Generalized Derivations and Left Ideals in Prime and Semiprime Rings

Abstract

Let 𝑅 be an associative ring, 𝜆 a nonzero left ideal of 𝑅, 𝑑∶𝑅→𝑅 a derivation and 𝐺∶𝑅→𝑅 a generalized derivation. In this paper, we study the following situations in prime and semiprime rings: (1) 𝐺(𝑥∘𝑦)=𝑎(xy±yx); (2) 𝐺[𝑥,𝑦]=𝑎(xy±yx); (3) 𝑑(𝑥)∘𝑑(𝑦)=𝑎(xy±yx); for all 𝑥,𝑦∈𝜆 and 𝑎∈{0,1,−1}.

1. Introduction

Throughout this paper, let 𝑅 be an associative ring, 𝜆 a left ideal of 𝑅, 𝑑 a derivation of 𝑅 and 𝐺 a generalized derivation of 𝑅. For any two elements 𝑥,𝑦∈𝑅, [𝑥,𝑦] will denote the commutator element 𝑥𝑦−𝑦𝑥 and 𝑥∘𝑦 denotes 𝑥𝑦+𝑦𝑥. We use extensively the following basic commutator identities: [𝑥𝑦,𝑧]=[𝑥,𝑧]𝑦+𝑥[𝑦,𝑧] and [𝑥,𝑦𝑧]=[𝑥,𝑦]𝑧+𝑦[𝑥,𝑧]. Recall that a ring 𝑅 is called prime, if for any 𝑎,𝑏∈𝑅, 𝑎𝑅𝑏=(0) implies that either 𝑎=0 or 𝑏=0 and is called semiprime if for any 𝑎∈𝑅, 𝑎𝑅𝑎=(0) implies 𝑎=0. An additive mapping 𝑑∶𝑅→𝑅 is said to be a derivation of 𝑅 if for any 𝑥,𝑦∈𝑅, 𝑑(𝑥𝑦)=𝑑(𝑥)𝑦+𝑥𝑑(𝑦) holds. The generalized derivation of 𝑅 is defined as an additive mapping 𝐺∶𝑅→𝑅 such that 𝐺(𝑥𝑦)=𝐺(𝑥)𝑦+𝑥𝑑(𝑦) holds for any 𝑥,𝑦∈𝑅, where 𝑑 is a derivation of 𝑅. So, every derivation is a generalized derivation, but the converse is not true in general. If 𝑑=0, then we have 𝐺(𝑥𝑦)=𝐺(𝑥)𝑦 for all 𝑥,𝑦∈𝑅, which is called a left multiplier mapping of 𝑅. Thus, generalized derivation generalizes both the concepts, derivation as well as left multiplier mapping of 𝑅.

In [1], Daif and Bell proved that if 𝑅 is a semiprime ring with a nonzero ideal 𝐼 and 𝑑 is a derivation of 𝑅 such that 𝑑([𝑥,𝑦])=±[𝑥,𝑦] for all 𝑥,𝑦∈𝐼, then 𝐼 is central ideal. In particular, if 𝐼=𝑅, then 𝑅 is commutative. Recently, Quadri et al. [2] have generalized this result replacing derivation 𝑑 with a generalized derivation in a prime ring 𝑅. More precisely, they obtained the following result.

Let 𝑅 be a prime ring and 𝐼 a nonzero ideal of 𝑅. If 𝑅 admits a generalized derivation 𝐹 associated with a nonzero derivation 𝑑 such that any one of the following holds: (i) 𝐹([𝑥,𝑦])=[𝑥,𝑦] for all 𝑥,𝑦∈𝐼; (ii) 𝐹([𝑥,𝑦])=−[𝑥,𝑦] for all 𝑥,𝑦∈𝐼; (iii) 𝐹(𝑥∘𝑦)=(𝑥∘𝑦) for all 𝑥,𝑦∈𝐼; (iv) 𝐹(𝑥∘𝑦)=−(𝑥∘𝑦) for all 𝑥,𝑦∈𝐼; then 𝑅 is commutative.

Recently in [3], the first author of this paper has studied all the results of [2] in semiprime ring. In the present paper, our aim is to discuss similar identities in a left sided ideal of a semiprime rings.

2. Main Results

Theorem 2.1. Let 𝑅 be a semiprime ring and 𝜆 a nonzero left ideal of 𝑅. If G is a generalized derivation of 𝑅 associated with a derivation 𝑑 of 𝑅 such that 𝐺(𝑥∘𝑦)=𝑎(𝑥±𝑦) for all 𝑥,𝑦∈𝜆, where 𝑎∈{0,1,−1}, then [𝜆,𝜆]𝑑(𝜆)=0.

