EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 15, No. 4, 2022, 2032-2042 ISSN 1307-5543 – ejpam.com Published by New York Business Global Reverse Derivations on δ-prime rings Iman Taha1,∗, Rohaidah Masri1, Ahmad Al Khalaf2, Rawdah Tarmizi1 1 Department of Mathematics, Faculty of Sciences and Mathematics, Sultan Idris Universiti, Tanjong Malim, Perak, Malaysia 2 Department of Mathematics and Statistics, Faculty of Sciences, Imam Mohammad Ibn Saud Islamic University, Riyadh, Riyadh, Saudi Arabia Abstract. In this paper, we generalized Posner’s theorem, then, Mayne’s theorem has been ex- tended to get a main result, and presented by the following theorem, if δ is a nonzero centralizing reverse derivations on a nonzero δ-ideal U of δ –prime ring R, then R is commutative. 2020 Mathematics Subject Classifications: 16W25, 16N60 Key Words and Phrases: Reverse derivation, δ-prime ring, δ-ideal 1. Introduction Throughout this paper, R assumed to be an associative ring with unity and the center Z(R), the commutator is defined as [u, v] = uv − vu, is also called a Lie commutator for elements u, v ∈ R, the symbol C(R) stand for the set of all commutator ideals generated by [u, v], the set annU = {r ∈ R : rU = 0} is the annihilator of U of R. The smallest positive integer n such that n · u = 0 for all u ∈ R is the characteristic of the ring R. We will employ some commutator properties like [uv,w] = u[v, w]+[u,w]v and [u, vw] = v[u,w] + [u, v]w ∀u, v, w ∈ R. Furthermore, we recall that R is called a prime ring if uRv = {0}, then either u = 0 or v = 0, and by analogy, R is called a semiprime ring, for u ∈ R, if uRu = {0}, then u = 0. A map F from R to R is said to be a centralizing on U if [u, F (u)] ∈ Z(R) for all u ∈ U . An additive map δ : R → R is called a derivation on R, if the condition δ(uv) = δ(u)v + uδ(v), ∀u, v ∈ R holds, while an additive map δ : R → R is said to be a reverse derivation on R if satisfies the rule δ(uv) = δ(v)u+ vδ(u), ∀u, v ∈ R. Furthermore, for a fixed element u ∈ R the additive map ∂u : R → R defined by, ∂u(v) = uv − vu, where v ∈ R is called a partial derivation generated by u ∈ R (∂u(U) = [u, U ] = {[u, t] : t ∈ U}). ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v15i4.4602 Email addresses: tfaith80gmail.com (I. Taha), ajalkalaf@imamu.edu.sa (A. Al Khalaf), rohaidah@fsmt.upsi.edu.my (R. Masri), rawdah@fsmt.upsi.edu.my (R. Tarmizi) https://www.ejpam.com 2032 © 2022 EJPAM All rights reserved. I. Taha et al. / Eur. J. Pure Appl. Math, 15 (4) (2022), 2032-2042 2033 Let ∆ be a subset of the set of all derivations D on R. An ideal U of R such that δ(U) ⊆ U is used to be called a δ- ideal of R. A ring R is called δ-prime if, for any two δ- ideals U, V of R, the condition UV = 0 infers that either U = 0 or V = 0, equivalently, we call a ring R that δ- semiprime, for U ⊆ D, if [U,U ] ̸= 0, implies U ̸= 0. Moreover, other terminologies are standard and they were considered as in [10],[11] and [12]. Posner’s First Theorem demonstrated that the