Discriminants of complete intersection space curves
aa r X i v : . [ c s . S C ] F e b Discriminants of complete intersection space curves
Laurent Busé
Université Côte d’Azur, Inria2004 route des Lucioles,06 902 Sophia Antipolis, France [email protected] Ibrahim Nonkané
Université Ouaga 2, IUFIC,12 BP 417 Ouagadougou 12, Burkina Faso [email protected]
ABSTRACT
In this paper, we develop a new approach to the discrimi-nant of a complete intersection curve in the 3-dimensionalprojective space. By relying on the resultant theory, we firstprove a new formula that allows us to define this discrimi-nant without ambiguity and over any commutative ring, inparticular in any characteristic. This formula also providesa new method for evaluating and computing this discrimi-nant efficiently, without the need to introduce new variablesas with the well-known Cayley trick. Then, we obtain newproperties and computational rules such as the covarianceand the invariance formulas. Finally, we show that our defi-nition of the discriminant satisfies to the expected geometricproperty and hence yields an effective smoothness criterionfor complete intersection space curves. Actually, we showthat in the generic setting, it is the defining equation of thediscriminant scheme if the ground ring is assumed to be aunique factorization domain.
1. INTRODUCTION
Discriminants are central mathematical objects that haveapplications in many fields. Let K be a field and supposegiven integers 1 ≤ c ≤ n and 1 ≤ d , . . . , d c . Let S be theset of all c -uples of homogeneous polynomials f , . . . , f c inthe polynomial ring K [ x , . . . , x n ] of degree d , . . . , d c respec-tively. Consider the subset D of S corresponding to those c -uples of homogeneous polynomials that define an algebraicsubvariety in P n − K which is not smooth and of codimension c . It is well-known that D is an irreducible hypersurfaceprovided d i ≥ i , or provided c = n (in which case D is nothing but the resultant variety) [8]. The discrim-inant polynomial is then usually defined as an equation of D . It is a homogeneous polynomial in the coefficients of eachpolynomial f i whose vanishing provides a smoothness crite-rion [8, 1]. This geometric approach to discriminants yieldsa beautiful theory with many remarkable results (e.g. [8]).However, whereas there are strong interests in computingwith discriminants (e.g. [7, 14, 13, 17, 4]), including in the . field of number theory, this approach is not tailored to de-velop the required formalism. For instance, having the dis-criminant defined up to a nonzero multiplicative constantis an important drawback, especially when computing overfields of positive characteristic. Another point is about thecomputation of discriminants: it is usually done by meansof the famous Cayley trick that requires to introduce newvariables, which has a bad effect on the computational cost.In some cases, there exist an alternative to the above geo-metric definition of discriminants. In the case c = n , whichcorresponds to resultants, there is a huge literature wherethe computational aspects are treated extensively. In par-ticular, a vast formalism is available and many formulas al-low to compute resultants, as for instance the well-knownMacaulay formula (e.g. [9, 10, 5]). When c < n the theorybecomes much more delicate. Nevertheless, for both cases c = 1 (hypersurfaces) and c = n − c = 1 goes back to Demazure[6, 7] and the case c = n − c, n ) = (2 , c, n ) = (2 ,
4) was the lastremaining case to complete the picture in P .Before going into further details, we provide an example toillustrate the contribution of this paper. The Clebsch cubicprojective surface is defined by the homogeneous polynomial f := 13 X i =1 x i − ( X i =1 x i ) ! ∈ Z [ x , x , x , x ] . y [3, Definiton 4.6] and the Macaulay formula, we get3 · Disc( f ) = Res( ∂ f , . . . , ∂ f ) = − × . Thus, Disc( f ) = − f := ax + x x + x + x + x ∈ Z [ a ][ x , x , x , x ] . The formula (4) we will prove in this paper allows to com-pute the discriminant of the intersection curve between theClebsch surface and these quadratic forms; we getDisc( f , f ) = 2 a (cid:0) a + 442944 a + 1163408 a +1303260 a + 416575 a + 238468 a − a − (cid:1) · (cid:0) a + 8208 a + 10656 a + 14069 a + 11134 a + 3176 (cid:1) . In characteristic 5, the Clebsch surface f = 0 is singular atthe point P = (1 : 1 : 1 : 1). So, if the surface defined bythe equation f = 0 goes through P then their intersectioncurve will be singular at P . In general, this is not the case.Indeed, we have thatDisc( f , f ) = 2 a (cid:0) a + 4 a + 3 a + 3 a + a + 2 (cid:1) · (cid:0) a + 3 a + a + 4 a + 4 a + 1 (cid:1) mod 5 . Now, if a is specialized to 5 b − f = 0 to go through P . Applying this special-ization the above formula, we obtainDisc( f , f | a =5 b − ) = 2 · (5 b − · · (cid:0) b − . . . (cid:1) (cid:0) b − . . . (cid:1) mod 5so that this discriminant now vanishes modulo 5 as expected.The paper is organized as follows. In Section 2 we prove anew formula, based on resultants, that is used to provide anew definition of the discriminant of a complete intersectionspace curve. Then, in Section 3 we give some propertiesand computational rules of this discriminant by relying onthe existing formalism of resultants. Finally, in Section 4we show that our definition is correct in the sense that itsatisfies to the expected geometric property, in particular ityields a universal and effective smoothness criterion whichis valid in arbitrary characteristic.In the sequel, we will rely heavily on the theory of re-sultants and its formalism, including the Macaulay formula.We refer the reader to [9] and [5, Chapter 3]. We will alsoassume some familiarity with the definition of discriminantsin the case c = n − § − ) andDisc( − ) respectively.
