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Journal of Taibah University for Science Volume 8, 2014 - Issue 4

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Original Article

Derivations and Jordan ideals in prime ringsFootnote
Peer review under responsibility of Taibah University.

Lahcen OukhtiteTaibah University, College of Sciences, Department of Mathematics, P.O. Box 2285, Madina, Saudi Arabia;Université Moulay Ismaïl, Faculté des Sciences et Techniques, Département de Mathématiques, Errachidia, MoroccoCorrespondence[email protected] [email protected]

Abdellah MamouniUniversité Moulay Ismaïl, Faculté des Sciences et Techniques, Département de Mathématiques, Errachidia, Morocco

Charef BeddaniTaibah University, College of Sciences, Department of Mathematics, P.O. Box 2285, Madina, Saudi Arabia

Pages 364-369 | Published online: 16 Apr 2018

Cite this article
https://doi.org/10.1016/j.jtusci.2014.04.004
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In this article

Abstract
1 Introduction
2 Preliminaries
3 Main results
Acknowledgements
Footnotes
References

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Abstract

The purpose of this paper is to study derivations satisfying certain differential identities on Jordan ideals of prime rings. Some well known results characterizing commutativity of prime rings by derivations have been generalized by using Jordan ideals. Moreover, we provide examples to show that our results cannot be extended to semi-prime rings.

MSC:

16W10
16W25
16U80

Keywords:

Prime-rings
Derivations
Jordan ideals
Commutativity

1 Introduction

The recent literature contains numerous results indicating how the global structure of a ring is often tightly connected to the behavior of special mappings defined on the ring. A well known result of Posner [Citation1] states that a prime ring must be commutative if it admits a nonzero centralizing derivation. This theorem has been generalized by a several authors in various directions. Moreover, Posner theorem has been extremely influential and at least indirectly it initiate the study of various notions. The most general and important one among them is the notion of a functional identity.

Recently, many authors have studied the commutativity of prime and semiprime rings admitting suitably constrained additive mappings, as automorphisms, derivations, skew derivations, and generalized derivations acting on appropriate subsets of the rings. Furthermore, obtained results are in general extensions of other ones previously proven just for the action of the considered mapping on the whole ring.

In the present paper we will study the structure of a prime ring R having derivations which satisfy suitable algebraic properties on a Jordan ideal J. More precisely, we will prove the following theorems:

Theorem 1

If R admits two derivations d₁ and d₂ such that

$[d_{1} (x), d_{2} (y)] = [x, y] for all x, y \in J,$ then R is commutative.

Theorem 2

Let d₁, d₂ and d₃ be nonzero derivations of R. If either

$d_{3} (y) d_{1} (x) - d_{2} (x) d_{3} (y) = 0 or d_{3} (y) d_{1} (x) - d_{2} (x) d_{3} (y) = [x, y]$ for all x, y ∈ J, then R is commutative and d₁ = d₂.

Theorem 3

Let d₁ and d₂ be derivations of R with d₁ is nonzero. If $[d_{2} (x), d_{1} (y)] = d_{1} [x, y] for all x, y \in J,$ then R is commutative.

Theorem 4

Let d₂ be a derivation of R. Then there exists no nonzero derivation d₁ such that

$d_{2} (x) \circ d_{1} (y) = d_{1} (x \circ y) for all x, y \in J .$

2 Preliminaries

Throughout this paper N will be an associative ring with center Z(R). Recall that R is prime (resp. semi-prime) it has the property that xRy = 0 (resp. xRx = 0) for x, y ∈ R implies x = 0 or y = 0 (resp. x = 0). An additive mapping d : R → R is called a derivation if d(xy) = d(x)y + xd(y) holds for all pairs x, y ∈ R. A mapping f : R → R is said to be strong commutativity preserving on a subset S of R if [f(x), f(y)] = [x, y] for all x, y ∈ S. As usual [x, y] = xy − yx and x ∘ y = xy + yx will denote the well-known Lie and Jordan products, respectively. Recalling that R is called 2-torsion free if 2x = 0 implies x = 0 for all x ∈ R. An additive subgroup J of R is a Jordan ideal if x ∘ r ∈ J for all x ∈ J and r ∈ R.

