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International Scholarly Research Notices / 2011 / Article

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Volume 2011 |Article ID 750382 | 5 pages | https://doi.org/10.5402/2011/750382

Generalized Derivations and Left Ideals in Prime and Semiprime Rings

Basudeb Dhara 1 and Atanu Pattanayak1

1Department of Mathematics, Belda College, Paschim Medinipur, Belda 721424, India

Academic Editor: A. Rapinchuk
Received19 May 2011
Accepted07 Jul 2011
Published15 Aug 2011

Abstract

Let 𝑅 be an associative ring, πœ† a nonzero left ideal of 𝑅, π‘‘βˆΆπ‘…β†’π‘… a derivation and πΊβˆΆπ‘…β†’π‘… a generalized derivation. In this paper, we study the following situations in prime and semiprime rings: (1) 𝐺(π‘₯βˆ˜π‘¦)=π‘Ž(xyΒ±yx); (2) 𝐺[π‘₯,𝑦]=π‘Ž(xyΒ±yx); (3) 𝑑(π‘₯)βˆ˜π‘‘(𝑦)=π‘Ž(xyΒ±yx); for all π‘₯,π‘¦βˆˆπœ† and π‘Žβˆˆ{0,1,βˆ’1}.

1. Introduction

Throughout this paper, let 𝑅 be an associative ring, πœ† a left ideal of 𝑅, 𝑑 a derivation of 𝑅 and 𝐺 a generalized derivation of 𝑅. For any two elements π‘₯,π‘¦βˆˆπ‘…, [π‘₯,𝑦] will denote the commutator element π‘₯π‘¦βˆ’π‘¦π‘₯ and π‘₯βˆ˜π‘¦ denotes π‘₯𝑦+𝑦π‘₯. We use extensively the following basic commutator identities: [π‘₯𝑦,𝑧]=[π‘₯,𝑧]𝑦+π‘₯[𝑦,𝑧] and [π‘₯,𝑦𝑧]=[π‘₯,𝑦]𝑧+𝑦[π‘₯,𝑧]. Recall that a ring 𝑅 is called prime, if for any π‘Ž,π‘βˆˆπ‘…, π‘Žπ‘…π‘=(0) implies that either π‘Ž=0 or 𝑏=0 and is called semiprime if for any π‘Žβˆˆπ‘…, π‘Žπ‘…π‘Ž=(0) implies π‘Ž=0. An additive mapping π‘‘βˆΆπ‘…β†’π‘… is said to be a derivation of 𝑅 if for any π‘₯,π‘¦βˆˆπ‘…, 𝑑(π‘₯𝑦)=𝑑(π‘₯)𝑦+π‘₯𝑑(𝑦) holds. The generalized derivation of 𝑅 is defined as an additive mapping πΊβˆΆπ‘…β†’π‘… such that 𝐺(π‘₯𝑦)=𝐺(π‘₯)𝑦+π‘₯𝑑(𝑦) holds for any π‘₯,π‘¦βˆˆπ‘…, 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(π‘₯𝑦)=𝐺(π‘₯)𝑦 for all π‘₯,π‘¦βˆˆπ‘…, 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([π‘₯,𝑦])=Β±[π‘₯,𝑦] for all π‘₯,π‘¦βˆˆπΌ, 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([π‘₯,𝑦])=[π‘₯,𝑦] for all π‘₯,π‘¦βˆˆπΌ; (ii) 𝐹([π‘₯,𝑦])=βˆ’[π‘₯,𝑦] for all π‘₯,π‘¦βˆˆπΌ; (iii) 𝐹(π‘₯βˆ˜π‘¦)=(π‘₯βˆ˜π‘¦) for all π‘₯,π‘¦βˆˆπΌ; (iv) 𝐹(π‘₯βˆ˜π‘¦)=βˆ’(π‘₯βˆ˜π‘¦) for all π‘₯,π‘¦βˆˆπΌ; 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 πœ† a nonzero left ideal of 𝑅. If G is a generalized derivation of 𝑅 associated with a derivation 𝑑 of 𝑅 such that 𝐺(π‘₯βˆ˜π‘¦)=π‘Ž(π‘₯±𝑦) for all π‘₯,π‘¦βˆˆπœ†, where π‘Žβˆˆ{0,1,βˆ’1}, then [πœ†,πœ†]𝑑(πœ†)=0.

