On the minimum number of high degree curves containing few points
aa r X i v : . [ m a t h . M G ] J a n ON THE MINIMUM NUMBER OF HIGH DEGREE CURVESCONTAINING FEW POINTS
MARIO HUICOCHEACONACYT/UAZ
Abstract.
Let d, n ∈ Z + and A be a nonempty finite subset of R . A curveof degree d is the zero set of a polynomial of degree d in R [ x, y ]; denote by C d the family of curves of degree d . For any C ∈ C d , we say that C isdetermined by A if for any C ∈ C d such that C ∩ A ⊇ C ∩ A , we have that C = C ; we denote by D d ( A ) the family of curves of degree d determined by A . Write O d,n ( A ) := { C ∈ D d ( A ) : | C ∩ A | ≤ n } . In this paper we statetwo Sylvester-Gallai type results. In the first one, we show that if there is no C ∈ C d containing A , then (cid:12)(cid:12)(cid:12)(cid:12) O d, d − d +42 ( A ) (cid:12)(cid:12)(cid:12)(cid:12) = Ω d (cid:16) | A | d (cid:17) ;moreover we give a construction which shows that this lower bound is the bestpossible. In the second main result of this paper, it is shown that if d ≥
3, thereis no C ∈ C d containing A and any subset B of A with | B | = n is contained inat most one C ∈ C d , then (cid:12)(cid:12)(cid:12)(cid:12) O d, n +1 − (cid:16) d +22 (cid:17) ( A ) (cid:12)(cid:12)(cid:12)(cid:12) = Ω n,d (cid:18) | A | (cid:16) d +22 (cid:17) − (cid:19) ;furthermore we show that this lower bound is not trivial. Introduction
In this paper R , Z , Z + , Z +0 denote the set of real numbers, integers, positiveintegers and nonnegative integers, respectively. For any set X , we denote by P ( X )the family of subsets of X and by P n ( X ) the family of subsets Y of X such that | Y | = n . For any n, m ∈ Z , we write [ n, m ] := { k ∈ Z : n ≤ k ≤ m } . Let d, n ∈ Z + .A curve of degree d is a subset C of R which is the zero set of a polynomial in R [ x, y ] of degree d ; we denote by C d the family of curves of degree d in R . For each A ∈ P ( R ), we say that C ∈ C d is determined by A if for any C ∈ C d satisfyingthat C ∩ A ⊇ C ∩ A , we have that C = C ; we denote by D d ( A ) the family ofelements of C d which are determined by A . We write O d,n ( A ) := { C ∈ D d ( A ) : | C ∩ A | ≤ n } , and O d ( A ) := O d, ( d +22 ) − ( A ) . Since there is always a curve of degree d passingthrough (cid:0) d +22 (cid:1) − C ∈ D d ( A ) satisfies that | C ∩ A | ≥ (cid:0) d +22 (cid:1) −
1, and therefore O d,n ( A ) = ∅ for all n < (cid:0) d +22 (cid:1) − Key words and phrases.
Sylvester-Gallai type results, plane curves, Veronese map.
Theorem 1.1.
Let A ∈ P (cid:0) R (cid:1) be finite. If A is not contained in a line, then O ( A ) = ∅ .Proof. See [3]. (cid:3)
Sylvester-Gallai theorem has opened a complete research field in discrete geom-etry, see for instance [3], [5], [8], [17]. An important Sylvester-Gallai problem is tobound |O ( A ) | in terms of | A | . A number of quantitative results have been found,see [6], [8], [12], [16]. One of these results is the Dirac-Motzkin conjecture whichwas proven by B. Green and T. Tao. Theorem 1.2.
There is an absolute constant c > with the following property.Let A ∈ P (cid:0) R (cid:1) be finite with | A | > c . If A is not contained in a line, then |O ( A ) | ≥ | A | . Proof.
See [8, Thm. 1.2]. (cid:3)
Another important problem is to find Sylvester-Gallai qualitative type resultsfor different geometric objects instead of lines (e.g. conics, circles, hyperplanes,etc.). For conics, this was done by J. Wiseman and P. Wilson.
Theorem 1.3.
Let A ∈ P (cid:0) R (cid:1) be finite. If A is not contained in a curve of degree , then O ( A ) = ∅ .Proof. See [19, Thm.]. (cid:3)
Different proofs of Theorem 1.3 were found later, see [4], [7]. However, untilthis paper, no Sylvester-Gallai type results have been obtained for curves of higherdegree. In particular, the following conjecture of Wiseman and Wilson remainsunproven.
Conjecture 1.4.
Let d ∈ Z + and A ∈ P (cid:0) R (cid:1) be finite. If A is not contained in acurve of degree d , then O d ( A ) = ∅ . In the last few years, a mix of the previous two Sivester-Gallai type problems havebeen studied. This means that given a nonempty finite subset A in R and a family F of geometric objects (e.g. circles, conics, hyperplanes, etc.), we want (underreasonable assumptions) to determine or at least bound the number of elements F of F such that F is determined by A and | F ∩ A | is small. This problem has beenstudied at least for lines, circles, conics and hyperplanes, see [1], [2], [4], [8], [11],[13], [14], [15]. However, there are no results for high degree curves (as we saidabove, to our best knowledge, neither qualitative Silvester-Gallai results exist forhigh degree curves). In this paper we are interested in this problem when F is thefamily of curves with a given degree. The first result of this paper is the next one. Theorem 1.5.
For each d ∈ Z + , there are c = c ( d ) , c = c ( d ) > with thefollowing property. Let A ∈ P (cid:0) R (cid:1) be finite with | A | > c . If A is not contained ina curve of degree d , then (cid:12)(cid:12)(cid:12) O d, d − d +42 ( A ) (cid:12)(cid:12)(cid:12) ≥ c | A | d . For d ∈ { , } , note that d − d +42 = (cid:0) d +22 (cid:1) −
1. Thus Theorem 1.5 can be seenas a generalization of Theorem 1.2 (although, in the proof of Theorem 1.5, we donot care about the best possible value of c ), and Theorem 1.5 implies Theorem 1.3for big enough sets. Moreover Theorem 1.5 is optimal as we see in the next result. N THE MINIMUM NUMBER OF HIGH DEGREE CURVES CONTAINING FEW POINTS 3
Theorem 1.6.
For each d, m ∈ Z such that m > max n d − d +42 , d +4 d o and d > , there is A ∈ P m (cid:0) R (cid:1) such that i) A is not contained in a curve of degree d . ii) (cid:12)(cid:12)(cid:12) O d, d − d +42 ( A ) (cid:12)(cid:12)(cid:12) ≤ (cid:0) | A |− ( d +12 ) d (cid:1) . Unfortunately, for d >
2, we have that d − d +42 ≥ (cid:0) d +22 (cid:1) so Theorem 1.5 doesnot provide information about O d,n ( A ) when (cid:0) d +22 (cid:1) − ≤ n < d − d +42 . The nextresult of this paper deals with this problem when A satisfies that each B ∈ P n ( A )is contained in at most one curve of degree d . Theorem 1.7.
For each d, n ∈ Z with d ≥ and n ≥ (cid:0) d +22 (cid:1) − , there are c = c ( n, d ) , c = c ( n, d ) > with the following property. Let A ∈ P (cid:0) R (cid:1) be finitesuch that for each B ∈ P n ( A ) , we have that B is contained in at most one curve ofdegree d . If | A | > c and A is not contained in a curve of degree d , then (cid:12)(cid:12)(cid:12) O d, n +1 − ( d +22 )( A ) (cid:12)(cid:12)(cid:12) ≥ c | A | ( d +22 ) − . The values of c , c , c and c can be determined explicitly (although they dependon some values of some auxiliary results). As a consequence of Theorem 1.7, if any B ∈ P ( d +22 ) − ( A ) is contained in at most one curve of degree d , then |O d ( A ) | =Ω (cid:16) | A | ( d +22 ) − (cid:17) so Theorem 1.7 implies a particular case of Conjecture 1.4. AlsoTheorem 1.7 is not trivial as we see in the next theorem (although we do not knowif the lower bound Ω (cid:16) | A | ( d +22 ) − (cid:17) is the best one). Theorem 1.8.
For each d, n, m ∈ Z + such that n ≥ (cid:0) d +22 (cid:1) − and m > n + 1 − (cid:0) d +22 (cid:1) , there is A ∈ P m (cid:0) R (cid:1) such that i) A is not contained in a curve of degree d . ii) Each B ∈ P n ( A ) is contained in at most one curve of degree d . iii) (cid:12)(cid:12)(cid:12) O d, n +1 − ( d +22 )( A ) (cid:12)(cid:12)(cid:12) ≤ (cid:0) | A |− ( d +22 ) − (cid:1) . We explain the main ideas in the proofs of Theorem 1.5 and Theorem 1.7.i) Let d ∈ Z + , A be a nonempty finite subset of R and ψ d be the d -Veronesemap. For each hyperplane H in R ( d +22 ) − , we have that ψ − d ( H ) ∈ C e forsome e ∈ [1 , d ]; conversely, for each C ∈ S de =1 C e , there is a hyperplane H in R ( d +22 ) − such that ψ − d ( H ) = C . Thus the problem of finding curvesdetermined by A which contain few points of A is (almost) equivalent tothe problem of finding hyperplanes generated by ψ d ( A ) which contain fewpoints of ψ d ( A ).ii) It is easier to deal with hyperplanes than to do it with curves. As wementioned above, there are already results that assure the existence ofseveral hyperplanes in R e which are generated but contain few points of agiven set S , see [1], [2], [14], [15]. Nonetheless, these results require thatany e points of S generate a hyperplane. In general, the set ψ d ( A ) wontsatisfy this condition in R ( d +22 ) − . Thus we cannot take advantage of [1],[2], [14], [15]. What we will do is to fix a (cid:0) d +22 (cid:1) − F and a 2-dimensional flat G such that F ∩ G = ∅ . Considering R ( d +22 ) − embedded into P ( d +22 ) − and the homogenization G h ∼ = P of G in P ( d +22 ) − , we will project each (cid:0) d +22 (cid:1) − K containing F into K ∩ G h (which is always a single point in the projective space G h ∼ = P ); this willinduce a map π F : R ( d +22 ) − \ F → P . In this way we take the problem ofhyperplanes containing F into a problem of lines.iii) Nevertheless, we will need very special flats F . To be able to say that if aline L contains few points of ( π F ◦ ψ d )( A ), then the curve ( π F ◦ ψ d ) − ( L )contains few points of A , we need that F satisfies certain conditions. Theflats F will be the affine hull of ψ d ( B ) where B is a subset of A whichsatisfies some technical conditions. The family of subsets B of A satisfyingthose assumptions will be denoted by N d ( A ). The longest and most tediouspart of this paper (which is Section 3) is to warranty that N d ( A ) is not verysmall, however this is the core of this article. The proof of this fact willdepend on the structure of A ; specifically, it depends on whether there is alow degree curve that contains several points of A or not.iv) Let F be the affine hull of ψ d ( B ) for some B ∈ N d ( A ) and denote by ϕ therestriction of π F ◦ ψ d to R \ ψ − d ( F ). In Section 4 we will prove that thereare k ≤ d and a finite collection (bounded in terms of d ) of Zariski closedsubsets { E i } i ∈ I of R such that • ϕ ( E i ) is a singleton for each i ∈ I . • For each a ∈ R \ ( ψ − d ( F ) ∪ S i ∈ I E i ), we have that ϕ ( a )
6∈ { ϕ ( E i ) : i ∈ I } . • For each a ∈ R \ ( ψ − d ( F ) ∪ S i ∈ I E i ), we have that | ϕ − ( ϕ ( a )) | ≤ k .v) With the map ϕ as in iv), any line L in R disjoint from { ϕ ( E i ) : i ∈ I } which contains exactly two points of ϕ (cid:0) A \ ψ − d ( F ) (cid:1) satisfies that ϕ − ( L )is an element of C d which contains few points of A and is determined by it.A result of T. Boys , C. Valculescu and F. de Zeeuw (see Lemma 2.11) willwarranty that there are several lines L as above, and therefore we will havea number of the desired curves for each B ∈ N d ( A ). This can be done foreach flat F generated by an element of N d ( A ). Thus, since |N d ( A ) | ≫ N d ( A ) are fundamental tools in the proofs of the mainresults. We introduce and state some properties of the families N d ( A ) in Section 3.Using the elements of the families N d ( A ), we will take the problem of finding curveswith few points into finding ordinary lines which avoid a finite set and this will bedone in Section 4. The proofs of the main results of this paper are completed inSection 5. 2. Preliminaries
In this section we state some notation and results that will be needed later.Let p ( x, y ) ∈ R [ x, y ] and d ∈ Z + . We denote by Z ( p ( x, y )) its zero set in R and by deg( p ( x, y )) its degree. We say that p ( x, y ) ∈ R [ x, y ] is irreducible if N THE MINIMUM NUMBER OF HIGH DEGREE CURVES CONTAINING FEW POINTS 5 deg( p ( x, y )) > p ( x, y ) = p ( x, y ) p ( x, y ), we get that p i ( x, y ) ∈ R for some i ∈ { , } . We say that Z ( p ( x, y )) is irreducible if p ( x, y ) isirreducible. Write C ≤ d := d [ e =1 C e R d [ x, y ] := { p ( x, y ) ∈ R [ x, y ] : deg( p ( x, y )) ∈ [1 , d ] } ;also, for technical reasons, we write C ≤ := ∅ . In R [ x, y ], we define the relation p ( x, y ) ∼ q ( x, y ) if there is r ∈ R \ { } such that p ( x, y ) = r · q ( x, y ), and we denoteby [ p ( x, y )] the class of p ( x, y ) and by R [ x, y ] / ∼ the set of classes. For any subset X of R [ x, y ], we write X/ ∼ := { [ p ( x, y )] ∈ R [ x, y ] / ∼ : [ p ( x, y )] ∩ X = ∅} . Note that for any p ( x, y ) , q ( x, y ) ∈ R [ x, y ] such that [ p ( x, y )] = [ q ( x, y )], we havethat deg( p ( x, y )) = deg( q ( x, y )) and Z ( p ( x, y )) = Z ( q ( x, y )); thus σ d : R d [ x, y ] / ∼−→ C ≤ d , σ d ([ p ( x, y )]) = Z ( p ( x, y )) . is well defined. Let p ( x, y ) ∈ R [ x, y ] be such that deg( p ( x, y )) > p ( x, y ) = r Q ni =1 p i ( x, y ) m i with m , m , . . . , m n ∈ Z + , r ∈ R ,[ p i ( x, y )] = [ p j ( x, y )] for all i, j ∈ [1 , n ] such that i = j , and p i ( x, y ) is irreducible foreach i ∈ [1 , n ]. Then the irreducible curves Z ( p ( x, y )) , Z ( p ( x, y )) , . . . , Z ( p n ( x, y ))are known as the irreducible components of Z ( p ( x, y )). The irreducible componentssatisfy that Z ( p ( x, y )) = n [ i =1 Z ( p i ( x, y )) = Z n Y i =1 p i ( x, y ) ! and deg( p ( x, y )) = n X i =1 m i deg( p i ( x, y )) ≥ n X i =1 deg( p i ( x, y )) . We will use a weak version of Bezout’s theorem.
Theorem 2.1.
Let d, e ∈ Z + , C ∈ C ≤ d and C ∈ C ≤ e . If C and C do not havean irreducible component in common, then | C ∩ C | ≤ de. Proof.
See [9, Ch. I.7]. (cid:3)
The next facts can be proven easily by the reader.
Remark 2.2.
Let d ∈ Z + and e, f ∈ [1 , d ] . i) For any C ∈ C ≤ e and C ∈ C ≤ f , notice that C ∪ C ∈ C ≤ e + f . ii) For any C ∈ C e and C ∈ C ≤ d such that C ⊆ C , there is C ∈ C ≤ d − e suchthat C = C ∪ C . iii) For any A ∈ P (cid:0) R (cid:1) such that | A | ≤ (cid:0) d +22 (cid:1) − , there is C ∈ C ≤ d such that A ⊆ C . For a curve C , it may happen that there exist p ( x, y ) , q ( x, y ) ∈ R [ x, y ] such that[ p ( x, y )] = [ q ( x, y )] but Z ( p ( x, y )) = C = Z ( q ( x, y )). The next lemma shows thatfor each C ∈ C ≤ d there are at most d d classes [ p ( x, y )] ∈ R d [ x, y ] / ∼ such that C = Z ( p ( x, y )). MARIO HUICOCHEA CONACYT/UAZ
Lemma 2.3.
Let d ∈ Z + and C ∈ C ≤ d with pairwise distinct irreducible components Z ( p ( x, y )) , Z ( p ( x, y )) , . . . , Z ( p n ( x, y )) . Then σ − d ( C ) = (" n Y i =1 p m i i ∈ R d [ x, y ] / ∼ : m , m , . . . , m n ∈ Z + , n X i =1 m i deg( p i ) ≤ d ) . Since the number of solutions ( m , m , . . . , m n ) ∈ Z + n of P ni =1 m i deg( p i ) ≤ d isbounded by d d , we get in particular that | σ − d ( C ) | ≤ d d . Proof.
See [10, Cor.7]. (cid:3)
Let d ∈ Z +0 and e ∈ [0 , d ]. A translation F of a vectorial subspace V of R d will be called a flat . We write dim F := dim V , and also if V is an e -dimensionalsubspace, we say that F is an e -flat; in particular, 1-flats are lines and d − e -flats in R d will be denoted by G e . For any subset S of R d , we denote by Fl( S ) the smallest flat (with respect to ⊆ ) which contains S and we write dim S := dim Fl( S ). If S = ∅ , we consider Fl( S ) = ∅ and dim S = − S = { s , s , . . . , s n } , we write Fl( s , s , . . . , s n ) := Fl( S ). The family of e -flats F in R d such that there is a subset R of S satisfying that F = Fl( R ) will be denotedby G e ( S ).A fundamental tool in this paper is the Veronese map. Write I d := n ( n, m ) ∈ Z +0 2 : n + m ∈ [1 , d ] o so | I d | = (cid:0) d +22 (cid:1) −
1. The d -Veronese map is themap ψ d : R −→ R ( d +22 ) − , ψ d ( a , a ) = ( a n a m ) ( n,m ) ∈ I d To avoid confusion, the ring of polynomials which corresponds to R will be de-noted by R [ x, y ] and the ring of polynomials which corresponds to R ( d +22 ) − will bedenoted by R [ z ( n,m ) ] ( n,m ) ∈ I d . There is a quite important relation between elementsof R d [ x, y ] / ∼ and hyperplanes in R ( d +22 ) − given by the next map τ d : R d [ x, y ] / ∼−→ G ( d +22 ) − ,τ d r (0 , + X ( n,m ) ∈ I d r ( n,m ) x n y m = Z r (0 , + X ( n,m ) ∈ I d r ( n,m ) z ( n,m ) The Veronese map has some well-known properties that we will need later. Theproof of the following facts can be found in standard algebraic geometry books, seefor instance [9, Ch. I], [18, Ch. 1].
