Radon’s construction and matrix relations generating syzygies

Let Πn\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\varPi _n$$\end{document} be the set of bivariate polynomials of degree not greater than n. A Πn\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\varPi _n$$\end{document}-correct set of nodes is a set such that the Lagrange interpolation problem with respect to these nodes has a unique solution. A maximal line of a Πn\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\varPi _n$$\end{document}-correct set is any line containing exactly n+1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$n+1$$\end{document} nodes. Syzygy matrices can be used to find linear factors of the fundamental polynomials and detect maximal lines. We suggest to use matrix relations in order to generate syzygies, identify linear factors of fundamental polynomials and detect maximal lines. We interpret our results in the important case of GC sets trying to shed some light on the Gasca-Maeztu conjecture.


Introduction
The Gasca-Maeztu conjecture [13] is still an outstanding problem in polynomial interpolation in two varialbles. So far it has been proven up to total degree n = 5 [14] with proofs of increasing complexity, for three and more variables not even its correct formulation is clear, cf. [11]. A promising approach, based on methods from algebraic geometry, was introduced in [12] and suggests to consider so-called syzygy matrices that capture information about the structure of the underlying interpolation problem. In this paper, which is a continuation of [8,9], we collect further information about these syzygy matrices and explicitly show their behavior in certain cases of a given defect. Even if this is not yet a proof of the conjecture we consider these results useful for further studying this connection between interpolation problems and the associated syzygy matrices.
We begin by recalling the setup we are considering in a more detailed way, using terminology from [9]. We will only consider the case of two variables in this paper. The bivariate Lagrange interpolation problem of degree n − 1 consists of finding a polynomial in Π n−1 , the space of bivariate polynomials of total degree not greater than n − 1, whose values coincide with prescribed data on a finite set Y ⊂ R 2 . If the Lagrange interpolation problem to data on Y in Π n−1 has a unique solution, we say that the set Y is Π n−1 -correct. The fact that a set is Π n−1 -correct can be revealed by inspecting I (Y ), the ideal of all polynomials vanishing on Y . A set Y with #Y = dim Π n−1 is Π n−1 -correct if and only if I (Y ) ∩ Π n−1 = {0}. The properties of the ideal of polynomials vanishing on a set can be read off specific bases of the ideal called H-bases and, if Y is Π n−1 -correct, H-bases of I (Y ) can be described as bases of the vector space I (Y ) ∩ Π n . Homogeneous relations between elements of a given H-basis are called syzygies. A matrix of maximal rank formed by first degree polynomials whose rows are syzygies of an H-basis for I (Y ) is called a syzygy matrix for the ideal. Many features of the interpolation problem can be interpreted in terms of syzygy matrices. Conversely, syzygy matrices can be used to derive properties of the corresponding H-basis, cf. [9,12].
The Berzolari-Radon construction extends a Π n−1 -correct set Y n−1 by adding n +1 nodes on a line W with W ∩ Y n−1 = ∅ and gives rise to a Π n -correct set Y n . The fundamental polynomials for the nodes in Y n \Y n−1 , i.e., the polynomials that vanish at all other nodes and take the value 1 at a given node, provide a particular H-basis h = (h 0 , . . . , h n ) of I (Y n−1 ), suitable for explicitly constructing syzygy matrices. Maximal lines in Y n are lines containing n +1 nodes. Any maximal line of Y n different from W is a linear factor of all elements of the H -basis h except one. Such maximal lines can be detected in the syzygy matrices as shown in [9,12].
In order to characterize first degree factors of the fundamental polynomials in h, the syzygy matrix relation Σh = 0 could be used. In this paper, we propose and investigate an alternative approach that deals with a more general relation of the form Sh = ge instead, where S ∈ Π (n+1)×(n+1) 1 is now a square matrix whose determinant coincides up to a nonzero constant with a nonzero ideal polynomial g ∈ I (Y ) ∩ Π n+1 and e = [1, . . . , 1] T ∈ R n+1 . Specifically, in Proposition 5 a relation of the form Sh = ge is derived with a matrix S = A with null diagonal entries and whose off diagonal elements a i j are affine polynomials that can be related to linear factors of the elements of h as shown in Theorem 2. In Theorem 1, we also identify maximal lines as the common off diagonal element in a column of A, which recovers the fundamental property of the syzygy matrices in [9].
