Results on Planar Near Rings

Transcription

1 Interntionl Mthemticl Forum, Vol. 9, 2014, no. 23, HIKARI Ltd, Results on Plnr Ner Rings Edurd Domi Deprtment of Mthemtics, University A.Xhuvni, Elbsn, Albni Copyright 2014 Edurd Domi. This is n open ccess rticle distributed under the Cretive Commons Attribution License, which permits unrestricted use, distribution, nd reproduction in ny medium, provided the originl work is properly cited. Abstrct In this pper we introduce some results for plnr - ner-rings nd plnr - ner-fields. Through new definition of -ner -fields we prove condition when plnr -ner-ring is plnr - ner field. Mthemtics Subject Clssifiction: 16Y30 Keywords: Plnr ner ring, plnr -ner ring, plnr - ner- field. 1. Introduction Plnrity is introduced in lgebr by Mrshll Hll in his prominent coordinstion of projective plne by plnr ternry rings [4]. At [6] J. L. Zemmer defined plnr nerfield s ner-ring in which the eqution x = bx + c is unique solution for every b. Michel Anshel nd Jmes R. Cly defined in [1] plnr ner-rings, which s expected, hve geometric interprettions. Here we will give concepts nd we will present sme uxiliry

2 1140 Edurd Domi propositions, which we will use further in the presenttion of the min results of the proceeding. Let consider M nd s two non empty sets. Let s consider M nd s two non-empty sets. Every mp of M x x M in M is clled - multipliction in M nd is denoted s (). The result of this multipliction for elements, b M nd is denoted b. According to Stynryn [2], - ner-ring is clssified ordered triple (M, +, () ) where M nd re non empty sets, + is n ddition in M, while () is - multipliction on M stisfying the following conditions: (i) (M, +) is group.( not necessrily belin) (ii) (, b, c,, ) M 3 x 2, (b)c = (bc). (iii) (, b, c,, ) M 3 x, ( + b)c = c + bc. An e element of -ner-ring M is clled identity element if for every M nd every we hve e = e =. Let (N, +) be ner ring. The reltion = m such tht = m b if only if x = bx for ll xn is n equivlence reltion nd is clled the reltion of equl multipliers in N [3]. Definition 1.1 [3] Triple (N, +, ) is plnr ner - ring when : 3 (i) The reltion of equl multipliers = m hs t lest three equivlence clsses N/ = m (ii) (, b, c, ) N 3 m b the eqution x = bx + c hs unique solution for xn.

3 Results on plnr ner rings 1141 Let M be -ner-ring. We define in the M the ech reltion = m such tht = m b then nd only then when x = bx, for ech x M nd for ech. Obviously, the reltion = m is n equivlence reltion. If = m b we will sy tht nd b re equl multipliers. The reltion = m we going to cll it reltion of the equl multipliers. Definition 1.2[5] - ner-rings (M, +, () ) is clled plnr if: (i) Reltion of the equl multipliers = m hs t lest three equivlence clsses, mening M/= m 3. (ii) For ny three elements, b, c of M such tht m b nd for ech, x = xb + c hs unique solution in M. Exmple 1.3 Let there be (M n (P), +) the group of mtrices of the order n with elements from field P, with hs t lest three elements. For ny subset of nonsingulr mtrices set of order n with elements from the field P we define in M n (P), - multipliction () such tht for ny two mtrices A, B of M n (P) nd for ech mtrix, we hve A B = A B, where, B re respectively the determinnts of mtrices, B. It is esy to be convinced tht (M n (P) +, () ) is -ner-ring. In this -ner-ring we hve A = m B if nd only if for ny X M n (P) X A = X B.

4 1142 Edurd Domi Tking nonsingulr mtrix X, we hve A = B. Since P hs t lest three elements we hve t lest three mtrices tht does not hve the sme determinnts, therefore M n (P)/= m 3. If A m B, mening A B, mtrix eqution X A = X B + C hs the unique solution the mtrix X = 1 C. A B Therefore, -ner-ring (M n (P), +, () ) is plnr. For plnr -ner-ring (M, +, () ) note A = { M = m 0} To mke it esy write the set M\A = M. For ech M nd for ech, the eqution x = x0 + = there is unique solution which is noted 1. It cn hppen tht 1 = 1 b lthough b. For M nd for ech note B = {b M 1 b = b}. Theorem 1.4[5] For ny plnr -ner-ring (M, +, () ) the following sttements re true: (i) M = A (, ) M B (ii) B M = B for ech M nd for ech M. Definition 1.5[4] A -ner-ring M is clled -ner -field if for every, the ner-ring M = (M, +, ) is ner-field. Let there be (M, +, () ) plnr -ner-ring. If for every, ner-ring (M, +, ) is unitry, then (M, +, () ) would be clled unitry plnr -ner-ring. If for ech, ner-ring (M, +, ) is ner-field, then (M, +, () ) will be clled plnr -ner-field.[5]

5 Results on plnr ner rings 1143 Proposition 1.6 [5] If (M, +, () ) is unitry plnr -ner-ring, then it is plnr -ner-field. 2. Results on plnr ner rings Theorem 2.1 Let be (M, +, () ) plnr -ner-ring: For ech nd for ech M, B is closed with respect to binry opertion of M for which x y = xy nd is group in reltion with the opertion induced in it. Proof.For M we hve 1 = = 1 (1 ) = (1 1 ). The eqution x = there is unique solution so we hve 1 1 =1. Thus, 1 B nd consequently 1 in the left identity of subgroup ( B, ). Assume tht is the unique solution to the eqution x =1, mening =1. Then (1 ) = 1 () = 1 1 = 1. From the iniquity of the solution of the eqution x = 1 from Theorem1.4 we hve 1 = so B Let there be b n element of the subgroup ( B, ). For n element b there is n element b B b such tht bb =1 b.

