Explicit factors of some iterated resultants and discriminants

In this paper, the result of applying iterative univariate resultant constructions to multivariate polynomials is analyzed. We consider the input polynomials as generic polynomials of a given degree and exhibit explicit decompositions into irreducible factors of several constructions involving two times iterated univariate resultants and discriminants over the integer universal ring of coefficients of the entry polynomials. Cases involving from two to four generic polynomials and resultants or discriminants in one of their variables are treated. The decompositions into irreducible factors we get are obtained by exploiting fundamental properties of the univariate resultants and discriminants and induction on the degree of the polynomials. As a consequence, each irreducible factor can be separately and explicitly computed in terms of a certain multivariate resultant. With this approach, we also obtain as direct corollaries some results conjectured by Collins and McCallum which correspond to the case of polynomials whose coefficients are themselves generic polynomials in other variables. Finally, a geometric interpretation of the algebraic factorization of the iterated discriminant of a single polynomial is detailled.


Introduction
Resultants provide an essential tool in constructive algebra and in equation solving, for projecting the solution of a polynomial system into a space of smaller dimension.In the univariate case, a well-known construction due to J.J. Sylvester (1840) consists in eliminating the monomials 1, X, . . ., X m+n−1 in the multiples (X i P (X)) 0≤i≤n−1 , (X j Q(X)) 0≤j≤m−1 , of two given polynomials P, Q of degree m and n, and in taking the determinant of the corresponding (m+n)×(m+n) matrix.Though the first resultant construction appeared probably in the work of E. Bézout [4] (see also Euler's work), and although contemporary to related works (Jacobi 1835, Richelot 1840, Cauchy 1840, . . .), this method remains well-known as Sylvester's resultant.It is nowadays a fundamental tool used in effective algebra to eliminate a variable between two polynomials.
It is a natural belief that equipped with such a tool which eliminates one variable at a time, one can iteratively eliminate several variables.This approach was actually exploited, for instance in [30], to deduce theoretical results (such as the existence of eliminant polynomials) in several variables.However, if we are interested in structural results as well as practical computations or complexity issues, this approach is far from being optimal.This explains the study and development of different types of multivariate resultants, including projective [23,18], anisotropic [18,19], toric [14,13,12], residual [19,7,5], determinantal [8] resultants.
Nevertheless, in some algorithms, such as in Cylindrical Algebraic Decomposition (CAD), an induction is applied on the dimension of the problems and iterated univariate resultants and subresultants are used at many steps of the algorithm [9,3].In its original paper on quantifier elimination for real closed fields by cylindrical algebraic decomposition [9], Collins used these iterated resultants in a geometric context and observed certain intriguing factorizations that "suggest some theorems" [9, p.178].The same year, 1975, Van der Waerden responded to these observations in a handwritten letter where he gave some intuitive hints for some of the phenomena noticed by Collins.More recently, McCallum [26,27] proved rigorously that certain iterated resultants have some irreducible factors, but he only showed the existence and did not give a way to compute them independently.As pointed out by Jean-Pierre Jouanolou, in 1868 and 1869 Olaus Henrici published two outstanding papers [16,17] addressing the decomposition of the discriminant of a discriminant.In particular, he gave the expected factorization of such a repeated discriminant.However, he did not prove the irreducibility of these factors which is a more difficult task.One of our goals in this paper is to give the decomposition into irreducible factors of these iterated resultant computations.
From a geometric point of view, we are interested in the solutions of equations depending on some parameters and by analyzing what happens "above", when we move these parameters.The number of solutions might change if we cross the set of points where a vertical line is tangent to the solution set (that is to the polar variety in the vertical direction).This polar variety projected in one dimension less, might have singularities where the number of solutions changes effectively or which are only due to the superposition of distinct points of this polar variety.These critical points of the polar variety of an algebraic surface f (x, y, z) = 0 in R 3 are used effectively in algorithms for computing the topology of the surface (see e.g.[28]).The projection of the polar curve (say on the (x, y)-plane) is obtained by a discriminant computation and the critical points of this projected curve are again computed by a discriminant.These values are then used to analyze where the topology of a plane section is changing in order to deduce the topology of the whole surface f (x, y, z) = 0. Similar projection tools are also implicitly used in higher dimension for the triangulation of hypersurfaces (see e.g.[15]), which leads in the algebraic context to univariate resultant computation (see e.g.[10,3]).
Another of our objectives is to show how these critical points corresponding generically to folds, double folds or pleats of the surface can be related to explicit factors in iterated resultant constructions.The main results of this paper are complete explicit factorizations of two times iterated univariate resultants and discriminants of generic polynomials of given degree.We actually give the decomposition of these iterated resultants over the integer universal ring of coefficients of polynomials of given degree.Such a formulation has the advantage to allow the pre-computation of a given factor.It has direct applications to the topological computation of algebraic surfaces, which was our starting point.
Our approach is based on the study of these iterated resultants in generic situations.Most of the interesting formulas are obtained by suitable specialization of this case.These specializations are performed using the formalism of the multivariate resultant as it has been originaly introduced and deeply developed by Jouanolou to whom this paper is dedicated in recognation of its outstanding contributions to resultant theory (e.g.[18,19,20]).It should be noticed that this approach can be pushed further to study more particular situations (corresponding to other type of multivariate resultants) that we did not consider in this paper, but which could be interesting for specific applications.
The paper is structured as follows.In the next section, we recall the definitions and main properties of resultants and discriminants that we will use.In section 3, we consider the computation of iterated resultants of 4 and then 3 polynomials.In section 4, we analyze the iterated computation of resultants of discriminants, first the resultant of the discriminants of two distinct polynomials P 1 , P 2 and next the resultant of the discriminant of P 1 and the resultant of P 1 , P 2 .We here extend the previous work [26] and prove some properties conjectured by Collins and McCallum.In section 5, we study the discriminant of a resultant, simplifying the proof and also extending some results of [26,27].These developments are used in cascade to provide, in section 6, the complete factorization into irreducible components of a discriminant of a discriminant for a generic polynomial, as conjectured in [27].These new results have direct corollaries for polynomials whose coefficients are themselves generic polynomials in other parameter variables, which we provide.

