For this purpose, we assume that we are given a birational parametrization of the surface that has the property of being transversal (this is a notion introduced in the paper, see Section 3 for the precise definition).Essentially, the idea of transversality is to assume that the multiplicity of the base points is minimal.Since, by definition (see Section 3) this multiplicity is introduced as a multiplicity of intersection of two algebraic curves, one indeed is requiring the transversality of the corresponding tangents.In this paper, we have not approached the problem of eliminating this hypothesis, and we leave it as future work in case it exists.
The general idea to solve the problem is as follows.We are given a birational parametrization P and let Q be the searched birational polynomial parametrization without base points; let us say, first of all, that throughout the paper we work projectively.Then, there exists a birational map, say S P , that relates both parametrizations as Q = P • S P .Then, taking into account that the base locus of S P and P are the same, that they coincide also in multiplicity, and applying some additional properties on base points stated in Subsection 3.1, we introduce a 2-dimensional linear system of curves, associated to an effective divisor generated by the base points of P.Then, using the transversality we prove that every basis of the linear system, composed with a suitable birational transformation, provides a reparametrization of P that yields to a polynomial parametrization with empty base locus.
The structure of the paper is as follows.In Section 2, we introduce the notation and we recall some definitions and properties on base points, essentially taken from [5].
In Section 3 we state some additional required properties on base points, we introduce the notion of transversality of a base locus, both for birational maps of the projective plane and for rational surface projective parametrizations.Moreover, we establish some fundamental properties that require the transversality.Section 4 is devoted to state the theoretical frame for solving the central problem treated in the paper.In Section 5, we derive the algorithm that is illustrated by means of some examples.We finish the paper with a section on conclusions.

Preliminary on Basic Points and Notation
In this section, we briefly recall some of the notions related to base points and we introduce some notation; for further results on this topic we refer to [5].We distinguish three subsections.In Subsection 2.1, the notation that will be used throughout the paper is introduced.The next subsection focuses on birational surface parametrizations, and the third subsection on birational maps of the projective plane.

Notation
Let, first of all, start fixing some notation.Throughout this paper, K is an algebraically closed field of characteristic zero.x = (x 1 , . . ., x 4 ), y = (y 1 , . . ., y 4 ) and t = (t 1 , t 2 , t 3 ).F is the algebraic closure of K( x , y ).In addition, P k (K) denotes the k-dimensional projective space, and G (P k (K)) is the set of all projective transformations of P k (K).
Furthermore, for a rational map where the non-zero m i are homogenous polynomial in h of the same degree, we denote by deg(M) the degree deg h (m i ), for m i non-zero, and by degMap(M) the degree of the map M; that is, the cardinality of the generic fiber of M (see e.g.[7]).
For L ∈ G (P k 2 (K)), and M ∈ G (P k 1 (K)) we denote the left composition and the right composition, respectively, by be homogeneous and non-zero, where L is a field extension of K. Then C (f ) denotes the projective plane curve defined by f over the algebraic closure of L.
Let C (f ), C (g) be two curves in P 2 (K).For A ∈ P 2 (K), we represent by mult A (C (f ), C (g)) the multiplicity of intersection of C (f ) and C (g) at A. Also, we denote by mult(A, C (f )) the multiplicity of C (f ) at A.
Finally, S ⊂ P 3 (K) represents a rational projective surface.