Remark 2.1

We shall use without explicit mention the fact that if J is a Jordan ideal of R, then 2[R, R]J ⊆ J and 2J[R, R] ⊆ J (see [2, Lemma 1]). Moreover, From [Citation3] we have 4jRj ⊂ J, 4j²R ⊂ J and 4Rj² ⊂ J for all j ∈ J.

In all that follows R will be a 2-torsion free prime ring and J a nonzero Jordan ideal of R. We leave the proofs of the following easy facts to the reader.

Fact 1.	=	If [a, x²] = 0 for all x ∈ J, then a ∈ Z(R).
Fact 2.	=	If J ⊂ Z(R), then R is commutative.
Fact 3.	=	If R is noncommutative such that $a [r, xy] b = 0 for all x, y \in J, and r \in R,$ then a = 0 or b = 0.

Lemma 2.2

If d is a derivation of R such that d(x²) = 0 for all x ∈ J, then d = 0.

Proof

Let us consider $R = R \times R^{0}$ where R⁰ denotes the opposite ring of R. It is known that $R$ equipped with the exchange involution *_ex, defined by *_ex(x, y) = (y, x), is a *_ex-prime ring (see [Citation4]). Moreover, if we set $J = J \times J$ , then $J$ is a *_ex-Jordan ideal of $R$ . Now let $D$ be the additive mapping defined on $R$ by $D (x, y) = (d (x), 0)$ . Clearly, $D$ is a nonzero derivation satisfying $D (u^{2}) = 0$ for all $u \in J$ . Applying [5, Lemma 3] we conclude that $D = 0$ and therefore d = 0. □

Proposition 2.3

If R admits derivations d₁ and d₂ such that d₁d₂(x) = 0 for all x ∈ J, then d₁ or d₂ is zero.

Proof

Assume that d₂ ≠ 0. We have

(1) $d_{1} d_{2} (x) = 0 for all x \in J .$ (1)

Replacing x by 4x²y in Equation(1)(1) $d_{1} d_{2} (x) = 0 for all x \in J .$ (1) , where y ∈ J, we obtain

(2) $d_{2} (x^{2}) d_{1} (y) + d_{1} (x^{2}) d_{2} (y) = 0 for all x, y \in J .$ (2)

Substituting 2[r, uv]y for y in Equation(2)(2) $d_{2} (x^{2}) d_{1} (y) + d_{1} (x^{2}) d_{2} (y) = 0 for all x, y \in J .$ (2) , where $u, v \in J, r \in R$ , we get

(3) $d_{2} (x^{2}) [r, uv] d_{1} (y) + d_{1} (x^{2}) [r, uv] d_{2} (y) = 0 for all u, v, x, y \in J, r \in R .$ (3)

Putting 4y²t instead of y in Equation(3)(3) $d_{2} (x^{2}) [r, uv] d_{1} (y) + d_{1} (x^{2}) [r, uv] d_{2} (y) = 0 for all u, v, x, y \in J, r \in R .$ (3) , where t ∈ R, we find that

(4) $d_{2} (x^{2}) [r, uv] y^{2} d_{1} (t) + d_{1} (x^{2}) [r, uv] y^{2} d_{2} (t) = 0 for all u, v, x, y \in J, r \in R .$ (4)

Writing d₂(z) instead of t in Equation(4)(4) $d_{2} (x^{2}) [r, uv] y^{2} d_{1} (t) + d_{1} (x^{2}) [r, uv] y^{2} d_{2} (t) = 0 for all u, v, x, y \in J, r \in R .$ (4) , where z ∈ J, we obtain

(5) $d_{1} (x^{2}) [r, uv] y^{2} d_{2}^{2} (z) = 0 for all u, v, x, y \in J_{1}, r \in R .$ (5)

Using Fact 3, because of d₂ ≠ 0, either R is commutative or d₁(x²) = 0 for all x ∈ J in which case d₁ = 0 by Lemma 2.2.

Assume that R is commutative; replacing x by 2x² in Equation(1)(1) $d_{1} d_{2} (x) = 0 for all x \in J .$ (1) , we get

(6) $2 d_{2} (x) d_{1} (x) = 0 for all x \in J .$ (6)

Hence for all x ∈ J we have d₁(x) = 0 or d₂(x) = 0 by using Brauer’s trick, we obtain d₁(J) = {0} or d₂(J) = {0}. Since d₂ ≠ 0, then d₁(J) = 0 and thus d₁ = 0. □

Proposition 2.4

If R admits derivations d₁ and d₂ such that d₁(d₂(x) − x) = 0 for all x ∈ J, then d₁ is zero.