Background material and notation
In this section we give the notation and quickly present the tools, as resultants and discriminants, that we will use all along this paper. 2.1.Resultants.
2.1.1.The univariate case.Let S be a commutative ring (with unity) and consider the two polynomials in S[X] where a i , b j ∈ S, and m and n are both positive integers.Their resultant (in degrees m, n) 1 that we will denote Res X (f, g), is defined as the determinant of the well-known Sylvester matrix We emphasize that the notation Res X (f, g) denotes the resultant of f and g with respect to the variable X as polynomials of their expected degree, which is here m and n respectively.It is important to keep this in mind since the closed 1 Notice that the dependence on the degrees m, n can be avoided if one considers homogeneous polynomials.
formulas we will prove in this paper are using this convention; see Remark 3.5 for an illustration.
This "eliminant polynomial" has a long history and many known properties.One can learn about it in many places in the literature, for instance [11,22]; see also [14, chapter 12] and [1] for a detailed exposition.In the sequel we will especially use the following properties: • Res X (f, g) belongs to the ideal (f, g) ⊂ S[X], • Res X (f, g) is homogeneous of degree n, resp.m, in the a i 's, resp. in the b j 's, • Res X (f, g) is homogeneous of degree mn if we set deg(a i ) := m − i and deg(b j ) := n − j for all i = 0, . . ., m and j = 0, . . ., n.We also recall the definition of the principal subresultant of f and g that we will use later on.It is defined as the determinant of the above Sylvester matrix where the two last lines and columns number n and n + m are erased; more precisely SRes (1) ).Note that the subresultants share a lot of properties with the resultants; we refer the interested reader to [1].
2.1.2.The multivariate case.All along this paper we will also use resultants of several homogeneous polynomials; we now quickly recall this notion.Although they are usually defined "geometrically" (as equations of certain hypersurfaces obtained by projection of an incidence variety), we will follow the formalism developed by Jouanolou [18] because it easily provides many properties of resultants.Suppose given an integer n ≥ 1, and a sequence of positive integers d 1 , . . ., d n .One considers the n "generic" homogeneous polynomials of degree d 1 , . . ., d n , respectively, in the variables X 1 , . . ., X n (all assumed to have weight 1) : where m := (X 1 , . . ., X n ) ⊂ C. It is naturally graded and it turns out that its degree zero graded part, denoted TF m (f 1 , . . ., f n ) 0 , is a principal ideal of U and has a unique generator, denoted Res X1:•••:Xn or simply Res, which satisfies (2.1) Res(X d1 1 , . . ., X dn n ) = 1.To define the resultant of any given n-uples of homogeneous polynomials in the variables X 1 , . . ., X n (and also to clarify the left side of the equality (2.1)) one proceeds as follows: Let S be a commutative ring.For all integers i ∈ {1, . . ., n}, suppose given a homogeneous polynomial of degree d i in the variables X 1 , . . ., X n and consider the morphism θ : U → S : U j,α → u j,α which corresponds to the specialization of the polynomials f i to the polynomials g i .Then, given an inertia form a ∈ TF m (f 1 , . . ., f n ) we set a(g 1 , . . ., g n ) := θ(a).In particular, the resultant of g 1 , . . ., g n is nothing but Also, if S = U and θ is the identity (i.e.g i = f i for all i), then we get a = a(f 1 , . . ., f n ); this clarifies the notation Res(f 1 , . . ., f n ) for the inertia form Res ∈ U.
Resultants have a lot of interesting properties.We recall the ones we will use in the sequel and refer the reader to [18, §5] for the proofs.Mention that many formulas are known to compute explicitly these resultants (e.g.[23], [20], [14], [11] and the reference therein).
Let S be any commutative ring and suppose given f 1 , . . ., f n homogeneous polynomials in the polynomial ring S[X 1 , X 2 , . . ., X n ] of positive degree d 1 , . . ., d n respectively.
In the following, we are going to consider resultant computation for eliminating a subset X i1 , . . ., X i k of the variables.Such a resultant, obtained by considering the homogenization of the polynomials with respect to this variable subset, will be denoted hereafter Res Xi 1 ,...,Xi k .With this notation, for homogeneous polynomials f 1 , . . ., f n ∈ S[X], we have 2.2.Discriminants.Given a polynomial where n ≥ 1 and S is any commutative ring, we recall that the discriminant of P , denoted Disc X (P ), satisfies the equality where ∂ X stands for the derivative with respect to the variable X, i.e. ∂/∂ X .Its properties follow immediately from the ones of the resultant.In particular, its degree in the coefficients of P is 2(d − 1).Note that in our notation the polynomial P is seen as a polynomial of its expected degree, here n, as we did for the resultant.Indeed, in (2.3) the considered resultants are in degrees (n, n − 1) and (n − 1, n) respectively.where ∂ i stands for ∂/∂ Xi for all i.It is an irreducible polynomial in U which is homogeneous of degree 3(d − 1) 2 in the U α 's.Note that we also have the following equality in U:
Finally, we also recall that being given two homogeneous polynomials P 1 and P 2 in the variables X 1 , X 2 , X 3 of degree d 1 , d 2 respectively, their discriminant is defined by the formula [21] Disc X1:X2:X3 (P is S-linear and for all integers k ≥ 0 we have ∂ X (X k ) = kX k−1 .For any integer i ≥ 1 we define ∂ i X as the composition of ∂ X with itself i times; for instance Now, suppose given an element a ∈ S and consider the S-linear map For instance, for all integers k ≥ 1 we have δ a (X k ) = k−1 j=0 a k−1−j X j .As above, for any integer i ≥ 1 we define δ i a as the composition of δ a with itself i times.It follows that for all integers n ≥ 1 (2.5) Indeed, we have P (X) = P (a) + (X − a)δ a (P )(X) by definition.Applying this formula to δ a (P ) we deduce (2.5) for n = 2.The general case is obtained by a similar induction.
Although S is only assumed to be a commutative ring, we still have the expected formula Lemma 2.2.For k ≥ 1, for any P ∈ S[X] and a ∈ S, we have the equality k! δ k a P (a) = ∂ k X P (a) in S. Proof.Notice that δ a P (X) is of degree deg(P ) − 1 in the variable X.Thus if d > deg(P ), we have δ d a (P ) = 0. We deduce that for any polynomial P ∈ S[X], By the binomial identity, for any n ∈ N, By identification of the coefficients in the basis (X − a) k of S[X], we deduce that for all k ≤ n, we have . By linearity, we also have this identity for any polynomial P ∈ S[X].
