International Journal of Algebra, Vol. 4, 2010, no. 2, S. Uma

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1 International Journal of Algebra, Vol. 4, 2010, no. 2, α 1, α 2 Near-Rings S. Uma Department of Mathematics Kumaraguru College of Technology Coimbatore, India psumapadma@yahoo.co.in R. Balakrishnan Department of Mathematics V.O.C. College Tuticorin, India T. Tamizh Chelvam Department of Mathematics Manonmaniam Sundaranar University Tirunelveli, India Abstract In this paper we introduce the notion of α 1,α 2 near rings and study some of their properties. Also we distinguish them by characterizing them separately. Mathematics Subject Classification: 16Y30 Keywords: α 1 near ring, α 2 near ring, sub directly irreducible, idempotents, nilpotent 1 Introduction Throughout this paper N stands for a right near ring (N, +,.) with at least two elements. 0 denotes the identity element of the group (N,+) and we write xy for x.y for any two elements x,y of N. If for each a in N, there exists b in N such that a = aba, then N is known as a regular near ring. Analogous to this concept we define α 1 and α 2 near rings. We distinguish them by furnishing examples and also obtain their complete characterizations. For notations and definitions we refer to Pilz[3].

2 72 S. Uma, R. Balakrishnan and T. Tamizh Chelvam 2 Preliminary Notes In this section, we review some of the basic facts in near rings which are used in subsequent sections. Definition 2.1 (Def. 9.4 in Pilz[3]) A near ring N is said to be weak commutative if xyz = xzy for every x,y,z in N. Definition 2.2 (Theorem 1.60 in Pilz[3]) N is sub directly irreducible if and only if the intersection of non zero ideals of N is non zero. Definition 2.3 (Def in Pilz[3])A sub near ring M of N is called invariant if MN M and NM M. Lemma 2.4 (Corollary 9.38 in [3]) Every sub directly irreducible zero symmetric near ring N without nonzero nilpotents is integral. Every non-zero idempotent is a right identity.( through out this paper E denotes the set of all idempotents of N). Lemma 2.5 If N is a zero symmetric near ring then for any ideal I of N, NI I and hence NIN I. Proof: n(n + i) - nn I for any i in I and n,n in N. Substituting n =0 and using the hypothesis that N is zero-symmetric we get NI I. Also I N I. Hence N I N IN I and the proof is complete. Lemma 2.6 If N is a sub commutative near ring and E 0 then idempotents are central. Proof: If N is a sub commutative near ring then Na = an for every a N. Let e E where E is the set of all idempotents of N as in Lemma 2.4. Now Ne = en implies for any n N ne = em and en = xe for some m, x in N ene = e(ne) =e(em) =em = ne (1) and ene =(en)e =(xe)e = xe = en (2) Now equations (1) and (2) imply en = ne and the desired result follows.

3 α 1, α 2 near rings 73 3 α 1,α 2 near rings In this section we define α 1 and α 2 near rings and give certain examples of these new concepts. Definition 3.1 Let N be a right near ring. If (1)for every a in N there exists x in N such that a = xax then we say N is an α 1 near ring. (2) for every a in N-{0} there exists x in N-{0} such that x = xax then we say Nisanα 2 near ring. Example 3.2 (a) Let (N,+) be the Klein s four group with multiplication defined as per Scheme 18, p. 408, Pilz[3]. 0 a b c a a a a a b 0 0 b b c a a c c Obviously (N, +,.) is a regular near ring. This near ring is α 1 (since b0b = 0, cac = a, bbb = b, ccc = c) as well as α 2 (since aaa = a, cbc = c, aca = a). (b) The near ring (N, +,.) defined on the Klein s four group N={ 0, a, b, c } where multiplication is defined as per scheme 2, p. 408, Pilz[3]. 0 a b c a 0 0 a a b 0 a b b c 0 a c c is an α 1 near ring (since bab = ab = a, bbb = b, ccc = c, aoa = 0). However it is neither α 2 (since there is no x in N - { 0 } such that xax = x) nor regular. (c) Near fields are α 1 near rings ( 1n1 = n for all n in N) and α 2 near rings ( n 1 nn 1 = n 1 ). (d) Every Boolean near ring is α 1 as well as α 2, since aaa = aa = a for every a in N. (e) The near ring (N, +,.) where (N,+) is the group of integers modulo 6 and. is defined as per scheme 34, p. 409 of Pilz[3], is not regular.

