On the dimension of spaces of algebraic curves passing through $n$-independent nodes

TL;DR

This paper investigates the maximum number of algebraic curves of bounded degree passing through a specific set of nodes with certain independence properties, providing a characterization of the extremal case and an application to a conjecture.

Contribution

It establishes an upper bound of four on the number of such curves and characterizes the structure of node sets when this bound is attained.

Findings

Maximum of four linearly independent curves of degree ≤ k passing through the nodes.

Characterization of node sets with exactly four such curves.

Application to the Gasca-Maeztu conjecture.

Abstract

Let a set of nodes $\mathcal X$ in plain be $n$-independent, i.e., each node has a fundamental polynomial of degree $n.$ Suppose also that $|\mathcal X|= d(n,k-2)+2,$ where $d(n,k-2) = (n+1)+n+\cdots+(n-k+4)$ and $\ k\le n-1.$ In this paper we prove that there can be at most $4$ linearly independent curves of degree less than or equal to $k$ passing through all the nodes of $\mathcal X.$ We provide a characterization of the case when there are exactly four such curves. Namely, we prove that then the set $\mathcal X$ has a very special construction: All its nodes but two belong to a (maximal) curve of degree $k-2.$ At the end, an important application to the Gasca-Maeztu conjecture is provided.

Equations

$${\Pi }_n=\left\{\sum _{i+j\le n}{a}_{ij}{x}^i{y}^j\right\}.$$

$$N:=N_n:=\dim {\Pi }_n=(1/2)(n+1)(n+2).$$

$$\mathcal{X}={\mathcal{X}}_s=\{(x_1,y_1),(x_2,y_2),\dots ,(x_s,y_s)\}.$$

$$p(x_i,y_i)=c_i,i=1,\dots ,s,$$

$$p(A)=1\ \ \text{and}\ \ p\big{|}_{{\mathcal{X}}\setminus\{A\}}=0,$$

$$p\in\Pi_{n}\ \ \text{and}\ \ p\big{|}_{{\mathcal{X}}_{N}}=0\quad\implies\quad p=0.$$

$${{\mathcal{P}}}_{n,{\mathcal{X}}}=\left\{p\in\Pi_{n}:p\big{|}_{\mathcal{X}}=0\right\}.$$

$$\dim {\mathcal{P}}_{n,\mathcal{X}}=N-|\mathcal{Y}|,$$

$$p\in {\Pi }_n\text{and}p{|}_{{\mathcal{X}}_{n+1}}=0⟹p=\ell r,$$

$$p\in {\Pi }_n\text{and}p{|}_{{\mathcal{X}}_d}=0⟹p=qr,$$

$${\mathcal{P}}_{k,\mathcal{X}}=\left\{p:p\in {\Pi }_k,p(A)=p(B)=0\right\}=\left\{q\mu :q\in {\Pi }_2,q(A)=q(B)=0\right\}.$$

$$\dim {\mathcal{P}}_{k,\mathcal{X}}=\dim \left\{q\in {\Pi }_2:q(A)=q(B)=0\right\}=\dim {\mathcal{P}}_{2,\{A,B\}}=6-2=4.$$

$$\mathcal{Y}=\mathcal{X}\cup \mathcal{A},$$

$$p{\ell }_{01}{\ell }_2\cdots {\ell }_{n-k}=q_{k-1}s.$$

$$p{\ell }_{01}=q_{k-1}\beta ,where\beta \in {\Pi }_2.$$

$$p\in\Pi_{k},\ p\big{|}_{{\mathcal{X}}}=0\ \implies\ p=q_{k-1}\ell,\ \hbox{where}\ \ell\in\Pi_{1}.$$

$$q_{k-1}={\ell }_{01}{q}_{k-2}where{q}_{k-2}\in {\Pi }_{k-2}.$$

$${\mathcal{X}}^{\prime \prime }:=\mathcal{X}\cup \{B_1,B_2,B_3\}\cup \mathcal{A}.$$

$${\sigma }_0{\ell }_0{\ell }_1{\ell }_2\dots {\ell }_{n-k-1}$$

$${\sigma }_0{\ell }_0{\ell }_1{\ell }_2\dots {\ell }_{n-k-1}=\sigma q,q\in {\Pi }_{n-k}.$$

$${\sigma }_0{\ell }_0=\sigma \ell ,where\ell \in {\Pi }_1.$$

$$\sigma ={\ell }_{0}r,r\in {\Pi }_{k-1}.$$

$$\sigma =\ell q,whereq\in {\Pi }_{k-1}.$$

$$p_{C_i}^{\star }=\mu r_i,r_i\in {\Pi }_3,i=1,2.$$

$$p_{C_i}^{\star }=\mu \ell {\beta }_i,i=1,2.$$

$$\mathcal{Y}\subset \mu .$$

$$p_D^{\star }={\mu }^{\prime }{q}^{\prime },q^{\prime }\in {\Pi }_3.$$

$$p_A^{\star }=\mu \ell q_A,where{q}_A\in {\Pi }_2.$$

Full text

At the end, an important application to the Gasca-Maeztu conjecture is provided.

