مجلة إبن الھیثم للعلوم الصرفة و التطبیقیة 2012 السنة 25 المجلد 1 العدد Ibn Al-Haitham Journal for Pure and Applied Science No. 1 Vol. 25 Year 2012 On Projective 3-Space Over Galois Field A. SH. Al-Mukhtar Department of Mathematics, College of Education Ibn-Al-Haitham ,University of Baghdad Received in : 11 May 2011 Accepted in :16 June 2011 Abstract The purpose of this paper is to give the definition of projective 3-space PG(3,q) over Galois field GF(q), q = pm for some prime number p and some integer m. Also, the definition of the plane in PG(3,q) is given and state the principle of duality . Moreover some theorems in PG(3,q) are proved. Keywords: plane, duality, Galois field. 1- Introduction, [1,2] A projective 3 – space PG(3,K) over a field K is a 3 – dimensional projective space which consists of points, lines and planes with the incidence relation between them. The projective 3 – space satisfies the following axioms: A. Any two distinct points are contained in a unique line. B. Any three distinct non-collinear points, also any line and point not on the line are contained in a unique plane. C. Any two distinct coplanar lines intersect in a unique point. D. Any line not on a given plane intersects the plane in a unique point. E. Any two distinct planes intersect in a unique line. A projective space PG(3,q) over Galois field GF(q), q = p m, for some prime number p and some integer m, is a 3 – dimensional projective space. Any point in PG(3,q) has the form of a quadrable (x1, x2, x3, x4), where x1, x2, x3, x4 are elements in GF(q) with the exception of the quadrable consisting of four zero elements. Two quadrables (x1, x2, x3, x4) and (y1, y2, y3, y4) represent the same point if there exists  in GF(q) \ {0} such that (x1, x2, x3, x4) =  (y1, y2, y3, y4), this is denoted by (x1, x2, x3, x4)  (y1, y2, y3, y4). Similarly, any plane in PG(3,q) has the form of a quadrable [x1, x2, x3, x4], where x1, x2, x3, x4 are distinct elements in GF(q) with the exception of the quadrable consisting of four zero elements. مجلة إبن الھیثم للعلوم الصرفة و التطبیقیة 2012 السنة 25 المجلد 1 العدد Ibn Al-Haitham Journal for Pure and Applied Science No. 1 Vol. 25 Year 2012 Two quadrables [x1, x2, x3, x4] and [y1, y2, y3, y4] represent the same plane if there exists  in GF(q) \ {0} such that [x1, x2, x3, x4] =  [y1, y2, y3, y4], this is denoted by [x1, x2, x3, x4]  [y1, y2, y3, y4].. Also a point P(x1, x2, x3, x4) is incident with the plane  [a1, a2, a3, a4] iff a1 x1 + a2 x2 + a3 x3 + a4 x4 = 0. Definition 1.1: [2] A plane  in PG(3,q) is the set of all points P(x1, x2, x3, x4) satisfying a linear equation u1 x1 + u2 x2 + u3 x3 + u4 x4 = 0. This plane is denoted by  [u1, u2, u3, u4]. It should be noted that if one takes another representation of P, say ( x1,  x2,  x3,  x4), then since u1  x1 + u2  x2 + u3  x3 + u4  x4 =  (u1 x1 + u2 x2 + u3 x3 + u4 x4), the definition of a plane is independent of the choice of representations of points on it. 2- Principle of Duality Definition 2.1: [3] For any S = PG(n,K), there is a dual space S*, whose points and primes (subspaces of dimensions (n – 1)) are respectively the primes and points of S. For any theorem true in S, there is an equivalent