2. DEFINITION AND FORMULA
Suppose given two positive integers d , d and considerthe generic homogeneous polynomials in the four variables x = ( x , x , x , x ) f := X | α | = d U ,α x α , f := X | α | = d U ,α x α . We denote by A = Z [ U i,α ; i = 1 , , | α | = d i ] the univer-sal ring of coefficients and we define the polynomial ring C = A [ x ]. The partial derivative of the polynomial f i with respect to the variable x j will be denoted by ∂ j f i . More-over, given four homogeneous polynomials p , p , p , p inthe variables x , the determinant of their Jacobian matrixwill be denoted by J ( p , p , p , p ) := det ( ∂ j p i ) i,j =1 ..., . Theorem Using the above notation, assume that d + d ≥ . Let l, m, n be three linear forms l ( x ) = X i =1 l i x i , m ( x ) = X i =1 m i x i , n ( x ) = X i =1 n i x i , and denote by A ′ the polynomial ring extension of A with thecoefficients l i ’s, m j ’s and n k ’s of the linear forms l, m, n .Then, there exists a unique polynomial in A , denoted by Disc( f , f ) and called the universal discriminant of f and f , which is independent of the coefficients of l, m, n and thatsatisfies to the following equality in A ′ : Res ( f , f , J ( f , f , l, m ) , J ( f , f , l, n )) = Disc( f , f ) · Res ( f , J ( f , l, m, n )) , f , J ( f , l, m, n )) Disc ( f , f , l ) . By convention, if d j = 1 we set Res ( f , J ( f , l, m, n )) , f , J ( f , l, m, n )) = J ( f j , l, m, n ) D j where D j = ( d + d − d j )( d + d − d j − . Given a commutative ring R and two homogeneous poly-nomials g := X | α | = d u ,α x α , g := X | α | = d u ,α x α in R [ x ] of degree d , d respectively, the map of rings ρ from A [ x ] to R [ x ] which sends U i,α to u i,α and leave each variable x i invariant, is called the specialization map of the universalpolynomials f , f to the polynomials g , g , as ρ ( f i ) = g i . Definition Suppose given a commutative ring R , twopositive integers d , d such that d + d ≥ and two homoge-neous polynomials g , g in R [ x ] of degree d , d respectively.Denoting by ρ the specialization map as above, we define thediscriminant of the polynomials g , g as Disc( g , g ) = Disc( ρ ( f ) , ρ ( f )) := ρ (Disc( f , f )) ∈ R. Proof of Theorem 1.
To prove the claimed formula,one can assume that A ′ is the universal ring of the coef-ficients of the polynomials f , f , l, m, n over the integers.Our first step is to show that Disc( f , f , l ) divides R := Res( f , f , J ( f , f , l, m ) , J ( f , f , l, n )) . For that purpose, denote by D the ideal of A [ x ] generatedby f , f and all the 3-minors of the Jacobian matrix ofthe polynomials f , f , l . We also define the ideal m = ( x )and we recall from [3, Theorem 3.23] that Disc( f , f , l ) is agenerator of the ideal of inertia forms of D , i.e. the ideal( D : m ∞ ) ∩ A = { p ∈ A such that ∃ ν ∈ N : m ν · p ⊂ D} . Now, from the similar characterization of the resultant bymeans of inertia forms [9, Proposition 2.3], we deduce thatthere exists an integer N such that m N · R ⊂ ( f , f , J ( f , f , l, m ) , J ( f , f , l, n )) ⊂ A [ x ] . But J ( f , f , l, m ) and J ( f , f , l, n ) belong to D , so we de-duce that R ∈ ( D : m ∞ ) ∩ A. It follows that R is an inertiaform of D and it is hence divisible by Disc( f , f , l ).ur second step is to prove that the resultant R := Res ( f , J ( f , l, m, n ) , f , J ( f , l, m, n ))divides R . For all i = 1 , . . . ,
4, we obviously have thatdet ∂ i f ∂ f ∂ f ∂ f ∂ f ∂ i f ∂ f ∂ f ∂ f ∂ f l i l l l l m i m m m m n i n n n n = 0 . By developing each of these determinants with respect totheir first column, we get the linear system l m n l m n l m n l m n J ( f , f , m, n ) − J ( f , f , l, n ) J ( f , f , l, m ) = ∂ f ∂ f ∂ f ∂ f J ( f , l, m, n ) − ∂ f ∂ f ∂ f ∂ f J ( f , l, m, n ) . The matrix of this linear system is nothing but the transposeof the Jacobian matrix of the polynomials l, m, n . Denote by∆ any of its 3-minor. Then, Cramer’s rules show that bothpolynomials ∆ · J ( f , f , l, m ) and ∆ · J ( f , f , l, n ) belongto the ideal generated by the polynomials J ( f , l, m, n ) and J ( f , l, m, n ). Therefore, the divisibility property of resul-tants [9, § R dividesRes( f , f , ∆ J ( f , f , l, m ) , ∆ J ( f , f , l, n )) = ∆ r · R , where r = 2 d d ( d + d − x . As it is well-known, ∆ is an irreducible polynomial,being the determinant of a matrix of indeterminates. There-fore, to conclude this second step we have to show that ∆does not divide R . For that purpose, we consider the spe-cialization η of the coefficients of f and f so that η ( f ) = d Y i =1 p i ( x ) , η ( f ) = d Y i =1 q i ( x ) , (1)where the p i ’s and q j ’s are generic linear forms; we add theircoefficients as new variables to A ′ . Using