As a consequence, if S is assumed to be a Q-algebra then (2.5) shows that for all integers n ≥ 1 If n is bigger than the degree of P , then we get the Taylor expansion formula We now fix a notation that we will use all along the paper.We will mainly manipulate polynomials in the three variables X 1 , X 2 and X 3 over a commutative ring S. Introducing a new indeterminate X 4 , for any polynomial P ∈ S[X 1 , X 2 , X 3 ] we set so that we have Observe that δ 3,4 (P ) is nothing but δ X4 (P ) ] and therefore that all the above definitions and formulas can be expressed with this notation.Therefore, we have The above definition can be extended to any variable.We denote by δ 2 i,j the corresponding operation, X i playing the role of X 3 , and X j the role of X 4 .If the indices i, j are omitted, we implicitly refer to the variables X 3 and X 4 .Notice that δ 3,4 P |X4=X3 = ∂ 3 P (X 3 ).Also, by convention we set δ i,i P = ∂ i P .
We will need the two following properties: for any polynomials L, Q, we have To prove (2.8), we remark that and divide by (X 4 − X 3 ) to get the formula.To prove (2.9), we substitute (2.6) in the previous relation and obtain 3,4 (Q), from which we deduce the relation (2.9).
Finally, mention that if S is assumed to be a Q-algebra then we have . Similarly, we have (2.10) Notice that we will very often omit variables X 1 and X 2 in the sequel, especially in the proofs to avoid to overload the computations and the text.2.4.A Bertini's lemma.The elimination of the variables X 1 , X 2 , X 3 , X 4 between polynomials constructed from the polynomials k = 1, . . ., r, where U denotes the universal coefficients ring Proof.Assume that R is not irreducible and can be splitted into the product of two factors in [x, a By sending x to 0 we observe that R(a i,j,α since R is irreducible so that we only have to rename each coefficient a i,j,α , 0) must be an invertible element in .But since R 1 and R 2 are homogeneous in the coefficients a (k) i,j,α this implies that either R 1 or R 2 is an invertible element in .

Resultant of resultants
In all this section, we suppose given four positive integers d 1 , d 2 , d 3 , d 4 and four homogeneous polynomials where U denotes the universal ring of coefficients ].We denote by X 4 a new indeterminate.Our first result is the most general situation of an iterated resultant with four different polynomials.
we have the following equality in U: Moreover, the above quantity is non-zero, irreducible and multi-homogeneous with respect to the set of coefficients (U Proof.First of all, we observe that the iterated resultant Res X2 (R 12 , R 34 ) and the resultant are both non-zero polynomials, for they both specialize to the quantity (−1) d1d2d3d4 if the polynomials 2 and X d4 3 respectively.Note also that the statement about the multi-degree of R follows from the homogeneity property of resultants.
To prove the irreducibility of R, we proceed by induction on the positive integer which is checked to be irreducible.Thus, we assume that R is irreducible up to a given integer p ≥ 4 and we will prove that R is irreducible if d = p + 1.To do this, first observe that one of the integers d 1 , d 2 , d 3 , d 4 must be greater or equal to 2. We can assume that d 1 ≥ 2 without loss of generality.Consider the specialization φ leaving invariant the polynomials P 2 , P 3 and P 4 and sending the polynomial P 1 to the product L 1 Q 1 where L 1 and Q 1 are both generic forms of respective degree 1 and d 1 − 1 ≥ 1.Then, by multiplicativity of resultants we have the equality whose right hand side is a product of two irreducible polynomials by our induction hypothesis.As the specialization φ is homogeneous (in terms of the coefficients of the P i 's, L 1 and Q 1 ), the number of irreducible factors of R can not decrease under the specialization φ and we deduce that R is the product of two irreducible polynomials R 1 and R 2 .But then, one of these two factors must depend on the coefficients of P 1 , say R 1 , and therefore φ(R 1 ) must depend on the coefficients of L 1 and Q 1 .This implies that R 2 is an invertible element in and consequently that R is irreducible.It remains to prove the claimed equality.To do this, we rewrite P k (1, X 2 , X 3 ), for all k = 1, . . ., 4, as and we then easily see from well-known properties of the Sylvester resultant that i,j ) and (U Of course, completely analogous results hold for R 34 , in particular After homogenization with the variable X 1 , it follows that there exists an integer N such that ) and P 4 (X 1 , X 2 , X 4 ) are not generic polynomials).It implies by the divisibility property of resultants, that R = Res(P 1 (X 3 ), P 2 (X 3 ), P 3 (X 4 ), P 4 (X 4 )) divides the quantity Since R is irreducible and since the second term in the right hand side of the above product does not depend on the coefficients of P 4 , we deduce that R divides Res X2 (R 12 , R 34 ).Now, from the degree properties of R 12 and R 13 we deduce that Res X2 (R 12 , R 34 ) is, similarly to R, multi-homogeneous with respect to the set of coefficients (U i,j ) i,j , (U i,j ) i,j , (U i,j ) i,j , (U i,j ) i,j of multi-degree This shows that Res X2 (R 12 , R 34 ) and R are equal up to multiplication by an invertible element in .To determine this invertible element, we take again the specialization sending P 1 to X d1 1 , P 2 to X d2 3 , P 3 to X d3 2 , P 4 to X d4 3 , and check that R specializes to (−1) d1d2d3d4 , as well as Res X2 (R 12 , R 34 ).
A specialization of this theorem gives the following result.
where x denotes a set of variables (x 1 , . . ., x n ) for some integer n ≥ 1 and S is any commutative ring, then the iterated resultant Res y (Res Moreover, if the polynomials f 1 , f 2 , f 3 and f 4 are sufficiently generic then this iterated resultant is irreducible and has exactly degree Proof.It is a corollary of Theorem 3.1.Indeed, the claimed equality is deduced from the equality given in Theorem 3.1 by substituting X 2 by y, X 3 by z, by seeing X 1 as the homogenization variable of the variables (X 2 , X 3 ) and by specializing each coefficient is specialized to a polynomial in x of degree at most d k we deduce the claimed degree bound as a consequence of the isobarity formula for resultants.