Recall that a GC set is a Π n -correct set all whose fundamental polynomials are products of first degree polynomials. Due to the relation between repeated elements in the syzygy matrix and maximal lines, syzygy matrices are a promising tool for attacking the aforementioned Gasca-Maeztu conjecture which can be rephrased as the fact that any GC set can be obtained as a Berzolari-Radon extension of a GC set of smaller degree. To illustrate this connection, we end the paper by considering known GC sets and relating them to the properties derived in Theorem 1 and Theorem 2.

H-bases and syzygy matrices of correct sets
Let Π be the space of bivariate polynomials and Π n be the subspace of Π consisting of all polynomials of total degree not greater than n.
has a unique solution in Π n for any f : Y → R.
From the definition, it follows that the number of elements of any Π n -correct set Y is #Y = dim Π n = n+2 2 . Let us denote by the ideal of polynomials vanishing on the set Y . We observe that a set Y with #Y = dim Π n is Π n -correct if and only if Π n ∩ I (Y ) = 0. Associated to each node y of a Π n -correct set Y , there exist Lagrange fundamental polynomials y,Y ∈ Π n such that y,Y (y) = 1 and y,Y ∈ I (Y \{y}). Since the interpolation error of a polynomial p always belongs to I (Y ), many interpolation formulae can be related to the representation of elements in the ideal I (Y ).
Definition 2 A finite sequence (h 0 , . . . , h n ) of elements in an ideal I of polynomials is called an H-basis for I if for any f ∈ I there exist polynomials g j ∈ Π deg f −deg h j , j = 0, . . . , n, such that f = n j=0 g j h j . In this paper we shall consider H-bases consisting of polynomials of degree n of I (Y ) with Y a Π n−1 -correct set. As a consequence of Lemma 1 and Theorem 3 of [9] we can state the following result.

. , h n ) is an H-basis of I (Y ) if and only if it is a basis of the vector space I (Y ) ∩ Π n .
Some properties of an H-basis can be related to their syzygies.
In Theorem 4 of [9], it was shown that for any H-basis of the ideal I (Y ) of a Π n−1correct set there exist a matrix in Π n×(n+1) 1 whose rows are formed by independent syzygies of degree at most one. This motivates the following definition. In Theorem 5 of [9], it was shown that any syzygy formed by polynomials in Π 1 is a linear combination of the rows of a particular affine syzygy matrix Σ and that the set of all affine syzygy matrices is of the form AΣ, where A ∈ R n×n is a nonsingular matrix. Since we only consider affine syzygy matrices here, we will sometimes drop the explicit mentioning of degree at most one and just speak of a syzygy matrix.
The linear combinations of the rows of a syzygy matrix Σ with entries σ jk , j = 0, . . . , n − 1, k = 0, . . . , n, generate a vector subspace of Π n+1 1 of dimension n. Hence, there exist polynomials s n0 , . . . , s nn ∈ Π 1 such that (s n0 , . . . , s nn ) does not belong to the subspace generated by the rows of Σ and hence n k=0 s nk h k =: g = 0. Now, we define s jk = s nk + σ jk , for j = 0, . . . , n − 1, k = 0, . . . , n, and find that where S [ j] is the matrix obtained from S by replacing the j-th column by e.
Proof Let M ∈ R n×(n+1) of rank n with Me = 0. Multiplying in the matrix relation Sh = ge by M we obtain M S[h 0 , . . . , h n ] T = 0. Since M S is a n × (n + 1) matrix whose rank over the field of rational functions is n, we deduce that Σ := M S is a syzygy matrix. Now letΣ be another syzygy matrix. By Theorem 5 of [9], there exist a nonsingular real matrix A ∈ R n×n such thatΣ = AΣ = AM S =M S, witĥ M := AM. SinceM is a matrix of rank n such thatMe = AMe = 0, the result follows.