Proof

We have

(7) $d_{1} (d_{2} (x) - x) = 0 for all x \in J .$ (7)

If R is commutative, replacing x by 2x² in Equation(7)(7) $d_{1} (d_{2} (x) - x) = 0 for all x \in J .$ (7) we get

(8) $d_{1} (x) d_{2} (x) = 0 for all x \in J .$ (8)

Since a commutative prime ring is an integral domain, then Eq. Equation(8)(8) $d_{1} (x) d_{2} (x) = 0 for all x \in J .$ (8) forces d₁ = 0 or d₂ = 0.

If d₂ = 0, then d₁(J) = {0} so that d₁ = 0.

Assume that R is noncommutative and d₂ ≠ 0.

Replacing x by 4x²y in Equation(7)(7) $d_{1} (d_{2} (x) - x) = 0 for all x \in J .$ (7) , where y ∈ J, we obtain

(9) $d_{2} (x^{2}) d_{1} (y) + d_{1} (x^{2}) d_{2} (y) = 0 for all x, y \in J .$ (9)

Substituting 2[r, uv]y for y in Equation(9)(9) $d_{2} (x^{2}) d_{1} (y) + d_{1} (x^{2}) d_{2} (y) = 0 for all x, y \in J .$ (9) , where $u, v \in J, r \in R$ , we get

(10) $d_{2} (x^{2}) [r, uv] d_{1} (y) + d_{1} (x^{2}) [r, uv] d_{2} (y) = 0 for all u, v, x, y \in J, r \in R .$ (10)

Replacing y by 4y²t, with t ∈ R in Equation(10)(10) $d_{2} (x^{2}) [r, uv] d_{1} (y) + d_{1} (x^{2}) [r, uv] d_{2} (y) = 0 for all u, v, x, y \in J, r \in R .$ (10) we obtain

(11) $d_{2} (x^{2}) [r, uv] y^{2} d_{1} (t) + d_{1} (x^{2}) [r, uv] y^{2} d_{2} (t) = 0 for all u, v, x, y \in J, r, t \in R .$ (11)

Putting t(d₂(z) − z) instead of t with z ∈ J in Equation(11)(11) $d_{2} (x^{2}) [r, uv] y^{2} d_{1} (t) + d_{1} (x^{2}) [r, uv] y^{2} d_{2} (t) = 0 for all u, v, x, y \in J, r, t \in R .$ (11) , we find that

(12) $d_{1} (x^{2}) [r, uv] y^{2} {td}_{2} (d_{2} (z) - z) = 0 for all u, v, x, y, z \in J, r, t \in R .$ (12)

In light of Fact 3, either $d_{2}^{2} (z) = d_{2} (z)$ for all z ∈ J or d₁(x²) = 0 for all x ∈ J.

If $d_{2}^{2} (z) = d_{2} (z)$ for all z ∈ J, then replacing z by 4z²u, with u ∈ J, we obtain d₂(z²)d₂(u) = 0 and replacing again u by 4uz² we get d₂(z²)Jd₂(z²) = 0 for all z ∈ J.

In view of [2, Lemma 2.6], the above relation yields that d₂(z²) = 0 for all z ∈ J, but in light of Lemma 2.2, this is impossible. Accordingly, d₂ = 0 and our hypothesis reduce to d₁(J) = {0} so that d₁ = 0. □

Proposition 2.5

If R admits derivations d₁ and d₂ such that d₁(x)d₂(y) = [x, y] for all x, y ∈ J, then either d₁ = 0 or d₂ = 0 and thus R is commutative.