If the polynomials f 1 , f 2 , f 3 and f 4 are sufficiently generic then it is clear that the degree bound is reached: this and the irreducibility statement is a consequence of Lemma 2.3.
The formula proved in Theorem 3.1 can be specialized to get the factorization of an iterated resultant which may occur quite often in practical situations (implicitization of a rational surface, projection of the intersection of two surfaces in projective space,. . .e.g.[24,6]).

Then, we have the equality
where the right hand side is a product of two irreducible polynomials in U.
The multi-degree computation and the irreducibility of Res(P 1 , P 2 , P 3 ) are known properties of resultants.The only point which requires a proof is the irreducibility of the factor (which is easily seen to be non-zero by a straightforward specialization) D := Res(P 1 (X 3 ), P 2 (X 4 ), P 3 (X 3 ), δ 3,4 (P 1 )(X 3 , X 4 )).
To do this, we proceed similarly to what we did in Theorem 3.
and a generic form Q 1 of degree d 1 − 1 ≥ 2. We call φ the map corresponding to this specialization.By (2.8), we have and we deduce after some manipulations on resultants that Either by our induction hypothesis or by Theorem 3.1, it turns out that the three resultants involved in the right hand side of the above computation are irreducible in U.So if D were reducible, say D = D 1 D 2 , then each factor should be homogeneous in the coefficients (U i,j ) i,j and hence φ(D 1 ) and φ(D 2 ) should be homogeneous in the coefficients of Q 1 and L 1 of the same degree.But As a consequence of this proposition, we get the following result which improves Theorem 3.1 and Theorem 3.2 in [26].
Corollary 3.4.Given three polynomials f k (x, y, z), k = 1, . . ., 3, of the form where x denotes a set of variables (x 1 , . . ., x n ) for some integer n ≥ 1 and S is any commutative ring, then the iterated resultant Res y (Res is of degree at most d 2 1 d 2 d 3 in x and we have Moreover, if the polynomials f 1 , f 2 , f 3 are sufficiently generic then this iterated resultant has exactly degree d 2 1 d 2 d 3 in x and both resultants on the right hand side of the above equality are distinct and irreducible.This corollary can be interpreted geometrically as follows.For simplicity we assume that x is a unique variable x and that f 1 , f 2 and f 3 are three polynomial in x, y, z.The resultant R 12 := Res z (f 1 , f 2 ) defines the projection of the intersection curve between the two surfaces {f 1 = 0} and {f 2 = 0}.Similarly, R 13 := Res z (f 1 , f 3 ) defines the projection of the intersection curve between the two surfaces {f 1 = 0} and {f 3 = 0}.Then the roots of Res y (R 12 , R 13 ) can be decomposed into two distinct sets: the set of roots x 0 such that there exists y 0 and z 0 such that f 1 (x 0 , y 0 , z 0 ) = f 2 (x 0 , y 0 , z 0 ) = f 3 (x 0 , y 0 , z 0 ), and the set of roots x 1 such that there exist two distinct points (x 1 , y 1 , z 1 ) and ( . The first set gives rise to the term Res x,y,z (f 1 , f 2 , f 3 ) in the factorization of the iterated resultant Res y (Res 12 , Res 13 ), and the second set of roots corresponds to the second factor.Remark 3.5.Before going further, it is a good point to emphasize the fact that all the formulas presented in this paper are universal in the sens that they remain true for any specialization of the coefficients of the given polynomials.It should be noticed that this is true if and only if we take the univariate resultants in their expecting degrees.For instance, in the above corollary, Res z (f 1 , f 2 ) denotes the resultant w.r.t. the variable z of the polynomials f 1 , f 2 which are seen as polynomials of degree d 1 and d 2 respectively (even if their degree is actually lower for a given specialization).If one does not take care of this point, then one looses the universal property of these formulas as in [26, §7] where it is observed that if f 1 := y 2 + z + x, f 2 := −y 2 + z and f 3 := y 2 + z then Res y,z (f 1 , f 2 , f 3 ) = 0 (for there is a base point at infinity) but Res y (Res z (f 1 , f 2 ), Res z (f 1 , f 3 )) = 4x 2 = 0, considering f 1 , f 2 and f 3 as polynomials in z of degree 1 and not 2.In such a case, formulas similar to the ones we proved above require the use of more sophisticated resultants which can take into account some particular structure of polynomial systems.For instance, in this example one can take into account the presence of the base point defined by the ideal (y 2 , w), where w is the homogenizing variable, by considering the residual resultant [5,7] (or the multi-homogeneous resultant which is the same here): we have the decomposition and we can check that the residual resultant equals 4x 2 (up to a sign).

Resultants of discriminants
In this section, we will give the factorization of three iterated resultants corresponding to two cases of a resultant of a discriminant and a resultant and to the case of a resultant of two discriminants.As we will see, the first case we are going to treat yields the two others by suitable specializations.
Then the following equality holds in U: Moreover, the iterated resultant Res X2 (D 1 , R 23 ) ∈ U is irreducible and multi-homogeneous with respect to the coefficients (U i,j ) i,j and (U Proof.By (2.3), we easily get that and hence, using the formula proved in Theorem 3.1 where we specialize the polynomials P 2 , P 3 and P 4 to the polynomials ∂ 3 P 1 , P 2 and P 3 respectively, we deduce that, in U, The classical multi-degree formula for resultants gives the claimed result concerning the multi-degree of the iterated resultant Res (1) 0,0 .We now assume that d ≥ 5.If d 3 ≥ 2 then we consider the specialization φ which sends P 3 to the product of the generic linear form L 3 and the generic homogeneous polynomial Q 3 of degree d 3 − 1 ≥ 1 and leave P 1 and P 2 invariant.We get, using the multiplicativity of resultants Therefore, by the induction hypothesis φ(D) = (U (1) 0,0 ) d2d3 R 1 R 2 where R 1 and R 2 are irreducible polynomials such that R 1 depends only on the coefficients of L 3 and R 2 depends only on the coefficients of Q 3 .Since φ is a homogeneous specialization, each irreducible factor of D which depends on P 3 must depends on L 3 and Q 3 , we deduce that D has only one irreducible factor depending on P 3 .Moreover, the irreducible factors of D which do not depend on P 3 are left invariant by φ so we deduce that D equals (U (1) 0,0 ) d2d3 times an irreducible factor and we are done.If d 2 ≥ 2 then we can argue exactly in the same way.
