n−Primary Submodules IBN AL- HAITHAM J. FOR PURE & APPL. SCI VOL.22 (2) 2009 Construction of Complete (k,n)-arcs in the Projective Plane PG(2,11) Over Galois Field GF(11), 3  n  11 A.T. Mahammad Department of Mathematics, College of Education Ibn-Al-Haitham , University of Baghdad Abstract The purpose of this work is to construct complete (k,n)-arcs in the projective 2-space PG(2,q) over Galois field GF(11) by adding some points of index zero to complete (k,n–1)- arcs 3  n  11. A (k,n)-arcs is a set of k points no n + 1 of which are collinear. A (k,n)-arcs is complete if it is not contained in a (k + 1,n)-arcs. Introduction Mayssa 2004 (4), constructed of complete (k,n)-arcs in PG(2,17) and Sawsan 2001 (6), showed the classification and construction of (k,n)-arcs from (k,m)-arcs in PG(2,q) m < n. And Ban, (8) showed the classification and construction of (k,4)-arc, k = 17, 18,…, 34, in PG(2,11). This paper is divided into two sections, section one consists of proving basic, theorems and giving some definitions of projective plane, (k,n)-arcs, maximal and complete arcs…ets. Section two consists of the projective plane of order eleven. The construction of complete (k,2)-arcs call it c1, c2, c3, …, c9 and the construction of complete (k,n)-arcs from complete (k,n – 1)-arcs in PG(2,11), where n = 3, 4, …, 9, 10 gave the points Pi and lines Li.in PG(2,11) are determined in the table (1,1). Section One 1.1 Definition "Projective Plane" (1) A projective plane PG(2,q) over Galois field GF(q) is a two-dimensional projective space, which consists of points and lines with incidence relation between them. In PG(2,q) there are q 2 + q + 1 points, and q 2 + q + 1 lines, every line contains 1 + q points and every point is on 1 + q lines, all these points in PG(2,q) have the form of a triple (a1,a2,a3) where a1, a2, a3  GF(q); such that (a1,a2,a3)  (0,0,0). Two points (a1,a2,a3) and (b1,b2,b3) represent the same point if there exists  GF(q)\{0}, such that (b1,b2,b3) =  (a1,a2,a3). There exists one point of the form (1,0,0). There exists q points of the form (x,1,0). There exists q 2 points of the form (x,y,1), similarly for the lines. A point 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 represented by (x1,x2,x3) is incident with the line represented by 1 2 3 a a a           . if (x1,x2,x3) 1 2 3 a a a           = 0  a1x1 + a2x2 + a3x3 = 0. The projective plane PG(2,q) satisfying the following axioms: 1. Any two distinct lines intersected in a unique point. 2. Any two distinct points are contained in a unique line. 3. There exists at least four points such that no three of them are collinear. IBN AL- HAITHAM J. FOR PURE & APPL. SCI VOL.22 (2) 2009 1.2 Definition (1) 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]. 1.3 Definition "Quadric" (1) A quadric Q in PG(n – 1,q) is a primal of order two, so Q is a quadric, then Q = V( F), where F is a quadric form, that is: n 2 2 ij i j 11 1 12 1 2 nn n i j i, j 1 F a x x a x a x x a x        1.4 Definition "Conics"(1) Let Q(2,q) be the set of quadrics in PG(2,q), that is the varieties V(F), where: 2 2 2 11 1 22 2 33 3 12 1 2 13 1 3 23 2 3 F a x a x a x a x x a x x a x x      If V(F) is non-singular, then quadric is conic. 1.5 Definition "(k,n)-arcs" A (k,n)-arc, K in PG(2,q) is a set of K points such that some line in PG(2,q) meets K in n points but such that no line meets K in more that n points, where n  2. A line L in PG(2,q) is an i-secant of a (k,n)-arc K if L  K = i. Let Ti denoted the total number of i-secants to K in PG(2,q). 0-secant is called an external line, a 1-secant is called a unisecant, a 2-secant is called a bisecant. 