the multiplicativityproperty of resultants, a straightforward computation yieldsthe following irreducible factorization formula η ( R ) = d Y i =1 J ( p i , l, m, n ) ! d ( d − · d Y i =1 J ( q i , l, m, n ) ! d ( d − · Y i,j,r,s Res( p i , p j , q r , q s ) (2)where the last product runs over the integers i, j = 1 , . . . , d ,with i < j and r, s = 1 , . . . , d with r < s . Since η (∆) = ∆and ∆ is not a factor in the above formula, we deduce that∆ does not divide R .The third step in this proof is to show that the discrimi-nant D ∞ := Disc( f , f , l ) and the resultant R are coprimepolynomials in A ′ . Since D ∞ is irreducible [3, Theorem3.23]), we have to show that it does not divide R . Consideragain the specialization η given by (1) and assume that D ∞ is a factor in R . Then, since D ∞ is independent on the coefficients of the linear forms m and n , η ( R ) must containsome factors that depend on the coefficient of l but not on m and n . However, the decomposition formula (2) showsthat η ( R ) contains only irreducible factors that do dependon three linear forms l, m, n , or on none of them. Therefore,we deduce that D ∞ does not divide R .To conclude this proof, we observe that the previous re-sults show that D ∞ R divides R . Moreover, straightforwardcomputations shows that D ∞ R and R are both homoge-neous polynomials with respect to the coefficients of l of thesame degree, and the same happens to be true with respectto the coefficients of m and n .To compute the discriminant it is much more efficient tospecialize the formula in Theorem 1 by giving to the lin-ear forms l, m, n some specific values, for instance a singlevariable. Consider the Jacobian matrix associated to thepolynomials f , f Jac( f , f ) := (cid:18) ∂ f ∂ f ∂ f ∂ f ∂ f ∂ f ∂ f ∂ f (cid:19) and its minors that we will denote by J i,j ( f , f ) := (cid:12)(cid:12)(cid:12)(cid:12) ∂ i f ∂ j f ∂ i f ∂ j f (cid:12)(cid:12)(cid:12)(cid:12) . (3)In the sequel, given a (homogeneous) polynomial p ( x ), forall j = 1 , . . . , p j the polynomial p inwhich the variable x j is set to zero. Corollary Suppose given a commutative ring R , twopositive integers d , d such that d + d ≥ and two homoge-neous polynomials g , g in R [ x ] of degree d , d respectively.Then, Res( g , g , J , ( g , g ) , J , ( g , g )) = ( − d d · Disc( g , g )Res( g , ∂ g , g , ∂ g )Disc (cid:0) g , g (cid:1) . (4) Proof.
Straightforward by applying the formula in The-orem 1 with l = x , m = x , and n = x . We notice thatDisc( f , f , x ) = ( − d d Disc (cid:16) f , f (cid:17) (5)by property of the discriminant of three homogeneous poly-nomials in four variables [3, Proposition 3.13].From a computational point of view the above formulaallows to compute the discriminant of any couple of homo-geneous polynomials g , g ∈ R [ x ] as a ratio of determinantssince all the other terms in (5) can be expressed as ratio ofdeterminants by means of the Macaulay formula. There isno need to introduce new variables as in the Cayley trick andthe formula is universal in the coefficients of the polynomialsover the integers.
3. PROPERTIES AND COMPUTATIONALRULES
In this section, we provide some properties and compu-tational rules of the discriminant Disc( f , f ) as defined inthe previous section. In particular, we give precise formu-las regarding the covariance and invariance properties. Wealso provide a detailed computation of a particular class ofcomplete intersection curves in order to illustrate how ourformalism allows to handle the discriminant and simplify itscomputation and evaluation over any ring of coefficients. Inwhat follows, R denotes a commutative ring. .1 First elementary properties From Theorem 1, it is clear that the discriminant Disc( f , f )is homogeneous with respect to the coefficients of f , respec-tively f and that these degrees can easily be computed. Asexpected, we recover the degrees of the usual geometric def-inition of discriminant (see [16, 15, 1]). Proposition 3 (Homogeneity).
The universal discrim-inant is homogeneous of degree δ i with respect to the coeffi-cient of f i where, setting e = d − and e = d − , δ = d (3 e + 2 e e + e ) , δ = d (3 e + 2 e e + e ) . Proof.
This is a straightforward computation from thedefining equality (see Theorem 1), since the degrees of resul-tants and discriminants of finitely many points are known(see [9, Proposition 2.3] and [3, Proposition 3.9]).
Proposition 4 (Permutation of the polynomials).
Let g , g ∈ R [ x , . . . , x ] be two homogeneous polynomialsof degree d and d respectively, then Disc ( g , g ) = Disc ( g , g ) . Proof.
This is a straightforward consequence of the sim-ilar property for resultants [9, § Proposition 5 (Elementary transformations).