Proof. If 𝐺(πœ†)=0, then for any π‘₯,π‘¦βˆˆπœ†, 0=𝐺(π‘₯𝑦)=𝐺(π‘₯)𝑦+π‘₯𝑑(𝑦)=π‘₯𝑑(𝑦),(2.1) that is, πœ†π‘‘(πœ†)=0. This gives our conclusion. So let 𝐺(πœ†)β‰ 0. Then by our assumption, we have 𝐺(π‘₯βˆ˜π‘¦)=π‘Ž(π‘₯±𝑦)(2.2) for all π‘₯,π‘¦βˆˆπœ†. Putting 𝑦=𝑦π‘₯, π‘₯βˆˆπœ†, we obtain that 𝐺((π‘₯βˆ˜π‘¦)π‘₯)=π‘Ž((π‘₯±𝑦)π‘₯). Since 𝐺 is a generalized derivation of 𝑅, this implies that 𝐺(π‘₯βˆ˜π‘¦)π‘₯+(π‘₯βˆ˜π‘¦)𝑑(π‘₯)=π‘Ž(π‘₯±𝑦)π‘₯. This gives by using (2.2) that (π‘₯βˆ˜π‘¦)𝑑(π‘₯)=0 for all π‘₯,𝑦,βˆˆπœ†. Now, we replace 𝑦 with 𝑧𝑦, where π‘§βˆˆπœ†, and then we get [][]0=𝑧(π‘₯βˆ˜π‘¦)𝑑(π‘₯)+π‘₯,𝑧𝑦𝑑(π‘₯)=π‘₯,𝑧𝑦𝑑(π‘₯)(2.3) for all π‘₯,𝑦,π‘§βˆˆπœ†. Since πœ† is a left ideal, it follows that [π‘₯,𝑧]𝑅𝑦𝑑(π‘₯)=0 for all π‘₯,𝑦,π‘§βˆˆπœ†. Since 𝑅 is semiprime, it must contain a family Ξ©={π‘ƒπ›ΌβˆΆπ›ΌβˆˆΞ›} of prime ideals such that β‹‚π›ΌβˆˆΞ›π‘ƒπ›Ό={0}. If 𝑃 is typical member of Ξ© and π‘₯βˆˆπœ†, we have either [π‘₯,πœ†]βŠ†π‘ƒ or πœ†π‘‘(π‘₯)βŠ†π‘ƒ.
For fixed 𝑃, the sets 𝑇1={π‘₯βˆˆπœ†βˆΆ[π‘₯,πœ†]βŠ†π‘ƒ} and 𝑇2={π‘₯βˆˆπœ†βˆΆπœ†π‘‘(π‘₯)βŠ†π‘ƒ} form two additive subgroups of πœ† such that 𝑇1⋃𝑇2=πœ†. Therefore, either 𝑇1=πœ† or 𝑇2=πœ†, that is, either [πœ†,πœ†]βŠ†π‘ƒ or πœ†π‘‘(πœ†)βŠ†π‘ƒ. Both of these two conditions together imply that [πœ†,πœ†]𝑑(πœ†)βŠ†π‘ƒ for any π‘ƒβˆˆΞ©. Therefore, β‹‚[πœ†,πœ†]𝑑(πœ†)βŠ†π›ΌβˆˆΞ›π‘ƒπ›Ό=0.

Corollary 2.2. Let 𝑅 be a prime ring and πœ† be a nonzero left ideal of 𝑅. If 𝑅 admits a generalized derivation 𝐺 associated with a derivation 𝑑 such that 𝐺(π‘₯βˆ˜π‘¦)=π‘Ž(π‘₯βˆ˜π‘¦) for all π‘₯,π‘¦βˆˆπœ†, where π‘Žβˆˆ{0,1,βˆ’1}, then one of the following holds: (i)πœ†π‘‘(πœ†)=0;(ii)𝑅 is commutative ring with char(𝑅)=2;(iii)𝑅 is commutative ring with char(𝑅)β‰ 2 and 𝐺(π‘₯)=π‘Žπ‘₯ for all π‘₯βˆˆπœ†.