Proof. If 𝐺(𝜆)=0, then for any 𝑥,𝑦∈𝜆, 0=𝐺(𝑥𝑦)=𝐺(𝑥)𝑦+𝑥𝑑(𝑦)=𝑥𝑑(𝑦),(2.1) that is, 𝜆𝑑(𝜆)=0. This gives our conclusion. So let 𝐺(𝜆)≠0. Then by our assumption, we have 𝐺(𝑥∘𝑦)=𝑎(𝑥±𝑦)(2.2) for all 𝑥,𝑦∈𝜆. Putting 𝑦=𝑦𝑥, 𝑥∈𝜆, we obtain that 𝐺((𝑥∘𝑦)𝑥)=𝑎((𝑥±𝑦)𝑥). Since 𝐺 is a generalized derivation of 𝑅, this implies that 𝐺(𝑥∘𝑦)𝑥+(𝑥∘𝑦)𝑑(𝑥)=𝑎(𝑥±𝑦)𝑥. This gives by using (2.2) that (𝑥∘𝑦)𝑑(𝑥)=0 for all 𝑥,𝑦,∈𝜆. Now, we replace 𝑦 with 𝑧𝑦, where 𝑧∈𝜆, and then we get 0=𝑧(𝑥∘𝑦)𝑑(𝑥)+[𝑥,𝑧]𝑦𝑑(𝑥)=[𝑥,𝑧]𝑦𝑑(𝑥)(2.3) for all 𝑥,𝑦,𝑧∈𝜆. Since 𝜆 is a left ideal, it follows that [𝑥,𝑧]𝑅𝑦𝑑(𝑥)=0 for all 𝑥,𝑦,𝑧∈𝜆. Since 𝑅 is semiprime, it must contain a family Ω={𝑃_𝛼∶𝛼∈Λ} of prime ideals such that ⋂_𝛼∈Λ𝑃_𝛼={0}. If 𝑃 is typical member of Ω and 𝑥∈𝜆, we have either [𝑥,𝜆]⊆𝑃 or 𝜆𝑑(𝑥)⊆𝑃.
For fixed 𝑃, the sets 𝑇₁={𝑥∈𝜆∶[𝑥,𝜆]⊆𝑃} and 𝑇₂={𝑥∈𝜆∶𝜆𝑑(𝑥)⊆𝑃} form two additive subgroups of 𝜆 such that 𝑇₁⋃𝑇₂=𝜆. Therefore, either 𝑇₁=𝜆 or 𝑇₂=𝜆, that is, either [𝜆,𝜆]⊆𝑃 or 𝜆𝑑(𝜆)⊆𝑃. Both of these two conditions together imply that [𝜆,𝜆]𝑑(𝜆)⊆𝑃 for any 𝑃∈Ω. Therefore, [𝜆,𝜆]𝑑(𝜆)⊆⋂_𝛼∈Λ𝑃_𝛼=0.

Corollary 2.2. Let 𝑅 be a prime ring and 𝜆 be a nonzero left ideal of 𝑅. If 𝑅 admits a generalized derivation 𝐺 associated with a derivation 𝑑 such that 𝐺(𝑥∘𝑦)=𝑎(𝑥∘𝑦) for all 𝑥,𝑦∈𝜆, where 𝑎∈{0,1,−1}, then one of the following holds: (i)𝜆𝑑(𝜆)=0;(ii)𝑅 is commutative ring with char(𝑅)=2;(iii)𝑅 is commutative ring with char(𝑅)≠2 and 𝐺(𝑥)=𝑎𝑥 for all 𝑥∈𝜆.

Proof. By Theorem 2.1, we have [𝜆,𝜆]𝑑(𝜆)=0. This gives 0=𝜆,𝜆²𝑑(𝜆)=[𝜆,𝜆]𝜆𝑑(𝜆)=[𝜆,𝜆]𝑅𝜆𝑑(𝜆).(2.4) Since 𝑅 is prime, either [𝜆,𝜆]=0 or 𝜆𝑑(𝜆)=0. Now 𝜆𝑑(𝜆)=0 gives our conclusion (i). So, let [𝜆,𝜆]=0 which gives 0=[𝜆,𝑅𝜆]=[𝜆,𝑅]𝜆 implying 0=[𝜆,𝑅]. Again, this gives 0=[𝑅𝜆,𝑅]=[𝑅,𝑅]𝜆. Since left annihilator of a left-sided ideal is zero, [𝑅,𝑅]=0, that is, 𝑅 is commutative. If char(𝑅)=2, we obtain conclusion (ii). So assume that char(𝑅)≠2. Then our assumption 𝐺(𝑥∘𝑦)=𝑎(𝑥∘𝑦) gives 2𝐺(𝑥𝑦)=2𝑎(𝑥𝑦) for all 𝑥,𝑦∈𝜆. Since char(𝑅)≠2, then 𝐺(𝑥𝑦)=𝑎(𝑥𝑦) for all 𝑥,𝑦∈𝜆. This gives for all 𝑥,𝑦∈𝜆, 0=𝐺(𝑥𝑦)−𝑎(𝑥𝑦)=𝐺(𝑥)𝑦+𝑥𝑑(𝑦)−𝑎𝑥𝑦=(𝐺(𝑥)−𝑎𝑥)𝑦+𝑥𝑑(𝑦).(2.5) Let 𝑟∈𝑅. Since 𝑅 is commutative, 𝑥𝑟∈𝜆. Put 𝑥=𝑥𝑟 in last result, we get 0=(𝐺(𝑥)−𝑎𝑥)𝑟𝑦+𝑥𝑟𝑑(𝑦)+𝑥𝑑(𝑟)𝑦={(𝐺(𝑥)−𝑎𝑥)𝑦+𝑥𝑑(𝑦)}𝑟+𝑑(𝑟)𝑦𝑥. Since 𝑅 is commutative, using (2.5), it yields 𝑑(𝑅)𝜆²={0}, implying 𝑑=0. Then, (2.5) implies (𝐺(𝑥)−𝑎𝑥)𝜆=0, which yields 𝐺(𝑥)−𝑎𝑥=0 for all 𝑥∈𝜆.