1.6 Definition "Complete (k,n)-arcs" (1) A (k,n)-arc in PG(2,q) is complete if there is no (k+1,n)-arc containing it. 1.7 Definition (1) A point N not in (k,n)-arc K is said to be has index i if there exists exactly i (2-secants) through N. Ci = Ni = the number of points of index i. 1.8 Definition "Maximal (k,n)-arcs" (2) A (k,n)-arc K in PG(2,q) is a maximal arc if k = (n – 1) q + n. 1.9 Theorem (2) Let M be a point of (k,2)-arc A in PG(2,q), then the number of unisecant through M is u = q + 2 – k. Proof: There exists exactly q + 1 lines through a point M in a(k,2)-arc A of PG(2,q), which are the bisecants and the unisecants of the arc. There exists exactly (k – 1) bisecants of the arc A through M and the other (k – 1) points of the arc, since the arc contains exactly k points. The number of unisecants through M is u, then u = q + 1 – (k – 1) = q + 1 – k + 1 = q + 2 – k. 1.10 Theorem (2) Let Ti be the number of the i-secants of a (k,n)-arc A in PG(2,q), then: (a) T2 = k (k – 1) / 2 (b) T1 = k u, u is the number of unisecants of each point of A. (c) T0 = q(q – 1) / 2 + u(u – 1) / 2. Proof (a): T2 = the number of bisecants of the (k,n)-arc A, the (k,n)-arc A contains k points, each two of them determine a bisecant line, so: T2 = k 2       = k! / (k – 2)!  2! = k(k – 1) / 2 IBN AL- HAITHAM J. FOR PURE & APPL. SCI VOL.22 (2) 2009 Proof (b): T1 = the number of unisecants to the (k,n)-arc A. By Theorem (1.6) there exists exactly u = q + 2 – k lines through any point M in (k,n)-arc A, since the number of points on (k,n)-arc is k. Then there exists ku = k(q + 2 – k) unisecants of the (k,n)-arc A. Proof (c): T0 be the number of the external lines to the (k,n)-arc A, then; T0 + T1 + T2 = q 2 + q + 1 represents all the lines in PG(2,q) then, T0 = q 2 + q + 1 – T1 – T2 from part (a) and (b) T0 = q 2 + q + 1 – k u – k(k – 1) / 2 Since, u = q + 2 – k  k = q + 2 – u, then T0 = q 2 + q + 1 – u (q + 2 – u) – (q + 2 – u)(q + 1 – u) / 2 T0 = 1 2 [2q 2 + 2q + 2 – 2u(q + 2 – u) – (q + 2 – u) (q + 1 – u)] T0 = 1 2 [2q 2 + 2q + 2 – 2uq – 4u + 2u 2 – q 2 – q + uq – 2q – 2 + 2u + uq + u – u 2 ] T0 = 1 2 [2q 2 + 2q – 4u + 2u 2 – q 2 – 3q + 3u – u 2 ] T0 = 1 2 [q 2 – q + u 2 – u] T0 = q (q – 1) / 2 + u(u – 1) / 2 1.11 Theorem (3) A (k,n)-arc A in PG(2,q) is complete if and only if C0 = 0. Proof:  Let A be a complete (k,n)-arc in PG(2,q) and suppose that C0  0, then  at least one point say N has an index zero and N  A. Then A  {N} is an arc in PG(2,q). Hence A  A  {N}. Which implies that the (k,n)-arc A is incomplete (contradicts the hypothesis).  suppose that C0 = 0 for the (k,n)-arc A then there are no points of index zero, for A, so the (k,n)-arc A is a complete. 1.12 Theorem (3) If a (k,n)-arc A is maximal arc in PG(2,q), then, (a) if n = q + 1, then A = PG(2,q) (b) if n = q, then A = PG(2,q) \ L , where L is line (c) if 2  n  q, then nq and the dual of the complements of (k,n)-arc A forms a (q(q + 1 – n) / n,q / n)-arc, also maximal. Proof (a): A (k,n)-arc A is a maximal in PG(2,q), then k = (n – 1)q + n, and if n = q + 1, then k = ((q + 1) – 1) q + (q + 1) = q 2 + q + 1 points A = (q 2 + q + 1,q + 1) = PG(2,q). Proof (b): When n = q, since A is a maximal arc, then A= (n + 1) q + n, A =(q – 1) q + q = q 2 PG(2,q) = q 2 + q + 1 PG(2,q) \ L = PG(2,q) – L = q 2 + q + 1– (q + 1) = q 2 = A. Then A = PG(2,q) \ L. IBN AL- HAITHAM J. FOR PURE & APPL. SCI VOL.22 (2) 2009 Proof (c): When 2  n  q, there exists a point M not in A, so the number of 0-secants through M is q / n, it follows that n / q. the dual of complement of (k,n)-arc A is (T0,q / n)-arc is maximal. Then (q(q+1 – n) / n,q / n)-arc is maximal. 1.13 Lemma (4) For a (k,n)-arc in PG(2,q), the following equation hold: 1. n i i 0 T   = q2 + q + 1 2. n i i 1 iT   = k (q +1) 3. n i i 2 i(i 1)T / 2   = k(k – 1) / 2 4. n i i 2 (i 1)p   = k – 1 Note: Ti denote the total number of i-secants to the arc in PG(2,q). 1.14 Theorem (5) A (k,n)-arc A In PG(2,q) is maximal if and only if every line in PG(2,q) is a 0-secant or n-secant. Proof:  Suppose that (k,n)-arc A is maximal arc in PG(2,q), then the result was proved in the theorem.  