Case of surface parametrizations
In this subsection, we consider a rational parametrization of the projective rational surface S , namely, where t = (t 1 , t 2 , t 3 ) and the p i are homogenous polynomials of the same degree such that gcd(p 1 , . . ., p 4 ) = 1.
Definition 1.A base point of P is an element A ∈ P 2 (K) such that p i (A) = 0 for every i ∈ {1, 2, 3, 4}.We denote by B(P) the set of base points of P. That is In order to deal with the base points of the parametrization, we introduce the following auxiliary polynomials: where x i , y i are new variables.We will work with the projective plane curves C (W i ) in P 3 (F).Similarly, for M = (M 1 : M 2 : M 3 ) ∈ G (P 3 (K)), we define, (2.3) Remark 1.Some times, we will need to specify the parametrization in the polynomials above.In those cases, we will write W P i or W M,P i instead of W i or W M i ; similarly, we may write C (W P 1 ) and C (W M,P
Using the multiplicity of intersection of these two curves, we define the multiplicity of a base point as follows.
Definition 2. The multiplicity of a base point A ∈ B(P) is mult A (C (W 1 ), C (W 2 )), that is, is the multiplicity of intersection at A of C (W 1 ) and C (W 2 ); we denote it by mult(A, B(P)) := mult A (C (W 1 ), C (W 2 )) (2.4) In addition, we define the multiplicity of the base points locus of P, denoted mult(B(P)), as mult(B(P)) := Note that, since gcd(p 1 , . . ., p 4 ) = 1, the set B(P) is either empty of finite.
For the convenience of the reader we recall here some parts of Proposition 2 in [5].
2. If A ∈ B(P), then the tangents to C (W L 1 ) at A (similarly to C (W L 2 )), with the corresponding multiplicities, are the factors in K where T i is the product of the tangents, counted with multiplicities, of C (L i (P)) at A, and where ǫ i = 1 if mult(A, C (L i (P)))) = min{mult(A, C (L i (P))) | i = 1, . . ., 4} and 0 otherwise.

Case of rational maps of P 2 (K)
In this subsection, let where gcd(s 1 , s 2 , s 3 ) = 1, be a dominant rational transformation of P 2 (K).
That is, the base points of S are the intersection points of the projective plane curves, C (s i ), defined over K by s i ( t ), i = 1, 2, 3. We denote by B(S) the set of base points of S.
We introduce the polynomials where x i , y j are new variables and we consider the curves C (V i ) over the field F; compare with (2.3).Similarly, for every L ∈ G (P 2 (K)) we introduce the polynomials Remark 2. Some times, we will need to specify the rational map in the polynomials above.In those cases, we will write V S i or V L,S i instead of V i or V L i ; similarly, we may write C (V S 1 ) and C (V L,S
As we did in Subsection 2.2, we have the following notion of multiplicity.
Definition 4. For A ∈ B(S), we define the multiplicity of intersection of A, and we denote it by mult(A, B(S)), as (2.9) In addition, we define the multiplicity of the base points locus of S, denoted mult(B(S)), as (note that, since gcd(s 1 , s 2 , s 3 ) = 1, B(S) is either finite or empty) Next result is a direct extension of Proposition 2 in [5] to the case of birational transformation of P 2 (K).
)), with the corresponding multiplicities, are the factors in K where T i is the product of the tangents, counted with multiplicities, of C (L i (S)) at A, and where and 0 otherwise.

Transversal Base Locus
In this section, we present some new results on base points that complement those in [5] and we introduce and analyze the notion of transversality in conexion with the base locus.