Theorem 2.3. Let 𝑅 be a semiprime ring, 𝜆 a nonzero left ideal of 𝑅 and 𝐺 a generalized derivation of 𝑅 associated with a derivation 𝑑 of 𝑅. If 𝐺[𝑥,𝑦]=𝑎(𝑥𝑦±𝑦𝑥) for all 𝑥,𝑦∈𝜆, then [𝜆,𝜆]𝑑(𝜆)=0.

Proof. If 𝐺(𝜆)=0, then 0=𝐺(𝜆²)=𝐺(𝜆)𝜆+𝜆𝑑(𝜆)=𝜆𝑑(𝜆) and hence, [𝜆,𝜆]𝑑(𝜆)=0, which is our conclusion. Assume next that 𝐺(𝜆)≠0. Then by our assumption, we have 𝐺[𝑥,𝑦]=𝑎(𝑥𝑦±𝑦𝑥)(2.6) for all 𝑥,𝑦∈𝜆. Put 𝑦=𝑦𝑥 and get 𝐺([𝑥,𝑦]𝑥)=𝑎(𝑥𝑦±𝑦𝑥)𝑥, that is, 𝐺([𝑥,𝑦])𝑥+[𝑥,𝑦]𝑑(𝑥)=𝑎(𝑥𝑦±𝑦𝑥)𝑥. Now using (2.6), the above relation yields [𝑥,𝑦]𝑑(𝑥)=0 for all 𝑥,𝑦∈𝜆. Putting 𝑦=𝑧𝑦, where 𝑧∈𝜆, we obtain that [𝑥,𝑧]𝑦𝑑(𝑥)=0, which is same as (2.3) in Theorem 2.1. By same argument of Theorem 2.1, we can conclude the result here.

Corollary 2.4. Let 𝑅 be a prime ring and 𝜆 a nonzero left ideal of 𝑅. If 𝐺 is a generalized derivation of 𝑅 associated with a derivation 𝑑 of 𝑅 such that 𝐺[𝑥,𝑦]=𝑎(𝑥𝑦±𝑦𝑥) for all 𝑥,𝑦∈𝜆, where 𝑎={0,1,−1}, then either 𝑅 is commutative or 𝜆𝑑(𝜆)=0 and one of the following holds: (i)𝜆[𝜆,𝜆]=0;(ii)𝐺(𝑥)=∓𝑎𝑥 for all 𝑥∈𝜆. In case 𝐺(𝑥)=−𝑎𝑥 for all 𝑥∈𝜆, with 𝑎≠0, then char(𝑅)=2.