composition of two nonzero derivations on a prime ring R, with charR ̸= 2, is not a derivation. The Second Theorem of Posner proved that if the nonzero derivation d on a prime ring R is a centralizing on R, then R is commutative [26]. Mayne generalized Posner’s theorem when a ring R has an automorphism or a nonzero centralizing derivation on some ideal U ̸= 0, concluding that R is commutative [22]. Likewise, some generalizations of these results with different ways for a prime and semiprime rings are condidered in [6, 7, 13, 14, 21–24, 26, 28, 29]. In general, they have showed commutativity of prime and semiprime rings admit centralizing derivations on specific subsets of R. Overall, Brešar and Vukman who started the researching on reverse derivation concept (see [8]), recently, Samman and Alyamani [27] came, with many properties of reverse derivations in prime (resp. semiprime) rings. 2. Preliminaries Many researchers studied the properties of Lie rings with derivations D of differentially simple, prime and semiprime rings (see for example [1–4], [14, 15], [16, 17] and [18, 28], where further references can be found for the widening in this field. Passman in [25], has investigated the commutative rings with semiprime Lie ring D. There are many papers in this line, as [9, 19, 20]. The following main lemmas that will be used to prove our new results, to which we shall refer, stated as in the following: Lemma 1. [26, Lemma 3] Let R be a prime ring, and d a derivation of R such that ad(a)− d(a)a = 0 for all a ∈ R. Then R is commutative. Lemma 2. [21, Theorem] Let R be a prime ring with a nontrivial centralizing automor- phism. Then, R is a commutative integral domain. Lemma 3. [22, Theorem] Let R be a prime ring and A ̸= {0} be an ideal of R. If R has a nontrivial automorphisim or derivation T such that uuT − uTu is in the center of R and uT is in U for every u in U , then R is commutative. Lemma 4. [4, Lemma 13] Let A ̸= 0 be a Lie δ-ideal of a δ-semiprime and subring of a ring R of charR ̸= 2. Then A ⊆ Z(R) or A contains a non-central δ-ideal of R. I. Taha et al. / Eur. J. Pure Appl. Math, 15 (4) (2022), 2032-2042 2034 Therefore the purpose of our research is to study the structure of reverse derivation on δ– prime ring and some properties of centralizing reverse derivations on nonzero δ– ideal of δ– prime ring. 3. Some properties and examples of Reverse Derivation In fact, it is not necessary that every derivation is a reverse derivation on a ring R or vice versa. Taking into consideration when the ring R is commutative, then the derivation and the reverse derivation are coincides. Therefore it is possible to define an example about this case. Example 1. Let R = {[ u v 0 0 ] : u, v ∈ T } , such that T is a ring with T 2 ̸= {0}. Let δ : R → R be an additive mapping defined as δ ([ u v 0 0 ]) = [ 0 v 0 0 ] : ∀u, v ∈ T. and δ̀ : R → R be an additive mapping defined as δ̀ ([ u v 0 0 ]) = [ 0 u 0 0 ] : ∀u, v ∈ T. It is easy to see that δ is a derivation, but not a