Key words: algebraic curves, $n$-independent nodes, maximal curves, Gasca-Maeztu conjecture.

Mathematics Subject Classification (2010): 41A05, 41A63.

1 Introduction

Denote the space of all bivariate polynomials of total degree $\leq n$ by $\Pi_{n}:$

[TABLE]

We have that

[TABLE]

Consider a set of $s$ distinct nodes

[TABLE]

The problem of finding a polynomial $p\in\Pi_{n}$ which satisfies the conditions

[TABLE]

is called interpolation problem.

A polynomial $p\in\Pi_{n}$ is called a fundamental polynomial for a node $A\in{\mathcal{X}}$ if

[TABLE]

where p\big{|}_{\mathcal{X}} means the restriction of $p$ on $\mathcal{X}.$ We denote the fundamental polynomial by $p_{A}^{\star}.$ Sometimes we call fundamental also a polynomial that vanishes at all nodes of ${\mathcal{X}}$ but one, since it is a nonzero constant times a fundamental polynomial.

Definition 1.1**.**

The interpolation problem with a set of nodes ${\mathcal{X}}_{s}$ and $\Pi_{n}$ is called $n$-poised if for any data $(c_{1},\ldots,c_{s})$ there is a unique polynomial $p\in\Pi_{n}$ satisfying the interpolation conditions (1.1).

A necessary condition of poisedness is $|{\mathcal{X}}_{s}|=s=N.$

Proposition 1.2**.**

A set of nodes ${\mathcal{X}}_{N}$ is $n$-poised if and only if

[TABLE]

Next, let us consider the concept of $n$-independence (see [3], [5]).

Definition 1.3**.**

A set of nodes ${\mathcal{X}}$ is called $n$-independent if all its nodes have $n$-fundamental polynomials. Otherwise, it is called $n$-dependent.

Fundamental polynomials are linearly independent. Therefore a necessary condition of $n$-independence of ${\mathcal{X}}_{s}$ is $s\leq N$.

2 Some properties of $n$-independent nodes

Let us start with the following simple

Lemma 2.1** (e.g., [6] Lemma 2.2).**

Suppose that a node set ${\mathcal{X}}$ is $n$-independent and a node $A\notin\mathcal{X}$ has $n$-fundamental polynomial with respect to the set ${\mathcal{X}}\cup\{A\}.$ Then the latter node set is $n$-independent, too.

Denote the distance between the points $A$ and $B$ by $\rho(A,B).$ Let us bring the following (see e.g., [4], [8])

Lemma 2.2**.**

Suppose that ${\mathcal{X}}_{s}=\{A_{i}\}_{i=1}^{s}$ is an $n$-independent set. Then there is a number $\epsilon>0$ such that any set ${\mathcal{X}}_{s}^{\prime}=\{A_{i}^{\prime}\}_{i=1}^{s},$ with the property that $\rho(A_{i},A_{i}^{\prime})<\epsilon,\ i=1,\ldots,s,$ is $n$-independent too.

Next result concerns the extension of $n$-independent sets

Lemma 2.3** (e.g., [5], Lemma 2.1).**

Any $n$-independent set ${\mathcal{X}}$ with $|{\mathcal{X}}|<N$ can be enlarged to an $n$-poised set.

In the sequel we will need the following modification of the above result.

Lemma 2.4**.**

Given $n$-independent sets ${\mathcal{X}}_{s_{i}},\ i=1,\ldots,m,$ where $|{\mathcal{X}}_{s_{i}}|={s_{i}}<N,$ a node $A,$ and any number $\epsilon>0.$ Then there is a node $A^{\prime},$ such that $\rho(A,A^{\prime})<\epsilon$ and each set ${\mathcal{X}}_{s_{i}}\cup\{A^{\prime}\},\ i=1,\ldots,m,$ is $n$-independent.

Proof.

