International Journal of Advanced and Applied Sciences, 5(5) 2018, Pages: 79-81 Contents lists available at Science-Gate International Journal of Advanced and Applied Sciences Journal homepage: http://www.science-gate.com/IJAAS.html The commutativity of prime rings with homoderivations E. F. Alharfie 1, *, N. M. Muthana 2 1Department of Mathematics, Tabuk University, Tabuk, Saudi Arabia 2Department of Mathematics, King Abdulaziz University, Jeddah, Saudi Arabia ARTICLE INFO ABSTRACT Article history: Let 푅 be a ring with center 푍(푅), and 퐼 be a nonzero left ideal. An additive Received 4 November 2017 mapping ℎ: 푅 → 푅 is called a homoderivation on 푅 if ℎ(푥푦) = ℎ(푥)ℎ(푦) + Received in revised form ℎ(푥)푦 + 푥ℎ(푦)for all 푥. 푦 ∈ 푅. In this paper, we prove the commutativity of R 19 February 2018 if any of the following conditions is satisfied for all 푥. 푦 ∈ 푅: (i) 푥ℎ(푦) ± Accepted 12 March 2018 푥푦 ∈ 푍(푅). (ii) 푥ℎ(푦) ± 푦푥 ∈ 푍(푅). (iii) 푥ℎ(푦) ± [푥. 푦] ∈ 푍(푅) (iv)[푥. 푦] ∈ 푍(푅)(v)[ℎ(푥)푦] ± 푥푦 푍(푅)and (vi) [ℎ(푥). 푦] ± 푦푥 ∈ 푍(푅). This result is in Keywords: the sprite of the well-known theorem of the commutativity of prime and Prime ring semiprime rings with derivations satisfying certain polynomial constraints. Homoderivation Also, we prove that the commutativity of prime ring on R, if R admits a Commutativity theorems nonzero homoderivation ℎ such that ℎ([푥. 푦]) = ±[푥. 푦] for all 푥. 푦 in a nonzero left ideal. © 2018 The Authors. Published by IASE. This is an open access article under the CC BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/). 1. Introduction Lemma 1.1 (Lemma 4): Let b and ab be in the center of a prime ring R. If b≠0, then a is in Z(R) *Throughout, 푅 denotes a ring with a center 푍(푅). (Mayne, 1984). We write [푥. 푦] 푓표푟 푥푦 − 푦푥 and is called the commutator. A ring 푅 is called prime if 푎푅푏 = 0 Remark 1.2 (Remark 3): Let R be a prime ring. If R implies 푎 = 0 표푟 푏 = 0 and is called semiprime if contains a nonzero commutative left ideal, then R is a commutative ring (Bresar, 1993). 푎푅푎 = 0 then 푎 = 0. A derivation on 푅 is an additive mapping 푑: 푅 → 푅 satisfying 푑(푥푦) = 푑(푥)푦 + 푥푑(푦) Lemma 1.3 (Lemma 1.1): Let R be a ring and 0≠I a for all 푥. 푦 ∈ 푅. El Sofy (2000) defined a right ideal of R. Suppose that a∈I such that an = 0 for homoderivation on 푅 to be an additive mapping ℎ a fixed integer n. Then R has a nonzero nilpotent from 푅 into itself such that ℎ(푥푦) = ℎ(푥)ℎ(푦) + ideal (Herstein, 1969). ℎ(푥)푦 + 푥ℎ(푦) for all 푥. 