Remark 2.4.
Let d ∈ Z + . i) The map ψ d is an isomorphism onto its image. ii) The map τ d is a bijection. Note that for any [ p ( x, y )] ∈ R d [ x, y ] / ∼ , wehave that ψ d ( Z ( p ( x, y ))) = τ d ([ p ( x, y )]) ∩ ψ d ( R ) . iii) For all e ∈ [1 , d ] and A ∈ P (cid:0) R (cid:1) , we have that dim ψ e ( A ) ≤ dim ψ d ( A ) . N THE MINIMUM NUMBER OF HIGH DEGREE CURVES CONTAINING FEW POINTS 7
A crucial part of this paper is to take the problem of finding curves determinedby A containing few points of A into the problem of finding hyperplanes in R ( d +22 ) − generated by ψ d ( A ) which contain few points of ψ d ( A ). The main tool to do thisis the following lemma. Lemma 2.5.
Let d ∈ Z + and A ∈ P (cid:0) R (cid:1) be such that there is no element of C ≤ d which contains A . Then τ d (cid:0) σ − d ( D d ( A )) (cid:1) = G ( d +22 ) − ( ψ d ( A )) . Proof.
See [10, Lemma 12]. (cid:3)
Let d ∈ Z + , p ( x, y ) ∈ R d [ x, y ] and A ∈ P (cid:0) R (cid:1) be finite. Since A is finite, thereis always a line L = Z ( rx + sy − t ) in R such that L ∩ A = ∅ . Therefore Z ( p ( x, y )) ∩ A = ( Z ( p ( x, y )) ∩ A ) ∪ ( L ∩ A )= Z (cid:16) p ( x, y ) · ( rx + sy − t ) d − deg( p ( x,y )) (cid:17) ∩ A. From this observation, we get the next fact.
Remark 2.6.
Let d ∈ Z + , p ( x, y ) ∈ R d [ x, y ] and A ∈ P (cid:0) R (cid:1) be finite. Then thereis q ( x, y ) ∈ R [ x, y ] such that deg( q ( x, y )) = d and Z ( p ( x, y )) ∩ A = Z ( q ( x, y )) ∩ A. Lemma 2.7.
Let e ∈ Z + , f ∈ Z +0 , F be a proper flat in R ( e +22 ) − and A ∈P ( f +22 ) (cid:0) R (cid:1) be such that A is not contained in an element of C ≤ f . i) If e ≥ f , then | ψ e ( A ) ∩ F | ≤ F . ii) If e < f , then | ψ e ( A ) ∩ F | ≤ (cid:0) f +22 (cid:1) − (cid:0) f − e +22 (cid:1) − (cid:0) e +22 (cid:1) + 2 + dim F . iii) If e < f , then dim ψ d ( A \ ψ − e ( F )) ≥ (cid:0) f − e +22 (cid:1) − for all d ≥ f − e .Proof. First we show i). If f = 0, then | A | = 1 so | ψ e ( A ) ∩ F | ≤ | ψ e ( A ) | = 1 ≤ F. Thus, from now on, we assume that f >
0. Since A is not contained in an elementof C ≤ f , Remark 2.4.ii implies that ψ f ( A ) cannot be contained in a hyperplane of R ( f +22 ) − and therefore(1) (cid:18) f + 22 (cid:19) − ψ f ( A ) . In so far as e ≥ f , Remark 2.4.iii implies that dim ψ f ( A ) ≤ dim ψ e ( A ) and thereby(2) dim ψ f ( A ) ≤ dim ψ e ( A ) ≤ | ψ e ( A ) | − | A | − . Since | A | = (cid:0) f +22 (cid:1) , we get from (1) and (2) that(3) dim ψ e ( A ) = | ψ e ( A ) | − | A | − . As a consequence of (3), for any S ∈ P ( ψ e ( A )), we get that dim S = | S | −
1; inparticular, dim ψ e ( A ) ∩ F = | ψ e ( A ) ∩ F | − , and hence | ψ e ( A ) ∩ F | = 1 + dim ψ e ( A ) ∩ F ≤ F, which completes the proof of i). MARIO HUICOCHEA CONACYT/UAZ
Now we prove ii). For any flat E in R ( e +22 ) − , write α ( E ) := (cid:0) e +22 (cid:1) − − dim E .In so far as F is a proper flat, note that α ( F ) ≥
0. The proof of ii) will be done byinduction on α ( F ). First suppose that α ( F ) = 0 so F is a hyperplane in R ( e +22 ) − .Then Remark 2.4.ii implies that ψ − e ( F ) ∈ C ≤ e . Trivially,(4) ψ − e ( F ) ∩ A ⊆ ψ − e ( F ) . We claim that(5) | ψ − e ( F ) ∩ A | ≤ (cid:18) f + 22 (cid:19) − (cid:18) f − e + 22 (cid:19) . Indeed, if (5) is false, then | A \ ψ − e ( F ) | = | A | − | ψ − e ( F ) ∩ A | < | A | − (cid:18) f + 22 (cid:19) + (cid:18) f − e + 22 (cid:19) = (cid:18) f − e + 22 (cid:19) , and therefore Remark 2.2.iii implies that there exists C ∈ C ≤ f − e such that(6) A \ ψ − e ( F ) ⊆ C. However, from (4) and (6), A = ( ψ − e ( F ) ∩ A ) ∪ ( A \ ψ − e ( F )) ⊆ ψ − e ( F ) ∪ C with ψ − e ( F ) ∪ C ∈ C ≤ f by Remark 2.2.i; this contradicts the assumption so (5) istrue. Therefore | F ∩ ψ e ( A ) | = | ψ − e ( F ) ∩ A |≤ (cid:18) f + 22 (cid:19) − (cid:18) f − e + 22 (cid:19) (cid:16) by (5) (cid:17) = (cid:18) f + 22 (cid:19) − (cid:18) f − e + 22 (cid:19) − α ( F ) (cid:16) since α ( F ) = 0 (cid:17) , and the basis of induction is complete. Now assume that the claim holds for allflats E such that 0 ≤ α ( E ) < α ( F ). Since A is not contained in an elementof C ≤ f and e < f , we get that A is not contained in an element of C ≤ e . ThusRemark 2.4.ii implies that ψ e ( A ) is not contained in a hyperplane and thereforedim ψ e ( A ) = (cid:0) e +22 (cid:1) −
1. In particular, this means that ψ e ( A ) \ F = ∅ , and then wecan fix a ∈ A \ ψ − e ( F ). Set E := Fl( F ∪ { ψ e ( a ) } ). Since ψ e ( a ) F , we have thatdim E = 1 + dim F , and therefore(7) α ( E ) + 1 = α ( F ) . In so far as ψ e ( a ) ∈ ψ e ( A ) \ F , we get that(8) | F ∩ ψ e ( A ) | + 1 ≤ | E ∩ ψ e ( A ) | . Then | F ∩ ψ e ( A ) | ≤ | E ∩ ψ e ( A ) | − (cid:16) by (8) (cid:17) ≤ (cid:18) f + 22 (cid:19) − (cid:18) f − e + 22 (cid:19) − α ( E ) − (cid:16) by induction (cid:17) = (cid:18) f + 22 (cid:19) − (cid:18) f − e + 22 (cid:19) − α ( F ) , (cid:16) by (7) (cid:17) and this completes the induction and the proof of ii). N THE MINIMUM NUMBER OF HIGH DEGREE CURVES CONTAINING FEW POINTS 9
Finally, we show iii). Since F is a proper flat, there is a hyperplane H whichcontains F . Remark 2.4.ii implies that ψ − e ( H ) ∈ C ≤ e . Since H ⊇ F ,(9) ψ − e ( F ) ∩ A ⊆ ψ − e ( H ) ∩ A ⊆ ψ − e ( H ) . We claim that(10) dim ψ f − e ( A \ ψ − e ( F )) ≥ (cid:18) f − e + 22 (cid:19) − . Indeed, if (10) does not hold, then dim ψ f − e ( A \ ψ − e ( F )) < (cid:0) f − e +22 (cid:1) −
1, and hencethere exists a hyperplane K in R ( f − e +22 ) − which contains ψ f − e ( A \ ψ − e ( F )) with ψ − f − e ( K ) ∈ C ≤ f − e by Remark 2.4.ii. Then(11) A \ ψ − e ( F ) ⊆ ψ − f − e ( K ) . Nonetheless, from (9) and (11), A = ( ψ − e ( F ) ∩ A ) ∪ ( A \ ψ − e ( F )) ⊆ ψ − e ( H ) ∪ ψ − f − e ( K ) , and Remark 2.2.i yields that ψ − e ( H ) ∪ ψ − f − e ( K ) ∈ C ≤ f ; this contradicts the as-sumption so (10) needs to be true. Now, in so far as d ≥ f − e , Remark 2.4.iii yieldsthat(12) dim ψ d ( A \ ψ − e ( F )) ≥ dim ψ f − e ( A \ ψ − e ( F )) . Then iii) is a consequence of (10) and (12). (cid:3)
Lemma 2.8.
Let e ∈ Z +0 and A ∈ P (cid:0) R (cid:1) be such that A is finite and it is notcontained in an element of C ≤ e . Denote by R the family of elements B ∈ P ( e +22 )( A ) such that B is not contained in an element of C ≤ e . Then |R| ≥ e +22 ) − | A | . Proof.
Let d ∈ Z + and T ∈ P (cid:0) R d (cid:1) be such that dim T = d . Set S d ( T ) := { R ∈ P d +1 ( T ) : dim R = d } . First we show that(13) |S d ( T ) | ≥ d | T | . We prove (13) by induction on d . If d = 1, then S ( T ) = { R ∈ P ( T ) : dim R = 1 } = P ( T )so |S ( T ) | = (cid:0) | T | (cid:1) ≥ | T | , and the basis of induction is complete. Assume that (13)holds for d − d . Since dim T = d , there exists S ∈ P d ( T ) suchthat dim S = d −
1; fix S ∈ P d ( T ) such that dim S = d − H := Fl( S ).We have two cases. • Assume that | T ∩ H | ≥ | T | . Fix t ∈ T \ H . Notice that for all R ∈ S d − ( T ∩ H ), we have that | R ∪ { t }| = | R | + 1 = d and dim R ∪ { t } = 1 + dim R = d so R ∪ { t } ∈ S d ( T ). This map S d − ( T ∩ H ) → S d ( T ) , R R ∪ { t } is injective so(14) |S d ( T ) | ≥ |S d − ( T ∩ H ) | . By induction, |S d − ( T ∩ H ) | ≥ d − | T ∩ H | so (14) leads to |S d ( T ) | ≥ |S d − ( T ∩ H ) | ≥ d − | T ∩ H | ≥ d | T | , and this completes the induction in this case. • Assume that | T ∩ H | < | T | . Note that for all t ∈ T \ H , we have that | S ∪ { t }| = | S | + 1 = d and dim S ∪ { t } = 1 + dim S = d so S ∪ { t } ∈ S d ( T ).The map T \ H → S d ( T ) , t S ∪ { t } is injective so |S d ( T ) | ≥ | T \ H | = | T | − | T ∩ H | > | T | ≥ d | T | , and this completes the induction.If e = 0, then C ≤ e = ∅ so R = {{ a } : a ∈ A } which means that |R| = | A | .Thus, from now on, assume that e >
0. Since A is not contained in an element of C ≤ e , Remark 2.4.ii implies that ψ e ( A ) is not contained in a hyperplane and thendim ψ e ( A ) = (cid:0) e +22 (cid:1) −
1. Applying (13) to ψ e ( A ) and (cid:0) e +22 (cid:1) −
1, we get that(15) (cid:12)(cid:12)(cid:12) S ( e +22 ) − ( ψ e ( A )) (cid:12)(cid:12)(cid:12) ≥ e +22 ) − | ψ e ( A ) | = 12( e +22 ) − | A | . For each R ∈ S ( e +22 ) − ( ψ e ( A )), we have that dim R = (cid:0) e +22 (cid:1) − R is not containedin a hyperplane, and hence, by Remark 2.4.ii, ψ − e ( R ) is not contained in an elementof C ≤ e . Thus the map S ( e +22 ) − ( ψ e ( A )) → R , R ψ − e ( R )is well defined and injective yielding that |R| ≥ (cid:12)(cid:12)(cid:12) S ( e +22 ) − ( ψ e ( A )) (cid:12)(cid:12)(cid:12) , and finally (15) implies the claim. (cid:3) Lemma 2.9.
Let d ∈ Z + , f ∈ [0 , d − , C ∈ C d − f and B ∈ P ( f +22 )( R \ C ) besuch that B is not contained in an element of C ≤ f . Take B ∈ P ( d +22 ) − ( R ) suchthat B ⊇ B and B ∩ C = B \ B . Then, for any e ∈ [ d − f, d ] and C ∈ C ≤ e suchthat C ⊇ C , we have that (16) | B ∩ C | < (cid:18) d + 22 (cid:19) − (cid:18) d − e + 22 (cid:19) − . Proof.
Since C ⊆ C , Remark 2.2.ii implies that there is C ∈ C ≤ e + f − d such that C = C ∪ C . In so far as B ∩ C = ∅ , notice that(17) B ∩ C = B ∩ ( C ∪ C ) = B ∩ C ⊆ C . If (16) is false, then(18) | B \ C | ≤ | B \ C | = | B | − | B ∩ C | ≤ (cid:18) d − e + 22 (cid:19) − . If e = d , then (18) implies that B ⊆ C = C ∪ C ; hence, inasmuch as B ∩ C = ∅ ,we get that B ⊆ C but this contradicts that B is not contained in an element of N THE MINIMUM NUMBER OF HIGH DEGREE CURVES CONTAINING FEW POINTS 11 C ≤ f . If e < d , then, by (18), Remark 2.2.iii yields the existence of C ∈ C ≤ d − e suchthat(19) B \ C ⊆ C . Since C ∈ C ≤ e + f − d and C ∈ C ≤ d − e , note that C ∪ C ∈ C ≤ f by Remark 2.2.i.Notice that B ⊆ C ∪ C by (17) and (19); however, this contradicts that B isnot contained in an element of C ≤ f and it proves (16). (cid:3) Lemma 2.10.
Let d ∈ Z + , e ∈ [1 , d − , C ∈ C e and B ∈ P ( R ) . i) If dim ψ d − e ( B \ C ) = (cid:0) d − e +22 (cid:1) − , then dim ψ d ( B ∪ C ) = (cid:0) d +22 (cid:1) − . ii) If dim ψ d − e ( B \ C ) = (cid:0) d − e +22 (cid:1) − , then dim ψ d ( B ∪ C ) = (cid:0) d +22 (cid:1) − .Proof. First we show i). Assume that dim ψ d ( B ∪ C ) < (cid:0) d +22 (cid:1) − ψ d ( B ∪ C )is contained in a hyperplane of R ( d +22 ) − . Then Remark 2.4.ii implies that thereis C ∈ C ≤ d such that B ∪ C ⊆ C ; in particular, C ⊆ C and then Remark2.2.ii yields the existence of C ∈ C ≤ d − e such that C = C ∪ C . In so far as B ∪ C ⊆ C = C ∪ C , we get that B \ C ⊆ C . Inasmuch as C ∈ C ≤ d − e ,Remark 2.4.ii implies that ψ d − e ( B \ C ) is contained in a hyperplane of R ( d − e +22 ) − and therefore dim ψ d − e ( B \ C ) < (cid:0) d − e +22 (cid:1) −
1, which proves i).Now we prove ii). Assume that dim ψ d ( B ∪ C ) = (cid:0) d +22 (cid:1) − ψ d − e ( B \ C ) = (cid:18) d − e + 22 (cid:19) − . We have to deal with two cases. • Suppose that dim ψ d ( B ∪ C ) > (cid:0) d +22 (cid:1) − ψ d ( B ∪ C ) = (cid:0) d +22 (cid:1) − ψ d − e ( B \ C ) = (cid:18) d − e + 22 (cid:19) − . Indeed, if (21) is false, there is a hyperplane of R ( d − e +22 ) − which contains ψ d − e ( B \ C ). Then, by Remark 2.4.ii, there is a curve C ′ ∈ C ≤ d − e suchthat B \ C ⊆ C ′ . Hence B ∪ C = ( B \ C ) ∪ C ⊆ C ′ ∪ C with C ′ ∪ C ∈ C ≤ d by Remark 2.2.i. This means that ψ d ( B ∪ C ) is containedin a hyperplane by Remark 2.4.ii, and thus dim ψ d ( B ∪ C ) < (cid:0) d +22 (cid:1) −
1. Thiscontradiction proves (21). • Suppose that dim ψ d ( B ∪ C ) < (cid:0) d +22 (cid:1) −
2. On the one hand, there are in-finitely many hyperplanes in R ( d +22 ) − containing ψ d ( B ∪ C ). On the otherhand, for each curve C ′ ∈ C ≤ d , there are only finitely many [ p ( x, y )] ∈ R d [ x, y ] / ∼ such that C ′ = Z ( p ( x, y )) by Lemma 2.3. Therefore we canchoose p ( x, y ) , p ( x, y ) ∈ R d [ x, y ] such that ψ d ( B ∪ C ) ⊆ τ d ([ p ( x, y )]), ψ d ( B ∪ C ) ⊆ τ d ([ p ( x, y )]) and Z ( p ( x, y )) = Z ( p ( x, y )); write C := Z ( p ( x, y )) and C := Z ( p ( x, y )). Since ψ d ( B ∪ C ) ⊆ τ d ([ p ( x, y )]), Re-mark 2.4.ii implies that B ∪ C ⊆ C . Then Remark 2.2.ii yields the existenceof C ∈ C ≤ d − e such that C = C ∪ C . Since B ∪ C ⊆ C = C ∪ C , weget that B \ C ⊆ C . In so far as C ∈ C ≤ d − e , Remark 2.4.ii impliesthat there is a hyperplane H in R ( d − e +22 ) − such that ψ − d − e ( H ) = C
32 MARIO HUICOCHEA CONACYT/UAZ and ψ d − e ( B \ C ) ⊆ H . Proceeding in the same way with C , there exist C ∈ C ≤ d − e and a hyperplane H in R ( d − e +22 ) − such that C = C ∪ C , ψ − d − e ( H ) = C and ψ d − e ( B \ C ) ⊆ H . Since C = C , notice that C = C and therefore H = H . Now, since ψ d − e ( B \ C ) ⊆ H ∩ H , we get that dim ψ d − e ( B \ C ) < (cid:0) d − e +22 (cid:1) − (cid:3) Lemma 2.11.