Let us use induction with respect to the number of sets: $m.$ Suppose that we have one set ${\mathcal{X}}_{s}.$ Since $s<N,$ there is a nonzero polynomial $p\in\Pi_{n}$ such that p\big{|}_{{\mathcal{X}}_{s}}=0. Now evidently there is a node $B\notin\mathcal{X},$ such that $\rho(A,B)<\epsilon$ and $p(B)\neq 0.$ Thus $p$ is an $n$-fundamental polynomial of the node $B$ with respect to the set ${\mathcal{X}}\cup\{B\}.$ Hence, in view of Lemma 2.1, the set ${\mathcal{X}}_{s}\cup\{B\}$ is $n$-independent. Then, assume that Lemma is true in the case of $m-1$ sets, i.e., there is a node $B,$ such that $\rho(A,B)<(1/2)\epsilon$ and each set ${\mathcal{X}}_{s_{i}}\cup\{B\},\ i=1,\ldots,m-1,$ is $n$-independent. In view of Lemma 2.2, there is a number $\epsilon^{\prime}<(1/2)\epsilon$ such that for any $C,$ with $\rho(C,B)<\epsilon^{\prime},$ each set ${\mathcal{X}}_{s_{i}}\cup\{C\},\ i=1,\ldots,m-1,$ is $n$-independent. Next, in view of first step of induction, there is a node $A^{\prime}$ such that $\rho(A^{\prime},B)<(1/2)\epsilon$ and the set ${\mathcal{X}}_{s_{m}}\cup\{A^{\prime}\}$ is $n$-independent. Now, it is easily seen that $A^{\prime}$ is a desirable node. ∎

Denote the linear space of polynomials of total degree at most $n$ vanishing on ${\mathcal{X}}$ by

[TABLE]

The following two propositions are well-known.

Proposition 2.5** (e.g., [5]).**

For any node set ${\mathcal{X}}$ we have that

[TABLE]

where ${\mathcal{Y}}$ is a maximal $n$-independent subset of ${\mathcal{X}}.$

Proposition 2.6**.**

Assume that $\ell$ is a line and ${\mathcal{X}}_{n+1}$ is any subset of $\ell$ containing $n+1$ points. Then we have that

[TABLE]

where $r\in\Pi_{n-1}.$

A plane algebraic curve is the zero set of some bivariate polynomial. To simplify notation, we shall use the same letter $p$, say, to denote the polynomial $p$ of degree $\geq 1$ and the curve given by the equation $p(x,y)=0$.

Set $d(n,k):=N_{n}-N_{n-k}=(1/2)k(2n+3-k).$ The following is a generalization of Proposition 2.6.

Proposition 2.7** ([9], Prop. 3.1).**

*Let $q$ be an algebraic curve of degree $k\leq n$ without multiple components. Then the following hold.

i) Any subset of $q$ containing more than $d(n,k)$ nodes is $n$-dependent.

ii) Any subset ${\mathcal{X}}_{d}$ of $q$ containing exactly $d=d(n,k)$ nodes is $n$-independent if and only if the following condition holds:*

[TABLE]

where $r\in\Pi_{n-k}.$

Thus, according to Proposition 2.7, i), at most $d(n,k)$ nodes of $\mathcal{X}$ can lie in the curve $q$ of degree $k\leq n$. This motivates the following definition.

Definition 2.8** ([9], Def. 3.1).**

Given an $n$-independent set of nodes $\mathcal{X}_{s},$ with $s\geq d(n,k).$ A curve of degree $k\leq n$ passing through $d(n,k)$ points of $\mathcal{X}_{s},$ is called maximal.

We say that a node $A$ of an $n$-poised set ${\mathcal{X}}$ uses a line $\ell$ if the latter divides the fundamental polynomial of $A,$ i.e., $p^{\star}_{A}=\ell q,$ for some $q\in\Pi_{n-1}.$

Next, we bring a characterization of maximal curves:

Proposition 2.9** ([9], Prop. 3.3).**

Let a node set ${\mathcal{X}}$ be n-poised. Then a polynomial $\mu$ of degree $k,\ k\leq n,$ is a maximal curve if and only if it is used by any node in ${\mathcal{X}}\setminus\mu.$

Proposition 2.10** ([8], Prop. 3.5).**

Assume that $\sigma$ is an algebraic curve of degree $k,$ without multiple components, and ${\mathcal{X}}_{s}\subset\sigma$ is any $n$-independent node set of cardinality $s,\ s<d(n,k).$ Then the set ${\mathcal{X}}_{s}$ can be extended to a maximal $n$-independent set ${\mathcal{X}}_{d}\subset\sigma$ of cardinality $d=d(n,k)$.

Finally, let us bring a well-known

Lemma 2.11**.**

Suppose that $m$ linearly independent curves pass through all the nodes of ${\mathcal{X}}.$ Then for any node $A\notin{\mathcal{X}}$ there are $m-1$ linearly independent curves, belonging to the linear span of given curves, passing through $A$ and all the nodes of ${\mathcal{X}}.$

3 Main result

Let us start with the following result from [7] (see also, [1]).

