# Generalized Derivations and Left Ideals in Prime and Semiprime Rings

Authors: Basudeb Dhara; Atanu Pattanayak
Journal: ISRN Algebra
 (2011)

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## Abstract

LetR
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}.


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## Body

## 1. Introduction

Throughout this paper, letR 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.LetRbe a prime ring andIa nonzero ideal ofR. IfRadmits a generalized derivationFassociated with a nonzero derivationdsuch that any one of the following holds: (i)F([x,y])=[x,y]for allx,y∈I; (ii)F([x,y])=-[x,y]for allx,y∈I; (iii)F(x∘y)=(x∘y)for allx,y∈I; (iv)F(x∘y)=-(x∘y)for allx,y∈I; thenRis 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.
LetR 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.
IfG(λ)=0, then for any x,y∈λ,
(2.1)0=G(xy)=G(x)y+xd(y)=xd(y),
that is, λd(λ)=0. This gives our conclusion. So let G(λ)≠0. Then by our assumption, we have
(2.2)G(x∘y)=a(x±y)
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
(2.3)0=z(x∘y)d(x)+[x,z]yd(x)=[x,z]yd(x)
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={Pα:α∈Λ} of prime ideals such that ⋂α∈ΛPα={0}. If P is typical member of Ω and x∈λ, we have either [x,λ]⊆P or λd(x)⊆P.
For fixedP, 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∈ΛPα=0.Corollary 2.2.
LetR 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 Theorem2.1, we have [λ,λ]d(λ)=0. This gives
(2.4)0=[λ,λ2]d(λ)=[λ,λ]λd(λ)=[λ,λ]Rλd(λ).
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,
(2.5)0=G(xy)-a(xy)=G(x)y+xd(y)-axy=(G(x)-ax)y+xd(y).
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.
LetR 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.
IfG(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
(2.6)G[x,y]=a(xy±yx)
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.
LetR 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 Theorem2.3, 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
(2.7)G([x,y]z+y[x,z])=a{(xy±yx)z∓y[x,z]}
for all x,y∈λ. Since G acts as a left multiplier map on λ, this implies
(2.8)G([x,y])z+G(y)[x,z]=a(xy±yx)z∓ay[x,z].
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.
LetR 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 allx,y∈λ,
(2.9)d(x)d(y)+d(y)d(x)=a(xy±yx).
Putting y=yx, we get
(2.10)d(x)(d(y)x+yd(x))+(d(y)x+yd(x))d(x)=a(xy±yx)x.
Using (2.9), we have
(2.11)d(x)yd(x)+d(y)[x,d(x)]+yd(x)2=0.
Put y=xy in (2.11), and get
(2.12)d(x)xyd(x)+{xd(y)+d(x)y}[x,d(x)]+xyd(x)2=0.
Left multiplying (2.11) by x and then subtracting from (2.12) yields
(2.13)[d(x),x]yd(x)+d(x)y[x,d(x)]=0
for all x,y∈λ. Put y=d(x)y in (2.13), we have
(2.14)[d(x),x]d(x)yd(x)+d(x)2y[x,d(x)]=0.
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).



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*Source: 750382-2011-08-15.xml*