EXPLICIT FORMULAS FOR THE MULTIVARIATE RESULTANT

CARLOS D’ ANDREA AND ALICIA DICKENSTEIN

Abstract. We present formulas for the multivariate resultant as a quotient of two determinants. They extend the classical Macaulay formulas, and involve matrices of considerably smaller size, whose non zero entries include coefficients of the given polynomials and coefficients of their Bezoutian. These formulas can also be viewed as an explicit computation of the morphisms and the determinant of a resultant complex.

1. Introduction

Given n homogeneous polynomials f_{1}, …, f_{n} in n variables over an algebraically closed field k with respective degrees d_{1}, …, d_{n}, the resultant Res_{d_{1},...,d_{n}}(f_{1}, …, f_{n}) is an irreducible polynomial in the coefficients of f_{1}, …, f_{n}, which vanishes whenever f_{1}, …, f_{n} have a common root in projective space. The study of resultants goes back to classical work of Sylvester, Bézout, Cayley, Macaulay and Dixon. The use of resultants as a computational tool for elimination of variables as well as a tool for the study of complexity aspects of polynomial system solving in the last decade, has renewed the interest in finding explicit formulas for their computation (cf. [1], [3], [4], [5],[14], [18],[20], [22],[23]).

By a determinantal formula it is meant a matrix whose entries are polynomials in the coefficients of f_{1}, …, f_{n} and whose determinant equals the resultant Res_{d_{1},...,d_{n}}(f_{1}, …, f_{n}). The interest on such a formula is the computation of the resultant, and so it is implicit that the entries should be algorithmically computed from the inputs. It is also meant that all non-zero entries have degree strictly less than the degree of the resultant.

In case all d_{i} have a common value d, all currently known determinantal formulas are listed by Weyman and Zelevinsky in [27]. This list is short: if d ≥ 2, there exist determinantal formulas for all d just for binary forms (given by the well known Sylvester matrix), ternary forms and quaternary forms; when n = 5, the only possible values for d are 2 and 3; finally, for n = 6, there exists a determinantal formula only for d = 2. We find similar strict restrictions on general n, d_{1}, …, d_{n} (cf. Lemma 5.3).

Both authors are supported by Universidad de Buenos Aires, grant TX094.

The second author is also supported by CONICET, Argentina, and the Wenner-Gren Foundation, Sweden.

Given d_{1}, …, d_{n}, denote t_{n} := ∑_{i=1}^{n}(d_{i}−1) the critical degree. Classical Macaulay formulas [21] describe the resultant Res_{d_{1},...,d_{n}}(f_{1}, …, f_{n}) as an explicit quotient of two determinants. These formulas involve a matrix of size at least the number of monomials in n variables of degree t_{n} + 1, and a submatrix of it.

Macaulay’s work has been revisited and sharpened by Jouanolou in [17], where he proposes for each t ≥ 0, a square matrix M_{t} of size

ρ (t) := \binom{t + n − 1}{n − 1} + i(t_{n} − t) (1)

whose determinant is a nontrivial multiple of Res_{d_{1},...,d_{n}}(f_{1}, …, f_{n}) (cf. [17], 3.11.19.7). Here, i(t_{n} − t) denotes the dimension of the k−vector space of elements of degree t_{n} −t in the ideal generated by a regular sequence of n polynomials with degrees d_{1}, …, d_{n}. Moreover, Jouanolou shows that the resultant may be computed as the ratio between the determinant of M_{t_{n}} and the determinant of one of its square submatrices. (cf. [17], Corollaire 3.9.7.7).

In this paper, we explicitly find the extraneous factor in Jouanolou’s formulation, i.e. the polynomial det(M_{t})/Res_{d_{1},...,d_{n}}(f_{1}, …, f_{n}), for all t ≥ 0 which again happens to be the determinant of a submatrix 𝔼_{t} of M_{t} for every t, and this allows us to present new resultant formulas à la Macaulay for the resultant, i.e. as a quotient of two determinants

(2) Res_{d_{1},...,d_{n}}(f_{1}, …, f_{n}) = \frac{det(M_{t})}{det(𝔼_{t})}.

For t > t_{n}, we recover Macaulay’s classical formulas. For t ≤ t_{n}, the size of the matrix M_{t} is considerably smaller.

In order to give explicit examples, we need to recall the definition of the Bezoutian associated with f_{1}, …, f_{n} (cf. [2], [16], [19], [25] and [17] under the name “Formes de Morley”). Let (f_{1}, …, f_{n}) be a sequence of generic homogeneous polynomials with respective degrees d_{1}, …, d_{n}

f_{i} := ∑_{|α_{i}|=d_{i}} a_{α_{i}}X^{α_{i}} ∈ A [X_{1}, …, X_{n}] ,

where A is the factorial domain A := ℤ [a_{α_{i}}]_{|α_{i}|=d_{i},i=1,...,n} .

Introduce two sets of n variables X, Y and for each pair (i, j) with 1 ≤ i, j ≤ n, write Δ_{ij}(X, Y) for the incremental quotient

\frac{f_{i}(Y_{1}, …, Y_{j−1}, X_{j}, …, X_{n}) − f_{i}(Y_{1}, …, Y_{j}, X_{j+1}, …, X_{n})}{X_{j} − Y_{j}}. (3)

Note that f_{i}(X) − f_{i}(Y) = ∑_{j=1}^{n} Δ_{ij}(X, Y)(X_{j} − Y_{j}).

The determinant

Δ(X, Y) := det(Δ_{ij}(X, Y))_{1≤i,j≤n} = ∑_{|γ|≤t_{n}} Δ_{γ} (X) .Y^{γ}. (4)

is a representative of the Bezoutian associated with (f_{1}, …, f_{n}) . It is a homogeneous polynomial in A [X, Y] of degree t_{n}.

Recall also that

deg Res_{d_{1},...,d_{n}}(f_{1}, …, f_{n}) = ∑_{i=1}^{n} d_{1} … d_{i−1} · d_{i} … d_{n}.

As a first example, let n = 3, (d_{1}, d_{2}, d_{3}) = (1, 1, 2) , and let

\begin{aligned}f_{1} & = & a_{1}X_{1} + a_{2}X_{2} + a_{3}X_{3} \\ f_{2} & = & b_{1}X_{1} + b_{2}X_{2} + b_{3}X_{3} \\ f_{3} & = & c_{1}X_{1}^{2} + c_{2}X_{2}^{2} + c_{3}X_{3}^{2} + c_{4}X_{1}X_{2} + c_{5}X_{1}X_{3} + c_{6}X_{2}X_{3} \end{aligned}

be generic polynomials of respective degrees 1, 1, 2. Here, t_{3} = 1. Macaulay’s classical matrix M_{2} looks as follows:

( \begin{matrix}a_{1} & 0 & 0 & 0 & 0 & c_{1} \\ 0 & a_{2} & 0 & b_{2} & 0 & c_{2} \\ 0 & 0 & a_{3} & 0 & b_{3} & c_{3} \\ a_{2} & a_{1} & 0 & b_{1} & 0 & c_{4} \\ a_{3} & 0 & a_{1} & 0 & b_{1} & c_{5} \\ 0 & a_{3} & a_{2} & b_{3} & b_{2} & c_{6} \end{matrix} )

and its determinant equals −a_{1}Res_{1,1,2}. The extraneous factor is the 1 × 1 minor formed by the element in the fourth row, second column.

On the other hand, because of Lemma 5.3, we can exhibit a determinantal formula for ±Res_{1,1,2}, and it is given by Proposition 5.6 for t = [\frac{t_{3}}{2} ] = 0 by the determinant of

( \begin{matrix}Δ_{(1,0,0)} & a_{1} & b_{1} \\ Δ_{(0,1,0)} & a_{2} & b_{2} \\ Δ_{(0,0,1)} & a_{3} & b_{3} \end{matrix} ) ,

where Δ_{γ} are coefficients of the Bezoutian (4). Explicitly, we have

Δ_{(1,0,0)} = c_{1}(a_{2}b_{3} − a_{3}b_{2}) − c_{4}(a_{1}b_{3} − a_{3}b_{1}) + c_{5}(a_{1}b_{2} − a_{2}b_{1}),
Δ_{(0,1,0)} = c_{6}(a_{1}b_{2} − a_{2}b_{1}) − c_{2}(a_{1}b_{3} − b_{1}a_{3})

and

Δ_{(0,0,1)} = c_{3}(a_{1}b_{2} − b_{1}a_{2}).

This is the matrix M_{0} corresponding to the linear transformation Ψ_{0} which is defined in (9).

Take now n = 4, and (d_{1}, d_{2}, d_{3}, d_{4}) = (1, 1, 2, 3) . The critical degree is 3. Macaulay’s classical matrix, M_{4}, has size 35 × 35. Because the degree of Res_{1,1,2,3} is 2 + 3 + 6 + 6 = 17, we know that its extraneous factor must be a minor of size 18 × 18. By Proposition 5.6, we can find the smallest possible matrix for t = 1 or t = 2. Set t = 2. We get the following 12 × 12 matrix

