International Journal of Pure and Applied Sciences and Technology

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1 Int. J. Pure Appl. Sci. Technol., 6(2) (2011), pp International Journal of Pure and Applied Sciences and Technology ISSN Available online at Research Paper Jordan Left Derivations on Semiprime Gamma Rings A.C. Paul 1 and Md. Mizanor Rahman 2,* 1 Department of Mathematics, Rajshahi University, Rajshahi, Bangladesh. 2 Department of Mathematics, Jagannath University, Dhaka, Bangladesh. * Corresponding author, (mizanorrahman@gmail.com) (Received: ; Accepted: ) Abstract: The object of this paper is to work on Jordan left derivations of semiprime gamma rings.let M be a 2 and 3-torsion free semiprime gamma ring. Let d and G be Jordan left derivations on M such that d 2 (M) = G (M). Then we prove that d(m) = 0. Keywords: Jordan left derivation, gamma ring, semiprime gamma ring. 1. Introduction The notion of a gamma ring was first introduced by N. Nobusawa (1964). Barnes (1966) generalized the concept of N. Nobusawa gamma ring and gave an concrete definition of a gamma ring. Y. Ceven (2002) defined a Jordan left derivation on gamma rings and showed that a Jordan left derivation on completely prime gamma rings is a left derivation. M. Soyturk (1994) investigated the commutativity of prime gamma rings with left and right derivations. He obtained some significant results on the commutativity of prime gamma rings of characteristic not equal to 2 and 3. M. Asci and S. Ceran (2007) obtained some commutativity results of prime gamma rings with left derivation. A.C. Paul and Amitabh Kumer Halder worked on Jordan left derivations of two torsion free ΓM- Modules. They prove that the existence of a nonzero Jordan left derivation on M into X implies M is commutative. They also prove that if M is a prime Γ-ring, then every Jordan left derivation is a left derivation. In this paper, we consider two Jordan left derivations d and G on a 2 and 3-torsion free semiprime Γ-ring with a condition d 2 (M) = G (M), and prove that d(m) = 0.

2 Int. J. Pure Appl. Sci. Technol., 6(2) (2011), Preliminaries Gamma ring : Let M and Γ be additive Abelian groups. M is called a Γ-ring if for any x, y, z M and α, β Γ, the following conditions are satisfied: (1) x α y M (2)(x+y) αz = x α z +y α z x(α+β)z=xαz+xβz xα(y+z)=xαy+xαz (3)(xαy)βz=xα(yβz) Prime gamma ring: A Γ-ring M is called prime if aγmγb = 0 (with a, b M) implies a = 0 or b=0 Semi prime gamma ring: A Γ-ring M is called semi prime if aγmγa = 0 (with a M) implies a = 0. A gamma ring M is n-torsion free if nx = 0 x = 0 for all x M. Derivation: Let M be a Γ-ring and let d:m M be an additive map. d is called a derivation if for any a, b M and α Γ. d(aαb) = d(a)αb + aαd(b) Jordan derivation: d is called a Jordan derivation if for any a M and α Γ, d(aαa)=d(a)αa+aαd(a). Left derivation: d is called a left derivation if for any a, b M and (aαb)=aαd(b)+bαd(a). α Γ, Jordan left derivation: d is called a Jordan left derivation if for any a M and α Γ, d(aαa)=2aαd(a) Let M be a Gamma-ring. An element x M is said to be nilpotent if (xα) n x = 0 for all x M and α Γ. We define the commutativity by [a, b] α = aαb- bαa. An example of a left derivation is given by Y. Ceven (2002) 3. Jordan left derivation of semiprime Γ-rings: We begin with the following lemmas: Lemma 3.1. Let M be a two torsion free Γ-ring and d a Jordan left derivation on M. Then for all a, b M and for all α Γ : (i) d(aαb + bαa) = 2aαd(b) + 2bαd(a) (ii) d(aαbαa) = aαaαd(b) + 3aαbαd(a) bαaαd(a) (iii) d(aαbαc + cαbαa) = aαcαd(b) + cαaαd(b) + 3aαbαd(c) + 3cαbαd(a) bαcαd(a) bαaαd(c). Proof : (i) is obtained by computing d((a + b) α (a + b)) (ii) From (i), d(aαb + bαa) = 2aαd(b) + 2bαd(a) Then replacing aαb + bαa for b, we have d(aαbαa + aαbαa) + 2aααd(b) + 4aαbαd(a) + 2aαbαd(a) 2bαaαd(a) (iii) is obtained replacing a + c for a in (ii)