Further results on base points
We start analyzing the rationality of the curve is a birational parametrization of C (V L 1 ).Proof.We start proving that for every where z i,j are undetermined coefficients satisfying that the determinant of the corresponding matrix is not zero.Furthermore, for L ∈ G (P 2 (K)), we denote by z L the coefficient list of L. We also introduce the polynomial We consider the birational extension R x : P 2 (F) be the open set where the R x is bijective; say that U x = P 2 (F) \ ∆.We express the close set ∆ as ∆ = ∆ 1 ∪ ∆ 2 where ∆ 1 is either empty or it is a union of finitely many curves, and ∆ 2 is either empty or finite many points.We fix our attention in ∆ 1 .Let f ( t ) be the defining polynomial of ∆ 1 .Let Z( z , t 1 , t 2 ) be the remainder of f when diving by V L 1 w.r.t t 3 .Note that R L does not divide f since R L is irreducible and depends on z .Hence Z is no zero.Let α( z ) be the numerator of a non-zero coefficient of Z w.r.t.{t 1 , t 2 } and let β( z ) the l.c.m. of the denominators of all coefficients of Z w.r.t.{t 1 , t 2 }.Then, we define Ω 1 as Let a, b ∈ C (R L ) ∩ U x be two different points, then by injectivity R x (C (R L )) contains at least two points, namely R x (a) and R x (b).In this situation, since Next Lemma analyzes the rationality of the curves C (L i (S)) where L = (L 1 : Proof.Let U be the open subset where R is a bijective map, and let {ρ j,1 t 1 + ρ j,2 t 2 + ρ j,3 t 3 } j=1,...,n be the linear forms defining the lines, if any, included in P 2 (K) \ U.Then, we take Ω 2 = ∩ n j=1 Σ j where In this situation, reasoning as in the last part of the proof of Lemma 3, we get that R(C The following lemma follows from Lemma 2. Then expressing L(S(A)) = 0 in terms of matrices, since L is invertible, we have that S(A) = L −1 (0, 0, 0) = (0, 0, 0).Thus, A ∈ B(S).
Lemma 6.There exists a non-empty Zariski open subset Proof.Let A ∈ B(S).Then, by Lemma 2 (2), we have that Let us assume w.l.o.g. that the minimum above is reached for i = 1.Then all (m A −1)order derivatives of the forms s i vanish at A, and there exists an m A -order partial derivative of s 1 not vanishing at A. Let us denote this partial derivative as ∂ m A .Now, let L be as in the proof of Lemma 3.Then, is a non-zero polynomial because ∂ m A s 1 (A) = 0. We then consider the open subset (see proof of Lemma 3 for the notation z L ) In this situation, we take Let us prove that Ω 3 satisfies the property in the statement of the lemma.Let L ∈ Ω 3 and A ∈ B(S).Let m A be as in (3.1).Then all partial derivatives of L i (S), of any order smaller than m A , vanishes at A. Moreover, since Remark 4. We note that the proofs of Lemma 5 and Lemma 6 are directly adaptable to the case of birational surface parametrizations.So, both lemmas hold if M ∈ G (P 3 (K)) and we replace S by the birational surface parametrizaion P and L S by M P := M • P.
In the following, we denote by Sing(D), the set of singularities of an algebraic plane curve D.
where T i is the product of the tangents, counted with multiplicities, to C (L i (S)) at A.

Let
A ∈ B(S), and let T i be the product of the tangents, counted with multiplicities, to ). ( 2) follows from Lemma 6 and Lemma 2.
(3) By (2) the tangents to C (V L 1 ) and to C (V L 2 ) at A are T 1 := x i T i and T 2 := y i T i , respectively.Since gcd(T 1 , T 2 , T 3 ) = 1, then T i is primitive, and hence gcd(T 1 , T 2 ) = 1.That is, C (V L 1 ) and C (V L 2 ) intersect transversally at A. From here, the results follows.

Transversality
We start introducing the notion of transversality for birational maps of P 2 (K).Definition 5. We say that S is transversal if either B(S) = ∅ or for every A ∈ B(S) it holds that (see In this case, we also say that the base locus of S is transversal. In the following lemma, we see that the transversality is invariant under left composition with elements in G (P 2 (K)).
Lemma 7. If S is transversal, then for every L ∈ G (P 2 (K)) it holds that L S is transversal.
The next lemma characterizes the transversality by means of the tangents of C (s i ) at the base points.Lemma 8.The following statements are equivalent 1. S is transversal.
2. For every A ∈ B(S) it holds that gcd(T 1 , T 2 , T 3 ) = 1, where T i is the product of the tangents, counted with multiplicities, to C (s i ) at A.
First of all, we observe that, because of Lemma 7, we may assume w.l.o.g. that Lemma 6 applies to S. So, by Definition 5, S is transversal if and only if for every and, by Definition 2, if and only if Furthermore, using Theorem 2.3.3 in [28], we have that if and only V 1 and V 2 intersect transversally at A i.e. if the curves have no common tangents at A which is equivalent to gcd(T 1 , T 2 , T 3 ) = 1.The proof finishes taking into account that, by Lemma 6 mult(A, In the last part of this section, we analyze the relationship of the transversality of a birational map of the projective plane and the transversality of a birational projective surface parametrization.For this purpose, first we introduce the notion of transversality for parametrizations.Definition 6.Let P be a birational surface parametrization of P 3 (K).We say that P is transversal if either B(P) = ∅ or for every A ∈ B(P) it holds that (see In this case, we say that the base locus of P is transversal.
We start with some technical lemmas.
Lemma 9.If P is transversal, then for every M ∈ G (P 2 (K)) it holds that M P is transversal.
Proof.The proof is analogous to the proof of Lemma 7.
Lemma 10.The following statements are equivalent 1. P is transversal.