Let g , g , h , h be four homogeneous polynomials in R [ x ] of de-gree d , d , d − d , d − d respectively. Then, Disc ( g , g + h g ) = Disc ( g + h g , g ) = Disc ( g , g ) . Proof.
This is a straightforward consequence of the in-variance of resultants under elementary transformations [9, § In this section, we give precise statements about two im-portant properties of the discriminant: its geometric covari-ance and its geometric invariance under linear change of vari-ables.
Proposition 6 (Covariance).
Suppose given two ho-mogeneous polynomials g , g in R [ x ] of the same degree d ≥ and a square matrix ϕ = ( u i,j ) i,j =1 , with coefficientsin R , then Disc( u , g + u , g , u , g + u , g ) =det( ϕ ) d ( d − Disc( g , g ) . Proof.
By definition, it is sufficient to prove this formulain the universal setting. For simplicity, we use the formula(4). Setting ˜ f := u , f + u , f and ˜ f = u , f + u , f ,we observe that J k,l ( ˜ f , ˜ f ) = det( ϕ ) J k,l ( f , f ) so thatRes( ˜ f , ˜ f , J , ( ˜ f , ˜ f ) , J , ( ˜ f , ˜ f ))= det( ϕ ) d ( d − Res( ˜ f , ˜ f , J , , J , ) . In addition, by the covariance of resultants [9, § f , ˜ f , J , , J , ) = det( ϕ ) d ( d − Res( f , f , J , , J , ) so that we deduce thatRes( ˜ f , ˜ f , J , ( ˜ f , ˜ f ) , J , ( ˜ f , ˜ f ))= det( ϕ ) d ( d − d − Res( f , f , J , , J , ) . The covariance of resultants also shows thatRes( ˜ f , ∂ ˜ f , ˜ f , ∂ ˜ f )= det( ϕ ) d ( d − d − Res( f , ∂ f , f , ∂ f )and the covariance property of discriminants of finitely manypoints [3, Proposition 3.18] yieldsDisc (cid:18) ˜ f , ˜ f (cid:19) = det( ϕ ) d ( d − Disc( f , f ) . From all these equalities and (4), we deduce the claimedformula.
Proposition 7 (Invariance).
Let g , g be two homo-geneous polynomials in R [ x ] of degree d i ≥ and let ϕ =( c i,j ) i,j be a square matrix with entries in R . For allhomogeneous polynomial g ∈ R [ x ] we set g ◦ ϕ ( x , x , x , x ) := g X j =1 c ,j x j , . . . , X j =1 c ,j x j ! . Then, we have that
Disc( g ◦ ϕ, g ◦ ϕ ) = det( ϕ ) D Disc( g , g ) where D = d d (cid:0) ( e + e ) − e e (cid:1) , e i = d i − , i = 1 , . Proof.
As always, to prove this formula we may assumethat we are in the universal setting, f and f being theuniversal homogeneous polynomials of degree d and d re-spectively. We will also denote by l, m, n three generic linearform and by φ the generic square matrix of size 4.Applying Theorem 1, we get the equalityRes( f ◦ ϕ, f ◦ ϕ, J ( f ◦ ϕ, f ◦ ϕ, l ◦ ϕ, m ◦ ϕ ) ,J ( f ◦ ϕ, f ◦ ϕ, l ◦ ϕ, n ◦ ϕ ))= Disc( f ◦ ϕ, f ◦ ϕ )Disc ( f ◦ ϕ, f ◦ ϕ, l ◦ ϕ ) Res( f ◦ ϕ,J ( f ◦ ϕ, l ◦ ϕ, m ◦ ϕ, n ◦ ϕ ) , f ◦ ϕ, J ( f ◦ ϕ, l ◦ ϕ, m ◦ ϕ, n ◦ ϕ ))(6)(observe that l ◦ ϕ, m ◦ ϕ, n ◦ ϕ are all linear forms in x ).Now, by [3, Proposition 3.27], we know thatDisc ( f ◦ ϕ, f ◦ ϕ, l ◦ ϕ ) = det( ϕ ) d d ( e + e ) Disc( f , f , l ) . Also, by the chain rule formula for the derivative of thecomposition of functions, we have the formulas J ( f ◦ ϕ, f ◦ ϕ, l ◦ ϕ, m ◦ ϕ )) = J ( f , f , l, m ) ◦ [ ϕ ] · det( ϕ ) J ( f ◦ ϕ, f ◦ ϕ, l ◦ ϕ, n ◦ ϕ )) = J ( f , f , l, n ) ◦ [ ϕ ] · det( ϕ ) J ( f i ◦ ϕ, l ◦ ϕ, m ◦ ϕ, n ◦ ϕ )) = J ( f i , l, m, n ) ◦ [ ϕ ] · det( ϕ )from we deduce, using the invariance of resultants [9, § f ◦ ϕ, J ( f ◦ ϕ, l ◦ ϕ, m ◦ ϕ, n ◦ ϕ ) , f ◦ ϕ,J ( f ◦ ϕ, l ◦ ϕ, m ◦ ϕ, n ◦ ϕ ))= det( ϕ ) d d e e Res( f , J ( f , l, m, n ) , f , J ( f , l, m, n ))ndRes( f ◦ ϕ, f ◦ ϕ, J ( f ◦ ϕ, f ◦ ϕ, l ◦ ϕ, m ◦ ϕ ) ,J ( f ◦ ϕ, f ◦ ϕ, l ◦ ϕ, n ◦ ϕ ))= det( ϕ ) d d ( e + e ) Res( f , f , J ( f , f , l, m ) , J ( f , f , l, n )) . From here, the claimed formula follows from the substitutionof the above equalities in (6) and the comparison with theformula given in Theorem 1.
