# Discriminants of complete intersection space curves

In this paper, we develop a new approach to the discrimi-nant of a complete intersection curve in the 3-dimensional projective space. By relying on the resultant theory, we first prove a new formula that allows us to define this discrimi-nant without ambiguity and over any commutative ring, in particular in any characteristic. This formula also provides a new method for evaluating and computing this discrimi-nant efficiently, without the need to introduce new variables as with the well-known Cayley trick. Then, we obtain new properties and computational rules such as the covariance and the invariance formulas. Finally, we show that our definition of the discriminant satisfies to the expected geometric property and hence yields an effective smoothness criterion for complete intersection space curves. Actually, we show that in the generic setting, it is the defining equation of the discriminant scheme if the ground ring is assumed to be a unique factorization domain.

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## 1 Introduction

Discriminants are central mathematical objects that have applications in many fields. Let be a field and suppose given integers and . Let be the set of all -uples of homogeneous polynomials in the polynomial ring of degree respectively. Consider the subset of corresponding to those -uples of homogeneous polynomials that define an algebraic subvariety in which is not smooth and of codimension . It is well-known that is an irreducible hypersurface provided for some , or provided (in which case is nothing but the resultant variety) [9]. The discriminant polynomial is then usually defined as an equation of . It is a homogeneous polynomial in the coefficients of each polynomial whose vanishing provides a smoothness criterion [9, 2]. This geometric approach to discriminants yields a beautiful theory with many remarkable results (e.g. [9]). However, whereas there are strong interests in computing with discriminants (e.g. [8, 15, 14, 18, 5]), including in the field of number theory, this approach is not tailored to develop the required formalism. For instance, having the discriminant defined up to a nonzero multiplicative constant is an important drawback, especially when computing over fields of positive characteristic. Another point is about the computation of discriminants: it is usually done by means of the famous Cayley trick that requires to introduce new variables, which has a bad effect on the computational cost.

In some cases, there exist an alternative to the above geometric definition of discriminants. In the case , which corresponds to resultants, there is a huge literature where the computational aspects are treated extensively. In particular, a vast formalism is available and many formulas allow to compute resultants, as for instance the well-known Macaulay formula (e.g. [10, 11, 6]). When the theory becomes much more delicate. Nevertheless, for both cases (hypersurfaces) and (finitely many points) discriminants can be defined rigorously and their formalism has been developed. The case goes back to Demazure [7, 8] and the case has been initiated by Krull [12, 13]. In both cases, the discriminant is defined by means of resultants, via a universal formula. This allows to develop the formalism, to obtain useful computational rules and also to compute it efficiently by taking advantage of the Macaulay formula for resultants; see [4] for more details.

The goal of this paper is to provide a similar treatment in the case . Our approach relies on the characterization of this discriminant by means of a universal formula where resultants and discriminants of finitely many points appear. As far as we know, this formula is new and provide the first (efficient) method to compute the discriminant of a complete intersection curve over any ring. In particular, we provide a closed formula that allows to compute it as a ratio of determinants. We emphasize that the computations are done in dimension at most 3, that is to say that there is no need to introduce new variables as with the Cayley trick. We mention that the problem of studying and computing discriminants goes back to the remarkable paper [17] of Sylvester in 1864. The case was the last remaining case to complete the picture in .

Before going into further details, we provide an example to illustrate the contribution of this paper. The Clebsch cubic projective surface is defined by the homogeneous polynomial

 f1:=13(4∑i=1x3i−(4∑i=1xi)3)∈Z[x1,x2,x3,x4].

By [4, Definiton 4.6] and the Macaulay formula, we get

 35⋅Disc(f1)=Res(∂1f1,…,∂4f1)=−35×5.

Thus, and we recover that the Clebsch surface is smooth except in characteristic 5. Now, consider the family of quadratic forms

 f2:=ax21+x1x2+x22+x23+x24∈Z[a][x1,x2,x3,x4].

The formula (4) we will prove in this paper allows to compute the discriminant of the intersection curve between the Clebsch surface and these quadratic forms; we get

 Disc(f1,f2)=2a(110592a7+442944a6+1163408a5+1303260a4+416575a3+238468a2−33924a−4448)⋅(3456a5+8208a4+10656a3+14069a2+11134a+3176)2.

