A NEW ALGORITHM FOR IMPLICITIZING A PARAMETRIC ALGEBRAIC SURFACE

Given a parametric representation of an algebraic projective surface S of the ordinary space we give a new algorithm for finding the implicit cartesian equation of S. The algorithm is based on finding a suitable finite number of points on S and computing, by linear algebra, the equation of the surface of least degree that passes through the points. AMS Subject Classification: 14Q10


Introduction
The problem of determining an implicit representation of an irreducible parametric algebraic surface in P 3 has been classically faced by elimination theory and can be solved by the computation of Gröbner bases or by the computation of resultants (see for example [2] and [4]).These methods have been implemented in various softwares of Computer Algebra.For doing these computations on the computer the surface has to be parametrically represented by polynomials with coefficients rational numbers (or more simply integers) and the result is an equation whose coefficients are rational numbers (or integers).Another possible representation of the coefficients of the parametric polynomials is in the set of integers modulo a prime number.In this paper, we introduce an alternative method that reconducts the implicitation of S to the computation of the equation of the surface that contains a suitable set of points on S. The algorithm is based on the fact that, if we find an appropriate set T of general points on S, a polynomial F vanishes on the polynomials that represent parametrically S if and only if F vanishes on T .This gives rise to a homogeneous linear system of equations.A non null solution of the system (which can be found very easily by Gauss reduction) gives the coefficients of the implicit equation of S.
Throughout this paper S will be an algebraic surface in the projective space P 3 over an algebraically closed field k.

Computing the Implicit Equation of a Parametric Surface by Points
Definition 2.1.We say that an irreducible (algebraic) surface S ⊂ P 3 is parametric if there exists a rational map Φ : given by homogeneous polynomials f i (t 0 , t 1 , t 2 ), i = 0, 1, 2, 3 of the same degree r whose image is a dense subset of V .We set Since S is a surface of P 3 it is also the set of solutions of a homogeneous polynomial F ∈ K[X 0 , ..., X 3 ] in 4 variables.The corresponding equation F = 0 is said to implicitize the parametric surface S and F is said to be the polynomial defining S (F is unique modulo constant terms of k).
Lemma 2.2.The degree of F is less or equal to r 2 .
Proof.Let F be the polynomial defining S. The degree of F is the maximum number of points of intersection with a line If we substitute the parametric equations of S in the equations of the line we get a system of two polynomials of degree r in the variables t 0 , t 1 , t 2 , t 3 which has, as solutions, the points of intersection of two plane curves of degree r.By Bezout theorem these points are at most r 2 , and we have the claim.

Definition 2.3. [5]
A set S = {P 1 , . . ., P N } ⊂ P 2 of N = n+2 2 points in the plane is in generic position if it is not contained in a projective curve of degree ≤ n.
Example 2.4.Six points that do not lie on a conic and ten points that do not lie on a cubic are in generic position.Lemma 2.5.Let S = {(x 0 , x 1 , x 2 , x 3 ) ∈ P 3 | x i = f i (t 0 , t 1 , t 2 ), i = 0, 1, 2, 3} be a parametric surface given by homogeneous polynomials of the same degree r.Let d be a fixed integer, and P 1 , . . ., P N be N = rd+2 2 points of P 2 in generic position such that Q i = Φ(P i ) form a set T of N points of P 3 .If F is a homogeneous polynomial of K[X 0 , ..., X 3 ] with degree d ′ ≤ d then F vanishes on S if and only if F vanishes on T .In particular S is not contained in a surface of degree ≤ d if and only if such is T .
Proof.Clearly, since T ⊂ S, a a polynomial vanishing on S vanishes on T .We prove the converse by contradiction.Let F be a homogeneous polynomial of degree d ′ vanishing on the set of points T and assume that there exists a point P ∈ P 2 such that F (Φ(P )) = 0. Then F (f 0 , f 1 , f 2 , f 3 ) is a non-zero polynomial of degree d ′ r ≤ dr vanishing on the N = rd+2 2 points P i and this contradicts the assumption of generic position.By Lemma 1.5 the computation of the equation that defines S is reconducted to the computation of the equation that vanishes on a finite set of points.Before starting the computation one has to find a set of points in generic position.Almost all sets of N = d+m m points S = {P 1 , . . ., P N } ⊂ P m are in generic position ( [3]).It follows that a random choice for the coordinates of the P i gives points in generic position.A systematic way of finding points in generic position in P m is given by the following two results.Proposition 2.6.Let d be a positive integer and a ir ∈ k, i = 0, . . ., d, r = 1, 2. Suppose a i 1 r = a i 2 r , for any i 1 = i 2 , and r = 1, 2. No curve of A 2 with degree ≤ d contains the N = d+2 2 affine points of the set S = {(a Proof.We have to prove that any polynomial F (X, Y ), of degree less then d, vanishing on all the points of S is null.By the hypotheses the polynomial G(X) = F (X, a 02 ), of degree less then d, has d + 1 roots a 01 , . . ., a 0d thus it is null.Hence F (X, Y ) = (X − a 02 )F 1 and F 1 has degree less then d − 1.By applying the same argument to the polynomial F 1 (X, a 12 ) and so on, we get F (X, Y ) = d n=0 (X 2 − a n2 F ′ so that F has degree greater then d or it is null.