Since h 0 , . . . , h n are nonzero polynomials in I (Y ) ∩ Π n and det S = 0, we deduce that Sh is a nonzero vector and then g is a nonzero polynomial in I (Y ) ∩ Π n+1 . Since all elements in S are polynomials in Π 1 , we have that det S ∈ Π n+1 and det S [ j] ∈ Π n , j = 0, . . . , n. By Cramer's rule, Since deg h j = n, j = 0, . . . , n, we have that is independent of j ∈ {0, . . . , n}. Taking into account that the h j form a basis of I (Y ) ∩ Π n , they have no common divisor and we deduce that g divides det S. Since det Since g ∈ I (Y ) ∩ Π n+1 is a nonzero polynomial, there are two possible cases in Proposition 2, deg g = n or deg g = n + 1.
If deg g = n, then g = give rise to syzygies consisting of first degree polynomials. Since all syzygies of first degree can be generated by the rows of a syzygy matrix as shown in Theorem 5 of [9], we deduce that a linear combination of the rows of S − ec T equals 0, that that is, the constant row vector c T is a linear combination of the rows of S. From this fact, it follows that det S ∈ Π n . Since g coincides with det S up to a constant factor, we have that deg det S = n and thus coincides with h j up to a constant factor. Therefore, If, on the other hand, the degrees of g and det S are exactly n + 1, then and no linear combination of the rows can be a constant vector. In the following result, we compare different relations Sh = ge.

H-bases generated by Berzolari-Radon extensions
All information about the interpolation problem is contained in the ideal I (Y ), hence also in any basis of the ideal. Since the construction we are considering proceeds by total degree, it is convenient to use an H-basis. We can construct an H-basis explicitly using interpolation properties of a Berzolari-Radon extension. From the H-basis we determine, in an elementary and explicit way, a syzygy matrix. These syzygy matrices describe the geometry of the node configuration as pointed out in [12].
Recall that a line is the zero set Z (w) := {x ∈ R 2 | w(x) = 0} of a first degree polynomial w ∈ Π 1 \Π 0 . In 1948, Radon [15] suggested a construction that extends a Π n−1 -correct set Y n−1 by adding n + 1 nodes t 0 , . . . , This construction is also attributed to Berzolari [1] due to which it is called the Berzolari-Radon construction nowadays.
Conversely, if Y n is a Π n -correct set and we remove a line Z (w) of n + 1 nodes, the resulting set Y n−1 := Y n \Z (w) is a Π n−1 -correct set. In Theorem 1 of [9] the Berzolari-Radon construction is reviewed providing explicit formulae for the fundamental polynomials of the extension. The fundamental polynomials of the extension can be used to describe an H-basis (h 0 , . . . , h n ) of the ideal I (Y n−1 ).
Let us revisit the construction of the syzygy matrices from [9], conveniently modified for our purposes. In the next result we construct an H-basis (h 0 , . . . , h n ) of the ideal I (Y n−1 ) and find a matrix relation for defining syzygies of (h 0 , . . . , h n ). To that end, we introduce some notation that we will use in the remainder of the paper.
These polynomials belong to Π n−1 and interpolate d j /w, j = 0, . . . , n + 1, and d n+1 /w 2 on Y n−1 , respectively. Moreover, set h := [h 0 , . . . , h n ] T and Finally, let B = [b jk ] j,k=0,...,n ∈ R (n+1)×(n+1) be the matrix whose entries are and with I denoting the identity matrix, and use S [ j] for the matrix obtained by replacing the j-th column of S by e, j = 0, . . . , n. Proof (a) The set Y n is Π n -correct and the Lagrange fundamental polynomials for the nodes in the set Y n can be written in the form (see Theorem 1 of [9]) By Theorem 2 of [9], the ( t j ,Y n ) j=0,...,n form a basis of the vector space I (Y n−1 ) ∩Π n and thus are an H-basis for the ideal I (Y n−1 ). Since we deduce that Since the polynomials g 0 , . . . , g n , g n+1 ∈ I (Y n ) ∩ Π n+1 are linearly independent, we deduce from Corollary 1 of [9] that (g 0 , . . . , g n , g n+1 ) is a basis of I (Y n ) ∩ Π n+1 and H-basis of I (Y n ).