In characteristic 5, the Clebsch surface is singular at the point . So, if the surface defined by the equation goes through then their intersection curve will be singular at . In general, this is not the case. Indeed, we have that

 Disc(f1,f2)=2a(2a7+4a6+3a5+3a2+a+2)⋅(a5+3a4+a3+4a2+4a+1)2mod5.

Now, if is specialized to then we force the surfaces defined by to go through . Applying this specialization the above formula, we obtain

 Disc(f1,f2|a=5b−4)=2⋅(5b−4)⋅5⋅(31250b7−…)(3125b5−…)2mod5

so that this discriminant now vanishes modulo 5 as expected.

The paper is organized as follows. In Section 2 we prove a new formula, based on resultants, that is used to provide a new definition of the discriminant of a complete intersection space curve. Then, in Section 3 we give some properties and computational rules of this discriminant by relying on the existing formalism of resultants. Finally, in Section 4 we show that our definition is correct in the sense that it satisfies to the expected geometric property, in particular it yields a universal and effective smoothness criterion which is valid in arbitrary characteristic.

In the sequel, we will rely heavily on the theory of resultants and its formalism, including the Macaulay formula. We refer the reader to [10] and [6, Chapter 3]. We will also assume some familiarity with the definition of discriminants in the case for which we refer the reader to [4, §3.1]. Resultants and discriminants will be denoted by and respectively.

## 2 Definition and formula

Suppose given two positive integers and consider the generic homogeneous polynomials in the four variables

 f1:=∑|α|=d1U1,αxα,  f2:=∑|α|=d2U2,αxα.

We denote by the universal ring of coefficients and we define the polynomial ring . The partial derivative of the polynomial with respect to the variable will be denoted by . Moreover, given four homogeneous polynomials in the variables , the determinant of their Jacobian matrix will be denoted by

###### Theorem 1

Using the above notation, assume that . Let be three linear forms

 l(x––)=4∑i=1lixi, m(x––)=4∑i=1mixi, n(x––)=4∑i=1nixi,

and denote by the polynomial ring extension of with the coefficients ’s, ’s and ’s of the linear forms . Then, there exists a unique polynomial in , denoted by and called the universal discriminant of and , which is independent of the coefficients of and that satisfies to the following equality in :

 Res(f1,f2,J(f1,f2,l,m),J(f1,f2,l,n))=Disc(f1,f2)⋅Res(f1,J(f1,l,m,n)),f2,J(f2,l,m,n))Disc(f1,f2,l).

By convention, if we set

 Res(f1,J(f1,l,m,n)),f2,J(f2,l,m,n))=J(fj,l,m,n)Dj

where .

Given a commutative ring and two homogeneous polynomials

 g1:=∑|α|=d1u1,αxα,  g2:=∑|α|=d2u2,αxα

in of degree , respectively, the map of rings from to which sends to and leave each variable invariant, is called the specialization map of the universal polynomials to the polynomials , as .

###### Definition 1

Suppose given a commutative ring , two positive integers such that and two homogeneous polynomials in of degree respectively. Denoting by the specialization map as above, we define the discriminant of the polynomials as

 Disc(g1,g2)=Disc(ρ(f1),ρ(f2)):=ρ(Disc(f1,f2))∈R.

[of Theorem 1] To prove the claimed formula, one can assume that is the universal ring of the coefficients of the polynomials over the integers.

Our first step is to show that divides

 R:=Res(f1,f2,J(f1,f2,l,m),J(f1,f2,l,n)).

For that purpose, denote by the ideal of generated by and all the 3-minors of the Jacobian matrix of the polynomials . We also define the ideal and we recall from [4, Theorem 3.23] that is a generator of the ideal of inertia forms of , i.e. the ideal

 (D:m∞)∩A={p∈A such that ∃ν∈N:mν⋅p⊂D}.

Now, from the similar characterization of the resultant by means of inertia forms [10, Proposition 2.3], we deduce that there exists an integer such that

 mN⋅R⊂(f1,f2,J(f1,f2,l,m),J(f1,f2,l,n))⊂A[x––].

But and belong to , so we deduce that It follows that is an inertia form of and it is hence divisible by .