Proof

Assume that d₂ ≠ 0, we have

(13) $d_{1} (x) d_{2} (y) = [x, y] for all x, y \in J .$ (13)

Replacing y by 4yz² in Equation(13)(13) $d_{1} (x) d_{2} (y) = [x, y] for all x, y \in J .$ (13) , where z ∈ J, we obtain

(14) $d_{1} (x) {yd}_{2} (z^{2}) = y [x, z^{2}] for all x, y, z \in J .$ (14)

Replacing y by 2[r, s]y in Equation(14)(14) $d_{1} (x) {yd}_{2} (z^{2}) = y [x, z^{2}] for all x, y, z \in J .$ (14) we get

(15) $d_{1} (x) [r, s] {yd}_{2} (z^{2}) = [r, s] y [x, z^{2}] for all x, y, z \in J, r, s \in R .$ (15)

Left multiplication of Equation(14)(14) $d_{1} (x) {yd}_{2} (z^{2}) = y [x, z^{2}] for all x, y, z \in J .$ (14) by [r, s] get

(16) $[r, s] d_{1} (x) {yd}_{2} (z^{2}) = [r, s] y [x, z^{2}] for all x, y, z \in J, r, s \in R .$ (16)

From Eqs. Equation(15)(15) $d_{1} (x) [r, s] {yd}_{2} (z^{2}) = [r, s] y [x, z^{2}] for all x, y, z \in J, r, s \in R .$ (15) and Equation(16)(16) $[r, s] d_{1} (x) {yd}_{2} (z^{2}) = [r, s] y [x, z^{2}] for all x, y, z \in J, r, s \in R .$ (16) we obtain [d₁(x), [r, s]]yd₂(z²) = 0 so that

(17) $[d_{1} (x), [r, s]] {Jd}_{2} (z^{2}) = 0 for all x, z \in J, r, s \in R .$ (17)

Since d₂ is nonzero, the primeness of R forces

(18) $[d_{1} (x), [r, s]] = 0 for all x \in J, r, s \in R .$ (18) so that d₁ is commuting, then [4, Theorem 2], forces that R is commutative and Eq. Equation(13)(13) $d_{1} (x) d_{2} (y) = [x, y] for all x, y \in J .$ (13) becomes

(19) $d_{1} (x) d_{2} (y) = 0 for all x, y \in J,$ (19)

which, because of d₂ ≠ 0, leads to d₁ = 0 so that [x, y] = 0 for all x, y ∈ J which implies that R is commutative. □

3 Main results

It is known that a 2-torsion free prime ring must be commutative if it admits a strong commutativity preserving derivation d, that is a derivation satisfying

$[d (x), d (y)] = [x, y] for all x, y \in R .$

Our aim in the following theorem is to generalize this result of Bell and Daif in two directions. First of all we will only assume that the commutativity condition is imposed on a Jordan ideal of R rather than on R. Secondly we will treat the case of two derivations instead of one derivation.

Theorem 3.1

If R admits derivations d₁ and d₂ such that [d₁(x), d₂(y)] = [x, y] for all x, y ∈ J, then R is commutative.

Proof

If d₁ = 0 or d₂ = 0, then our hypothesis becomes [x, y] = 0 for all x, y ∈ J, so that R is commutative by [2, Lemma 2.7] together with Fact 2.

Now assume that d₁ and d₂ are nonzero derivations such that

(20) $[d_{1} (x), d_{2} (y)] = [x, y] for all x, y \in J .$ (20)

Replacing x by 4xz² in Equation(20)(20) $[d_{1} (x), d_{2} (y)] = [x, y] for all x, y \in J .$ (20) , where z ∈ J, we get

(21) $d_{1} (x) [z^{2}, d_{2} (y)] + [x, d_{2} (y)] d_{1} (z^{2}) = 0 for all x, y, z \in J .$ (21)

Again, replacing x by 2x[r, uv] in Equation(21)(21) $d_{1} (x) [z^{2}, d_{2} (y)] + [x, d_{2} (y)] d_{1} (z^{2}) = 0 for all x, y, z \in J .$ (21) , where $u, v \in J, r \in R$ , we obtain

(22) $d_{1} (x) [r, uv] [z^{2}, d_{2} (y)] + [x, d_{2} (y)] [r, uv] d_{1} (z^{2}) = 0 for all u, v, x, y, z \in J, r \in R .$ (22)

Putting 4tx² for x in Equation(22)(22) $d_{1} (x) [r, uv] [z^{2}, d_{2} (y)] + [x, d_{2} (y)] [r, uv] d_{1} (z^{2}) = 0 for all u, v, x, y, z \in J, r \in R .$ (22) , where t ∈ R, we find that

(23) $d_{1} (t) x^{2} [r, uv] [z^{2}, d_{2} (y)] + [t, d_{2} (y)] x^{2} [r, uv] d_{1} (z^{2}) = 0$ (23) for all $u, v, x, y, z \in J$ and r, t ∈ R.