Theorem 3.1** ([7], Thm. 1).**

Assume that ${\mathcal{X}}$ is an $n$-independent set of $d(n,k-1)+2$ nodes lying in a curve of degree $k$ with $k\leq n.$ Then the curve is determined uniquely by these nodes.

Next result in this series is the following

Theorem 3.2** ([8], Thm. 4.2).**

Assume that ${\mathcal{X}}$ is an $n$-independent set of $d(n,k-1)+1$ nodes with $k\leq n-1.$ Then two different curves of degree $k$ pass through all the nodes of ${\mathcal{X}}$ if and only if all the nodes of ${\mathcal{X}}$ but one lie in a maximal curve of degree $k-1.$

Now let us present the main result of this paper:

Theorem 3.3**.**

Assume that ${\mathcal{X}}$ is an $n$-independent set of $d(n,k-2)+2$ nodes with $k\leq n-1.$ Then four linearly independent curves of degree less than or equal to $k$ pass through all the nodes of ${\mathcal{X}}$ if and only if all the nodes of ${\mathcal{X}}$ but two lie in a maximal curve of degree $k-2.$

Let us mention that the inverse implication here is evident. Indeed, assume that $d(n,k-2)$ nodes of ${\mathcal{X}}$ are located in a curve $\mu$ of degree $k-2$. Therefore the curve $\mu$ is maximal and the remaining two nodes of ${\mathcal{X}},$ denoted by $A$ and $B,$ are outside of it: $A,B\notin\mu.$ Hence we have that

[TABLE]

Thus we readily get that

[TABLE]

In the last equality we use the fact that any two nodes are $2$-independent. We get also that there can be at most $4$ linearly independent curves of degree $\leq k$ passing through all the nodes of $\mathcal{X}.$

Before starting the proof of Theorem 3.3 let us present two lemmas.

Lemma 3.4**.**

Assume that ${\mathcal{X}}$ is an $n$-independent node set and a node $A\in\mathcal{X}$ has an $n$-fundamental polynomial $p^{\star}_{A}$ such that $p^{\star}_{A}(A^{\prime})\neq 0.$ Then we can replace the node $A$ with $A^{\prime}$ such that the resulted set ${\mathcal{X}}^{\prime}:={\mathcal{X}}\cup\{A^{\prime}\}\setminus\{A\}$ is again an $n$-independent. In particular, such replacement can be done in the following two cases:

i) If a node $A\in\mathcal{X}$ belongs to several components of $\sigma$ then we can replace it with a node $A^{\prime},$ which belongs only to one component of $\sigma;$

ii) If a curve $q$ is not a component of an $n$-fundamental polynomial $p^{\star}_{A}$ then we can replace the node $A$ with a node $A^{\prime}$ lying in $q.$

Proof.

Indeed, notice that $p^{\star}_{A}(A^{\prime})\neq 0$ means that $p^{\star}_{A}$ is a fundamental polynomial also for the node $A^{\prime}$ with respect to the set ${\mathcal{X}}^{\prime}.$ Next, for (i) note that a fundamental polynomial of a node $A$ differs from [math] in a neighborhood of $A.$ Finally, for (ii) note that $q$ is not a component of $p^{\star}_{A}$ means that there is a point $A^{\prime}\in q$ such that $p^{\star}_{A}(A^{\prime})\neq 0.$∎

Lemma 3.5**.**

Assume that the hypotheses of Theorem 3.3 hold and assume additionally that there is a curve $q_{k-1}\in\Pi_{k-1}$ passing through all the nodes of ${\mathcal{X}}.$ Then all the nodes of ${\mathcal{X}}$ but two lie in a maximal curve $\mu$ of degree $k-2.$

Proof.

First note that the curve $q_{k-1}$ is of exact degree $k-1,$ since it passes through more than $d(n,k-2)$ $n$-independent nodes. This implies also that $q_{k-1}$ has no multiple component. Therefore we can extend the set ${\mathcal{X}}$ till a maximal $n$-independent set ${\mathcal{Y}}\subset q_{k-1},$ by adding $n-k+1$ nodes, i.e.,

[TABLE]

where ${\mathcal{A}}=\{A_{0},\ldots,A_{n-k}\}.$

In view of Lemma 3.4, i), we may suppose that the nodes from ${\mathcal{A}}$ are not intersection points of the components of the curve $q_{k-1}.$

Next, we are going to prove that these $n-k+1$ nodes are collinear together with $m\geq 2$ nodes from ${\mathcal{X}}.$

To this end denote the line through the nodes $A_{0}$ and $A_{1}$ by $\ell_{01}.$ Then for each $i=2\ldots,n-k,$ choose a line $\ell_{i}$ passing through the node $A_{i}$ which is not a component of $q_{k-1}.$ We require also that each line passes through only one of the mentioned nodes and therefore the lines are distinct.