2.
For every A ∈ B(P) it holds that gcd(T 1 , . . ., T 4 ) = 1, where T i is the product of the tangents, counted with multiplicities, to C (p i ) at A.
Proof.The proof is analogous to the proof of Lemma 8.
The following lemma focusses on the behavior of the base points of P when right composing with elements in G (P 2 (K)).

If P is transversal then
We consider the curves C (W P i ) and C (W P L i ) (see Remark 1), and we note that C (W P L i ) is the transformation of C (W i ) under the birational transformation L −1 of P 2 (K).Now, (2) and (3) follow from Definition 2, and (4) from Lemma 10.
Next results analyze the base loci of birational surface parametrizations assuming that there exists one of them with empty base locus.Proof.Since Q = P(S) and B(Q) = ∅, by Theorem 11 in [5] we get that B( L S S) = B( L P P) for L S in a certain open subset of G (P 2 (K)) and L P in a certain open subset of G (P 3 (K)).Now, using Lemma 5, and Remark 4 one concludes the proof of statement (1).Statement (2) follows from Theorem 11 in [5], taking into account that Q is birational.
We start observing that because of Lemma 12 one has that B(S) = B(P).Now, let us consider L ∈ G (P 2 (K)) and M ∈ G (P and, in consequence, B(Q * ) = ∅.Moreover, Q * and P * parametrize the same surface.Furthermore, by Lemma 7, S * is transversal.Thus, S * , P * , Q * satisfy the hypotheses of the lemma.On the other hand, by Lemma 5 and Remark 4, we have that B(S * ) = B(S) = B(P) = B(P * ).Furthermore, by Lemmas 1, 2 we have that mult(A, )).Therefore, by Lemma 6 and Remark 4, we can assume w.lo.g. that for every A ∈ B(S) = B(P) it holds that Now, let A ∈ B(P) and let m := mult(A, C (V 1 )).We can assume w.l.o.g that A = (0 : 0 : 1).Let T i denote the product of the tangents to s i at A. Also, let deg(S) = s deg(P) = p, and deg(Q) = q Then, by (3.2), we may write: where g j,i (t 1 , t 2 ) are homogeneous forms of degree j.In addition, let q i be expressed as where F j,i (t 1 , t 2 ) are homogeneous forms of degree j.Then Using this expression and (3.3) can be expressed as + (terms of degree in t 3 strictly smaller than q(s − m)) . (3.5) Now, let us prove that H i is not identically zero.We first observe that H i = q i (T 1 , T 2 , T 3 ) for i ∈ {1, 2, 3, 4}.We also note that if there exists i ∈ {1, 2, 3, 4} such that H i = 0, by (3.2), it must happen that for all i ∈ {1, 2, 3, 4} it holds that H i = 0. Let H 1 be zero.Then, T = (T 1 : T 2 : T 3 ) / ∈ P 2 (K), because otherwise T ∈ B(Q) and B(Q) = ∅.Therefore, T is a curve parametrization.Thus, if H i = 0, then T parametrizes a common component of the four curves C (p i ).But this implies that gcd(q 1 , q 2 , q 3 , q 4 ) = 1 which is a contradiction.Thus, by (3.2), mult(A, We finish this section stating the relationship between the transversality of S and P under the assumption that P(S −1 ) does not have base points.
Theorem 1.Let P and Q be two birational projective parametrizations of the same surface S such that Q(S) = P and B(Q) = ∅.Then, S is transversal if and only if P is transversal.

Therefore, mult(A, B(S))
Thus, S is transversal if and only if P is transversal.