So it only remains to consider the case where d 2 = d 3 = 1.Let ψ be the specialization which sends P 1 to the product of the generic linear form L 1 := aX 1 +bX 2 +cX 3 times the generic homogeneous polynomial Q 1 of degree d 1 −1 and leaves P 2 and P 3 invariant.By the basic properties of resultants and our induction hypothesis we get: × R where R is an irreducible polynomial which does not depend on the coefficients of L 1 and does depend on the coefficients of Q 1 , P 2 of P 3 ; in particular it has degree 2(d 1 − 2) in the coefficients of Q 1 .Observe that Res(L 1 (X 3 ), Q 1 (X 3 ), P 2 (X 4 ), P 3 (X 4 )) is irreducible by theorem 3.1 and has degree d 1 − 1 in the coefficients of L 1 and 1 in the coefficients of Q 1 .Since ψ is an homogeneous specialization, each irreducible factor of D which depends on P 1 must depend on L 1 and Q 1 and have the same degree with respect to the coefficients of these two polynomials.From this property and the above computation we deduce that D has only one irreducible factor that depends on P 1 .Moreover, since the others irreducible factors of D are left invariant by ψ we deduce that D equals (U (1) 0,0 ) d1 times an irreducible polynomial.
Corollary 4.2.Given three polynomials f k (x, y, z), k = 1, 2, 3, of the form where x denotes a set of variables (x 1 , . . ., x n ) for some integer n ≥ 1 and S is any commutative ring, then the iterated resultant Res y (Disc Moreover, if the polynomials f 1 , f 2 , f 3 are sufficiently generic then this iterated resultant has exactly degree d 1 (d 1 − 1)d 2 d 3 in x and the iterated resultant We can now specialize the formula of Proposition 4.1 to get the factorization of two kinds of iterated resultants: the resultant of two discriminants and the resultant of a discriminant of a polynomial f and a resultant of f and another polynomial.We begin with the simplest one.
Then the following equality holds in U: Moreover, the resultant Res X2 (D 1 , D 2 ) ∈ U is irreducible and bi-homogeneous with respect to the coefficients (U (1) i,j ) i,j and (U Proof.We set D := (U and hence, using the formula proved in Proposition 4.1, where we specialize the polynomial P 3 to the polynomial ∂ 3 P 2 , we deduce that, in U, D is equal to The classical multi-degree formula for resultants gives the claimed result concerning the bi-degree of the iterated resultant Res X2 (D 1 , D 2 ).To prove its irreducibility, we proceed by induction on the integer First, if d = 4, that is to say d 1 = d 2 = 2, then we check by hand (or with a computer) that D equals an irreducible polynomial in U times the factor (U We now assume that d ≥ 5. Without loss of generality we can also assume that d 1 ≥ d 2 (since the problem is completely symmetric in P 1 and P 2 ) and hence that d 1 ≥ 3. Consider the specialization φ which sends P 1 to the product of the generic linear form L 1 := aX 1 + bX 2 + cX 3 times the generic homogeneous polynomial Q 1 of degree d 1 − 1 and leave P 2 invariant.Using properties of resultants we get: Using our inductive hypothesis and Proposition 4.1 we deduce that where R 1 is an irreducible polynomial of degree d 2 (d 2 − 1)(d 1 − 1) in the coefficients of L 1 and d 2 (d 2 − 1) in the coefficients of Q 1 , and R 2 is an irreducible polynomial independent of the coefficients of L 1 and of degree 2d 2 (d 2 − 1)(d 1 − 2) in the coefficients of Q 1 .Note that we already know that (U (1) 0,0 ) d2(d2−1) (U (2) 0,0 ) d1(d1−1) is a factor of D.Moreover, since φ is an homogeneous specialization, each remaining irreducible factor of D which depends on P 1 must depend on L 1 and Q 1 with the same degree, so we deduce that R 2  1 R 2 comes from the same irreducible factor of D and we conclude that D equals (U (1) 0,0 ) d2(d2−1) (U (2) 0,0 ) d1(d1−1) times an irreducible polynomial in U.
where d 1 , d 2 ≥ 2, x denotes a set of variables (x 1 , . . ., x n ) for some integer n ≥ 1 and S is a commutative ring, then the iterated resultant Res y (Disc z (f 1 ), Disc z (f 2 )) ∈ S[x] is of degree at most d 1 (d 1 − 1)d 2 (d 2 − 1) in x and we have (a Moreover, if the polynomials f 1 , f 2 , f 3 are sufficiently generic then this iterated resultant has exactly degree d 1 (d 1 − 1)d 2 (d 1 − 1) in x and is irreducible.

Resultant of a discriminant and a resultant sharing one polynomial.
We now turn to the second specialization of Proposition 4.1 which is a little more intricate than the previous one.Note that this iterated resultant has also been studied in [26,Theorem 3.3].In order to improve the previous analysis, we begin with two technical results.We recall that we sometimes omit the variables X 1 , X 2 , i.e. we note P (X 3 ) instead of P (X 1 , X 2 , X 3 ) to not overload the text.
Proof.We will denote by R the above resultant.We first use a geometric argument to justify that R is the product of a certain power of the coefficient U (1) 0,0 and a certain power of an irreducible polynomial that we will denote T(P 1 , P 2 ) ∈ U.