reverse derivation, while δ̀ is both a derivation and a reverse derivation. Now, if the ring R is commutative, then δ(uv) = δ(vu). if δ is a derivation on a ring R, so δ(uv) = δ(u)v + uδ(v) and δ(vu) = δ(v)u+ vδ(u). This means δ(u)v + uδ(v) = δ(v)u+ vδ(u). Thus, δ is a reverse derivation too. Lemma 5. Let R be a δ-prime ring, U ̸= {0} a δ-ideal of R, which δ ̸= 0 a reverse derivation on a R. If δ(U) = 0, then δ(R) = 0. proof. Since U is δ-ideal of R, then RU ⊆ U and UR ⊆ U. I. Taha et al. / Eur. J. Pure Appl. Math, 15 (4) (2022), 2032-2042 2035 So δ(RU) ⊆ δ(U) = 0 and δ(UR) ⊆ δ(U) = 0, then δ(RU) = δ(UR) = 0. Since δ is a reverse derivation then, δ(RU) = δ(U)R+ Uδ(R) = 0 and δ(UR) = δ(R)U +Rδ(U) = 0, this means Uδ(R) = δ(R)U = 0. Thus, we deduce that δ(R) ⊆ annU , but U is a δ-ideal, hence δ(R) = 0. Lemma 6. Let δ ̸= 0 be a reverse derivation on a δ-ring R and let U ̸= {0} be δ-ideal of R. If [δ(u), u] = 0 ∀u ∈ U, (1) then R is commutative. proof. Now, let linearize the identity (1) on U , then we have for all u, v ∈ U 0 = [δ(u+ v), u+ v] = [δ(u), u] + [δ(u), v]+ [δ(v), u] + [δ(v), v] = [δ(u), v] + [δ(v), u]. [δ(u), v] = [u, δ(v)]. (2) Write vδ(u) instead of δ(u) in (2), then we get [u, δ(v)] = [vδ(u), v] = v[δ(u), v] + [v, v]δ(u) = v[δ(u), v] = v[u, δ(v)] = vuδ(v)− vδ(v)u = vuδ(v)− δ(v)vu = [vu, δ(v)] = I. Taha et al. / Eur. J. Pure Appl. Math, 15 (4) (2022), 2032-2042 2036 [δ(vu), v] = [δ(u)v + uδ(v), v] = [δ(u)v, v] + [uδ(v), v] = [δ(u), v]v + [u, v]δ(v) [v, u]δ(v) = 0. (3) Now, replace u by uv in (3), then, [v, uv]δ(v) = 0, [v, u]vδ(v) = 0, thus [v, u]Uδ(v) = 0. Hence, since R is a δ-prime ring, either δ(u) = 0, then u = 0 and this contradicts the assumption, or [v, u] = 0 for all u, v ∈ U , therefore U is commutative. So we get UC(R) = 0, then C(R) = 0, hence R is commutative. 4. Reverse Derivation on δ- Ideal Through this section, we will prove several lemmas arriving to the extension of lemma (1), presented by the main theorem. Lemma 7. Let R be a δ-prime ring and let δ ̸= 0 be a reverse derivation on R. If u[δn(u), R] = 0, ∀u ∈ R or [δn(v), R]u = 0, ∀u, v ∈ R, n ∈ Z+. Then either u = 0 or v ∈ Z(R). proof. Assume u, v ∈ R and n ∈ Z+, then from the assumption u[δn(u), R] = 0, this equivalents to u∂δn(v)(R) = 0, (4) then 0 = u∂δn(v)(ab) = u∂δn(v)(b)a+ ub∂δn(v)(a), ∀a, b ∈ R Now from (4) ub∂δn(v)(a) = 0, I. Taha et al. / Eur. J. Pure Appl. Math, 15 (4) (2022), 2032-2042 2037 This means uR[δn(v), a] = 0. Consequently, uRδk([δn(v), a]) = 0. What forces that u = 0 or [δn(v), a] = 0. Hence v ∈ Z(R). Lemma 8. Let R be a δ-prime ring and let U ̸= {0} be a right δ-ideal, which δ is a reverse derivation on R. If U is commutative, then R is commutative. proof. Assume that u ∈ U . Since U is commutative, then ∂u(U) = [u, U ] = 0. Now by lemma (5), ∂u(R) = 0, then u ∈ Z(R),∀u ∈ U , hence U ⊆ Z(R). Thus C(R) = 0, this means R is commutative. Lemma 9. Let R be a δ-prime ring and δ ̸= 0 a reverse derivation on R. If [v, δn(u)v] ∈ Z(R),∀u, v ̸= 0 ∈ R,n ∈ Z+, then u ∈ Z(R). proof. Assume a ∈ R, then we get 0 = [δn(u)v, a] = δn(u)[v, a] + [δn(u), a]v = [δn(u), a]v. Then, by lemma (7), u ∈ Z(R). Lemma 10. Let R be a δ-prime ring of charR ̸= 2 and let U be a δ-ideal of R, which δ ̸= 0 is. If [a, δ(a)] ∈ Z(R) ∀a ∈ U, (5) then [a, δ(a)] = 0. proof. Assume that a, b ∈ U . Now by linearlizing the identity (5), we see [a+ b, δ(a+ b)] = [a, δ(a)] + [a, δ(b)] + [b, δ(a)] + [b, δ(b)], = [a, δ(b)] + [b, δ(a)], then, from (5) [a, δ(b)] + [b, δ(a)] ∈ Z(R). (6) Now replace b by a2 in (6), we have [a, δ(a2)] + [a2, δ(a)] = [a, δ(a)a+ aδ(a)] + [a2, δ(a)]. I. Taha et al. / Eur. J. Pure Appl. Math, 15 (4) (2022), 2032-2042 2038 Then, [a, δ(a2)] + [a2, δ(a)] = 4a[a, δ(a)] ∈ Z(R). Hence, a[a, δ(a)] ∈ Z(R). (7) Therefore, [a[a, δ(a)], δ(a)] = 0 a[a, δ(a)]δ(a)− δ(a)a[a, δ(a)] = 0 [a, δ(a)](aδ(a)− δ(a)a) = 0 [a, δ(a)]2 = 0. From (5), δ([a, δ(a)]) ∈ Z(R) and δ([a, δ(a)]) = [δ(a), δ(a)] + [a, δ2(a)] = [a, δ2(a)]. We obtain, δ([a, δ2(a)]) ∈ Z(R). Now δ([a, δ2(a)]) = [δ(a), δ2(a)] + [a, δ3(a)] = [a, δ3(a)]. Then, [a, δ3(a)] ∈ Z(R). (8) Now, using the induction on a number n, so we have [a, δn(a)] ∈ Z(R). (9) replacing b in (6) by aδn(a), we have [a, δ(aδn(a))] + [aδn(a), δ(a)] ∈ Z(R). I. Taha et al. / Eur. J. Pure Appl. Math, 15 (4) (2022), 2032-2042 2039 Then, [a, δ(aδn(a))] + [aδn(a), δ(a)] = [a, δn+1(a)a+ δn(a)δ(a)] + [aδn(a), δ(a)] = [a, δn+1(a)a] + [a, δn(a)δ(a)]− [δ(a), aδn(a)] = [a, δn+1(a)]a+ δn(a)[a, δ(a)] + [a, δn(a)]δ(a)− a[δ(a), δn(a)]− [δ(a), a]δn(a) = S. Then, 0 = [S, δn(a)] =[[a, δn+1(a)]a, δn(a)]+ (10) [δn(a)[a, δ(a)], δn(a)]+ [[a, δn(a)]δ(a), δn(a)]− [a[δ(a), δn(a)], δn(a)]− [[δ(a), a]δn(a), δn(a)]. Now substituting instead b in (6) by a2δn(a), then we get [a, δ(a2δn(a))] + [a2δn(a), δ(a)] ∈ Z(R). Then, [a, δ(a2δn(a))] + [a2δn(a), δ(a)] = [a, δ(a)aδn(a))] + [a, aδ(a)δn(a)] + [a, a2δn+1(a)]− [δ(a), a2δn(a)] = 4[a, δ(a)]aδn(a) + [a, δn(a)]aδ(a) + [a, δn(a)]δ(a)a+ [a, δn+1(a)]a2 − [δ(a), δn(a)]a2 = M. Now, multiply M by [a, δn(a)] and in view of (10), w have [a, δn(a)]2aδ(a) + [a, δn(a)]2δ(a)a− [δ(a), δn(a)][a, δn(a)]a2. Then, by continue the processes, we obtain [a, δn(a)]3δ(a) = 0, and so [a, δn+1(a)]4δ(a) = 0, REFERENCES 2040 then, [a, δn+1(a)]4R = 0, This means B = ∞∑ n=1 ∑ a∈U [a, δn(a)]R is a sum of nilpotent ideals and so U is a nil ideal, then B = 0. this means [a, δ(a)] = 0. Lemma 11. Let R be δ- prime ring of charR ̸= 2 and let δ be a reverse derivation, such that [a, δ(a)] ∈ Z(R),∀a ∈ R. Then R is commutative. proof. Recall that U = [R,R] is a Lie ideal of a prime ring R. Moreover, δ([R,R]) ⊆ [R,R]. Since every ideal in δ- prime ring is a δ- ideal, Now, if U = [R,R] is commutative, then, C(R) is a nil ideal (see [5], Lemma 1.7). Hence C(R) = 0 and R is commutative. Therefore, by using lemma (4), U = [R,R] contains a δ- ideal of R. Thus, by (6), this implies [a, δ(a)] ∈ Z(R), ∀a ∈ R, this gives δ(U) ∈ Z(R), then for all a ∈ U , ([a, δ(a)] = 0, and based on lemma (8) R will be commutative. 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