Substituting d₂(y)t for t in Equation(23)(23) $d_{1} (t) x^{2} [r, uv] [z^{2}, d_{2} (y)] + [t, d_{2} (y)] x^{2} [r, uv] d_{1} (z^{2}) = 0$ (23) , we get d₁d₂(y)tx²[r, uv][z², d₂(y)] = 0 and therefore

(24) $d_{1} d_{2} (y) {Rx}^{2} [r, uv] [z^{2}, d_{2} (y)] = 0 for all u, v, x, y, z \in J, r \in R .$ (24)

Since R is a prime ring, we obtain d₁d₂(y) = 0 or x²[r, uv][z², d₂(y)] = 0 for all $u, v, x, y, z \in J, r \in R$ . By using Brauer’s trick, we obtain d₁d₂(y) = 0 for all y ∈ J or x²[r, uv][z², d₂(y)] = 0 for all $u, v, x, y, z \in J, r \in R$ .

Since d₁ and d₂ are nonzero, hence Proposition 2.3 forces

(25) $x^{2} [r, uv] [z^{2}, d_{2} (y)] = 0 for all u, v, x, y, z \in J, r \in R .$ (25)

In view of Fact 3, Eq. Equation(25)(25) $x^{2} [r, uv] [z^{2}, d_{2} (y)] = 0 for all u, v, x, y, z \in J, r \in R .$ (25) together with J ≠ {0}, yield R is commutative or [z², d₂(y)] = 0 for all y, z ∈ J, in which case, Fact 1 forces d₂(J) ⊆ Z(R). Hence, d₂ is centralizing on J and [4, Theorem 2], assures that R is commutative. □

In [Citation6] Herstein proved that a 2-torsion free prime ring must be commutative if it admits a nonzero derivation d which satisfies [d(x), d(y)] = 0 for all x, y ∈ R. Using the proof of Theorem 1, the following corollary gives a more general version of Herstein’s result.

Corollary 3.2

If R admits nonzero derivations d₁ and d₂ such that [d₁(x), d₂(y)] = 0 for all x, y ∈ J, then R is commutative.

Proof

Assume that

(26) $[d_{1} (x), d_{2} (y)] = 0 for all x, y \in J .$ (26)

Replacing x by 4xz² in Equation(26)(26) $[d_{1} (x), d_{2} (y)] = 0 for all x, y \in J .$ (26) , where z ∈ J, we get

(27) $d_{1} (x) [z^{2}, d_{2} (y)] + [x, d_{2} (y)] d_{1} (z^{2}) = 0 for all x, y, z \in J .$ (27)

Since Eq. Equation(27)(27) $d_{1} (x) [z^{2}, d_{2} (y)] + [x, d_{2} (y)] d_{1} (z^{2}) = 0 for all x, y, z \in J .$ (27) is the same as Eq. Equation(21)(21) $d_{1} (x) [z^{2}, d_{2} (y)] + [x, d_{2} (y)] d_{1} (z^{2}) = 0 for all x, y, z \in J .$ (21) , reasoning as in the proof of Theorem 3.1, we conclude that R is commutative. □

Motivating by Herstein’s result [Citation6], F.W. Niu in [Citation7] proved that a 2-torsion free prime ring R admitting nonzero derivations d₁, d₂ and d₃ such that ″d₃(y)d₁(x) = d₂(x)d₃(y) for all x, y ∈ R″ must be commutative. However, this result is less precise because in this case necessarily d₁ = d₂. Our aim in the following theorem is to extend this result to Jordan ideals in a more general form.

Theorem 3.3

Let d₁, d₂ and d₃ be nonzero derivations of R. If either d₃(y)d₁(x) = d₂(x)d₃(y) or d₃(y)d₁(x) − d₂(x)d₃(y) = [x, y] for all x, y ∈ J, then R is commutative and d₁ = d₂.