We rewrite the polynomial P 1 as where the c i 's are homogeneous polynomials in U[X 1 , X 2 ] of degree i respectively; we have Embedding into the algebraic closure Q of Q, the variety defined by the equation R = 0 is the projection of the incidence variety 3,4 P 1 (x 3 , x 4 ) = P 2 (x 4 ) = 0} (where N Q denotes the affine space whose coordinates are the N := (d 1 + 2)(d 1 + 1) + (d 2 + 2)(d 2 + 1) indeterminate coefficients, over Q) by the canonical projection on the second factor Consider the canonical projection of W onto the first factor π 1 : W → È 3 , which is surjective, and denote by D the line in P 3 which is defined by the equations X 1 = 0 and X 2 = 0. On the one hand, we observe that for all x ∈ P 3 \ D the fiber π −1 1 (x) is a linear space of codimension 4 in N ; therefore the algebraic closure of W |P 3 \D is an irreducible variety of dimension N − 1 in È 3 × N whose projection by π 2 gives an irreducible hypersurface in N .We denote by T(P 1 , P 2 ) a defining equation of this irreducible hypersurface.On the other hand, for all x ∈ D the fiber π −1 1 (x) is always included in the linear space of equation U 0,0 = 0 (it is actually exactly this linear space if x = (0 : 0 : 1 : 0) and the linear space U (1) are constants that we have to determine (note that T(P 1 , P 2 ) can be chosen in U because we know that the variety defined by R = 0 can be defined by an equation in U since it is a resultant of polynomials in U).
Now, we will prove by induction on the integer First, by Lemma 4.5 and using the computation (4.2), we note that we always have We check by hand (or with a computer) that the induction hypothesis (4.3) is true for d = 4, i.e. d 1 = 3 and d 2 = 1 and we assume that d ≥ 5.If d 2 ≥ 2 we specialize P 2 to the product of two generic homogeneous polynomials, say P ′ 2 and P ′′ 2 ; then R(P 1 , P 2 ) specializes, by multiplicativity of the resultants, to (U (1) 0,0 ) d1d2 R(P 1 , P ′ 2 )R(P 1 , P ′′ 2 ).Since each irreducible factor of R(P 1 , P 2 ) must depend on P ′ 2 and P ′′ 2 and since we already know that c(d 1 , d 2 ) ≥ d 1 d 2 , we deduce that R satisfies (4.3) for all couple (d 1 , d 2 ) such that d 2 ≥ 2.
We now turn to the case where d 1 ≥ 4 and d 2 = 1.Consider the homogeneous specialization φ which sends P 1 to X 3 Q 1 where Q 1 is the generic homogeneous polynomial of degree d 1 − 1.Using the properties of resultants we get And since, by (2.9), δ Moreover, (2.6) implies that δ So finally, we deduce that Since T(Q 1 , P 2 ) is irreducible (by our induction hypothesis), it follows that r(d 1 , d 2 ) must equal 1 for each irreducible factor of the above specialization must appear to a power which is a multiple of r(d 1 , d 2 ).Also, since P 2 (X 1 , X 2 , 0) is a generic linear form in X 1 and X 2 it turns out that Res(Q is, up to a linear change of coordinates, a discriminant of a univariate polynomial: it equals U (1) 0,0 times an irreducible polynomial in U.Moreover, it is easy to see that U )) (for this resultant does not vanish under this condition).Therefore, we deduce that a(d 1 , d 2 ) ≤ d 1 d 2 and then that a(d 1 , d 2 ) = d 1 d 2 .Finally, the three resultants in the above specialization formula are clearly primitive 2 (either by the induction hypothesis for the last one or either because it stays primitive after the change of coordinate induced by the linear polynomial P 2 ) and it follows that c(d 1 , d 2 ) = 1.
The formula on the degree is obtained as a direct consequence of the known degree formula for the resultants of several homogeneous polynomials.Proposition 4.7.Suppose that d 1 ≥ 2. We set Then the following equality holds in U: where the irreducible polynomial T(P 1 , P 2 ) has been defined in Lemma 4.6 for Moreover, Res X1:X2:X3 (P 1 , P 2 , ∂ 3 P 1 ) ∈ U is irreducible and the iterated resultant ). Proof.First, the classical multi-degree formula for resultants gives the claimed result for the bi-degree of the iterated resultant Res X2 (D 1 , R 12 ).By Lemma 4.5, we have 3,4 P 1 ) and by Lemma 4.6, we know that X3,X4 P 1 ) = (U (1) 0,0 ) d1d2 T(P 1 , P 2 ), where T(P 1 , P 2 ) is an irreducible polynomial, which implies the claimed formula.
We proceed as we already did several times: by induction on the integer d := We check by hand that R is irreducible if d 1 = 2 and d 2 = 1 and suppose that d ≥ 4. If d 2 ≥ 2 then one specializes P 2 to a product of two generic forms and we conclude using the multiplicativity property of the resultant.Otherwise, we have d 1 ≥ 3 and one specializes P 1 to L 1 Q 1 where Q 1 is the generic homogeneous polynomial of degree d 1 − 1 and L 1 := aX 1 + bX 2 + cX 3 is the generic linear form; this sends R to where Res(Q 1 , P 2 , ∂ 3 Q 1 ) is irreducible by our induction hypothesis and also where Res(Q 1 , P 2 , L 1 ) is irreducible for it is the resultant of three generic polynomials.
Examining the degrees in the coefficients of Q 1 and L 1 of the above factors and using the fact that each irreducible factor of R must specializes to a polynomial having the same degree in the coefficients of Q 1 and L 1 , we deduce that R is irreducible.
A specialization of this proposition gives the following result which covers and precises [26,Theorem 3.3].
Corollary 4.8.Given two polynomials f k (x, y, z), k = 1, 2, of the form where d 1 , d 2 ≥ 2, x denotes a set of variables (x 1 , . . ., x n ) for some integer n ≥ 1 and S is any commutative ring, then the iterated resultant is of degree at most d 2 1 d 2 (d 1 − 1) in x and we have where we recall that, in S[x], we have Moreover, if the polynomials f 1 , f 2 , f 3 are sufficiently generic then this iterated resultant has exactly degree

Discriminant of a resultant
In this section, we suppose given two positive integers d 1 , d 2 ≥ 2 and two homogeneous polynomials where, as usual, U denotes the universal ring of coefficients [U i,j ].We will hereafter focus on the factorization of a discriminant of a resultant.
