يمفضاءات جزئیة في مجموعات من ستو GF(q)حول حقل كالوا PG(2,q)سقاطي ا میساء جلیل محمد جامعة بغداد ،ابن الھیثم -كلیة التربیة ،قسم الریاضیات خالصةال حــول PG(2,q)جزئیـة فــي المسـتوي االسـقاطي لفضـاءات مجموعـات انـواع مــن تتــم درسـفـي هـذا البحـث الـبعض مـن خـالل االمثلـة مـع العالقـات التـي تـربط بـین هـذه المجموعـات بعضـها وبیـان بعـض GF(q). احقـل كـالو .والمبرهنات IHJPAS Sets of Subspaces of a Projective Plane PG(2,q) Over Galois Field GF(q) M. J. Mahammad Departme nt of Mathematics,College of Education Ibn-Al-Haitham , Unive rsity of Baghdad Abstract In this thesis, some sets of subsp aces of p rojective p lane PG(2,q) over Galois field GF(q) and the relations between them by some theorems and examp les can be shown. 1. Introduction A recurring them of this work is the characterization of algebraic varieties in PG(2,q) as finite sets of p oints with certain combinatorial p rop erties of . this work. Section one which contains some definitions of nucleus p oint, t-fold nucleus p oint, Blocking set, t-fold Blocking set, Unital set, (k,n)-arc, flag, st rong representive sy st em and, the set of ty p e (0,1,2,q + 1). Section two, contains some theorems about these subsets and the relation between them and some examp les which about some of these subsets. 2.1 De fini tion "Projective Plane" [2] A p rojective p lane PG(2,q) over Galois field GF(q) is a two-dimensional p rojective sp ace, which consists of p oints and lines with relation between them, in PG(2,q) there are q 2 + q + 1 p oints, and q 2 + q + 1 lines, every line contains 1 + q p oints and every p oint is on 1 + q lines, any p oint in PG(2,q) has the form of a trip le (a1,a2,a3) where a1, a2, a3  GF(q); such that (a1,a2,a3)  (0,0,0). Two p oints (a1,a2,a3) and (b1,b2,b3) represent t he same point if there exists  GF(q)\{0}, such that (b1,b2,b3) =  (a1,a2,a3). Similarly any line in PG(2,q) has the form of a trip le [a1,a2,a3], where a1,a2,a3  GF(q); such that [a1,a2,a3] ≠ [0,0,0]. Two lines [a1,a2,a3] and [b1,b2,b3] represent the same line if there exists  GF(q)\{0}, such that [b1,b2,b3] =  [a1,a2,a3]. There exists one p oint of the form (1,0,0). There exist q p oints of the form (x,1,0). There exist q 2 p oints of the form (x,y ,1). A p oint p (x1,x2,x3) is incident with the line L[a1,a2,a3] if and only if a1x1 + a2x2 + a3x3 = 0, i.e. a point rep resented by (x1,x2,x3) and the line rep resented by 1 2 3 a a a           , then (x1,x2,x3) 1 2 3 a a a           = 0  a1x1 +a2x2 + a3x3=0. Any p rojective plane PG(2,q) satisfies the following axioms: 1. Any two dist inct lines are intersected in a unique p oint. 2. Any two dist inct p oints are contained in a unique line. 3. There exist at least four p oints such that no t hree of them are collinear. 2.2 De fini tion "Blocking S et"[2] A blocking set B of PG(2,q) is a set of p oints intersecting every line of PG(2,q) in at least one point, so B is blocking set if and only if PG(2,q)\B is blocking set. IHJPAS 2.3 De fini tion "Minimal Blocking S et" [2] A Blocking set B is called minimal in PG(2,q) when no p rop er subset of it is st ill a blocking set such that,  p  B, B\{p} is not a blocking set. 2.4 De fini tion "Nucle i S et"[2] Let S be a set in PG(2,q), let p be a point in PG(2,q) and p  S, p is