Now suppose that $p\in\Pi_{k}$ vanishes on ${\mathcal{X}}.$ Consider the polynomial $r=p\ell_{01}\ell_{2}\cdots\ell_{n-k}.$ We have that $r\in\Pi_{n}$ and $r$ vanishes on the node set ${\mathcal{Y}},$ which is a maximal $n$-independent set in the curve $q_{k-1}.$ Therefore we obtain that $r=q_{k-1}s,$ where $s\in\Pi_{n-k+1}.$ Thus we have that

[TABLE]

The lines $\ell_{i},\ i=2,\ldots,n-k,$ are not components of $q_{k-2}.$ Therefore they are components of the polynomial $s.$ Thus we obtain that

[TABLE]

Now let us verify that $\ell_{01}$ is a component of $q_{k-1}.$ Indeed, otherwise it is a component of the conic $\beta$ and we get that

[TABLE]

Therefore we get $\dim{\mathcal{P}}_{k,\mathcal{X}}=3,$ which contradicts the hypothesis.

Thus we conclude that

[TABLE]

Then let us show that all the nodes of ${\mathcal{A}}$ belong to $\ell_{01}.$ Suppose conversely that a node from ${\mathcal{A}},$ say $A_{2},$ does not belong to the line $\ell_{01}.$ Then in the same way as in the case of the line $\ell_{01}$ we get that $\ell_{02}$ is a component of $q_{k-1}.$ Thus the node $A_{0}$ is an intersection point of two components of $q_{k-1},$ i.e., $\ell_{01}$ and $\ell_{02},$ which contradicts our assumption.

We have that the curve $q_{k-2}$ passes through at most $d(n,k-2)$ nodes from ${\mathcal{X}}.$ Hence, in view of the relation (3.1), we get that at least $2$ nodes from ${\mathcal{X}}$ belong to the line $\ell_{01}.$

Next we will show that exactly $2$ nodes from ${\mathcal{X}\cap}\ell_{01}$ do not belong to $q_{k-2}.$ This will prove Lemma since, in view of (3.1), we have that ${\mathcal{X}\setminus}\ell_{01}\subset q_{k-2}$

Assume by way of contradiction that at least $3$ nodes from from ${\mathcal{X}\cap}\ell_{01}$ do not belong to $q_{k-2}.$

Let us verify that then one can move the node $A_{0},$ from $\ell_{01}$ to a new location $A_{0}^{\prime}\in q_{k-2}\setminus\ell_{01},$ such that the resulted set ${\mathcal{Y}}={\mathcal{X}}\cup{\mathcal{A}}$ remains $n$-independent. This will contradict the proved fact ${\mathcal{A}}\subset\ell_{01}.$

In view of Lemma 3.4, ii), for this we need to find an $n$-fundamental polynomial of $A_{0}$ for which $q_{k-2}$ is not a component. Let us show that any fundamental polynomial of $A_{0}$ has this property. Indeed, suppose conversely that for an $n$-fundamental polynomial $p^{\star}_{A_{0}}\in\Pi_{n}$ the curve $q_{k-2}$ is a component, i.e., $p^{\star}_{A_{0}}=q_{k-2}r,$ where $r\in\Pi_{n-k+2}.$ We get from here that $r$ vanishes at $\geq 3+(n-k+1)-1=n-k+3$ nodes in $\ell.$ Therefore, in view of Proposition 2.6, $r$ vanishes at all the points of $\ell_{01}$ including $A_{0},$ which is a contradiction. ∎

Now we are in a position to present

Proof of Theorem 3.3.

Recall that it remains to prove the direct implication. Let $\sigma_{1},\ldots,\sigma_{4},$ be the four curves of degree $\leq k$ that pass through all the nodes of the $n$-independent set ${\mathcal{X}}$ with $|{\mathcal{X}}|=d(n,k-2)+2.$

Consider three collinear nodes $B_{1},B_{2},B_{3}\notin{\mathcal{X}}$ such that the following two conditions are satisfied:

$(i)$ The set ${\mathcal{X}}\cup\{B_{1},B_{2},B_{3}\}$ is $n$-independent;

$(ii)$ The line through $B_{1},B_{2},B_{3},$ does not pass through any node from ${\mathcal{X}}.$

Let us verify that one can find such nodes $B_{1},B_{2},B_{3},$ or the conclusion of Theorem 3.3 holds. Indeed, in view of Lemma 2.3, we can start by choosing some two nodes $B_{i}^{\prime},\ i=1,2,$ such that

$(i)^{\prime}$ The set $\mathcal{X}\cup\{B_{1}’,B_{2}’\}$ is $n$-independent.