For any n ∈ Z + , there are c = c ( n ) , c = c ( n ) ∈ R with thefollowing property. For all T ∈ P n (cid:0) R (cid:1) and S ∈ P (cid:0) R (cid:1) such that | S | > c and S \ T is not collinear, we have that |{ L ∈ O ( S ) : L ∩ T = ∅}| ≥ | S | − c . Proof.
See [4, Lemma 2.5]. (cid:3) Families N d ( A )The purpose of this section is to define and prove some properties of the families N d ( A ) which will be fundamental objects in the proof of the main results of thispaper.Let d ∈ Z + and A ∈ P ( R ). We denote by N d ( A ) the family of subsets B ∈P ( d +22 ) − ( A ) which have following four properties:i) dim ψ d ( B ) = (cid:0) d +22 (cid:1) − e ∈ [1 , d −
1] and C ∈ C e , we have that | B ∩ C | < (cid:0) d +22 (cid:1) − (cid:0) d − e +22 (cid:1) .iii) For all e ∈ [1 , d −
1] and C ∈ C e such that | B ∩ C | = (cid:0) d +22 (cid:1) − (cid:0) d − e +22 (cid:1) − ψ d − e ( B \ C ) = (cid:0) d − e +22 (cid:1) − e ∈ [1 , d −
1] and C ∈ C e such that | B ∩ C | < (cid:0) d +22 (cid:1) − (cid:0) d − e +22 (cid:1) − ψ d − e ( B \ C ) > (cid:0) d − e +22 (cid:1) − d ∈ Z + , e ∈ [1 , d − A, B, C ∈ P ( R ) and D ∈ P ( B ), write V e ( D ) := Fl( ψ e ( D )) W e ( B, D ) := Fl( ψ d − e ( B \ ψ − e ( V e ( D )))) α e ( D ) := (cid:18) e + 22 (cid:19) − − dim V e ( D ) β e ( B, D ) := (cid:18) d − e + 22 (cid:19) − − dim W e ( B, D ) γ e ( B, D ) := | V e ( D ) ∩ ψ e ( B ) | µ e ( B, D ) := α e ( D ) < α e ( D ) + γ e ( B, D ) + (cid:0) d − e +22 (cid:1) if α e ( D ) ≥ . N THE MINIMUM NUMBER OF HIGH DEGREE CURVES CONTAINING FEW POINTS 13 τ e ( B, D ) := { α e ( D ) , β e ( B, D ) } < γ e ( B, D ) > (cid:0) d +22 (cid:1) − (cid:0) d − e +22 (cid:1) − α e ( D ) + β e ( B, D ) + | B | + 2 if min { α e ( D ) , β e ( B, D ) } ≥ γ e ( B, D ) = (cid:0) d +22 (cid:1) − (cid:0) d − e +22 (cid:1) − α e ( D ) + β e ( B, D ) + | B | + 3 if min { α e ( D ) , β e ( B, D ) } ≥ γ e ( B, D ) < (cid:0) d +22 (cid:1) − (cid:0) d − e +22 (cid:1) − .U e ( B, D ) := ψ − d ( V d ( B )) if α e ( D ) < ψ − d ( V d ( B )) ∪ ψ − e ( V e ( D )) if β e ( B, D ) < ≤ α e ( D ); ψ − d ( V d ( B )) ∪ ψ − e ( V e ( D )) ∪ ψ − d − e ( W e ( B, D )) if β e ( B, D ) , α e ( D ) ≥ .I ( B, C ) := { ( f, E ) ∈ [1 , d − × P ( B ) : C * U f ( B, E ) }N d ( A, B, C ) := { E ∈ N d ( A ) : B ∈ P ( E ) and E ∩ C = E \ B } . The following facts are direct consequences of the definitions.
Remark 3.1.
Let d ∈ Z + , e ∈ [1 , d − , B ∈ P (cid:0) R (cid:1) and D ∈ P ( B ) . i) Note that B ⊆ (cid:0) B ∩ ψ − e ( V e ( D )) (cid:1) ∪ (cid:0) B ∩ ψ − d − e ( W e ( B, D )) (cid:1) ; in particular, | B | ≤ (cid:12)(cid:12) B ∩ ψ − e ( V e ( D )) (cid:12)(cid:12) + (cid:12)(cid:12) B ∩ ψ − d − e ( W e ( B, D )) (cid:12)(cid:12) . ii) Let B ∈ P (cid:0) R (cid:1) , B , D ∈ P ( B ) and D ∈ P ( B ) . If V e ( D ) ⊆ V e ( D ) = R ( e +22 ) − and W e ( B , D ) = W e ( B , D ) , then U e ( B , D ) ⊆ U e ( B , D ) . iii) Let B ∈ P (cid:0) R (cid:1) , B , D ∈ P ( B ) and D ∈ P ( B ) . If V e ( D ) = V e ( D ) ,then W e ( B , D ) ⊆ W e ( B , D ) and U e ( B , D ) ⊆ U e ( B , D ) . iv) If min { α e ( D ) , β e ( B, D ) } ≥ , then τ e ( B, D ) ≤ α e ( D ) + β e ( B, D ) + | B | + 3 . The main results of this section are Lemma 3.5 and Lemma 3.9. Their proofsare rather technical so, for the sake of comprehension, we sketch them before westate the auxiliary results that we need to prove them. We will say that A ∈ P ( R )is d -regular if | A ∩ C | ≤ d +8 | A | for all C ∈ C ≤ d . • We start with Lemma 3.5. Assume that A is d -regular and take B := ∅ .Given b = ( b , b , . . . , b ( d +22 ) − ) ∈ A ( d +22 ) − , we will construct recursivelya chain B ( B ( . . . ( B ( d +22 ) − with B i +1 = B i ∪ { b i +1 } for each i ∈ [0 , (cid:0) d +22 (cid:1) − b ∈ A ( d +22 ) − , we get B ( d +22 ) − ∈N d ( A, ∅ , R ). On the one hand, for each i ∈ [0 , (cid:0) d +22 (cid:1) − b i +1 is not ina forbidden subset U i of R , then B ( d +22 ) − ∈ N d ( A, ∅ , R ); here we will use Lemma 3.3 and Lemma 3.4. On the other hand, since A is d -regular, foreach i ∈ [0 , (cid:0) d +22 (cid:1) − U i and A isvery small so B ( d +22 ) − ∈ N d ( A, ∅ , R ) almost always. • Now we sketch the proof of Lemma 3.9. Given b =( b ( f +22 ) +1 , b ( f +22 ) +2 , . . . , b ( d +22 ) − ) ∈ ( A ∩ C )( d +22 ) − ( f +22 ) − , we will con-struct recursively a chain B ( B ( . . . ( B ( d +22 ) − ( f +22 ) − with B i +1 = B i ∪ { b ( f +22 ) + i +1 } for each i ∈ [0 , (cid:0) d +22 (cid:1) − (cid:0) f +22 (cid:1) − b ∈ ( A ∩ C )( d +22 ) − ( f +22 ) − , we get B ( d +22 ) − ∈ N d ( A, B , C ). Asin the sketch of Lemma 3.5, we need that b ( f +22 ) + i +1 is not in a forbiddensubset U i of R for each i ∈ [0 , (cid:0) d +22 (cid:1) − (cid:0) f +22 (cid:1) − A ∩ U i is very small because of Bezout’s theorem (applied to some curvesthat intersect C ).The next properties will shorten the proofs of some auxiliary results Lemma 3.2.
Let e, d ∈ Z + with e ∈ [1 , d − , B ∈ P (cid:0) R (cid:1) and D ∈ P ( B ) . i) If α e ( D ) ≥ , then there is C ∈ C ≤ e such that ψ − e ( V e ( D )) ⊆ C ii) If β e ( B, D ) ≥ − , then there is C ∈ C ≤ d − e such that ψ − d − e ( W d − e ( D )) ⊆ C iii) If | B | ≤ (cid:0) d +22 (cid:1) − , then there is C ∈ C ≤ d such that ψ − d ( V d ( B )) ⊆ C . iv) If | B | ≤ (cid:0) d +22 (cid:1) − , then there are C ∈ C ≤ e , C ∈ C ≤ d − e and C ∈ C ≤ d suchthat U e ( B, D ) ⊆ C ∪ C ∪ C .Proof. First we show i). In so far as α e ( D ) ≥
0, notice that V e ( D ) is a proper flatin R ( e +32 ) − . Hence there exists a hyperplane H containing V e ( D ). Remark 2.4.iiimplies that ψ − e ( H ) is in C ≤ e and it satisfies that ψ − e ( V e ( D )) ⊆ ψ − e ( H ), and thisproves i).We prove ii). Since β e ( B, D ) ≥ − W e ( B, D ) is a proper flat in R ( d − e +22 ) − .This implies that there exists a hyperplane H containing W e ( B, D ). Remark 2.4.iiyields that ψ − d − e ( H ) is in C ≤ d − e and it satisfies that ψ − d − e ( W e ( B, D )) ⊆ ψ − d − e ( H ),which proves ii).To prove iii), note that since | B | ≤ (cid:0) d +22 (cid:1) − V d ( B ) = dim Fl( ψ d ( B )) ≤ | ψ d ( B ) | − | B | − ≤ (cid:18) d + 22 (cid:19) − . Thus there is a hyperplane H containing V d ( B ), and then, by Remark 2.4.ii, ψ − d ( H )is in C ≤ d and it satisfies that ψ − d ( V d ( B )) ⊆ ψ − d ( H ).Finally, notice that iv) is a consequence of i), ii) and iii). (cid:3) Lemma 3.3.
Let d ∈ Z + , A, C ∈ P (cid:0) R (cid:1) and B ∈ P ( A ) , and write U := S ( e,D ) ∈ I ( B,C ) U e ( B, D ) . Assume that A \ U = ∅ and max { τ e ( B, D ) , µ e ( B, D ) } < (cid:0) d +22 (cid:1) for all ( e, D ) ∈ I ( B, C ) . Then, for all b ∈ A \ U and ( e, D ) ∈ I ( B ∪ { b } , C ) , max { τ e ( B ∪ { b } , D ) , µ e ( B ∪ { b } , D ) } < (cid:18) d + 22 (cid:19) . N THE MINIMUM NUMBER OF HIGH DEGREE CURVES CONTAINING FEW POINTS 15
Proof.
Fix b ∈ A \ U and ( e, D ) ∈ I ( B ∪ { b } , C ). If α e ( D ) <
0, then τ e ( B ∪ { b } , D )and µ e ( B ∪ { b } , D ) are zero so we assume from now on that(22) α e ( D ) ≥ . We divide the proof of the claim into two cases. • First assume that b D . Then D is a subset of B and Remark 3.1.iiiimplies that U e ( B, D ) ⊆ U e ( B ∪ { b } , D ); this implies that ( e, D ) ∈ I ( B, C )since ( e, D ) ∈ I ( B ∪ { b } , C ). The inclusion ( e, D ) ∈ I ( B, C ) leads to(23) max { τ e ( B, D ) , µ e ( B, D ) } < (cid:18) d + 22 (cid:19) . Because ( e, D ) ∈ I ( B, C ), we have that b U e ( B, D ). Notice that (22)yields that ψ − e ( V e ( D )) ⊆ U e ( B, D ) so b ψ − e ( V e ( D )) and thereby γ e ( B ∪ { b } , D ) = | V e ( D ) ∩ ψ e ( B ∪ { b } ) | = | V e ( D ) ∩ ψ e ( B ) | = γ e ( B, D ) . (24) Thus µ e ( B ∪ { b } , D ) = α e ( D ) + γ e ( B ∪ { b } , D ) + (cid:18) d − e + 22 (cid:19) (cid:16) by (22) (cid:17) = α e ( D ) + γ e ( B, D ) + (cid:18) d − e + 22 (cid:19) (cid:16) by (24) (cid:17) = µ e ( B, D ) (cid:16) by (22) (cid:17) < (cid:18) d + 22 (cid:19) . (cid:16) by (23) (cid:17) (25) If β e ( B, D ) <
0, then dim W e ( B, D ) > (cid:0) d − e +22 (cid:1) − W e ( B ∪ { b } , D ) ≥ dim W e ( B, D ) > (cid:18) d − e + 22 (cid:19) − , and therefore β e ( B ∪ { b } , D ) <
0; thus, if β e ( B, D ) <
0, we have that τ e ( B ∪ { b } , D ) = 0, and then by (25) we are done. From now on, weassume that(26) β e ( B, D ) ≥ . From (22) and (26), note that α e ( D ) , β e ( B, D ) ≥ b U e ( B, D ),notice that b ψ − e ( V e ( D )) and b ψ − d − e ( W e ( B, D )); this leads to ψ d − e ( b ) ∈ W e ( B ∪ { b } , D ) \ W e ( B, D ), and then W e ( B, D ) is a flat properly con-tained in W e ( B ∪ { b } , D ) implying thatdim W e ( B ∪ { b } , D ) ≥ W e ( B, D ) , and therefore(27) β e ( B, D ) ≥ β e ( B ∪ { b } , D ) + 1 . If β e ( B ∪ { b } , D ) < γ e ( B ∪ { b } , D ) > (cid:0) d +22 (cid:1) − (cid:0) d − e +22 (cid:1) −
1, then τ e ( B ∪ { b } , D ) = 0, and we are done by (25). Thus we assume that this isnot the case, and then (24) and (27) lead to β e ( B, D ) − ≥ β e ( B ∪ { b } , D ) ≥ γ e ( B, D ) = γ e ( B ∪ { b } , D ) ≤ (cid:18) d + 22 (cid:19) − (cid:18) d − e + 22 (cid:19) − . (29) If γ e ( B ∪ { b } , D ) = (cid:0) d +22 (cid:1) − (cid:0) d − e +22 (cid:1) −
1, then (29) leads to(30) γ e ( B, D ) = γ e ( B ∪ { b } , D ) = (cid:18) d + 22 (cid:19) − (cid:18) d − e + 22 (cid:19) − , and hence τ e ( B ∪ { b } , D ) = α e ( D ) + β e ( B ∪ { b } , D ) + | B ∪ { b }| + 2 (cid:16) by (22),(28),(30) (cid:17) ≤ α e ( D ) + β e ( B, D ) + | B | + 2 (cid:16) by (28) (cid:17) = τ e ( B, D ) . (cid:16) by (22),(28),(30) (cid:17) If γ e ( B ∪ { b } , D ) < (cid:0) d +22 (cid:1) − (cid:0) d − e +22 (cid:1) −
1, then (29) leads to(31) γ e ( B, D ) = γ e ( B ∪ { b } , D ) < (cid:18) d + 22 (cid:19) − (cid:18) d − e + 22 (cid:19) − , and hence τ e ( B ∪ { b } , D ) = α e ( D ) + β e ( B ∪ { b } , D ) + | B ∪ { b }| + 3 (cid:16) by (22),(28),(31) (cid:17) ≤ α e ( D ) + β e ( B, D ) + | B | + 3 (cid:16) by (28) (cid:17) = τ e ( B, D ) . (cid:16) by (22),(28),(31) (cid:17) Thus, in any case, τ e ( B ∪ { b } , D ) ≤ τ e ( B, D ), and then (23) implies that(32) τ e ( B ∪ { b } , D ) ≤ τ e ( B, D ) < (cid:18) d + 22 (cid:19) , The claim follows from (25) and (32). • Assume that b ∈ D . Since D \ { b } ⊆ D , we have that(33) V e ( D \ { b } ) ⊆ V e ( D ) , and then(34) α e ( D \ { b } ) ≥ α e ( D ) . Now we claim that(35) ( V e ( D ) \ V e ( D \ { b } )) ∩ ψ e ( B ) = ∅ . Indeed, if (35) is false, then there is a ∈ B such that ψ e ( a ) ∈ V e ( D ) \ V e ( D \{ b } ). Sincedim V e ( D ) − dim V e ( D \ { b } ) ≤ | D | − | D \ { b }| = 1 , we have that the flat generated by V e ( D \ { b } ) and ψ e ( a ) has to be V e ( D ),but this flat is precisely Fl( V e ( D \ { b } ) ∪ { ψ e ( a ) } ) = V e (( D \ { b } ) ∪ { a } ) so(36) V e ( D ) = V e (( D \ { b } ) ∪ { a } ) . From (36), we can apply Remark 3.1.iii to get that(37) U e ( B, ( D \ { b } ) ∪ { a } ) ⊆ U e ( B ∪ { b } , D ) . Since ( e, D ) ∈ I ( B ∪ { b } , C ), we have that C * U e ( B ∪ { b } , D ). From (37),we get that C * U e ( B, ( D \ { b } ) ∪ { a } ) and hence ( e, ( D \ { b } ) ∪ { a } ) ∈ I ( B, C ). Thus, in so far as b U , we get that b U e ( B, ( D \ { b } ) ∪ { a } ).From (22) and (36), note that α e (( D \{ b } ) ∪{ a } ) ≥ ψ − e ( V e (( D \{ b } ) ∪ N THE MINIMUM NUMBER OF HIGH DEGREE CURVES CONTAINING FEW POINTS 17 { a } )) ⊆ U e ( B, ( D \{ b } ) ∪{ a } ) and hence b ψ − e ( V e (( D \{ b } ) ∪{ a } )). Thismeans that ψ e ( b ) V e (( D \ { b } ) ∪ { a } ); however, b ∈ D so ψ e ( b ) ∈ V e ( D )but this contradicts (36) and proves (35). Now, from (35),( ψ − e ( V e ( D )) \ ψ − e ( V e ( D \ { b } ))) ∩ B = ∅ , and thus B \ ψ − e ( V e ( D \ { b } )) = ( B ∪ { b } ) \ ψ − e ( V e ( D )). This gives(38) W e ( B, D \ { b } ) = W e ( B ∪ { b } , D ) , and therefore(39) β e ( B, D \ { b } ) = β e ( B ∪ { b } , D ) . From (22), (33) and (38), we can apply Remark 3.1.ii and we get that(40) U e ( B, D \ { b } ) ⊆ U e ( B ∪ { b } , D ) . Since ( e, D ) ∈ I ( B ∪{ b } , C ), we have that C * U e ( B ∪{ b } , D ). Thus, from(40), we obtain that C * U e ( B, D \ { b } ) and hence ( e, D \ { b } ) ∈ I ( B, C ).This implies by assumption that(41) max { τ e ( B, D \ { b } ) , µ e ( B, D \ { b } ) } < (cid:18) d + 22 (cid:19) . Since ( e, D \ { b } ) ∈ I ( B, C ) and b U , we get that b U e ( B, D \ { b } ).From (22) and (34), we have that α e ( D \ { b } ) ≥ ψ − e ( V e ( D \ { b } )) ⊆ U e ( B, D \ { b } ) and hence b ψ − e ( V e ( D \ { b } )). This means that(42) ψ e ( b ) V e ( D \ { b } ) , and thereby V e ( D \ { b } ) is a flat properly contained in V e ( D ); in particular,(43) α e ( D \ { b } ) ≥ α e ( D ) + 1 . Then γ e ( B ∪ { b } , D ) = | V e ( D ) ∩ ψ e ( B ∪ { b } ) | = | ( V e ( D ) \ V e ( D \ { b } )) ∩ ψ e ( B ∪ { b } ) | + | V e ( D \ { b } ) ∩ ψ e ( B ∪ { b } ) | =1 + | V e ( D \ { b } ) ∩ ψ e ( B ∪ { b } ) | (cid:16) by (35) (cid:17) =1 + | V e ( D \ { b } ) ∩ ψ e ( B ) | (cid:16) by (42) (cid:17) =1 + γ e ( B, D \ { b } ) . (44) Then µ e ( B ∪ { b } , D ) = α e ( D ) + γ e ( B ∪ { b } , D ) + (cid:18) d − e + 22 (cid:19) (cid:16) by (22) (cid:17) ≤ α e ( D \ { b } ) + γ e ( B, D \ { b } ) + (cid:18) d − e + 22 (cid:19) (cid:16) by (43),(44) (cid:17) = µ e ( B, D \ { b } ) (cid:16) by (22),(43) (cid:17) < (cid:18) d + 22 (cid:19) . (cid:16) by (41) (cid:17) (45) If β e ( B ∪ { b } , D ) < γ e ( B ∪ { b } , D ) > (cid:0) d +22 (cid:1) − (cid:0) d − e +22 (cid:1) −
1, then τ e ( B ∪ { b } , D ) = 0, and we are done by (45). Thus suppose that theseinequalities are not true and then (39) and (44) give β e ( B, D \ { b } ) = β e ( B ∪ { b } , D ) ≥ γ e ( B, D \ { b } ) + 1 = γ e ( B ∪ { b } , D ) ≤ (cid:18) d + 22 (cid:19) − (cid:18) d − e + 22 (cid:19) − . (47) If γ e ( B ∪ { b } , D ) = (cid:0) d +22 (cid:1) − (cid:0) d − e +22 (cid:1) −
1, then (47) leads to(48) γ e ( B, D \ { b } ) + 1 = γ e ( B ∪ { b } , D ) = (cid:18) d + 22 (cid:19) − (cid:18) d − e + 22 (cid:19) − , and hence τ e ( B ∪ { b } , D ) = α e ( D ) + β e ( B ∪ { b } , D ) + | B ∪ { b }| + 2 (cid:16) by (22),(46),(48) (cid:17) ≤ α e ( D \ { b } ) + β e ( B, D \ { b } ) + | B | + 2 (cid:16) by (43),(46) (cid:17) ≤ τ e ( B, D \ { b } ) . (cid:16) by (22),(46),(48) (cid:17) If γ e ( B ∪ { b } , D ) < (cid:0) d +22 (cid:1) − (cid:0) d − e +22 (cid:1) −
1, then (47) leads to(49) γ e ( B, D \ { b } ) + 1 = γ e ( B ∪ { b } , D ) < (cid:18) d + 22 (cid:19) − (cid:18) d − e + 22 (cid:19) − , and hence τ e ( B ∪ { b } , D ) = α e ( D ) + β e ( B ∪ { b } , D ) + | B ∪ { b }| + 3 (cid:16) by (22),(46),(49) (cid:17) ≤ α e ( D \ { b } ) + β e ( B, D \ { b } ) + | B | + 3 (cid:16) by (43),(46) (cid:17) = τ e ( B, D \ { b } ) . (cid:16) by (22),(46),(49) (cid:17) In any case, τ e ( B ∪ { b } , D ) ≤ τ e ( B, D \ { b } ), and then (41) leads to(50) τ e ( B ∪ { b } , D ) ≤ τ e ( B, D \ { b } ) < (cid:18) d + 22 (cid:19) . The claim follows from (45) and (50). (cid:3)
Lemma 3.4.
