# Generalized Derivations on Prime Near Rings

Authors: Asma Ali; Howard E. Bell; Phool Miyan
Journal: International Journal of Mathematics and Mathematical Sciences
 (2013)

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## Abstract

LetN be a near ring. An additive mapping f:N→N is said to be a
right generalized (resp., left generalized) derivation with associated derivation d on N if f(xy)=f(x)y+xd(y) (resp., f(xy)=d(x)y+xf(y)) for all x,y∈N. A mapping
f:N→N is said to be a generalized derivation with associated derivation d on N if
f is both a right generalized and a left generalized derivation with associated derivation
d on N. The purpose of the present paper is to prove some theorems in the setting of
a semigroup ideal of a 3-prime near ring admitting a generalized derivation, thereby
extending some known results on derivations.


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## Body

## 1. Introduction

Throughout the paper,N denotes a zero-symmetric left near ring with multiplicative center Z, and for any pair of elements x,y∈N, [x,y] denotes the commutator xy-yx, while the symbol (x,y) denotes the additive commutator x+y-x-y. A near ring N is called zero-symmetric if 0x=0, for all x∈N (recall that left distributivity yields that x0=0). The near ring N is said to be 3-prime if xNy={0} for x,y∈N implies that x=0 or y=0. A near ring N is called 2-torsion-free if (N,+) has no element of order 2. A nonempty subset A of N is called a semigroup right (resp., semigroup left) ideal if AN⊆A (resp., NA⊆A), and if A is both a semigroup right ideal and a semigroup left ideal, it is called a semigroup ideal. An additive mapping d:N→N is a derivation on N if d(xy)=d(x)y+xd(y), for all x,y∈N. An additive mapping f:N→N is said to be a right (resp., left) generalized derivation with associated derivation d if f(xy)=f(x)y+xd(y) (resp., f(xy)=d(x)y+xf(y)), for all x,y∈N, and f is said to be a generalized derivation with associated derivation d on N if it is both a right generalized derivation and a left generalized derivation on N with associated derivation d. (Note that this definition differs from the one given by Hvala in [1]; his generalized derivations are our right generalized derivations.) Every derivation on N is a generalized derivation.In the case of rings, generalized derivations have received significant attention in recent years. We prove some theorems in the setting of a semigroup ideal of a 3-prime near ring admitting a generalized derivation and thereby extend some known results [2, Theorem 2.1], [3, Theorem 3.1], [4, Theorem 3], and [5, Theorem 3.3].


