Two Remarks on Blocking Sets and Nuclei in Planes of Prime Order

Transcription

1 Designs, Codes and Cryptography, 10, 9 39 (1997) c 1997 Kluwer Academic Publishers, Boston. Manufactured in The Netherlands. Two Remarks on Blocking Sets and Nuclei in Planes of Prime Order ANDRÁS GÁCS Department of Computer Science, Eötvös Loránd University, Budapest H-1088, Budapest, Múzeum krt HUNGARY PÉTER SZIKLAI Department of Computer Science, Eötvös Loránd University, Budapest H-1088, Budapest, Múzeum krt HUNGARY TAMÁS SZŐNYI Department of Computer Science, Eötvös Loránd University, Budapest H-1088, Budapest, Múzeum krt HUNGARY Communicated by: D. Jungnickel Received February 15, 1995; Revised October, 1995; Accepted October 10, 1995 Abstract. In this paper we characterize a sporadic non-rédei type blocking set of PG(, 7) having minimum cardinality, and derive an upper bound for the number of nuclei of sets in PG(, q) having less than q + 1 points. Our methods involve polynomials over finite fields, and work mainly for planes of prime order. Keywords: blocking set, nucleus, polynomials. 1. Introduction In this paper we study two problems concerning blocking sets and nuclei in planes of prime order. The two objects are in close relation, see Blokhuis [3], Bruen [8]. The problems considered here are not directly related, but the methods used in the proofs are similar: polynomials over finite fields are used. The first problem is about blocking sets. A point-set B in a projective plane is called a blocking set if it intersects every line but contains no line. A blocking set is called minimal (irreducible) if it is minimal subject to set inclusion. For a blocking set (or more generally, for any pointset) a line is called an i-secant, if it intersects the blocking set in exactly i points. Instead of 1-secant also the term tangent will be used. So a blocking set is minimal if it has a tangent at each of its points. It is easy to prove that a line intersects a blocking set B in at most B q points. If there is a line L with B L = B q, then the blocking set is said to be of Rédei type (or maximal type). For the size of a blocking set of PG(, p) (the Desarguesian plane of order p), p a prime, Blokhuis [] recently proved that B 3(p+1)/. Blokhuis proof is algebraic; combinatorial arguments only give B q+ q+1 (Bruen [7]), but they work for any plane of order q. There are blocking sets of Rédei type satisfying B =3(q+1)/ (see [9, Chap. 13.4], [1, Par. 36] and [4]). For planes of prime order Lovász and Schrijver [10] characterized blocking sets of Rédei

2 30 GÁCS, SZIKLAI AND SZŐNYI type satisfying B =3(q+1)/. There is only one example known of a blocking set of size 3(q + 1)/ which is not of Rédei type. The affine plane of order 3 can be embedded in PG(, 7) (as the points of inflexion of a non-singular cubic). The 1 lines of this affine plane cover each point of PG(, 7), hence in the dual plane we obtain a blocking set of size 1 = 3(7 + 1)/. The lines of the dual plane corresponding to the points of the original affine plane intersect the blocking set in 4 = points, so this is a blocking set of almost Rédei type. The first part of this paper is devoted to the characterization of this particular blocking set. Our main result in this direction is the following: if B is a blocking set of size 3(p + 1)/ in PG(, p), p prime, and there is a line intersecting B in (p + 1)/ points, then p 7. Moreover, for p < 7 the blocking sets are of Rédei type (with respect to another line) and for p = 7 there is a unique example being isomorphic to the one described in the previous paragraph. Our methods also yield a new proof of Lovász, Schrijver s characterization of minimum cardinality blocking sets of Rédei type. The second part of this paper is about nuclei. The nucleus of a (q + 1)-set S was first defined by Mazzocca as a point out of S such that every line through it meets S in exactly one point. Blokhuis, Wilbrink [5] proved that a (q + 1)-set cannot have more than q 1 nuclei if it is not a line. Recently, Blokhuis [3] generalized the notion of a nucleus for sets of size greater than q + 1 in the following way: the point P / S is a nucleus of S,ifevery line through P intersects S. He proved that the number of nuclei