Proof. By Theorem 2.1, we have [πœ†,πœ†]𝑑(πœ†)=0. This gives ξ€Ί0=πœ†,πœ†2𝑑[][](πœ†)=πœ†,πœ†πœ†π‘‘(πœ†)=πœ†,πœ†π‘…πœ†π‘‘(πœ†).(2.4) Since 𝑅 is prime, either [πœ†,πœ†]=0 or πœ†π‘‘(πœ†)=0. Now πœ†π‘‘(πœ†)=0 gives our conclusion (i). So, let [πœ†,πœ†]=0 which gives 0=[πœ†,π‘…πœ†]=[πœ†,𝑅]πœ† implying 0=[πœ†,𝑅]. Again, this gives 0=[π‘…πœ†,𝑅]=[𝑅,𝑅]πœ†. 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)β‰ 2. Then our assumption 𝐺(π‘₯βˆ˜π‘¦)=π‘Ž(π‘₯βˆ˜π‘¦) gives 2𝐺(π‘₯𝑦)=2π‘Ž(π‘₯𝑦) for all π‘₯,π‘¦βˆˆπœ†. Since char(𝑅)β‰ 2, then 𝐺(π‘₯𝑦)=π‘Ž(π‘₯𝑦) for all π‘₯,π‘¦βˆˆπœ†. This gives for all π‘₯,π‘¦βˆˆπœ†, 0=𝐺(π‘₯𝑦)βˆ’π‘Ž(π‘₯𝑦)=𝐺(π‘₯)𝑦+π‘₯𝑑(𝑦)βˆ’π‘Žπ‘₯𝑦=(𝐺(π‘₯)βˆ’π‘Žπ‘₯)𝑦+π‘₯𝑑(𝑦).(2.5) Let π‘Ÿβˆˆπ‘…. Since 𝑅 is commutative, π‘₯π‘Ÿβˆˆπœ†. Put π‘₯=π‘₯π‘Ÿ in last result, we get 0=(𝐺(π‘₯)βˆ’π‘Žπ‘₯)π‘Ÿπ‘¦+π‘₯π‘Ÿπ‘‘(𝑦)+π‘₯𝑑(π‘Ÿ)𝑦={(𝐺(π‘₯)βˆ’π‘Žπ‘₯)𝑦+π‘₯𝑑(𝑦)}π‘Ÿ+𝑑(π‘Ÿ)𝑦π‘₯. Since 𝑅 is commutative, using (2.5), it yields 𝑑(𝑅)πœ†2={0}, implying 𝑑=0. Then, (2.5) implies (𝐺(π‘₯)βˆ’π‘Žπ‘₯)πœ†=0, which yields 𝐺(π‘₯)βˆ’π‘Žπ‘₯=0 for all π‘₯βˆˆπœ†.

Theorem 2.3. Let 𝑅 be a semiprime ring, πœ† a nonzero left ideal of 𝑅 and 𝐺 a generalized derivation of 𝑅 associated with a derivation 𝑑 of 𝑅. If 𝐺[π‘₯,𝑦]=π‘Ž(π‘₯𝑦±𝑦π‘₯) for all π‘₯,π‘¦βˆˆπœ†, then [πœ†,πœ†]𝑑(πœ†)=0.