## 2. Preliminary Results

We begin with several lemmas, most of which have been proved elsewhere.Lemma 1 (see [3, Lemma 1.3]).
LetN be a 3-prime near ring and d be a nonzero derivation on N.(i)
IfU is a nonzero semigroup right ideal or a nonzero semigroup left ideal of N, then d(U)≠{0}.(ii)
IfU is a nonzero semigroup right ideal of N and x is an element of N which centralizes U, then x∈Z.Lemma 2 (see [3, Lemma 1.2]).
LetN be a 3-prime near ring.(i)
Ifz∈Z∖{0}, then z is not a zero divisor.(ii)
IfZ∖{0} contains an element z for which z+z∈Z, then (N,+) is abelian.(iii)
Ifz∈Z∖{0} and x is an element of N such that xz∈Z, then x∈Z.Lemma 3 (see [3, Lemmas 1.3 and 1.4]).
LetN be a 3-prime near ring and U be a nonzero semigroup ideal of N. Let d be a nonzero derivation on N.(i)
Ifx,y∈N and xUy={0}, then x=0 or y=0.(ii)
Ifx∈N and xU={0} or Ux={0}, then x=0.(iii)
Ifx∈N and d(U)x={0} or xd(U)={0}, then x=0.Lemma 4 (see [3, Lemma 1.5]).
IfN is a 3-prime near ring and Z contains a nonzero semigroup left ideal or a semigroup right ideal, then N is a commutative ring.Lemma 5.
Iff is a generalized derivation on N with associated derivation d, then (d(x)y+xf(y))z= d(x)yz+xf(y)z, for all x,y,z∈N.Proof.
We prove only (ii), since (i) is proved in [2]. For all x,y,z∈N we have f((xy)z)= f(xy)z+xyd(z)=(d(x)y+xf(y))z+xyd(z) and f(x(yz))=d(x)yz+xf(yz)= d(x)yz+xf(f(y)z+ yd(z))= d(x)yz+xf(y)z+xyd(z). Comparing the two expressions for f(xyz) gives (ii).Lemma 6.
LetN be a 3-prime near ring and f a generalized derivation with associated derivation d.(i)
f(x)y+xd(y)=xd(y)+f(x)y for all x,y∈N.(ii)
d(x)y+xf(y)=xf(y)+d(x)y for all x,y∈N.Proof.
(i)f(x(y+y))=f(x)(y+y)+xd(y+y)=f(x)y+f(x)y+xd(y)+xd(y), and f(xy+xy)= f(x)y+xd(y)+f(x)y+xd(y). Comparing these two equations gives the desired result.
(ii) Again, calculatef(x(y+y)) and f(xy+xy) and compare.Lemma 7.
LetN be a 3-prime near ring and f a generalized derivation with associated derivation d. Then f(Z)⊆Z.Proof.
Letz∈Z and x∈N. Then f(zx)=f(xz); that is, f(z)x+zd(x)=d(x)z+xf(z). Applying Lemma 6(i), we get zd(x)+f(z)x=d(x)z+xf(z). It follows that f(z)x=xf(z) for all x∈N, so f(z)∈Z.Lemma 8.
LetN be a 3-prime near ring and U a nonzero semigroup ideal of N. If f is a nonzero right generalized derivation of N with associated derivation d, then f(U)≠{0}.Proof.
Supposef(U)={0}. Then f(ux)=f(u)x+ud(x)=0=ud(x) for all u∈U and x∈N, and it follows by Lemma 3(ii) that d=0. Therefore f(xu)=f(x)u=0 for all u∈U and x∈N, and another appeal to Lemma 3(ii) gives f=0, which is a contradiction.Lemma 9 (see [2, Theorem 4.2]).
LetN be a 3-prime near ring with a nonzero right generalized derivation f with associated derivation d. If f(N)⊆Z then (N,+) is abelian. Moreover, if N is 2-torsion-free, then N is a commutative ring.Lemma 10 (see [2, Theorem 4.1]).
LetN be a 2-torsion-free 3-prime near ring and f a nonzero generalized derivation on N with associated derivation d. If [f(N),f(N)]={0}, then N is a commutative ring.