called nucleus p oint of a set S if every line in PG(2,q) through p intersects S exactly one, the set of nucleus p oints of S called nuclei set and denoted by N(S). By the following examp le we exp lain the definition: In a p rojective p lane PG(2,4), let S = {5,6,8,9,10,12,13,14,15,16,17,18,21} and let N(S) = {1,2,3,4} then  P  N(S); P is nucleus p oint since:through p oint 1, t here are 5 lines which are [1 0 0], [0 0 1],[2 0 1], [0 3 1],[3 3 1] such that each one of them intersects S in one p oint which are {5,9,10,14,18}, resp ectively. Similarly for the other p oints {2,3,4}. 2.5 De fini tion "t-Fold nucleus Point"[4] A p oint p  PG(2,q) is a t-fold nucleus p oint of a set S  PG(2,q) if p  S and every line through p meets S at least t p oints of S. 2.6 De fini tion "Uni tal S et" [1] A unital set PG(2,q) of a square order q is a set U of (q q +1) p oints such that each line in PG(2,q) meets U either one or q +1 p oints, i.e., every line in PG(2,q) is a tangent or a secant of U if contain (1) p oint or ( q +1) p oints of line. By the following examp le we exp lain the definition: Let U = {1,6,7,10,11,16,17,18,19} be a set in PG(2,4) and U contains 9 p oints. U is a unital set since: U contains 4 4 +1 = 9 p oints and every and line in PG(2,4) meets U in 1 or 3 p oints, as shown in the table (1,2). 2.7 De fini tion "(0,1,2,q + 1)-S et"[1] A set of p oints in PG(2,q) is called of ty p e (0,1,2,q + 1) if every line in PG(2,q) meets the set in 0,1,2 or q + 1 points. 2.8 De fini tion "n-secant"[2] A line L in PG(2,q) is an i-secant of a (K,n)-arc K if :  K∩ L = i , i=0,1,2,......,n. 2.9 De fini tion "(k,n)-arc" [2] A (k,n)-arc in PG(2,q) is a set S of k p oints with p rop erty that every line i contains at most n p oints of S, a (k,n)-arc S is called comp lete arc if it is not contained in a (k + 1,n)-arc. A (k,n)-arc in PG(2,q) is maximal-arc if every line in PG(2,q) is a zero secant or an n-secant of the (k,n)-arc. 2.10 De fini tion "Flag"[1] A flag in PG(2,q) is an incident p oint-line p air; Flag = {(p i,Lj ); pi  Lj , i,j=1,2,…,q 2 +q+1}. 2.11 De fini tion "S trong Representi ve S yste m" [1] A set S = {(p 1,L1),(p 2,L2),…,(p s,Ls)} of flags p i  Lj  i = j is a st rong representive sy st em. The members of st rong represent sy st em that S  q q +1, with equality if and only if S consists of incident p oint-t angent p airs of a unital, we denoted p oints of S by P(S) and the lines of S by L(S), P(S) and L(S) are called sp acialy and the others are called ordinary S is called maximal if is not p art of Larger-strong represent. 2.12 Definition "Complete Nuclei Set" "New" Let N(S) be a set of all nucleus p oints of a set S in PG(2,q), N(S), is called comp lete nuclei set if N(S) = PG(2,q) \ S. 3. The Relation Be tween The S ets S ubspace of PG(2q) Over GF(q) This section contains theorems to show that some relations between the : Blocking set ,(k,n)-arc, unital set, nuclei set, st rong representive sy st em and the set of ty p e (0,1,2,q+1). IHJPAS The orem Let N(S) be comp lete, then S is a line. Proof: For every p oint p in N(S), p is a nucleus p oint of S, then every line in PG(2,q) through p meets S in exactly one point. Since there exists q + 1 lines through p , then there exists at least q + 1 p oints