Suppose every line in PG(2,q) is a 0-secant or n-secant. If T1 = T2 = T3 = … = Tn – 1 = 0, then n i i 1 iT   = k (q +1) (by Lemma (1.13), (2)) T1 +2 T2 + … + (n – 1) Tn – 1 + n Tn = k (q +1) n Tn = k (q +1) …[1] n i i 2 i(i 1)T / 2   = k(k – 1) / 2 (Lemma (1.13), (3)) T2 + 3T3 + … + n(n – 1) Tn / 2 = k(k – 1) / 2 n(n – 1) Tn / 2 = k(k – 1) / 2 n(n – 1) Tn = k(k – 1) …[2] From equation [1], we get: n Tn / k = q + 1 …[3] From equation [2], we get: n Tn / k = (k – 1) / (n – 1) …[4] From equations [3] and [4], we get (k – 1) / (n – 1) = q + 1  (k – 1) = (q + 1) (n – 1)  (k – 1) = (n – 1) q + (n – 1)  k = (n – 1)q + n (k,n)-arc A is maximal arc (by definition 1.5) Section Two The projective plane PG(2,11) contains 133 points, 133 lines, every line contains 12 points and every points is on 12 points. The points and lines of PG(2,11) are shown in table (1,1). 2.1 The Construction of (k,2)-arc in PG(2,11) (2) Let A = (1,2,13,25) be the set of unit and reference points in PG(2,11) as in the table (1,1) such that: IBN AL- HAITHAM J. FOR PURE & APPL. SCI VOL.22 (2) 2009 1 = (1,0,0), 2 = (0,1,0), 13 = (0,0,1), 25 = (1,1,1), A is (4,2)-arc, since no three points of A are collinear, the points of A are the vertices of a quadrangle whose sides are the lines. L1 = [1,2] = {1,2,3,4,5,6,7,8,9,10,11,12} L2 = [1,13] = {1,13,14,15,16,17,18,19,20,21,22,23} L3 = [1,25] = {1,24,25,26,27,28,29,30,31,32,33,34} L4 = [2,13]= {2,13,24,35,46,57,68,79,90,101,112,123} L5 = [2,25]= {2,14,25,36,47,58,69,80,91,102,113,124} L6 = [13,25] = {3,13,25,37,49,61,73,85,97,109,121,133} The diagonal points of A are the points {3,14,24} where, L1  L6 = 3; L2 L5 =14; L3  L4 = 24. Which are the intersection of pairs of the opposite sides, then there are 61 points on the sides of the quadrangle, four of them are points of the arc A and three of them are the diagonal points of A, so there are 72 points not on the sides of quadrangle which are the points of index zero for A, these points are: 38, 39, 40, 41, 42, 43, 44, 45, 48, 50, 51, 52, 53, 54, 55, 56, 59, 60, 62, 63, 64, 65, 66, 67, 70, 71, 72, 74, 75, 76, 77, 78, 81, 82, 83, 84, 86, 87, 88, 89, 92, 93, 94, 95, 96, 98, 99, 100, 103, 104, 105, 106, 107, 108, 110, 111, 114, 115, 116, 117, 118, 119, 120, 122, 125, 126, 127, 128, 129, 130, 131, 132. Hence A is incomplete (4,2)-arc. 2.2 The Conics in PG(2,11) Through the Reference and Unit Points (1) The general equation of the conic is: 2 2 2 1 1 2 2 3 3 4 1 2 5 1 3 6 2 3 F a x a x a x a x x a x x a x x 0       …[1] By substituting the points of the arc A in [1], then: 1 = (1,0,0) implies that a1 = 0, 2 = (0,1,0), then a2 =0, 13 = (0,0,1), then a3 = 0, 25 = (1,1,1), then a4 + a5 + a6 = 0. Hence, from equation [1] a4 x1 x2 + a5 x1 x3 + a6 x2 x3 = 0 …[2] If a4 = 0, then a5 x1 x3 + a6 x2 x3 = 0, and hence x3(a5 x1 + a6 x2) = 0, then x3 =0 or a5 x1 + a6 x2 = 0, which is a pair of lines, then the conic is degenerated, therefore for a4  0, similarly a5  0 and a6  0. Dividing equation [2] by a4, one can get: 5 6 1 2 1 3 2 3 4 4 a a x x x x x x 0 a a    , then x1 x2 +  x1 x3 +  x2 x3 = 0 …[3] where 5 6 4 4 a a , a a    , so that 1 +  +  = 0 (mod.11)  = – (1 + ), then [3] can be written as:x1 x2 +  x1 x3 – (1 + ) x2 x3 = 0 where   0 and   10 for if  = 0 or  = 10, then degenerated conics, can be obtained thus  = 1, 2, 3, 4, 5, 6, 7, 8, 9. 2.3 The Equation and the Points of the Conics of PG(2,11) Through the Reference and Unit Points (1) 1. If  = 1, then the equation of the conic C1 is x1 x2 + x1 x3 + 9 x2 x3 = 0, the points of C1 are: {1, 2, 13, 25, 40, 53, 63, 77, 87, 100, 104, 116}, which is a complete (12,2)-arc, since there are no points of index zero for C1. 2. If  = 2, then the equation of the conic C2 is x1 x2 + 2x1 x3 + 8 x2 x3 = 0, the points of C2 are: {1, 2, 13, 25, 42, 50, 59, 78, 84, 96, 110, 131}, which is a complete (12,2)-arc, since there are no points of index zero for C2. 3. If  = 3, then the equation of the conic C3 is x1 x2 + 3x1 x3 + 7 x2 x3 = 0, the points of C3 are: {1, 2, 13, 25, 41, 48, 64, 76, 89, 95, 115, 132}, which is a complete (12,2)-arc, since there are no points of index zero for C3. IBN AL- HAITHAM J. FOR PURE & APPL. SCI VOL.22 (2) 2009 4. If  = 4, then the equation of the conic C4 is x1 x2 + 4x1 x3 + 6 x2 x3 = 0, the points of C4 are: {1, 2, 13, 25, 44, 56, 65, 72, 82, 108, 118, 125}, which is a complete (12,2)-arc, since there are no points of index zero for C4. 