Proper Polynomial Reparametrization
In this section, we deal with the central problem of the paper, namely, the determination, if they exist, of proper (i.e.birational) polynomial parametrizations of rational surfaces.For this purpose, we distinguish several subsections.In the first subsection, we fix the general assumptions and we propose our strategy.In the second subsection, we perform the theoretical analysis, and in the last subsection we prove the existence of a linear subspace, computable from the input data, and containing the solution to the problem.
We start recalling what we mean with a polynomial projective parametrization.We say that a projective parametrization is polynomial if its dehomogenization w.r.t. the fourth component, taking t i = 1 for some i ∈ {1, 2, 3}, is polynomial; note that the fourth component of a polynomial projective parametrization has to be a power of t i for some i ∈ {1, 2, 3}.Clearly, a similar reasoning is applicable w.r.t.other dehomogenizations.On the other hand, we say that a parametrization is almost polynomial if its fourth component is the power of a linear form.
The important fact is that a rational surface admits a birational polynomial parametrization if and only if it admits a birational almost polynomial parametrization.Furthermore, if we have an almost polynomial parametrization, and its fourth component is a power of the linear form L * 3 ( t ), we may consider two additional linear forms L * 1 , L * 2 such that L * = (L * 1 : L * 2 : L * 3 ) ∈ G (P 2 (K)) and then the composition of the almost polynomial parametrization with (L * ) −1 is a polynomial parametrization of the same surface.

General assumptions and strategy
In our analysis we have two main assumptions.Se assume that the rational surface S admits a polynomial birational parametrization with empty base locus.Throughout the rest of the paper, let us fix one of these parametrizations and denote it by Q; that is, with q i homogenous polynomials of the same degree such that gcd(q 1 , . . ., q 4 ) = 1, is a proper polynomial parametrization of S satisfying that B(Q) = ∅.Note that, by Corollary 6 in [5], the degree of S is then the square of a natural number.Moreover, we introduce a second assumption.We assume that we are given a transversal birational parametrization of S .Throughout the rest of the paper, let us fix P as a transversal proper parametrization of S , and let P be expressed as in (2.1).
Our goal is to reach Q, or more precisely an almost polynomial parametrization of S , from P. For this purpose, first we observe that, since both P and Q are birational, they are related by means of a birational map of P 2 (K), say S P .More precisely, S P := Q −1 • P. In the following, we represent S P as where gcd(s 1 , s 2 , s 3 ) = 1.Note that, because of Theorem 1, since P is transversal, then S P is transversal.In addition, let R P := S −1 P ( t ) = P −1 • Q.In the sequel, we represent R P as where gcd(r 1 , r 2 , r 3 ) = 1.
So, in order to derive Q from P it would be sufficient to determine S P , and hence R P , because Q = P(R P ).Furthermore if, instead of determining S P , we obtain L S P := L • S P , for some L ∈ G (P 2 (K)), then instead of Q we get which is almost polynomial, and hence solves the problem.Taking into account this fact we make the following two considerations: 1. We can assume w.l.o.g. that B(P) = ∅.Indeed, if B(P) = ∅, by Theorem 10 and Corollary 9 in [5], we get that B(S P ) = ∅.Furthermore, by Corollary 7 in [5], we obtained that deg(S P ) = 1.Thus, using that Q is indeed polynomial, we get that the fourth component of P is the power of a linear form, and therefore the input parametrization P would be already almost polynomial, and hence the problem would be solved.
2. We can assume w.l.o.g. that S P satisfies whatever property reachable by means of a left composition with elements in G (P 2 (K)), as for instance those stated in Lemmas 3, or 4, or 6.In particular, by Lemma 7, the transversality is preserved.
In other words, in the set R of all birational transformations of P 2 (K), we consider the equivalence relation ∼, defined as S ∼ S * if there exists L ∈ G (P 2 (K)) such that L • S = S * , and we work with the equivalence classes in R/ ∼.
Therefore, our strategy will be to find a birational map M of P 2 (K) such that P(M −1 ) is almost polynomial.For this purpose, we will see that it is enough to determine a dominant rational transformation M of P 2 (K) (later, we will prove that such a transformation is indeed birational) such that 1. deg(M) = deg(S P ).

∀ A ∈ B(M) it holds that mult(A, B(M)) = mult(A, B(S P )).
The difficulty is that both M and S P are unknown.Nevertheless, by Corollary 10 and Theorem 3 in [5], we have that deg(S P ) = deg(P) deg(S ) .
Note that deg(P) is given and deg(S ) can be determined by applying, for instance, the formulas in [14] (see also [13]).On the other hand, taking into account Lemma 12, we can achieve our goal by focusing on P.More precisely, we reformulate the above conditions into the equivalent following conditions.
Conditions 1.We say that a rational dominant map for all A ∈ B(P).
In the following subsections, we will see that rational dominant maps satisfying Conditions 1 provide an answer to the polynomiality problem.