Then, according to Lemma 2.2, for some positive $\epsilon$ all the nodes in $\epsilon$ neighborhoods of $B_{i}^{\prime},\ i=1,2,$ satisfy $(i)^{\prime}.$ Thus, from this neighborhoods we can choose the nodes $B_{i},\ i=1,2,$ respectively, such that the line through them: $\ell_{0}$ does not pass through any node from ${\mathcal{X}}.$

Now, let us esablish existence of the node $B_{3},$ satisfying the condition $(i).$

Assume, by way of contradiction, that there is no such node $B_{3}\in\ell_{0}.$

This, in view of Lemma 2.1, means that any polynomial $p\in\Pi_{n}$ vanishing on ${\mathcal{X}}\cup\{B_{1},B_{2},\}$ vanishes identically on $\ell_{0}.$ By using Lemma 2.11 we may choose a such polynomial $p$ from the linear span of linearly independent curves $\sigma_{i},i=1,2,3.$ Then we get that p\in\Pi_{k},\ p\big{|}_{\ell_{0}}=0. Thus we have $p=\ell_{0}q,$ where $q\in\Pi_{k-1}.$ Now, in view of $(ii)$ we readily deduce that the curve $q$ of degree $\leq k-1$ passes through all the nodes of $\mathcal{X}.$ Thus the proof of Theorem is completed in view of Lemma 3.5.

Next, in view of Proposition 2.11, there is a curve of degree at most $k,$ denoted by $\sigma,$ which passes through all the nodes of ${\mathcal{X}}^{\prime}:={\mathcal{X}}\cup\{B_{1},B_{2},B_{3}\}.$

Now notice that the curve $\sigma$ passes through more than $d(n,k-2)$ nodes and therefore its degree equals either to $k-1$ or $k.$ By taking into account Lemma 3.5 we may assume that the degree of the curve $\sigma$ equals to $k.$ Evidently we may assume also that $\sigma$ has no multiple component.

Therefore, by using Proposition 2.10, we can extend the set ${\mathcal{X}}^{\prime}$ till a maximal $n$-independent set ${\mathcal{X}}^{\prime\prime}\subset\sigma.$ To this end, in view of $|{\mathcal{X}}^{\prime\prime}|=d(n,k)$, we need to add a set of $d(n,k)-(d(n,k-2)+2)-3=2(n-k)$ nodes to ${\mathcal{X}}^{\prime}$, denoted by $\mathcal{A}:=\{A_{1},\ldots,A_{2(n-k)}\}.$ Thus we have that

[TABLE]

Thus the curve $\sigma$ becomes maximal with respect to this set. In view of Lemma 3.4, i), we require that each node of $\mathcal{A}$ belongs only to one component of the curve $\sigma.$

Then, by using Lemma 2.11, we get a curve $\sigma_{0}$ of degree at most $k,$ different from $\sigma$ that passes through all the nodes of ${\mathcal{X}}$ and two more nodes: $A_{0},A_{0}^{\prime}\in\mathcal{A},$ which will be described below.

In the sequel we intend to divide the set of nodes $\mathcal{A}$ into $n-k$ pairs such that the lines $\ell_{1},\ldots,\ell_{n-k-1}$ through the first $n-k-1$ pairs from them, respectively, are not components of $\sigma.$ The last pair is specified as $A_{0},A_{0}^{\prime}.$

Before establishing the mentioned division of $\mathcal{A}$ let us verify how we can finish the proof by using it.

Recall that $\ell_{0}$ is the line through the triple of the nodes $B_{1},B_{2},B_{3}.$ Notice that the following polynomial

[TABLE]

of degree $n$ vanishes at all the $d(n,k)$ nodes of ${\mathcal{X}}^{\prime\prime}\subset\sigma.$ Consequently, according to Proposition 2.7, $\sigma$ divides this polynomial:

[TABLE]

The distinct lines $\ell_{1},\ell_{2},\dots,\ell_{n-k-1},$ do not divide the polynomial $\sigma\in\Pi_{k}$, therefore all they have to divide $q\in\Pi_{n-k}.$ Therefore, we get from (3.2):

[TABLE]

Since the curves $\sigma$ and $\sigma_{0}$ are different the lines $\ell_{0}$ and $\ell$ also are different. Therefore the line $\ell_{0}$ has to divide $\sigma\in\Pi_{k}$:

[TABLE]

In view of above condition (iii) the line $\ell_{0}$ does not pass through any node of $\mathcal{X}.$ Therefore the curve $r$ of degree $k-1$ passes through all the nodes of $\mathcal{X}.$ Thus the proof of Theorem is completed in view of Lemma 3.5.