5. If  = 5, then the equation of the conic C5 is x1 x2 + 5x1 x3 + 5 x2 x3 = 0, the points of C5 are: {1, 2, 13, 25, 43, 51, 67, 71, 99, 103, 119, 127}, which is a complete (12,2)-arc, since there are no point of index zero for C5. 6. If  = 6, then the equation of the conic C6 is x1 x2 + 6x1 x3 + 4x2 x3 = 0, the points of C6 are: {1, 2, 13, 25, 45, 62, 88, 98, 105, 114, 126}, which is a complete (12,2)-arc, since there are no points of index zero for C6. 7. If  = 7, then the equation of the conic C7 is x1 x2 + 7x1 x3 + 3x2 x3 = 0, the points of C7 are: {1, 2, 13, 25, 38, 55, 75, 81, 94, 106, 122, 129}, which is a complete (12,2)-arc, since there are no points of index zero for C7. 8. If  = 8, then the equation of the conic C8 is x1 x2 + 8x1 x3 + 2x2 x3 = 0, the points of C8 are: {1, 2, 13, 25, 39, 60, 74, 86, 92, 111, 120, 128}, which is a complete (12,2)-arc, since there are no points of index zero for C8. 9. If  = 9, then the equation of the conic C9 is x1 x2 + 9x1 x3 + 1x2 x3 = 0, the points of C9 are: {1, 2, 13, 25, 54, 66, 70, 83, 93, 107, 117, 130}, which is a complete (12,2)-arc, since there are no points of index zero for C9. Thus there are nine complete (12,2)-arcs (conics) in PG(2,11) through the reference and the unit points. Hence each arc is a maximum arc, since contains (12) points. 2.4 The Construction of Complete (k,n)-arcs in PG(2,11) (2) 1. The construction of complete arcs of degree 3 In 2.3, we found nine complete (k,2)-arcs which are C1, C2, C3, …, C9, so the complete arcs of degree 3 can be constructed from some complete arcs of degree 2, say C1, C1 = {1, 2, 13, 25, 40, 53, 63, 77, 87, 100, 104, 116}. C1 is not complete (k,3)-arc, since there exist some points of index zero for C1 which are {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 54, 55, 56, 57, 58, 59, 60, 61, 62, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 78, 79, 80, 81,82, 83,84, 85, 86, 88, 89, 90, 91, 92, 93, 94, 95, 96, 98, 99, 101, 102, 103, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 128, 129, 130, 131, 132, 133}, one can add to C1 seven points of index zero which are:{12, 14, 45, 49, 57, 70, 128}, then it can be obtained a complete (19,3)-arc, H1 = {1, 2, 12, 13, 14, 25, 40, 45, 49, 53, 57, 63, 70, 77, 87, 100, 104, 116, 128} since each point not in H1 is on at least one 3-secant and H1 intersect each line in at most 3 points, thus C0 = 0, since there are no points of index zero for H1. Similarly one can find complete arcs of degree 3 from C2, C3, …, C9, by adding some points of index zero to each one of them, call them: H2, H3, …, H9. 2. The construction of complete arcs of degree 4 One will try to construct complete arcs of degree 4 from the complete arcs of degree 3, taken the complete (19,3)-arc: H1 = {1, 2, 12, 13, 14, 25, 40, 45, 49, 53, 57, 63, 70, 77, 87, 100, 104, 116, 128}, since there exist some points of index zero for H1 which are {3, 4, 5, 6, 7, 8, 9, 10, 11, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 41, 42, 43, 44, 46, 47, 48, 50, 51, 52, 54, 55, 56, 57, 58, 59, 60, 61, 62, 64, 65, 66, 67, 68, 69, 71, 72, 73, 74, 75, 76, 78, 79, 80, 81,82, 83,84, 85, 86, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 101, 102, 103, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 129, 130, 131, 132, 133}. the arc H1 is incomplete (19,4)-arc, one can add to H1 eight of these points which are:{10, 23, 32, 38, 47, 84, 90, 105}, then it can be obtained a complete (27,4)-arc S1, S1 = {1, 2, 10, 12, 13, 14, 23, 25, 32, 38, 40, 45, 47, 49, 53, 57, 63, 70, 77, 84, 87, 90, 100, 104, 105, 116, 128}, S1 is a complete (27,4)-arc, since every point not on S1 is on at least one 4-secant, there are no points of index zero for S1 intersect each line in at most 4 points. Similarly one can find IBN AL- HAITHAM J. FOR PURE & APPL. SCI VOL.22 (2) 2009 complete arcs of degree 4 from by adding some points of index zero to H2, H3, …, H9 to obtain complete arcs of degree 4, call them S1, S2, …, S9. 