A small minimal blocking set in PG(n,p^t), spanning a (t-1)-space, is linear
aa r X i v : . [ m a t h . C O ] O c t A small minimal blocking set in PG( n, p t ),spanning a ( t − Peter Sziklai Geertrui Van de VoordeJanuary 1, 2019
Abstract
In this paper, we show that a small minimal blocking set withexponent e in PG( n, p t ), p prime, spanning a ( t/e − F p e -linear set, provided that p > t/e ) −
11. As a corol-lary, we get that all small minimal blocking sets in PG( n, p t ), p prime, p > t −
11, spanning a ( t − F p -linear, henceconfirming the linearity conjecture for blocking sets in this particularcase. Keywords:
Blocking set, linearity conjecture, linear set
In this section, we introduce the necessary background and notation. If V isa vector space, then we denote the corresponding projective space by PG( V ).If V has dimension n + 1 over the finite field F q , with q elements, q = p t , p prime, then we also write V as V( n + 1 , q ) and PG( V ) as PG( n, q ).A blocking set in PG( n, q ) is a set B of points such that every hyperplaneof PG( n, q ) contains at least one point of B . Such a blocking set is some-times called a 1 -blocking set , or a blocking set with respect to hyperplanes . Ablocking set B is called small if | B | < q + 1) / minimal if no propersubset of B is a blocking set.A point set S in PG( V ), where V = V( n + 1 , p t ), is called F q -linear ifthere exists a subset U of V that forms an F q -vector space for some F q ⊂ F p t ,such that S = B ( U ), where B ( U ) := {h u i F pt : u ∈ U \ { }} .
1e have a one-to-one correspondence between the points of PG( n, q h ) andthe elements of a Desarguesian ( h − D of PG( h ( n + 1) − , q ). Thisgives us a different view on linear sets; namely, an F q -linear set is a set S ofpoints of PG( n, q h ) for which there exists a subspace π in PG( h ( n + 1) − , q )such that the points of S correspond to the elements of D that have a non-empty intersection with π . We identify the elements of D with the points ofPG( n, q h ), so we can view B ( π ) as a subset of D , i.e. B ( π ) = { R ∈ D| R ∩ π = ∅} . For more information on this approach to linear sets, we refer to [7].The linearity conjecture for blocking sets (see [13]) states that(LC) All small minimal blocking sets in PG( n, q ) are linear sets.Up to our knowledge, this is the complete list of cases in which the lin-earity conjecture for blocking sets in PG( n, p t ), p prime, with respect tohyperplanes, has been proven. • t = 1 (for n = 2, see [2]; for n >
2, see [5]) • t = 2 (for n = 2, see [11]; for n >
2, see [10]) • t = 3 (for n = 2, see [8]; for n >
2, see [10]) • B is of R´edei-type, i.e., there is a hyperplane meeting B in | B | − p t points (for n = 2, see [1, 3]; for n >
2, see [9]) • dim h B i = t (see [12]).In this paper, we show that if dim h B i = t −
1, and the characteristicof the field is sufficiently large, B is a linear set, as a corollary of the maintheorem. Main Theorem.
A small minimal blocking set B in PG( n, q ) , with exponent e , q = p t , p prime, q := p e , q ≥ , t/e = h , spanning an ( h − -dimensionalspace is an F q -linear set. A subspace clearly meets an F p -linear set in 1 mod p or 0 points. Thefollowing theorem shows that for a small minimal blocking set, the sameholds. 2 heorem 1. [12, Theorem 2.7] If B is a small minimal blocking set in PG( n, p t ) , p prime, then B intersects every subspace of PG( n, p t ) in mod p or points. From this theorem, we get that every small minimal blocking set B inPG( n, p t ), p prime, has an exponent e ≥
1, which is the largest integer forwhich every hyperplane intersects B in 1 mod p e points. The following theorem by Sziklai characterises the intersection of particularlines with a small minimal blocking set as a linear set.