of a (q + k)-set is at most k(q 1) and used this bound to derive bounds for blocking sets in the affine plane AG(, q). Our aim is to extend the definition of nuclei to sets having less than q + 1 points and bound the number of nuclei. Let S be a pointset of size q + 1 k. The point P / S is a nucleus if the lines through P intersect S in at most one point. For planes of prime order we prove (see Theorem 3..) that the number of nuclei of a set of size q + 1 k is smaller than (k + 1)(q 1), which is the symmetric version of Blokhuis result [3].. Blocking Sets Let B be a minimal blocking set in PG(, q), B =q+kand suppose that there is a line L with B L =k 1. The line L will play the role of the line at infinity; denote by D the set B L and by U the set B \ L AG(, q). We use cartesian coordinates in AG(, q) with points denoted by (x, y). Infinite points can be identified with slopes. The infinite point (m) corresponds to the parallel class of lines with equation mx + b y = 0. The aim of this Section is to characterize blocking sets of almost Rédei type (i.e. a blocking set B for which there exists a line intersecting it in one point less than B q, the maximal possible intersection of a blocking set and a line). More precisely, the following theorem will be proved. THEOREM.1 Let q = p be a prime and B a blocking set of size p + k = 3(p + 1)/ in PG(, p). Moreover, suppose that there is a line intersecting B in exactly k 1 = p+1 points. Then p 7, and for p < 7, B is of Rédei type. For p = 7 there is a unique blocking set satisfying the conditions of the theorem.

3 TWO REMARKS ON BLOCKING SETS AND NUCLEI IN PLANES OF PRIME ORDER 31 Let us begin the proof with the following lemma, which holds for arbitrary q. LEMMA. Let B be a minimal blocking set in PG(, q) with B =q+k and suppose that there is a line L intersecting B in exactly k 1 points. Then there is a point A / B such that every line joining A to a point of L \ B contains two points of B. Hence k (q + 3)/. Proof. We use the notation introduced above. Consider a point (direction) y of L \ B. The lines through (y) should be blocked by U, hence from (y) we see at least one point of U on each line with slope y, and there is just one -secant to U with slope y. Let U ={(a i,b i ) :i=1,...q +1}, and define the polynomial q+1 q+1 H(x, y) = (x + ya i b i ) = h j (y)x q+1 j. (1) i=1 Let H y (x) := H(x, y). Note that deg(h j ) j. Let us suppose for the proof of this lemma that ( ) D. Consider a direction (y) / D. Since there is at least one point of U on each line with slope y, H y (x) is divisible by x q x for such y s. For (y) / Dthe quotient H y (x)/(x q x) = x+h 1 (y) is linear, x+ya b, say. Let us interpret this fact geometrically: if the point A : (a, b) is joined to the points of L \ B, then these lines contain at least two (and hence exactly two) points of U. IfAwere a point of B then we could delete it, hence B would not be minimal. If A / B then q + k = B k 1+(q+ k)=q+3 k, whence k (q + 3)/. Remark. The Lemma can also be proved using the method of Proof of Proposition of Blokhuis [1]. The present proof follows the ideas of [14]. The more general situation when there is a line intersecting B in slightly less than B qpoints, is studied in [6]. From now on we concentrate on the extremal case of Lemma. and for the sake of simplicity deal with planes of prime order. In other words, suppose q = p is a prime and k = (p + 3)/. In this case, by Lemma., the -secants of B through points of L \ B meet in a common point A(a, b). By a suitable linear transformation, one can achieve the following situation: 1. a = b = 0,. a p+1 = 0, and finally that 3. {(a i, b i ) : i = 1,...,p}={(x,f(x)) : x GF(p)}, where f (x) = c p 1 x p c x +1 is a polynomial over GF(p). In other words, condition 3 just means that ( ) / D. Note that these conditions yield deg(h j ) j 1 for j p, since the coefficient of y j in h j is the j-th elementary symmetric polynomial in the a i s, which is 0 by conditions and 3. Also note that b p+1 0 by condition 1 and b p+1 1 by condition 3. On the other hand, by the proof of Lemma., H y (x) = x p+1 x whenever y / D, y, so we have at least (p 1)/ roots for each h j ( j p ). This means that h 1 (y) = h (y) = =hp 1(y)=0 identically.