## 3. Main Results

The theorems that we prove in this section extend the results proved in [2, Theorems 2.1 and 3.1], [3, Theorems 2.1, 3.1, and 3.3], and [5, Theorem 3.3].Theorem 11.
LetN be a 3-prime near ring and U be a nonzero semigroup ideal of N. Let f be a nonzero right generalized derivation with associated derivation d. If f(U)⊆Z, then (N,+) is abelian. Moreover, if N is 2-torsion-free, then N is a commutative ring.Proof.
We begin by showing that(N,+) is abelian, which by Lemma 2(ii) is accomplished by producing z∈Z∖{0} such that z+z∈Z. Let a be an element of U such that f(a)≠0. Then for all x∈N, ax∈U and ax+ax=a(x+x)∈U, so that f(ax)∈Z and f(ax)+f(ax)∈Z; hence we need only to show that there exists x∈N such that f(ax)≠0. Suppose that this is not the case, so that f((ax)a)=0= f(ax)a+axd(a)= axd(a), for all x∈N. By Lemma 3(i) either a=0 or d(a)=0.
Ifd(a)=0, then f(xa)=f(x)a+xd(a);thatis,f(xa)=f(x)a∈Z, for all x∈N. Thus [f(u)a,y]=0, for all y∈N, and u∈U. This implies that f(u)[a,y]=0, for all u∈U and y∈N and Lemma 2(i) gives a∈Z. Thus 0=f(ax)=f(xa)=f(x)a, for all x∈N. Replacing x by u∈U, we have f(U)a=0, and by Lemmas 2(i) and 8, we get that a=0. Thus, we have a contradiction.
To complete the proof, we show that ifN is 2-torsion-free, then N is commutative.
Consider first the cased=0. This implies that f(ux)=f(u)x∈Z for all u∈U and x∈N. By Lemma 8, we have u∈U such that f(u)∈Z∖{0}, so N is commutative by Lemma 2(iii).
Now consider the cased≠0. Let c∈Z∖{0}. This implies that if x∈U, f(xc)=f(x)c+xd(c)∈Z. Thus (f(x)c+xd(c))y=y(f(x)c+xd(c)) for all x,y∈U and c∈Z. Therefore by Lemma 5(i), f(x)cy+xd(c)y=yf(x)c+yxd(c) for all x,y∈U and c∈Z. Since d(c)∈Z and f(x)∈Z, we obtain d(c)[x,y]=0, for all x,y∈U and c∈Z. Let d(Z)≠0. Choosing c such that d(c)≠0 and noting that d(c) is not a zero divisor, we have [x,y]=0 for all x,y∈U. By Lemma 1(ii), U⊆Z; hence N is commutative by Lemma 4.
The remaining case isd≠0 and d(Z)={0}. Suppose we can show that U∩Z≠{0}. Taking z∈(U∩Z)∖{0} and x∈N, we have f(xz)=f(x)z∈Z; therefore f(N)⊆Z by Lemma 2(iii) and N is commutative by Lemma 9.
Assume, then, thatU∩Z={0}. For each u∈U, f(u2)=f(u)u+ud(u)=u(f(u)+d(u))∈U∩Z, so f(u2)=0. Thus, for all u∈U and x∈N, f(u2x)=f(u2)x+u2d(x)=u2d(x)∈U∩Z, so u2d(x)=0, and by Lemma 3(iii)  u2=0. Since f(xu)=f(x)u+xd(u)∈Z for all u∈U and x∈N, we have (f(x)u+xd(u))u=u(f(x)u+xd(u)), and right multiplying by u gives uxd(u)u=0. Consequently d(u)uNd(u)u={0}, so that d(u)u=0 for all u∈U. Since u2=0, d(u2)=d(u)u+ud(u)=0 for all u∈U, so f(u)u=0 for all u∈U. But by Lemma 8, there exists u0∈U for which f(u0)≠0; and since f(u0)∈Z, we get u0=0—a contradiction. Therefore U∩Z≠{0} as required.Theorem 12.
LetN be a 3-prime near ring with a nonzero generalized derivation f with associated nonzero derivation d. Let U be a nonzero semigroup ideal of N. If [f(U),f(U)]=0, then (N,+) is abelian.Proof.
