International Journal of Pure and Applied Mathematics Volume 80 No. 1 2012, 1-15 ISSN: 1311-8080 (printed version) url: http://www.ijpam.eu PA ijpam.eu RELATION BETWEEN PROPERTIES OF TERNARY SEMIRINGS AND PROJECTIVE PLANES Flamure Ibraimi 1, Alit Ibraimi 2 Department of Mathematics State University of Tetovo Tetovo, MACEDONIA Abstract: The purpose of the present paper is to study the concept of semiring in projective plane. We introduce the properties of ternary semirings and we show that projective plane with these properties is a desarguesian projective plane. AMS Subject Classification: 51E1 Key Words: projective plane, theorem of desargues, ternary semiring 1. Introduction Ternary semirings are one of the generalized structures of semirings introduced by Dutta and Kar [1]. D.H. Lehmer [3] introduced the notion of ternary algebraic system called triplexes which turn out to be commutative ternary groups. Some works on ternary semiring may be found in [2], [4], [5], [6], [7], [8], [9] and [10]. Main purpose in this paper is to give the relation between properties of ternary semirings and projective plane. By coordinatization of projective plane we define addition and multiplication from the set of all points P P : PIl, P J to the set of points on l. Then by introducing the properties of ternary semirings, we show that projective plane with these properties is a desarguesian projective plane. Received: April 22, 2011 Correspondence author c 2012 Academic Publications, Ltd. url: www.acadpubl.eu
2 F. Ibraimi, A. Ibraimi 2. Preliminaries Definition 1. Incidence structure or incidence geometry is called a tripple Σ = (P, B, I) of sets P, B, I which satisfy conditions: P B = ϕ, I P B (P, B are disjunct set and I an binary relation between the elements of P and B). Definition 2. Theincidence structurep = (P,B,I) iscalled theprojective plane if and only if: 1. Any two distinct points are incident with a unique line. 2. Any two distinct lines are incident with a unique point. 3. There exist four points such that no three are incident with one line. Any set of points which satisfies conditions (i) and (ii) is called a closed configuration, and any closed configuration which does not satisfy condition (iii) is called a degenerate plane. Lemma 1. Any projective plane contains a quadrilateral. Theorem 1. Let P be a finite projective plane. There exists a positive integer n 2 such that: 1. Each line contains exactly n+1 points; 2. Each point is on exactly n+1 lines; 3. Plane P contains n 2 +n+1 points and n 2 +n+1 lines. Definition 3. Ordered triple (A,+ ) is said to be a ternary semiring if the following condition are satisfied: 1. (A, +) is commutative semigroup, 2. (abc)de = a(bcd)e = ab(cde), 3. (a+b)cd = acd +bcd, 4. a(b+c)d = abd +acd, 5. ab(c+d) = abc +abd,
RELATION BETWEEN PROPERTIES OF... 3 Figure 1 for all a, b, c, d, e A. Definition 4. Let (A, + )be a ternary semiring. If there exists an element 0 Asuch that 0+x = xand 0xy = x0y = xy0 =0 for all x, y A, then 0 is called the zero element or simply the zero of the ternary semiring (A,+ ). In this case we say that Sis a ternary semiring with zero. Coordinatization. An ordered quadruple (OXY I) of distinct point such that no three are incident whith one line will be called a coordinate system. In this coordinate system (OXYI) we define the following elements l = OI,l 1 = OY,l 2 = OX,l = XY (see Figure 1). Now define set A = {P P PIl, P J. For any point P not incident with the line l we can define A = PY lb = PX l and the ordered pair (AB) to be the coordinate of the point P(AB). Also we should mention that the incidence structure (P, B, I) is to be Desarguesian projective plane. Proposition 1. There exists a one to one correspondence between P = {P P P and A A. In addition, we define a function λ A+B from the set A to the set of points on l. Let (O,X,Y,I) be a coordinate system with associated lines l 1,l 2,l 3,l and the set A. For any two points AB A we define a B = (XB l 1 ) J. Now λ A+B = [(YA a B ) X] l, AB A (see Figure 2). So, λ A+B = A + B, AB A. Corollary 1. Let B A, a B = (XB l 1 ) J and C(X Y ), C / l. Then CIa B if and only if Y = X +B.