Proof. By the Theorem 2.3, we may conclude that [𝜆,𝜆]𝑑(𝜆)=0. Then by same argument as given in Corollary 2.2, we obtain that either 𝑅 is commutative or 𝜆𝑑(𝜆)=0. Let 𝑅 be noncommutative, then for any 𝑥,𝑦∈𝜆, we have 𝐺(𝑥𝑦)=𝐺(𝑥)𝑦+𝑥𝑑(𝑦)=𝐺(𝑥)𝑦, that is, 𝐺 acts as a left multiplier map on 𝜆. Then for any 𝑥,𝑦,𝑧∈𝜆, replacing 𝑦 with 𝑦𝑧 in our hypothesis 𝐺[𝑥,𝑦]=𝑎(𝑥𝑦±𝑦𝑥), we have 𝐺([𝑥,𝑦]𝑧+𝑦[𝑥,𝑧])=𝑎{(𝑥𝑦±𝑦𝑥)𝑧∓𝑦[𝑥,𝑧]}(2.7) for all 𝑥,𝑦∈𝜆. Since 𝐺 acts as a left multiplier map on 𝜆, this implies 𝐺([𝑥,𝑦])𝑧+𝐺(𝑦)[𝑥,𝑧]=𝑎(𝑥𝑦±𝑦𝑥)𝑧∓𝑎𝑦[𝑥,𝑧].(2.8) By using 𝐺[𝑥,𝑦]=𝑎(𝑥𝑦±𝑦𝑥), it gives 𝐺(𝑦)[𝑥,𝑧]=∓𝑎𝑦[𝑥,𝑧], that is, (𝐺(𝑦)±𝑎𝑦)[𝑥,𝑧]=0 for all 𝑥,𝑦∈𝜆. Replacing 𝑦 with 𝑦𝑢, where 𝑢∈𝜆, we find that (𝐺(𝑦)±𝑎𝑦)𝑢[𝑥,𝑧]=0, which gives (𝐺(𝑦)±𝑎𝑦)𝑅𝜆[𝑥,𝑧]=0. Since 𝑅 is prime, either 𝜆[𝜆,𝜆]=0 or 𝐺(𝑦)=∓𝑎𝑦 for all 𝑦∈𝜆. When 𝐺(𝑦)=−𝑎𝑦, our assumption 𝐺[𝑥,𝑦]=𝑎(𝑥𝑦+𝑦𝑥) implies −𝑎[𝑥,𝑦]=𝑎(𝑥𝑦+𝑦𝑥) for all 𝑥,𝑦∈𝜆. This implies 2𝑎𝑥𝑦=0, that is, 2𝑎𝑅𝜆²=0 implies char(𝑅)=2, unless 𝑎=0.

Theorem 2.5. Let 𝑅 be a semiprime ring, 𝜆 a nonzero left ideal of 𝑅 and 𝑑 a derivation of 𝑅. If 𝑑(𝑥)∘𝑑(𝑦)=𝑎(𝑥𝑦±𝑦𝑥) for all 𝑥,𝑦∈𝜆, where 𝑎∈{0,1,−1}, then 𝜆[𝑥,𝑑(𝑥)]₂=0. In case 𝜆=𝑅 and 𝑅 is 2-torsion free, 𝑑 maps 𝑅 into its center.

Proof. We have for all 𝑥,𝑦∈𝜆, 𝑑(𝑥)𝑑(𝑦)+𝑑(𝑦)𝑑(𝑥)=𝑎(𝑥𝑦±𝑦𝑥).(2.9) Putting 𝑦=𝑦𝑥, we get 𝑑(𝑥)(𝑑(𝑦)𝑥+𝑦𝑑(𝑥))+(𝑑(𝑦)𝑥+𝑦𝑑(𝑥))𝑑(𝑥)=𝑎(𝑥𝑦±𝑦𝑥)𝑥.(2.10) Using (2.9), we have 𝑑(𝑥)𝑦𝑑(𝑥)+𝑑(𝑦)[𝑥,𝑑(𝑥)]+𝑦𝑑(𝑥)²=0.(2.11) Put 𝑦=𝑥𝑦 in (2.11), and get 𝑑(𝑥)𝑥𝑦𝑑(𝑥)+{𝑥𝑑(𝑦)+𝑑(𝑥)𝑦}[𝑥,𝑑(𝑥)]+𝑥𝑦𝑑(𝑥)²=0.(2.12) Left multiplying (2.11) by 𝑥 and then subtracting from (2.12) yields [𝑑(𝑥),𝑥]𝑦𝑑(𝑥)+𝑑(𝑥)𝑦[𝑥,𝑑(𝑥)]=0(2.13) for all 𝑥,𝑦∈𝜆. Put 𝑦=𝑑(𝑥)𝑦 in (2.13), we have [𝑑(𝑥),𝑥]𝑑(𝑥)𝑦𝑑(𝑥)+𝑑(𝑥)²𝑦[𝑥,𝑑(𝑥)]=0.(2.14) Left multiplying (2.13) by 𝑑(𝑥) and then subtracting from (2.14), we get [𝑥,𝑑(𝑥)]₂𝑦𝑑(𝑥)=0 for all 𝑥,𝑦∈𝜆. This implies [𝑥,𝑑(𝑥)]₂𝑦[𝑥,𝑑(𝑥)]₂=0 and hence, 𝑦[𝑥,𝑑(𝑥)]₂𝑅𝑦[𝑥,𝑑(𝑥)]₂=0 for all 𝑥,𝑦∈𝜆. Since 𝑅 is semiprime, 𝜆[𝑥,𝑑(𝑥)]₂=0. In case 𝜆=𝑅, [𝑥,𝑑(𝑥)]₂=0 for all 𝑥∈𝑅, and then by [4], 𝑑(𝑅)⊆𝑍(𝑅).