Our second step is to prove that the resultant

 R0:=Res(f1,J(f1,l,m,n),f2,J(f2,l,m,n))

divides . For all , we obviously have that

 det⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝∂if1∂1f1∂2f1∂3f1∂4f1∂if2∂1f2∂2f2∂3f2∂4f2lil1l2l3l4mim1m2m3m4nin1n2n3n4⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠=0.

By developing each of these determinants with respect to their first column, we get the linear system

 ⎛⎜ ⎜ ⎜⎝l1m1n1l2m2n2l3m3n3l4m4n4⎞⎟ ⎟ ⎟⎠⎛⎜⎝J(f1,f2,m,n)−J(f1,f2,l,n)J(f1,f2,l,m)⎞⎟⎠=⎛⎜ ⎜ ⎜⎝∂1f2∂2f2∂3f2∂4f2⎞⎟ ⎟ ⎟⎠J(f1,l,m,n)−⎛⎜ ⎜ ⎜⎝∂1f1∂2f1∂3f1∂4f1⎞⎟ ⎟ ⎟⎠J(f2,l,m,n).

The matrix of this linear system is nothing but the transpose of the Jacobian matrix of the polynomials . Denote by any of its 3-minor. Then, Cramer’s rules show that both polynomials and belong to the ideal generated by the polynomials and . Therefore, the divisibility property of resultants [10, §5.6] implies that divides

 Res(f1,f2,ΔJ(f1,f2,l,m),ΔJ(f1,f2,l,n))=Δr⋅R,

where ; observe that is independent of . As it is well-known, is an irreducible polynomial, being the determinant of a matrix of indeterminates. Therefore, to conclude this second step we have to show that does not divide . For that purpose, we consider the specialization of the coefficients of and so that

 η(f1)=d1∏i=1pi(x––),  η(f2)=d2∏i=1qi(x––), (1)

where the ’s and ’s are generic linear forms; we add their coefficients as new variables to . Using the multiplicativity property of resultants, a straightforward computation yields the following irreducible factorization formula

 η(R0)=(d1∏i=1J(pi,l,m,n))d2(d2−1)⋅(d2∏i=1J(qi,l,m,n))d1(d1−1)⋅∏i,j,r,sRes(pi,pj,qr,qs)4 (2)

where the last product runs over the integers , with and with . Since and is not a factor in the above formula, we deduce that does not divide .

The third step in this proof is to show that the discriminant and the resultant are coprime polynomials in . Since is irreducible [4, Theorem 3.23]), we have to show that it does not divide . Consider again the specialization given by (1) and assume that is a factor in . Then, since is independent on the coefficients of the linear forms and , must contain some factors that depend on the coefficient of but not on and . However, the decomposition formula (2) shows that contains only irreducible factors that do depend on three linear forms , or on none of them. Therefore, we deduce that does not divide .

To conclude this proof, we observe that the previous results show that divides . Moreover, straightforward computations shows that and are both homogeneous polynomials with respect to the coefficients of of the same degree, and the same happens to be true with respect to the coefficients of and .

To compute the discriminant it is much more efficient to specialize the formula in Theorem 1 by giving to the linear forms some specific values, for instance a single variable. Consider the Jacobian matrix associated to the polynomials

and its minors that we will denote by

 Ji,j(f1,f2):=∣∣∣∂if1∂jf1∂if2∂jf2∣∣∣. (3)

In the sequel, given a (homogeneous) polynomial , for all we will denote by the polynomial in which the variable is set to zero.

###### Corollary 2

Suppose given a commutative ring , two positive integers such that and two homogeneous polynomials in of degree respectively. Then,

 Res(g1,g2,J1,2(g1,g2),J2,3(g1,g2))=(−1)d1d2⋅Disc(g1,g2)Res(g1,∂2g1,g2,∂2g2)Disc(¯¯¯¯¯g14,¯¯¯¯¯g24). (4)

Straightforward by applying the formula in Theorem 1 with , , and . We notice that

 Disc(f1,f2,x4)=(−1)d1d2Disc(¯¯¯f41,¯¯¯f42) (5)

by property of the discriminant of three homogeneous polynomials in four variables [4, Proposition 3.13].

From a computational point of view the above formula allows to compute the discriminant of any couple of homogeneous polynomials as a ratio of determinants since all the other terms in (5) can be expressed as ratio of determinants by means of the Macaulay formula. There is no need to introduce new variables as in the Cayley trick and the formula is universal in the coefficients of the polynomials over the integers.