Δ_{(2,0,0,0)}^{1} Δ_{(2,0,0,0)}^{2} Δ_{(2,0,0,0)}^{3} Δ_{(2,0,0,0)}^{4} a_{1} 0 0 0 0 0 0 c_{1}
Δ_{(0,2,0,0)}^{1} Δ_{(0,2,0,0)}^{2} Δ_{(0,2,0,0)}^{3} Δ_{(0,2,0,0)}^{4} 0 a_{2} 0 0 b_{2} 0 0 c_{2}
Δ_{(0,0,2,0)}^{1} Δ_{(0,0,2,0)}^{2} Δ_{(0,0,2,0)}^{3} Δ_{(0,0,2,0)}^{4} 0 0 a_{3} 0 0 b_{3} 0 c_{3}
Δ_{(0,0,0,2)}^{1} Δ_{(0,0,0,2)}^{2} Δ_{(0,0,0,2)}^{3} Δ_{(0,0,0,2)}^{4} 0 0 0 a_{4} 0 0 b_{4} c_{4}
Δ_{(1,1,0,0)}^{1} Δ_{(1,1,0,0)}^{2} Δ_{(1,1,0,0)}^{3} Δ_{(1,1,0,0)}^{4} a_{2} a_{1} 0 0 b_{1} 0 0 c_{5}
Δ_{(1,0,1,0)}^{1} Δ_{(1,0,1,0)}^{2} Δ_{(1,0,1,0)}^{3} Δ_{(1,0,1,0)}^{4} a_{3} 0 a_{1} 0 0 b_{1} 0 c_{6}
Δ_{(1,0,0,1)}^{1} Δ_{(1,0,0,1)}^{2} Δ_{(1,0,0,1)}^{3} Δ_{(1,0,0,1)}^{4} a_{4} 0 0 a_{1} 0 0 b_{1} c_{7}
Δ_{(0,1,1,0)}^{1} Δ_{(0,1,1,0)}^{2} Δ_{(0,1,1,0)}^{3} Δ_{(0,1,1,0)}^{4} 0 a_{3} a_{2} 0 b_{3} b_{2} 0 c_{8}
Δ_{(0,1,0,1)}^{1} Δ_{(0,1,0,1)}^{2} Δ_{(0,1,0,1)}^{3} Δ_{(0,1,0,1)}^{4} 0 a_{4} 0 a_{2} b_{4} 0 b_{2} c_{9}
Δ_{(0,0,1,1)}^{1} Δ_{(0,0,1,1)}^{2} Δ_{(0,0,1,1)}^{3} Δ_{(0,0,1,1)}^{4} 0 0 a_{4} a_{3} 0 b_{4} b_{3} c_{10}
a_{1}  a_{2}  a_{3}  a_{4}  0 0 0 0 0 0 0 0
b_{1}  b_{2}  b_{3}  b_{4}  0 0 0 0 0 0 0 0

where

f_{1} =  a_{1}X_{1} + a_{2}X_{2} + a_{3}X_{3} + a_{4}X_{4}
f_{2} =  b_{1}X_{1} + b_{2}X_{2} + b_{3}X_{3} + b_{4}X_{4}
f_{3} = c_{1}X_{1}^{2} + c_{2}X_{2}^{2} + c_{3}X_{3}^{2} + c_{4}X_{4}^{2} + c_{5}X_{1}X_{2} + c_{6}X_{1}X_{3}
+c_{7}X_{1}X_{4} + c_{8}X_{2}X_{3} + c_{9}X_{2}X_{4} + c_{10}X_{3}X_{4},

f_{4} is a homogeneous generic polynomial of degree 3 in four variables, and for each γ, |γ| = 2, we write

Δ_{γ}(X) = ∑_{j=1}^{4} Δ_{γ}^{j}X_{j},

which has degree 1 in the coefficients of each f_{i}, i = 1, …, 4. The determinant of this matrix is actually ±a_{1}Res_{1,1,2,3}. Here, the extraneous factor is the minor 1 × 1 of the matrix obtained by taking the element in the fifth row, sixth column.

In the following table, we display the minimal size of the matrices M_{t} and the size of classical Macaulay matrix for several values of n, d_{1}, …, d_{n}.

n  (d_{1}, …, d_{n})  min size  classical 2  (10, 70)  70  80 2  (150, 200)  200  350 3  (1, 1, 2)  3  6 3  (1, 2, 5)  14  28 3  (2, 2, 6)  21  45 4  (1, 1, 2, 3))  12  35 4  (2, 2, 5, 5)  94  364 4  (2, 3, 4, 5)  90  364 5  (4, 4, 4, 4, 4)  670  4845 7 (2, 3, 3, 3, 3, 3, 3)  2373  38760 10  (3, 3, …, 3)  175803  14307150 20  (2, 2, …, 2)  39875264 131282408400 .

We give in section 4 an estimate for the ratio between these sizes. However, it should be noted that the number of coefficients of the Bezoutian that one needs to compute increases when the size of the matrix M_{t} decreases. We refer to [16] and [24] for complexity considerations on the computation of Bezoutians. In particular, this computation can be well parallelized. Also, the particular structure of the matrix and the coefficients could be used to improve the complexity estimates; this problem is studied for n = 2 and n = 3 in [11].

Our approach combines Macaulay’s original ideas [21], expanded by Jouanolou in [17], with the expression for the resultant as the determinant of a Koszul complex inspired by the work of Cayley [7]. We also use the work [9], [10] of Chardin on homogeneous subresultants, where a Macaulay style formula for subresultants is presented. In fact, we show that the proposed determinants are explicit non-zero minors of a bigger matrix which corresponds to one of the morphisms in a Koszul resultant complex which in general has many non zero terms, and whose determinant is Res_{d_{1},...,d_{n}}(f_{1}, …, f_{n}) (cf. Theorem 5.1). These are the complexes considered in [27],[15] in the equal degree case, built from the spectral sequence associated with a twisted Koszul complex at the level of sheaves.

We give explicit expressions for the morphisms in these complexes in terms of the Bezoutian associated with f_{1}, …, f_{n} for degrees under critical degree, addressing in this manner a problem raised by Weyman and Zelevinsky in [27] (cf. also [15, 13.1.C].

In the last sections, we show that different classical formulas can be viewed as special cases of the determinantal formulas that we present here (cf. [15],[27]). In particular, we also recover in this setting the “affine” Dixon formulas considered in [14] and we classify in particular all such determinantal formulas.

2. Notations and some preliminary statements

Let S_{u} denote the A-free module generated by the monomials in A[X] with degree u. If u < 0, then we set S_{u} = 0. Define also the following free submodules E^{t,j} ⊆ S^{t,j} ⊆ S_{t−d_{j}}, for all j = 1, …, n :

S^{t,j} := ⟨X^{γ}, |γ| = t − d_{j}, γ_{1} < d_{1}, …, γ_{j−1} < d_{j−1}⟩ (5)

(6) E^{t,j} := ⟨X^{γ} ∈ S^{t,j}, there exists i ≠ j : γ_{i} ≥ d_{i}⟩ .

Note that E^{t,n} = 0, and S^{t,1} = S_{t−d_{1}} ∀t ∈ ℕ_{0}.

Let j_{u} : S_{u} → S_{u}^{∗} be the isomorphism associated with the monomial bases in S_{u} and denote by T_{γ} := j_{u}(X^{γ}) the elements in the dual basis. Convention. All spaces that we will consider have a monomial basis, or a dual monomial basis. We shall suppose all these bases have a fixed order. This will allow us to define matrices “in the monomial bases”, with no ambiguity.

Let ψ_{1,t} be the A-linear map

ψ_{1,t} : S_{t_{n}−t}^{∗} → S_{t}

which sends

T_{γ} ↦ Δ_{γ} (X) , (7)

where the polynomial Δ_{γ} (X) is defined in (4). Let Δ_{t} denote the matrix of ψ_{1,t} in the monomial bases.

Lemma 2.1. For suitable orders of the monomial bases in S_{t} and S_{t_{n}−t}, we have that

tΔ_{t} = Δ_{t_{n}−t}.

Proof. It holds that Δ (X, Y) = Δ (Y, X) by the symmetry property of Bezoutians (cf. [17, 3.11.8]). This implies that

∑_{|γ|=t_{n}−t} Δ_{γ} (X) Y^{γ} = ∑_{|λ|=t} Δ_{λ} (Y) X^{λ} = ∑_{|γ|=t_{n}−t, |λ|=t} c_{γλ}X^{λ}Y^{γ},

with c_{γλ} ∈ A. It is easy to see that if Δ_{t} = (c_{γλ})_{|γ|=t_{n}−t, |λ|=t} then Δ_{t_{n}−t} = (c_{γλ})_{|λ|=t, |γ|=t_{n}−t} .

Let us consider also the Sylvester linear map ψ_{2,t} :

\begin{aligned}ψ_{2,t} : & S^{t,1} & ⊕ & ⋯ & ⊕ & S^{t,n} & → & S_{t} \\ & ( g_{1} & , & … & , & g_{n}) & ↦ & ∑_{i=1}^{n} g_{i}f_{i}, \end{aligned} (8)

and denote by D_{t} its matrix in the monomial bases. As usual, ψ_{2,t_{n}−t}^{∗} denotes the dual mapping of (8) in degree t_{n} − t.

Denote

Ψ_{t} : S_{t_{n}−t}^{∗} ⊕ (S^{t,1} ⊕ ⋯ ⊕ S^{t,n}) → S_{t} ⊕ (S^{t_{n}−t,1} ⊕ ⋯ ⊕ S^{t_{n}−t,n})^{∗} (9)

the A-morphism defined by

(T , g ) ↦ (ψ_{1,t}(T) + ψ_{2,t}(g), ψ_{2,t_{n}−t}^{∗} (T)) , (10)

and call M_{t} the matrix of Ψ_{t} in the monomial bases.

Denote also by E_{t} the submatrix of M_{t} whose columns are indexed by the monomials in E^{t,1} ∪ ⋯ ∪ E^{t,n−1}, and whose rows are indexed by the monomials X^{γ} in S_{t} for which there exist two different indices i, j such that γ_{i} ≥ d_{i}, γ_{j} ≥ d_{j}. With these choices it is not difficult to see that M_{t} and E_{t} (when defined) are square matrices.

Remark 2.2. Observe that E_{t} is actually a submatrix of D_{t}. In fact, E_{t} is transposed of the square submatrix named E (t) in [10], and whose determinant is denoted by Δ (n, t) in [21, Th. 6].

Lemma 2.3. M_{t} is a square matrix of size ρ(t), where ρ is the function defined in (1).

Proof. The assignment which sends a monomial m in S^{t,i} to x_{i}^{d_{i}} · m injects the union of the monomial bases in each S^{t,i} onto the monomials of degree t which are divisible by some x_{i}^{d_{i}}. It is easy to see that the cardinality of the set of complementary monomials of degree t is precisely H_{d}(t), where H_{d}(t) denotes the dimension of the t-graded piece of the quotient of the polynomial ring over k by the ideal generated by a regular sequence of homogeneous polynomials with degrees d_{1}, …, d_{n} (cf. [17, 3.9.2]). Moreover, using the assignment (γ_{1}, …, γ_{n}) ↦ (d_{1} − 1 − γ_{1}, …, d_{n} − 1 − γ_{n}) , it follows that

H_{d}(t) = H_{d}(t_{n} − t). (11)

We can compute explicitly this Hilbert function by the following formula (cf. [21, §2]):

\begin{matrix}∏ & {n \atop i=1} & (1 − Y^{d_{i}}) \\ (1 − Y) & n \end{matrix} = ∑_{t=0}^{∞} H_{d}(t).Y^{t}. (12)

Then,

rk (S^{t,1} ⊕ ⋯ ⊕ S^{t,n}) = rk S_{t} − H_{d}(t).