6 1144 Edurd Domi Now we will prove tht becuse B = B b. Let c be n element of B B, which is not empty b b B B. Then 1 c = c = 1 b c. Thus, 1 b nd 1 b re the solutions of the eqution xc = c nd hence from the iniquity of the solution of this eqution we hve 1 =1 b. From this equlities nd definitions of sets B, B b we obtin the equlity B = B b. So, for every b B we hve bb = 1 =1 b, which shows tht ech element b of the subgroup ( B, ) hs left inverse with respect to his left identity1. Thus, subgroup ( B, ) is group. Proposition 2.2 For ech two elements, c of M the mp : 1 c x is n isomorphism. B B c such tht (x) = Proof. By Theorem 1.4 (ii) the definition of the mp : B Bc is correct becuse 1 c x x 1c x is the element of the group B. For ny two elements x, y of c B the equlities re true: (x y) = 1 c (x y) = 1 c (xy) = (1 c x)y = [(1 c x) 1 c ]y = (1 c x)(1 c y) = (x) (y), which implies tht the mp is homomorphism of the group ( B, ) in group ( B, ). c For every b B c we hve 1 c b = b. becuse of this equlity we hve: (1 1 c ) = 1 (1 c b) = 1 b, obtining tht the elements1 1 c, 1 re solutions of eqution xb = 1 (1 c b), which hs unique solution. Thus, we hve 1 1 c =1. In the sme wy we cn prove even the equlity 1 c 1 =1 c.

7 Results on plnr ner rings 1145 If 1 c x = 1 c y for every two elements x, y of B, then the equlities re true: 1 (1 c x) = (1 1 c )x = 1 x = x = 1 (1 c x) = 1 (1 c y) = (1 1 c )y = 1 y = y, which proved tht the homomorphism is monomorphism. For ech element y B c element1 y B Since the equlities re true: (1 y) = 1 c (1 y) = (1 c 1 )y = 1 y = y, monomorphism is epimorphism nd consequently is isomorphism of group ( B, ) in group ( B c, ). Proposition 2.3 For ech M nd for ech element, 1 is - right identity element, ( d M, d1 = d). Proof. For ech element d M we tke the eqution x1 = d1. This eqution hs solution n element d nd in the sme wy the element d1 is its solution becuse the equlities re true: (d1 ) 1 = d( 1 1 c ) = d 1. From the iniquity of the solution of the eqution, obtin tht for every d M we hve d 1 = d, mening the element d M is -right identity. Proposition 2.4 If (M, +, () ) is plnr -ner-ring such tht (, b) M 2, = m b = b nd (,, b) (M) 2, 1 = 1 b,

8 1146 Edurd Domi (M, +, () ) is plnr -ner-field. Proof. For three elements, b, c of M, where b nd for ech eqution x = bx + c hs only one solution. If 0, then m 0 nd consequently A = {0}. Since for ech (, b) (M) 2, 1 =1 b, for ny two elements, b different from zero of M, we hve B = B b = M. Thus, M = M\ {0} = M* = B nd therefore M therefore is closed with respect binry opertion of M for which x y = xy nd forms the group in reltion with the opertion inducted in it. Therefore, for ech ner-ring (M, +, ) is plnr by definition 1.5 mening (M, +, () ) is plnr -ner-field by Proposition 1.6 If -ner-ring (M, +, () ) hs no zero divisor, then it would cll tht -ner-ring is integrl domin. If these -ner-ring (M, +, () ) is plnr, then we would cll it tht plnr -nerring is integrl domin. By the Proposition 2.4 we obtin: Corollry 2.8 If (M, +, () ) plnr -ner-ring is integrl domin such s References (, b) M 2, = m b = b, then it is plnr -ner-field. [1] Anchel, M., Cly, R., Plnrity in lgebric system, Bul.Amer.Mth. Soc., 74 (1968). [2] Celestin, C.F., Ner ring. Some developments linked to semigroups nd, [3] Cly, J. R., Ner rings, Geneses nd Appliction, Oxford University Press [4] Domi, E., Petro, P., G -Ner Fields nd Their Chrcteriztions by Qusi idels, Interntionl Mthemticl Forum, 5 (2010), No

9 Results on plnr ner rings 1147 [5] Domi, E., Petro, P., Plnr -ner-ring, AJNTS 2 (2010). [6] Hll, M. Jr., Projective plne, Trns.Amer.Mth.Soc. 54 (1943). [7] Pilz, G., Ner rings. The theory nd its pplictions. North. Hollnd, Amsterdm. Revised edition, 1984 [8] Zemmer, J, L., Ner fields, plnr nd nonplnr, The Mth. Student, 31 (1964). Received: My 11, 2014