Theorem 2. [13, Corollary 5.2]
Let B be a small minimal blocking set withexponent e in PG( n, q ) , q = p t , p prime. If for a certain line L , | L ∩ B | = p e + 1 , then F p e is a subfield of F q and L ∩ B is a subline PG(1 , p e ) . Using the 1 mod p -result (Theorem 1), it is not too hard to derive anupper bound on the size of a small minimal blocking set in PG( n, q ) as donein [14]. This bound is a weaker version of the bound in Corollary 5.2 of [13]. Lemma 3. [14, Lemma 1]
The size of a small minimal blocking set B withexponent e in PG( n, q h ) , q := p e ≥ , p prime, is at most q h + q h − + q h − +3 q h − . In this paper, we will make use of the fact that we can find lower boundson the number of secant lines to a small minimal blocking set. In the nextlemma, one considers the number of ( q + 1)-secants to the blocking set B ,which will give a linear intersection with the blocking set by Theorem 2. Lemma 4. [14, Lemma 4]
A point of a small minimal blocking set B withexponent e in PG( n, q h ) , q := p e ≥ , p prime, lying on a ( q + 1) -secant,lies on at least q h − − q h − + 1 ( q + 1) -secants. For the proof of Lemma 7, we will make use of the concept of pointexponents of a blocking set and the well-known fact that the projection of asmall minimal blocking set is a small minimal blocking set.
Lemma 5. [12, Corollary 3.2]
Let n ≥ . The projection of a small minimalblocking set in PG( n, q ) , from a point Q / ∈ B onto a hyperplane skew to Q ,is a small minimal blocking set in PG( n − , q ) . The exponent e P of a point P of a small minimal blocking set B is thelargest number for which every line through P meets B in 1 mod p e P or 0points. The following lemma is essentially due to Blokhuis.3 emma 6. (See [4, Lemma 2.4(1)]) If B is a small minimal blocking set in PG(2 , q ) , q = p t , p prime, with | B | = q + κ , and P is a point with exponent e P , then the number of secants to B through P , is at least ( q − κ + 1) /p e P + 1 . Lemma 7.
A point P with exponent e P = 2 e of a small minimal blockingset B with exponent e in PG( n, q h ) , q := p e ≥ , p prime, lies on at least q h − − q h − − q h − − q h − + 1 secant lines to B .Proof. If n = 2, Lemma 3, together with Lemma 6, shows that the numberof secant lines to B is at least ( q h − q h − − q h − − q h − + 1) /q + 1 ≥ q h − − q h − − q h − − q h − + 1.If n >
2, then let L be a line through P , meeting B in q + 1 points.There exists a point Q , not on B , lying only on tangent lines to B . Let ˜ B be the projection of B from Q onto a hyperplane through L . By Lemma 5,˜ B is a small minimal blocking set in PG( n − , q ). It is clear that every linethrough P meets ˜ B in 1 mod q or 0 points, and that there is a line, namely L , meeting ˜ B in 1 + q points, so e P = 2 e in the blocking set ˜ B . It followsthat the number of secant lines through a point P with exponent 2 e to B isat least the number of secant lines through the point P with exponent 2 e to˜ B in PG( n − , q h ). Continuing this process, we see that this number is atleast the number of secant lines through the point P with exponent 2 e in asmall minimal blocking set ˜ B in PG(2 , q h ), and the statement follows. In the following lemma, we will distinguish planes according to their inter-section size with a small minimal blocking set. We will call a plane with q + q + 1 non-collinear points of B a good plane, while all other planes willbe called bad . Note that also planes meeting B in only points on a line, orskew to B are called bad . The following lemma shows that good planes meeta small minimal blocking set in a linear set. Lemma 8. If Π is a plane of PG( n, q ) containing at least non-collinearpoints of a small minimal blocking set B in PG( n, q ) , with exponent e , q = p t , p prime, q := p e , then(i) q + q + 1 ≤ | B ∩ Π | . (ii) If | B ∩ Π | = q + q + 1 , then B ∩ Π is F q -linear.(iii) If | B ∩ Π | > q + q + 1 , then | B ∩ Π | ≥ q + q + 1 . roof. (i) By Lemma 1, every line meets B in 1 mod q or 0 points. Sincewe find 3 non-collinear points, it is easy to see that | B ∩ Π | ≥ q + q + 1.(ii) From the previous argument, we easily see that if | B ∩ Π | = q + q +1,then every line in Π contains 0, 1 or q + 1 points of B . Suppose that thereexist two ( q + 1)-secants that meet in a point, not in B , then the number ofpoints in Π ∩ B is at least q + q + 1 + q . Hence, every two ( q + 1)-secantsmeet in a point of B . Moreover, through two points of B ∩ Π, there is aunique ( q + 1)-secant, so B meets Π in an F q -subplane.