Similarly,

rk (S^{t_{n}−t,1} ⊕ ⋯ ⊕ S^{t_{n}−t,n})^{∗} = rk (S_{t_{n}−t})^{∗} − H_{d}(t_{n} − t).

Therefore, M_{t} is square of size rk S_{t} − H_{d}(t_{n} − t) + rk S_{t_{n}−t}. Since i(t_{n} − t) = rk S_{t_{n}−t} − H_{d}(t_{n} − t), the size of M_{t} equals rk S_{t} + i(t_{n} − t) = ρ(t).

Remark 2.4. Ordering properly the monomial bases, M_{t} is the transpose of the matrix which appears in [17, 3.11.19.7]. It has the following structure:

[ \begin{matrix}Δ_{t} & D_{t} \\ ^{t}D_{t_{n}−t} & 0 \end{matrix} ] . (13)

Remark 2.5. Because ψ_{2,t} = 0 if and only if t < min{d_{i}}, we have that Ψ_{t} = ψ_{2,t} + ψ_{1,t} if t > t_{n} − min{d_{i}}, and Ψ_{t} = ψ_{2,t} if t > t_{n}.

Finally, denote 𝔼_{t} the square submatrix of M_{t} which has the following structure:

𝔼_{t} = [ \begin{matrix}∗ & E_{t} \\ ^{t}E_{t_{n}−t} & 0 \end{matrix} ] . (14)

It is clear from the definition that det(𝔼_{t}) = ± det(E_{t}) det(E_{t_{n}−t}).

Remark 2.6. Dualizing (10) and using lemma 2.1 with a careful inspection at (13) and (14), we have that ordering properly their rows and columns,

tM_{t} = M_{t_{n}−t} and ^{t}𝔼_{t} = 𝔼_{t_{n}−t}.

3. Generalized Macaulay formulas

We can extend the map ψ_{2,t} in (8) to the direct sum of all homogeneous polynomials with degrees t − d_{1}, …, t − d_{n}, and the map ψ_{2,t_{n}−t} to the direct sum of all homogeneous polynomials with degrees t_{n} − t − d_{1}, …, t_{n} − t − d_{n}, to get a map

Ψ̃_{t} : (S_{t_{n}−t})^{∗}⊕(S_{t−d_{1}} ⊕ ⋯ ⊕ S_{t−d_{n}}) → S_{t}⊕(S_{t_{n}−t−d_{1}} ⊕ ⋯ ⊕ S_{t_{n}−t−d_{n}})^{∗} .

We can thus see the matrix M_{t} of Ψ_{t} in (9) as a choice of a square submatrix of Ψ̃_{t}. We will show that its determinant is a non zero minor of maximal size.

Proposition 3.1. Let M′_{t} be a square matrix over A of the form

M′_{t} := [ \begin{matrix}Δ_{t} & F_{t} \\ ^{t}F_{t_{n}−t} & 0 \end{matrix} ] (15)

where F_{t} has i(t) columns and corresponds to a restriction of the map

\begin{matrix}S_{t−d_{1}} ⊕ ⋯ ⊕ S_{t−d_{n}} & → & S_{t} \\ (g_{1}, …, g_{n}) & ↦ & ∑_{i=1}^{n} g_{i} f_{i}; \end{matrix}

and similarly for F_{t_{n}−t} in degree t_{n} − t. Then, det(M′_{t}) is a multiple of Res_{d_{1},...,d_{n}}(f_{1}, …, f_{n}) (probably zero).

Proof. It is enough to mimic for the matrix M′_{t} the proof performed by Jouanolou in [17, Prop. 3.11.19.10] to show that the determinant of the matrix M′_{t} is an inertia form of the ideal ⟨f_{1}, …, f_{n}⟩ (i.e. a multiple of the resultant). We include this proof for the convenience of the reader.

Let N := ∑_{i=1}^{n} #{α_{i} ∈ ℕ^{n} : |α_{i}| = d_{i}}. Given an algebraically closed field k, and a = (a_{α_{i}})_{|α_{i}|=d_{i}, i=1,... ,n}, a point in k^{N}, we denote by f_{1}(a), …, f_{n}(a) ∈ k[X] the polynomials obtained from f_{1}, …, f_{n} when the coefficients are specialized to a, and similarly for the coefficients of the Bezoutian. Because of the irreducibility of Res_{d_{1},...,d_{n}}(f_{1}, …, f_{n}), it is enough to show that for all a ∈ k^{N} such that f_{1}(a), …, f_{n}(a) have a non trivial solution in k^{n}, the determinant of the specialized matrix M′_{t}(a) is equal to 0.

Suppose that this is case, and let (p_{1}, …, p_{n}) be a non trivial solution. Without loss of generality, we can suppose p_{1} ≠ 0. One of the rows of M′_{t}(a) is indexed by X_{1}^{t}. Replace all the elements in that row as follows:

1. if the element belongs to a column indexed by a monomial X^{γ}, |γ| = t_{n} − t, then replace it with Δ_{γ}(a);

2. if it belongs to a column indexed by a monomial X^{γ} ∈ S_{t−d_{i}},

replace it with X^{γ} f_{i}(a). It is easy to check that, the determinant of the modified matrix is equal to X_{1}^{t} det(M′_{t}(a)). Now, we claim that under the specialization X_{i} ↦ p_{i}, the determinant of the modified matrix will be equal to zero if and only if det(M′_{t}(a)) = 0.

In order to prove this, we will show that the following submatrix of size (i(t_{n} − t) + 1) × \binom{n+t−1}{n−1} has rank less or equal than i(t_{n} − t) :

\left[{Δ_{γ_{1}}(a)(p)  …  Δ_{γ_{s}}(a)(p) \atop ^{t}F_{t_{n}−t}(a)}\right] .

This, combined with a Laplace expansion of the determinant of the modified matrix, gives the desired result.

If the rank of the block [^{t}F_{t_{n}−t}(a)] is less than i(t_{n} − t), then the claim follows straightforwardly. Suppose this is not the case. Then the family {X^{γ} f_{i}(a), X^{γ} ∈ S_{t_{n}−t−d_{i}}} is a basis of the piece of degree t_{n} − t of the generated ideal I(a) := ⟨f_{1}(a), …, f_{n}(a)⟩. We will show that in this case the polynomial ∑_{|γ|=t_{n}−t} Δ_{γ}(a)(p)X^{γ} belongs to I(a), which proves the claim.

Because of (3) and (4), the polynomial (X_{1} − Y_{1}) Δ(a)(X, Y) lies in the ideal ⟨f_{1}(a)(X)−f_{1}(a)(Y), …, f_{n}(a)(X)−f_{n}(a)(Y)⟩. Specializing Y_{i} ↦ p_{i}, we deduce that (X_{1} − p_{1}) ∑_{j=0}^{t_{n}}(∑_{|γ|=j} Δ_{γ}(a)(p) X^{γ}) is in the graded ideal I(a). This, combined with the fact that p_{1} ≠ 0, proves that ∑_{|γ|=j} Δ_{γ}(a)(p) X^{γ} ∈ I(a) for all j.

In particular, Res_{d_{1},...,d_{n}}(f_{1}, …, f_{n}) divides det(M_{t}). We describe the extraneous factor explicitly in the following theorem, which is the main result in this section. Before stating it, we set the following convention: if the matrix 𝔼_{t} is indexed by an empty set, we define det (𝔼_{t}) = 1.

Theorem 3.2. For any t ≥ 0, det (M_{t}) ≠ 0 and det(𝔼_{t}) ≠ 0. Moreover, we have the following formula à la Macaulay :

Res_{d_{1},...,d_{n}}(f_{1}, …, f_{n}) = ±\frac{det(M_{t})}{det(𝔼_{t})} .

For the proof of Theorem 3.2, we will need the following auxiliary lemma. Let D_{t} and E_{t} be the matrices defined in §2 before Lemma 2.3.

Lemma 3.3. Let t ≥ 0, and Λ a ring which contains A. Suppose we have a square matrix M with coefficients in Λ which has the following structure:

M = [ \begin{matrix}M_{1} & D_{t} \\ M_{2} & 0 \end{matrix} ] ,

where M_{1}, M_{2} are rectangular matrices. Then, there exists an element m ∈ Λ such that

det (M) = m . det (E_{t})

Proof. D_{t} is square if and only if t > t_{n}. (cf. [21, §3]). In this case,

det(M) = ± det(M_{2}) det(D_{t});

because of Macaulay’s formula (cf. [21, Th. 5]), we have that the right hand side equals

± det(M_{2}) det(E_{t})Res_{d_{1},...,d_{n}}(f_{1}, …, f_{n}),

and the conclusion follows easily.

Suppose now 0 ≤ t ≤ t_{n}. As in the introduction, let i(t) denote the dimension of the k-vector space of elements of degree t in the ideal generated by a regular sequence of n polynomials with degrees d_{1}, …, d_{n}. Then D_{t} has i(t)+H_{d} (t) rows and i(t) columns, and there is a bijection between the family F of H_{d} (t) monomials of degree t, and the maximal minors m_{F} of D_{t}. Namely, m_{F} is the determinant of the square submatrix made by avoiding all rows indexed by monomials in F.

It is not hard to check that m_{F} is the determinant ϕ_{F}^{∗} which is used in [10], for computing the subresultant associated with the family {X^{γ}}_{γ∈F}.

Now, using the generalized Macaulay’s formula for the subresultant (cf. [10]), we have that

m_{F} = ± det(E_{t}).Δ_{F}^{t},

where Δ_{F}^{t} is the subresultant associated with the family F. It is a polynomial in A which vanishes under a specialization of the coefficients f_{1}(a), …, f_{n}(a) if and only if the family {X^{γ}}_{γ∈F} fails to be a basis of the t− graded piece of the quotient k[X_{1}, …, X_{n}]/⟨f_{1}(a), …, f_{n}(a)⟩ (cf. [9]).

Let m_{F}^{c} be the complementary minor of m_{F} in M (i.e. the determinant of the square submatrix of M which is made by deleting all rows and columns that appear in m_{F}). By the Laplace expansion of the determinant, we have that

det (M) = ∑_{F} s_{F} · m_{F} · m_{F}^{c} = det (E_{t}) (∑_{F} s_{F} · m_{F}^{c} · Δ_{F}^{t})

with s_{F} = ±1. Setting m = ∑_{F} s_{F} · m_{F}^{c} · Δ_{F}^{t} ∈ Λ, we have the desired result.