Theoretical analysis
We start this analysis with some technical lemmas.For this purpose, S , Q, P, S P , R P are as in the previous subsection.We recall that Q(S P ) = P, B(Q) = ∅, R P = S −1 P , P is transversal, and hence S P is also transversal.Moreover, by Lemma 12, S P satisfies Conditions 1. Furthermore, in the sequel, let with gcd(s 1 , s 2 , s 3 ) = 1, be dominant rational map of P 2 (K) satisfying Conditions 1.
Proof.We use the notation introduced in Lemma 3. We take L ∈ G (P 3 (K)) such that 1. C (V L 1 ) is rational (see (2.8) for the definition of V L 1 constructed from M, and Lemma 3 for the existence of L).
In addition, we consider a projective transformation N( t ) in the parameters t such that deg Lemma 15.Let M be a rational dominant map of P 2 (K).If M satisfies Conditions 1, then M is birational.
Moreover, the second equality follows as in the previous reasoning, taking into account that S satisfies Condition 1, and hence B(S P ) = B(S) = B(P).
(2) follows taking into account that the multiplicity of a point on a curve, as well as the multiplicity of intersection, does not change under projective transformations.
and such that ℘ is uniquely determined by B(P).
Proof.We first observe that we can assume w.l.o.g. that no base point of P is on the line at infinity x 3 = 0. Indeed, let L ∈ G (P 2 (K)) be such that B(P) is contained in the affine plane x 3 = 1.We consider In addition, because of Lemma 16, S * satisfies the hypothesis of the theorem.
Let C (V 1 ) denote the curve associated to S as in (2.7).By Lemma 3, taking L in the corresponding open subset of G (P 2 (K)), we have that ) then the affine parametrization ρ( x , h 1 ) := (w 1 ( x , h 1 , 1)/w 3 ( x , h 1 , 1), w 2 ( x , h 1 , 1)/w 3 ( x , h 1 , 1)) is affinely surjective (see [21] and [28]).Now, let A = (a 1 : a 2 : 1) ∈ B(P).By Lemma 2, P ∈ C (V 1 ).We observe that, by taking L in the open subset of Lemma 6, we may assume that We consider the polynomial Since the affine parametrization has been taken surjective, we have that and that for every root t 0 of g A it holds that ρ(t 0 ) = (a 1 , a 2 ).We write w i as On the other hand, we express s i as where deg(T i,j ) = j, j ∈ {m A , . . ., deg(S)}, and In other words, g A divides s i (W).Now, for B = (b 1 : b 2 : 1) ∈ B(P), with A = B, it holds that gcd(g A , g B ) = 1, since otherwise there would exist a root t 0 of gcd(g A , g B ), and this implies that ) which is a contradiction.Therefore, we have that We observe that the factor defined by the base points does not depend on i.Thus, since s i (W) does depend on i, we get that f i is the factor depending on i.Furthermore, For the next theorem, we recall that S P = (s 1 : s 2 : s 3 ) with gcd(s 1 , s 2 , s 3 ) = 1; see (4.2).Theorem 3. Let S be transversal.If i ∈ {1, 2, 3} then s i (R) = Z i ( t ) ℘( t ) where Z i is a linear form, deg(℘( t )) = mult(B(P)) and such that ℘ is uniquely determined by B(P).
Proof.We first observe that we can assume w.l.o.g. that no base point of P is on the line at infinity x 3 = 0. Indeed, let L ∈ G (P 2 (K)) be such that B(P) is contained in the affine plane x 3 = 1.We consider S * := S L P = (s * 1 : s * 2 : s * 3 ), S * := S L , and In addition, because of Lemma 16, S * and S * satisfy the hypotheses of the theorem.