Let d ∈ Z + , A, C ∈ P (cid:0) R (cid:1) and B ∈ P ( A \ C ) . Take B ∈P ( d +22 ) − ( A ) which satisfies the following properties. i) dim ψ d ( B ) = (cid:0) d +22 (cid:1) − . ii) B ⊇ B . iii) B ∩ C = B \ B . iv) For all ( e, D ) ∈ I ( B, C ) , we have that max { τ e ( B, D ) , µ e ( B, D ) } < (cid:0) d +22 (cid:1) . v) For all ( e, D ) ∈ ([1 , d − × P ( B )) \ I ( B, C ) with α e ( D ) ≥ , we have that γ e ( B, D ) < (cid:0) d +22 (cid:1) − (cid:0) d − e +22 (cid:1) and β e ( B, D ) < .Then B ∈ N d ( A, B , C ) .Proof. Because of the assumptions i), ii) and iii), it suffices to show that B satisfiesthe conditions ii), iii) and iv) of the definition of N d ( A ). First we show that for all N THE MINIMUM NUMBER OF HIGH DEGREE CURVES CONTAINING FEW POINTS 19 e ∈ [1 , d −
1] and C ∈ C e , we have that(51) | C ∩ B | < (cid:18) d + 22 (cid:19) − (cid:18) d − e + 22 (cid:19) . Fix p ( x, y ) ∈ R e [ x, y ] such that C = Z ( p ( x, y )), and write H := τ e ([ p ( x, y )]) and D := B ∩ ψ − e ( H ). Hence V e ( D ) ⊆ H and Remark 2.4.ii implies that C = ψ − e ( H )so(52) | C ∩ B | = | H ∩ ψ e ( B ) | = | V e ( D ) ∩ ψ e ( B ) | = γ e ( B, D ) . In so far as dim V e ( D ) ≤ dim H = (cid:0) e +22 (cid:1) −
2, we get that α e ( D ) ≥
0, and therefore(53) µ e ( B, D ) = α e ( D ) + γ e ( B, D ) + (cid:18) d − e + 22 (cid:19) . If ( e, D ) ∈ I ( B, C ), then µ e ( B, D ) < (cid:0) d +22 (cid:1) by iv). Insomuch as α e ( D ) ≥
0, weobtain from (53) that γ e ( B, D ) < (cid:0) d +22 (cid:1) − (cid:0) d − e +22 (cid:1) . If ( e, D ) I ( B, C ), then γ e ( B, D ) < (cid:0) d +22 (cid:1) − (cid:0) d − e +22 (cid:1) by v). Hence, in any case,(54) γ e ( B, D ) < (cid:18) d + 22 (cid:19) − (cid:18) d − e + 22 (cid:19) , and thus (51) follows from (52) and (54).Now take e ∈ [1 , d −
1] and C ∈ C e such that(55) | C ∩ B | = (cid:18) d + 22 (cid:19) − (cid:18) d − e + 22 (cid:19) − , and we will show that(56) dim ψ d − e ( B \ C ) = (cid:18) d − e + 22 (cid:19) − . Since | B \ C | = | B | − | C ∩ B | = (cid:18) d − e + 22 (cid:19) − , we have that(57) dim ψ d − e ( B \ C ) ≤ | B \ C | − (cid:18) d − e + 22 (cid:19) − . Fix p ( x, y ) ∈ R e [ x, y ] such that C = Z ( p ( x, y )), and write H := τ e ([ p ( x, y )]) and D := B ∩ ψ − e ( H ). Hence V e ( D ) ⊆ H and Remark 2.4.ii implies that ψ e ( C ) = H ∩ ψ e ( R ) so | C ∩ B | = | ψ e ( C ) ∩ ψ e ( B ) | = | H ∩ ψ e ( B ) | = | V e ( D ) ∩ ψ e ( B ) | = γ e ( B, D ) , and then (55) implies that(58) γ e ( B, D ) = (cid:18) d + 22 (cid:19) − (cid:18) d − e + 22 (cid:19) − . In so far as dim V e ( D ) ≤ dim H = (cid:0) e +22 (cid:1) −
2, we get that(59) α e ( D ) ≥ . Since D = B ∩ ψ − e ( H ) = B ∩ C , we have that B \ ψ − e ( V e ( D )) = B \ C so W e ( B, D ) = Fl( ψ d − e ( B \ ψ − e ( V e ( D ))) = Fl( ψ d − e ( B \ C )) , and then(60) dim W e ( B, D ) = dim ψ d − e ( B \ C ) . From (57) and (60),(61) β e ( B, D ) ≥ . From (58), (59) and (61),(62) τ e ( B, D ) = α e ( D ) + β e ( B, D ) + | B | + 2 . From (59) and (61), min { α e ( D ) , β e ( B, D ) } ≥
0; then v) implies that ( e, D ) ∈ I ( B, C ). Thus iv) implies that τ e ( B, D ) < (cid:0) d +22 (cid:1) , and then (59) and (62) gives β e ( B, D ) <
1. Now, from (61), β e ( B, D ) = 0 and then (60) yields (56).Now take e ∈ [1 , d −
1] and C ∈ C e such that(63) | C ∩ B | < (cid:18) d + 22 (cid:19) − (cid:18) d − e + 22 (cid:19) − , and we will prove that(64) dim ψ d − e ( B \ C ) > (cid:18) d − e + 22 (cid:19) − . Fix p ( x, y ) ∈ R e [ x, y ] such that C = Z ( p ( x, y )), and write H := τ e ([ p ( x, y )]) and D := B ∩ ψ − e ( H ). Hence V e ( D ) ⊆ H and Remark 2.4.ii implies that ψ e ( C ) = H ∩ ψ e ( R ) so | C ∩ B | = | ψ e ( C ) ∩ ψ e ( B ) | = | H ∩ ψ e ( B ) | = | V e ( D ) ∩ ψ e ( B ) | = γ e ( B, D ) , and thus (63) gives(65) γ e ( B, D ) < (cid:18) d + 22 (cid:19) − (cid:18) d − e + 22 (cid:19) − . Since dim V e ( D ) ≤ dim H = (cid:0) e +22 (cid:1) −
2, we get that(66) α e ( D ) ≥ . Inasmuch as D = B ∩ ψ − e ( H ) = B ∩ C , notice that B \ ψ − e ( V e ( D )) = B \ C so W e ( B, D ) = Fl( ψ d − e ( B \ ψ − e ( V e ( D ))) = Fl( ψ d − e ( B \ C )) , and thus(67) dim W e ( B, D ) = dim ψ d − e ( B \ C ) . We claim that(68) β e ( B, D ) < . Indeed, if (68) is false, then β e ( B, D ) ≥ { α e ( D ) , β e ( B, D ) } ≥ e, D ) ∈ I ( B, C ) and thereby τ e ( B, D ) < (cid:0) d +22 (cid:1) by iv). However, (65), (66) and β e ( B, D ) ≥ τ e ( B, D ) = α e ( D ) + β e ( B, D ) + | B | + 3 ≥ (cid:18) d + 22 (cid:19) , which is impossible. This means that (68) is true, and thereby (64) is a consequenceof (67) and (68). (cid:3) Recall that A ∈ P ( R ) is d -regular if | A ∩ C | < d +8 | A | for all C ∈ C ≤ d . N THE MINIMUM NUMBER OF HIGH DEGREE CURVES CONTAINING FEW POINTS 21
Lemma 3.5.
Let d ∈ Z with d > and A be d -regular. Then |N d ( A, ∅ , R ) | ≥ d +3 ! | A | ( d +22 ) − . Proof.
Set g := (cid:0) d +22 (cid:1) −
3. Throughout this proof, for any b ∈ A g , its i -th entrywill be denoted by b i , i.e. b = ( b , b , . . . , b g ). For any b ∈ A g , write B ( b ) := ∅ .Now, for all i ∈ [1 , g ] and b ∈ A g , set B i ( b ) := { b , b , . . . , b i } U i ( b ) := [ ( e,D ) ∈ I ( B i − ( b ) , R ) U e ( B i − ( b ) , D ) . For all i ∈ [1 , g ] and b ∈ A g , notice that | B i ( b ) | ≤ g < (cid:0) d +22 (cid:1) −
1. Then Lemma3.2.iv implies that for all e ∈ [1 , d −
1] and D ∈ P (cid:0) B i ( b ) (cid:1) , we get that U e (cid:0) B i ( b ) , D (cid:1) is contained in the union of 3 curves in C ≤ d ; in particular, U e (cid:0) B i ( b ) , D (cid:1) cannotcontain R , and therefore(69) I (cid:0) B i ( b ) , R (cid:1) = [1 , d − × P (cid:0) B i ( b ) (cid:1) . For each i ∈ [1 , g ], write R i := (cid:8) b ∈ A g : b i ∈ U i ( b ) (cid:9) . The first step in the proofis to show that for all i ∈ [1 , g ],(70) | R i | ≤ g | A | g . For each i ∈ [1 , g ] , E ∈ P ([1 , i ]) and b ∈ A g , set E ( b ) := { b j : j ∈ E } . Noticethat P (cid:0) B i ( b ) (cid:1) = { E ( b ) : E ∈ P ([1 , i ]) } . Now, for each i ∈ [1 , g ] , E ∈ P ([1 , i ])and e ∈ [1 , d − R i,e,E := (cid:8) b ∈ A g : b i ∈ U e ( B i − ( b ) , E ( b )) (cid:9) . From (69),notice that for all i ∈ [1 , g ], we get that R i = S E ∈P ([1 ,i ]) S d − e =1 R i,e,E and then(71) | R i | ≤ X E ∈P ([1 ,i ]) d − X e =1 | R i,e,E | . For each i ∈ [1 , g ] , E ∈ P ([1 , i ]) and b ∈ A g , there exist C , C , C ∈ C ≤ d such that U e ( B i − ( b ) , E ( b )) ⊆ C ∪ C ∪ C by Lemma 3.2.iv; then A ∩ U e ( B i − ( b ) , E ( b )) ⊆ A ∩ ( C ∪ C ∪ C ) , and, since A is d -regular,(72) | A ∩ U e ( B i − ( b ) , E ( b )) | ≤ X j =1 | A ∩ C i | < d +8 | A | ≤ d +6 | A | . Furthermore, in so far as g = (cid:0) d +22 (cid:1) − < d +2 , we have that dg g +1 ≤ d +6 .Hence, inasmuch as the i -th entry of b ∈ R i,e,E has to be in A ∩ U e ( B i − ( b ) , E ( b )),(72) leads to(73) | R i,e,E | ≤ | A | g − | A ∩ U e ( B i − ( b ) , E ( b )) | ≤ d +6 | A | g ≤ gd g +1 | A | g . Since |P ([1 , i ]) × [1 , d − | ≤ d g , we have that (71) and (73) imply (70). Set T := (cid:8) b ∈ A g : b i U i ( b ) for all i ∈ [1 , g ] (cid:9) so T = A g \ S gi =1 R i . From(70),(74) | T | = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) A g \ g [ i =1 R i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≥ | A | g − g X i =1 | R i | ≥ | A | g . The next step is to show that the map φ : T −→ P ( A ) , φ ( b ) = B g ( b )satisfies that(75) φ ( T ) ⊆ N d ( A, ∅ , R ) . We claim that for all b ∈ T , i ∈ [1 , g ] and ( e, D ) ∈ [1 , d − × P (cid:0) B i ( b ) (cid:1) , we getthat(76) max (cid:8) τ e ( B i ( b ) , D ) , µ e ( B i ( b ) , D ) (cid:9) < (cid:18) d + 22 (cid:19) . We show (76) by induction on i ∈ [1 , g ]. For i = 1, notice that B ( b ) = { b } andthen α e ( D ) ≤ (cid:0) e +22 (cid:1) − , β e ( B ( b ) , D ) ≤ (cid:0) d − e +22 (cid:1) − γ e ( B ( b ) , D ) ≤ (cid:8) τ e ( B ( b ) , D ) , µ e ( B ( b ) , D ) (cid:9) ≤ (cid:18) e + 22 (cid:19) + (cid:18) d − e + 22 (cid:19) − < (cid:18) d + 22 (cid:19) , and then (76) follows in this case. Assume that (76) holds for i − ≥ i . Since b ∈ T , note that b i ∈ A \ U i ( b ). By induction,max (cid:8) τ e ( B i − ( b ) , D ) , µ e ( B i − ( b ) , D ) (cid:9) < (cid:18) d + 22 (cid:19) for all ( e, D ) ∈ [1 , d − × P (cid:0) B i − ( b ) (cid:1) = I (cid:0) B i − ( b ) , R (cid:1) so B i − ( b ) and b i ∈ A \ U i ( b ) satisfy the assumptions of Lemma 3.3, and hence this lemma implies that B i − ( b ) ∪ { b i } = B i ( b ) satisfies (76) completing the induction. In particular, (76)holds for i = g so(77) max (cid:8) τ e ( B g ( b ) , D ) , µ e ( B g ( b ) , D ) (cid:9) < (cid:18) d + 22 (cid:19) for all b ∈ T and ( e, D ) ∈ [1 , d − × P (cid:0) B g ( b ) (cid:1) . For all b ∈ T , we get that b i U i ( b ); in particular, b i ψ − d ( V d ( B i − ( b ))). This means that the flats V d ( B ( b )) ( V d ( B ( b )) ( . . . ( V d ( B g ( b )) are contained properly so(78) dim V d ( B g ( b )) ≥ g − . On the other hand,dim V d ( B g ( b )) = dim Fl( ψ d ( B g ( b ))) ≤ | ψ d ( B g ( b )) | − | B g ( b ) | − g − ψ d ( B g ( b )) = dim V d ( B g ( b )) = g − (cid:18) d + 22 (cid:19) − . We want to apply Lemma 3.4 with B = B g ( b ), B = ∅ and C = R , and this canbe done because its assumptions are satisfied (indeed: i) holds by (79); ii) and iii)are trivial; iv) holds by (69) and (77); v) never happens by (69)). Hence Lemma3.4 implies that B g ( b ) ∈ N d ( A, ∅ , R ) and it proves (75). N THE MINIMUM NUMBER OF HIGH DEGREE CURVES CONTAINING FEW POINTS 23
To conclude the proof, note that for each B ∈ φ ( T ), we have that φ − ( B ) ⊆{ ( b , . . . , b g ) ∈ A g : { b , . . . , b g } = B } . Therefore, for all B ∈ N d ( A, ∅ , R ), wehave that | φ − ( B ) | ≤ g !, and then (74) and (75) yield that(80) |N d ( A, ∅ , R ) | ≥ g ! | T | ≥ g !2 | A | g . In so far as g = (cid:0) d +22 (cid:1) − < d +2 , we have that g !2 ≤ d +3 ! and then (80) impliesthe claim of the lemma. (cid:3) Lemma 3.6.