We now give the proof of Theorem 3.2.

Proof. In [21] it is shown that det (E_{t}) ≠ 0, ∀t ≥ 0. This implies that det (𝔼_{t}) ≠ 0. In order to prove that det (𝔼_{t}) = det(E_{t}) det(E_{t_{n}−t}) divides det(M_{t}), we use the following trick: consider the ring B := ℤ [b_{α_{i}}]_{|α_{i}|=d_{i},i=1,...,n} , where b_{α_{i}} are new variables, and the polynomials

f_{b,i} := ∑_{|α_{i}|=d_{i}} b_{α_{i}}X^{α_{i}} ∈ B [X_{1}, …, X_{n}] .

Let D_{t}^{b} the matrix of the linear transformation ψ_{2,t}^{b} determined by the formula (8) but associated with the sequence f_{b,1}, …, f_{b,n} instead of f_{1}, …, f_{n}. Set Λ := ℤ [a_{α_{i}}, b_{α_{i}}], and consider the matrix M(a, b) with coefficients in Λ given by

M(a, b) = [ \begin{matrix}Δ_{t} & D_{t} \\ ^{t}D_{t_{n}−t}^{b} & 0 \end{matrix} ] .

It is easy to see that M(a, a) = M_{t}, and because of Lemma 3.3, we have that det (E_{t}) divides det (M(a, b)) in Λ. Transposing M(a, b) and using a symmetry argument, again by the same lemma, we can conclude that det (E_{t_{n}−t}^{b}) divides det (M(a, b)) in Λ, where E_{t_{n}−t}^{b} has the obvious meaning.

The ring Λ is a factorial domain and det(E_{t}) and det(E_{t_{n}−t}^{b}) have no common factors in Λ because they depend on different sets of variables. So, we have

det (M(a, b)) = p(a, b) det(E_{t}) det(E_{t_{n}−t}^{b})

for some p ∈ Λ. Now, specialize b_{α_{i}} ↦ a_{α_{i}}. The fact that det (M_{t}) is a multiple of the resultant has been proved in Proposition 3.1 (see also [17, Prop. 3.11.19.21]) for 0 ≤ t ≤ t_{n}, and in [21] for t > t_{n}. On the other side, since Res_{d_{1},...,d_{n}}(f_{1}, …, f_{n}) is irreducible and depends on all the coefficients of f_{1}, …, f_{n} while det(E_{t}) and det(E_{t_{n}−t}) do not depend on the coefficients of f_{n}, we conclude that Res_{d_{1},...,d_{n}}(f_{1}, …, f_{n}) divides p(a, a). Moreover, the following lemma shows that they have the same degree. Then, their ratio is a rational number λ. We can see that λ = ±1, considering the specialized family X_{1}^{d_{1}}, …, X_{n}^{d_{n}}.

Lemma 3.4. For each i = 1, …, n the degree deg_{(a_{α_{i}})} (M_{t}) of M_{t} in the coefficients of f_{i} equals

deg_{(a_{α_{i}})} (Res_{d_{1},...,d_{n}}(f_{1}, …, f_{n})) + deg_{(a_{α_{i}})} (E_{t}) + deg_{(a_{α_{i}})} (E_{t_{n}−t}) =
d_{1} … d_{i−1}. d_{i+1} … d_{n} + deg_{(a_{α_{i}})} (E_{t}) + deg_{(a_{α_{i}})} (E_{t_{n}−t})

Proof. Set J_{u}(i) := {X^{γ} ∈ S_{u}, γ_{i} ≥ d_{i}, γ_{j} < d_{j} ∀j ≠ i}, u = t, t_{n} − t. From the definitions of ψ_{2,t} and E_{t}, it is easy to check that, if δ_{t} is a maximal minor of D_{t},

deg_{(a_{α_{i}})} (δ_{t}) − deg_{(a_{α_{i}})} (E_{t}) = #J_{t}(i).

Using Laplace expansion, it is easy to see that det (M_{t}) may be expanded as follows

det (M_{t}) = ∑_{δ_{t},δ_{tn−t}} s_{δ} · m_{δ} · δ_{t} · δ_{t_{n}−t}

where s_{δ} = ±1, δ_{t_{n}−t} is a maximal minor of ^{t}D_{t_{n}−t} and m_{δ} is a minor of size H_{d}(t) in Δ_{t}.

As each entry of Δ_{t} has degree 1 in the coefficients of f_{i}, the lemma will be proved if we show that

#J_{t}(i) + #J_{t_{n}−t}(i) + H_{d}(t) = d_{1} … d_{i−1}. d_{i+1} … d_{n}. (16)

Now, as already observed in the proof of Lemma 2.3, H_{d}(t) can be computed as the cardinality of the following set:

H_{d,t} := {X^{γ} ∈ S_{t}, γ_{j} < d_{j} ∀j}, (17) and d_{1} … d_{i−1}. d_{i+1} … d_{n} is the cardinality of

Γ_{i} := {X_{1}^{γ_{1}} … X_{i−1}^{γ_{i−1}}X_{i+1}^{γ_{i+1}} … X_{n}^{γ_{n}} , γ_{j} < d_{j} ∀j}.

In order to prove (16) it is enough to exhibit a bijection between Γ_{i} and the disjoint union J_{t}(i) ⋃ J_{t_{n}−t}(i) ⋃ H_{d,t}. This is actually a disjoint union for all t, unless t_{n} − t = t. But what follows shows that the bijection is well defined even in this case.

Let Xγ̂ ∈ Γ_{i}, γ̂ = (γ_{1}, …, γ_{i−1}, γ_{i+1}, …, γ_{n}) with γ_{j} < d_{j} ∀j ≠ i. If |γ̂| ≤ t, then there exists a unique γ_{i} such that γ := (γ_{1}, …, γ_{n}) ∈ ℕ_{0}^{n} verifies |γ| = t. If γ_{i} < d_{i}, then we send Xγ̂ to X^{γ} ∈ H_{d,t}. Otherwise, we send it to X^{γ} ∈ J_{t}(i).

If |γ̂| > t, let γ̂^{∗} denote the multiindex

(d_{1} − 1 − γ_{1}, …, d_{i−1} − 1 − γ_{i−1}, d_{i+1} − 1 − γ_{i+1}, …, d_{n} − 1 − γ_{n}) .

Then, |γ̂^{∗}| < t_{n}−t, and there exists a unique γ_{i} such that the multiindex γ defined by

(d_{1} − 1 − γ_{1}, …, d_{i−1} − 1 − γ_{i−1}, γ_{i}, d_{i+1} − 1 − γ_{i+1}, …, d_{n} − 1 − γ_{n})

has degree t_{n} − t. We can send Xγ̂ to X^{γ} ∈ J_{t_{n}−t}(i) provided that γ_{i} ≥ d_{i}. Suppose this last statement does not happen, this implies that the monomial with exponent

γ^{∗} := (γ_{1}, …, γ_{i−1}, d_{i} − 1 − γ_{i}, γ_{i+1}, …, d_{n})

has degree t contradicting the fact that |γ̂| > t.

With these rules, it is straightforward to check that we obtain the desired bijection.

Changing the order of the sequence (f_{1}, …, f_{n}), and applying Theorem (3.2), we deduce that

Corollary 3.5. Res_{d_{1},...,d_{n}}(f_{1}, …, f_{n}) = gcd{maximal minors of Ψ̃_{t}}.

4. Estimating the size of M_{t}

We have, for each integer t ≥ 0, a matrix M_{t} of size ρ(t), where ρ was defined in (1), whose determinant is a nontrivial multiple of the resultant, and such that, moreover, its extraneous factor is a minor of it. We want to know which is the smallest matrix we can have.

We can write ρ as

ρ(t) = \binom{n + t − 1}{n − 1} + \binom{n + t_{n} − t − 1}{n − 1} − H_{d}(t_{n} − t).

It is straightforward to check that \binom{n+t−1}{n−1}+\binom{n+t_{n}−t−1}{n−1} is the restriction to the integers of a polynomial ϕ(t) in a real variable t, symmetric with respect to \frac{t_{n}}{2} (i.e. ϕ(\frac{t_{n}}{2} + t) = ϕ(\frac{t_{n}}{2} − t) for all t). Moreover, ϕ reaches its minimum over [0, t_{n}] at t = \frac{t_{n}}{2} . Since

ρ(t) = ϕ(t) − H_{d}(t) = ϕ(t_{n} − t) − H_{d}(t_{n} − t) = ρ(t_{n} − t), (18)

in order to study the behaviour of ρ we need to understand how H_{d}(t) varies with t. We denote as usual the integer part of a real number x by the symbol [x] .

Proposition 4.1. H_{d}(t) is non decreasing on (the integer points of ) the interval [0, [\frac{t_{n}}{2} ]] .

Proof. We will prove this result by induction on n. The case n = 1 is obvious since t_{1} = d−1 and H_{d}(t) = 1 for any t = 0, …, d−1. Suppose then that the statement holds for n variables and set

d̂:= (d_{1}, …, d_{n+1}) ∈ ℕ_{0}^{n+1},
d := (d_{1}, …, d_{n}).

Let t < t + 1 ≤ [\frac{t_{n+1}}{2} ] . We want to see that φ(t) := Hd̂(t + 1) − Hd̂(t) is non negative. Recall from (17) that, for every t ∈ ℕ_{0}, Hd̂(t) equals the cardinality of the set

{γ ∈ ℕ_{0}^{n+1} : |γ| = t, 0 ≤ γ_{i} ≤ d_{i} − 1, i = 1, …, n + 1}.

Then, it can also be computed as

∑_{j=0}^{d_{n+1}−1} #{γ̂ ∈ ℕ_{0}^{n} : |γ̂| = t − j, 0 ≤ γ̂_{i} ≤ d_{i} − 1, i = 1, …, n},

which gives the equality Hd̂(t) = ∑_{j=0}^{d_{n+1}−1} H_{d}(t − j). It follows that φ(t) = H_{d}(t + 1) − H_{d}(t + 1 − d_{n+1}).