4 3 GÁCS, SZIKLAI AND SZŐNYI We need an algebraic lemma which will help us to translate information about the polynomial H(x, y) into information about the polynomial f (x). It resembles the Newton Waring Girard formulae relating power-sums and elementary symmetric polynomials. First of all some notation: for a 1,...,a t ;b 1,...,b t GF(p), σ k,l (a 1,...,a t ;b 1,...,b t )will denote the sum of all products of type a i1 a ik b j1 b jl, where {i 1,...,i k }and { j 1,..., j l } are disjoint subsets of {1,...,t} of cardinality k and l respectively. Note that this is essentially (( 1) l times) the coefficient of x t k l y k in the polynomial H(x, y) = t i=1 (x + a i y b i ) which we are using with t = p + 1. Note also that σ 0,l (a 1,...,a t ;b 1,...,b t )= σ l (b 1,...,b t ) is the l-th elementary symmetric polynomial of the b i -s and similarly σ k,0 (a 1,...,a t ;b 1,...,b t )=σ k (a 1,...,a t ). LEMMA.3 Let 1 s p be a fixed integer, a 1,...,a t ;b 1,...,b t GF(p). If σ k,l (a 1,...,a t ;b 1,...,b t ) = 0 for all k, l 0, 1 k + l s, then t i=1 a i k l b i = 0 for all k, l 0, 1 k + l s. (We define 0 0 = 1, this may occur in the power sum whenever k or l is zero.) Proof. For simplicity we shall write σ k,l instead of σ k,l (a 1,...a t ;b 1,...,b t ). For k = 0, by the fundamental theorem of symmetric polynomials, bi l is a polynomial of σ 0,1,σ 0,,...,and σ 0,l. By choosing b 1 = =b t =0 one sees that the constant term of this polynomial is zero, so by the conditions of the lemma, i bl i = 0. The l = 0 case is similar. In the general case we use induction on k + l. By the above case one may assume that k, l > 0 are fixed integers. Introduce the following function: N(m, n) = ai m bi n a i 1 a ik m b j1 b jl n i;i 1,...,i k m ; j 1,..., j l n all distinct where 0 m k, 0 n l. (For any other values of m and n we define N(m, n) to be zero.) It is easy to see that N(1, 0) = N(0, 1) = k!l!σ k,l = 0 and we want to prove N(k, l) = 0. By the induction hypothesis i i 1,...,i k m ; j 1,..., j l n ai m bi n = (ai m 1 bi n 1 + +aj m l n b n j l n ), for any 1 m + n < k + l. From this it follows for all 1 m + n < k + l that N(m, n) = a i1 b jl n ai m bi n i 1,...,i k m ; j 1,..., j l n i i 1,..., j l n all distinct = a i1 b jl n (ai m 1 bi n 1 + +aj m l n b n j l n ) i 1,...,i k m ; j 1,..., j l n all distinct = (k m)n(m +1,n) (l n)n(m,n +1). For m = k or n = l the recursion still works, since N(k + 1, n) and N(m, l + 1) were defined to be zero. Using the above recursion starting from N(k, l) and going downward, we get N(k 1, l) = N(k, l), N(k, l) =!N(k, l) and in general it s easily seen by induction on k+l m n,

5 TWO REMARKS ON BLOCKING SETS AND NUCLEI IN PLANES OF PRIME ORDER 33 that for 1 m + n k + l ( 1)k+l m n N(m, n) = N(k, l). (k + l m n)! This implies that N(k, l) = ( 1)k+l 1 N(1, 0) = 0, so the proof is complete. (k+l 1)! Lemma.3 will be used in conjunction with the following easily proved proposition: PROPOSITION.4 For any prime power q and polynomial f (x) = c q 1 x q 1 + +c 0 over GF(q) we have x GF(q) f(x) = c q 1. Now we are ready to prove Theorem.1. Proof of Theorem.1: Recall that the affine part of our blocking set consists of the point (0, b p+1 ) (where b p+1 0, 1), plus the graph of a polynomial f (x) = c p 1 x p c x +1. The conditions of Lemma.3 are satisfied with s = (p 1)/ and t = p + 1. Since a p+1 = 0, it follows that for any k + l (p 1)/, { 0, if k > 0; x k f(x) l = b l p+1, if k = 0. x GF(p) Using this