(iii) By Theorem 1, if there is a line L of Π containing more than ( q + 1)points of B , then | L ∩ B | ≥ q + 1, and | Π ∩ B | ≥ q + q + 1. So from nowon, we may assume that every line meets B in 0, 1 or q + 1 points. If there isan F q -subplane strictly contained in Π ∩ B , then clearly | B ∩ Π | ≥ q + q + 1,so we may assume that there is no F q -subplane contained in Π ∩ B .Let L be a ( q + 1)-secant in Π, let P be a point of B ∩ L , let Q be a pointof B \ L and let M be the line P Q . From Theorem 2, we know that L ∩ B and M ∩ B are sublines over F q . These sublines define a unique F q -subplaneΠ . Let N be a line, not through P , through a point of L ∩ B , say R andof M ∩ B , say R . Let N be another line, not through R or R , meeting L in a point R of B and M in a point R of B . If T is the intersection pointof N and N , then T belongs to the subplane Π .Now suppose that T is a point of B , then N meets B in a subline,containing 3 points of the subline Π ∩ N . This implies that the subline N ∩ B is completely contained in B . The same holds for the subline N ∩ B ,and repeating the same argument, for every subline through T meeting L and M in points of B , different from P . Again repeating the same argument, fora point T ′ = T on N , not on L or M , yields that Π is contained in B , acontradiction. This implies that the q − B on the line N , noton L or M , are different from the q − B on the line N , not on L or M . Varying N and N over all lines meeting L and M in points of B ,we get that there are at least q ( q −
1) + 2 q + 1 points in B ∩ Π.To avoid abundant notation, we continue with the following hypothesison B . B is a small minimal blocking set in PG( n, q ), with exponent e , q = p t , p prime, q := p e , t/e = h , spanning an ( h − -dimensional space . Lemma 9.
A plane of
PG( n, q ) contains at most q + q + q + 1 points of B .Proof. Suppose there exists a plane Π with more than q + q + q + 1 pointsof B , then, by Theorem 1, | Π ∩ B | ≥ q + q + 2 q + 1. We prove by induction5hat, for all 2 ≤ k ≤ h −
1, there is a k -space, containing at least ( q k +20 − / ( q −
1) + q k − points of B . The case k = 2 is already settled, so supposethere is a j -space Π j , j < h −
1, containing at least ( q j +20 − / ( q −
1) + q j − points of B . Since B spans an ( h − j < h −
1, there is a point Q in B , not in Π j . Because a line containing two points of B contains at least q + 1 points of B , this implies that |h Q, Π j i ∩ B | ≥ ( q j +30 − / ( q −
1) + q j .By induction, we obtain that B contains at least ( q h +10 − / ( q −
1) + q h − points, a contradiction, since | B | ≤ q h + q h − + q h − + 3 q h − . Lemma 10.
Let L be a ( q + 1) -secant to B . Then either L lies on at least q h − − q h − + 1 good planes, or L lies on bad planes only. In the latter case,all planes with points of B contain at least q + q + 1 points of B outside of L .Proof. Let Q be a point on L , not on B . We project B from Q onto ahyperplane H , not through Q , and denote the image of this projection by ˜ B .Let P be the point L ∩ H . It follows from Lemma 5, that ˜ B is a small minimalblocking set. Since every subspace meets B in 1 mod q or 0 points, everysubspace meets ˜ B in 1 mod q or 0 points. Suppose that P has exponent e P = 1, then it follows from Lemma 4 that P lies on at least q h − − q h − + 1( q + 1)-secants. This means that there are at least q h − − q h − + 1 planesthrough L containing at least q + q + 1 points of B , which implies that | B | ≥ q ( q h − − q h − +1), a contradiction since | B | ≤ q h + q h − + q h − +3 q h − by Lemma 3.If P has exponent e P at least 4, we get that the planes through L whichcontain a point of B , not on L , contain at least q + q + 1 points of B , whichis impossible by Lemma 9. We conclude that P has exponent e P = 2 or e P = 3. If P has exponent e P = 3, then every plane through L that containsa point of B not on L , contains at least q + q + 1 points, and hence, allplanes through L are bad.Finally, if P has exponent 2, we know from Lemma 7 that there are atleast s = q h − − q h − − q h − − q h − + 1 secant lines through P , which impliesthat there are at least s planes through L containing a point of B outside of L . Suppose t of the s planes are bad, then, using Lemma 8(iii), B contains atleast t (2 q ) + ( s − t )( q ) + q + 1 points. If we put t = 3 q h − − q h − − q h − + 1,we get a contradiction since | B | ≤ q h + q h − + q h − + 3 q h − by Lemma 3. Lemma 11.