theorem true in S*. In particular, if T is a theorem in S stated in terms of points, primes and incidence, the same theorem is true in S* and gives a dual theorem T* in S by interchanging "point" and "prime" whenever they occur. In PG(3,K) point and plane are dual, where as the dual of a line is a line. Theorem 2.2: The points of PG(3,q) have unique forms which are (1,0,0,0), (x,1,0,0), (x, y,1,0), (x, y, z,1) for all x, y, z in GF(q). Proof : Let P(x1, x2, x3, x4); x1; x2, x3, x4GF(q) be any point in PG(3,q), then either x40 or x4=0. If x4  0, then P(x1, x2, x3, x4)  31 2 4 4 4 ( , , ,1) xx x x x x , where 1 4  x x x , 2 4  x y x , 3 4  x z x . If x4 = 0, then either x3  0 or x3 = 0. If x3  0, then P(x1, x2, x3, 0)  1 2 3 3 ( , ,1,0) x x x x , where 1 3  x x x , 2 3  x y x . If x3 = 0, then either x2  0 or x2 = 0. If x2  0, then P(x1, x2, 0, 0)  1 2 ( ,1,0,0) x x = P(x, 1, 0, 0), where 1 2  x x x . مجلة إبن الھیثم للعلوم الصرفة و التطبیقیة 2012 السنة 25 المجلد 1 العدد Ibn Al-Haitham Journal for Pure and Applied Science No. 1 Vol. 25 Year 2012 If x2 = 0, then x1  0 and P(x1, 0, 0, 0)  1 1 ( ,0,0, 0) x x = P(1, 0, 0, 0). Similarly, one can prove the dual of theorem 1. Theorem 2.3: The planes of PG(3,q) have unique forms which are [1,0,0,0], [x,1,0,0], [x, y,1,0], [x, y, z,1] for all x, y, z in GF(q). Theorem 2.4: [1] Every line in PG(3,q) contains exactly q + 1 points. Theorem 2.5: [1] Every point in PG(3,q) is on exactly q + 1 lines. Theorem 2.6: [1] Every plane in PG(3,q) contains exactly q 2 + q + 1 points (lines). Theorem 2.7: [1] Every point in PG(3,q) is on exactly q 2 + q + 1 p lanes. Theorem 2.8: There exist q 3 + q 2 + q + 1 points in PG(3,q). Proof : From theorem 1, the points of PG(3,9) have unique forms which are (1,0,0,0), (x,1,0,0), (x, y,1,0), (x, y, z,1) for all x, y, z in GF(q). It is clear that there exists one point of the form (1,0,0,0). There exist q points of the form (x,1,0,0). There exist q 2 points of the form (x, y,1,0). There exist q 3 points of the form (x, y, z,1). Similarly, one can prove the dual of theorem 2.8. Theorem 2.9: There exist q 3 + q 2 + q + 1 p lanes in PG(3,q). Theorem 2.10: Any two p lanes in PG(3,q) intersect in exactly q + 1 points. Proof : By axiom E, since any two planes intersect in a unique line and each line in PG(3,q) contains exactly q + 1 points, then any two p lanes intersect in exactly q + 1 points. Theorem 2.11: مجلة إبن الھیثم للعلوم الصرفة و التطبیقیة 2012 السنة 25 المجلد 1 العدد Ibn Al-Haitham Journal for Pure and Applied Science No. 1 Vol. 25 Year 2012 Any line in PG(3,q) is on exactly q +1 planes. Proof : Let l be any line in PG(3,q) and m be another line in PG(3,q) not coplanar with l . m contains exactly q + 1 points. By axiom B, l determines a unique plane with any point of m. Hence there exist q + 1 planes through l . If there exists another plane through l , then this plane intersects m in another point which is a contradiction. Hence l is on exactly q + 1 planes. Theorem 2.12: Any two points in PG(3,q) are on exactly q + 1 planes. Proof : Since any two points determine a unique line and by theorem 10, then every line is on exactly q + 1 planes. Theorem 2.13: There exist (q 2 + 1) (q 2 + q + 