Proof

(i) Assume that

(28) $d_{3} (y) d_{1} (x) - d_{2} (x) d_{3} (y) = 0 for all x, y \in J .$ (28)

Replacing x by 4xz² in Equation(28)(28) $d_{3} (y) d_{1} (x) - d_{2} (x) d_{3} (y) = 0 for all x, y \in J .$ (28) , where z ∈ J, we get

(29) $d_{2} (x) [d_{3} (y), z^{2}] + [d_{3} (y), x] d_{1} (z^{2}) = 0 for all x, y, z \in J .$ (29)

Substituting 2x[r, uv] for x in Equation(29)(29) $d_{2} (x) [d_{3} (y), z^{2}] + [d_{3} (y), x] d_{1} (z^{2}) = 0 for all x, y, z \in J .$ (29) , where $u, v \in J, r \in R$ , we obtain

(30) $d_{2} (x) [r, uv] [d_{3} (y), z^{2}] + [d_{3} (y), x] [r, uv] d_{1} (z^{2}) = 0 for all u, v, x, y, z \in J, r \in R .$ (30)

Replacing x by 4tx² in Equation(30)(30) $d_{2} (x) [r, uv] [d_{3} (y), z^{2}] + [d_{3} (y), x] [r, uv] d_{1} (z^{2}) = 0 for all u, v, x, y, z \in J, r \in R .$ (30) , where t ∈ R, we find that

(31) $d_{2} (t) x^{2} [r, uv] [d_{3} (y), z^{2}] + [d_{3} (y), t] x^{2} [r, uv] d_{1} (z^{2}) = 0$ (31) for all $u, v, x, y, z \in J$ and r, t ∈ R.

Substituting d₃(y)t for t in Equation(31)(31) $d_{2} (t) x^{2} [r, uv] [d_{3} (y), z^{2}] + [d_{3} (y), t] x^{2} [r, uv] d_{1} (z^{2}) = 0$ (31) , then we have d₂d₃(y)tx²[r, uv][d₃(y), z²] = 0 and therefore

(32) $d_{2} d_{3} (y) {Rx}^{2} [r, uv] [d_{3} (y), z^{2}] = 0 for all u, v, x, y, z \in J, r \in R .$ (32)

As Eq. Equation(32)(32) $d_{2} d_{3} (y) {Rx}^{2} [r, uv] [d_{3} (y), z^{2}] = 0 for all u, v, x, y, z \in J, r \in R .$ (32) is the same as Eq. Equation(24)(24) $d_{1} d_{2} (y) {Rx}^{2} [r, uv] [z^{2}, d_{2} (y)] = 0 for all u, v, x, y, z \in J, r \in R .$ (24) then we conclude that R is commutative. Thus Equation(28)(28) $d_{3} (y) d_{1} (x) - d_{2} (x) d_{3} (y) = 0 for all x, y \in J .$ (28) becomes d₃(y)(d₁(x) − d₂(x)) = 0 so that

(33) $d_{3} (y) R (d_{1} (x) - d_{2} (x)) = 0 for all x, y \in J .$ (33)

Since d₃ is nonzero we get d₂ = d₁.

(ii) Assume that d₃(y)d₁(x) − d₂(x)d₃(y) = [x, y] for all x, y ∈ J If d₃ = 0 then our theorem is trivial by [2, Lemma 2.7] together with Fact 2.

Hence one can suppose that d₃ ≠ 0.

Assume that d₁ and d₂ are nonzero derivations such that

(34) $d_{3} (y) d_{1} (x) - d_{2} (x) d_{3} (y) = [x, y] for all x, y \in J .$ (34)

Replacing x by 4xz² in Equation(34)(34) $d_{3} (y) d_{1} (x) - d_{2} (x) d_{3} (y) = [x, y] for all x, y \in J .$ (34) , where z ∈ J, we get

(35) $d_{2} (x) [d_{3} (y), z^{2}] + [d_{3} (y), x] d_{1} (z^{2}) = 0 for all x, y, z \in J .$ (35)

Since Eq. Equation(35)(35) $d_{2} (x) [d_{3} (y), z^{2}] + [d_{3} (y), x] d_{1} (z^{2}) = 0 for all x, y, z \in J .$ (35) is the same as Eq. Equation(29)(29) $d_{2} (x) [d_{3} (y), z^{2}] + [d_{3} (y), x] d_{1} (z^{2}) = 0 for all x, y, z \in J .$ (29) , reasoning as in (i), we conclude that R is commutative. Moreover, our hypothesis becomes

(36) $d_{3} (y) R (d_{1} (x) - d_{2} (x)) = 0 for all x, y \in J$ (36)

which, in light of d₃ ≠ 0, leads to d₁ = d₂. □

M.N. Daif in [Citation8] proved that if R a semiprime ring, U a nonzero right ideal of R and d a nonzero U^*-derivation (i.e. [d(y), d(x)] = d[x, y] for all x, y ∈ U), then d(U) centralizes [U, U].