(c) We note that (v − v j )h j ∈ I (Y n ) ∩ Π n+1 , j = 0, . . . , n, are linearly independent polynomials since the homogeneous leading forms of h 0 , . . . , h n are linearly independent. By (b), there exist constants b jk such that Since 1, u, v form a basis of Π 1 , we can interpret (8) as a polynomial in v and w and compare the coefficients of v n+1 in both sides of (8), leading to b j,n+1 = 1, j = 0, . . . , n. Consequently, by (6), and thus Sh = g n+1 e.
Let us now describe a formula for elements b jk of the matrix B. Using (3) and (4) we have Comparing this with (9), we find that Due to (7) we have Hence, substituting x = t k into formula (10) and noticing that w(t k ) = 0, we obtain which is the expression given in (5). In particular b j j = 0, j = 0, . . . , n, and B has zero diagonal entries.
(d) From Proposition 2, it follows that there exists a constant c such that g n+1 = c det S and h j = c det S [ j] , j = 0, . . . , n. Expanding det S and ordering the expansion by powers of w, we get Expanding the determinant of S [ j] into powers of w, we deduce that where B [ j] is the matrix obtained from B, replacing its j-th row by e.
We can now obtain syzygies in a similar way to Sect. 3 of [9]. Forming the difference of any two equations in (9), Let Σ (i) be the matrix whose rows are the syzygies σ ji , j = 0, . . . , n. Observe that the matrix Σ (i) is the matrix obtained from S = v I − V − w B by subtracting the i-th row from all the other rows. By Proposition 2, Σ (i) is a syzygy matrix of (h 0 , . . . , h n ),

be the matrix obtained by removing the j-th column in Σ (i) . From Proposition 4 (d), it follows that
and, using Cramer's rule, we can deduce h j = (−1) i− j det Σ (i) j (see also Theorem 4 of [9]).
In general, we can apply Proposition 2 to obtain syzygy matrices M S multiplying any constant matrix M ∈ R n×(n+1) of rank n with Me = 0 by the matrix S = Let us now consider a modification of the relation Sh = g n+1 e of Proposition 4 (c). For this purpose, we define the crucial matrix that we will consider for the rest of the paper and whose entries can be regarded as equations of lines related with the set Y n of nodes.

Definition 6 Under the hypotheses of Definition 5, we define
whose entries are the polynomials that are either zero or have degree 1. ( (c) we have for any k = 0, 1, . . . , n, that Proof (a) Using (15) and the relations (9) we find that and so we have From (16), it follows that coefficient of v n+1 in the expansion of g in terms of v and w is n. From (14), the coefficient of v n+1 in det(−A) is det(I − ee T ) = −n. Applying Proposition 2 and comparing terms in v n+1 we deduce that g = − det(−A) = (−1) n det A. Moreover, setting w = 0 in (16) and noting that d n+1 does not vanish identically on this line, allows us to conclude that g = 0.
(c) Differentiating (16), substituting x = t k and recalling that w(t k ) = 0, we get Now we differentiate (3) and obtain, using (5), Observe that the matrix relation −Ah = −ge can be obtained from Sh = g n+1 e using The syzygy matrices Σ ( j) , j = k, are the matrices obtained from S = v I − V − w B by subtracting the j-th row from all other rows. Since the differences between the rows of S and the corresponding of −A are the same, the syzygy matrices Σ ( j) can be obtained either from the matrix S or −A by subtracting the j-th row from all the other rows. For instance, choosing j = 0, we obtain the syzygy matrix a 01 a 02 − a 12 · · · a 0n − a 1n −a 20 a 01 − a 21 a 02 · · · a 0n − a 2n . . .