with l = 1, k = 0, 1,...,(p 3)/, one gets c p 1 = b p+1, c p = =cp+1 = 0. For p = 3 there is just one possibility: f (x) = x + 1, b 4 =, which yields a blocking set of Rédei type: for example the line y = x + contains 3 points of B. For p 5 we can use our equation also with l =, k = 0 to obtain b p+1 = (c p 1 + c p 1 +c p 1 ), i.e. c p 1 = b p+1. For p = 5 there are four possibilities for f : f (x) = x 4 + x + 1, f (x) = x 4 x + 1, f (x) = 3x 4 + x + 1or f(x)=3x 4 x +1, while b p+1 =,, 3 and 3, respectively. It is easy to check that these lead to blocking sets, but they are all of Rédei type with respect to the line y =, y =, y = 1 and y = 1, respectively. For p 7 our equation with l =, k = 1,,...,(p 5)/ can be used to derive = =c =0. Now for k = 0, l = 3weget c p 3 b 3 p+1 = (c3 p 1 +3c p 1 +3c p 1c p 1 +3c p 1 +3c p 1), i.e. (using c p 1 = c p 1 )c p 1 (c p 1 +1)=0. Since c p 1 = b p+1 0, c p 1 = 1 follows. If p > 7 then our equation with l = 4, k = 0gives b 4 p+1 = (c4 p 1 +4c3 p 1 +6c p 1 c p 1 +6c p 1 +1c p 1c p 1 + c 4 p 1 + 4c p 1 + 6c p 1 ).

6 34 GÁCS, SZIKLAI AND SZŐNYI Putting 1 in place of c p 1 and b p+1 ; and in place of c p 1, we get a contradiction, showing that p > 7 cannot occur. For p = 7 there are two possibilities: f (x) = x 6 +3x 3 +1or f(x)= x 6 3x 3 +1 with b 8 = 1in both cases. It is easy to check, that they really give rise to blocking sets with the desired properties. It is also obvious, that they are essentially the same: the reflection in the line y = 0 takes one blocking set to the other. This remark completes the classification. Since there is just one blocking set with the prescribed properties, the blocking set described in the proof of Theorem.1 has to be isomorphic to the one described in the Introduction. Remark. Lemma.3 can also be used to classify all minimal blocking sets of Rédei type in PG(, p), p prime (cf. [10]). Similarly to the case just worked out, now U ={(a i,b i ): i =1,...,p}is the affine part of B, while the part at infinity consists of those y L for which H(x, y) := p i=1 (x a i y + b i ) x p x. We will show that B p+ p+3 implies B =p+ p+3 and after a suitable linear transformation U can be written as {(x, x p+1 ) : x GF(p)}. After linear transformation U can be written as U ={(x,f(x)) : x GF(p)}, where f (x) = x s + c s x s + +c x is a polynomial over GF(p).(Fors=3this means f (x) = x 3.) Using Lemma.3, since H(x, y) = x p x for at least p 3 y-s, x GF(p) xk f(x) l = 0 for all k + l p 3 l =, k = 1,,..., f(x)= x p+1 p+1. Putting l = 1, k = 0, 1,...yields s follows indeed..ifs= p+1, then from Finally we prove that s p 1 leads to contradiction. There are integers l, k, s.t. ls+k = p 1, 0 k < s. Since s p 1, l. A simple calculation shows that for s 3, l + k p 3 automatically holds, so x k f (x) l = 0 which is a contradiction (it should be 1 times the coefficent of x s, i.e. 1.) s = is a contradiction, since f (x) = x leads to a blocking set containing p points from the line at infinity. 3. Nuclei of Sets Definition 3.1. Given a set of points S in the projective plane PG(,q), q = p e, q >, p a prime, the point P is a nucleus of S if each of the lines through P contains at most one point of S. (It implies that P is not in S.) Note that if S > q + 1 then there exists no nucleus of S in this sense, since in this case at least one of the q + 1 lines through P would contain at least two points of S. This makes the notion of nucleus a whole : first it was introduced by Mazzocca for S =q+1 (see [5]), then extended for S q+1 by Blokhuis [3]. Let S =q+1 r,r 0, and n the number of infinite points in S (n implies that there are no ideal nuclei of S). Sometimes we shall write a = q + 1 r n for the