4 F. Ibraimi, A. Ibraimi Figure 2 Figure 3 Now, also in (OXYI) we define another function δ AB from the set A to the setofpointsonl byδ AB = [(YB m A ) J] l,m A = [(XA YI) O], AB A (see Figure 3). So, δ AB = A B, AB A. Corollary 2. Let A A, m A = [(XA YI) O] and C(X Y ), C / l. Then CIm A if and only if Y = AX. Theorem 1. 3. Relation between Properties of Ternary Semirings and Projective Planes 1. (A+B)+C = A+(B +C), ABC A. 2. A+B = B +A, AB A.
RELATION BETWEEN PROPERTIES OF... 5 Figure 4 3. A+O = O+A = A, A A. 4. (A B) C = A (B C), ABC A. Proof. 1. To proof that (A + B) + C = A + (B + C), ABC A, let ABC O and A C. In the coordinate system (OXYI) { let ABC l. To define S 1 = A + B we determine XB l 1 = B } {,XC l 1 = C } then from the points B,C and Jwe determine the lines a B = B J,a C = C J, YA a B = {A 1 },XA 1 l = {S 1 (see Figure 4). Now, YB a C = {B 1 }, B 1 X l = {S 2 define the point S 2 = B + C and a S2 = (XS 2 l{ 1 ) J } Since S 1 = A+B we define S 3 = (A+B)+C with YS 1 a C = S 1,XS 1 l = {S 3, S 1 (A + B(A+B) + C) also we define A 1 S 1 l = {J 1. Let S 4 = A + (B + C) then we define S 2 = B + C, YA a S2 = {A 2 }XA 2 l = {S 4. Point S 4 S 3 and A 2 (AA + (B +C)). We need to prove the collinearity of the points A 2, S 1 and X. By the conclusion of the Theorem of Desargues, we might look for such triangles perspective from a point to obtain the necessary collinearity. WehavetrianglesYBXandS 1 JA 1perspectivefromthepoint
6 F. Ibraimi, A. Ibraimi Figure 5 S 1. Thus, YB S 1 J = {B 1}, BX JA 1 = {B and YX S 1 A 1 = {J 1 }, from the theorem of Desargues are collinear. Now we consider triangles E JB 1 and YA 1 J 1 perspective from B. So, the points E J YA 1 = {A 2, JB 1 A 1 J 1 = {S 1 and E B 1 YJ 1 = {X are collinear, based on the Theorem of Desargues. This completes the proof of part (i). 2. For A + B = B + A, AB A, let AB O and A B while for A = B = O the result is immediate. Let ABC incident with the linel. To define S = A + B, we have to determine the point XB l 1 = {B, B J = a B, YA a B = {A 2 } then XA 2 l = {S and A 2 = (AA+B) (see Figure 5). For S 1 = B+A we define the line a A = (XA OY) J and point YB a A = {B 1 }, than XB 1 l = {S 1. Since S S 1 we notice A+B = B+A. Now we can prove collinearity of the point A 2,B 1,Xusing the theorem of Desargues. Let us consider triangles A AA 1 and B BB 2 perspective from the point O for which we have A A B B = {X, A A 1 B B 2 = {J, AA 1 BB 2 = {Y. Thus, XJY are collinear point. Now, we choose triangles A B J and ABY perspective from O, so that A B AB = {O, A J AY = {A 1, B J BY = {B 2 } are collinear. Finally, since the points A 2 (AA+B), B 1 (BB +A) and X are collinear then A+B = B +A. 3. A + O = O + A = A, A A. If XA OY = {A, A { J = a A and OY a A = A } then A X l = {A} (see Figure 6).
RELATION BETWEEN PROPERTIES OF... 7 Figure 6 Therefore, O +A = A A A. In similar way we can see that A+O = A, A A. 4. Let ABC A and ABC O to prove that (A B) C = A (B C). Initially, to determine P 1 = A B we introduce points XA YI = {A, OA = m A, YB m A = {B 1 the pointb 1 (BA B) then XB 1 l = {P 1. (fig.7). { } If P 2 = (A B) C, we define P 1 YI = P 1 OP 1 = m P 1, YC m P1 = {C 2 }, C 2 (C,(A B) C), { then C 2 X l = {P 2 }. Now, if P 3 = B C we have XB YI = B }, OB = m B, YC m B = {C 1 }, B 1 (C,B C) then, XC 1 l = {P 3 }. Next, let define the point P 4 = A (B C). We introduce YP 3 m A = {P},P(BCA (B C)) then, PX l = {P 4. Also from thefig. 7, we can see that P 4 P 2. We needto prove thecollinearity of the points C 2,P,X. Now, it s necessary to look for triangles to be used in the theorem of Desargues to establish this collinearity. We have trianglesb P 1 C 1andBB 1 P 3 perspectivefromx. Thus,B P 1 BB 1 = {Y, P 1 C 1 B 1 P 3 = {F and B C 1 BP 3 = {O by the theorem of Desargues are collinear points. Then, by the triangles OP 1 B 1 and YC 1 P 3 perspective from F we have collinearity of the points OP 1 YC 1= {C 2, OB 1 YP 3 = {P and P 1 B 1 C 1 P 3 = {X. This completes the proof of (iii). Theorem 2. 1. (A B C) D E = A (B C D) E = A B (C D E), ABC,DE A.