1) lines in PG(3,q). Proof : In PG(3,q), there exist q 3 + q 2 + q + 1 planes, and each plane contains exactly q 2 + q + 1 lines, then the numbers of lines is equal to (q 3 + q2 + q + 1)( q2 + q + 1), but each line is on q + 1 planes, then there exist exactly 3 2 2 2 2( 1)( 1) ( 1)( 1) ( 1)           q q q q q q q q q lines in PG(3,q). Now, some theorems on projective 3-space PG(3,q) can be proved. Theorem 2.14: Four distinct points A(x1, x2, x3, x4), B(y1, y2, y3, y4), C(z1, z2, z3, z4), and D(w1, w2, w3, w4) are coplanar iff 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 0   x x x x y y y y z z z z w w w w Proof : Let  [u1, u2, u3, u4] be a plane containing the points A, B, C, D, then x1 u1 + x2 u2 + x3 u3 + x4 u4 = 0 y1 u1 + y2 u2 + y3 u3 + y4 u4 = 0 z1 u1 + z2 u2 + z3 u3 + z4 u4 = 0 w1 u1 + w2 u2 + w3 u3 + w4 u4 = 0 It is known from the linear algebra that this system of equations have non zero solutions for u1,u2, u3,u4 iff  = 0. Thus the necessary and sufficient conditions for four points to be coplanar that  = 0. مجلة إبن الھیثم للعلوم الصرفة و التطبیقیة 2012 السنة 25 المجلد 1 العدد Ibn Al-Haitham Journal for Pure and Applied Science No. 1 Vol. 25 Year 2012 Corollary 2.15: If four distinct points in PG(3,q) A(x1, x2, x3, x4), B(y1, y2, y3, y4), C(z1, z2, z3, z4), and D(w1, w2, w3, w4) are collinear, then  = 0. This follows from theorem 2.14 and the incidence of these points on a line of some plane. From the principle of duality , one can prove: Theorem 2.16: Four distinct planes in PG(3,q) A[x1, x2, x3, x4], B[y1, y2, y3, y4], C[z1, z2, z3, z4], and D[w1, w2, w3, w4] are concurrent (intersecting in one point) iff 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 0   x x x x y y y y z z z z w w w w Theorem 2.17: The equation of the plane determined by three distinct points A(y 1, y2, y3, y4), B(z1, z2, z3, z4), and C(w1, w2, w3, w4) is 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 2 3 4 3 1 4 1 2 4 3 2 1 2 3 4 1 3 1 4 2 1 2 4 3 3 2 1 4 2 3 4 3 1 4 1 2 4 3 2 1 0      x x x x y y y y z z z z w w w w y y y y y y y y y y y y z z z x z z z x z z z x z z z x w w w w w w w w w w w w where (x1, x2, x3, x4) be any variable point on the plane, and it’s coordinates are: 2 3 4 3 1 4 1 2 4 3 2 1 2 3 4 3 1 4 1 2 4 3 2 1 2 3 4 3 1 4 1 2 4 3 2 1 , , ,          y y y y y y y y y y y y z z z z z z z z z z z z w w w w w w w w w w w w Similarly, one can prove the dual of this theorem. Theorem 2.18: The equation of the point determined by three distinct planes (non-collinear) in PG(3,q) a[y1, y2, y3, y4], b[z1, z2, z3, z4], and c[w1, w2, w3, w4] is 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4  x x x x y y y y z z z z w w w w مجلة إبن الھیثم للعلوم الصرفة و التطبیقیة 2012 السنة 25 المجلد 1 العدد Ibn Al-Haitham Journal for Pure and Applied Science No. 1 Vol. 25 Year 2012 2 3 4 3 1 4 1 2 4 3 2 1 2 3 4 1 3 1 4 2 1 2 4 3 3 2 1 4 2 3 4 3 1 4 1 2 4 3 2 1 0    y y y y y y y y y y y y z z z x z z z x z z z x z z z x w w w w w w w w w w w w where [x1, x2, x3, x4] be any variable plane passing through the point, and it’s coordinates are: 2 3 4 3 1 4 1 2 4 3 2 1 2 3 4 3 1 4 1 2 4 3 2 1 2 3 4 3 1 4 1 2 4 3 2 1 , , ,           y y y y y y y y y y y y z z z z