Maximal lines
Maximal lines, introduced by de Boor in [11], are a key tool in the description of Π n -correct sets that can be obtained from the Berzolari-Radon construction. Since no line in the plane can contain more than n +1 nodes of a Π n -correct set, maximality has to be understood in the sense that such a maximal lines contains the maximal number of nodes possible among all lines in the plane.

Definition 7
A maximal line Z (w) of a Π n -correct set Y is defined by the property that #(Z (w) ∩ Y ) = n + 1. The set of maximal lines of a Π n -correct set Y will be denoted by M(Y ).
The main structural advantage of the syzygy matrices is that they enable the detection of maximal lines, see Theorem 6 of [9]. The following result shows that maximal lines can even be directly detected in the matrix A. Since divides g and all terms in l = j,k a jl h l but not h k , we deduce that divides a jk . Therefore a jk = for all j = k. So all off diagonal elements of the k-th column of A coincide with . (b) Assume that all off diagonal elements of the k-th column of A coincide, that is By Proposition 2, h j coincides up to a constant with the determinant of a matrix S [ j] , whose k-th column is (x)(e−e j ). So, divides h j for all j = k. For any y ∈ Y n−1 \L, take n y (x) a first degree polynomial such that n y (y) = 0 and n y (t k ) = 0. Then n y y,Y n−1 ∈ Π n ∩ I (Y n−1 ) and since (h 0 , . . . , h n ) is a basis of Π n ∩ I (Y n−1 ), we can write n y y,Y n−1 = n j=0 c j h j .
Evaluating at t k and taking into account (11), we conclude that c k = 0 and, since divides h j for all j = k and not n y we conclude that divides all polynomials y,Y n−1 , y ∈ Y n−1 \L. From the fact that y,Y n−1 , y ∈ Y n−1 \L are linearly independent polynomials in Π n−2 := { p : p ∈ Π n−2 }, we deduce that #(Y n−1 \L) ≤ dim Π n−2 and #(Y n−1 ∩ L) ≥ n, which implies that L is a maximal line. (c) By (a), any maximal line passing through some node in T is a factor of det A. So, the product of all maximal lines passing through a node in T divides det A.
The zero set of each fundamental polynomial t j ,Y n is an algebraic curve H j := Z (h j ) determined by the polynomial h j . If the gradient ∇h j (y) = 0 at some point y ∈ H j , then ∇h j (y)(· − y) = 0 is the equation of the tangent line of H j at y ∈ H j . If is a linear factor of h j , then the line Z ( ) is a component of the algebraic curve H j . Therefore, if L is a component of H j passing through y ∈ H j , then coincides with ∇h j (y)(· − y) up to a nonzero constant factor. Proposition 5 can be used to relate the elements of the matrix A ∈ Π (n+1)×(n+1)  Proof (a) Assume that a first degree polynomial is a factor of g, that is, g = q with q ∈ Π n . Clearly w is not a factor of g because w divides g − nd but not d. Since q can vanish at most at n nodes of T = {t 0 , . . . , t n } and g ∈ Π n+1 ∩ I (Y n ) vanishes at T , we conclude that there exists some k such that (t k ) = 0 and q(t k ) = 0. Differentiating, we have that ∇g(t k ) = q(t k )∇ . From Proposition 5 (c) we conclude that and j =k a jk are first degree polynomials vanishing at the same point t k with proportional gradients. So we deduce that coincides, up to a constant factor, with j =k a jk . (b) The proof is completely analogous. Assume that h j = q j , with q j ∈ Π n−1 . Since h j vanishes at T \{t j } and q j can vanish at most at n − 1 nodes of T \{t j }, we deduce that there exists k = j such that (t k ) = 0 and q j (t k ) = 0. Differentiating and using Proposition 5 (c), we deduce that ∇ and ∇a jk are both proportional to ∇h j (t k ) and we deduce that coincides, up to a constant factor with a jk .