Assume thatx∈N is such that [x,f(U)]=[x+x,f(U)]=0. For all u,v∈U such that u+v∈U, [x+x,f(u+v)]=0.
This implies that(1)(x+x)f(u+v)=f(u+v)(x+x),(x+x)f(u)+(x+x)f(v)=f(u+v)x+f(u+v)x,f(u)(x+x)+f(v)(x+x)=xf(u+v)+xf(u+v),f(u)x+f(u)x+f(v)x+f(v)x=xf(u)+xf(v)+xf(u)+xf(v),xf(u)+xf(u)+xf(v)+xf(v)=xf(u)+xf(v)+xf(u)+xf(v),x(f(u)+f(v)-f(u)-f(v))=0,xf(u+v-u-v)=0.
This equation may be restated as xf(c)=0, where c=(u,v).
Leta,b∈U. Then ab∈U and ab+ab=a(b+b)∈U, so [f(ab),f(U)]={0}=[f(ab)+f(ab),f(U)], and by the argument in the previous paragraph, f(ab)f(c)=0. We now have f(U2)f(u,v)=0 for all u,v∈U such that u+v∈U. Taking w∈U2 and x∈N, we get f(xw)f((u,v))=(d(x)w+ xf(w))f((u,v))=0=d(x)wf((u,v)), and since U2 is a nonzero semigroup ideal by Lemma 3(ii) and d≠0, Lemma 3(i) gives
(2)f((u,v))=0∀u,v∈Usuchthatu+v∈U.
Take u=ry and v=rz, where r∈U and y,z∈N, so that u+v=ry+rz=r(y+z)∈U. By (2) we have
(3)f((ry,rz))=f(r(y,z))=0∀r∈U,y,z∈N.
Replacing r by rw, w∈U, we obtain f(r(wy,wz))=0=d(r)(w(y,z))+rf((wy,wz)); so by (3) d(r)U(y,z)=0 for all r∈U and y,z∈N. It follows immediately by Lemmas 1(i) and 3(i) that (y,z)=0 for all y,z∈N; that is, (N,+) is abelian.Theorem 13.
LetN be a 2-torsion-free 3-prime near ring and U be a nonzero semigroup ideal of N. If f is a nonzero generalized derivation with associated derivation d such that [f(U),f(U)]=0, then N is a commutative ring if it satisfies one of the following: (i)  d(Z)≠{0}; (ii) U∩Z≠{0}; (iii)d=0 and f(Z)≠{0}.Proof.
(i) Leta∈N centralizes f(U), and let z∈Z such that d(z)≠0. Then a centralizes f(uz) for all u∈U, so that a(f(u)z+ud(z))=(f(u)z+ud(z))a and aud(z)=ud(z)a. Since d(z)∈Z∖{0}, d(z)[a,u]=0=[a,u] for all u∈U. Therefore a centralizes U, and by Lemma 1(ii), a∈Z. Since f(U) centralizes f(U), f(U)⊆Z and our result follows by Theorem 11.
(ii) We may assumed(Z)={0}. Let z∈(U∩Z)∖{0}. Then for all x,y∈N, f(xz)=f(x)z and f(yz)=f(y)z commute; hence z2[f(x),f(y)]=0=[f(x),f(y)]. Our result now follows from Lemma 10.
(iii) Letu,v∈U and z∈Z such that f(z)≠0. Then [f(zu),f(u)]=0=[f(z)u,f(v)], and since f(z)∈Z∖{0}, f(z)[u,f(u)]=0=[u,f(v)]. Thus f(U) centralizes U, and by Lemma 1(ii), f(U)⊆Z. Our result now follows by Theorem 11.We have already observed that iff is a generalized derivation with d=0, then f(x)y=xf(y) for all x,y∈N. For 3-prime near rings, we have the following converse.Theorem 14.
LetN be a 3-prime near ring and U be a nonzero semigroup ideal of N. If f is a nonzero right generalized derivation of N with associated derivation d and f(x)y=xf(y), for all x,y∈U, then d=0.Proof.
We are given thatf(x)y=xf(y) for all x,y∈U. Substituting yz for y, we get f(x)yz=xf(yz)=x(f(y)z+yd(z)) for all x,y,z∈U. It follows that xyd(z)=0 for all x,y,z∈U; that is, xUd(z)={0} for all x,z∈U. By Lemma 3(i), d(U)=0, and hence d=0 by Lemma 1(i).