## 3 Properties and computational rules

In this section, we provide some properties and computational rules of the discriminant as defined in the previous section. In particular, we give precise formulas regarding the covariance and invariance properties. We also provide a detailed computation of a particular class of complete intersection curves in order to illustrate how our formalism allows to handle the discriminant and simplify its computation and evaluation over any ring of coefficients. In what follows, denotes a commutative ring.

### 3.1 First elementary properties

From Theorem 1, it is clear that the discriminant is homogeneous with respect to the coefficients of , respectively and that these degrees can easily be computed. As expected, we recover the degrees of the usual geometric definition of discriminant (see [17, 16, 2]).

###### Proposition 3 (Homogeneity)

The universal discriminant is homogeneous of degree with respect to the coefficient of where, setting and ,

 δ1=d2(3e21+2e1e2+e22),  δ2=d1(3e22+2e1e2+e21).

This is a straightforward computation from the defining equality (see Theorem 1), since the degrees of resultants and discriminants of finitely many points are known (see [10, Proposition 2.3] and [4, Proposition 3.9]).

###### Proposition 4 (Permutation of the polynomials)

Let be two homogeneous polynomials of degree and respectively, then

 Disc(g2,g1)=Disc(g1,g2).

This is a straightforward consequence of the similar property for resultants [10, §5.8] and discriminants of finitely many points [4, Proposition 3.12 i)].

###### Proposition 5 (Elementary transformations)

Let be four homogeneous polynomials in of degree respectively. Then,

 Disc(g1,g2+h2g1)=Disc(g1+h1g2,g2)=Disc(g1,g2).

This is a straightforward consequence of the invariance of resultants under elementary transformations [10, §5.9] and the invariance of discriminants of finitely many points under elementary transformations [4, Proposition 3.12].

### 3.2 Covariance and invariance

In this section, we give precise statements about two important properties of the discriminant: its geometric covariance and its geometric invariance under linear change of variables.

###### Proposition 6 (Covariance)

Suppose given two homogeneous polynomials in of the same degree and a square matrix with coefficients in , then

 Disc(u1,1g1+u1,2g2,u2,1g1+u2,2g2)=det(φ)6d(d−1)2Disc(g1,g2).

By definition, it is sufficient to prove this formula in the universal setting. For simplicity, we use the formula (4). Setting and , we observe that so that

 Res(~f1,~f2,J1,2(~f1,~f2),J2,3(~f1,~f2))=det(φ)4d2(d−1)Res(~f1,~f2,J1,2,J2,3).

In addition, by the covariance of resultants [10, §5.11],

 Res(~f1,~f2,J1,2,J2,3)=det(φ)4d(d−1)2Res(f1,f2,J1,2,J2,3)

so that we deduce that

 Res(~f1,~f2,J1,2(~f1,~f2),J2,3(~f1,~f2))=det(φ)4d(d−1)(2d−1)Res(f1,f2,J1,2,J2,3).

The covariance of resultants also shows that

 Res(~f1,∂2~f1,~f2,∂2~f2)=det(φ)d(d−1)(2d−1)Res(f1,∂2f1,f2,∂2f2)

and the covariance property of discriminants of finitely many points [4, Proposition 3.18] yields

 Disc(¯¯¯¯¯¯~f14,¯¯¯¯¯¯~f24)=det(φ)3d(d−1)Disc(¯¯¯f41,¯¯¯f42).

From all these equalities and (4), we deduce the claimed formula.

###### Proposition 7 (Invariance)

Let be two homogeneous polynomials in of degree and let be a square matrix with entries in . For all homogeneous polynomial we set

 g∘φ(x1,x2,x3,x4):=g(4∑j=1c1,jxj,…,4∑j=1c4,jxj).

Then, we have that

where , , .

As always, to prove this formula we may assume that we are in the universal setting, and being the universal homogeneous polynomials of degree and respectively. We will also denote by three generic linear form and by the generic square matrix of size .