A point P of B lying on a ( q + 1) -secant, lies on at most one ( q + 1) -secant L that lies on only bad planes.Proof. Let P be a point of B lying on a ( q + 1)-secant and let L be a linethrough P that only lies on bad planes. From Lemma 9 and Lemma 10, we6et that q + q + 1 ≤ | Π ∩ B | ≤ q + q + q + 1 for all planes Π through L ,containing points of B outside of L .By Lemma 3, | B | ≤ q h + q h − + q h − + 3 q h − , so there are at most q h − + 2 q h − planes through L containing points of B outside of L . Since P lies on at least q h − − q h − + 1 ( q + 1)-secants, there are at least twoplanes Π and Π containing at least q − q + 1 ( q + 1)-secants through P . Suppose that L ′ is a ( q + 1)-secant through P , different from L , lying ononly bad planes. At least one of the planes Π , Π , say Π , does not contain L ′ . We will now show that for all k ≤ h −
2, there exists a k -space through Π ,not containing L ′ , containing at least q k − q k − ( q + 1)-secants through P .For k = 2, the statement is true, hence, suppose it holds for all k ≤ j < h − ′ be a j -space through Π , not containing L ′ and containing at least q j − q j − ( q + 1)-secants through P .Let | Π ′ ∩ B | = A , then a ( j + 1)-space Π ′′ through Π ′ , containing a pointof B , not in Π ′ , contains at least ( q − A + 1 points of B , not in Π ′ . Wesee that the number of ( j + 1)-spaces containing a point of B , not in Π ′ , ismaximal if the number of points in Π ′ is minimal. Since | B ∩ Π | ≥ q + q +1, | B ∩ Π ′ | ≥ ( q + q + 1) q j − + 1. This implies that the number of points of B in such a ( j + 1)-space, outside of Π ′ is at least q j +10 − q j + q j − − q j − + q .Since | B | ≤ q h + q h − + q h − + 3 q h − , the number of such ( j + 1)-spaces is atmost q h − j − + 2 q h − j − + 4 q h − j − . At most ( q j +10 − / ( q −
1) ( q + 1)-secantsthrough P lie in Π ′ . Suppose that all ( j + 1)-spaces through Π ′ , exceptpossibly h Π ′ , L i , contain at most q j − q j − ( q + 1)-secants through P , notin Π ′ , then the number of ( q + 1)-secants through P is at most( q h − j − + 2 q h − j − + 4 q h − j − − q j − q j − ) + ( q j +10 − / ( q − , a contradiction if j < h −
2, since there are at least q h − − q h − + 1 ( q + 1)-secants through P . We may conclude, by induction, that there exists an( h − ′′ , not through L ′ , that contains at least q h − − q h − ( q + 1)-secants through P . Since L ′ does not lie in Π ′′ , this implies that there areat least q h − − q h − different planes through L ′ that each have at least q points outside of L , a contradiction since | B | ≤ q h + q h − + q h − + 3 q h − .This implies that there is at most one line through P that lies on only badplanes. Lemma 12.
Assume h > and q > h − . Denote the ( q + 1) -secants,not lying on only bad planes, through a point P of B that lies on at least ne ( q + 1) -secant, by L , . . . , L s . Let x be a point of the spread elementcorresponding to P in PG( h ( n + 1) − , q ) and let ℓ i be the line through x such that B ( ℓ i ) = L i ∩ B . Let L = { ℓ , . . . , ℓ s } , then hLi has dimension h .Proof. From Lemma 4 and Lemma 11 we get that s is at least q h − − q h − +1 − q h − − q h − . From Lemmas 8(ii) and 10, we get that through everyline L i , i = 1 , . . . , s , there are at least q h − − q h − + 1 good planes, sayΠ ij , j = 1 , . . . , t , such that B ∩ Π ij = B ( π ij ), for a plane π ij through ℓ i .Denote the set of planes { π ij , ≤ i ≤ s, ≤ j ≤ t } by V , and the set of lines { ℓ , . . . , ℓ s } by L .A fixed plane π ij of V , say π , contains q + 1 lines of L , say ℓ , . . . , ℓ q +1 .The lines ℓ , . . . , ℓ q +1 lie on a set of at least ( q +1)( q h − − q h − )+1 differentplanes of V . On these planes, there lies a set P of at least ( q + 1)( q h − − q h − ) q different points y , . . . , y u , not in π , such that B ( y i ) ⊂ B .We claim that B ( y r ) = B ( y ′ r ) implies that y r = y ′ r for y r and y ′ r in P ( ∗ ). We know that y r lies on π ij and y ′ r lies on π i ′ j ′ for some i, i ′ , j, j ′ . Since B ( π ij ) = B ∩ Π ij and B ( π i ′ j ′ ) = B ∩ Π i ′ j ′ , the lines hB ( xy r ) i and hB ( xy ′ r ) i are( q + 1)-secants to B . Since we assume that B ( y r ) = B ( y ′ r ), these ( q + 1)-secants coincide. Moreover, B ( xy r ) ⊂ B and B ( xy ′ r ) ⊂ B , so xy r and xy ′ r aretransversal lines through the same regulus, which forces y r = y ′ r . This provesour claim, hence, different points of the point set P give rise to differentpoints of B .We will prove that, for all 2 ≤ k ≤ h there exists a k -space through x with at least q k − − (5 k − q k − lines of L . The existence of π provesthis statement for k = 2. Assume, by induction, that there exists a j -spacethrough x , say ν , where j < h −
1, containing at least q j − − (5 j − q j − lines of L .We will now count the number of couples ( ℓ ∈ L contained in ν , r apoint, not in ν with h r, ℓ i ∈ V ). The number of lines of L in ν is at least q j − − (5 j − q j − , the number of points r / ∈ ν with h r, ℓ i ∈ V for somefixed ℓ , is at least ( q h − − q h − ) q − ( q j +10 − / ( q − r with h r, ℓ i ∈ V , is by ( ∗ ) at most | B | , hence, the number of points r / ∈ ν with h r, ℓ i ∈ V is at most | B | − ( q j − − (5 j − q j − ) q − r , lying on (say) X different planes h r, ℓ i of V with X ≥ ( q j − − (5 j − q j − )( q h − q h − − ( q j +10 − / ( q − q h + q h − + q h − + 3 q h − − q j + (5 j − q j − − . This last expression is larger than q j − − (5( j + 1) − q j − , if h >
3, for all j ≤ h − j +1-space h r, ν i , contains at least ( q j − − (5( j +1) − q j − ) q + 1 lines of L , hence, by induction, we find an h -dimensional-spacethrough x containing at least q h − − (5 h − q h − lines of L .Suppose now that there is a line of L , say ℓ s , not in this h -space ξ . ByLemma 10, there are at least q h − − q h − planes through ℓ s , giving rise to( q h − − q h − )( q − q ) points z , which are not contained in ξ , such that B ( z ) ⊂ B . By ( ∗ ), and the fact that there are at least ( q h − − (5 h − q h − ) q + 1points y in ξ such that B ( y ) ⊂ B , we get that | B | ≥ q h + q h − + q h − + 3 q h − ,a contradiction.This shows that the dimension of hLi is h .We now use the following theorem, which is an extension of [11, Remark3.3]. Theorem 13. [6, Corollary 1]
A blocking set of size smaller than q in PG( n, q ) is uniquely reducible to a minimal blocking set. Main Theorem.
A small minimal blocking set B in PG( n, q ) , with exponent e , q = p t , p prime, q := p e , q ≥ , t/e = h , spanning an ( h − -dimensionalspace is an F q -linear set.Proof. As seen in Lemma 12, there exists an h -dimensional space ξ in PG(( n +1) h − , q ), such that |B ( ξ ) ∩ B | ≥ q h − q h − + 1. Define ˜ B to be the unionof B ( ξ ) and B and recall that B ( ξ ) is a small minimal F q -linear block-ing set in PG( n, q ). Clearly, ˜ B is a blocking set, and its size is equal to | B | + |B ( ξ ) | − | B ∩ B ( ξ ) | . Hence, | ˜ B | is at most ( q h +10 − / ( q −
1) + q h + q h − + q h − +3 q h − − ( q h − q h − +1) < q h . Theorem 13 shows that B = B ( ξ ),so we may conclude that B is an F q -linear set.By the fact that the exponent of a small minimal blocking set in PG( n, q )is at least one (see Theorem 1), we get the following corollary. Corollary 14.
All small minimal blocking sets in
PG( n, p t ) , p prime, p > t − spanning a ( t − -space, are F p -linear. Acknowledgment:
This research was conducted while the second au-thor was visiting the Department of Computer Science at E¨otv¨os Lor´andUniversity in Budapest. The author thanks all members of the Finite Ge-ometry group for their hospitality during her stay.The first author acknowledges the partial support of the grants OTKAT-49662, K-81310, T-67867, CNK-77780, ERC, Bolyai and T ´AMOP.The authors thank the anonymous referees for their valuable comments.9 eferences [1] S. Ball. The number of directions determined by a function over a finitefield.
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