in S. If there exists another p oint R in S, then there existS another line through P and R in S then there exists another line through P and R, which is a contradiction since there exists exactly q + 1 lines through P. Hence S contains exactly q + 1 p oints P1, …, Pq + 1 , sup p ose these p oints are not collinear, then  at least one point, say Pq + 1, not collinear with two p oints say P1 and P2, the line PPq + 1 intersects the line P1P2 in one p oint, say Pq +2, Pq +2 is on the line P1P2 , then Pq +2 is not on S hence the line PPq + 1 , since the line PPq + 1 intersects S in only one point Pq + 1. Pq + 2 is not in S, then any line through it intersects S in one p oint but the line P1P2 which p assing Pq +2 intersects S in two p oints which is a contradiction. Hence the p oints P1, P2, …, Pq + 1 are collinear, similarly for any three p oints P1, P2, …, Pq + 1 are collinear. The orem [1] "without prove" Let B be a blocking in PG(2,q), t hen for every P PG(2,q) \B, p is t-fold nucleus p oint, if and only if B is a t-fold blocking set, 2  t  q. Proof: sup p ose that B is a blocking set and every P PG(2,q) \B, P is a t-fold blocking nucleus p oint of B, then every line in PG(2,q) through P meets B in at least t p oints, but from definition of blocking set B can not contain any line, so every line in PG(2,q) meets B in at least t p oints, then B is a t-fold blocking set. Conversely , sup p ose B is a t-fold blocking set then every line in PG(2,q) meets B in at least t p oints, then for every p oint P PG(2,q) \B and every line through P intersects B in at least t- p oints, then for every P PG(2,q) \B , P is a t-fold nucleus p oint. The orem In PG(2,q), q = p 2 , every unital set is a (qp +1,p +1)-arc, q  3. Proof: Let U be a unital set in PG(2,q), q = p 2 then every line in PG(2,q), q = p 2 intersectS U in either 1 or p + 1 p oints and U is a set of qp + 1 p oints and there are no p + 2 p oints are collinear then U is (p q+1,p +1)-arc. The converse is not true as shown by the following examp le: In p rojective p lan PG(2,4), let K = {1,2,3,4,7,8,11,12,13,15,16,21}, K is an (12,4)-arc, then K is not a unital set since: 1. Every unital in PG(2,q) contains (9) p oints, but (k,4)-arc contains (12) p oints. 2. Some line in PG(2,4) meets (k,4)-arc in 4 p oints, but every line in PG(2,4) most meets every unital set in either 1 or 3 points. 3. The orem In PG(2,q), q = p 2 every unital set is a blocking set. Proof: let U be a unital set in PG(2,q), q = p 2 then U contains qp +1 p oints and every line in PG(2,q), q = p 2 intersects U in either 1 or p + 1 p oints so every line in PG(2,q), q = p 2 meets U, but U dose not contain any line; then U is a blocking set. The converse is not true and showed by the following examp le; In p rojective p lan PG(2,4), let B = {1,2,5,8,9,10,12,13,14,15,16,21}, then B is blocking set but it is not a unital set since: 1. Every unital in PG(2,4) contains (9) p oints, but B contains (12) p oints. 