3
be the open subset in lemma 6 applied to S and S P , respectively.Taking L ∈ Ω S 3 ∩ Ω S P 3 (note that G (P 2 (K)) is irreducible and hence the previous intersection is non-empty), we may assume that We consider the polynomial g A = gcd(w 1 − a 1 w 3 , w 2 − a 2 w 3 ).Reasoning as in the Proof of Theorem 2 we get that deg h (g A ) = m A (4.12) and that for every root t 0 of g A it holds that ρ(t 0 ) = (a 1 , a 2 ).We write w i as w i = g A • w * i + a i w 3 for i =∈ {1, 2}.On the other hand, by (4.10), (4.11), we have that mult(A, C (s i )) = m A .Therefore, we can express s i as where deg(T i,j ) = j, j ∈ {m A , . . ., deg(S P )}, and In other words, g A divides s i (W).Now, for B = (b 1 : b 2 : 1) ∈ B(P), with A = B, reasoning as in the proof of Theorem 2, it holds that gcd(g A , g B ) = 1.Therefore, we have that In this situation, let us introduce the notation t * := (t 3 , 0, −t 1 , t 1 , t 2 ) and t * * = (t 3 , 0, −t 1 , φ −1 (t 1 , t 2 )).Then, for i ∈ {1, 2, 3}, we have that This concludes the proof.
Corollary 2. If S is transversal, there exists L ∈ G (P 2 (K)) such that S = L S P .
Corollary 3. The following statements are equivalent 1. S is transversal.
2. There exists L ∈ G (P 2 (K)) such that S = L S P .

The Solution Space
In this subsection we introduce a linear projective variety containing the solution to our problem and we show how to compute it.We start identifying the set of all projective curves, including multiple component curves, of a fixed degree d, with the projective space (see [9], [28] or [32] for further details) More precisely, we identify the projective curves of degree d with the forms in K[ t ] of degree d, up to multiplication by non-zero K-elements.Now, these forms are identified with the elements in V d corresponding to their coefficients, after fixing an order of the monomials.By abuse of notation, we will refer to the elements in V d by either their tuple of coefficients, or the associated form, or the corresponding curve.
Similarly, one has the next lemma Lemma 18.If M is a birational transformation of P 2 (K) and L ∈ G (P 2 (K)) then L(M) = L( L M).
Furthermore, the following theorem holds Theorem 4. Let M be a birational transformation of P 2 (K) and let {n 1 , n 2 , n 3 } be a basis of L(M) and N := (n 1 : n 2 : n 3 ).There exists Since {m 1 , m 2 , m 3 } is also a basis of L(M), one has that L := ( λ 1,j t j : λ 2,j t j : λ 3,j t j ) ∈ G (P 2 (K))) and N = L • M. Remark 5. Observe that, by Theorem 4, all bases of L(M) generate birational maps of P 2 (K).Corollary 4. Let M be a birational transformation of P 2 (K).The following statements are equivalent 1. M is transversal.
Proof.It follows from Theorem 4 and Lemma 7.
In the following results we analyze the bases of L(S P ).So, S, R, P, Q and S are as the in previous subsections.Proof.By Corollary 3, there exists L ∈ G (P 2 (K)) such that M = L S P .Now, by Lemma 18, L(S P ) = L(M).Taking into account that {m 1 , m 2 , m 3 } are linearly independent, we get the result.
The previous results show that the solution to our problem lies in L(S P ).However, knowing L(S P ) implies knowing S P , which is essentially our goal.In the following, we see how to achieve L(S P ) by simply knowing B(S P ) and the base point multiplicities; note that, under the hypotheses of this section, this information is given by P. Definition 7. Let M be a birational transformation of P 2 (K).We define the linear system of base points of M, and we denote it by L (M), as the linear system of curves, of degree deg(M), Observe that L (M) is the deg(M)-linear system associated to the effective divisor A∈B(M) mult(A, B(M)) • A Remark 6.We observe that if M satisfies Conditions 1, in particular S P , then L (M) is the deg(S P )-degree linear system generated by the effective divisor The following lemma is a direct consequence of the definition above.Lemma 19.Let M be a birational transformation of P Next lemma relates the L(M) and L (M).Lemma 20.If M is a transversal birational map of P 2 (K), then L(M) ⊂ L (M).
Proof.Let M = (m 1 : m 2 : m 3 ).By Lemmas 17 and 20, we have that dim(L (M)) ≥ 2. Let us assume that dim(L (M)) = k > 2 and let {n 1 , . . ., n k+1 } be a basis of L (M) where Now, we take three points in P 2 (K) that will be crucial later.For their construction, we first consider an open Zariski subset Σ ⊂ P 2 (K) where M : Σ → M(Σ) ⊂ P 2 (K) is a bijective map.Then, M(Σ) is a constructible set of P 2 (K) (see e.g.Theorem 3.16 in [7]).Thus, P 2 (K) \ M(Σ) can only contain finitely many lines.On the other hand, we consider the open subset Ω 2 ⊂ G (P 2 (K)) introduced in Lemma 4, and we take L = (L 1 : and since M(B 1 ), M(B 2 ), M(B 3 ) are not collinear we get that a 1 = a 2 = a 3 = 0 that is a contradiction.Therefore, C (f ) and C (g) do not share components.Thus, by Bézout's theorem the number of intersections of C (f ) and C (g), properly counted, is deg(M) 2 .In addition, we oberve that f ∈ L(M) ⊂ L (M) (see Lemma 20 Proof.By Lemma 20, L(M) ⊂ L (M).Thus, using Lemmas 17 and 21, we get the result.
transversality of parametrizations.In addition, we observe that we require to the input parametrization to be proper (i.e.birational).This can be checked for instance using the algorithms in [12].