z z z z z z z z w w w w w w w w w w w w Notation 2.19: If v is the vector with components (a1, a2, a3, a4), then the symbol P(v) means that the coordinates of the point P are (a1, a2, a3, a4) in a projective 3–space S = PG(3,K). Definition 2.20:[3] The points Pi(vi), with i = 1, …, m are linearly dependent or independent according as the vectors vi are linearly dependent or independent. Definition 2.21:[3] If the points P1, P2, , Pm are linearly dependent, then at least one of the ci’s of the equation 1 ( ) 0    m i i i i c v is not equal to zero, say c1, then P1 = 1 1 c ( c2 P2 + c3 P3 +  + cm Pm ). The point P1 is then said to be a linear combination of the points P2, P3, , Pm. This definition may be dualized by replacing the word "point" by the word "plane", and the geometric meaning of linear dependence of points or planes may now be given. Theorem 2.22: Two points (planes) in PG(3,q) are linearly dependent iff they coincide. Proof : Let P and Q be any two points. If P and Q are linearly dependent, then there exist c1 and c2 such that (c1, c2)  (0,0), c1 P + c2 Q = . If c1 = 0, then c2 Q = . This implies c2 = 0, since Q  (0,0,0). Then c1  0 and similarly c2  0, 2 1 c Q c    . This means that P and Q coincide. If P and Q are coincide, then there exist c1 0, c2  0 s.t . c1 P = c2 Q. Hence, c1 P  c2 Q =  and thus P and Q are linearly dependent. Theorem 2.23: Four points in PG(3,q) are linearly dependent iff they are coplanar. Proof : مجلة إبن الھیثم للعلوم الصرفة و التطبیقیة 2012 السنة 25 المجلد 1 العدد Ibn Al-Haitham Journal for Pure and Applied Science No. 1 Vol. 25 Year 2012 Let A(x1, x2, x3, x4), B(y1, y2, y3, y4), C(z1, z2, z3, z4), and D(w1, w2, w3, w4) be any four points in S. If A, B, C, D are linearly dependent, then there exist c1, c2, c3 and c4 in K such that (c1, c2, c3, c4)  (0,0,0,0) and c1 A+ c2 B+ c3 C + c4 D =  c1 A + c2 B + c3 C + c4 D = c1 (x1, x2, x3, x4) + c2 (y1, y2, y3, y4) + c3 (z1, z2, z3, z4) + c4 (w1, w2, w3, w4) = (0,0,0,0) c1 x1 + c2 y1 + c3 z1 + c4 w1 = 0 c1 x2 + c2 y2 + c3 z2 + c4 w2 = 0 c1 x3 + c2 y3 + c3 z3 + c4 w3 = 0 c1 x4 + c2 y4 + c3 z4 + c4 w4 = 0 (1) This system has non zero solutions for c1, c2, c3, c4 iff 1 2 3 41 1 1 1 1 2 3 42 2 2 2 1 2 3 43 3 3 3 1 2 3 44 4 4 4 0    x x x xx y z w y y y yx y z w z z z zx y z w w w w wx y z w by theorem 2.14 the points A, B, C, D are coplanar. Conversely, if the points A, B, C, D are coplanar, then 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 0   x x x x y y y y z z z z w w w w , then 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 0 x y z w x y z w x y z w x y z w . So the system (1) of equations has non zero solutions for c1, c2, c3, c4. Thus A, B, C, D are linearly dependent. Theorem 2.24: Any five points (planes) in PG(3,q) in S are linearly dependent. Proof : Let A(a1, a2, a3, a4), B(b1, b2, b3, b4), C(c1, c2, c3, c4), D(d1, d2, d3, d4) and E(e1, e2, e3, e4) be any five points in S. Let a A + b B + c C + d D + e E =  a (a1,a2,a3,a4) + b (b1,b2,b3,b4) + c (c1,c2,c3,c4) + d (d1,d2,d3,d4) + e (e1,e2,e3,e4) =  a a1 + b b1 + c c1 + d d1 + e e1 = 0 a a2 + b b2 + c c2 + d d2 + e e2 = 0 a a3 + b b3 + c c3 + d d3 + e e3 = 0 a a4 + b b4 + c c4 + d d4 + e e4 = 0 This system