2. Some line in PG(2,4) meets (k,4)-arc in 4 p oints, but every line most meets every unital set in either 1 or 3 p oints. IHJPAS The orem Let U be a unital set in PG(2,q 2 ) then every p oint P  U ,P is either nucleus p oint or (q+1)-fold nucleus p oint. Proof: Let U is a unital set, t hen every line meets U in either 1 or q + 1 p oint, for every p oint P, every line through P meets U in either 1 or q + 1 p oints, then P is either nucleus or q + 1-fold nucleus p oint. The orem Let S = {(p i,Li); p i  B, B is minimal blocking set; Li PG(2,q) i} then S is a st rong representive sy st em. Proof: Let S = {(p i,Lj ); pi  B,  i ;Lj  PG(2,q) } t o p rove that S is a strong representive sy st em. Sup p ose that S is not st rong representive sy st em then  P0,P1  B s.t .  L  PG(2,q) such that (P0,L), (P1,L)  S, then the line L in PG(2,q) intersects B in P0, P1  B \ {P0} or B \{P1} is blocking set, which is contradiction since B is a minimal blocking set and  L  PG(2,q), L  B = P or L is t angent t o B in P. The orem Every maximal (k,2)-arc in PG(2,q) with no 0-secant is a set of t y p e (0,1,2,q+1)-set. Proof: It is clear that the maximal (k,2)-arc with no 0-secant mean that every line in PG(2,q) is a 2- secant of (k,2)-arc, so every line intersects maximal (k,2)-arc in two p oints then the maximal (k,2)-arc is a set of t y p e (0,1,2,q+1)-set. 4. Conclusi on and Recommandation In this research, we took of subsp aces of PG(2,q) like Blocking, Nuclei, Unital, (0,1,2,q+1) –set, strong representive sy st em and complete nuclei set as new definitions. Then we found some relations between these subsets and exp lain them by theorems like; if N(S) is comp lete then S is line,  p  PG(2,q)\B , p is t-fold nucleus, if and only if B is t-fold blocking set, every unital set is (qp +1,p +1)-arc, q  3, pu, u is unital set then p is either nucleus or q+1- fold nucleus p oint and other relations. Some of definitions were exp lained by examp les and tables like unital set and nucleus p oint. So as some theorems, this relation will lead to make new sets of subsp aces included at this p rojective sp ace or others. Re ferences 1. Susan Barwich, and Gary Ebert, (2008). "Unitals in Projective Plane", DOI:10.1007/978- 0-387-76366-8-1,© Sp ringer Science + Business M edia, LLc 2. Blokhuis,A.; Hischfed,J.W.P. ; JungnickD.and Thas, J .F. (2001), "Finite Geometries", Kluwer Academic, Publishers. 3. Beutel Sp acher, A.and de clerck, F. (1993), "Finite Geometry and Combinatorics", Combridge University Press. 4. Lamb, J.D.and Preece, D.A. (1999), "Surreys in Combinatorics", combridge University Press. IHJPAS Table (1.1):The points and lines of PG(2,4) Table (1.2) :The points and Line s of Untial Se t i P i Li 1 1 0 0 1 2 3 4 5 2 0 1 0 2 6 10 14 18 3 1 1 0 4 6 12 17 19 4 2 1 0 5 7 12 14 21 5 3 1 0 5 8 11 17 18 6 0 0 1 1 18 19 20 21 7 1 0 1 2 7 11 15 19 8 2 0 1 1 6 7 8 9 9 3 0 1 2 8 12 16 20 10 0 1 1 5 6 13 15 20 11 1 1 1 3 9 12 15 18 12 2 1 1 5 9 10 16 19 13 3 1 1 4 7 13 16 18 14 0 2 1 3 8 13 14 19 15 1 2 1 3 7 10 17 20 16 2 2 1 4 8 10 15 21 17 3 2 1 4 9 11 14 20 18 0 3 1 1 14 15 16 17 19 1 3 1 2 9 13 17 21 20 2 3 1 3 6 11 16 21 21 3 3 1 1 10 11 12 13 i P i Li 1 1 0 0 1 2 0 1 0 6 10 18 3 1 1 0 6 17 19 4 2 1 0 7 5 3 1 0 11 17 18 6 0 0 1 1 18 19 7 1 0 1 7 11 19 8 2 0 1 1 6 7 9 3 0 1 16 10 0 1 1 6 11 1 1 1 18 12 2 1 1 10 16 19 13 3 1 1 7 16 18 14 0 2 1 19 15 1 2 1 7 10 17 16 2 2 1 10 17 3 2 1 11 18 0 3 1 1 16 17 19 1 3 1 17 20 2 3 1 6 11 16 21 3 3 1 1 10 11 IHJPAS Graph: This graph shows that a set S contains exactly q+1 points IHJPAS