Let d, f ∈ Z be such that d ≥ and f ∈ [0 , d − , and B ∈ P ( f +22 ) (cid:0) R (cid:1) be such that B is not contained in an element of C ≤ f . Then, for all e ∈ [1 , d − and D ∈ P ( B ) , max { τ e ( B, D ) , µ e ( B, D ) } < (cid:18) d + 22 (cid:19) . Proof.
Fix e ∈ [1 , d −
1] and D ∈ P ( B ). If α e ( D ) <
0, then τ e ( B, D ) = µ e ( B, D ) =0. Hence, from now on, we assume that(81) α e ( D ) ≥ . First we show that(82) µ e ( B, D ) < (cid:18) d + 22 (cid:19) . If e ≥ f , then Lemma 2.7.i leads to | ψ e ( B ) ∩ V e ( D ) | ≤ V e ( D ) , and hence | ψ e ( B ) ∩ V e ( D ) | + α e ( D ) ≤ V e ( D ) + α e ( D )= (cid:18) e + 22 (cid:19) − ≤ (cid:18) d + 12 (cid:19) − (cid:18) d − e + 12 (cid:19) . (cid:16) since d − ≥ e (cid:17) If e < f , then Lemma 2.7.ii leads to | ψ e ( B ) ∩ V e ( D ) | ≤ (cid:18) f + 22 (cid:19) − (cid:18) f − e + 22 (cid:19) − (cid:18) e + 22 (cid:19) + 2 + dim V e ( D )so | ψ e ( B ) ∩ V e ( D ) | + α e ( D ) ≤ (cid:18) f + 22 (cid:19) − (cid:18) f − e + 22 (cid:19) ≤ (cid:18) d + 12 (cid:19) − (cid:18) d − e + 12 (cid:19) . (cid:16) since d − ≥ f (cid:17) Thus, in any case,(83) γ e ( B, D ) + α e ( D ) = | ψ e ( B ) ∩ V e ( D ) | + α e ( D ) ≤ (cid:18) d + 12 (cid:19) − (cid:18) d − e + 12 (cid:19) . We get (82) as follows. µ e ( B, D ) = γ e ( B, D ) + α e ( D ) + (cid:18) d − e + 22 (cid:19) (cid:16) by (81) (cid:17) = (cid:18) d + 12 (cid:19) − (cid:18) d − e + 12 (cid:19) + (cid:18) d − e + 22 (cid:19) (cid:16) by (83) (cid:17) < (cid:18) d + 22 (cid:19) . Now we prove that(84) τ e ( B, D ) < (cid:18) d + 22 (cid:19) . If β e ( B, D ) <
0, then τ e ( B, D ) = 0 and in this case we are done by (82). Thus weassume that(85) β e ( B, D ) ≥ . Remark 3.1.i gives(86) (cid:12)(cid:12) B ∩ ψ − e ( V e ( D )) (cid:12)(cid:12) + (cid:12)(cid:12) B ∩ ψ − d − e ( W e ( B, D )) (cid:12)(cid:12) ≥ | B | = (cid:18) f + 22 (cid:19) . From (81) and (85), Remark 3.1.iv yields that(87) τ e ( B, D ) ≤ α e ( D ) + β e ( B, D ) + (cid:18) f + 22 (cid:19) + 3 . We divide the conclusion of the proof of (84) into four cases. • Assume that e ≥ f and d − e ≥ f . We apply Lemma 2.7.i to V e ( D ) and W e ( B, D ) to get α e ( D ) ≤ (cid:18) e + 22 (cid:19) − − | ψ e ( B ) ∩ V e ( D ) | (88) β e ( B, D ) ≤ (cid:18) d − e + 22 (cid:19) − − | ψ d − e ( B ) ∩ W e ( B, D ) | . (89) From (86), (88) and (89), we get that(90) α e ( D ) + β e ( B, D ) ≤ (cid:18) e + 22 (cid:19) + (cid:18) d − e + 22 (cid:19) − − (cid:18) f + 22 (cid:19) . Hence τ e ( B, D ) ≤ α e ( D ) + β e ( B, D ) + (cid:18) f + 22 (cid:19) + 3 (cid:16) by (87) (cid:17) ≤ (cid:18) e + 22 (cid:19) + (cid:18) d − e + 22 (cid:19) (cid:16) by (90) (cid:17) < (cid:18) d + 22 (cid:19) . (cid:16) since d ≥ (cid:17) N THE MINIMUM NUMBER OF HIGH DEGREE CURVES CONTAINING FEW POINTS 25 • Assume that e ≥ f and d − e < f . We apply Lemma 2.7.i to V e ( D ) andLemma 2.7.ii to W e ( B, D ) so α e ( D ) ≤ (cid:18) e + 22 (cid:19) − − | ψ e ( B ) ∩ V e ( D ) | (91) β e ( B, D ) ≤ (cid:18) f + 22 (cid:19) − (cid:18) f − ( d − e ) + 22 (cid:19) − − | ψ d − e ( B ) ∩ W e ( B, D ) | . (92) We have two subcases. ⋆ Suppose that e = d −
1. Then β e ( B, D ) = (cid:18) (cid:19) − − dim W e ( B, D ) = − dim W e ( B, D )so (85) implies that(93) β e ( B, D ) = dim W e ( B, D ) = 0 . Since dim W e ( B, D ) = 0, we have that W e ( B, D ) is a point and hence(86) leads to(94) (cid:12)(cid:12) B ∩ ψ − e ( V e ( D )) (cid:12)(cid:12) ≥ (cid:18) f + 22 (cid:19) − . From (91), (93) and (94), we get that(95) α e ( D ) + β e ( B, D ) = α e ( D ) ≤ (cid:18) e + 22 (cid:19) − (cid:18) f + 22 (cid:19) . Hence τ e ( B, D ) ≤ α e ( D ) + β e ( B, D ) + (cid:18) f + 22 (cid:19) + 3 (cid:16) by (87) (cid:17) ≤ (cid:18) e + 22 (cid:19) + 3 (cid:16) by (95) (cid:17) < (cid:18) d + 22 (cid:19) . (cid:16) since e + 1 = d ≥ (cid:17) ⋆ Suppose that e < d −
1. From (86), (91) and (92), we get that(96) α e ( D ) + β e ( B, D ) ≤ (cid:18) e + 22 (cid:19) − (cid:18) f − ( d − e ) + 22 (cid:19) − . Thus τ e ( B, D ) ≤ α e ( D ) + β e ( B, D ) + (cid:18) f + 22 (cid:19) + 3 (cid:16) by (87) (cid:17) ≤ (cid:18) e + 22 (cid:19) − (cid:18) f − ( d − e ) + 22 (cid:19) + (cid:18) f + 22 (cid:19) + 1 (cid:16) by (96) (cid:17) < (cid:18) d + 22 (cid:19) . (cid:16) since f, e + 1 ≤ d − (cid:17) • Assume that e < f and d − e ≥ f . We apply Lemma 2.7.ii to V e ( D ) andLemma 2.7.i to W e ( B, D ) so α e ( D ) ≤ (cid:18) f + 22 (cid:19) − (cid:18) f − e + 22 (cid:19) − | ψ e ( B ) ∩ V e ( D ) | (97) β e ( B, D ) ≤ (cid:18) d − e + 22 (cid:19) − − | ψ d − e ( B ) ∩ W e ( B, D ) | . (98) We have three subcases. ⋆ Suppose that e = 1 and α e ( D ) >
0. Since e = 1,(99) α e ( D ) = (cid:18) (cid:19) − − dim V e ( D ) = 1 − dim V e ( D )so the assumption α e ( D ) > α e ( D ) − V e ( D ) = 0 . Since dim V e ( D ) = 0, we have that V e ( D ) is just a point and thereby(86) gives(101) (cid:12)(cid:12) B ∩ ψ − d − e ( W e ( B, D )) (cid:12)(cid:12) ≥ (cid:18) f + 22 (cid:19) − . From (98), (100) and (101), we get that(102) α e ( D ) + β e ( B, D ) = 1 + β e ( B, D ) ≤ (cid:18) d − e + 22 (cid:19) − (cid:18) f + 22 (cid:19) . Hence τ e ( B, D ) ≤ α e ( D ) + β e ( B, D ) + (cid:18) f + 22 (cid:19) + 3 (cid:16) by (87) (cid:17) ≤ (cid:18) d − e + 22 (cid:19) + 3 (cid:16) by (102) (cid:17) < (cid:18) d + 22 (cid:19) . (cid:16) since d ≥ (cid:17) ⋆ Suppose that e = 1 and α e ( D ) = 0. Proceeding as in (99), we concludethat(103) 1 + α e ( D ) = dim V e ( D ) = 1 . If γ e ( B, D ) > (cid:0) d +22 (cid:1) − (cid:0) d − e +22 (cid:1) − d , then τ e ( B, D ) = 0. Thus weassume that γ e ( B, D ) ≤ d . If γ e ( B, D ) = d , then (86) implies that | ψ d − e ( B ) ∩ W e ( B, D ) | ≥ (cid:18) f + 22 (cid:19) − γ e ( B, D ) = (cid:18) f + 22 (cid:19) − d, and then (98) and (103) lead to τ e ( B, D ) = α e ( D ) + β e ( B, D ) + (cid:18) f + 22 (cid:19) + 2 ≤ (cid:18) d − e + 22 (cid:19) + d. If γ e ( B, D ) ≤ d −
1, then (86) gives | ψ d − e ( B ) ∩ W e ( B, D ) | ≥ (cid:18) f + 22 (cid:19) − γ e ( B, D ) ≥ (cid:18) f + 22 (cid:19) − d + 1 , N THE MINIMUM NUMBER OF HIGH DEGREE CURVES CONTAINING FEW POINTS 27 and then (98) and (103) imply that τ e ( B, D ) = α e ( D ) + β e ( B, D ) + (cid:18) f + 22 (cid:19) + 3 ≤ (cid:18) d − e + 22 (cid:19) + d. In any case, τ e ( B, D ) ≤ (cid:18) d + 12 (cid:19) + d < (cid:18) d + 22 (cid:19) .⋆ Suppose that e >
1. From (86), (97) and (98), we get that(104) α e ( D ) + β e ( B, D ) ≤ (cid:18) d − e + 22 (cid:19) − (cid:18) f − e + 22 (cid:19) − . Thus τ e ( B, D ) ≤ α e ( D ) + β e ( B, D ) + (cid:18) f + 22 (cid:19) + 3 (cid:16) by (87) (cid:17) ≤ (cid:18) d − e + 22 (cid:19) − (cid:18) f − e + 22 (cid:19) + (cid:18) f + 22 (cid:19) + 1 (cid:16) by (104) (cid:17) < (cid:18) d + 22 (cid:19) . (cid:16) since e > (cid:17) • Assume that e < f and d − e < f . The definitions of W e ( B, D ) and V e ( D ) lead to W e ( B, D ) ⊇ ψ d − e ( B \ ψ − e ( V e ( D ))) and V e ( D ) ⊇ ψ e ( B \ ψ − d − e ( W e ( B, D ))). Then we get that dim W e ( B, D ) ≥ dim ψ d − e ( B \ ψ − e ( V e ( D ))) and dim V e ( D ) ≥ dim ψ e ( B \ ψ − d − e ( W e ( B, D ))).Hence, as a consequence of Lemma 2.7.iii applied to V e ( D ) and W e ( B, D ),we get that α e ( D ) ≤ (cid:18) e + 22 (cid:19) − − (cid:18) f − ( d − e ) + 22 (cid:19) (105) β e ( B, D ) ≤ (cid:18) d − e + 22 (cid:19) − (cid:18) f − e + 22 (cid:19) − . (106) Using that (cid:0) g +22 (cid:1) = P g +1 i =1 i for any g ∈ Z +0 , it is proven easily that(107) (cid:18) e + 22 (cid:19) + (cid:18) d − e + 22 (cid:19) + (cid:18) f + 22 (cid:19) − (cid:18) f − ( d − e ) + 22 (cid:19) − (cid:18) f − e + 22 (cid:19) < (cid:18) d + 22 (cid:19) . Hence τ e ( B, D ) ≤ α e ( D ) + β e ( B, D ) + (cid:18) f + 22 (cid:19) + 3 (cid:16) by (87) (cid:17) ≤ (cid:18) e + 22 (cid:19) + (cid:18) d − e + 22 (cid:19) + (cid:18) f + 22 (cid:19) − (cid:18) f − ( d − e ) + 22 (cid:19) − (cid:18) f − e + 22 (cid:19) (cid:16) by (105),(106) (cid:17) < (cid:18) d + 22 (cid:19) . (cid:16) by (107) (cid:17) This concludes the proof of (84). The claim is a consequence of (82) and (84). (cid:3)
Lemma 3.7.
Let d ∈ Z + , f ∈ [0 , d − , C ∈ C d − f be irreducible and B ∈P ( f +22 )( R \ C ) be such that B is not contained in an element of C ≤ f . Take B ∈ P ( d +22 ) − ( R ) such that B ⊇ B and B ∩ C = B \ B , and also take ( e, D ) ∈ ([1 , d − × P ( B )) \ I ( B, C ) such that α e ( D ) ≥ . Then β e ( B, D ) < and γ e ( B, D ) < (cid:0) d +22 (cid:1) − (cid:0) d − e +22 (cid:1) − .Proof. Since α e ( D ) ≥ U e ( B, D ) = ψ − d ( V d ( B )) ∪ ψ − e ( V e ( D )) if β e ( B, D ) < ψ − d ( V d ( B )) ∪ ψ − e ( V e ( D )) ∪ ψ − d − e ( W e ( B, D )) if β e ( B, D ) ≥ . In so far as | B | < (cid:0) d +22 (cid:1) −
1, Lemma 3.2.iii implies that there is C ∈ C ≤ d suchthat ψ − d ( V d ( B )) ⊆ C . Since α e ( D ) ≥
0, Lemma 3.2.i yields the existence of C ∈ C ≤ d such that ψ − e ( V e ( D )) ⊆ C . Now, if β e ( B, D ) <
0, write C = ∅ , and if β e ( B, D ) ≥
0, we fix C ∈ C ≤ d − e such that ψ − d − e ( W e ( B, D )) ⊆ C (which exists byLemma 3.2.ii). Thus, in any case,(108) U e ( B, D ) ⊆ C ∪ C ∪ C . In so far as ( e, D ) I ( B, C ), note that C ⊆ U e ( B, D ). From (108), we have that C is an irreducible component of C or C ∪ C . We claim that(109) C ⊆ C ∪ C . If (109) is false, then C is a component of C . Lemma 2.9 applied to C and C gives(110) | B ∩ C | < (cid:18) d + 22 (cid:19) − (cid:18) d − d + 22 (cid:19) − (cid:18) d + 22 (cid:19) − . However, B ⊆ B ∩ ψ − d ( V d ( B )) ⊆ B ∩ C ;hence | B ∩ C | = | B | = (cid:0) d +22 (cid:1) − β e ( B, D ) < . If (111) is false, then C = ∅ . Since C ∈ C ≤ e and C ∈ C ≤ d − e , notice that C ∪ C ∈ C ≤ d . From (109), we can apply Lemma 2.9 to C and C ∪ C so(112) | B ∩ ( C ∪ C ) | < (cid:18) d + 22 (cid:19) − . From Remark 3.1.i, B ⊆ ψ − e ( V e ( D )) ∪ ψ − d − e ( W e ( B, D )) ⊆ C ∪ C . Hence | B ∩ ( C ∪ C ) | = | B | = (cid:0) d +22 (cid:1) − γ e ( B, D ) < (cid:18) d + 22 (cid:19) − (cid:18) d − e + 22 (cid:19) − . From (111), we get that C = ∅ . From (109), C ⊆ C . Thus, since C is irreducible, e ≥ d − f . Applying Lemma 2.9 to C and C ,(114) | B ∩ C | < (cid:18) d + 22 (cid:19) − (cid:18) d − e + 22 (cid:19) − . N THE MINIMUM NUMBER OF HIGH DEGREE CURVES CONTAINING FEW POINTS 29
From (114), γ e ( B, D ) = | B ∩ ψ − e ( V e ( D )) | ≤ | B ∩ C | < (cid:18) d + 22 (cid:19) − (cid:18) d − e + 22 (cid:19) − , and this proves (113). The claim of the lemma is a consequence of (111) and(113). (cid:3) Lemma 3.8.
Let d ∈ Z + , f ∈ [0 , d − , C ∈ C ≤ d be irreducible and B ∈ P ( R ) be such that | B | ≤ (cid:0) d +22 (cid:1) − . Then, for all ( e, D ) ∈ I ( B, C ) , | C ∩ U e ( B, D ) | ≤ d . Proof.
Fix p ( x, y ) ∈ R d [ x, y ] such that C = Z ( p ( x, y )). Write F := V d ( B ) , F := V e ( D ) and F := W e ( B, D ), and also set d := d , d := e and d := d − e .Depending on the values of α e ( D ) and β e ( B, D ), there is a subset I of [1 ,
3] such that U e ( B, D ) = S i ∈ I ψ − d i ( F i ) with F i a proper flat in R ( di +22 ) − for each i ∈ I (here weuse the assumption | B | ≤ (cid:0) d +22 (cid:1) − F ≤ | B | − < (cid:0) d +22 (cid:1) − e, D ) ∈ I ( B, C ), we get that C * U e ( B, D ) = [ i ∈ I ψ − d i ( F i ) . Thus C * ψ − d i ( F i ) and therefore ψ d i ( C ) * F i for each i ∈ I . Since F i is a properflat in R ( di +22 ) − and ψ d i ( C ) * F i , there exists a hyperplane H i in R ( di +22 ) − suchthat H i ⊇ F i and ψ d i ( C ) * H i . For each i ∈ I , set C i := ψ − d i ( H i ) and notethat C is not a component of C i . Thus Theorem 2.1 applied to each intersection | C ∩ C i | yields | C ∩ U e ( B, D ) | ≤ X i ∈ I | C ∩ C i | ≤ d ( d + d + d ) = 2 d , and this concludes the proof. (cid:3) Lemma 3.9.