8 F. Ibraimi, A. Ibraimi Figure 7 2. A B O = A O B = O A B = O, AB A. 3. (A+B) C D = A C D +B C D, ABC,D A. 4. A (B +C) D = A B D +A C D, ABC,D A. 5. A B (C +D) = A B C +A B D, ABC,D A. Proof. 1. For (A B C) D E = A B (C D E), ABC,DE A, { let ABC,DE OI. To define P 1 = A B we determine XA YI = A }, OA = m A, YB m A = {B 1 } then XB 1 l = {P 1 }. (fig.8). Now, if we determine P 2 = (A B) C we havexp 1 YI = {P 1 }, P 1 = m P 1, YC m P1 = {C 1 },where XC 1 l = {P 2 }. { Now, for the point P 3 = D E we introduce XD YI = D }, OD = m D, YE m D = {E 1 } and XE 1 l = {P 3 }. Finally, to define P = (A B C) D E we have XP 2 YI = {P 2 }, OP 2 = m P 2, YP 3 m P2 =
RELATION BETWEEN PROPERTIES OF... 9 Figure 8 {P 3 }and then XP 3 l = {P, P 3 (D E(A B C) D E). To proof (i) we have to define also the point P = A B (C D E). Using the point P 1 we define P 4 = C D with XC YI = {C }, OC = m C, YD m C = {D 1 } than, D 1 X l = {P 4. Also we denote P 3 = m C YP 3. Now, if P 5 = (C D) E we have, XP 4 YI = {P 4 }, OP 4 = m P 4, YE m P4 = {E 2 } than, E 2 X l = {P 5. Finally, for P = A B (C { D E), since OP 1 = m P 1, YP 5 m P1 = {P 5 } than, XP 5 l = P } where P 5 (C D E,A B (C D E)) We can see that P P. For this part we have to prove the collinearity of the point P 3,P 5,X. Let consider triangles CC 1 P 5 and C P 2 P 3 perspective from J. Hence, from the theorem of Desargues the point CC 1 C P 2 = {Y, C 1P 5 P 2 P 3 = {M and CP 5 C P 3 = {O are collinear. Now, we have triangles OC 1P 2 and YP 5 P 3 perspective from M. Thus, OC 1 YP 5 = {P 5, OP 2 YP 3 = {P 3 } and C 1 P 2 P 5P 3 = {X are collinear point. This, completes the first part of (i). The proof of the second part we will continuous based on case (iii) of Theorem 1. So, let A (B C D) E, ABC,DE A denote in the form (A G) E = A (G E) where G = B C D. To define
10 F. Ibraimi, A. Ibraimi Figure 9 P 1 = B C we determine XB YI = {B }, OB = m B, YC m B = {C 1 }, C 1 X l = {P 1 (see Figure 9). Now, wedefineg = B C DwithXP 1 YI ={P 1 }, OP 1 =m P 1,YD m P1 = {D 1 andd 1 X l = {G}. Next, forp 2 = A GwehaveXA YI ={A }, OA =m A, YG m A = {G } thenxg l = {P 2. IfwedefineP 3 = (A G) E, then XP 2 YI ={P 2 }, OP 2 =m P 2, YE m P2 = {E 1 and XE 1 l = {P 3. Next, if XG YI ={G }, OG =m G, YE m G = {E 2, then XE 2 l = {P 4 where P 4 = G E. Now, finally to define P 5 = A(G E) we have YP 4 m A = {P 4, XP 4 l = {P 5 and P 5 P 3. Finally, based on the Theorem of Desargues wehave toproofthat thepoint E 1,P 4,X arecollinear. From triangles GG P 4 and G P 2 E 2 perspective from X we have GG G P 2 ={Y}, G P 4 P 2 E 2= {M 1 andgp 4 G E 2 = {O}collinear. Todeterminecollinearity ofthepointe 1,P 4 andxwehavetriangles OG P 2 andyp 4E 2 perspective from M 1 where OG YP 4 = {P 4, G P 2 P 4E 2 = {X} and OP 2 YE 2 = {E 1. This completes the proof of (i). 2. A B O = A O B = O A B = O, AB A. If we define P = A B then from figure 10, we can see that hold A B O = O.