Observe that, in view of the above argument, we may assume that any line component of the curve $\sigma$ passes through at least a node from $\mathcal{X}.$

Next let us establish the above mentioned division of the node set $\mathcal{A}.$

Recall that each node of $\mathcal{A}$ belongs only to one component of the curve $\sigma.$ Therefore the line components of $\sigma$ do not intersect at the nodes of $\mathcal{A}.$

By using induction on $m$ one can prove easily the following

Lemma 3.6**.**

Suppose that a finite set of lines ${\mathcal{L}}$ and $2m$ nodes lying in these lines are given, such that no node is an intersetion point of two lines. Then one can divide the node set into $m$ pairs such that no pair belongs to the same line from ${\mathcal{L}}$ if and only if each line from ${\mathcal{L}}$ contains no more than $m$ nodes.

Thus the above mentioned division of the node set $\mathcal{A}$ into $n-k$ pairs is possible if and only if no $n-k$ nodes of ${\mathcal{A}}_{0}:=\mathcal{A}\setminus\{A_{0},A_{0}^{\prime}\}$ are located in a line component, where the nodes $A_{0},A_{0}^{\prime}$ are those associated with the curve $\sigma_{0}.$ Observe also that we may associate any two nodes of $\mathcal{A}$ with $\sigma_{0}.$

Now notice that, in view of $|\mathcal{A}|=2(n-k),$ there can be at most two undesirable line components for the set $\mathcal{A},$ i.e., lines containing each $n-k$ nodes from it. This case can be settled easily, by associating with $\sigma_{0}$ one node from each of the two components.

Next suppose that there is only one undesirable line component with $n-k$ or $n-k+1$ nodes. In this case one or two nodes from this line we associate with $\sigma_{0},$ respectively.

Finally consider the case of one undesirable line component $\ell$ with $s\geq n-k+2$ nodes. We have that

[TABLE]

Now we will move $s-n+k-1$ nodes, one by one, from $\ell$ to the other component $q$ such that the set ${\mathcal{X}}^{\prime\prime}:={\mathcal{X}}\cup\{B_{1},B_{2},B_{3}\}\cup\mathcal{A}$ remains $n$-independent. Again, in view of Lemma 3.4, i), we require that each moved node belongs only to one component of the curve $\sigma.$

To establish each described movement, in view of Lemma 3.4, ii), it suffices to prove that during this process each node $A\in\ell\cap\mathcal{A},$ has no $n$-fundamental polynomial for which the curve $q$ is a component. Suppose conversely that $p^{\star}_{A}=qr,\ r\in\Pi_{n-k+1}.$ Now, we have that $r$ vanishes at $\geq n-k+1$ nodes in $\ell\cap\mathcal{A}\setminus\{A\},$ and at least at a node from $\ell\cap\mathcal{X}$ mentioned above. Thus $r$ together with $p^{\star}_{A}$ vanishes at the whole line $\ell,$ including the node $A,$ which is a contradiction. It remains to note that there will be no more undesirable line, except $\ell,$ in the resulted set $\mathcal{A},$ after the described movement of the nodes, since we finish by keeping exactly $n-k+1$ nodes in $\ell\cap\mathcal{A}.$

∎

4 An application to the Gasca-Maeztu conjecture

Recall that a node $A\in{\mathcal{X}}$ uses a line $\ell$ means that $\ell$ is a factor of the fundamental polynomial $p=p^{\star}_{A},$ i.e., $p=\ell r,$ for some $r\in\Pi_{n-1}.$

A $GC_{n}$-set in plane is an $n$-poised set of nodes where the fundamental polynomial of each node is a product of $n$ linear factors.

The Gasca-Maeztu conjecture states that any $GC_{n}$-set possesses a subset of $n+1$ collinear nodes.

It was proved in [2] that any line passing through exactly $2$ nodes of a $GC_{n}$-set ${\mathcal{X}}$ can be used at most by one node from ${\mathcal{X}}.$

It was proved in [7] that any used line passing through exactly $3$ nodes of a $GC_{n}$-set ${\mathcal{X}}$ can be used either by exactly one or three nodes from ${\mathcal{X}}.$

Below we consider the case of lines passing through exactly $4$ nodes.