3. The construction of complete arcs of degree 5 In the same method in 1 and 2, one can construct complete arcs of degree 5 by adding some points of index zero to complete arcs of degree 4, for example by taking S1, and the points of index zero for S1 : {3, 4, 5, 6, 7, 8, 9, 11, 15, 16, 17, 18, 19, 20, 21, 22, 24, 26, 27, 28, 29, 30, 31, 33, 34, 35, 36, 37, 39, 41, 42, 43, 44, 46, 48, 50, 51, 52, 54, 55, 56, 58, 59, 60, 61, 62, 64, 65, 66, 67, 68, 69, 71, 72, 73, 74, 75, 76, 78, 79, 80, 81, 82, 83, 85, 86, 88, 89, 91, 92, 93, 94, 95, 96, 97, 98, 99, 101, 102, 103, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 129, 130, 131, 132, 133}, by adding to S1 nine of these points which are: {8, 22, 27, 43, 56, 62, 74, 85,112 }, so one can get a complete arc of degree 5 call M1 , M1 ={1, 2, 8, 10, 12, 13, 14, 22, 23, 25, 27, 32, 38, 40, 43, 45, 47, 49, 53, 56, 57, 62, 63, 70, 74, 77, 84, 85, 87, 90, 100, 104, 105, 112, 116, 128}, M1 is complete (36,5)- arc, since there are no point of index zero; i.e. C0 = 0, so every points not in M1 is on at least one 5-secant, and M1 intersects each line in at most 5 points, Similarly one can find complete arcs of degree 5 by adding some point of index zero to : S2, S3, …, S9 , to obtain complete arcs of degree 5, call them, M1, M3, ….,M9. 4. The construction of complete arcs of degree 6 Complete arcs of degree 6 can be obtained from the complete arcs of degree 5 by adding some points of index zero, for example, one takes the (36,6)-arc, The points of index zero for M1 are:{3, 4, 5, 6, 7, 9, 11, 15, 16, 17, 18, 19, 20, 21, 24, 26, 28, 29, 30, 31, 33, 34, 35, 36, 37, 39, 41, 42, 44, 46, 48, 50, 51, 52, 54, 55, 58, 59, 60, 61, 64, 65, 66, 67, 68, 69, 71, 72, 73, 75, 76, 78, 79, 80, 81, 82, 83, 86, 88, 89, 91, 92, 93, 94, 95, 96, 97, 98, 99, 101, 102, 103, 106, 107, 108, 109, 110, 111, 113, 114, 115, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 129, 130, 131, 132, 133}, and M1 ={1, 2, 8, 10, 12, 13, 14, 22, 23, 25, 27, 32, 38, 40, 43, 45, 47, 49, 53, 56, 57, 62, 63, 70, 74, 77, 84, 85, 87, 90, 100, 104, 105, 112, 116, 128 }, by adding to M1 eleven of these points which are {6, 30, 54, 67, 69, 75, 79, 92, 93, 107, 120}, so we have N1 ={1, 2, 6, 8, 10, 12, 13, 14, 22, 23, 25, 27, 30, 32, 38, 40, 43, 45, 47, 49, 53, 54, 56, 57, 62, 63, 67, 69, 70, 74, 75, 77, 79, 84, 85, 87, 90, 92, 93, 100, 104, 105, 107, 112, 116, 120, 128}, then N1 is complete (47,6)-arc, since There are no points of index zero for N1. Similarly one can construct complete arcs of degree 6 by adding some points of index zero to M2 , M3, …, M9, then complete of degree 6 can be obtained,and call them N2, N3, …, N9. 5. The construction of complete arcs of degree 7 Complete arcs of degree 7 can be constructed from the complete arcs of degree 6, one can take the (47,6)-arc, N1 is complete arc of degree 7, since there exist some points of index zero which are:{3, 4, 5, 7, 9, 11, 15, 16, 17, 18, 19, 20, 21, 24, 26, 28, 29, 31, 33, 34, 35, 36, 37, 39, 41, 42, 44, 46, 48, 50, 51, 52, 55, 58, 59, 60, 61, 64, 65, 66, 68, 71, 72, 73, 76, 78, 80, 81, 82, 83, 86, 88, 89, 91, 94, 95, 96, 97, 98, 99, 101, 102, 103, 106, 108, 109, 110, 111, 113, 114, 115, 117, 118, 119, 121, 122, 123, 124, 125, 126, 127, 129, 130, 131, 132, 133}. By adding to N1 eleven of these points which are:{5, 21, 51, 58, 61, 64, 82, 83, 111, 117, 121}, then K1 ={1, 2, 5, 6, 8, 10, 12, 13, 14, 21, 22, 23, 25, 27, 30, 32, 38, 40, 43, 45, 47, 49, 51, 53, 54, 56, 57, 58, 61, 62, 63, 64, 67, 69, 70, 74, 75, 77, 79, 82, 83, 84, 85, 87, 90, 92, 93, 100, 104, 105, 107, 111, 112, 116, 117, 120, 121, 128} is a complete (58,7)-arc, since there are no points of index zero, thus every point not in K1 is on at least one 7-secant and K1 intersects each line in at most 7 points. Similarly , constructed arcs of degree 7can be contructed from N2, N3, …, N9, call them K2, K3, …, K9. IBN AL- HAITHAM J. FOR PURE & APPL. SCI VOL.22 (2) 2009 6. The construction of complete arcs of degree 8 Complete arcs of degree 8 can be constructed from the complete arcs of degree 7, one can take the (58,7)-arc, k1 is complete (58,7)-arc, since there exist some points of index zero which are:{3, 4, 5, 7, 9, 11, 15, 16, 17, 18, 19, 20, 24, 26, 28, 29, 31, 33, 34, 35, 36, 37, 