Proof. If 𝐺(πœ†)=0, then 0=𝐺(πœ†2)=𝐺(πœ†)πœ†+πœ†π‘‘(πœ†)=πœ†π‘‘(πœ†) and hence, [πœ†,πœ†]𝑑(πœ†)=0, which is our conclusion. Assume next that 𝐺(πœ†)β‰ 0. Then by our assumption, we have 𝐺[]π‘₯,𝑦=π‘Ž(π‘₯𝑦±𝑦π‘₯)(2.6) for all π‘₯,π‘¦βˆˆπœ†. Put 𝑦=𝑦π‘₯ and get 𝐺([π‘₯,𝑦]π‘₯)=π‘Ž(π‘₯𝑦±𝑦π‘₯)π‘₯, that is, 𝐺([π‘₯,𝑦])π‘₯+[π‘₯,𝑦]𝑑(π‘₯)=π‘Ž(π‘₯𝑦±𝑦π‘₯)π‘₯. Now using (2.6), the above relation yields [π‘₯,𝑦]𝑑(π‘₯)=0 for all π‘₯,π‘¦βˆˆπœ†. Putting 𝑦=𝑧𝑦, where π‘§βˆˆπœ†, we obtain that [π‘₯,𝑧]𝑦𝑑(π‘₯)=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 πœ† a nonzero left ideal of 𝑅. If 𝐺 is a generalized derivation of 𝑅 associated with a derivation 𝑑 of 𝑅 such that 𝐺[π‘₯,𝑦]=π‘Ž(π‘₯𝑦±𝑦π‘₯) for all π‘₯,π‘¦βˆˆπœ†, where π‘Ž={0,1,βˆ’1}, then either 𝑅 is commutative or πœ†π‘‘(πœ†)=0 and one of the following holds: (i)πœ†[πœ†,πœ†]=0;(ii)𝐺(π‘₯)=βˆ“π‘Žπ‘₯ for all π‘₯βˆˆπœ†. In case 𝐺(π‘₯)=βˆ’π‘Žπ‘₯ for all π‘₯βˆˆπœ†, with π‘Žβ‰ 0, then char(𝑅)=2.

Proof. By the Theorem 2.3, we may conclude that [πœ†,πœ†]𝑑(πœ†)=0. Then by same argument as given in Corollary 2.2, we obtain that either 𝑅 is commutative or πœ†π‘‘(πœ†)=0. Let 𝑅 be noncommutative, then for any π‘₯,π‘¦βˆˆπœ†, we have 𝐺(π‘₯𝑦)=𝐺(π‘₯)𝑦+π‘₯𝑑(𝑦)=𝐺(π‘₯)𝑦, that is, 𝐺 acts as a left multiplier map on πœ†. Then for any π‘₯,𝑦,π‘§βˆˆπœ†, replacing 𝑦 with 𝑦𝑧 in our hypothesis 𝐺[π‘₯,𝑦]=π‘Ž(π‘₯𝑦±𝑦π‘₯), we have [][][]}𝐺(π‘₯,𝑦𝑧+𝑦π‘₯,𝑧)=π‘Ž{(π‘₯𝑦±𝑦π‘₯)π‘§βˆ“π‘¦π‘₯,𝑧(2.7) for all π‘₯,π‘¦βˆˆπœ†. Since 𝐺 acts as a left multiplier map on πœ†, this implies [][][].𝐺(π‘₯,𝑦)𝑧+𝐺(𝑦)π‘₯,𝑧=π‘Ž(π‘₯𝑦±𝑦π‘₯)π‘§βˆ“π‘Žπ‘¦π‘₯,𝑧(2.8) By using 𝐺[π‘₯,𝑦]=π‘Ž(π‘₯𝑦±𝑦π‘₯), it gives 𝐺(𝑦)[π‘₯,𝑧]=βˆ“π‘Žπ‘¦[π‘₯,𝑧], that is, (𝐺(𝑦)Β±π‘Žπ‘¦)[π‘₯,𝑧]=0 for all π‘₯,π‘¦βˆˆπœ†. Replacing 𝑦 with 𝑦𝑒, where π‘’βˆˆπœ†, we find that (𝐺(𝑦)Β±π‘Žπ‘¦)𝑒[π‘₯,𝑧]=0, which gives (𝐺(𝑦)Β±π‘Žπ‘¦)π‘…πœ†[π‘₯,𝑧]=0. Since 𝑅 is prime, either πœ†[πœ†,πœ†]=0 or 𝐺(𝑦)=βˆ“π‘Žπ‘¦ for all π‘¦βˆˆπœ†. When 𝐺(𝑦)=βˆ’π‘Žπ‘¦, our assumption 𝐺[π‘₯,𝑦]=π‘Ž(π‘₯𝑦+𝑦π‘₯) implies βˆ’π‘Ž[π‘₯,𝑦]=π‘Ž(π‘₯𝑦+𝑦π‘₯) for all π‘₯,π‘¦βˆˆπœ†. This implies 2π‘Žπ‘₯𝑦=0, that is, 2π‘Žπ‘…πœ†2=0 implies char(𝑅)=2, unless π‘Ž=0.