(c) Assume that h j is a product of first degree polynomials. Since w cannot be a factor of h j and h j vanishes at T \{t j }, we deduce that there exists exactly one factor jk vanishing at t k , k = j and we can assume without loss of generality that By (b), jk coincides up to a constant with a jk . Then we have h j = c k = j a jk .
Comparing the coefficients of v n , we deduce that c = 1.
Theorem 2 (c) can also be proven using formula (12). If h j = k = j jk is a product of first degree polynomials, we can assume without loss of generality that jk (t k ) = 0 and hence there exists β jk such that jk = v − v k − β jk w. Comparing (12) with the expansion into powers of w in the product we deduce that β jk = b jk and jk = a jk , for all k = j. So h j = k = j a jk .
As a consequence of the preceding theorems, we can characterize maximal lines of an Π n−1 -correct set in terms of its ideal. Proof Suppose that Y n−1 contains a maximal line . Let t ∈ L\Y n−1 . By the Berzolari-Radon construction, we can choose a line W containing t 0 = t and no node of Y n−1 to construct Y n = Y n−1 ∪ {t 0 , . . . , t n }. By Theorem 1 (a), divides h 1 , . . . , h n and, by Proposition 4 (a) and the fact that (t) = 0, these polynomials are a basis of the vector space For the converse, we again make a Berzolari-Radon completion with t 0 = t and use Proposition 5 (c) and Theorem 1 (b) to conclude that L is a maximal line for Y n−1 .
Let us derive some consequences of Theorem 2. By (13), we know that h 0 = det Σ (0) 0 , where is the matrix obtained from S = v I − V − w B by subtracting the first row from the other rows and removing the first column. Denoting by the matrix obtained from B by subtracting the first row from the other rows and removing the first column, we can write If h 0 is the product of first degree polynomials, we get from Theorem 2 (c) that 0 is the product of its diagonal entries for all v, w. From this fact, it follows that the eigenvalues of the matrix

GC sets and factors of the fundamental polynomials
The geometric characterization introduced in [10] identifies Π n -correct sets Y in the plane such that all the Lagrange fundamental polynomials y,Y ∈ Π n , y ∈ Y , can be expressed as products of first degree polynomials.
such that for each y ∈ Y , there exists a set Γ y,Y of n lines containing all points in Y but y, that is, From the definition, it follows that a GC n set is Π n -correct. If Y n is a GC n set and a maximal line W ∈ M(Y n ) is removed, the remaining set Y n−1 = Y n \W is a GC n−1 -set and all the Lagrange polynomials t j ,Y n (x), t j ∈ Y n ∩ W , j = 0, . . . , n, are products of first degree polynomials, and so are the H-basis polynomials h 0 , . . . , h n .
In [13], Gasca and Maeztu conjectured that each GC n set contains at least one maximal line. So far, his conjecture has only been proved for degrees n ≤ 5 [2,4,14].
If the conjecture would hold for any degree, then each GC n -set Y n could be regarded as a Berzolari-Radon extension of a GC n−1 -set Y n−1 by adding n + 1 nodes on a line W with W ∩ Y n−1 = ∅.
The concept of defect introduced in [3,5,6] can be used to classify Π n -correct sets and consequently GC n sets.

Definition 9
The defect of a Π n -correct set Y , with n ≥ 1 is the number n+2−#M(Y ).
From the definition and the properties of Π n -correct sets, if follows that the defect of a Π n -correct set is a nonnegative integer less than or equal to n + 2.