3 Int. J. Pure Appl. Sci. Technol., 6(2) (2011), Lemma 3.2: Let M be a 2-torsion free and 3-torsion free Γ-ring, and d: M M a Jordan left derivations. If d([d(x), x] α, x] α )= 0 holds for all x M and α Γ, then [d(x), x] α α d(x)= 0 is fulfilled for all x Γ and α Γ. Proof: By lemma 3.1, we have 0 = d([d(x), x] α,,x] α ) = d(d(x)αxαx+xαxαd(x)) - 2d (xαd(x)αx) = 4 d(x)αxαd(x)+2xαxαd 2 (x) - 2xαxαd 2 (x) - 6xαd(x) αd(x) + 2xαd(x) αd(x) = 6 [d(x),x] α α d(x) 6 [d(x),x] α α d(x)= 0 Since M is 2 and 3-torsion free, [d(x),x] α α d(x)= Theorem: Let M be a 2-torsion free and 3-torsion free semiprime Γ-ring. If d and G of M into M are Jordan left derivations such that d 2 (M) = G(M), then d = 0. Proof : Let x M then xαx M putting xαx for x in d 2 (x) = G(x). (1) d(xαd(x)) = xαg(x), x M Let us prove that for all x M and α Γ. (2) d(d(x) αx) = 2d(x) αd(x) + xαg(x) Using lemma 3.1, we have d(d(x) αx + xαd(x))= 2d(x) αd(x) + 2xαd 2 (x) Using lemma 3.1 (i) we obtain d(d(x) αx) = 2d(x) αd(x) + 2xαd 2 (x) d(xαd(x)) = 2d(x) αd(x) + 2xαG(x) xαg(x) = 2d(x) αd(x) + xαg(x) d(d(x) αx xαd(x)) = 2d(x) αd(x) (3) d[d(x), x] α = 2d(x) αd(x) Linearing (3), we have d([d(x + y), x + y] α ) = 2d(x + y) αd(x + y) d([d(x) + d(y), x + y] α ) = 2(d(x) + d(y)) α (d(x) + d(y)) d([d(x), x] α + [d(y), x] α + [d(x), y] α + [d(y), y] α ) = 2[d(x)αd(x) + d(x)αd(y) + d(y)αd(x) + d(y) αd(y)] Using (3),we get d([d(y), x] α + [d(x), y] α ) = 2d(x)αd(y) + 2d(y) αd(x) Putting in the above relation y = xαx, using lemma 3.1 and (3), we obtain 0 = d([d(x), xαx]α + [d(xαx), x] α ) 2d(x) αd(xαx) 2d(xαx) αd(x) = d([d(x), x] α αx + xα [d(x), x] α ) + 2d(xα [d(x), x] α ) 4d(x) αxαd(x) 4xα d(x) αd(x) = 2[d(x), x] α αd(x) + 2xαd([d(x), x)] α ) + 2d(xα [d(x), x] α ) 4d(x) αxαd(x) 4xα d(x) αd(x) = 2d(x) αxαd(x) 2xαd(x) αd(x)+4xαd(x) αd(x) + 2d(xα [d(x), x] α ) 4d(x) αx d(x) 4xαd(x) = 2d(x) αxα d(x) 2xα d(x) αd(x) + 2d(xα [d(x), x] α )