Corollary The discriminant is invariant under per-mutation of the variables x . Proof.
It follows from Proposition 7 since D is even. Given a plane curve, we prove that its discriminant asdefined in Section 2, is compatible with its discriminant asa plane hypersurface [3, § Lemma Let g be a homogeneous polynomial in R [ x ] ofdegree d ≥ . Then, for all i = 1 , . . . , we have that Disc( g, x i ) = Disc (cid:16) g i (cid:17) . Proof.
By definition, it is sufficient to prove this equalityin the case where g is replaced by the generic homogeneouspolynomial f of degree d . We apply Theorem 1 with l = x r , m = x s , n = x t that are chosen so that { x i , x r , x s , x t } = { x , x , x , x } as sets. We obtain the equality R := Res( f, x i , ± ∂ t f, ± ∂ s f ) = Disc( f, x i )Disc ( f, x i , x r ) . Since the degree of f and one of its partial derivative areconsecutive integers, their product is always an even integer.It follows by standard properties of resultants that R doesnot depend on the sign of its entry polynomials, nor on theirorder, nor on the reduction of the variables, so that we have R = Res( f i , ∂ t f i , ∂ s f i ) = Res( f i , ∂ s f i , ∂ t f i ) . Now, by property of discriminants, in particular (5) and itsinvariance under permutation of variables [3, Proposition3.12], we haveDisc ( f, x i , x r ) = ( − d Disc (cid:16) f i , x r (cid:17) = Disc (cid:16) f i,r (cid:17) . Finally, [3, Proposition 4.7] shows thatDisc (cid:16) f i,r (cid:17) Disc( f i ) = Res (cid:16) f i , ∂ r f i , ∂ s f i (cid:17) and the claimed equality is proved. Proposition
Let g ∈ R [ x ] be a homogeneous polyno-mial of degree d ≥ and l = P i =1 l i x i be a linear form in R [ x ] . Then, for all i = 1 , . . . , we have that l d ( d − i Disc( g, l )= Disc (cid:0) g ( l i x − δ i l ( x ) , · · · , l i x − δ i l ( x )) (cid:1) where δ ji stands for the Kronecker symbol. Proof.
We assume that we are in the generic setting,which is sufficient to prove this corollary. Consider the lin-ear change of coordinates given by the matrix ϕ i defined asfollows: its i th row is the vector ( − l − l − l − l ) T and its other rows are filled with zeros except on the diagonalwhere we put − l i . Then, it is not hard to check that g ◦ ϕ i ◦ ϕ i ( x ) = g ( l i · x ) = l di g ( x ) . Therefore, by Proposition 3 we obtainDisc( g ◦ ϕ i ◦ ϕ i , l ) = Disc( l di g, l ) = l d ( d − i Disc( g, l ) . (7)On the other hand, since l = x i ◦ ϕ ( x ), Proposition 7 yieldsDisc( g ◦ ϕ i ◦ ϕ i , l ) = det( ϕ i ) d ( d − Disc( g ◦ ϕ i , x i )= l d ( d − i Disc( g ◦ ϕ i , x i )(notice that d ( d −
1) is even and det( ϕ i ) = − l i ). Then,using Lemma 9 we deduce thatDisc( g ◦ ϕ i ◦ ϕ i , l ) = l d ( d − i Disc( g ◦ ϕ ii ) . Compared with (7), this latter equality shows that l d ( d − i Disc( g, l ) = Disc( g ◦ ϕ ii )since l i is not a zero divisor in the universal ring of coeffi-cients. Finally, to conclude we observe thatDisc( g ◦ ϕ ii )= Disc ( g ( − l i x , · · · , l ( x ) − l i x i , · · · , − l i x ))= ( − d ( d − Disc ( g ( l i x , · · · , l i x i − l ( x ) , · · · , l i x ))where the last equality follows from the homogeneity of thediscriminant of a single polynomial [3, Proposition 4.7]. In order to illustrate the gain we obtain with the new for-malism we are developing, we give an explicit decompositionof the discriminant of a particular family of complete inter-section space curves that are drawn on a generalized cylinderwhose base is an arbitrary algebraic plane curve.
Proposition
Suppose given an element u ∈ R andtwo homogeneous polynomials f, g ∈ R [ x , x , x ] of degree d and d respectively. If d + d ≥ then Disc( ux d + f ( x , x , x ) , g ( x , x , x )) =( − d d d d ( d + d − u d [( d + d − − ( d − d − · Disc( f, g ) d − Disc( g ) d . Proof.