If t + 1 ≤ [\frac{t_{n}}{2} ] , we deduce that φ(t) ≥ 0 by inductive hypothesis. Suppose then that t + 1 is in the range [\frac{t_{n}}{2} ] < t + 1 ≤ [\frac{t_{n+1}}{2} ] . As H_{d}(t+1) = H_{d}(t_{n}−t−1), it is enough to show that t_{n}−t−1 ≥ t+1−d_{n+1} and t_{n}−t−1 ≤ [\frac{t_{n}}{2} ] , which can be easily checked, and the result follows again by inductive hypothesis.

Corollary 4.2. The size ρ (t) of the matrix M_{t} is minimal over ℕ_{0} when t = [\frac{t_{n}}{2} ] .

Proof. By (18), ρ has a maximum at [\frac{t_{n}}{2} ] over [0, t_{n}] because ϕ has a maximum and H_{d} has a minimum. If t > t_{n}, we have that ρ (t) = \binom{n+t−1}{n−1}. For t in this range, it is easy to check that ρ (t_{n}) = \binom{n+t_{n}−1}{n−1} − 1 < ρ (t) . Then, ρ(t) > ρ(t_{n}) ≥ ρ ([\frac{t_{n}}{2} ]) .

Remark 4.3. Note that when t_{n} is odd, ρ([\frac{t_{n}}{2} ]) = ρ([\frac{t_{n}}{2} ] + 1), and then the size of M_{t} is also minimal for t = [\frac{t_{n}}{2} ] + 1 in this case.

Denote p := ^{\frac{∑_{i=1}^{n} d_{i}}{n}} the average value of the degrees, and set q := \frac{p+1}{2p}. Note that except in the linear case when all d_{i} = 1, it holds that p > 1 and q < 1.

Proposition 4.4. Assume p > 1. The ratio between the size of the smallest matrix M_{t} and the classical Macaulay matrix M_{t_{n}+1} can be bounded by

\frac{ρ ([t_{n}/2])}{ρ (t_{n} + 1)} ≤ 2 q^{n−1}.

In particular, it tends to zero exponentially in n when the number of variables tends to infinity and p remains bigger that a constant c > 1..

Proof. When t_{n} is even, t_{n} − [t_{n}/2] = [t_{n}/2] and when t_{n} is odd, t_{n} − [t_{n}/2] = [t_{n}/2] + 1. In both cases,

\frac{ρ ([t_{n}/2])}{ρ (t_{n} + 1)} ≤ \frac{2 \binom{n+[t_{n}/2]}{n−1}}{\binom{n+t_{n}}{n−1}} = 2 \begin{matrix}([t_{n}/2] + n) & … & ([t_{n}/2] + 2) \\ (t_{n} + n) & … & (t_{n} + 2) \end{matrix} =

= 2 (\frac{[t_{n}/2] + n}{t_{n} + n}) (\frac{[t_{n}/2] + n − 1}{t_{n} + n − 1}) … (\frac{[t_{n}/2] + 2}{t_{n} + 2}) ≤

≤ 2 (\frac{[t_{n}/2] + n}{t_{n} + n})^{n−1} .

Since t_{n} = np − n, we deduce that

\frac{[t_{n}/2] + n}{t_{n} + n} ≤ \frac{\frac{np}{2} + \frac{n}{2}}{np} = \frac{1}{2} + \frac{1}{2p} = q,

as wanted.

5. Resultant complexes

In this section we consider Weyman’s complexes (cf. [27], [15]) and we make explicit the morphisms in these complexes, which lead to polynomial expressions for the resultant via determinantal formulas in the cases described in Lemma 5.3.

We will consider a complex which is a “coupling” of the Koszul complex K^{•}(t; f_{1}, …, f_{n}) associated with f_{1}, …, f_{n} in degree t and the dual of the Koszul complex K^{•}(t_{n} − t, f_{1}, …, f_{n})^{∗} associated with f_{1}, …, f_{n} in degree t_{n} −t. This complex arises from the spectral sequence derived from the Koszul complex of sheaves on ℙ^{n−1} associated with f_{1}, …, f_{n} twisted by O_{ℙ^{n−1}}(t). Here, O_{ℙ^{n−1}}(t) denotes as usual the t-twist of the sheaf of regular functions over the (n − 1)-projective space ℙ^{n−1} (see for instance [15, p. 34]). Its space of global sections can be identified with the space of homogeneous polynomials in n variables of degree t. We make explicit in terms of the Bezoutian the map ∂_{0} (see (10) below) produced by cohomology obstructions. In fact, the non-trivial contribution is given in terms of the mapping ψ_{1,t} defined in (7).

Precisely, let K^{•}(t; f_{1}, …, f_{n}) denote the complex

{0 ⟶ K(t)^{−n} ⟶^{δ_{−(n−1)}} … ⟶^{δ_{−1}} K(t)^{−1} ⟶^{δ_{0}} K(t)^{0} }, (19)

where

K(t)^{−j} = ⊕_{i_{1}<···<i_{j}} S_{t−d_{i_{1}}−···−d_{i_{j}}}

and δ_{−j} are the standard Koszul morphisms.

Similarly, let K^{•}(t_{n} − t; f_{1}, …, f_{n})^{∗} denote the complex

{K(t_{n} − t)^{0} ⟶^{δ_{0}^{∗}} K(t_{n} − t)^{1} ⟶^{δ_{1}^{∗}} … ⟶^{δ_{n}^{∗}} K(t_{n} − t)^{n} }, (20)

where

K(t_{n} − t)^{j} = ⊕_{i_{1}<···<i_{j}} S_{t_{n}−t−d_{i_{1}}−···−d_{i_{j}}}^{∗}

and δ_{j}^{∗} are the duals of the standard Koszul morphisms. Note that in fact K(t_{n} − t)^{n} = 0 for any t ≥ 0.

Now, define C^{•}(t; f_{1}, …, f_{n}) to be the following coupled complex

{0 ⟶ C^{−n} ⟶^{∂_{−(n−1)}} … ⟶^{∂_{−1}} C^{−1} ⟶^{∂_{0}} C^{0} ⟶^{∂_{1}} … ⟶^{∂_{n−1}} C^{n−1} ⟶ 0 }, (21)

where

\begin{aligned}C^{−j} & = & K(t)^{−j}, & j = 2, …, n \\ C^{j} & = & K(t_{n} − t)^{j+1}, & j = 1, …, n − 1 \\ C^{−1} & = & K(t_{n} − t)^{0} ⊕ K(t)^{−1} \\ C^{0} & = & K(t)^{0} ⊕ K(t_{n} − t)^{1} \end{aligned} (22)

and the morphisms are defined by

\begin{aligned}∂_{−j} & = & δ_{−j}, & j = 2, …, n − 1 \\ ∂_{j} & = & δ_{j}^{∗}, & j = 2, …, n − 1 \\ ∂_{−1} & = & 0 ⊕ δ_{−1} \\ ∂_{0} & = & (ψ_{1,t} + δ_{0}) ⊕ δ_{0}^{∗} \\ ∂_{1} & = & 0 + δ_{1}^{∗} \end{aligned} (23)

More explicitly, ∂_{0}(T, (g_{1}, …, g_{n})) = (ψ_{1,t}(T) + δ_{0}(g_{1}, …, g_{n}), δ_{0}^{∗}(T)) and ∂_{1}(h, (T_{1}, …, T_{n})) = δ_{1}^{∗}(T_{1}, …, T_{n}). Observe that ∂_{0} is precisely the mapping we called Ψ̃_{t} in the previous section.

As in the proof of Proposition 3.1, given an algebraically closed field k, and a = (a_{α_{i}})_{|α_{i}|=d_{i}, i=1,... ,n}, a point in k^{N}, we denote by f_{1}(a), …, f_{n}(a) the polynomials ∈ k[X] obtained from f_{1}, …, f_{n} when the coefficients are specialized to a. For any particular choice of coefficients in (21) we get a complex of k-vector spaces. We will denote the specialized modules and morphisms by K(t)^{1}(a), δ_{0}(a), etc. Let D denote the determinant (cf. [15, Appendix A], [12]) of the complex of A-modules (21) with respect to the monomial bases of the A-modules C^{ℓ}. This is an element in the field of fractions of A.

We now state the main result in this section.

Theorem 5.1. The complex (21) is generically exact, and for each specialization of the coefficients it is exact if and only if the resultant does not vanish. For any positive integer t we have that

D = Res_{d_{1},...,d_{n}}(f_{1}, …, f_{n}), (24)

and moreover, D equals the greatest common divisor of all maximal minors of a matrix representing the A-module map ∂_{0}.

Proof. For t > t_{n}, we get the Koszul complex in degree t, and so the specialized complex at a point a ∈ k^{N} is exact if and only if f_{1}(a), …, f_{n}(a) is a regular sequence, i.e. if and only if the resultant does not vanish. The fact that the determinant of this complex equals the resultant goes back to ideas of Cayley; for a proof see [12], [15] or [8].

Suppose 0 ≤ t ≤ t_{n}. Since δ_{0} ∘ δ_{−1} = δ_{1}^{∗} ∘ δ_{0}^{∗} = 0, it is easy to see that (21) is a complex.

Set

U := {a = (a_{α_{i}}) ∈ k^{N}, i = 1, …, n, |α_{i}| = d_{i} : det(M_{t}(a)) ≠ 0}.

Note that the open set U is non void because the vector of coefficients of {X_{1}^{d_{1}}, …, X_{n}^{d_{n}}} lies in U, since in this case det M_{t} = ±1. For any choice of homogeneous polynomials f_{1}(a), …, f_{n}(a) ∈ k[X] with respective degrees d_{1}, …, d_{n} and coefficients a in U, the resultant does not vanish by Theorem 3.2 and then the specialized Koszul complexes in (19) and (20) are exact.

Then, the dimension dim Im(δ_{0}(a)) of the image of δ_{0}(a) equals i(t) = dim < f_{1}(a), …, f_{n}(a) >_{t} . Similarly, dim(ker(δ_{0}^{∗}(a)) = i(t_{n} − t). Therefore,

dim ker(∂_{0}(a)) ≥ dim Im(∂_{−1}(a)) = dim Im(δ_{−1}(a)) =
= dim ker(δ_{0}(a)) = dim K(t)^{−1}(a) − i(t).

On the other side, the fact that M_{t}(a) is non singular of size ρ(t) implies that

dim ker(∂_{0}(a)) ≤ dim C^{−1}(a) − ρ(t) =
= dim K(t)^{−1}(a) + dim K(t_{n} − t)^{0}(a) − ρ(t) =
= ImK(t)^{−1}(a) + dim S_{t_{n}−t}(a) − ρ(t) =
= dim K(t)^{−1}(a) − i(t).