Our next aim is to consider a more general situation as follows. $[d_{2} (y), d_{1} (x)] = d_{1} [y, x]$ for all x, y in a subset S of R. In fact, the following theorem proves that if R is prime and J a Jordan ideal, then the above condition constitute a commutativity criterion. However, this result cannot be extended to semiprime rings (see our example).

Theorem 3.4

Let d₁ and d₂ be derivations of R with d₁ is nonzero. If [d₂(y), d₁(x)] = d₁[y, x] for all x, y ∈ J, then R is commutative.

Proof

If d₂ = 0, then our hypothesis becomes d₁[y, x] = 0 for all x, y ∈ J. Hence R is commutative by [9, Theorem 2.10]. Now assume that d₂ is a nonzero derivation such that

(37) $d_{2} (y) d_{1} (x) - d_{1} (x) d_{2} (y) = d_{1} [y, x] for all x, y \in J .$ (37)

Replacing x by 4xz² in Equation(37)(37) $d_{2} (y) d_{1} (x) - d_{1} (x) d_{2} (y) = d_{1} [y, x] for all x, y \in J .$ (37) , where z ∈ J, we get

(38) $d_{1} (x) [d_{2} (y) - y, z^{2}] + [d_{2} (y) - y, x] d_{1} (z^{2}) = 0 for all x, y, z \in J .$ (38)

Again, replacing x by 2x[r, uv] in Equation(38)(38) $d_{1} (x) [d_{2} (y) - y, z^{2}] + [d_{2} (y) - y, x] d_{1} (z^{2}) = 0 for all x, y, z \in J .$ (38) , where $u, v \in J$ and r ∈ R, we obtain

(39) $d_{1} (x) [r, uv] [d_{2} (y) - y, z^{2}] + [d_{2} (y) - y, x] [r, uv] d_{1} (z^{2}) = 0,$ (39) for all $u, v, x, y, z \in J$ and r ∈ R. Putting 4tx² for x in Equation(39)(39) $d_{1} (x) [r, uv] [d_{2} (y) - y, z^{2}] + [d_{2} (y) - y, x] [r, uv] d_{1} (z^{2}) = 0,$ (39) , where t ∈ R, we find that

(40) $d_{1} (t) x^{2} [r, uv] [d_{2} (y) - y, z^{2}] + [d_{2} (y) - y, t] x^{2} [r, uv] d_{1} (z^{2}) = 0$ (40) for all $u, v, x, y, z \in J$ and r, t ∈ R. Substituting (d₂(y) − y)t for t in Equation(40)(40) $d_{1} (t) x^{2} [r, uv] [d_{2} (y) - y, z^{2}] + [d_{2} (y) - y, t] x^{2} [r, uv] d_{1} (z^{2}) = 0$ (40) , then we have

$d_{1} (d_{2} (y) - y) {tx}^{2} [r, uv] [d_{2} (y) - y, z^{2}] = 0$ and therefore

$d_{1} (d_{2} (y) - y) {Rx}^{2} [r, uv] [d_{2} (y) - y, z^{2}] = 0 for all u, v, x, y, z \in J, r \in R$ which leads to d₁(d₂(y) − y) = 0 or [d₂(y) − y, z²] = 0. Since d₁ ≠ 0 Lemma 2.4 forces [d₂(y) − y, z²] = 0 and Fact 3 implies that d₂(y) − y ∈ Z(R) for all y ∈ J. Hence d₂ is centralizing on J and [4, Theorem 2] assures that R is commutative. □

As an application of Theorem 4, if we take d₂ = 0 we obtain Theorem 2.10 of [Citation9] as follows:

Corollary 3.5

If R admits a nonzero derivation d such that d([x, y]) = 0 for all x, y ∈ J, then R is commutative.

We now consider differential identities involving anti-commutators instead of commutators. Our result is of a different kind.

Theorem 3.6

Let d₂ be a derivation of R. Then there exists no nonzero derivation d₁ such that d₂(y) ∘ d₁(x) = d₁(y ∘ x) for all x, y ∈ J.