RELATION BETWEEN PROPERTIES OF... 11 Figure 10 Similarly, A O B = O A B = O, AB A. 3. Let (A+B) C D = A C D +B C D, ABC,D A, AB O and CD OI. First to define S 1 = A+B we have XB OY = {B }, B J = a B, YA a B = {A 1 and XA 1 l = {S 1 (see Figure 11). Now we can determine P 1 = C D with XC YI = {C, OC = m C, YD m C = {D 1 then, XD 1 l = {P 1. Let note M = (A+B) C D. To determine this point in l we should define XS 1 YI = {S { } 1, OS 1 = m S 1, YP 1 m S1 = P 1 andthenxp 1 l = {M. So, thepointp 1 (CD,(A+B) C D). AlsowehavetodefinethesumS = A C D+B C D, ABC,D A. Therefore, first we define point P 2 = A C D as follows, XA YI = { } {A, OA = m A, YP 1 m A = P 1 then, XP 1 l = {P 2. Now, if { } P 3 = B C D and XB YI = {B, OB = m B, YP 1 m B = then XP 1 l = {P 3. Finally, we can define sum S = A C D+B C D by XP 3 YO = {P 3, P 3 J = a P 3 YP 2 a P3 = {P 2 then, XP 2 l = {S where we have S M. Point P 2 (A C D,A C D+B C D). Thus, based on theorem of Desargues, we have to proof that the points O,A 1 andp 2 arecollinear. Fromtriangles B A 2AandP 3 P 2 P 2 perspectivefrom O we have B A 2 P 3 P 2 = {X, A 2A P 2 P 2 = {Y and B A P 3 P 2 = {K. Points X,Y and K are collinear by the Desargues theorem. Now, from trianglesb JP 3 andayp 2 perspectivefromk wehaveb P 3 AP 2 = {O, P 1
12 F. Ibraimi, A. Ibraimi Figure 11 B J AY = {A 1 and JP 3 YP 2 = {P 2, so that the points O,A 1 and P 2 are collinear. 4. A (B +C) D = A B D+A C D, ABC,D A. First we define S 1 = B +C and we have XC OY = {C, C J = a C, YB a C = {B 1 then, XB 1 l = {S 1. (fig.12). Now, we determine P 1 = A (B +C) then XA YI = {A, OA = m A, YS 1 m A = {S 1 and XS 1 l = {P 1. Next for P 2 = A (B +C) D we have XP 1 YI = {P 1, OP 1 = m P 1 YD m P1 = {D 1, XD 1 l = {P 2. In coordinate system wehave to definealso thesum S = A B D+A C D. Therefore, ifp 3 = A B D firstwedefinep 3 = A B withyb m A = {B 2, XB 2 l = {P 3. Now, XP 3 YI = {P 3, OP 3 = m P 3, YD m P = 3 {D 2 },XD 2 l = {P 3. To define P 4 = A C D, we determine first P 4 = A C and we have YC m A = {C 1, XC 1 l = {P 4. Now, XP 4 YI = {P 4, OP 4 = m P 4, YD m P 4 = {D 3, XD 3 l = {P 4. Finally, we define sum S = A B D +A C D and we have XP 4 YO = {G, GJ = a P4, YP 3 a P4 = {F}, FX l = {S where S P 2 We need to prove the collinearity of the points F,B 1 and O. From triangles BB 3 C and P 3 F 1 G perspective from O we have BB 3 P 3 F 1 = {Y, B 3 C F 1 G = {X,
RELATION BETWEEN PROPERTIES OF... 13 Figure 12 BC P 3 G = {K. We can see that the point Y,X,K are collinear. Let consider triangles YP 3 B and JGC perspective from K. Since, YP 3 JG = {F, P 3 B GC = {O, YB JC = {B 1 we can see that these points are collinear. This completes the proof of (iii). 5. Let A B (C +D) = A B C + A B D, ABC,D A and A,B O,I,CD O. Let define P 1 = A B, then XA YI = {A, OA = m A, YB m A = {B 1 and XB 1 l = {P 1 }, B 1 (BA B). (fig.13). Now to determine S 1 = C + D we have XD YO = {D, D J = a D, YC a D = {C 1 and then XC 1 l = {S 1, C 1 (CC +D). Next, we define P 2 = A B (C +D)and we have XP 1 YI = {P 1, OP 1 = m P 1, YS 1 m P1 = {S 1, XS 1 l = {P 2. On the other side if S 2 = A B C + A B D then we define P 3 = A B C and P 4 = A B D. For P 3 we determine YC m P1 = {C 2, XC 2 l = {P 3 so, C 2 (C,A B C). For P 4 we determine YD m P1 = {D 1, XD 1 l = {P 4 so, D 1 (D,A B D). Now, for S 2 = A B C + A B D we have XP 4 YO = {P 4, P 4 J = a P 4, YP 3 a P4 = {P 3, XP 3 l = {S 2 and P 3 (A B CA B C+A B D). Finally, wecanseethats 2 P 2. Alsowe determine thepoints C 2 l = {H}, P 4 H YC = {Q}, D H YC = {C 3,