Therefore, dim Im(∂_{−1}(a)) = dim ker(∂_{0}(a)) and the complex is exact at level −1.

In a similar way, we can check that the complex is exact at level 0, and so the full specialized complex (21) is exact when the coefficients a lie in U.

In order to compute the determinant of the complex in this case, we can make suitable choices of monomial subsets in each term of the complex starting from the index sets that define M_{t}(a) to the left and to the right. Then,

D(a) = \frac{det M_{t}(a)}{p_{1}(a) · p_{2}(a)},

where p_{1}(a) (resp. p_{2}(a)) is a quotient of product of minors of the morphisms on the left (resp. on the right).

Taking into account (19) and (20), it follows from [10] that

p_{1}(a) = det(E_{t}(a)), p_{2}(a) = det(E_{t_{n}−t}(a)),

and so by Theorem 3.2 we have

D(a) = Res_{d_{1},...,d_{n}}(f_{1}, …, f_{n})(a)\frac{det(𝔼_{t})(a)}{det(E_{t}(a)) det(E_{t_{n}−t}(a))} =

= Res_{d_{1},...,d_{n}}(f_{1}, …, f_{n})(a)

for all families of homogeneous polynomials with coefficients a in the dense open set U, and since D and the resultant are rational functions, this implies (24), as wanted. Moreover, it follows that the complex is exact if and only if the resultant does not vanish.

The fact that Res_{d_{1},...,d_{n}}(f_{1}, …, f_{n}) is the greatest common divisor of all maximal minors of the matrix representing ∂_{0} has been proved in Corollary 3.5.

We remark that from the statement of Theorem 5.1 plus a close look at the map at level 0, it is not hard to deduce that for a given specialization of f_{1}, …, f_{n} in k with non vanishing resultant, the specialized polynomials Δ_{γ}(a), |γ| = t_{n} −t generate the quotient of the polynomial ring k[X] by the ideal I(a) = ⟨f_{1}(a), …, f_{n}(a)⟩ in degree t. We can instead use the known dualizing properties of the Bezoutian in case the polynomials define a regular sequence, to provide an alternative proof of Theorem 5.1. This is a consequence of Proposition 5.2 below. We refer to [17],[[19], Appendix F], [25] and [26] for the relation between the Bezoutian and the residue (i.e. an associated trace) and we simply recall the properties that we will use.

Assume Res_{d_{1},...,d_{n}} (f_{1}(a), …, f_{n}(a)) is different from zero. This implies that f_{1}(a), …, f_{n}(a) is a regular sequence and their zero locus consists of the single point 0 ∈ k^{n}. Then, there exists a dualizing klinear operator

R_{0} : k[Y]/⟨f_{1}(a)(Y), …, f_{n}(a)(Y)⟩ ⟶ k,

called the residue or trace operator, which verifies 1. h(X) = R_{0} (h(Y) Δ(a)(X, Y)) in the quotient ring k[X]/I(a).

2. If h is homogeneous of degree t with t ≠ t_{n}, R_{0}(h) = 0 Then, for every polynomial h(X) ∈ k[X] of degree t, it holds that

h(X) = ∑_{|γ|=t_{n}−t} R_{0} (h(Y) Y^{γ}) Δ_{γ}(a)(X) mod I(a), (25)

where Δ(a)(X, Y) = ∑_{|γ|=t_{n}−t} Δ_{γ}(a)(X)Y^{γ} as in (4). As a consequence, the family {Δ_{γ}(a)(X)}_{|γ|=t_{n}−t}, (resp. |γ| = t) generates the graded piece of the quotient in degree t (resp. t_{n} − t). Moreover, it is easy to verify that for any choice of polynomials p_{i}(X, Y), q_{i}(X, Y) ∈ k[X, Y], i = 1, …, n, the polynomial Δ̃_{a}(X, Y) defined by

Δ̃_{a}(X, Y) := Δ(a)(X, Y) + ∑_{i=1}^{n} p_{i}(X, Y) f_{i}(a)(X) + q_{i}(X, Y) f_{i}(a)(Y). (26)

has the same dualizing properties as Δ(a)(X, Y).

We are ready to prove a kind of “converse” to Proposition 3.1.

Proposition 5.2. If Res_{d_{1},...,d_{n}} (f_{1}(a), …, f_{n}(a)) ≠ 0, it is possible to extract a square submatrix M′_{t} of Ψ̃_{t} as in (15) such that det (M′_{t}(a)) ≠ 0.

Proof. Since f_{1}(a)(X), …, f_{n}(a)(X) is a regular sequence in k[X], the dimensions of the graded pieces of the quotient k[X]/I(a) in degrees t and t_{n} − t are i(t) and i(t_{n} − t) respectively.

We can then choose blocks F_{t} and F_{t_{n}−t} as in (15) such that F_{t}(a) and F_{t_{n}−t}(a) have maximal rank. Suppose without loss of generality

that the blocks F_{t} and F_{t_{n}−t} have respectively the form [ \begin{matrix}Q_{t} \\ R_{t} \end{matrix} ] and [ \begin{matrix}Q_{t_{n}−t} \\ R_{t_{n}−t} \end{matrix} ], where Q_{t}(a) and Q_{t_{n}−t}(a) are square invertible matrices of

maximal size. We are going to prove that, with this choice, the matrix M′_{t}(a) is invertible.

Our specialized matrix will look as follows:

M′_{t}(a) = [ \begin{matrix}Δ_{t}(a) & \begin{matrix}Q_{t}(a) \\ R_{t}(a) \end{matrix} \\ ^{t}Q_{t_{n}−t}(a) ^{t}R_{t_{n}−t}(a) & 0 \end{matrix} ] .

Applying linear operations in the rows and columns of M′_{t}(a), it can be transformed into:

[ \begin{matrix}0 & 0 & Q_{t}(a) \\ 0 & Δ̃_{t,a} & R_{t}(a) \\ ^{t}Q_{t_{n}−t}(a) & ^{t}R_{t_{n}−t}(a) & 0 \end{matrix} ] ,

where the block [Δ̃_{t,a}] is square and of size H_{d}(t).

But it is easy to check that this Δ̃_{t,a} corresponds to the components in degree t of another Bezoutian Δ̃_{a}(X, Y) (in the sense of (26)). This is due to the fact that each of the linear operations performed on M′_{t}(a), when applied to the block Δ_{t,a}, can be read as a polynomial combination of f_{i}(a)(X) and f_{i}(a)(Y) applied to the bezoutian Δ(a)(X, Y).

Using the fact that the polynomials Δ̃_{γ,a}(X) read in the columns of Δ̃_{t,a} generate the quotient in degree t_{n} − t and they are as many as its dimension, we deduce that they are a basis and so

det ( \begin{matrix}0 & Δ̃_{t} \\ ^{t}Q_{t_{n}−t}(a) & ^{t}R_{t_{n}−t}(a) \end{matrix} ) ≠ 0,

which completes the proof of the claim.

We could then avoid the consideration of the open set U in the proof of Theorem 5.1, and use Proposition 5.2 to show directly that the complex is exact outside the zero locus of the resultant. In fact, this is not surprising since for all specializations such that the resultant is non zero, the residue operator defines a natural duality between the t-graded piece of the the quotient of the ring of polynomials with coefficients in k by the ideal I(a) and the t_{n} − t graded piece of the quotient, and we can read dual residue bases in the Bezoutian.

We characterize now those data n, d_{1}, …, d_{n} for which we get a determinantal formula.

Lemma 5.3. Suppose d_{1} ≤ d_{2} ≤ ⋯ ≤ d_{n}. The determinant of the resultant complex provides a determinantal formula for the resultant Res_{d_{1},...,d_{n}}(f_{1}, …, f_{n}) if and only if the following inequality is verified

d_{3} + ⋯ + d_{n} − n < d_{1} + d_{2} − 1. (27)

Moreover, when (27) holds, there exists a determinantal formula given by the resultant complex for each t such that

d_{3} + ⋯ + d_{n} − n < t < d_{1} + d_{2}. (28)

Remark 5.4. When all d_{i} have a common value d, (27) reads

(n − 2)d < 2d + n − 1,

which is true for any d for n ≤ 4, for d = 1, 2, 3 in case n = 5, for d = 1, 2 in case n = 6, and never happens for n > 7 unless d = 1, as we quoted in the introduction.

Proof. The determinant of the resultant complex provides a determinantal formula for Res_{d_{1},...,d_{n}}(f_{1}, …, f_{n}) precisely when C^{−2} = C_{1} = 0. This is respectively equivalent to the inequalities

t < d_{1} + d_{2}

and

t_{n} − t = d_{1} + … d_{n} − n − t < d_{1} + d_{2},

from which the lemma follows easily. We have decreased the right hand side of (27) by a unit in order to allow for a natural number t satisfying (28).

Corollary 5.5. For all n ≥ 7 there exists a determinantal formula only if d_{1} = d_{2} = d_{3} = 1 and n − 3 ≤ d_{4} + ⋯ + d_{n} < n, which forces all d_{i} to be 1 or at most, all of them equal 1 except for two of them which equal 2, or all of them equal 1 except for one of them which equals 3.

The proof of the corollary follows easily from the inequality (27). In any case, if a determinantal formula exists, we have a determinantal formula for t = [t_{n}/2], as the following proposition shows.

Proposition 5.6. If a determinantal formula given by the resultant complex exists, then M_{[t_{n}/2]} is square and of the smallest possible size ρ([\frac{t_{n}}{2} ]).

Proof. In order to prove that M_{[t_{n}/2]} is square, we need to check by Lemma 5.3 that

d_{3} + ⋯ + d_{n} − n < [\frac{t_{n}}{2} ] < d_{1} + d_{2}. (29)

If there exists a determinantal formula, then the inequalily (27) holds, from which it is straightforward to verify that

d_{3} + ⋯ + d_{n} − n < \frac{t_{n}}{2} < d_{1} + d_{2}.

To see that in fact (29) holds, it is enough to check that

d_{3} + ⋯ + d_{n} − n + 1/2 ≠ \frac{t_{n}}{2} = \frac{d_{1} + ⋯ + d_{n} − n}{2}.