39, 41, 42, 44, 46, 48, 50, 52, 55, 59, 60, 61, 64, 65, 66, 68, 71, 72, 73, 76, 78, 80, 81, 86, 88, 89, 91, 92, 94, 95, 96, 97, 98, 99, 101, 102, 103, 106, 108, 109, 110, 113, 114, 115, 118, 119, 122, 123, 124, 125, 126, 127, 129, 130, 131, 132, 133}. By adding to k1 thirteen of these points which are:{3, 16, 24, 26, 28, 35, 37, 41, 48, 59, 78, 98, 125}, to obtain a complete (71,8)-arc L1 and L1 ={1, 2, 3, 5, 6, 8, 10, 12, 13, 14, 16, 21, 22, 23, 24, 25, 26, 27, 28, 30, 32, 35, 37, 38, 40, 41, 43, 45, 46, 47, 49, 51, 53, 54, 56, 57, 58, 59, 61, 62, 63, 64, 67, 69, 70, 74, 75, 77, 79, 82, 83, 84, 85, 87, 90, 92, 93, 98, 100, 104, 105, 107, 111, 112, 116, 117, 120, 121, 125, 128} is a complete (71,8)-arc, since there are no points of index zero, thus every point on L1 is on at least one 8-secant and L1 intersects any line in at most 8 points. Similarly arcs of degree 8 can be constructed from K2, K3, …, K9, call them L2, L3, …, L9. 7. The construction of complete arcs of degree 9 Complete arcs of degree 9 can be constructed from the complete arcs of degree 8, the complete (71,8)-arc L1 is taken, L1 is in complete (71,9)-arc, the points of index zero of L1 are:{4, 7, 9, 11, 15, 17, 18, 19, 20, 29, 34, 36, 39, 42, 44, 46, 50, 52, 55, 60, 65, 66, 68, 71, 72, 73, 76, 80, 81, 86, 88, 89, 91, 94, 95, 96, 97, 99, 101, 102, 103, 106, 108, 109, 110, 113, 114, 115, 118, 119, 122, 123, 124, 126, 127, 129, 130, 131, 132, 133}. By adding to L1 twelve of these points which are:{4, 15, 29, 36, 44, 52, 65, 71, 80, 88, 119, 133}, then a complete (83,9)-arc call it O1 is obtained (83,9)-arc and O1 ={1, 2, 3, 4, 5, 6, 8, 10, 12, 13, 14, 15, 16, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 32, 35, 36, 37, 38, 40, 41, 43, 44, 45, 47, 48, 49, 51, 52, 53, 54, 56, 57, 58, 59, 61, 62, 63, 64, 65, 67, 69, 70, 74, 75, 77, 78, 79, 80, 82, 83, 84, 85, 87, 88, 90, 92, 93, 98, 100, 104, 105, 107, 111, 112, 116, 117, 119, 120, 121, 125, 128, 133} is a complete (83,9)-arc, since there are no points of index zero, thus every point on O1 is on at least one 9-secant and O1 intersects any line in at most 9 points. In the same way complete arcs of degree 9 can be obtained from arcs of degree 8, L2, L3, …, L9, call them O2, O3, …, O9. 8. The construction of complete arcs of degree 10 Complete arcs of degree 10 can be constructed from the complete arcs of degree 9 as the following: The complete arc of degree 9, O1 is complete (83,10)-arc, since there exist some points of index zero for O1 which are:{7, 9, 11, 17, 18, 19, 20, 31, 33, 34, 39, 42, 44, 46, 50, 55, 60, 66, 68, 72, 73, 76, 81, 86, 89, 91, 94, 95, 96, 97, 99, 101, 102, 103, 106, 108, 109, 110, 113, 114, 115, 118, 122, 123, 124, 126, 127, 129, 130, 131, 132}. Twelve of these points are added to O1 which are:{9, 17, 31, 42, 46, 73, 86, 95, 96, 99, 103, 113}, then a complete (95,10)-arc call it B1, is obtained B1 ={1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 13, 14, 15, 16, 17, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 35, 36, 37, 38, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 51, 52, 53, 54, 56, 57, 58, 59, 61, 62, 63, 64, 65, 67, 69, 70, 71, 73, 74, 75, 77, 78, 79, 80, 82, 83, 84, 85, 86, 87, 88, 90, 92, 93, 95, 96, 98, 99, 100, 103, 104, 105, 107, 111, 112, 113, 116, 117, 119, 120, 121, 125, 128, 133} is a complete (95,10)-arc, since there are no points of index zero, i.e. C0 = 0. Similarly complete arcs of degree 10 can be constructed, call it B2, B3, …, B9 from O2, O3, …, O9. 