If Y n−1 is a GC n−1 set of defect d, then we can choose a line W with Y n−1 ∩ W = ∅ and a subset T ⊂ W of n + 1 points containing all intersections K j ∩ W , j = 0, . . . , n − d, where K 0 , . . . , K n−d are the maximal lines of Y n−1 . By Theorem 1, all off diagonal elements of the j-th column of A coincide with a first degree polynomial k j for j = 0, . . . , n − d. If the Gasca-Maeztu, conjecture holds then M(Y n−1 ) = ∅, d < n + 1, and all off diagonal elements in the first column of A coincide.
On the other hand, if Y n is a GC n set and W is a maximal line of Y n , then Y n−1 = Y n \W is a GC n−1 set. Let T = Y n ∩ W . All maximal lines of Y n except W are maximal lines of Y n−1 intersecting W at some point in T , leading to matrices A such that some of their columns have coincident off diagonal elements.
Let us illustrate some properties described in Theorem 1 and Theorem 2 with all known GC n configurations.
Defect 0. Assume that Y n is a GC n set of defect 0. Let K 0 , . . . , K n , K n+1 be the maximal lines of Y n . Choose w a linear polynomial such that K n+1 = Z (w) and v a first degree polynomial such that 1, v, w form a basis of Π 1 . Take T = Y n ∩ Z (w), t j = K j ∩ W , j = 0, . . . , n, and Y n−1 = Y n \W . Since all off diagonal elements of the j-column of A are equal to k j , j = 0, . . . , n, we have that If we multiply A[h 0 , . . . , h n ], we see that all components of the product are equal to g = nk 0 · · · k n . So we find that k 0 · · · k n divides det A. Defect 1. Assume that Y n is a GC n set of defect 1, n ≥ 2. Let K 0 , . . . , K n the maximal lines of Y n . Choose w a linear polynomial such that K n = Z (w) and v a first degree polynomial such that 1, v, w form a basis of Π 1 . Take T = Y n ∩ Z (w), with t j = K j ∩ Z (w), j = 0, . . . , n − 1, t n ∈ Z (w)\{t 0 , . . . , t n−1 } and Y n−1 = Y n \Z (w). Then we have that The elements of the last column a jn are first degree polynomials corresponding to the line joining t n and the node in K j ∩ Y n−1 \ l = j K l . Since the defect is 1, not all elements in the last column of A coincide and there exists at least two linearly independent polynomials vanishing at t n among a jn , j = 0, . . . , n−1. All components of the product A[h 0 , . . . , h n ] are equal to g = k 0 · · · k n−1 n−1 j=0 a jn and k 0 · · · k n−1 divides det A. We remark that the linear factor n−1 j=0 a jn not corresponding to a maximal line can be directly obtained from Theorem 2 (a).

Defect 2.
In this case Y n is a GC n set of defect 2, n ≥ 3, and K 0 , . . . , K n−1 are their maximal lines. Choose w a linear polynomial such that K n−1 = Z (w) and v a first degree polynomial such that 1, v, w form a basis of Π 1 . Take T = Y n ∩ Z (w), t j = K j ∩ Z (w), j = 0, . . . , n − 2, and Y n−1 = Y n \Z (w). Let z ∈ Y n−1 \ n−1 l=0 K l be the unique node in Y n−1 not belonging to any maximal line and denote by L n−1 and L n the lines joining z with t n−1 and t n respectively. Then, it can be deduced that the sets Γ t,Y from Definition 8 are of the form where the union of the two lines A j,n−1 ∪ A j,n should contain t n−1 , t n , the node z and the two nodes in K j \ i = j K i , j = 0, . . . , n − 2. Since at least one of them should contain z, we deduce that at least one of the lines L n−1 or L n belong to Γ t j ,Y n for j = 0, . . . , n − 2, that is, one of the following three situations can happen: either only A j,n−1 = L n−1 , or only A j,n = L n or both at the same time, A j,n−1 = L n−1 , A j,n = L n . We can reorder the nodes in such a way that L n is used by the nodes t j with j = 0, . . . , m−1 but not L n−1 and L n−1 is used by the nodes t j with j = m, . . . , n−2. At least one node t j , j ∈ {m, . . . , n −2} does not use L n and so we can assume without