Because of the space limitation, we will only givethe main lines to prove this formula. First, we notice that itis sufficient to assume that we are in the universal setting,that is to say to assume that the coefficients of f, g and u are indeterminates over the integers.Set f = ux d + f and f = g . By Corollary 2, we havethatRes( f , f , J , , J , ) = ( − d d Disc ( f , f ) · Res( f , ∂ f , f , ∂ f )Disc( f, g ) . (8)Applying Laplace’s formula [9, § f , ∂ f , f , ∂ f ) = u d ( d − d − Res( g, ∂ f, ∂ g ) d and substituting this equality in (8), we deduce thatRes( f , f , J , , J , ) = ( − d d u d ( d − d − · Disc ( f , f ) Res( g, ∂ f, ∂ g ) d Disc( f, g ) . (9)ow, applying again Laplace’s formula we get thatRes( f , f , J , , J , ) (10)= u d ( d + d − Res( g, J , ( f, g ) , J , ( f, g )) d . In order to compute R := Res( g, J , ( f, g ) , J , ( f, g )), wefirst observe thatRes( g, x J , ( f, g ) , J , ( f, g )) = Res( g, x , J , ( f, g )) R by multiplicativity of resultants. From the definition of theJacobian determinants we have x J , ( f, g ) = d f∂ g − d g∂ f + x J , ( f, g )and we deduce thatRes( g, x J , ( f, g ) , J , ( f, g )) = Res( g, d f∂ g, J , ( f, g ))= d d ( d + d − Res( g, f, J , ( f, g ))Res( g, ∂ g, ∂ f ) · Res( g, ∂ g, ∂ g ) . But from the rule of permutation of polynomials for resul-tants [9, § g, f, J , ( f, g )) = ( − d + d Res( f, g, J , ( f, g ))= Disc( f, g )Res( f , g ) . Similarly, from the rule of permutations of polynomials andthe definition of discriminants of hypersurfaces [3, Definition4.6], we have Res( g, ∂ g, ∂ g ) = Disc( g )Disc( g ) and henceRes( g, x , J , ( f, g )) R = ( − d + d d d ( d + d − · Res( g, ∂ g, ∂ f )Disc( f, g )Disc( g )Res( f , g )Disc( g ) . (11)Now, it remains to compute R ′ = Res( g, x , J , ( f, g )) . Onthe one hand we haveRes( g, x , x J , ( f, g )) = Res( g, x , x ) R ′ . (12)On the other hand,Res ( g, x , x J , ( f, g ))= ( − d d Res (cid:16) g , x J , (cid:16) f , g (cid:17)(cid:17) (13)and since x J , ( f , g ) = d f ∂ g − d g ∂ f by the Eulerformula, we deduce thatRes( g , x J , ( f , g )) (14)= ( − d d d Res( g , f )Disc( g )Res( g , X ) . Finally, since Res( g , x ) = ( − d Res( g, x , x ) the com-parison of (12), (13) and (14) shows that R ′ = d d Disc( g )Res( f , g ) . Now, coming back to the factor R , we deduce from (11) that R = ( − d + d d d ( d + d − Res( g, ∂ g, ∂ f )Disc( f, g )Disc( g )and hence from (10) thatRes( f , f , J , , J , ) = ( − d ( d − d d d ( d + d − · u d ( d + d − Res( g, ∂ g, ∂ f ) d Disc( f, g ) d Disc( g ) d . Finally, we deduce from (3) that( − d d u d ( d − d − Disc ( f , f ) · Res( g, ∂ f, ∂ g ) d Disc( f, g )= ( − d ( d − d d d ( d + d − u d ( d + d − · Res( g, ∂ g, ∂ f ) d Disc( f, g ) d Disc( g ) d and the claimed formula follows.
4. THE GEOMETRIC PROPERTY
The aim of this section is to show that the discriminantDisc( f , f ) defined in Definition 1 satisfies to the expectedgeometric property, namely that its vanishing correspondsto the existence of a singular point on the curve intersectionof the two surfaces of equations f = 0 and f = 0 in P .We start by recalling the precise meaning of this geometricproperty as we will work over coefficient rings which are notnecessarily fields.Let k be a commutative ring. We consider the universalsetting over k , i.e. we suppose given two positive integers d , d and we consider the (generic) homogeneous polyno-mials in the four variables x = ( x , x , x , x ) f := X | α | = d U ,α x α , f := X | α | = d U ,α x α that are polynomials in k C = k A [ x ], where k A = k [ U i,α ; i =1 , , | α | = d i ] is the universal ring of coefficients over thebase ring k . If there is no possible confusion, we will omitthe subscript k in the above notation.We define the ideal m = ( x ) ⊂ C generated by the vari-ables x , the ideal J ⊂ C generated by all the 2-minorsof the Jacobian matrix of f and f , and the ideal D =( f , f ) + J ⊂ C . Thus, using notation (3), we have that D = ( f , f , J , , J , , J , , J , , J , , J , ) ⊂ C. The quotient ring B := C/ D is a graded ring with respectto the variables x . As such, it gives rise to the projec-tive scheme Proj( B ) ⊂ P A that corresponds to the points( u i,α ) i,α × P ∈ Spec( A ) × P k such that the correspondingpolynomials f , f and all the 2-minors of their Jacobian ma-trix vanish simultaneously at P . The canonical projection ofProj( B ) onto Spec( A ) is a closed subscheme ∆ of Spec( A )whose support is precisely what is commonly called the dis-criminant locus . By definition, the defining ideal of ∆ is theideal P = T m ( D ) ∩ A where T m ( D ) = ker C π −→ Y i =1 B x i ! ⊂ C = { p ∈ C such that ∃ ν ∈ N : m ν · p ⊂ D} is the so-called ideal of inertia forms – the notation B x i denotes the localization of B with respect to the variable x i and π is the product of the canonical quotient maps.In what follows, we will show that k Disc( f , f ), as definedby Definition 1, is a generator of k P if k is a UFD, so thatit satisfies to the expected geometric property. Before goinginto the details, we recall the following important and well-known result (see e.g. [1, 8]): if k is a field, then the reducedscheme of k ∆ is an irreducible hypersurface, i.e. the radicalof k P is a principal and prime ideal, so that it is generatedby an irreducible polynomial D k ∈ k A . This polynomial isot unique; it is unique up to multiplication by a nonzeroelement in k . In addition, D k is homogeneous of degree δ i (see Proposition 3 for the definition of δ i ) with respect tothe coefficients of f i .We begin with some preliminary results on the Jacobianminors and the ideal J they generate. Lemma
For any j ∈ { , . . . , } we have that X k ∈{ ,..., } ,k = j x k J k,j ∈ ( d f , d f ) . Proof.