9. Them construction of complete arcs of degree 11 Complete arcs of degree 11 can be constructed from complete arcs of degree 10. The complete arcs of degree 10 B1 is taken. B1 is in complete (95,11)-arc, since there exist some points of index zero for B1 which are:{7, 11, 18, 19, 20, 33, 34, 39, 50, 55, 60, 66, 68, 72, 76, 81, 89, 91, 94, 97, 101, 102, 106, 108, 109, 110, 114, 115, 118, 122, 123, 124, 126, 127, 129, 130, 131, 132}, by adding to B1 (26) points of these points which are :{11, 19, 20, IBN AL- HAITHAM J. FOR PURE & APPL. SCI VOL.22 (2) 2009 33, 50, 66, 68, 89, 91, 94, 96, 101, 106, 108, 109, 110, 114, 115, 122, 124, 126, 127, 129, 130, 131, 132}, so we get a complete (121,11)-arc, call it Z1 = {1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 35, 36, 37, 38, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 56, 57, 58, 59, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 77, 78, 79, 80, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 98, 99, 100, 101, 103, 104, 105, 106, 107, 108, 109, 110, 112, 113, 114, 115, 116, 117, 119, 120, 122, 124, 125, 126, 127, 128, 129, 130, 131, 132, 133}, The Z1 is complete (121,11)-arc, since There are no point of index zero ,i.e. Co = 0. Similarly complete arcs of degree 11, Z2, Z3,…,.Z9 can be constructed from complete arcs of degree 10. Refrence 1. Hirschfeld, J.W.P., (1979), Oxford University press, Oxford. 2. Mayssa, G. M., (2004), M.Sc. Thesis, Baghdad University, College of Education Ibn-Al- Haitham, Iraq. 3. Aziz, S.M., (2001), M.Sc. Thesis, Mosul University. 4. Sawsan, J. K., (2001), M.Sc. Thesis, University of Baghdad, College of Education Ibn- Al-Haitham, Iraq. (، رسالة ماجستير، جامعة الموصل، العراق.1002بان عبد الكريم، ) .5 6. Rashad, (1999), M.Sc. Thesis, University of Baghdad, College of Education Ibn-Al- Haitham Table :(1,1) of the points and lines of PG(2,11) i Pi Li 1 1 0 0 2 13 24 35 46 57 68 79 90 101 112 123 2 0 1 0 1 13 14 15 16 17 18 19 20 21 22 23 3 1 1 0 12 13 34 44 54 64 74 84 94 104 114 124 4 2 1 0 7 13 29 45 50 66 71 87 92 108 113 129 5 3 1 0 9 13 31 38 56 63 70 88 95 102 120 127 6 4 1 0 10 13 32 40 48 67 75 83 91 110 118 126 7 5 1 0 4 13 26 39 52 65 78 80 93 106 119 132 8 6 1 0 11 13 33 42 51 60 69 89 98 107 116 125 9 7 1 0 5 13 27 41 55 58 72 86 100 103 117 131 10 8 1 0 6 13 28 43 47 62 77 81 96 111 115 130 11 9 1 0 8 13 30 36 53 59 76 82 99 105 122 128 12 10 1 0 3 13 25 37 49 61 73 85 97 109 121 133 13 0 0 1 1 2 3 4 5 6 7 8 9 10 11 12 14 1 0 1 2 23 34 45 56 67 78 89 100 111 122 133 15 2 0 1 2 18 29 40 51 62 73 84 95 106 117 128 16 3 0 1 2 20 31 42 53 64 75 86 97 108 119 130 17 4 0 1 2 21 32 43 54 65 76 87 98 109 120 131 18 5 0 1 2 15 26 37 48 59 70 81 92 103 114 125 19 6 0 1 2 22 33 44 55 66 77 88 99 110 121 132 20 7 0 1 2 16 27 38 49 60 71 82 93 104 115 126 21 8 0 1 2 17 28 39 50 61 72 83 94 105 116 127 22 9 0 1 2 19 30 41 52 63 74 85 96 107 118 129 23 10 0 1 2 14 25 36 47 58 69 80 91 102 113 124 24 0 1 1 1 123 124 125 126 127 128 129 130 131 132 133 25 1 1 1 12 23 33 43 53 63 73 83 93 103 113 123 26 2 1 1 7 18 34 39 55 60 76 81 97 102 118 123 27 3 1 1 9 20 27 45 52 59 77 84 91 109 116 123 28 4 1 1 10 21 29 37 56 64 72 80 99 107 115 123 29 5 1 1 4 15 28 41 54 67 69 82 95 108 121 123 30 6 1 1 11 22 31 40 49 58 78 87 96 105 114 123 31 7 1 1 5 16 30 44 47 61 75 89 92 106 120 123 32 8 1 1 6 17 32 36 51 66 70 85 100 104 119 123 33 9 1 1 8 19 25 42 48 65 71 88 94 111 117 123 34 10 1 1 3 14 26 38 50 62 74 86 98 110 122 123 35 0 2 1 1 68 69 70 71 72 73 74 75 76 77 78 36 1 2 1 11 23 32 41 50 59 68 88 97 106 115 124 37 2 2 1 12 18 28 38 48 58 68 89 99 109 119 129 38 3 2 1 5 20 34 37 51 65 68 82 96 110 113 127 39 4 2 1 7 21 26 42 47 63 68 84 100 105 121 126 40 5 2 1 6 15 30 45 49 64 68 83 98 102 117 132 41 6 2 1 9 22 29 36 54 61 68 86 93 111 118 125 42 7 2 1 8 16 33 39 56 62 68 85 91 108 114 131 43 8 2 1 10 17 25 44 52 60 68 87 95 103 122 130 44 9 2 1 3 19 31 43 55 67 68 80 92 104 116 128 45 10 2 1 4 14 27 40 53 66 68 81 94 107 120 133 46 0 3 1 1 90 91 92 93 94 95 96 97 98 99 100 47 1 3 1 10 23 31 39 47 66 74 82 90 109 117 125 48 2 3 1 6 18 33 37 52 67 71 86 90 105 120 124 49 3 3 1 12 20 30 40 50 60 70 80 90 111 121 131 50 4 3 1 4 21 34 36 49 62 75 88 90 103 116 129 51 5 3 1 8 15 32 38 55 61 78 84 90 107 113 130 52 6 3 1 7 22 27 43 48 64 69 85 90 106 122 127 53 7 3 1 11 16 25 45 54 63 72 81 90 110 119 128 54 8 3 1 3 17 29 41 53 65 77 89 90 102 114 126 55 9 3 1 9 19 26 44 51 58 76 83 90 108 115 133 56 10 3 1 5 14 28 42 56 59 73 87 90 104 118 132 57 0 4 1 1 101 102 103 104 105 106 107 108 109 110 111 58 1 4 1 9 23 30 37 55 62 69 87 94 101 119 126 59 2 4 1 11 18 27 36 56 65 74 83 92 101 121 130 60 3 4 1 8 20 26 43 