7 TWO REMARKS ON BLOCKING SETS AND NUCLEI IN PLANES OF PRIME ORDER 35 number of the affine points of S. The main result of this Section is the following bound on the number of nuclei. THEOREM 3.. If S is not contained in a line and ( q+1 n) r+1 0 (mod p), then N(S) (r+1)(q 1). Remark. This bound is sharp as the following example shows: take q r points on a line and one point off the line. Note that this configuration is a degenerate projective subplane. Remark. The number theoretic condition on the binomial coefficient can be dropped in many cases, cf. Corollaries 3.5, 3.6. and Theorem 3.8. In order to prove Theorem 3. some lemmas are required. Let F be a field containing all the (q + 1)-st roots of unity, E ={α F:α q+1 =1}, E =q+1. Let A be a subset of E with A =q+1 r. For any set H E denote by σ i (H) the i-th elementary symmetric polynomial of the elements of H if H i 1; σ 0 (H) = 1; and if i > H or i < 0 let σ i (H) = 0. The next lemma gives a relation among the elementary symmetric polynomials of A using the fact that its elements are mutually distinct. LEMMA 3.3. Let A E, A =q+1 r, and let σ i = σ i (A), σ i Proof. r σ k σ r+1 k = 0. We have X q+1 1 = a A(X a) α (E \ A) (X α) = (X q+1 r σ 1 X q r + +( 1) k σ k X q+1 r k +( 1) q+1 r σ q+1 r ) (X r σ 1 Xr 1 + = σ i (E \ A). Then +( 1) k σk Xr k...+( 1) r σr ). The coefficient of X q+1 u is (for u = 1,,...,q) 0=( 1) u u σk σ u k. (u) From (u), (u = 1,...,q)σu can be computed explicitly in terms of the σ j s(0 j u 1): u 1 σu = σk σ u k. (u )

8 36 GÁCS, SZIKLAI AND SZŐNYI Finally, as σr+1 = 0, we have r σk σ r+1 k = 0, (r + 1 ) completing the proof of the Lemma. Note that this is the relation among the σ i (A)s of the smallest possible degree. LEMMA 3.4 For j 1 j ( )( ) a + k 1 a ( 1) k = 0. k j k Proof. Since (x + 1) a = ( ) a + i 1 ( 1) i x i, i i=0 and (x + 1) a = i=0 ( ) a x i, i it follows 1 = (x + 1) a (x + 1) a = x j j ( )( ) a + k 1 a ( 1) k, k j k which immediately gives the Lemma. Now we are ready to prove the main theorem of this Section. Proof of Theorem 3.: We can associate with the affine points of the plane the elements of GF(q ), and with the infinite points the (q + 1)-st roots of unity. If U is a point of PG(, q) then the corresponding element of GF(q ) will be denoted by u (the same letter). The points X, Y AG(,q) and the infinite point I are collinear iff for the corresponding elements (x y) q 1 = i. Without loss of generality one can suppose that S contains n infinite points, and also that n < S, since S is not contained in a line. Therefore S ={s 1,s,...,s q+1 r n,ε 1,ε,...,ε n }where {s 1,...,s q+1 r n,ε 1,...,ε n } GF(q ) and ε q+1 i = 1 for i = 1,...,n(the s i s correspond to the affine, the ε i s to the infinite points). An affine point is a nucleus if and only if for the corresponding x GF(q )the elements (x s 1 ) q 1,(x s ) q 1,...,(x s q+1 r n ) q 1,ε 1,ε,...,ε n are pairwise distinct (q +1)-