Let d ∈ Z be such that d ≥ , f ∈ [0 , d − , C ∈ C d − f be irreducibleand A ∈ P ( R ) be such that | A ∩ C | ≥ d d +22 ) and A is not contained in anelement of C ≤ d . For all B ∈ P ( f +22 )( A \ C ) such that B is not contained in anelement of C ≤ f , we have that |N d ( A, B , C ) | ≥ d +3 ! | A ∩ C | ( d +22 ) − − ( f +22 ) . Proof.
Write g := (cid:0) d +22 (cid:1) − − (cid:0) f +22 (cid:1) , A := A ∩ C and B := n b , b , . . . , b ( f +22 ) o .In this proof, for any b ∈ A g , its i -th entry will be denoted by b ( f +22 ) + i , i.e. b = (cid:16) b ( f +22 ) +1 , b ( f +22 ) +2 , . . . , b ( d +22 ) − (cid:17) . For any b ∈ A g , write B ( b ) := B . Forall i ∈ [1 , g ] and b ∈ A g , write B i ( b ) := B ∪ n b ( f +22 ) +1 , b ( f +22 ) +2 , . . . , b ( f +22 ) + i o = n b , b , . . . , b ( f +22 ) + i o U i ( b ) := [ ( e,D ) ∈ I ( B i − ( b ) ,C ) U e ( B i − ( b ) , D ) . For all b ∈ A g , i ∈ [1 , g ] and ( e, D ) ∈ I ( B i − ( b ) , C ), we have that C * U e ( B i − ( b ) , D ). Thus, applying Lemma 3.8 to C and U e ( B i − ( b ) , D ), we getthat(115) | C ∩ U e ( B i − ( b ) , D ) | ≤ d . For each i ∈ [1 , g ], write R i := n b ∈ A g : b ( f +22 ) + i ∈ U i ( b ) o . First we show thatfor all i ∈ [1 , g ],(116) | R i | ≤ d d +22 ) − | A | g − . For each i ∈ [1 , g ] , E ∈ P (cid:16)h , (cid:0) f +22 (cid:1) + i i(cid:17) and b ∈ A g , set E ( b ) := { b j : j ∈ E } . Notice that P (cid:0) B i ( b ) (cid:1) = n E ( b ) : E ∈ P (cid:16)h , (cid:0) f +22 (cid:1) + i i(cid:17)o . Now, for each i ∈ [1 , g ] , E ∈ P (cid:16)h , (cid:0) f +22 (cid:1) + i i(cid:17) and e ∈ [1 , d − R i,e,E := n b ∈ A g : b ( f +22 ) + i ∈ U e ( B i − ( b ) , E ( b )) and ( e, E ( b )) ∈ I ( B i − ( b ) , C ) o This definition gives that for all i ∈ [1 , g ], R i ⊆ d − [ e =1 [ E ∈P ([ , ( f +22 ) + i ]) R i,e,E and thereby(117) | R i | ≤ X E ∈P ([ , ( f +22 ) + i ]) d − X e =1 | R i,e,E | . For each i ∈ [1 , g ] , E ∈ P (cid:16)h , (cid:0) f +22 (cid:1) + i i(cid:17) and e ∈ [1 , d − b ∈ R i,e,E then b ( f +22 ) + i ∈ U e ( B i − ( b ) , E ( b )) ∩ A ⊆ U e ( B i − ( b ) , E ( b )) ∩ C ; thus, by (115), b ∈ R i,e,E has at most 2 d possible values in its i -th entry and therefore(118) | R i,e,E | ≤ d | A | g − . Since (cid:12)(cid:12)(cid:12) P (cid:16)h , (cid:0) f +22 (cid:1) + i i(cid:17) × [1 , d − (cid:12)(cid:12)(cid:12) ≤ d d +22 ) − , we have that (117) and (118)imply (116).Set T := n b ∈ A g : b ( f +22 ) + i U i ( b ) for all i ∈ [1 , g ] o so T = A g \ S gi =1 R i .From (116),(119) g X i =1 | R i | ≤ gd d +22 ) − | A | g − ≤ d d +22 ) − | A | g − . Since | A | ≥ d d +22 ) by assumption, we get from (119) that(120) | T | = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) A g \ g [ i =1 R i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≥ | A | g − g X i =1 | R i | ≥ | A | g − d d +22 ) − | A | g − ≥ | A | g . The second step in the proof is to show that the map φ : T −→ P ( A ) , φ ( b ) = B g ( b ) N THE MINIMUM NUMBER OF HIGH DEGREE CURVES CONTAINING FEW POINTS 31 satisfies that(121) φ ( T ) ⊆ N d ( A, B , C ) . We claim that for all b ∈ T , i ∈ [0 , g ] and ( e, D ) ∈ I ( B i ( b ) , C ), we have that(122) max (cid:8) τ e ( B i ( b ) , D ) , µ e ( B i ( b ) , D ) (cid:9) < (cid:18) d + 22 (cid:19) . We show (122) by induction on i ∈ [0 , g ]. For i = 0, we apply Lemma 3.6 to B = B ( b ). Assume that (122) holds for i − ≥ i . In so faras b ∈ T , note that b i ∈ A \ U i ( b ). By induction,max (cid:8) τ e ( B i − ( b ) , D ) , µ e ( B i − ( b ) , D ) (cid:9) < (cid:18) d + 22 (cid:19) for all ( e, D ) ∈ I (cid:0) B i − ( b ) , C (cid:1) so B i − ( b ) and b i ∈ A \ U i ( b ) satisfy the assump-tions of Lemma 3.3, and hence this lemma implies that B i − ( b ) ∪ { b i } = B i ( b )satisfies (122) completing the induction. In particular, (122) is true for i = g so,for all b ∈ T and ( e, D ) ∈ I (cid:0) B g ( b ) , C (cid:1) ,(123) max (cid:8) τ e ( B g ( b ) , D ) , µ e ( B g ( b ) , D ) (cid:9) < (cid:18) d + 22 (cid:19) . Now, for all ( e, D ) ∈ ([1 , d − × P ( B g ( b ))) \ I (cid:0) B g ( b ) , C (cid:1) such that α e ( D ) ≥ β e ( B, D ) < γ e ( B, D ) < (cid:18) d + 22 (cid:19) − (cid:18) d − e + 22 (cid:19) − . (125)For all b ∈ T and i ∈ [1 , g ], we get that b ( f +22 ) + i U i ( b ); in particular, b ( f +22 ) + i ψ − d ( V d ( B i − ( b ))). Then the flats V d ( B ( b )) ( V d ( B ( b )) ( . . . ( V d ( B g ( b )) arecontained properly so(126) dim V d ( B g ( b )) ≥ g + dim V d ( B ( b )) . Since B = B ( b ) is not contained in an element of C ≤ f , we get that ψ f ( B ( b ))is not contained in a hyperplane of R ( f +22 ) − so dim ψ f ( B ( b )) = (cid:0) f +22 (cid:1) −
1. FromRemark 2.4.iii, we get that dim ψ d ( B ( b )) ≥ dim ψ f ( B ( b )) sodim V d ( B ( b )) = dim ψ d ( B ( b )) ≥ dim ψ f ( B ( b )) = (cid:18) f + 22 (cid:19) − , and then (126) gives(127) dim V d ( B g ( b )) ≥ (cid:18) d + 22 (cid:19) − . On the other hand,dim V d ( B g ( b )) = dim Fl( ψ d ( B g ( b ))) ≤ | ψ d ( B g ( b )) |− | B g ( b ) |− (cid:18) d + 22 (cid:19) − ψ d ( B g ( b )) = dim V d ( B g ( b )) = (cid:18) d + 22 (cid:19) − . We can apply Lemma 3.4 with B = B g ( b ), B and C because its assumptions aresatisfied (indeed: i) holds by (128); ii) and iii) hold by the construction of B g ( b );iv) holds by (123); v) holds by (124) and (125)). Hence Lemma 3.4 implies that B g ( b ) ∈ N d ( A, B , C ) and this shows (121).Finally, for each B ∈ N d ( A, B , C ), | φ − ( B ) | ≤ g ! which is the number ofpermutations of the elements in B \ B . Therefore (120) and (121) give(129) |N d ( A, B , C ) | ≥ g ! | T | ≥ g !2 | A | g . Since g ≤ (cid:0) d +22 (cid:1) − < d +2 , we have that g !2 ≤ d +3 ! and hence (129) completesthe proof. (cid:3) From curves to lines
In this section we use the elements of the families N d ( A ) to generate flats thatwill be used to construct hyperprojections which are used to transform the problemof finding curves of degree d with few points into a problem of finding ordinary lineswhich avoid a finite set.Let d ∈ Z + , e ∈ [1 , d −
1] and B ∈ P ( R ). Write C d,e ( B ) := (cid:26) C ∈ C e : | C ∩ B | = (cid:18) d + 22 (cid:19) − (cid:18) d − e + 22 (cid:19) − (cid:27) C d ( B ) := d − [ f =1 C d,f ( B ) . Lemma 4.1.
Let d ∈ Z + and e ∈ [1 , d − . For any B ∈ N d ( R ) , |C d,e ( B ) | < d +2 . Proof.
Since B ∈ N d ( R ), we have that for any f ∈ [1 , d −
1] and C ∈ C f , | C ∩ B | < (cid:18) d + 22 (cid:19) − (cid:18) d − f + 22 (cid:19) ≤ (cid:18) d + 22 (cid:19) − | B | ;in particular, this means that there is no element C ∈ C ≤ e which contains B . Thus,from Remark 2.4.ii, no hyperplane in R ( e +22 ) − can contain ψ e ( B ) and therefore(130) dim ψ e ( B ) = (cid:18) e + 22 (cid:19) − . Set the map φ : C d,e ( B ) −→ P ( d +22 ) − ( d − e +22 ) − ( B ) , φ ( C ) = C ∩ B We will show that φ is injective. Take C , C ∈ C d,e ( B ) such that C ∩ B = C ∩ B .Write F := Fl( ψ e ( C ∩ B )) = Fl( ψ e ( C ∩ B )). Because C , C ∈ C ≤ e , we have thatFl( ψ e ( C )) and Fl( ψ e ( C )) are contained in hyperplanes by Remark 2.4.ii. For i ∈ { , } ,(131) F ⊆ Fl( ψ e ( C i ∩ B )) ⊆ Fl( ψ e ( C i ))so F is a proper flat in R ( e +22 ) − . We claim that F is a hyperplane. If F is nota hyperplane, then (130) implies that we can choose a hyperplane H in R ( e +22 ) − such that H ⊇ F and(132) | H ∩ ψ e ( B ) | > | F ∩ ψ e ( B ) | . N THE MINIMUM NUMBER OF HIGH DEGREE CURVES CONTAINING FEW POINTS 33
However, (cid:18) d + 22 (cid:19) − (cid:18) d − e + 22 (cid:19) − ≥ | ψ − e ( H ) ∩ B | (cid:16) since B ∈ N d ( R ) (cid:17) = | H ∩ ψ e ( B ) | > | F ∩ ψ e ( B ) | (cid:16) by (132) (cid:17) ≥ | C ∩ B | = (cid:18) d + 22 (cid:19) − (cid:18) d − e + 22 (cid:19) − , (cid:16) since C ∈ C d,e ( B ) (cid:17) and this contradiction shows that F is a hyperplane. Hence, since F is a hyperplane,(131) leads to Fl( ψ e ( C )) = F = Fl( ψ e ( C )) , and then Remark 2.4.ii yields that C = C implying that φ is injective. In so faras φ is injective,(133) |C d,e ( B ) | ≤ (cid:12)(cid:12)(cid:12) P ( d +22 ) − ( d − e +22 ) − ( B ) (cid:12)(cid:12)(cid:12) ≤ |P ( B ) | = 2 | B | = 2( d +22 ) − . Using that (cid:0) d +22 (cid:1) − < d +2 , the claim follows from (133). (cid:3) Let d ∈ Z + . As usual, P d := ( R d +1 \ { } ) / ∼ with y ∼ z if there is r ∈ R suchthat y = r · z . We will use the embedding R d −→ P d , ( z , z , . . . , z d ) [1 : z : z : . . . : z d ]so that R d can be seen as a subset of P d . For any flat F in R d defined by a familyof linear equations n r ,i + P dj =1 r j,i x j o i ∈ I (i.e. F = T i ∈ I Z (cid:16) r ,i + P dj =1 r j,i x j (cid:17) ),its homogenization will be the subset of elements [ z : z : . . . : z d ] ∈ P d such that P dj =0 r j,i z j = 0 for all i ∈ I , and we will denote it by F h . The next standard factsabout homogenization can be found in [9, Sec.I.2]. Remark 4.2.
Let d ∈ Z + , e ∈ [1 , d ] and F be an e -dimensional flat in R d . i) Then F h is an e -dimensional linear variety in P d , and F h ∼ = P e . ii) Let f ∈ [1 , d ] and G be an f -dimensional flat in R d . If e + f − d ≥ ,then F h ∩ G h = ∅ and F h ∩ G h is a g -dimensional linear variety in P d with g ≥ e + f − d . Let d ∈ Z + , e ∈ [0 , d −
1] and F be an e -dimensional flat in R d . Take G a d − e − R d such that F ∩ G = ∅ . For any z ∈ R d \ F , noticethat dim Fl( F ∪ { z } ) = e + 1 and dim Fl( F ∪ { z } ) ∩ G ≤ F ∩ G = ∅ . Onthe other hand, Remark 4.2.ii implies that Fl( F ∪ { z } ) h ∩ G h is a linear variety ofdimension at least 0. Therefore Fl( F ∪ { z } ) h ∩ G h is exactly a point. Identifying G h with P d − e − , we define the hyperprojection centered in F as the map π F : R d \ F −→ P d − e − , { π F ( z ) } = Fl( F ∪ { z } ) h ∩ P d − e − . An easy consequence of the the definition of π F is the following remark. Remark 4.3.
Let d ∈ Z + , e ∈ [0 , d − and F be an e -dimensional flat in R d . Forany f ∈ [ e + 1 , d ] , there is an bijective relation between the f -flats containing F andthe f − e − -dimensional linear varieties in P d − e − given by H π F ( H \ F ) . Let B ∈ N d ( R ). Recall that V d ( A ) = Fl( ψ d ( A )) for any A ∈ P ( R ). Write D d ( B ) := ψ − d ( V d ( B )) E d ( B ) := D d ( B ) ∪ [ C ∈C d ( B ) ψ − d ( V d ( C ∪ B )) , and the map ϕ d,B : R \ D d ( B ) −→ P , ϕ d,B ( a ) = π V d ( B ) ( ψ d ( a ));in other words, ϕ d,B = π V d ( B ) ◦ ψ d | R \ D d ( B ) . The main properties of ϕ d,B are provenin the next lemma. Lemma 4.4.