But if the equality holds, we would have that d_{3} + ⋯ + d_{n} = d_{1} + d_{2} + n − 1, which is a contradiction. According to Corollary 4.2, we also know that M_{[t_{n}/2]} has the smallest possible size.

6. Dixon formulas

We prove in this section that “affine” Dixon formulas can in fact be recovered in this setting. We first recall classical Dixon formulas to compute the resultant of three bivariate affine polynomials of degree d. We will make a slight change of notation in what follows. The input affine polynomials (having monomials of degree at most d in two variables (X_{1}, X_{2})) will be denoted f_{1}, f_{2}, f_{3} and we will use capital letters F_{1}, F_{2}, F_{3} to denote the homogeneous polynomials in three variables given by their respective homogenizations (with homogeneizing variable X_{3}). Dixon (cf. [13]) proposed the following determinantal formula to compute the resultant Res_{d,d,d}(f_{1}, f_{2}, f_{3}) = Res_{d,d,d}(F_{1}, F_{2}, F_{3}):

Let Bez(X_{1}, X_{2}, Y_{1}, Y_{2}) denote the polynomial obtained by dividing the following determinant by (X_{1} − Y_{1})(X_{2} − Y_{2}):

det ( \begin{matrix}f_{1}(X_{1}, X_{2}) & f_{2}(X_{1}, X_{2}) & f_{3}(X_{1}, X_{2}) \\ f_{1}(Y_{1}, X_{2}) & f_{2}(Y_{1}, X_{2}) & f_{3}(Y_{1}, X_{2}) \\ f_{1}(Y_{1}, Y_{2}) & f_{2}(Y_{1}, Y_{2}) & f_{3}(Y_{1}, Y_{2}). \end{matrix} )

Note that by performing row operations we have that Bez(X_{1}, X_{2}, Y_{1}, Y_{2}) equals the determinant of the matrix

det ( \begin{matrix}Δ_{11} & Δ_{21} & Δ_{31} \\ Δ_{12} & Δ_{22} & Δ_{32} \\ f_{1}(Y_{1}, Y_{2}) & f_{2}(Y_{1}, Y_{2}) & f_{3}(Y_{1}, Y_{2}), \end{matrix} )

where Δ_{ij} are as in (3). Write

Bez(X_{1}, X_{2}, Y_{1}, Y_{2}) = ∑_{|β|≤2d−2} B_{β}(X_{1}, X_{2})Y_{1}^{β_{1}}Y_{2}^{β_{2}}.

Set A := ℤ[a], where a denotes one indeterminate for each coefficient of f_{1}, f_{2}, f_{3}. Let S denote the free module over A with basis B given by all monomials in two variables of degree less or equal than d − 2, which has an obvious isomorphism with the free module S′ over A with basis B′ given by all monomials in three variables of degree equal to d − 2. The monomial basis of all polynomials in two variables of degree less or equal than 2d − 2 will be denoted by C.

Let M be the square matrix of size 2d^{2}−d whose columns are indexed by C and whose rows contain consecutively the expansion in the basis C of m · f_{1}, of m · f_{2}, and of m · f_{3}, where m runs in the three cases over B, and finally, the expansion in the basis C of all B_{β}, |β| ≤ d − 1. Then, Dixon’s formula says that

Res_{d,d,d}(f_{1}, f_{2}, f_{3}) = ± det M.

Here, d_{1} = d_{2} = d_{3} = d and n = 3, so that (27) holds and by (28) there is a determinantal formula for each t such that d − 3 < t < 2d. So, one possible choice is t = 2d − 2. Then, t_{3} − t = d − 1 < d, which implies < F_{1}, F_{2}, F_{3} >_{t_{3}−t}= 0. Also, t − d = d − 2 < d, and therefore S^{t,i} = S′, for all i = 1, 2, 3.

Let Δ(X_{1}, X_{2}, X_{3}, Y_{1}, Y_{2}, Y_{3}) = ∑_{|γ|≤3d−3} Δ_{γ}(X)Y^{γ} be the Bezoutian associated with the homogeneous polynomials F_{1}, F_{2}, F_{3}. We know that Res_{d,d,d}(F_{1}, F_{2}, F_{3}) = ± det M_{2d−2}. In this case, the transposed matrix tM_{2d−2} is a square matrix of the same size as M, and it is obvious that their 3d(d − 1)/2 first rows coincide (if the columns are ordered conveniently). According to (7), the last (d + 1)d/2 rows of ^{t}M_{2d−2} contain the expansion in the basis B′ of all Δ_{γ} , |γ| = d − 1.

Proposition 6.1. The “affine” matrix M and the “homogeneous” matrix ^{t}M_{2d−2} coincide.

Proof. Denote P (X_{1}, X_{2}, X_{3}, Y_{1}, Y_{2}, t) the homogeneous polynomial of degree 3d − 2 in 6 variables obtained by dividing the following determinant by (X_{1} − Y_{1})(X_{2} − Y_{2}) :

det ( \begin{matrix}Δ_{1,1}(F) & Δ_{2,1}(F) & Δ_{3,1}(F) \\ Δ_{1,2}(F) & Δ_{2,2}(F) & Δ_{3,2}(F) \\ F_{1}(Y_{1}, Y_{2}, t) & F_{2}(Y_{1}, Y_{2}, t) & F_{3}(Y_{1}, Y_{2}, t) \end{matrix} ) ,

where

Δ_{i,1}(F) := F_{i}(X_{1}, X_{2}, X_{3}) − F_{i}(Y_{1}, Y_{2}, X_{3}), i = 1, 2, 3

and

Δ_{i,2}(F) := F_{i}(Y_{1}, X_{2}, X_{3}) − F_{i}(Y_{1}, Y_{2}, X_{3}), i = 1, 2, 3.

It is easy to check that

(X_{3} − Y_{3})Δ(X, Y) = P (X_{1}, X_{2}, X_{3}, Y_{1}, Y_{2}, X_{3}) − P (X_{1}, X_{2}, X_{3}, Y_{1}, Y_{2}, Y_{3}) (30)

and that

P (X_{1}, X_{2}, 1, Y_{1}, Y_{2}, 1) = Bez(X_{1}, X_{2}, Y_{1}, Y_{2}) (31)

We are looking for the elements in Bez(X_{1}, X_{2}, Y_{1}, Y_{2}) of degree less or equal than d − 1 in the variables Y_{1}, Y_{2}. But it is easy to check that deg_{y} (P (X_{1}, X_{2}, X_{3}, Y_{1}, Y_{2}, Y_{3})) ≥ d. This, combined with the equality given in (30), implies that, for each 1 ≤ j ≤ d − 1 :

X_{3} ∑_{|γ|=j} Δ_{γ}(X)Y^{γ} − Y_{3} ∑_{|γ|=j−1} Δ_{γ}(X)Y^{γ}

is equal to the piece of degree j in the variables Y_{i} of the polynomial P (X_{1}, X_{2}, X_{3}, Y_{1}, Y_{2}, X_{3}).

Besides, this polynomial does not depend on Y_{3}, so the following formula holds for every pair γ, γ̃ such that γ = γ̃ + (0, 0, k), |γ| = j :

X_{3}^{k}Δ_{γ}(X) = Δγ̃(X). (32)

This allows us to compute Δ_{γ}(X) for every |γ| = d − 1, in terms of the homogeneization of B_{(γ_{1},γ_{2})}. From equation (32), the claim follows straightforwardly.

We conclude that Dixon’s formula can be viewed as a particular case of the determinantal expressions that we addressed. Moreover, Proposition 6.1 can be extended to any number of variables and all Dixon matrices as in [14, §3.5] can be recovered in degrees t such that ψ_{2,t_{n}−t}^{∗} = 0, i.e. such that t_{n} ≥ t > t_{n} − min{d_{1}, …, d_{n}}. As we have seen, all one can hope in general is the explicit quotient formula we give in Theorem 3.2. In fact, we have the following consequence of Lemma 5.3

Lemma 6.2. There exists a determinantal Dixon formula if and only if n = 2, or n = 3 and d_{1} = d_{2} = d_{3}, i.e. in the case considered by Dixon.

Proof. Assume d_{1} ≤ d_{2} ≤ ⋯ ≤ d_{n}. If inequality (28) is verified for t > t_{n} − d_{1}, we deduce that

(n − 2)d_{1} − n ≤ d_{3} + ⋯ + d_{n} − n < d_{1} − 2, (33)

and so (n − 3)d_{1} < n − 2. This equality cannot hold for any natural number d_{1} unless n ≤ 3. It is easy to check that for n = 2 there exist a determinantal Dixon formula for any value of d_{1}, d_{2}. In case n = 3,

(33) implies that d_{3} < d_{1} + 1. Then, d_{1} = d_{2} = d_{3}, as claimed.

7. Other known formulas and some extensions

We can recognize other well known determinantal formulas for resultants in this setting.

7.1. Polynomials in one variable. Let

f_{1}(x) = ∑_{j=0}^{d_{1}} a_{j}x^{j} , f_{2}(x) = ∑_{j=0}^{d_{2}} b_{j}x^{j}

be generic univariate polynomials (or their homogenizations in two variables) of degrees d_{1} ≤ d_{2}. In this case, inequality (28) is verified for all t = 0, …, d_{1} + d_{2} − 1 and so we have a determinantal formula for all such t. Here, t_{2} = d_{1} + d_{2} − 2. When t = d_{1} + d_{2} − 1 = t_{2} + 1 we have the classical Sylvester formula.

Assume d_{1} = d_{2} = d and write

\frac{f_{1}(x)f_{2}(y) − f_{1}(y)f_{2}(x)}{x − y} = ∑_{i,j=0}^{d−1} c_{ij}x^{i}y^{j}.

Then, the classical Bézout formula for the resultant between f_{1} and f_{2} says that

Res_{d,d}(f_{1}, f_{2}) = det(c_{ij}).

It is easy to see that we obtain precisely this formulation for t = d − 1. For other values of t we get formulas interpolating between Sylvester and Bézout as in [15, Ch. 12], even in case d_{1} ≠ d_{2}. It is easy to check that the smallest possible matrix has size d_{2}.

Suppose for example that d_{1} = 1, d_{2} = 2. In this case, [t_{2}/2] = [1/2] = 0, and M_{0} is a 2 × 2 matrix representing a map from S_{1}^{∗} to S_{0} ⊕ S_{0}^{∗}, whose determinant equals the resultant

Res_{1,2}(f_{1}, f_{2}) = a_{1}^{2}b_{0} − a_{0}a_{1}b_{1} + b_{2}a_{0}^{2}.