49 66 72 89 95 101 118 124 61 4 4 1 12 21 31 41 51 61 71 81 91 101 122 132 62 5 4 1 10 15 34 42 50 58 77 85 93 101 120 128 63 6 4 1 5 22 25 39 53 67 70 84 98 101 115 129 64 7 4 1 3 16 28 40 52 64 76 88 100 101 113 125 65 8 4 1 7 17 33 38 54 59 75 80 96 101 117 133 66 9 4 1 4 19 32 45 47 60 73 86 99 101 114 127 67 10 4 1 6 14 29 44 48 63 78 82 97 101 116 131 68 0 5 1 1 35 36 37 38 39 40 41 42 43 44 45 69 1 5 1 8 23 29 35 52 58 75 81 98 104 121 127 70 2 5 1 5 18 32 35 49 63 77 80 94 108 122 125 71 3 5 1 4 20 33 35 48 61 74 87 100 102 115 128 72 4 5 1 9 21 28 35 53 60 78 85 92 110 117 124 73 5 5 1 12 15 25 35 56 66 76 86 96 106 116 126 74 6 5 1 3 22 34 35 47 59 71 83 95 107 119 131 75 7 5 1 6 16 31 35 50 65 69 84 99 103 118 133 76 8 5 1 11 17 26 35 55 64 73 82 91 111 120 129 77 9 5 1 10 19 27 35 54 62 70 89 97 105 113 132 78 10 5 1 7 14 30 35 51 67 72 88 93 109 114 130 79 0 6 1 1 112 113 114 115 116 117 118 119 120 121 122 80 1 6 1 7 23 28 44 49 65 70 86 91 107 112 128 81 2 6 1 10 18 26 45 53 61 69 88 96 104 112 131 82 3 6 1 11 20 29 38 47 67 76 85 94 103 112 132 83 4 6 1 6 21 25 40 55 59 74 89 93 108 112 127 84 5 6 1 3 15 27 39 51 63 75 87 99 111 112 124 85 6 6 1 12 22 32 42 52 62 72 82 92 102 112 133 86 7 6 1 9 16 34 41 48 66 73 80 98 105 132 130 87 8 6 1 4 17 30 43 56 58 71 84 97 110 112 125 88 9 6 1 5 19 33 36 50 64 78 81 95 109 112 126 89 10 6 1 8 14 31 37 54 60 77 83 100 106 112 129 90 0 7 1 1 46 47 48 49 50 51 52 53 54 55 56 91 1 7 1 6 23 27 42 46 61 76 80 95 110 114 129 92 2 7 1 4 18 31 44 46 59 72 85 98 111 113 126 93 3 7 1 7 20 25 41 46 62 78 83 99 104 120 125 94 4 7 1 3 21 33 45 46 58 70 82 94 106 118 130 95 5 7 1 5 15 29 43 46 60 74 88 91 105 119 133 96 6 7 1 10 22 30 38 46 65 73 81 100 108 116 124 97 7 7 1 12 16 26 36 46 67 77 87 97 107 117 127 98 8 7 1 8 17 34 40 46 63 69 86 92 109 115 132 99 9 7 1 11 19 28 37 46 66 75 84 93 102 122 131 100 10 7 1 9 14 32 39 46 64 71 89 96 103 121 128 101 0 8 1 1 57 58 59 60 61 62 63 64 65 66 67 102 1 8 1 5 23 26 40 54 57 71 85 99 102 116 130 103 2 8 1 9 18 25 43 50 57 75 82 100 107 114 132 104 3 8 1 3 20 32 44 56 57 69 81 93 105 117 129 105 4 8 1 11 21 30 39 48 57 77 86 95 104 113 133 106 5 8 1 7 15 31 36 52 57 73 89 94 110 115 131 107 6 8 1 8 22 28 45 51 57 74 80 97 103 120 126 108 7 8 1 4 16 29 42 55 57 70 83 96 109 122 124 109 8 8 1 12 17 27 37 47 57 78 88 98 108 118 128 110 9 8 1 6 19 34 38 53 57 72 87 91 106 121 125 111 10 8 1 10 14 33 41 49 57 76 84 92 11 119 127 112 0 9 1 1 79 80 81 82 83 84 85 86 87 88 89 113 1 9 1 4 23 25 38 51 64 77 79 92 105 118 131 114 2 9 1 3 18 30 42 54 66 78 79 91 103 115 127 115 3 9 1 10 20 28 36 55 63 71 79 98 106 114 133 116 4 9 1 8 21 27 44 50 67 73 79 96 102 119 125 117 5 9 1 9 15 33 40 47 65 72 79 97 104 122 129 118 6 9 1 6 22 26 41 56 60 75 79 94 109 113 128 119 7 9 1 7 16 32 37 53 58 74 79 95 111 116 132 12 8 9 1 5 17 31 45 48 62 76 79 93 107 121 124 121 9 9 1 12 19 29 39 49 59 69 79 100 110 120 130 122 10 9 1 11 14 34 43 52 61 70 79 99 108 117 126 123 0 10 1 1 24 25 26 27 28 29 30 31 32 33 34 124 1 10 1 3 23 24 36 48 60 72 84 96 108 120 132 125 2 10 1 8 18 24 41 47 64 70 87 93 110 116 133 126 3 10 1 6 20 24 39 54 58 73 88 92 107 122 126 127 4 10 1 5 21 24 38 52 66 69 83 97 111 114 128 128 5 10 1 11 15 24 44 53 62 71 80 100 109 118 127 129 6 10 1 4 22 24 37 50 63 76 89 91 104 117 130 130 7 10 1 10 16 24 43 51 59 78 86 94 102 121 129 131 8 10 1 9 17 24 42 49 67 74 81 99 106 113 131 132 9 10 1 7 19 24 40 56 61 77 82 98 103 119 124 133 10 10 1 12 14 24 45 55 65 75 85 95 105 115 125 2002( 2) 22المجلد مجلة ابن الهيثم للعلوم الصرفة والتطبيقية حول PG(2,11)في المستوي االسقاطي (k,n)إنشاء األقواس الكاملة  n  11 3حيث ان GF(11)حقل كالوا علي طالب محمد جامعة بغداد ،ابن الهيثم -كلية التربية،قسم الرياضيات الخالصة في الفضاء االسقاطي الثناايي (k,n)ان الغاية االساسية من هذا البحث هو ايجاد قوس كامل PG(2,q) حااول حقاال كااالوGF(11) وذلااب بواسااطة اضااافة بقاا النقاااط دليى ااا سااف الاا القااوس . n  11 3حيث (k,n – 1)الكامل عى استقامة واحدة. n + 1من النقاط ليس هنالب kهو مجموعة (k,n)القوس .(k+1,n)هو قوس ال يمكن ان يكون محتوى في القوس (k,n)القوس الكامل