푦 ∈ 푅. The only additive map which is both derivation and homoderivation on Lemma 1.4 (Corollary 2.5): Let R be a prime ring of prime ring is the zero map. If 푆 ⊆ 푅, then a mapping characteristic not 2 and I a nonzero left ideal. If R 푓: 푅 → 푅 preserves 푆 푓 푓(푆) ⊆S. A mapping 푓: 푅 → admits a nonzero homoderivation h which is 푅 is said to be zero-power valued on 푆 if 푓 centralizing on I, then R is commutative. preserves 푆 and if for each 푥 ∈ 푆, there exists a positive integer 푛(푥) > 1 such that 푓푛(푥) = 0 (El 2. On the commutative conditions Sofy, 2000). Ashraf and Rehman (2001) had shown that if 푅 is a prime ring, 퐼 an ideal of 푅 and 푑: 푅 → 푅 Theorem 2.1: Let R be a prime ring of characteristic is a derivation of 푅, then 푅 is a commutative ring if not 2, and I be a nonzero left ideal in R. If h is a and only if 푅 satisfies any one of the properties nonzero homderivation which is zero-power valued on I. Then, for all x, y∈I, the following conditions are 푑(푥푦) ± 푥푦 ∈ 푍(푅). 푑(푥푦) + 푥푦 ∈ 푍(푅). 푑(푥푦) ± 푦푥 ∈ equivalent: 푍(푅). 푑(푥푦) + 푦푥 ∈ 푍(푅). 푑(푥)푑(푦) ± 푥푦 ∈ 푍(푅) 푎푛푑 푑(푥)푑(푦) + 푥푦 ∈ 푍(푅) 푓표푟 푎푙푙 푥. 푦 ∈ 퐼. i. 푥ℎ(푦) ± 푥푦 ∈ 푍(푅) ii. 푥ℎ(푦) ± 푦푥 ∈ 푍(푅) Motivated by these results, we prove a similar iii. 푥ℎ(푦) ± [푥. 푦] ∈ 푍(푅) result regarding homoderivations. To achieve our iv. ℎ(푦)푥 ± [푥. 푦] ∈ 푍(푅) aim, we will use the following lemma. v. [ℎ(푥). 푦] ± 푥푦 ∈ 푍(푅) vi. [ℎ(푥). 푦] ± 푦푥 ∈ 푍(푅) vii. 푅 is commutative. * Corresponding Author. Email Address: [email protected] (E. F. Alharfie) Proof: If (vii) holds then all other conditions are https://doi.org/10.21833/ijaas.2018.05.010 true. To prove (i) ⟹ (vii). By hypothesis, we have 2313-626X/© 2018 The Authors. Published by IASE. This is an open access article under the CC BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/) 푥ℎ(푦) ± 푥푦 ∈ 푍(푅) 푓표푟 푎푙푙 푥, 푦 ∈ 퐼. (1) 79 E. F. Alharfie, N. M. Muthana /International Journal of Advanced and Applied Sciences, 5(5) 2018, Pages: 79-81 Replacing x by yx in (1), we get 푦(푥ℎ(푦) ± 푥푦) ∈ (ℎ(푦)푥 ± [푥. 푦])푦 ∈ 푍(푅) 푍(푅). By Lemma 1.1, 푦 ∈ 푍(푅) 표푟 푥ℎ(푦) ± 푥푦 = 0. If Since h(y)x ± [x. y] ∈ Z(R), we get (h(y)x ± y∈Z(R) for all y∈I hence I⊂Z(R). Therefore I is [x. y])y = y(h(y)x ± [x. y]). So y(h(y)x ± [x. y]) ∈ commutative. By Remark 1.2, R is commutative. If Z(R). By Lemma 1.1, we have y ∈ Z(R) or h(y)x ± [x. y] = 0. If y ∈ Z(R) for all y ∈ I. I ⊆ Z(R). Then I is 푥ℎ(푦) ± 푥푦 = 0 푓표푟 푎푙푙 푥, 푦 ∈ 퐼. (2) commutative, by Remark 1.2, R is commutative. If Replace y by yx in (2), we have x(h(y) ±y)h(x)=0. ℎ(푦)푥 ± [푥. 