4 Int. J. Pure Appl. Sci. Technol., 6(2) (2011), (4) d(xα [d(x), x] α ) = d(x) αxαd(x) + xαd(x) αd(x), for x M and α Γ. Let us prove the identity (5) d([d(x), x] α αx) = d(x) αx αd(x) + xαd(x) αd(x), x M and α Γ. Using lemma 3.1 and 3.2 we have d([d(x), x] α αx + xα [d(x), x] α ) = 2[d(x), x]αd(x) + 2xαd([d(x), x] α ) = 2[d(x), x] α αd(x) + 4xα d(x) αd(x) Now applying (4) we obtain d([d(x), x] α αx )= 2[d(x), x] α αd(x) + 4(xαd(x) αd(x) d(x)α [d(x), x] α ) =2[d(x), x] α αd(x)+4xα d(x) d(x) αxαd(x) xαd(x) αd(x) = d(x) αxαd(x) + xαd(x) αd(x), x M which completes the proof (5). From (4) and (5) we obtain (6) d([[d(x), x]α, x] α ) = 0, x M From (6) and lemma 4 it follows (7) [d(x), x] α αd(x) = 0, x M Using (7), (3) and lemma 3.1 we obtain d(d(x) α [d(x), x] α ) = d(d(x) α [d(x), x] α + [d(x), x] α αd(x)) = 2d(x)αd([d(x), x] α ) + 2[d(x), x] α αg(x) (8) d(d(x) α [d(x), x] α ) = 4d(x) αd(x) αd(x)+2[d(x), x] α αg(x), x M and α Γ. Let us prove the relation (9) d(d(x)α [d(x), x] α ) = 6[d(x), x] α αg(x), x M and α Γ. Using (7) and lemma 3.1 we obtain 0 = d[(d(x), x] α αd(x)) = d(d(x) αxαd(x)) d(xαd(x) αd(x)) = d(x) αd(x) αd(x) αd(x) + 3d(x) αxαg(x) xαd(x) αg(x) d(xαd(x)αd(x)) (xαd(x) αd(x)) = d(x) αd(x) αd(x)+3d(x) αxα G(x) xα d(x) αg(x), x M and α Γ. Now we have (10) d(d(x)2αx+ xαd(x) αd(x))= 2d(x) αd(x) αd(x)+ 4xαd(x) αg(x), x M and α Γ. From the above relation and (10) it follows (11) d(d(x) αd(x) αx) = d(x) αd(x) αd(x) + 5xαd(x) α G(x) 3d(x) αx αg(x), x M and α Γ. From (10) and (11) we obtain d([d(x) αd(x), x] α ) = 6[x, d(x)] α αg(x) according to (7) 6[x, d(x)] α αg(x) = d([d(x) αd(x), x] α ) = d([d(x), x] α αd(x) + d(x) α [d(x), x] α )

5 Int. J. Pure Appl. Sci. Technol., 6(2) (2011), = d(d(x) α [d(x), x] α ) Which completes the proof of (9). Combining (8) with (9) we arrive at (12) d(x) αd(x) αd(x) + 2[d(x), x] α αg(x) = 0, for x M and α Γ. Now starting from (7) and using lemma 3.1, we obtain 0 = d(d(x) α [d(x), x] α αd(x)) = d(x) αd(x) αd([d(x), x] α ) + 3d(x) α [d(x), x] α αd 2 (x) [d(x), x] α αd(x) αd 2 (x) = 2d(x) αd(x) αd(x) αd(x) + 3d(x) α [d(x), x] α αg(x) (13) 2d(x) αd(x) αd(x) αd(x) + 3d(x) α [d(x), x] α αg(x) = 0, x M and α Γ. 4d(x) αd(x) αd(x) αd(x) + 6d(x) α [d(x), x] α αg(x) = 0 d(x) αd(x) αd(x) αd(x) + 3d(x) αd(x) αd(x) αd(x) + 6d(x) α [d(x), x] α αg(x) = 0 d(x) αd(x) αd(x) αd(x) + 3d(x)α (d(x) αd(x) αd(x) + 2[d(x), x] α αg(x)) = 0 (d(x)α) 3 d(x) = 0 by using 12. We have seen that d(x) is a nilpotent element of M. But since semiprime Γ-ring contains no non-zero nilpotent elements, we have d(x) = 0, for all x M. Which completes the proof of the theorem. References [1] M. Asci and S. Ceran, The commutativity in prime gamma rings with left derivation, Internat. Math. Forum, 2(3), 2007, [2] W.E. Barnes, On the Γ-rings of Nobusawa, Pacific J. Math. 18 (1966), , [3] Y. Ceven, Jordan left derivations on completely prime Γ-ring, C.U. Fen-Edebiyat Fakultesi Fen Bilimlere Dergisi, 23(2), 2002, [4] N. Nobusawa, On the generalization of the ring theory, Osaka J. Math., 1(1964), [5] A.C. Paul and Amitabh Kumar Halder, Jordan left derivations, J. of Mathematical Sciences, 13 (2009), [6] M. Soyturk, The commutativity of prime gamma rings with derivation, Turk. J. Math. 18 (1999),

On Commutativity of Completely Prime Gamma-Rings

On Commutativity of Completely Prime Gamma-Rings alaysian Journal of athematical Sciences 7(2): 283-295 (2013) ALAYSIAN JOURNAL OF ATHEATICAL SCIENCES Journal homepage: http://einspem.upm.edu.my/journal 1,2 I. S. Rakhimov, 3* Kalyan Kumar Dey and 3 Akhil