Proof.Introduce a new indeterminate U and set where the a i 's and the b j 's are polynomials in U[X 2 , U ]. Expanding the resultant with respect to its two last rows, we get where ∆ i (i = 1, 2, 3) are polynomials in the a i 's and b j 's.Taking the derivative with respect to the variable X 2 , we deduce that (5.1) Now, it is easy to check that we have Moreover, by invariance property (change of bases formula) and the same is true for the subresultant SRes X3 (P 1 , P 2 ).Therefore we deduce that (5.1) is nothing but the claimed equality by substituting X 3 by U .This lemma implies the following factorization.Proposition 5.2.In U, we have the equality

This iterated resultant is bi-homogeneous with respect to the sets of coefficients
and recall that, by definition, we have R 0 Disc X1:X2:X3 (P 1 , P 2 ) = Res(P 1 , P 2 , J 1 ).Setting and hence to deduce that Denoting D the above quantity, we have and finally Now, since two consecutive integers are always relatively prime, we deduce that c(d 1 , d 2 ) = ±1 ∈ Z.To determine exactly this integer, we compute the other side of the claimed equality.For simplicity, we will consider the specialization ψ which is similar to the specialization φ with in addition A = 0. We have It is not hard to see from the definition of the subresultant that SRes (1) and also to compute 1) .
At this point, the factorization given in the above proposition is not complete since we only know that one factor is irreducible.The following result shows that the second factor is not irreducible, but is the square of an irreducible polynomial and moreover that it can be interpreted as a particular iterated resultant itself.
where the a i 's and the b j 's are polynomials in U[X 1 , X 2 , U ].The subresultant SRes (1) , determinant which remains unchanged if we add, for all i = 1, . . ., d 1 + d 2 − 3, the line number i times to the last line which then becomes of the form where R 1 and R 2 are polynomials in U[X 1 , X 2 , X 3 , X 4 , U ].It follows that, by developping this determinant with respect to the last line, SRes ) and by invariance of the subresultant under the change of coordinates U ← X 3 + U we deduce that SRes which implies, after a substitution of U by X 3 and a suitable use of the divisibility property of the resultants, that divides Res X1:X2:X3 (P 1 , P 2 , SRes X3 (P 1 , P 2 )).An easy computation shows that these two resultants have the same bi-degree w.r.t. the coefficients of P 1 and P 2 ; therefore they are equal up to sign in U (we have already seen that Res X1:X2:X3 (P 1 , P 2 , SRes X3 (P 1 , P 2 )) is a primitive polynomial in U through a particular specialization).To determine the sign we consider again the specialization for which it is easy to see that both resultants then specialize to 1.
The following result can be seen as the main explanation of [26, theorem 3.4 It is bi-homogeneous with respect to the set of coefficients (U i,j ) i,j and(U i,j ) i,j of bi-degree Proof.Let us denote by R the above resultant.Embedding into the algebraic closure of É, the variety defined by the equation R = 0 is the projection of the incidence variety i,j ) ∈ È 3 × N such that P 1 = δ 3,4 (P 1 ) = P 2 = δ 3,4 (P 2 ) = 0} (where N is the number of indeterminate coefficients) by the canonical projection on the second factor π 2 : W ⊂ È 3 × N → N .But for a generic point (a (k) i,j ) ∈ N such that R = 0, we have at least two pre-image in W since if (x 1 : x 2 : x 3 : x 4 ) is such pre-image then (x 1 : x 2 : x 4 : x 3 ) is also a pre-image (which is generically different).It follows that the co-restriction of π 2 to the variety R = 0 has degree at least 2 and hence that, in U, where p is a positive integer, c(d) is also a positive integer but may depend on d := (d 1 , d 2 ) and r 1 (d), . . ., r p (d) are positive integers greater or equal to 2 and may also depend on d.Note that we know by Lemma 5.3 that c(d) = 1.To determine the other quantities we will proceed by induction on d 1 + d 2 ≥ 4 (remember that d 1 ≥ 2 and d 2 ≥ 2).First, we can check by hand (or with a computer) that the claim is true if d 1 = d 2 = 2: R is the square of an irreducible polynomial in U of bi-degree (3,3).Now, assume that d 1 + d 2 > 4; without loss of generality one may assume that d 1 ≥ 3. Consider the homogeneous specialization φ which sends P 1 to the product L 1 Q 1 where Q 1 is generic homogeneous polynomials of degree d 1 − 1 ≥ 2 and L 1 := aX 1 + bX 2 + cX 3 is a generic linear form.We have, using obvious notations, (5.2) since exchanging the role of X 3 and X 4 , we have Res(L 1 (X 3 ), Q 1 (X 4 ), P 2 (X 3 ), δ 3,4 (P 2 )) = Res(L 1 (X 4 ), Q 1 (X 3 ), P 2 (X 4 ), δ 3,4 (P 2 )) = Res(L 1 (X 4 ), Q 1 (X 3 ), P 2 (X 3 ), δ 3,4 (P 2 )) = Res(Q 1 (X 3 ), L 1 (X 4 ), P 2 (X 3 ), δ 3,4 (P 2 )).
Gathering the results of this section, we obtain the full factorization of the discriminant of a resultant.Theorem 5.6.Introducing a new indeterminate X 4 , we have in U: where Disc X1:X2:X3 (P 1 , P 2 ) and D(P 1 , P 2 ) are irreducible polynomials in U.This iterated resultant is bi-homogeneous with respect to the sets of coefficients U (1) i,j and U (2) i,j respectively of bi-degree As usual, we can specialize this result to obtain the following: Corollary 5.7.Given two polynomials f k (x, y, z), k = 1, 2, of the form where x denotes a set of variables (x 1 , . . ., x n ) for some integer n ≥ 1 and S is any commutative ring, the iterated resultant Disc y (Res in x and can be factorized, up to sign, as Moreover, if the polynomials f 1 and f 2 are sufficiently generic, then this iterated resultant has exactly degree d 1 d 2 (d 1 d 2 − 1) in x and the two terms in the right hand

Discriminant of a discriminant
In this section, we are interested in analyzing two iterated discriminants.Before going into the algebraic study, let us consider the problem from a geometric point of view.Suppose we are given an implicit surface f (x, y, z) = 0. Computing the discriminant of f in z consists in projecting the apparent contour (or polar) curve in the z direction, which is defined by f (x, y, z) = 0, ∂ z f (x, y, z) = 0. Computing the discriminant in y of this discriminant in z of f consists in computing the position of lines parallel to the y axis, which are tangent to the projected curve.