4 74 S. Uma, R. Balakrishnan and T. Tamizh Chelvam It is neither α 1 ( there is no x in N such that x3x = 3 ) nor α 2 ( there is no x in N-{0} such that x3x = x). (f) We consider the near ring defined on the Klein s four group N={ 0, a, b, c } where. satisfies the following table as per scheme 21, p.408 of Pilz[3].. 0 a b c a a a a a b 0 0 b 0 c a a c a This is an α 2 near ring as it satisfies ana = a for all n in N-{ 0 }. But it is not regular. It is worth noting that it is not an α 1 near ring.(since xcx c for any x in N). Remark 3.3 From the above examples it is clear that the two concepts - α 1 near ring and α 2 near ring - are different from each other, one does not imply the other in general and that these concepts do not imply regularity. 4 α 1 near ring In this section we study some of the important properties of α 1 near rings and give a complete characterization of such near rings. Proposition 4.1 In an α 1 near ring for every a in N there exists some x in N such that (i) a 2 x=xa 2 (ii) a = x n a x n for all n 1 Proof : (i)let N be an α 1 near ring and a N. Then there exists x in N such that a = xax. This implies xa 2 = (xa)a = xa(xax) = (xax)ax = a(ax) = a 2 x. Hence x a 2 = a 2 x for every a in N and the result follows. (ii)xax = x(xax)x = x 2 a x 2 =... = x n a x n for all n 1.

5 α 1, α 2 near rings 75 Remark 4.2 Example 3.2(b) shows that an α 1 near ring need not be regular. But the converse is established as in the following proposition. Proposition 4.3 Let N be a regular near ring. If N is weak commutative ( as in Def. 2.1) then N is an α 1 near ring. Proof : Since N is regular for every a N, there exists b N such that aba = a. Let x = ab then xax = (ab)a(ab) = (aba)ab = a(ab) = aba [since N is weak commutative]. Consequently xax = a for every a in N and N becomes an α 1 near ring. Proposition 4.4 Let N be a zero symmetric weak commutative α 1 ring. Then for any a,b in N ab = 0 implies ba = 0 near Proof : Suppose ab = 0 for a,b N. Since N is an α 1 near ring there exists x,y N such that xax = a and yby = b. Also since N is weak commutative ba = (yby)(xax) = yb(yxa)x = yb(yax)x = y(bya)x 2 = y(bay)x 2 = (yba)yx 2 = (yab) yx 2 = y(ab)yx 2 =y0yx 2 [since ab = 0] = 0. Proposition 4.5 Homomorphic image of an α 1 near ring is also an α 1 near ring. Proof : The proof is straight forward. Proposition 4.6 If I is an ideal of the α 1 near ring N then N/I is also an α 1 near ring. Proof : The function φ: N N/I be defined by φ(x) = I + x is an epimorphism. The rest of the proof is taken care of by proposition 4.5. Proposition 4.7 Every α 1 near ring N is isomorphic to a sub direct product of sub directly irreducible α 1 near rings. Proof : By Theorem 1.62, p. 26 of Pilz[3], N is isomorphic to a sub direct product of sub directly irreducible near rings N i s and each N i is a homomorphic image of N under the projection map π i. Now the desired result follows from proposition 4.5. We furnish below a characterization of α 1 near ring. Theorem 4.8 Nisanα 1 near ring if and only if every a in N can be written as a = u + v where u N 0 and v N c and u=x 0 (nx + m) - x 0 a x c,v=x 0 a x c + x c,x=x 0 + x c N 0 N c where x 0,n N 0 and x c,m N c. Further more u N 0 and v N c.

6 76 S. Uma, R. Balakrishnan and T. Tamizh Chelvam Proof : For the only if part, let a N. Since N is α 1 there exists x in N such that a = xax also by using Peirce decomposition we can write a = n + m, and x = x 0 + x c where x N, n, x 0 N 0 and m, x c N c. Now a = (x 0 + x c ) (n + m) x =(x 0 + x c ) (nx + mx ) =(x 0 + x c ) (nx + m), since m N c = x 0 (nx + m)+ x c (nx + m ) = x 0 (nx + m)+ x c, since x c N c = x 0 (nx + m) - x 0 a x c + x 0 a x c + x c, = u + v where u = x 0 (nx + m )- x 0 ax c and v = x 0 ax c + x c Now u 0 = [x 0 (nx + m) - x 0 a x c ]0 = x 0 (nx + m)0 - x 0 a x c 0 = x 0 (nx0 + m0) - x 0 a x c 0 = x 0 (nx0 + m) - x 0 a x c, Since m,x c N c = x 0 (nx c +mx c )- x 0 a x c, since x0= x c and m N c =0. Therefore u N 0. And v0 = [x 0 a x c + x c ]0 = x 0 a x c 0+x c 0 = x 0 a x c + x c, since x c N c =v. Therefore v N c. Thus a = u + v where u N 0 and v N c. This completes the proof of the only if part. For the if part we assume for every a in N with a = u + v where u N 0, v N c with u = x 0 [nx + m ] - x 0 a x c and v = x 0 a x c + x c where x = x 0 + x c, x 0,n N 0 and x c,m N c We shall show that N is an α 1 near ring. Nowa=u+v = x 0 (nx + m) - x 0 a x c + x 0 a x c + x c = x 0 (nx + mx) + x c, since m N c = x 0 (n + m)x + x c, = x 0 ax+x c ax, since x c N c =(x 0 + x c )ax =xax. Thus for every a in N, a = xax for some x in N. i.e. N is an α 1 near ring. 5 α 2 near ring Throughout this section N denotes an α 2 near ring. And N =N-{0} In this section we study some of the important properties and a characterization of α 2 near rings.