If we write f_{1}(x) = 0x^{2} + a_{1}x + a_{0} and we use the classical Bezout formula for d = 2, we would also get a 2 × 2 matrix but whose determinant equals b_{2} · Res_{1,2}(f_{1}, f_{2}). The exponent 1 in b_{2} is precisely the difference d_{2} − d_{1}.

7.2. Sylvester formula for three ternary quadrics. Suppose that n = 3, d_{1} = d_{2} = d_{3} = 2 and 2 ≠ 0. Let J denote the Jacobian determinant associated with the homogeneous polynomials f_{1}, f_{2}, f_{3}. A beautiful classical formula due to Sylvester says that the resultant Res_{2,2,2}(f_{1}, f_{2}, f_{3}) can be obtained as 1/512 times the determinant of the 6 × 6 matrix whose columns are indexed by the monomials in 3 variables of degree 2 and whose rows correspond to the expansion in this monomial basis of f_{1}, f_{2}, f_{3}, \frac{∂J}{∂X_{1}} , \frac{∂J}{∂X_{2}} and \frac{∂J}{∂X_{3}} . In this case, [t_{3}/2] = [3/2] = 1, and by Lemma 5.3 we have a determinantal formula in this degree since 2 − 3 < 1 < 4. From Euler equations

2f_{i}(X) = ∑_{j=1}^{3} X_{j} \frac{∂f_{i}(X)}{∂X_{j}},

we can write

2 (f_{i}(X) − f_{i}(Y)) = ∑_{j=1}^{3} (X_{j} \frac{∂f_{i}(X)}{∂X_{j}} − Y_{j} \frac{∂f_{i}(Y)}{∂Y_{j}})
= ∑_{j=1}^{3} (X_{j} − Y_{j})\frac{∂f_{i}(X)}{∂X_{j}} + Y_{j} (\frac{∂f_{i}(X)}{∂X_{j}} − \frac{∂f_{i}(Y)}{∂Y_{j}})
= ∑_{j=1}^{3} ((X_{j} − Y_{j}) \frac{∂f_{i}(X)}{∂X_{j}} + Y_{j} ∑_{l=1}^{3} \frac{∂^{2}f_{i}(X)}{∂X_{j}∂X_{l}}(X_{l} − Y_{l})) .

Because of (26), we can compute the Bezoutian using

Δ_{ij}(X, Y) := \frac{1}{2} (\frac{∂f_{i}(X)}{∂X_{j}} + ∑_{l=1}^{3} \frac{∂^{2}f_{i}(X)}{∂X_{l}∂X_{j}}Y_{l}) .

Using this formulation, it is not difficult to see that we can recover Sylvester formula from the equality Res_{2,2,2}(f_{1}, f_{2}, f_{3}) = det M_{1}.

7.3. Jacobian formulations. When t = t_{n}, one has H_{d}(t) = 1, and via the canonical identification of S_{0}^{∗} with A, the complex (22) reduces to the following modified Koszul Complex:

0 ⟶ K(t)^{−n} ⟶^{δ_{−(n−1)}} … ⟶^{δ_{−1}} A ⊕ K(t)^{−1} ⟶^{δ_{0}} K(t)^{0} ⟶0, (34)

where δ_{0} is the following map:

\begin{matrix}A & ⊕ & S_{t_{n}−d_{1}} ⊕ ⋯ ⊕ S_{t_{n}−d_{n}} & → & S_{t_{n}} \\ (λ & , & g_{1}, …, g_{n}) & ↦ & λ Δ_{0} + ∑_{i=1}^{n} g_{i} f_{i}, \end{matrix}

and Δ_{0} := Δ(X, 0). As a corollary of Theorem 5.1 we get that, for every specialization of the coefficients, Δ_{0} is a non-zero element of the quotient if the resultant does not vanish.

Assume that the characteristic of k does not divide the product d_{1} … d_{n}. It is a well-known fact that the jacobian determinant J of the sequence (f_{1}, …, f_{n}) is another non-zero element of degree t_{n}, which is a non-zero element of the quotient whenever the resultant does not vanish (cf. for instance [25]). In fact, one can easily check that

J = d_{1} … d_{n} Δ_{0} mod ⟨f_{1}, …, f_{n}⟩ . (35)

In [6], the same complex is considered in a more general toric setting, but using J instead of Δ_{0} . Because of (35), we can recover their results in the homogeneous case.

Theorem 7.1. Consider the modified complex (34) with J instead of Δ_{0}. Then, for every specialization of the coefficients, the complex is exact if and only if the resultant does not vanish. Moreover, the determinant of the complex equals d_{1} … d_{n} Res_{d_{1},...,d_{n}}(f_{1}, …, f_{n}).

We can also replace Δ_{0} by J in Macaulay’s Formula (Theorem 3.2), and have the following result:

Theorem 7.2. Consider the square submatrix M̃_{t_{n}} which is extracted from the matrix of δ_{0} in the monomial bases, choosing the same rows and columns of M_{t_{n}}. Then, det(M̃_{t_{n}}) ≠ 0, and we have the following formula à la Macaulay:

d_{1} … d_{n} Res_{d_{1},...,d_{n}}(f_{1}, …, f_{n}) = \frac{det(M̃_{t_{n}})}{det(E_{t_{n}})}.

We end the paper by addressing two natural questions that arise:

7.4. Different choices of monomial bases. Following Macaulay’s original ideas, one can show that there is some flexibility in the choice of the monomial bases defining S^{t,i} in order to get other non-zero minors, of Ψ̃_{t}, i.e different square matrices M′_{t} whose determinants are nonzero multiples of Res_{d_{1},...,d_{n}}(f_{1}, …, f_{n}) with different extraneous factors det(E′_{t}), det(E′_{t_{n}−t}), for appropiate square submatrices E′_{t}, E′_{t_{n}−t} of M′_{t}. Besides the obvious choices coming from a permutation in the indices of the variables, other choices can be made as follows.

For any i = 1, …, n, set d̂_{i} := (d_{1}, …, d_{i−1}, d_{i+1}, …, d_{n}) and define Hd̂_{i}(t) for any positive integer t by the equality

\begin{matrix}∏ & j≠i & (1 − Y^{d_{j}}) \\ (1 − Y) & n−1 \end{matrix} = ∑_{t=0}^{∞} Hd̂_{i}(t).Y^{t}.

For each t ∈ ℕ_{0}, set also Λ_{t} := {X^{γ} ∈ S_{t} : γ_{j} < d_{j}, j = 1, …, n}. We then have the following result:

Proposition 7.3. Let M′_{t} a square submatrix of Ψ̃_{t} of size ρ(t). Denote its blocks as in (15). Suppose that, for each i = 1, …, n, the block F_{t} has exactly Hd̂_{i}(t − d_{i}) of its columns corresponding to f_{i} in common with the matrix D_{t} defined in (8) and, also, the block F_{t_{n}−t} shares exactly Hd̂_{i}(t_{n} − t − d_{i}) columns corresponding to f_{i} with D_{t_{n}−t}. Then, if det (M′_{t}) is not identically zero, the resultant Res_{d_{1},...,d_{n}}(f_{1}, …, f_{n}) can be computed as the ratio \frac{det(M′_{t})}{det(𝔼′_{t})}, where 𝔼′_{t} is made by joining two submatrices E′_{t} of F_{t} and E′_{t_{n}−t} of F_{t_{n}−t}. These submatrices are obtained by omitting the columns in common with D_{t} (resp. D_{t_{n}−t}) and the rows indexed by all common monomials in D_{t} (resp. D_{t_{n}−t}) and all monomials in Λ_{t} (resp. Λ_{t_{n}−t}).

We omit the proof which is rather technical, and based in [21, 6a], and [9],[10].

7.5. Zeroes at infinity. Given a non-homogeneous system of polynomial equations f̃_{1}, …, f̃_{n} in n − 1 variables with respective degrees d_{1}, …, d_{n}, we can homogenize these polynomials and consider the resultant Res_{d_{1},...,d_{n}}(f_{1}, …, f_{n}) associated with their respective homogenizations f_{1}, …, f_{n}. However, this resultant may vanish due to common zeros of f_{1}, …, f_{n} at infinity in projective space ℙ^{n−1} even when there is no affine common root to f̃_{1}, = ⋯ = f̃_{n} = 0. We can in this case extend Canny’s construction [4] of the Generalised Characteristic Polynomial (GCP) for classical Macaulay’s matrices to the matrices M_{t} for any natural number t. In fact, when we specialize f_{i} to X_{i}^{d_{i}} for all i = 1, …, n, the Bezoutian is given by

∑_{j_{1}=0}^{d_{1}−1} ⋯ ∑_{j_{n}=0}^{d_{n}−1} X_{1}^{d_{1}−1−j_{1}} ⋯ X_{n}^{d_{n}−1−j_{n}} Y_{1}^{j_{1}} ⋯ Y_{n}^{j_{n}},

and the specialized matrix M_{t}(e) of M_{t} has a single non zero entry on each row and column which is equal to 1, so that det(M_{t}(e)) = ±1. We order the columns in such a way that M_{t}(e) is the identity matrix. With this convention, define the polynomial C_{t}(s) by

C_{t}(s) := \frac{Charpoly (M_{t})(s)}{Charpoly (𝔼_{t})(s)},

where s denotes a new variable and Charpoly means characteristic polynomial. We then have by the previous observation that

C_{t}(s) = Res_{d_{1},...,d_{n}}(f_{1} − s x_{1}^{d_{1}}, …, f_{n} − s x_{1}^{d_{n}}).

Moreover, this implies that C_{t}(s) coincides with Canny’s GCP C(s), but involves matrices of smaller size. Canny’s considerations on how to compute more efficiently the GCP also hold in this case. Of course, it is in general much better to find a way to construct “tailored” resultants for polynomials with a generic structure which is not dense, as in the case of sparse polynomial systems ([14], [15]).

Acknowledgements: We are grateful to J. Fernández Bonder, G. Massaccesi and J.M. Rojas for helpful suggestions. We are also grateful to David Cox for his thorough reading of the manuscript.

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Author’s addresses: Departamento de Matemática, F.C.E y N., UBA, (1428) Buenos Aires, Argentina.

cdandrea@dm.uba.ar                      alidick@dm.uba.ar