Applying Theorem 1, we get the equality

 Res(f1∘φ,f2∘φ,J(f1∘φ,f2∘φ,l∘φ,m∘φ),J(f1∘φ,f2∘φ,l∘φ,n∘φ))=Disc(f1∘φ,f2∘φ)Disc(f1∘φ,f2∘φ,l∘φ)Res(f1∘φ,J(f1∘φ,l∘φ,m∘φ,n∘φ),f2∘φ,J(f2∘φ,l∘φ,m∘φ,n∘φ)) (6)

(observe that are all linear forms in ). Now, by [4, Proposition 3.27], we know that

 Disc(f1∘φ,f2∘φ,l∘φ)=det(φ)d1d2(e1+e2)Disc(f1,f2,l).

Also, by the chain rule formula for the derivative of the composition of functions, we have the formulas

 J(f1∘φ,f2∘φ,l∘φ,m∘φ)) =J(f1,f2,l,m)∘[φ]⋅det(φ) J(f1∘φ,f2∘φ,l∘φ,n∘φ)) =J(f1,f2,l,n)∘[φ]⋅det(φ) J(fi∘φ,l∘φ,m∘φ,n∘φ)) =J(fi,l,m,n)∘[φ]⋅det(φ)

from we deduce, using the invariance of resultants [10, §5.13] and their homogeneity, that

 Res(f1∘φ,J(f1∘φ,l∘φ,m∘φ,n∘φ),f2∘φ,J(f2∘φ,l∘φ,m∘φ,n∘φ))=det(φ)d1d2e1e2Res(f1,J(f1,l,m,n),f2,J(f2,l,m,n))

and

 Res(f1∘φ,f2∘φ,J(f1∘φ,f2∘φ,l∘φ,m∘φ),J(f1∘φ,f2∘φ,l∘φ,n∘φ))=det(φ)d1d2(e1+e2)2Res(f1,f2,J(f1,f2,l,m),J(f1,f2,l,n)).

From here, the claimed formula follows from the substitution of the above equalities in (6) and the comparison with the formula given in Theorem 1.

###### Corollary 8

The discriminant is invariant under permutation of the variables .

It follows from Proposition 7 since is even.

### 3.3 Discriminant of a plane curve

Given a plane curve, we prove that its discriminant as defined in Section 2, is compatible with its discriminant as a plane hypersurface [4, §4.2].

###### Lemma 9

Let be a homogeneous polynomial in of degree . Then, for all we have that

 Disc(g,xi)=Disc(¯¯¯gi).

By definition, it is sufficient to prove this equality in the case where is replaced by the generic homogeneous polynomial of degree . We apply Theorem 1 with , , that are chosen so that as sets. We obtain the equality

 R:=Res(f,xi,±∂tf,±∂sf)=Disc(f,xi)Disc(f,xi,xr).

Since the degree of and one of its partial derivative are consecutive integers, their product is always an even integer. It follows by standard properties of resultants that does not depend on the sign of its entry polynomials, nor on their order, nor on the reduction of the variables, so that we have

 R=Res(¯¯¯fi,∂t¯¯¯fi,∂s¯¯¯fi)=Res(¯¯¯fi,∂s¯¯¯fi,∂t¯¯¯fi).

Now, by property of discriminants, in particular (5) and its invariance under permutation of variables [4, Proposition 3.12], we have

 Disc(f,xi,xr)=(−1)dDisc(¯¯¯fi,xr)=Disc(¯¯¯fi,r).

Finally, [4, Proposition 4.7] shows that

 Disc(¯¯¯fi,r)Disc(¯¯¯fi)=Res(¯¯¯fi,∂r¯¯¯fi,∂s¯¯¯fi)

and the claimed equality is proved.

###### Proposition 10

Let be a homogeneous polynomial of degree and be a linear form in . Then, for all we have that

 l2d(d−1)2iDisc(g,l)=Disc(g(lix1−δ1il(x––),⋯,lix4−δ4il(x––)))

where stands for the Kronecker symbol.

We assume that we are in the generic setting, which is sufficient to prove this corollary. Consider the linear change of coordinates given by the matrix defined as follows: its

row is the vector

and its other rows are filled with zeros except on the diagonal where we put . Then, it is not hard to check that

 g∘φi∘φi(x––)=g(l2i⋅–x)=l2dig(x––).

Therefore, by Proposition 3 we obtain

 Disc(g∘φi∘φi,l)=Disc(l2dig,l)=l6d(d−1)2iDisc(g,l). (7)

On the other hand, since , Proposition 7 yields

 Disc(g∘φi∘φi,l)=det(φi)d(d−1)2Disc(g∘φi,xi)=l