We illustrate it by some explicit computations, with the polynomial Its discriminant in z is a polynomial of degree 12 and the discriminant in y of this discriminant can be factorized as: Figures 3 1 and 2 illustrate the situation, where we represent the surface f = 0 and the projection of its apparent contour (the x-direction is pointing to the top of the image and z-direction to the left).
The first factor of degree 10 has 4 real roots corresponding to the smooth points of the surface with a tangent plane orthogonal to the x-direction.The second factor of multiplicity 2 corresponds to points of the polar variety which project in the (x, y)-plane onto the same point.Geometrically speaking, we have a double folding of the surface in the z-direction above these values.There are two such real points in our example.The last factor of multiplicity 3 corresponds to cusp points on the discriminant curve, which are the projection of a "fronce" or a pleat of the surface.There are 4 such real points.Notice that the branches of the discriminant curve between two of these cusp points and one of the double folding points form a very tiny loop, which is difficult to observe at this scale.
These phenomena can be explained from an algebraic point of view, as we will see hereafter.They have also been analyzed from a singularity theory point of view.A well-know result in singularity theory, due to H. Whitney (see [31,25,2]) asserts that the singularities of the projection of a generic surface onto a plane are of 3 types: • a regular point on the contour curve corresponding to a fold of the surface, • a cusp corresponding to a pleat, • a double point corresponding to the projection of two transversal folds.These are stable singularities, which remain by a small perturbation of the surface or of the direction of projection.For a more complete analysis of the singularities of the apparent contour, see also [29].
We are now going to analyze the algebraic side of these geometric properties for generic polynomials of a given degree.Suppose given a positive integer d ≥ 3 and a homogeneous polynomial where U is the universal coefficients ring [U i,j ].In this section we will study the discriminant of the discriminant of P .Here is the first factorization we can get as a specialization of the iterated resultant we studied in the previous section.
Proposition 6.1.We have the equality in U: Proof.This is essentially a specialization of the formula given in Proposition 5.6, which yields in our case the equality Res(P, δ 3,4 P, ∂ 3 P, δ 3,4 ∂ 3 P ).
Again by Euler identity, the relation (2.4) yields and by substitution in (6.2) and simplification by Res(∂ 2 P, ∂ 3 P, X 1 ) and d d−2 , we get that U 0,0 Disc(P, ∂ 3 P ) = Disc(P )Res(P, since Disc X3 (P (1, X 2 , X 3 )) is of degree d(d − 1) in X 2 and the discriminant of a polynomial of degree D is homogeneous of degree 2 (D − 1) in its coefficients.The claimed formula then follows immediately.
Concerning the degree, observe that Disc X3 (P (1, X 2 , X 3 )) has coefficients of degree 2 (d − 1) in the coefficients of P .The discriminant of a polynomial of degree D being homogeneous of degree 2 (D − 1) in its coefficients, we obtain that Disc X2 (Disc In the factorization of Disc X2 (Disc X3 (P )) given in this proposition, we only know that the factor Disc(P ) is known to be irreducible in U.The remaining of this section is devoted to the study of the full factorization of the two other factors.We begin with the study of the factor appearing in Proposition 6.1 which corresponds to the resultant of P , its first and second derivatives with respect to X 3 .a(d) ≥ 2 for all d ≥ 3, that is to say that U 2 0,0 divides Res(P, ∂ 3 P, ∂ It is clear that ∂ 2 3 P ∈ (X 1 , X 2 , U 0,0 X 3 ) and that Moreover, X 2 1 ∂ 2 1 P + 2X 1 X 2 ∂ 1 ∂ 2 P + X 2 2 ∂ 2 2 P ∈ (X 1 , X 2 ) 2 ⊂ (X 1 , X 2 , U 0,0 X 3 ) 2 .Therefore, the divisibility property of resultants implies that Res(X 1 , X 2 , U 0,0 X 3 ) 2×1×1 divides R, and since Res(X 1 , X 2 , U 0,0 X 3 ) = U 0,0 , we are done.

Date:
February 1, 2008.This work was first presented at the conference in honor of Jean-Pierre Jouanolou, held at Luminy, Marseille, 15-19 th May 2006.

4. 1 .
Resultant of a discriminant and a resultant.Proposition 4.1.Assume that d 1 ≥ 2 and set X2 (D 1 , R 23 ).Now, we proceed by induction on the integer d := d 1 + d 2 + d 3 ≥ 4 (remember d 1 ≥ 2, d 2 ≥ 1 and d 3 ≥ 1) to prove the irreducibility of the iterated resultant Res X3 (D 1 , R 23 ).First, if d = 4, that is to say d 1 = 2 and d 2 = d 3 = 1, then we check by hand that D equals an irreducible polynomial in U times U

Figure 1 .
Figure 1.A quartic surface and the projection of a polar curve.

Figure 2 .
Figure 2. The polar curve of the quartic surface projected on a plane.
Notations and derivatives.Let S be a commutative ring and suppose given a polynomial P (X) ∈ S[X].We denote by ∂ X P the formal derivative of P with respect to the variable X.More precisely, the map r], yields polynomials in U for which we are going to give irreducible factorizations.As in geometric applications, this computation is applied with the coefficients U 1, that is, by induction on the integer d := d 1 + d 2 + d 3 ≥ 4. We can check by hand (or with a computer) that D is an irreducible polynomial in U if (d 1 , d 2 , d 3 ) = (2, 1, 1).We thus assume that D is irreducible up to a given integer p ≥ 4 and we will prove that D is irreducible ifd = p + 1.If d 2 ≥ 2 (resp.d 3 ≥2) then we can specialize P 2 (resp.P 3 ) as a product of a generic linear form and a generic form of degree d 2 − 1 ≥ 1 (resp.d 3 − 1 ≥ 1) and conclude, exactly as we did in Theorem 3.1, that D is then irreducible.Otherwise, then d 1 ≥ 3 and we specialize P 1 to the product of a generic linear form ]. Introducing a new indeterminate X 4 , there exists a non-zero irreducible polynomial D(P 1 , P 2 ) in U such that Res X1:•••:X4 (P 1 , δ 3,4 (P 1 ), P 2 , δ 3,4 (P 2 )) = D(P 1 , P 2 ) 2 .