Algorithm 1 Transversality of a Parametrization
Require: A rational proper projective parametrization P( t ) of an algebraic surface S .return "P is not transversal".4: end if 5: if ∀ A ∈ B(P), gcd(T 1 , T 2 , T 3 , T 4 ) = 1, where T i is the product of the tangents, counted with multiplicities, to C (p i ) at A, then 6: return "P is transversal" else return "P is not transversal".7: end if Observe that Step 2 in Algorithm 1 provides a first direct filter to detect some non-transversal parametrizations, and Step 5 applies the characterization in Lemma 10.This justifies the next theorem Theorem 6. Algorithm 1 is correct.
The following algorithm is the central algorithm of the paper.

3:
Compute the inverse L −1 of L and return "P(L −1 ) is a rational proper polynomial parametrization of S ". 4: end if 5: Apply Algorithm 1 to check whether P is transversal.In the affirmative case go to the next Step.Otherwise return "Algorithm 2 is not applicable".10: Compute Q( t ) = (q 1 : q 2 : q 3 : q 4 ), where Q( t ) = P(R( t )).
11: if q 4 ( t ) is of the form (a 1 t 1 + a 2 t 2 + a 3 t 3 ) deg(Q) , then 12: return "Q(L −1 ) (where L is as in Step 2) is a rational proper polynomial parametrization of S " else return "S does not admit a polynomial proper parametrization with empty base locus".13: end if Theorem 7. Algorithm 2 is correct.
Proof.For the correctness of the first steps (1-4) we refer to the preamble in Section 4 where the almost polynomial parametrizations are treated.For the rest of the steps, we use the notation introduced in Section 4 and we assume the hypotheses there, namely, Q(S P ) = P and B(Q) = ∅.Since P is transversal, by Theorem 1, we have that S P is transversal.Now, by Theorem 5, L = L(S P ).In this situation, by Theorem 4, S = L S P for some L ∈ G (P 2 (K)).Therefore, P(R) has to be almost polynomial, and hence the last step generates a polynomial parametrization.If the fourth component of Q, namely q 4 is not the power of a linear form, then B(Q) = ∅.Remark 7. Let us comment some consequences and computational aspects involved in the execution of the previous algorithms.locus, and hence an algorithm to avoid the two complications mentioned above, if it is possible.For this purpose, we have had to introduce, and indeed impose, the notion of transversal base locus.This notion directly affects to the transversality of the tangents at the base points of the algebraic plane curves V i or W i (see (2.7) and (2.3)).This, somehow, implies that in general one may expect transversality in the input.In any case, we do deal here with the non-transversal case and we leave it as an open problem.We think that using the ideas, pointed out by J. Schicho in [20], on blowing up the base locus, one might transform the given problem (via a finite sequence of Cremone transformations and projective transformations) into the case of transversality.

Lemma 12 .
Let P and Q be two birational projective parametrizations of the same surface S such that Q(S) = P and B(Q) = ∅.It holds that 1. B(S) = B(P).