Let d ∈ Z + and B ∈ N d ( R ) . i) For all e ∈ [1 , d − and C ∈ C d,e ( B ) , we have that | ϕ d,B ( C \ D d ( B )) | = 1 . ii) For all a ∈ R \ E d ( B ) , we get that ϕ d,B ( a ) ϕ d,B ( E d ( B ) \ D d ( B )) . iii) For all a ∈ R \ E d ( B ) , we have that (cid:12)(cid:12)(cid:12) ϕ − d,B ( ϕ d,B ( a )) (cid:12)(cid:12)(cid:12) ≤ d − | D d ( B ) | .Proof. Since B ∈ N d ( R ), notice that for all e ∈ [1 , d −
1] and C ∈ C d,e ( B ),(134) dim ψ d − e ( B \ C ) = (cid:18) d − e + 22 (cid:19) − H in R ( d − e +22 ) − such that ψ d − e ( B \ C ) ⊆ H and therefore B \ C ⊆ ψ − d − e ( H ). In so far as ψ − d − e ( H ) ∈ C ≤ d − e and C ∈ C e ,we get that ψ − d − e ( H ) ∪ C ∈ C ≤ d by Remark 2.2.i. Thus, insomuch as B ∪ C = ( B \ C ) ∪ C ⊆ ψ − d − e ( H ) ∪ C, Remark 2.4.ii implies that ψ d ( B ∪ C ) is contained in hyperplane K of R ( d +22 ) − suchthat ψ − d ( K ) = ψ − d − e ( H ) ∪ C ; in particular,(135) dim ψ d ( B ∪ C ) ≤ (cid:18) d + 22 (cid:19) − . First we prove i). To show i), it suffices to prove that(136) dim ψ d ( B ∪ C ) = (cid:18) d + 22 (cid:19) − π V d ( B ) projects (cid:0) d +22 (cid:1) − V d ( B ) = Fl( ψ d ( B )) into pointsof P . From (134), V d − e ( B \ C ) is a proper flat in R ( d − e +22 ) − , and from (135), V d ( B ∪ C ) is a proper flat in R ( d +22 ) − ; therefore R \ ( ψ − d − e ( V d − e ( B \ C )) ∪ ψ − d ( V d ( B ∪ C ))) = ∅ . Fix a ∈ R \ ( ψ − d − e ( V d − e ( B \ C )) ∪ ψ − d ( V d ( B ∪ C ))). Since a ψ − d − e ( V d − e ( B \ C )),we get from (134) that(137) dim ψ d − e (( B ∪ { a } ) \ C ) = dim ψ d − e ( { a } ∪ ( B \ C )) = (cid:18) d − e + 22 (cid:19) − . Now we apply Lemma 2.10.ii to B ∪ { a } so (137) yields(138) dim ψ d (( B ∪ { a } ) ∪ C ) = (cid:18) d + 22 (cid:19) − . In so far as a ψ − d ( V d ( B ∪ C )), (138) leads to (136). N THE MINIMUM NUMBER OF HIGH DEGREE CURVES CONTAINING FEW POINTS 35
We prove ii) by contradiction. Assume that there is a ∈ R \ E d ( B ) such that ϕ d,B ( a ) ∈ ϕ d,B ( E d ( B ) \ D d ( B ))). Therefore there are e ∈ [1 , d −
1] and C ∈ C d,e ( B )such that ϕ d,B ( a ) = ϕ d,B ( C \ D d ( B ))). This equality means that ψ d ( a ) is in the (cid:0) d +22 (cid:1) − V d ( B ∪ C ) = Fl( ψ d ( B ∪ C )) so a ∈ ψ − d ( V d ( B ∪ C )) ⊆ E d ( B ) , which contradicts the assumption.Finally, we show iii) by contradiction. Assume that there is a ∈ R \ E d ( B ) suchthat(139) (cid:12)(cid:12)(cid:12) ϕ − d,B ( ϕ d,B ( a )) (cid:12)(cid:12)(cid:12) + | D d ( B ) | > d . Since B ∈ N d ( R ), we have that dim ψ d ( B ) = (cid:0) d +22 (cid:1) −
4. Now, in so far as a D d ( B ) = ψ − d (Fl( ψ d ( B ))), we get that(140) dim ψ d ( { a } ∪ B ) = (cid:18) d + 22 (cid:19) − V d ( { a }∪ B ) = Fl( ψ d ( { a }∪ B )) is a (cid:0) d +22 (cid:1) − ϕ d,B ( a ) by π V d ( B ) .Since V d ( { a } ∪ B ) is a (cid:0) d +22 (cid:1) − H and H distinct hyperplanes in R ( d +22 ) − such that H ∩ H = V d ( { a }∪ B ); write C := ψ − d ( H ) and C := ψ − d ( H )so that(141) C ∩ C = ψ − d ( H ∩ H ) = ψ − d ( V d ( { a } ∪ B )) , and therefore(142) | C ∩ C | = | ψ − d ( V d ( { a } ∪ B )) | ≥ (cid:12)(cid:12)(cid:12) ϕ − d,B ( ϕ d,B ( a )) (cid:12)(cid:12)(cid:12) + | D d ( B ) | . On the other hand, Remark 2.4.ii leads to C , C ∈ C ≤ d . We claim that C and C share an irreducible component. If this claim were false, then Theorem 2.1 wouldgive | C ∩ C | ≤ d ; however, this contradicts the inequality | C ∩ C | > d which isa consequence of (139) and (142). Therefore there is an irreducible curve C ∈ C e for some e ∈ [1 , d −
1] such that C ⊆ C ∩ C . From (141), C ⊆ ψ − d ( V d ( { a } ∪ B ))so ψ d ( C ) ⊆ V d ( { a } ∪ B ) = Fl( ψ d ( { a } ∪ B )); this and (140) yield(143) dim ψ d ( { a } ∪ B ∪ C ) = dim ψ d ( { a } ∪ B ) = (cid:18) d + 22 (cid:19) − . Since B ∈ N d ( R ), notice that | B ∩ C | ≤ (cid:18) d + 22 (cid:19) − (cid:18) d − e + 22 (cid:19) − . Thus we have two cases. • If | B ∩ C | = (cid:0) d +22 (cid:1) − (cid:0) d − e +22 (cid:1) −
1, then C ∈ C d,e ( B ). From (136), dim ψ d ( B ∪ C ) = (cid:0) d +22 (cid:1) −
3; however, since a ψ − d ( V d ( B ∪ C )) (because a ∈ R \ E d ( B )), we get thatdim ψ d ( { a } ∪ B ∪ C ) > dim ψ d ( B ∪ C ) = (cid:18) d + 22 (cid:19) − • If | B ∩ C | < (cid:0) d +22 (cid:1) − (cid:0) d − e +22 (cid:1) −
1, then (because B ∈ N d ( R )) we have that(144) dim ψ d − e ( B \ C ) > (cid:18) d − e + 22 (cid:19) − . Using (144) and Lemma 2.10, we conclude that dim ψ d ( B ∪ C ) > (cid:0) d +22 (cid:1) − (cid:18) d + 22 (cid:19) − ψ d ( { a } ∪ B ∪ C ) ≥ dim ψ d ( B ∪ C ) > (cid:18) d + 22 (cid:19) − , which is impossible.In any case, we reached a contradiction and this concludes the proof of iii). (cid:3) Let d, n ∈ Z + , A ∈ P ( R ) be finite and B ∈ N d ( A ). Set O d,n ( A, B ) := { C ∈ O d,n ( A ) : B ⊆ C } δ d ( A, B ) := max a ∈ A \ E d ( B ) (cid:12)(cid:12)(cid:12) ϕ − d,B ( ϕ d,B ( a )) (cid:12)(cid:12)(cid:12) . From Lemma 4.4, δ d ( A, B ) exists and δ d ( A, B ) ≤ d − | D d ( B ) | ≤ d − | B | = d − d + 42 . For any m ∈ Z + , let c = c ( m ) and c = c ( m ) be as in Lemma 2.11 and set c ( d ) := d · max ≤ m ≤ d d +2 c ( m ) c ( d ) := d · max ≤ m ≤ d d +2 c ( m ) . Then next two lemmas are the main results of this section.
Lemma 4.5.
Let d ∈ Z + and A ∈ P ( R ) be finite, d -regular and such that | A | ≥ d max { c ( d ) , c ( d ) , } . Write n := 2 δ d ( A, B ) + | D d ( B ) | . Then, for any B ∈N d ( A, ∅ , R ) , |O d,n ( A, B ) | ≥ d d +2 | A | . Proof.
Since A is finite, we can apply a linear automorphism in P to assume that ϕ d,B ( A \ D d ( B )) ∩ ( P \ R ) = ∅ ; thus we assume from now on that ϕ d,B ( A \ D d ( B )) ⊆ R . Write S := ϕ d,B ( A \ E d ( B )) , T := ϕ d,B ( E d ( B ) \ D d ( B )) and L := { L ∈ O ( S ) : L ∩ T = ∅} . From Lemma 4.4.ii, we have that S \ T = S . From Lemma 4.1, wehave that |C d,e ( B ) | < d +2 for all e ∈ [1 , d −
1] so |C d ( B ) | < d d +2 . Thus Lemma4.4.i and the previous inequality lead to(145) | T | ≤ |C d ( B ) | < d d +2 . For each C ∈ C d ( B ), fix a hyperplane H C in R ( d +22 ) − such that H C ⊇ V d ( C ∪ B );since ψ − d ( H C ) ∈ C ≤ d and A is d -regular,(146) (cid:12)(cid:12) A ∩ ψ − d ( H C ) (cid:12)(cid:12) ≤ d +8 | A | . From (145) and (146), X C ∈C d ( B ) (cid:12)(cid:12) A ∩ ψ − d ( H C ) (cid:12)(cid:12) ≤ d d +2 d +8 | A | ≤ | A | , N THE MINIMUM NUMBER OF HIGH DEGREE CURVES CONTAINING FEW POINTS 37 and therefore(147) | A \ E d ( B ) | = | A | − | A ∩ E d ( B ) | ≥ | A | − X C ∈C d ( B ) (cid:12)(cid:12) A ∩ ψ − d ( H C ) (cid:12)(cid:12) ≥ | A | . Since δ d ( A, B ) ≤ d by Lemma 4.4.iii, we get from (147) that(148) | S | ≥ δ d ( A, B ) | A \ E d ( B ) | ≥ d | A \ E d ( B ) | ≥ d | A | . On the one hand, S is not collinear because if S is contained in a line L , then A \ E d ( B ) is contained in ϕ − d,B ( L ) which is in C ≤ d , and (147) would contradict the d -regularity of A . On the other hand, (148) implies | S | > c ( d ). Thus we can applyLemma 2.11 to S and T , and we obtain that(149) |L| ≥ | S | − c ( d ) . From (148) and (149),(150) |L| ≥ d | A | . Denote by H the family of hyperplanes in R ( d +22 ) − generated by ψ d ( A ) and definethe maps η : L −→ H , η ( L ) = π − V d ( B ) ( L ) η : η ( L ) −→ C ≤ d , η ( η ( L )) = ϕ − d,B ( L ) . We will show that(151) η ( η ( L )) ⊆ O d,n ( A, B ) . For each L ∈ L , L is a line generated by elements of S = ϕ d,B ( A \ D d ( B )). Hence η ( L ) is a hyperplane generated by elements of ψ d ( A ), and then, by Lemma 2.5, η ( η ( L )) = ϕ − d,B ( L ) = ψ − d ( η ( L )) is an element of C ≤ d determined by A (i.e. η ( η ( L )) ∈ D d ( A )). For any L ∈ L , we have that L ∩ S = { ϕ d,B ( a ) , ϕ d,B ( a ) } forsome a , a ∈ A \ E d ( B ). In so far as η ( η ( L )) = ϕ − d,B ( L ), Lemma 4.4 leads to(152) η ( η ( L )) ∩ A = ϕ − d,B ( L ) ∩ A = (cid:16) ϕ − d,B ( ϕ d,B ( a )) ∪ ϕ − d,B ( ϕ d,B ( a )) ∪ D d ( B ) (cid:17) ∩ A. We conclude from (152) that | η ( η ( L )) ∩ A | ≤ | ϕ − d,B ( ϕ d,B ( a )) | + | ϕ − d,B ( ϕ d,B ( a )) | + | D d ( B ) | ≤ n, and therefore η ( η ( L )) ∈ O d,n ( A, B ) proving (151). From Remark 4.3, η is in-jective. From Lemma 2.3, for each C ∈ C ≤ d , there are at most d d hyperplanes in R ( d +22 ) − such that C = ψ − d ( H ) so | η − ( C ) | ≤ d d for all C ∈ C ≤ d . The previoustwo statements give | η ( η ( L )) | ≥ d d |L| , and then (151) yields that(153) |O d,n ( A, B ) | ≥ | η ( η ( L )) | ≥ d d |L| . Then the lemma follows from (150) and (153). (cid:3)
Lemma 4.6.
Let d ∈ Z + , f ∈ [0 , d − , C ∈ C d − f be irreducible, A ∈ P ( R ) befinite such that | A ∩ C | ≥ d d +8 d +8 max { c ( d ) , c ( d ) , } , and B ∈ P ( f +22 )( A \ C ) be such that B is not contained in an element of C ≤ f . Write n := 2 δ d ( A, B ) + | D d ( B ) | . i) For any B ∈ N d ( A, B , C ) , we get that |O d,n ( A, B ) | ≥ d d +2 | A ∩ C | . ii) Assume that | A \ C | ≤ d | A ∩ C | . Then, for any B ∈ N d ( A, B , C ) , weget that (cid:12)(cid:12) O d,d ( A, B ) (cid:12)(cid:12) ≥ d d +8 | A ∩ C | .Proof. Write S := ϕ d,B ( A \ E d ( B )) , S := ϕ d,B (( A ∩ C ) \ E d ( B )) and T := ϕ d,B ( E d ( B ) \ D d ( B )). Denote by H the family of hyperplanes in R ( d +22 ) − gen-erated by ψ d ( A ), denote by L the family of lines generated by ϕ d,B ( A \ D d ( B )) anddefine the maps η : L −→ H , η ( L ) = π − V d ( B ) ( L ) η : η ( L ) −→ C ≤ d , η ( η ( L )) = ψ − d ( η ( L )) . As in the first part of Lemma 4.5, we may assume that S is contained in R andwe have that(154) | T | ≤ |C d ( B ) | < d d +2 . The next step is to show that for any hyperplane H in R ( d +22 ) − containing V d ( B ),(155) | H ∩ ψ d ( C ) | ≤ d . Indeed, write C := ψ − d ( H ). Note that B ⊆ ψ − d ( V d ( B )) ⊆ ψ − d ( H ) = C so | B ∩ C | = | B | = (cid:0) d +22 (cid:1) −
3. Thus C is not a component of C because otherwiseLemma 2.9 applied to C and C leads to | B ∩ C | < (cid:0) d +22 (cid:1) −
3. Thereby Theorem2.1 applied to C and C gives | ψ d ( C ) ∩ H | = | C ∩ C | ≤ d , which shows (155). Remark 4.3 implies that for any line L ∈ R , there is a hyper-plane H in R ( d +22 ) − containing V d ( B ) such that π V d ( B ) ( H ) = L . Then (155) yieldsthat for any line L in R ,(156) | L ∩ ϕ d,B ( C \ D d ( B )) | ≤ d . The next step is to prove that(157) | S | ≥ d (cid:16) | A ∩ C | − d d +3 (cid:17) ≥ d | A ∩ C | . Denote by L T the family of lines generated by T . From (154),(158) |L T | ≤ (cid:18) | T | (cid:19) < d d +3 . For any L ∈ L T , the hyperplane π − V d ( B ) ( L ) contains V d ( B ) so (155) leads to(159) | π − V d ( B ) ( L ) ∩ ψ d ( C ) | ≤ d . N THE MINIMUM NUMBER OF HIGH DEGREE CURVES CONTAINING FEW POINTS 39
From (158) and (159),(160) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ψ d ( A ∩ C ) ∩ [ L ∈L T π − V d ( B ) ( L ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ X L ∈L T | ψ d ( C ) ∩ π − V d ( B ) ( L ) | < d d +3 . Since T is contained S L ∈L T L , we have that E d ( B ) is contained in ψ − d (cid:16)S L ∈L T π − V d ( B ) ( L ) (cid:17) . Thus (160) leads to(161) | ( A ∩ C ) \ E d ( B ) | ≥ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ( A ∩ C ) \ ψ − d [ L ∈L T π − V d ( B ) ( L ) !(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≥ | A ∩ C | − d d +3 . From Lemma 4.4.iii, each element in S = ϕ d,B (( A ∩ C ) \ E d ( B )) has at most δ d ( A, B ) ≤ d elements in its preimage so (161) leads to (157).Now we show i). Set L := { L ∈ O ( S ) : L ∩ T = ∅} . On the one hand, S is not collinear because if this is the case, then S is also collinear and then π − V d ( B ) ( S ) is contained in a hyperplane; however, (157) yields | S | > d , and thiswould contradict (155). On the other hand, (157) implies | S | > c ( d ). Thus we canapply Lemma 2.11 to S and T , and we obtain that(162) |L | ≥ | S | − c ( d ) . From (157) and (162),(163) |L | ≥ d | A ∩ C | − c ( d ) ≥ d | A ∩ C | Proceeding as in the last part of Lemma 4.5, it is concluded that η ( η ( L )) ⊆ O d,n ( A, B )so(164) |O d,n ( A, B ) | ≥ | η ( η ( L )) | ≥ d d |L | . Then i) is a straight consequence of (163) and (164).Finally, we show ii). Set S := ϕ d,B (( A \ C ) \ E d ( B )), and notice that(165) | S | ≤ | A \ C | ≤ d | A ∩ C | . Denote by L the family of lines L generated by S such that L ∩ ( T ∪ S ) = ∅ .For any s ∈ S , write L ( s ) := { L ∈ L : s ∈ L } . Since S ⊆ ϕ d,B ( C \ E d ( B )) ⊆ π V d ( B ) ( ψ d ( C )), we have that for each t ∈ S ∩ L , there is z ∈ ψ d ( C ) ∩ π − V d ( B ) ( L )such that π V d ( B ) ( z ) = t . Then (155) gives(166) | L ∩ S | ≤ | π − V d ( B ) ( L ∩ S ) | ≤ | π − V d ( B ) ( L ) ∩ ψ d ( C ) | ≤ d . On the one hand, (166) implies that for each L ∈ L , there are at most d elements s ∈ S such that L ∈ L ( s ) so(167) |L | ≥ d X s ∈ S |L ( s ) | . On the the other hand, (166) implies that for each s ∈ S , there are at least | S |− d lines generated by s and other element of S ; notice that at most | S | + | T | of theselines pass through an element of S ∪ T so (154), (157) and (165) lead to(168) |L ( s ) | ≥ | S | − d − | S | − | T | ≥ d | A ∩ C | . From (157), (167) and (168), we can bound |L | below as follows(169) |L | ≥ d X s ∈ S |L ( s ) | ≥ d | S || A ∩ C | ≥ d | A ∩ C | . We will show that(170) η ( η ( L )) ⊆ O d,d ( A, B ) . For any L ∈ L , notice that L ∩ S = L ∩ S insomuch as L ∩ ( S ∪ T ) = ∅ ; consideringthe preimages of π V d ( B ) , we get that η ( L ) ∩ ψ d ( A ) = η ( L ) ∩ π − V d ( B ) ( S ). As in(166), | η ( η ( L )) ∩ A | = | η ( L ) ∩ ψ d ( A ) | ≤ d , and this proves (170). From Remark 4.3, η is injective. From Lemma 2.3, for each C ∈ C ≤ d , there are at most d d hyperplanes in R ( d +22 ) − such that C = ψ − d ( H ) so | η − ( C ) | ≤ d d for all C ∈ C ≤ d . From this and (170),(171) (cid:12)(cid:12) O d,d ( A, B ) (cid:12)(cid:12) ≥ | η ( η ( L )) | ≥ d d |L | . Then ii) is a consequence of (169) and (171). (cid:3) Proofs of the main results
We conclude the proofs of the main results in this section.
Proof. (Theorem 1.5).