Corollary 4.1**.**

Let ${\mathcal{X}}$ be an $n$-poised set of nodes and $\ell$ be a line which passes through exactly $4$ nodes. Suppose $\ell$ is used by at least four nodes from ${\mathcal{X}}.$ Then it is used by exactly six nodes from ${\mathcal{X}}.$ Moreover, if it is used by six nodes, then they form a $2$-poised set. Furthermore, in the latter case, if ${\mathcal{X}}$ is a $GC_{n}$ set then the six nodes form a $GC_{2}$ set.

Proof.

Assume that $\ell\cap{\mathcal{X}}=\{A_{1},\ldots,A_{4}\}=:\mathcal{A}.$ Assume also that the four nodes in $\mathcal{B}:=\{B_{1},\ldots,B_{4}\}\in{\mathcal{X}}$ use the line $\ell:$ $p_{B_{i}}^{\star}=\ell\,q_{i},\ i=1,\ldots,4,$ where $q_{i}\in\Pi_{n-1}.$

The polynomials $q_{1},\ldots,q_{4},$ vanish at $N-8$ nodes of the set ${\mathcal{X}}^{\prime}:={\mathcal{X}}\setminus(\mathcal{A}\cup\mathcal{B}).$ Hence through these $N-8=d(n,n-3)+2$ nodes pass four linearly independent curves of degree $n-1.$ By Theorem 3.3 there exists a maximal curve $\mu$ of degree $n-3$ passing through $N-10$ nodes of ${\mathcal{X}}^{\prime}$ and the remaining two nodes denoted by $C_{1},C_{2},$ are outside of it. Now, according to Proposition 2.9, the nodes $C_{1},C_{2},$ use $\mu:$

[TABLE]

These polynomials $r_{i}$ have to vanish at the four nodes of $\mathcal{A}\subset\ell.$ Hence $q_{i}=\ell\beta_{i},\ i=1,2,$ with $\beta_{i}\in\Pi_{2}.$ Therefore, the nodes $C_{1},C_{2}$ use the line ${\ell}:$

[TABLE]

Hence if four nodes in $\mathcal{B}\subset{\mathcal{X}}$ use the line ${\ell}$ then there exist two more nodes $C_{1},C_{2}\in{\mathcal{X}}$ using it and all the nodes of ${\mathcal{Y}}:={\mathcal{X}}\setminus(\mathcal{A}\cup\mathcal{B}\cup\{C_{1},C_{2}\})$ lie in a maximal curve $\mu$ of degree $n-3:$

[TABLE]

Next, let us show that there is no seventh node using ${\ell}$. Assume by way of contradiction that except of the six nodes in $\mathcal{S}:=\{B_{1},\ldots,B_{4},C_{1},C_{2}\},$ there is a seventh node $D$ using ${\ell}.$ Of course we have that $D\in{\mathcal{Y}}.$

Then we have that the four nodes $B_{1},B_{2},B_{3}$ and $D$ are using $\ell$ therefore, as was proved above, there exist two more nodes $E_{1},E_{2}\in{\mathcal{X}}$ (which may coincide or not with $B_{4}$ or $C_{1},C_{2}$) using it and all the nodes of ${\mathcal{Y}}^{\prime}:={\mathcal{X}}\setminus(\mathcal{A}\cup\{B_{1},B_{2},B_{3},D,E_{1},E_{2}\})$ lie in a maximal curve $\mu^{\prime}$ of degree $n-3.$

We have also that

[TABLE]

Now, notice that both the curves $\mu$ and $\mu^{\prime}$ pass through all the nodes of the set ${\mathcal{Z}}:={\mathcal{X}}\setminus(\mathcal{A}\cup\mathcal{B}\cup\{C_{1},C_{2},D,E_{1},E_{2},\})$ with $|{\mathcal{Z}}|\geq N-13.$

Then, we get from Theorem 3.1, with $k=n-4$, that $N-13=d(n,n-4)+2$ nodes determine the curve of degree $n-3$ passing through them uniquely. Thus $\mu$ and $\mu^{\prime}$ coincide.

Therefore, in view of (4.1) and (4.2), $p_{D}^{\star}$ vanishes at all the nodes of ${\mathcal{Y}}$, which is a contradiction since $D\in{\mathcal{Y}}.$

Now, let us verify the last “moreover” statement. Suppose the six nodes in $\mathcal{S}\subset{\mathcal{X}}$ use the line ${\ell}.$ Then, as we obtained earlier, the nodes ${\mathcal{Y}}:={\mathcal{X}}\setminus(\mathcal{A}\cup\mathcal{S})$ are located in a maximal curve $\mu$ of degree $n-3.$ Therefore the fundamental polynomial of each $A\in\mathcal{S}$ uses $\mu$ and hence $\ell:$

[TABLE]

It is easily seen that $q_{A}$ is a $2$-fundamental polynomial of $A\in\mathcal{S}.$ ∎