14 F. Ibraimi, A. Ibraimi Figure 13 C 2 J P 4 H = {R and C 2 J YS 1 = {S 1. We have to proof collinearity of the points S 1,R and X. From triangles YJC 2 and D DH perspective from O we have YJ D D = {X, JC 2 DH = {S 2 and YC 2 D H = {C 3. Therefore, X,S 2 C 3 are collinear. From triangles C 3 XC 1 and HDJ perspective from D we have C 3 X HD = {S 2, XC 1 DJ = {S 1 and C 3 C 1 HJ = Y. Points Y,S 2,S 1are collinear. Now with triangles YJC 2 and P 4 P 4H perspective from O we have collinearity of the points YJ P 4 P 4 = {X, JC 2 P 4 H = {R and YC 2 P 4 H = {Q. From triangles YP 3C 2 and P 4JH perspective from O we have YP 3 P 4 J = {P 3, P 3C 2 JH = {X and YC 2 P 4H = {Q. Hence X,P 3,Q are collinear. And finally, from triangles D 1HP 4 and YS 1 J perspective from D we have collinearity of the points D 1 H YS 1 = {S 1, HP 4 S 1 J = {R and D 1P 4 YJ = {X} This completes the proof of (iv). From Theorem 1, (i), (iii), (iv) and Theorem 2 (i), (iii), (iv), (v), we obtain the following proposition. Proposition 2. Projective plane with elements from ternary semiring is a desarguesian plane.
RELATION BETWEEN PROPERTIES OF... 15 References [1] T.K. Dutta, S. Kar, On regular ternary semirings, Advances in Algebra, Proceedings of the ICM Satellite Conference in Algebra and Related Topics, World Scientific (2003), 343-355. [2] T.K. Dutta, S. Kar, A note on regular ternary semirings, Kyungpook Math. J., 46 (2006), 357-365. [3] D.H. Lehmer, A ternary analogue of Abelian groups, American Journal of Mathematics, 59 (1932), 329-338. [4] Kavikumar, Azme Khamis, Young Bae Jun, Fuzzy bi-ideals in ternary semirings, International Journal of Computational and Mathematical Sciences, 3, No. 4 (2009), 164-168. [5] T.K. Dutta, S. Kar, On the Jacobson radical of a ternary semiring, Southeast Asian Bulletin of Mathematics, 28, No. 1 (2004), 1-13. [6] T.K. Dutta, S. Kar, A note on the Jacobson radicals of ternary semirings, Southeast Asian Bulletin of Mathematics, 29, No. 2 (2005), 321-331. [7] T.K. Dutta, S. Kar, Two types of Jacobson radicals of ternary semirings, Southeast Asian Bulletin of Mathematics, 29, No. 4 (2005), 677-687. [8] T.K. Dutta, S. Kar, On matrix ternary semirings, International Journal of Mathematics and Analysis, 1, No. 1 (2006), 97-111. [9] S. Kar, On quasi-ideals and bi-ideals of ternary semirings, International Journal of Mathematics and Mathematical Sciences, 2005, No. 18 (2005), 3015-3023. [10] S. Kar, On structure space of ternary semirings, Southeast Asian Bulletin of Mathematics, 31 (2007), 547-555. [11] D. Hughes, F. Piper, Projective Planes Sringer Verlag, New York (1973). [12] S. Bilaniuk, A Problem Course on Projective Planes, Trent University, Ontario, Canada (2003).
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