From Theorem 1.2, the claim holds for d = 1. From [11,Thm 1.3], the statement is true for d = 2. Thus we assume that d ≥ c = d d +9 d +16 max { c ( d ) , c ( d ) , } and c = d d +6 (2 d )4 d +16 (2 d +3 )! satisfy the desired properties. Set c := (cid:0) d − d +42 ( d +22 ) − (cid:1) and c := ( d d +8 ) d c c . For all B ∈ N d ( A ), Lemma 4.4.iii implies that δ d ( A, B ) + (cid:18) d + 22 (cid:19) − δ d ( A, B ) + | B | ≤ δ d ( A, B ) + | D d ( B ) | ≤ d so 2 δ d ( A, B ) + | D d ( B ) | ≤ d − d +42 . Therefore(172) O d, δ d ( A,B )+ | D d ( B ) | ( A, B ) ⊆ O d, d − d +42 ( A, B ) . For all C ∈ O d, d − d +42 ( A ), we have that | A ∩ C | ≤ d − d +42 ; therefore there areat most c subsets B ∈ N d ( A ) such that C ∈ O d, d − d +42 ( A, B ), and this yields (cid:12)(cid:12)(cid:12) O d, d − d +42 ( A ) (cid:12)(cid:12)(cid:12) ≥ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) [ B ∈N d ( A ) O d, d − d +42 ( A, B ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≥ c X B ∈N d ( A ) (cid:12)(cid:12)(cid:12) O d, d − d +42 ( A, B ) (cid:12)(cid:12)(cid:12) . (173) N THE MINIMUM NUMBER OF HIGH DEGREE CURVES CONTAINING FEW POINTS 41
Since A is not contained in an element of C d , then A is not contained in anelement of C ≤ d by Remark 2.6. The conclusion of the proof is divided into twocases.i) Assume that A is d -regular. On the one hand, Lemma 3.5 gives(174) |N d ( A ) | ≥ |N d ( A, ∅ , R ) | ≥ d +3 ! | A | ( d +22 ) − . Lemma 4.5 and (172) imply that for all B ∈ N d ( A, ∅ , R ),(175) (cid:12)(cid:12)(cid:12) O d, d − d +42 ( A, B ) (cid:12)(cid:12)(cid:12) ≥ d d +2 | A | . Then (173), (174) and (175) yield (cid:12)(cid:12)(cid:12) O d, d − d +42 ( A ) (cid:12)(cid:12)(cid:12) ≥ c d d +2 (2 d +3 !) | A | ( d +22 ) − . ii) Assume that A is not d -regular. Then there is C ∈ C ≤ d such that | A ∩ C | ≥ d +8 | A | . Moreover, since C has at most d irreducible components, thereare f ∈ [0 , d −
1] and C ∈ C d − f irreducible such that(176) | A ∩ C | ≥ d d +8 | A | . Denote by R the family of subsets B ∈ P ( f +22 )( A \ C ) such that B is notcontained in an element of C ≤ f . Lemma 2.8 gives(177) |R| ≥ f +22 ) − | A \ C | . For all B ∈ R , Lemma 3.9 leads to(178) |N d ( A, B , C ) | ≥ d +3 ! | A ∩ C | ( d +22 ) − − ( f +22 ) ≥ d +3 ! | A ∩ C | d − . We have two subcases. ⋆ Assume that | A \ C | ≤ d | A ∩ C | . Hence Lemma 4.6.ii implies thatfor all B ∈ R and B ∈ N d ( A, B , C ),(179) (cid:12)(cid:12)(cid:12) O d, d − d +42 ( A, B ) (cid:12)(cid:12)(cid:12) ≥ (cid:12)(cid:12) O d,d ( A, B ) (cid:12)(cid:12) ≥ d d +8 | A ∩ C | . Then (cid:12)(cid:12)(cid:12) O d, d − d +42 ( A ) (cid:12)(cid:12)(cid:12) ≥ c X B ∈N d ( A ) (cid:12)(cid:12)(cid:12) O d, d − d +42 ( A, B ) (cid:12)(cid:12)(cid:12) (cid:16) by (173) (cid:17) ≥ c X B ∈R X B ∈N d ( A,B ,C ) (cid:12)(cid:12)(cid:12) O d, d − d +42 ( A, B ) (cid:12)(cid:12)(cid:12) ≥ c | A ∩ C | d , (cid:16) by (178), (179) (cid:17) and the claim holds by (176). ⋆ Assume that | A \ C | > d | A ∩ C | so that (177) leads to(180) |R| ≥ f +22 ) − | A \ C | > f +22 ) +1 d | A ∩ C | . From Lemma 4.6.i, we have that for all B ∈ R and B ∈ N d ( A, B , C ),(181) (cid:12)(cid:12)(cid:12) O d, d − d +42 ( A, B ) (cid:12)(cid:12)(cid:12) ≥ d d +2 | A ∩ C | . Thus (cid:12)(cid:12)(cid:12) O d, d − d +42 ( A ) (cid:12)(cid:12)(cid:12) ≥ c X B ∈N d ( A ) (cid:12)(cid:12)(cid:12) O d, d − d +42 ( A, B ) (cid:12)(cid:12)(cid:12) (cid:16) by (173) (cid:17) ≥ c X B ∈R X B ∈N d ( A,B ,C ) (cid:12)(cid:12)(cid:12) O d, d − d +42 ( A, B ) (cid:12)(cid:12)(cid:12) ≥ c | A ∩ C | d , (cid:16) by (178), (180), (181) (cid:17) and the claim holds by (176).Therefore in any case c and c work. (cid:3) We prove Theorem 1.6.
Proof. (Theorem 1.6).
Take a line L in R , B ∈ P ( d +12 )( R \ L ) such that there isno element of C ≤ d − which contains B , and B ∈ P m − ( d +12 )( L ). Set A := B ∪ B .Theorem 2.1 implies that for any C ∈ C ≤ d such that L is not a component of C ,(182) | L ∩ C | ≤ d. Now assume that A is contained in a curve C ∈ C ≤ d . Since d < m − (cid:0) d +12 (cid:1) = | A ∩ L | , (182) implies that C contains L . However, if C contains L , then B = A \ L ⊆ C \ L , and therefore there is an element of C ≤ d − which contains B contradicting the assumption. Thus A is not contained in a curve C ∈ C ≤ d , andthis shows i).Set η : O d, d − d +42 ( A ) −→ P d ( A ∩ L ) , η ( C ) = A ∩ L ∩ C. We show that η is well defined. Since | A ∩ L | = m − (cid:0) d +12 (cid:1) > d − d +42 − (cid:0) d +12 (cid:1) , wehave that C cannot contain L for any C ∈ O d, d − d +42 ( A ); hence (182) yields that | A ∩ L ∩ C | ≤ | L ∩ C | ≤ d . On the other hand, for any C ∈ O d, d − d +42 ( A ), C isdetermined by A so | A ∩ C | ≥ (cid:0) d +22 (cid:1) −
1, and hence | A ∩ L ∩ C | = | A ∩ C | − | ( A ∩ C ) \ L | ≥ (cid:18) d + 22 (cid:19) − − (cid:18) d + 12 (cid:19) = d. Thus η is well defined. Finally, η in injective. Indeed, take C , C ∈ O d, d − d +42 ( A )such that η ( C ) = η ( C ). Since C and C are determined by A , note that | A ∩ C | , | A ∩ C | ≥ (cid:0) d +22 (cid:1) − | ( A ∩ C ) \ L | , | ( A ∩ C ) \ L | ≥ (cid:0) d +12 (cid:1) . Thereby( A ∩ C ) \ L = ( A ∩ C ) \ L = B , and the equality η ( C ) = η ( C ) implies that A ∩ C = A ∩ C . Since C and C are determined by A , the previous equalityyields C = C . Finally, because η is injective, (cid:12)(cid:12)(cid:12) O d, d − d +42 ( A ) (cid:12)(cid:12)(cid:12) ≤ (cid:18) | A ∩ L | d (cid:19) = (cid:18) | A | − (cid:0) d +12 (cid:1) d (cid:19) , and this shows ii). (cid:3) We prove Theorem 1.7.
N THE MINIMUM NUMBER OF HIGH DEGREE CURVES CONTAINING FEW POINTS 43
Proof. (Theorem 1.7).
We prove that c = d d +9 d +16 max { c ( d ) , c ( d ) , } and c = d d +6 (2 d )4 d +16 (2 d +3 )! work. Write c := (cid:0) n +1 − ( d +22 )( d +22 ) − (cid:1) and c := c d d +2 (2 d +3 )!. Let B ∈ N d ( A ). Take a ∈ A such that δ d ( A, B ) = (cid:12)(cid:12)(cid:12) ϕ − d,B ( ϕ d,B ( a )) (cid:12)(cid:12)(cid:12) . Since π − V d ( B ) ( ϕ d,B ( a )) is a (cid:0) d +22 (cid:1) − R ( d +22 ) − , we havethat for any s ∈ ψ d ( A ) \ π − V d ( B ) ( ϕ d,B ( a )), the flat H s generated by π − V d ( B ) ( ϕ d,B ( a ))and s is a hyperplane. Then, by Lemma 2.5, ψ − d ( H s ) is an element of C d con-taining ϕ − d,B ( ϕ d,B ( a )) ∪ D d ( B ). Now, since A is not contained in an element of C d , Remark 2.6 implies that A is not contained in an element of C ≤ d . Therefore,by Remark 2.4.ii, ψ d ( A ) is not contained in a hyperplane of R ( d +22 ) − . Thus, since π − V d ( B ) ( ϕ d,B ( a )) is a (cid:0) d +22 (cid:1) − s , s ∈ ψ d ( A ) \ π − V d ( B ) ( ϕ d,B ( a )) and ψ − d ( H s ) = ψ − d ( H s ). Since ϕ − d,B ( ϕ d,B ( a )) ∪ D d ( B ) is contained in ψ − d ( H s ) and ψ − d ( H s ) and they are in C d , the assumption on n yields that | ϕ − d,B ( ϕ d,B ( a )) ∪ D d ( B ) | < n , and hence(183) δ d ( A, B ) + | B | ≤ δ d ( A, B ) + | D d ( B ) | = | ϕ − d,B ( ϕ d,B ( a )) ∪ D d ( B ) | < n. From (183), note that 2 δ d ( A, B ) + | D d ( B ) | ≤ n + 1 − (cid:0) d +22 (cid:1) , and hence(184) O d, δ d ( A,B )+ | D d ( B ) | ( A, B ) ⊆ O d, n +1 − ( d +22 )( A, B ) . For all C ∈ O d, n +1 − ( d +22 )( A ), we have that | A ∩ C | ≤ n + 1 − (cid:0) d +22 (cid:1) ; thus there areat most c subsets B ∈ N d ( A ) such that C ∈ O d, n +1 − ( d +22 )( A, B ), and this yields (cid:12)(cid:12)(cid:12) O d, n +1 − ( d +22 )( A ) (cid:12)(cid:12)(cid:12) ≥ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) [ B ∈N d ( A ) O d, n +1 − ( d +22 )( A, B ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≥ c X B ∈N d ( A ) (cid:12)(cid:12)(cid:12) O d, n +1 − ( d +22 )( A, B ) (cid:12)(cid:12)(cid:12) . (185)If A is d -regular, then we proceed exactly as in Case i) of Theorem 1.5 to concludethat (cid:12)(cid:12)(cid:12) O d, n +1 − ( d +22 )( A ) (cid:12)(cid:12)(cid:12) ≥ c d d +2 (2 d +3 !) | A | ( d +22 ) − . From now on, we assume that A is not regular. This means that there is C ∈ C ≤ d such that | A ∩ C | ≥ d +8 | A | . In so far as, C has at most d irreducible components,there are f ∈ [0 , d −
1] and C ∈ C d − f such that(186) | A ∩ C | ≥ d d +8 | A | . Notice that f = 0; otherwise, f > B ∈ P n ( A ∩ C ) and any curve C ∈ C f , we get a curve C ∪ C ∈ C ≤ d such that B ⊆ C ∪ C contradicting theassumption about n . Then, since f = 0, Lemma 3.9 imples that for all b ∈ A \ C ,(187) |N d ( A, { b } , C ) | ≥ d +3 ! | A ∩ C | ( d +22 ) − − ( f +22 ) = 12 d +3 ! | A ∩ C | ( d +22 ) − . From Lemma 4.6.i, we have that for all b ∈ A \ C and B ∈ N d ( A, { b } , C ),(188) (cid:12)(cid:12)(cid:12) O d, n +1 − ( d +22 )( A, B ) (cid:12)(cid:12)(cid:12) ≥ d d +2 | A ∩ C | . Hence (cid:12)(cid:12)(cid:12) O d, n +1 − ( d +22 )( A ) (cid:12)(cid:12)(cid:12) ≥ c X B ∈N d ( A ) (cid:12)(cid:12)(cid:12) O d, n +1 − ( d +22 )( A, B ) (cid:12)(cid:12)(cid:12) (cid:16) by (185) (cid:17) ≥ c X b ∈ A \ C X B ∈N d ( A, { b } ,C ) (cid:12)(cid:12)(cid:12) O d, n +1 − ( d +22 )( A, B ) (cid:12)(cid:12)(cid:12) ≥ c | A ∩ C | ( d +22 ) − , (cid:16) by (187), (188) (cid:17) and the claim holds by (186). (cid:3) Finally, we complete the proof of Theorem 1.8.
Proof. (Theorem 1.8).
Let C ∈ C d be irreducible and H be a hyperplane in R ( d +22 ) − such that C = ψ − d ( H ). Choose a ∈ R \ C . We construct recursivelya set S ∈ P m − ( ψ d ( R ) ∩ H ) such that dim R = | R | − R ∈ P ( S ) withdim R < (cid:0) d +22 (cid:1) −
2. Take s ∈ ψ d ( R ) ∩ H and write S := { s } . Now assume thatfor some i ∈ [1 , m − S i ∈ P i ( ψ d ( R ) ∩ H ) such thatdim R = | R | − R ∈ P ( S i ) with dim R < (cid:0) d +22 (cid:1) −
2. Let F i be the collectionof all flats F generated by the subsets of S i such that dim F < (cid:0) d +22 (cid:1) −
2. Since S i isfinite, F i is finite. On the other hand, for each F ∈ F i , there exists a hyperplane G in R ( d +22 ) − such that G = H and F ⊆ G ∩ H ; in particular, C is not a componentof ψ − d ( G ). Applying Theorem 2.1 to the curves ψ − d ( G ) and ψ − d ( H ) = C , wehave that | ψ − d ( G ) ∩ ψ − d ( H ) | ≤ d and thus(189) | F ∩ ( ψ d ( R ) ∩ H ) | ≤ | G ∩ ( ψ d ( R ) ∩ H ) | = | ψ − d ( G ) ∩ ψ − d ( H ) | ≤ d . From (189), we have that (cid:0)S F ∈F i F (cid:1) ∩ (cid:0) ψ d ( R ) ∩ H (cid:1) is finite. Since ψ d ( C ) ⊆ ψ d ( R ) ∩ H is not finite, ( ψ d ( R ) ∩ H ) \ S F ∈F i F = ∅ and we choose s i +1 inthis difference. Make S i +1 := S i ∪ { s i +1 } , and notice that S i +1 ⊆ ψ d ( R ) ∩ H , | S i +1 | = i + 1 and dim R = | R | − R ∈ P ( S i +1 ) with dim R < (cid:0) d +22 (cid:1) − s i +1 S F ∈F i F ). In this way we construct S , S , . . . , S m − and S := S m − has the desired properties. Set A := { a } ∪ ψ − d ( S ).Theorem 2.1 implies that for any C ∈ C ≤ d such that C * C ,(190) | C ∩ C | ≤ d . Since d < m − | A ∩ C | , if A is contained in a curve C ∈ C ≤ d , then C iscontained in C by (190). Nevertheless, in so far as C ∈ C d and C ∈ C ≤ d , we havethat C = C but this is impossible since a C . This proves i).Take B ∈ P n ( A ) and write B := B ∩ C . We claim thatdim ψ d ( B ) ≥ (cid:0) d +22 (cid:1) − B = B (cid:0) d +22 (cid:1) − B = B . (191)Indeed, if B = B , then | ψ d ( B ) | = | B | ≥ (cid:0) d +22 (cid:1) −
1; thus, if dim ψ d ( B ) < (cid:0) d +22 (cid:1) − ψ d ( B ) < (cid:18) d + 22 (cid:19) − ≤ | ψ d ( B ) | − , N THE MINIMUM NUMBER OF HIGH DEGREE CURVES CONTAINING FEW POINTS 45 which is impossible by the construction of S = ψ d ( A ∩ C ). If B = B , then B = B ∪ { a } so | ψ d ( B ) | = | B | − ≥ (cid:0) d +22 (cid:1) −
2; if dim ψ d ( B ) < (cid:0) d +22 (cid:1) −
3, thendim ψ d ( B ) < (cid:18) d + 22 (cid:19) − ≤ | ψ d ( B ) | − , which is impossible by the construction of S , and this proves (191). Now, if B = B ,then (191) implies that there is at most one hyperplane which contains ψ d ( B );hence, by Remark 2.4.ii, there is at most one element C ∈ C ≤ d such that B ⊆ C .If B = B , then (191) leads to dim ψ d ( B ) ≥ (cid:0) d +22 (cid:1) −
3. Since a C , we havethat ψ d ( a ) H , and since H ⊇ ψ d ( B ), we conclude that dim ψ d ( B ) ≥ ψ d ( B ) ≥ (cid:0) d +22 (cid:1) −
2. As in the previous case, we conclude that there is at mostone element C ∈ C ≤ d such that B ⊆ C , and this completes the proof of ii).Finally we show iii). Set η : O d, n +1 − ( d +22 )( A ) −→ P ( d +22 ) − ( A ∩ C ) , η ( C ) = A ∩ C ∩ C. We show that η is well defined. Take C ∈ O d, n +1 − ( d +22 )( A ). Since | A ∩ C | = m − > n + 1 − (cid:0) d +22 (cid:1) −
1, we have that C cannot contain C . Since C isdetermined by A , we have that | A ∩ C | ≥ (cid:0) d +22 (cid:1) −
1. We prove by contradictionthat(192) | A ∩ C ∩ C | ≤ (cid:18) d + 22 (cid:19) − . Assume that(193) | ψ d ( A ∩ C ∩ C ) | = | A ∩ C ∩ C | ≥ (cid:18) d + 22 (cid:19) − . By the construction of S , (193) yields that(194) dim ψ d ( A ∩ C ∩ C ) ≥ (cid:18) d + 22 (cid:19) − ψ d ( C );however, since C does not contain C , we have that ψ d ( C ∩ C ) is contained in twodifferent hyperplanes of R ( d +22 ) − and thereforedim ψ d ( A ∩ C ∩ C ) ≤ dim ψ d ( C ∩ C ) ≤ (cid:18) d + 22 (cid:19) − , which contradicts (194) and proves (192). On the other hand, since C is determinedby A , | A ∩ C | ≥ (cid:0) d +22 (cid:1) − | A ∩ C ∩ C | ≥ | A ∩ C | − |{ a }| ≥ (cid:18) d + 22 (cid:19) − η is well defined; furthermore these inequalities force that | A ∩ C ∩ C | = | A ∩ C | − |{ a }| so a ∈ C for all C ∈ O d, n +1 − ( d +22 )( A ). Hence, for any C , C ∈ O d, n +1 − ( d +22 )( A )such that A ∩ C ∩ C = A ∩ C ∩ C , we get that A ∩ C = A ∩ C . In so far as C and C are determined by A , we conclude that C = C . This yields that η isinjective so |O d, n +1 − ( d +22 )( A ) | ≤ (cid:18) | A ∩ C | (cid:0) d +22 (cid:1) − (cid:19) = (cid:18) | A | − (cid:0) d +22 (cid:1) − (cid:19) concluding the proof of iii). (cid:3) References [1] S. Ball,
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