9 TWO REMARKS ON BLOCKING SETS AND NUCLEI IN PLANES OF PRIME ORDER 37 st roots of unity. Lemma 3.3 can be used (with A ={(x s 1 ) q 1,(x s ) q 1,...,(x s q+1 r n ) q 1,ε 1,ε,...,ε n }) to deduce that ( ) r σk σ r+1 k ((x s 1 ) q 1,(x s ) q 1,...,(x s q+1 r n ) q 1,ε 1,ε,...,ε n )=0 when x is a nucleus. All the nuclei are roots of this polynomial, hence the number of nuclei is bounded by the degree of the polynomial, if it is not identically zero. It remains to show that the polynomial is of degree (r + 1)(q 1), and prove that the coefficient of x (r+1)(q 1) is non-zero. For u = 1,...,r u 1 σu = σj σ u j, (u ) from which by induction follows, substituting (x s 1 ) q 1,...,(x s q+1 r n ) q 1,ε 1,...,ε n, that deg(σi ) = i(q 1), considering σ i as a polynomial in x. Let z i denote the coefficient of x i(q 1) in σ i ((x s 1 ) q 1,...,(x s q+1 r n ) q 1,ε 1,...,ε n ). Let a = q + 1 r n, then for i = 1,...,r i 1 ( ) a z i = z j. i j On the other hand, by Lemma 3.4 for i = 1,...,r ( ) a+i 1 z i =( 1) i. i Finally, the leading coefficient of the polynomial on the left-hand side of equation (r + 1 ) is r ( ) a r ( )( ) a+j 1 z j = ( 1) j a r + 1 j j r + 1 j and using Lemma 3.4 again, one sees that this is ( ) q + 1 n ( 1) r+1 0, r + 1 by the assumption of Theorem 3.. So the polynomial is of degree (r + 1)(q 1) exactly, it is not the zero polynomial, hence it has at most (r + 1)(q 1) roots, and this is an upper bound for the number of nuclei.

10 38 GÁCS, SZIKLAI AND SZŐNYI COROLLARY 3.5 If q = p is prime, then the condition for the binomial coefficient is satisfied automatically, hence for a set S of p + 1 r non-collinear points the number of nuclei is at most (r + 1)(p 1) in PG(, p). COROLLARY 3.6 If there exists a line L intersecting S in exactly two points (i.e. n = ), then the condition for the binomial coefficient is satisfied automatically, hence for a set S of q + 1 r non-collinear points the number of nuclei is at most (r + 1)(q 1) in PG(, q). The next proposition characterizes small sets having many nuclei. THEOREM 3.7 If N(S) > 1 (q(q 1) +r(r 1)) then S (with the lines containing at least two points of S) is a (possibly degenerate) subplane. In particular, if S =q +1 r, S < q and S has exactly (r + 1)(q 1) nuclei then S is a degenerate subplane. Proof. Suppose S is not a (possibly degenerate) projective plane. Then S determines more lines than the number of its points (this is the classical theorem of de Bruijn and Erdős, see e.g. [11]), so at least q + r lines. In the plane k lines cover at least k(q + 1) ( k ) points, so the lines of S cover at least 1 (q + 3q + r r + ) points, hence the number of nuclei is at most q + q (q + 3q + r r + ) = 1 (q(q 1) + r(r 1)). But if r > q q then (r + 1)(q 1) > 1 (q(q 1)+r(r 1)), contradiction. An easy calculation shows that a proper subplane can not have so many nuclei. Remark. The results of this Section can be extended to spaces instead of planes and also to multiple nuclei, see [13]. THEOREM 3.8 If q = p, p > prime, S =p (i.e. r = 1) then in Theorem 3. the condition on the binomial coefficient can be eliminated. Proof. We may assume that for every secant l: S l 0,1 (modp) (otherwise l can be chosen as the ideal line), and S l p. First we show that every secant is either a p or a (p + 1) secant. Suppose that l is a k secant, k p, A a point out of it. Then the union of lines through A meeting S l contains at least p(p 1) + 1 > p points of S, a contradiction. Now we state that there are either p or p + 1 secants through each point of S. Indeed, suppose there are at most p 1 through an A S, then the number of points of S on them is at most (p 1)p + 1 < p, contradiction. Suppose there are at least p + secants through A, then the number of points of S on them is at least (p + )(p 1) + 1 > p, contradiction.