Theorem 2.5. Let 𝑅 be a semiprime ring, πœ† a nonzero left ideal of 𝑅 and 𝑑 a derivation of 𝑅. If 𝑑(π‘₯)βˆ˜π‘‘(𝑦)=π‘Ž(π‘₯𝑦±𝑦π‘₯) for all π‘₯,π‘¦βˆˆπœ†, where π‘Žβˆˆ{0,1,βˆ’1}, then πœ†[π‘₯,𝑑(π‘₯)]2=0. In case πœ†=𝑅 and 𝑅 is 2-torsion free, 𝑑 maps 𝑅 into its center.

Proof. We have for all π‘₯,π‘¦βˆˆπœ†, 𝑑(π‘₯)𝑑(𝑦)+𝑑(𝑦)𝑑(π‘₯)=π‘Ž(π‘₯𝑦±𝑦π‘₯).(2.9) Putting 𝑦=𝑦π‘₯, we get 𝑑(π‘₯)(𝑑(𝑦)π‘₯+𝑦𝑑(π‘₯))+(𝑑(𝑦)π‘₯+𝑦𝑑(π‘₯))𝑑(π‘₯)=π‘Ž(π‘₯𝑦±𝑦π‘₯)π‘₯.(2.10) Using (2.9), we have []𝑑(π‘₯)𝑦𝑑(π‘₯)+𝑑(𝑦)π‘₯,𝑑(π‘₯)+𝑦𝑑(π‘₯)2=0.(2.11) Put 𝑦=π‘₯𝑦 in (2.11), and get []𝑑(π‘₯)π‘₯𝑦𝑑(π‘₯)+{π‘₯𝑑(𝑦)+𝑑(π‘₯)𝑦}π‘₯,𝑑(π‘₯)+π‘₯𝑦𝑑(π‘₯)2=0.(2.12) Left multiplying (2.11) by π‘₯ and then subtracting from (2.12) yields [][]𝑑(π‘₯),π‘₯𝑦𝑑(π‘₯)+𝑑(π‘₯)𝑦π‘₯,𝑑(π‘₯)=0(2.13) for all π‘₯,π‘¦βˆˆπœ†. Put 𝑦=𝑑(π‘₯)𝑦 in (2.13), we have []𝑑(π‘₯),π‘₯𝑑(π‘₯)𝑦𝑑(π‘₯)+𝑑(π‘₯)2𝑦[]π‘₯,𝑑(π‘₯)=0.(2.14) Left multiplying (2.13) by 𝑑(π‘₯) and then subtracting from (2.14), we get [π‘₯,𝑑(π‘₯)]2𝑦𝑑(π‘₯)=0 for all π‘₯,π‘¦βˆˆπœ†. This implies [π‘₯,𝑑(π‘₯)]2𝑦[π‘₯,𝑑(π‘₯)]2=0 and hence, 𝑦[π‘₯,𝑑(π‘₯)]2𝑅𝑦[π‘₯,𝑑(π‘₯)]2=0 for all π‘₯,π‘¦βˆˆπœ†. Since 𝑅 is semiprime, πœ†[π‘₯,𝑑(π‘₯)]2=0. In case πœ†=𝑅, [π‘₯,𝑑(π‘₯)]2=0 for all π‘₯βˆˆπ‘…, and then by [4], 𝑑(𝑅)βŠ†π‘(𝑅).

References

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Copyright © 2011 Basudeb Dhara and Atanu Pattanayak. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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