Proof

Suppose there exists a nonzero derivation d₁ such that

(41) $d_{2} (y) d_{1} (x) + d_{1} (x) d_{2} (y) - d_{1} (x \circ y) = 0 for all x, y \in J .$ (41)

Replacing x by 4xz² in Equation(41)(41) $d_{2} (y) d_{1} (x) + d_{1} (x) d_{2} (y) - d_{1} (x \circ y) = 0 for all x, y \in J .$ (41) we get

(42) $d_{1} (x) [z^{2}, d_{2} (y) + y] - [x, d_{2} (y)] d_{1} (z^{2}) + {xd}_{1} (y \circ z^{2}) - (x \circ y) d_{1} (z^{2}) - {xd}_{1} [y, z^{2}] = 0,$ (42) for all x, y, z ∈ J. Substituting 2x[r, uv] for x in Equation(42)(42) $d_{1} (x) [z^{2}, d_{2} (y) + y] - [x, d_{2} (y)] d_{1} (z^{2}) + {xd}_{1} (y \circ z^{2}) - (x \circ y) d_{1} (z^{2}) - {xd}_{1} [y, z^{2}] = 0,$ (42) we obtain

(43) $d_{1} (x) [r, uv] [z^{2}, d_{2} (y) + y] - [x, d_{2} (y) + y] [r, uv] d_{1} (z^{2}) = 0,$ (43) for all $u, v, x, y, z \in J$ and r ∈ R. Reasoning as in Eq. Equation(39)(39) $d_{1} (x) [r, uv] [d_{2} (y) - y, z^{2}] + [d_{2} (y) - y, x] [r, uv] d_{1} (z^{2}) = 0,$ (39) , we arrive at R is commutative and our hypothesis becomes

(44) $d_{1} (x) d_{2} (y) - d_{1} (x) y - {xd}_{1} (y) = 0 for all x, y \in J .$ (44)

Replacing x by 2xz where z ∈ J we obtain

(45) $d_{1} (x) {zd}_{2} (y) - d_{1} (z) xy = 0 for all x, y, z \in J .$ (45)

Writing 2yu instead of y where u ∈ J, we obtain

(46) $d_{1} (x) {zyd}_{2} (u) = 0 for all u, x, y, z \in J .$ (46)

Which because of d₁ ≠ 0 yields d₂ = 0. Therefore our hypothesis becomes d₁(x ∘ y) = 0 for all x, y ∈ J and thus

(47) $d_{1} (x^{2}) = 0 for all x \in J .$ (47)

Applying Lemma 2.2 the last equation implies that d₁ = 0 which contradicts our hypothesis. Consequently, d₁ = 0. □

The following example proves that, Theorems 3.1, 3.3 and 3.4 cannot be extended to semiprime ring.

Example 3.7

Let R be a noncommutative prime ring and consider de semiprime ring $R = R \times Q [X]$ . It is straightforward to verify that J = {0} × Q[X] is a nonzero Jordan ideal of R and d₁(r, P(X)) = d₂(r, P(X)) = d₃(r, P(X)) = (0, P′(X)) are nonzero derivations of $R$ such that

$[d_{1} (x), d_{2} (y)] = [x, y], d_{3} (y) d_{1} (x) = d_{2} (x) d_{3} (y),$ $d_{3} (y) d_{1} (x) - d_{2} (x) d_{3} (y) = [x, y], [d_{2} (y), d_{1} (x)] = d_{1} [y, x]$ for all x, y ∈ J but $R$ is not commutative.

Acknowledgments

The first and the third authors are deeply indebted to the team work at the Deanship of the Scientific Research, Taibah University, for their valuable help and critical guidance and for facilitating administrative procedures. This research work was supported by a grant No. (6051/1435).

Notes

Peer review under responsibility of Taibah University.

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Derivations and Jordan ideals in prime ringsFootnote
Peer review under responsibility of Taibah University.

Abstract

1 Introduction

2 Preliminaries

3 Main results

Acknowledgments

References

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Derivations and Jordan ideals in prime ringsFootnotePeer review under responsibility of Taibah University.

Abstract

1 Introduction

2 Preliminaries

3 Main results

Acknowledgments

Notes

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Derivations and Jordan ideals in prime ringsFootnote
Peer review under responsibility of Taibah University.