Using the Euler formula, we have that (cid:12)(cid:12)(cid:12)(cid:12) d f ∂ j f d f ∂ j f (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) P i =1 x i ∂ i f ∂ j f P i =1 x i ∂ i f ∂ j f (cid:12)(cid:12)(cid:12)(cid:12) = X k ∈{ ,..., } ,k = j x k (cid:12)(cid:12)(cid:12)(cid:12) ∂ k f ∂ j f ∂ k f ∂ j f (cid:12)(cid:12)(cid:12)(cid:12) and the claim follows. Lemma
For any integer j ∈ { , } and any triple ofdistinct integers i , i , i in { , , , } we have that J i ,i .∂ i f j − J i ,i .∂ i f j + J i ,i .∂ i f j = 0 . Proof.
Develop the determinant of the 3-minor corre-sponding to the columns i , i , i in the Jacobian matrix of f , f and f j . Lemma If k is a domain, then for all i = 1 , . . . , theideal J x i ⊂ k C x i is a prime ideal. Proof.
For simplicity, we will assume that i = 4, theother cases being similar. In order to emphasize some par-ticular coefficients of f and f we rewrite them as follows: f i = U i, x d i + x d i − ( U i, x + U i, x + U i, x ) + h i , i = 1 , . We consider the A -algebra morphism η : C [ x − ] → C [ x − ] : U i,j
7→ − ∂ j f i /x d i − which leaves invariant all the variables x and all the coeffi-cients of f , f , except the U i,j ’s. As η ( ∂ j f i ) = − U i,j x d i − , η is surjective. Moreover, setting U = (cid:18) U , U , U , U , U , U , (cid:19) and denoting by U r,s the 2-minor of U corresponding to thecolumn number r, s , we have that η ( J r,s ) = x d + d − U r,s ,r, s ∈ { , , } , r = s. Considering the map η induced by η , η : C [ x − ] → C [ x − ] / ( U , , U , , U , ) , we deduce that ( J , , J , , J , ) .C [ x − ] ⊂ ker( η ) . Actually,this inclusion is an equality. Indeed, if p ∈ ker( η ) then η ( p ( U i,j )) = p (cid:16) − ∂ j f i /x d i − (cid:17) ∈ ( U , , U , , U , ) . (15)But since η ( − ∂ j f i /x d i − ) = U i,j , applying again η to (15)we deduce that p ( U i,j ) ∈ η ( U , , U , , U , ) = ( J , , J , , J , ) .C [ x − ] . It follows that η induces a graded isomorphism C x / J x ∼ −→ C [ x − ] / ( U , , U , , U , ) . (16)From here, if k is a domain then the ideal generated by the2-minors of U is a prime ideal (see [2, Theorem 2.10]) andhence C x / J x is a domain. The above lemma is the key result to deduce the followingproperties of the ideal of inertia forms T m ( D ). Proposition If k is a domain then B x i is a domainfor all i = 1 , . . . , . Proof.
We prove the claim for i = 4, the other casesbeing similar. Let p , p be two polynomials in C so that p p = 0 in B x , i.e. p p belongs to the ideal D up tomultiplication by a power of x . Using this fact and Lemma12, we deduce that there exists an integer ν such that x ν p p ∈ ( f , f , J , , J , , J , ) . (17)In order to emphasize the leading coefficients of f and f with respect to the variable x , we rewrite them as f i = U i, x d i + q i , i = 1 , . Denote by C the polynomial ring C in which the variables(coefficients) U , , U , are removed and consider the surjec-tive graded morphism ρ : C [ x − ] → C [ x − ] : U i,
7→ − q i /x d i which leaves invariant all the variables x and all the coeffi-cients of f , f , except U , , U , . It induces an isomorphism ρ : C x / ( f , f ) x ∼ −→ C [ x − ] . Now, by (17) we deduce that ρ ( p ) ρ ( p ) belongs to the ideal( J , , J , , J , ) .C [ x − ] . Therefore, using Lemma 14 we de-duce that either ρ ( p ) or ρ ( p ) belongs to this ideal, say ρ ( p ). This implies that there exists an integer µ such that x µ p ∈ ( f , f , J , , J , , J , ) ⊂ D . In turns, this implies precisely that p = 0 in B x , whichconcludes the proof. Corollary
For all i = 1 , . . . , we have that T m ( D ) = ker (cid:16) C π i −→ B x i (cid:17) = { p ∈ C such that ∃ ν ∈ N : x νi · p ∈ D} . Thus, both T m ( D ) and P are prime ideals if k is a domain. Proof.