of 4 linear homogeneous equations in 5 unknowns a, b, c, d, e has non trivial solutions since 4 < 5. Then A, B, C, D, E are linearly dependent. Theorem 2.25: مجلة إبن الھیثم للعلوم الصرفة و التطبیقیة 2012 السنة 25 المجلد 1 العدد Ibn Al-Haitham Journal for Pure and Applied Science No. 1 Vol. 25 Year 2012 In PG(3,q) if P1, P2, , Pm are linearly independent points while P1, P2, , Pm + 1 are linearly dependent, then the coordinates of the points may be chosen so that P1 + P2 +  + Pm = Pm + 1. Proof : Since the points P1, P2, , Pm + 1 are linearly dependent, constants c1, c2, , cm + 1  0, 0, , 0 exist such that c1 P1(v1) + c2 P2(v2) +  + cm Pm(vm) + cm + 1 P m + 1(v m + 1) = . Now, cm + 1  0, for otherwise the points P1, P2, , Pm would be dependent contrary to hypothesis. The equation may, therefore, be solved for Pm + 1 giving Pm + 1 = m 1 1 c   [ c1 P1(v1) +  + cm Pm(vm) ] = k1 P1(v1) +  + km Pm(vm) = P1(k1 v1) +  + Pm(km vm) where 1   i i m c k c , i = 1, , m or dropping the symbols ki vi , Pm + 1=P1+ P2++Pm. Theorem 2.26: In PG(3,q) a point D is on the plane determined by three distinct points A, B, C iff D is a linear combination of A, B, C. Proof : If D is on the plane determined by three distinct points, then A, B, C, D are coplanar. By theorem (5), they are linearly dependent, there exist constants a, b, c, d such that not all of them are zero and a A + b B + c C + d D = . If d = 0, then a A + b B + c C = , which implies that a = b = c = 0, since A, B, C are linearly independent, which is a contradiction. Since any three noncollinear points in the plane are linearly independent, [3]. So d  0, and then D = ( )  a d A + ( )  b d B + ( )  c d C Thus D is a linear combination of A, B, C. Suppose D is a linear combination of A, B, C, then there exist constants c1, c2, c3 not all of them are zero such that: D = c1 A + c2 B + c3 C, which implies c1 A + c2 B + c3 C + (–1) D = , then it follows that A, B, C, D are linearly dependent. By theorem (5), the points A, B, C, D are coplanar. References 1. Al-Mukhtar, A.Sh. (2008) Complete Arcs and Surfaces in three Dimensional Projective Space Over Galois Field, Ph.D. Thesis, University of Technology, Iraq. 2. Kirdar,M.S. and Al-Mukhtar, A.Sh. (2009) Engineering and Technology Journal, On Projective 3-Space, Vol.27(8): مجلة إبن الھیثم للعلوم الصرفة و التطبیقیة 2012 السنة 25 المجلد 1 العدد Ibn Al-Haitham Journal for Pure and Applied Science No. 1 Vol. 25 Year 2012 3. Hirschfeld, J. W. P. (1998) Projective Geometries Over Finite Fields, Second Edition, Oxford University Press. مجلة إبن الھیثم للعلوم الصرفة و التطبیقیة 2012 السنة 25 المجلد 1 العدد Ibn Al-Haitham Journal for Pure and Applied Science No. 1 Vol. 25 Year 2012 حول الفضاء الثالثي االسقاطي حول حقل كالوا آمال شهاب المختار جامعة بغداد، ابن الهیثم -كلیة التربیة، قسم الریاضیات 2011 حزیران 16: قبل البحث في 2011 آیار 11:استلم البحث في الخالصة ــقاطي ــاء تعریـــف الفــــضاء الثالثـــي االســ ــث هـــو إعطــ ــالوا PG(3,q)الغـــرض مــــن هـــذا البحــ ـــل كــ ، GF(q) فـــي حقـ q = p m لبعض قیم ، p و m اذ ان،p عدد أولي و mكذلك تقدیم تعریف المستوي في . عدد صحیحPG(3,q) ونص .PG(3,q)مبدأ الثنائیة وبرهنت بعض المبرهنات في . مستوي ، مبدأ الثنائیة ، حقل كالوا:الكلمات المفتاحیة