푦] = 0 푓표푟 푎푙푙 푥. 푦 ∈ 퐼. (8) Since h is zero-power valued on I, there exists an integer n(y)>1 such that hn(y)(y) = 0 for all y∈I. Replace y by xy in (8), h(x)(h(y) ± y)x = 0. Since Replacing y by h is zero-power valued on I, so h(x)yx = 0 for all x. y ∈ I . Then we get h(x)RIx = 0. 푦 − ℎ(푦) + ℎ2(푦) + ⋯ + (−1)푛(푦)−1ℎ푛(푦)−1(푦) By primeness of R we have h(x) = 0 since Ix ≠ 0 for all x ∈ I. in the last relation. We have 푥푦ℎ(푥) = 0 for all 푥, 푦 ∈ If h(x) = 0 for all x ∈ I , we have [x. y] = 0 by (8), 퐼. Hence, 푥푅퐼ℎ(푥) = 0 for all 푥 ∈ 퐼. But I ≠ 0. So then I is commutative. By Remark 1.2, R is Ih(x) = 0 for all 푥 ∈ 퐼. Hence the Eq. 2 implies x2 = 0 commutative. for all 푥 ∈ 퐼. By Lemma 1.3, R has a nonzero nilpotent ideal which contradict that R is prime ring. To prove (v) ⟶ (vii). By hypothesis, we have To prove (ii) → (vii). By hypothesis, we have [ℎ(푥). 푦] ± 푥푦 ∈ 푍(푅) 푓표푟 푎푙푙 푥. 푦 ∈ 퐼. (9) 푥ℎ(푦) ± 푦푥 ∈ 푍(푅) 푓표푟 푎푙푙 푥. 푦 ∈ 퐼. (3) Replace y by yh(x) in (8) for all x. y ∈ I, we have, Replace x by yx in (3), we have y(xh(y) ± yx) ∈ ([ℎ(푥). 푦] ± 푥푦)ℎ(푥) ∈ 푍(푅) 푓표푟 푎푙푙 푥. 푦 ∈ 퐼. Z(R). By Lemma 1.1, y ∈ Z(R) or xh(y) ± yx = 0, If By Lemma 1.1 either h(x) ∈ Z(R) or [h(x). y] + xy = y ∈ Z(R) for all y ∈ I. hence I ⊆ Z(R). Therefore I is 0 for all x. y ∈ I. commutative. By Remark 1.2, R is commutative. If If h(x) ∈ Z(R) for all x ∈ I. then [h(x). x] = 0. By 푥ℎ(푦) ± 푦푥 = 0 푓표푟 푎푙푙 푥. 푦 ∈ 퐼. (4) Lemma 1.4, R is commutative. If Replace y by xy in (4) we have xh(x)(h(y) ± y) = [ℎ(푥). 푦] + 푥푦 = 0 푓표푟 푎푙푙 푥. 푦 ∈ 퐼. (10) 0 for all x. y ∈ I. Since h is zero-power valued on I , so Replace y by xy in (10) xh(x)y = 0 for all x. y ∈ I. Hence, xh(x)RI = 0 for all x ∈ I ⋅ By primeness of R, we get xh(x) = 0 for all x ∈ 2 [ℎ(푥). 푥푦] + 푥푥푦 = 0 푓표푟 푎푙푙 푥. 푦 ∈ 퐼. I. So, by (3) we get x = 0 for all x ∈ I. By Lemma 푥([ℎ(푥). 푦] + 푥푦) + [ℎ(푥). 푥]푦 = 0 푓표푟 푎푙푙 푥. 푦 ∈ 퐼. 1.3, this is contradiction. To prove (iii) → (vii). By [ℎ(푥). 푥]푦 = 0 푓표푟 푎푙푙 푥. 푦 ∈ 퐼. hypothesis, we have [ℎ(푥). 푥]푅퐼 = 0 푓표푟 푎푙푙 푥. 푦 ∈ 퐼. 푥ℎ(푦) ± [푥. 푦] ∈ 푍(푅) 푓표푟 푎푙푙 푥. 푦 ∈ 퐼. (5) By primeness of R and I ≠ 0, we have [h(x). x] = 0 for all x ∈ I. By Lemma 1.4, R is commutative.