11 TWO REMARKS ON BLOCKING SETS AND NUCLEI IN PLANES OF PRIME ORDER 39 Next we show that there exists at most one point, through which there are only p secants. Suppose that A, B S have this property, then the secants through any of them together with the line AB contain at most (p 1) + (p + 1) <p points of S. If there are p + 1 secants through every point of S then all of them are p secants, and S is an affine Baer subplane, which has only N(S) =p pnuclei. If through A S there are only p secants, then (p 1) of them are (p + 1) secants and one of them, say l, isap secant. Let B S \ l, and l a secant through A, different from l and AB. Consider the p + 1 lines through B and the points of S l. Each of them has at least p points from S, so in particular their intersection with l must belong to S (since there are only p secants through A). So p = l S =p +1, contradiction. Acknowledgment The authors acknowledge the financial support of OTKA T 1430, OTKA F-01630, OTKA T and OTKA grants. References 1. A. Blokhuis, Characterization of seminuclear sets in a finite projective plane, J. of Geometry, Vol. 40 (1991) pp A. Blokhuis, On the size of a blocking set in PG(, p), Combinatorica, Vol. 14 (1994) pp A. Blokhuis, On nuclei and affine blocking sets, J. Comb. Theory A, Vol. 67 (1994) pp A. Blokhuis, Blocking sets in desarguesian planes, Combinatorics: Paul Erdős is Eighty, Vol., János Bolyai Mathematical Society, Budapest (1996), pp A. Blokhuis and H. A. Wilbrink, A characterization of exterior lines of certain sets of points in PG(, q), Geom. Dedicata, Vol. 3 (1987) pp A. Blokhuis, R. Pellikaan and T. Szőnyi, Blocking sets of almost Rédei type, to appear in J. Comb. Theory A. 7. A. A. Bruen, Baer subplanes and blocking sets, Bull. Amer. Math. Soc., Vol. 76 (1970) pp A. A. Bruen, Nuclei of sets of q + 1 points in PG(, q) and blocking sets of Rédei type, J. Comb. Theory A, Vol. 55 (1990) pp J. W. P. Hirschfeld, Projective Geometries over Finite Fields, Oxford University Press, Oxford (1979). 10. L. Lovász and A. Schrijver, Remarks on a theorem of Rédei, Studia Scient. Math. Hungar., Vol. 16 (1981) pp K. Metsch, Linear spaces with few lines, Springer Lecture Notes in Math., (1991) p L. Rédei, Lückenhafte Polynome über endlichen Körpern, Birkhäuser Verlag, Basel (1970) (English translation: Lacunary Polynomials over Finite Fields, North Holland, Amsterdam, 1973). 13. P. Sziklai, Nuclei of Point Sets in PG(n,q), to appear in Discr. Math. 14. T. Szőnyi, On the number of directions determined by a set of points in an affine Galois plane, J. Comb. Theory A, vol. 74 (1996), pp