Using Proposition 15, the proof of [3, Corollary3.21] applies verbatim to show that x i is not a zero divisor in B x j for all i = j . From here, we deduce that the canonicalmaps B x i → B x i x j , i = j , are all injectives maps and hencethe claimed equalities follow.We are now ready to prove the main result of this section. Theorem If k is a UFD then k Disc( f , f ) is a gen-erator of k P . It is hence an irreducible polynomial in k A . Proof.
First, let K be a field. From the geometric prop-erty we recalled previously, we know that the radical of K P isgenerated by an irreducible polynomial D K . Using Corollary16, we deduce that D K is actually a generator of K P .Now, assume that k is a domain and take again the nota-tion of Theorem 1. The resultantRes ( f , f , J ( f , f , l, m ) , J ( f , f , l, n ))is an inertia form of its four input polynomials and hence, bydeveloping the Jacobian determinants, we see that it belongsto k P · k A ′ . Therefore, Theorem 1 shows thatRes ( f , J ( f , l, m, n )) , f , J ( f , l, m, n )) · Disc( f , f )Disc ( f , f , l ) ∈ k P · k A ′ . (18)e claim that Disc( f , f , l ) does not belong to k P . Indeed,assume the contrary. By extension to the fraction field K of k , we would have that the square-free part of K Disc( f , f , l )belongs to the prime ideal K P . But K P is generated by D K sowe get a contradiction since K Disc( f , f , l ) is homogeneousof degree d ( e + 2 e ) < δ with respect to the coefficientsof f , and similarly with respect to the coefficients of f [3,Proposition 3.9]. With a similar argument, we also get thatthe resultant Res ( f , J ( f , l, m, n ) , f , J ( f , l, m, n )) does notbelong to k P since it is homogeneous of degree d (2 e e + e )with respect to the coefficients of f , and similarly with re-spect to the coefficients of f . Finally, as k P is a prime idealwe deduce from (18) that k Disc( f , f ) belongs to k P (recallthat k is here assumed to be a domain).As a first consequence, since k Disc( f , f ) and D k are ho-mogeneous polynomials of the same degree with respect tothe coefficients of each f i , we conclude that this theorem isproved if k is a assumed to be a field.Let p ∈ Z P be an inertia form and set N := d d ( e + e ). Our next aim is to show that Z Disc( f , f ) divides p N .For that purpose, using the definition of inertia forms andLemma 12, we deduce that there exists an integer ν suchthat x ν · p ∈ ( f , f , J , , J , , J , ) . Then, Lemma 13 showsthat ∂ f i · J , ∈ ( J , , J , ) so that we get that x ν · p · ∂ f i ∈ ( f , f , J , , J , ) . Now, by the divisibility property of resultants [9, § f , f , J , , J , ) divides R := Res( f , f , J , , x ν · p · ∂ f i ) . Applying computational rules of resultants and choosing i =1 we get that R = p N Res (cid:16) f , f , J , (cid:17) ν · Res( f , ∂ f , f , ∂ f )Res( f , f , ∂ f , ∂ f )where, in addition,Res (cid:16) f , f , J , (cid:17) = Disc (cid:16) f , f (cid:17) Res (cid:16) f , , f , (cid:17) by the definition of discriminants of finitely many points [3,Definition 3.5]. Combining the above equalities and using(4), we deduce that Disc( f , f ) divides the product p N Disc (cid:16) f , f (cid:17) ν − Res (cid:16) f , , f , (cid:17) ν · Res( f , f , ∂ f , ∂ f ) . (19)With a similar degree inspection as above and after exten-sion to Q , we deduce that Q Disc( f , f ), which is an irre-ducible polynomial, cannot divide the discriminant and thetwo resultants in (19). Then, we claim that the discrimi-nant and the two resultants in (19) are primitive polynomi-als. This is a known property for the first two ones. Forthe third one, namely Res( f , f , ∂ f , ∂ f ), we argue byspecialization: for instance,Res( f , Ux d , ∂ f , ∂ f ) = U d e Res (cid:16) f , ∂ f , ∂ f (cid:17) d = U d e Disc (cid:16) f (cid:17) d Disc (cid:16) f , (cid:17) d is a primitive polynomial since both discriminants on theright side are known to be primitive polynomials. Therefore,we conclude that Z Disc( f , f ) divides p N . Finally, from what we proved we deduce that k P andthe ideal generated by k Disc( f , f ) have the same radicals.Since we assume that k is a UFD, k P is prime and we de-duce that there exist an irreducible polynomial P k , an in-vertible element c ∈ k and a positive integer r such that k Disc( f , f ) = c · P rk . By extension to K we deduce imme-diately that r = 1, which concludes the proof.The above theorem shows that Z Disc( f , f ) is a primitiveand irreducible polynomial in k A . It also shows that the dis-criminant formula we gave provides an effective smoothnesscriterion (as the criterion in [7, p. 3] that applies verbatim).
5. ACKNOWLEDGMENTS
Part of this work was done while the second author wasvisiting IMSP at Benin, supported by the DAAD. Both au-thors warmly thank the ICTP for its hospitality and are verygrateful to Alicia Dickenstein, Marie-Fran¸coise Roy and Fer-nando Rodrigues Villegas for their continued support.
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