7 α 1, α 2 near rings 77 Proposition 5.1 In an α 2 near ring E { 0 }. Proof : Let N be an α 2 near ring. So for every a in N there exists an x in N such that xax = x. Here (ax) 2 = ax(ax) = a(xax) = ax and (xa) 2 = xa(xa) = (xax)a = xa. i.e. ax and xa are idempotents and hence E { 0 }. As in proposition 4.3, we have the following result. Proposition 5.2 Every regular near ring is an α 2 near ring. Proof : Let N be a regular near ring. Hence for every a in N there exist a,b in N such that aba = a. Let x = bab. Now xax = (bab)a(bab) = b(aba)bab = b(a)bab = b(aba)b = bab = x. This proves that every regular near ring is an α 2 near ring. Remark 5.3 Example 3.2(f) confirms that the converse of the above result is not true. Remark 5.4 Proposition 4.5 shows that a homomorphic image of an α 1 near ring is an α 1 near ring. It is worth noting that a homomorphic image of an α 2 near ring need not be so, however it is preserved under isomorphisms. Proof : Let (N, +) be the Klein s four group with N={ 0, a, b, c } and + satisfying the following table + 0 a b c 0 0 a b c a a 0 c b b b c 0 a c c b a 0 We consider two near rings (N, +, *) where * satisfies the following table as per scheme 23, p.408 of [3]. * 0 a b c a a a a a b b b b b c c c c c and (N, +,.) where. is defined as per scheme 22, p.408 of Pilz[3].

8 78 S. Uma, R. Balakrishnan and T. Tamizh Chelvam. 0 a b c a a a a a b c a a a a The mapping f : N N defined as f(0) = 0, f(a) = a, f(b) = 0, f(c) = a is a homomorphism. Here (N, +, *) is an α 2 near ring but (N, +,.) is not. This example is an evidence for the Remark 5.4. Definition 5.5 A sub near ring M of a near ring N is called an α 2 sub near ring if for every a in M there exists an x in M such that xax = x. Proposition 5.6 Let N be an α 2 near ring then (i)every invariant subgroup M of N ( as in Def. 2.3) is also an α 2 sub near ring. (ii)every ideal I of a zero symmetric α 2 near ring N is also an α 2 sub near ring. Proof : (i)let a M. Since N is an α 2 near ring there exists x in N such that x = xax. Now M is an invariant subgroup of N implies xax M. i.e. x M. Consequently M is an α 2 sub near ring. (ii) Let I be an ideal of the zero symmetric α 2 near ring N. Let a I. Since N is an α 2 near ring there exist x N such that xax = x. Now xax N I N I [ By Lemma 2.5 ]. The desired result now follows. We characterize α 2 near ring in the following theorem. Theorem 5.7 A near ring N is α 2 if and only if every N-subgroup contains a non zero idempotent. Proof : Suppose N be an α 2 near ring. If M is a N- subgroup and a M then there exists x N such that xax = x 0. As in the proof of Proposition 5.1 we see that xa M is an idempotent and xa 0. For the if part let a N. Clearly Na is an N-subgroup of N. Now, by hypothesis Na contains a non zero idempotent e. Then e = ba for some b in N. Suppose x = bab. Clearly then x 0. We have xax = bab a (bab) = ba ba ba b = e e e b = e b =(ba)b = x and this completes the proof. We conclude our discussion with the following theorem.

9 α 1, α 2 near rings 79 Theorem 5.8 Let N be a sub commutative α 2 near ring. Then the following are equivalent (i)n be an integral near ring. (ii)n is a near field. (iii) N is a zero symmetric reduced sub directly irreducible near ring Proof : (i) (ii) Suppose N be an α 2 near ring. Then by Proposition 5.2 E {0} and by Lemma 2.4 each non zero idempotent is a right identity of N. Since N is sub commutative, Lemma 2.6 demands that ef = fe for any two non zero idempotents e,f of N. Also f = fe = ef = e. Thus N has a unique non-zero idempotent e. It follows that e is the identity of N. Again since N is α 2 for every a in N there exists x N such that x = xax. Further ax and xa are idempotent (by Proposition 5.1). Consequently ax = xa = e and N becomes a near field. (ii) (iii) is obvious. (iii) (i) follows from Lemma 2.4. References [1] Dheena.P, A note on a paper of Lee, Journal of Indian Math. Soc. 53 (1988) [2] Henry E. Heartherly, Regular Near Rings, Journal of Indian Math. Soc. 38 (1974) [3] Pilz Günter, Near-Rings, North Holland, Amsterdam, [4] S.Suryanarayanan and N. Ganesan, Stable and Pseudo stable Nearrings, Indian J.Pure and Appl. Math. 19(12) December 1988, Received: August, 2009