## 4. Generalized Derivations Acting as a Homomorphism or an Antihomomorphism

In [4], Bell and Kappe proved that if R is a semiprime ring and d is a derivation on R which is either an endomorphism or an antiendomorphism on R, then d=0. Of course, derivations which are not endomorphisms or antiendomorphisms on R may behave as such on certain subsets of R; for example, any derivation d behaves as the zero endomorphism on the subring C consisting of all constants (i.e., the elements x for which d(x)=0). In fact in a semiprime ring R, d may behave as an endomorphism on a proper ideal of R. However as noted in [4], the behaviour of d is somewhat restricted in the case of a prime ring. Recently the authors in [6] considered (θ,ϕ)-derivation d acting as a homomorphism or an antihomomorphism on a nonzero Lie ideal of a prime ring and concluded that d=0. In this section we establish similar results in the setting of a semigroup ideal of a 3-prime near ring admitting a generalized derivation.Theorem 15.
LetN be a 3-prime near ring and U be a nonzero semigroup ideal of N. Let f be a nonzero generalized derivation on N with associated derivation d. If f acts as a homomorphism on U, then f is the identity map on N and d=0.Proof.
By the hypothesis(4)f(xy)=d(x)y+xf(y)=f(x)f(y)∀x,y∈U.
Replacing y by yz in the above relation, we get
(5)f(xyz)=d(x)yz+xf(yz)∀x,y,z∈U,
or
(6)f(xy)f(z)=d(x)yz+x(d(y)z+yf(z))∀x,y,z∈U.
This implies that
(7)(d(x)y+xf(y))f(z)=d(x)yz+xd(y)z+xyf(z)∀x,y,z∈U.
Using Lemma 5(ii), we get
(8)d(x)yf(z)+xf(y)f(z)=d(x)yz+xd(y)z+xyf(z)∀x,y,z∈U,
or
(9)d(x)yf(z)+xf(yz)=d(x)yz+xd(y)z+xyf(z)∀x,y,z∈U.
This implies that
(10)d(x)yf(z)+xd(y)z+xyf(z)=d(x)yz+xd(y)z+xyf(z)∀x,y,z∈U.
That is,
(11)d(x)yf(z)=d(x)yz∀x,y,z∈U.
Therefore
(12)d(x)y(f(z)-z)=0∀x,y,z∈U,
which implies that
(13)d(x)U(f(z)-z)={0}∀x,z∈U.
It follows by Lemma 3(i) that either d(U)=0 or f(z)=z for all z∈U.
In fact, as we now show, both of these conditions hold.
Suppose thatf(u)=u for all u∈U. Then for all u∈U and x∈N, f(xu)=xu=d(x)u+ xf(u)= d(x)u+xu; hence d(x)U={0} for all x∈N, and thus d=0.
On the other hand, suppose thatd(U)={0}, so that d=0. Then for all x,y∈U, f(xy)=f(x)y= f(x)f(y), so that f(x)(y-f(y))=0. Replacing y by zy, z∈N, and noting that f(zy)=zf(y), we see that f(x)N(y-f(y))={0} for all x,y∈U. Therefore, f(U)={0} or f is the identity on U. But f(U)={0} contradicts Lemma 8, so f is the identity on U.
We now know thatf is the identity on U and f(xy)=xf(y) for all x,y∈N. Consequently, f(ux)=ux=uf(x) for all u∈U and x∈N, so that U(x-f(x))={0} for all x∈N. It follows that f is the identity on N.Theorem 16.
LetN be a 3-prime near ring and U be a nonzero semigroup ideal of N. If f is a nonzero generalized derivation on N with associated derivation d. If f acts as an antihomomorphism on U, then d=0, f is the identity map on N, and N is a commutative ring.Proof.
We begin by showing thatd=0 if and only if f is the identity map on N.
Clearly iff is the identity map on N, xd(y)=0 for all x,y∈N, and hence d=0.
Conversely, assume thatd=0, in which case f(xy)=f(x)y=xf(y) for all x,y∈N. It follows that for any x,y,z∈U,
(14)f(yxz)=f(z)f(yx)=f(z)yf(x)=f(zy)f(x)=zf(y)f(x)=zf(xy).
On the other hand,
(15)f(yxz)=f(xz)f(y)=f(x)zf(y)=f(x)f(zy)=f(x)f(y)f(z)=f(yx)f(z)=f(y)xf(z)=f(y)f(xz)=f(y)f(x)z=f(xy)z.
Comparing (14) and (15) shows that f(U2) centralizes U, so that f(U2)⊆Z by Lemma 1(ii).
NowU2 is a nonzero semigroup ideal by Lemma 3(ii); hence f(U2)≠0 by Lemma 8. Choosing x,y∈U such that f(xy)≠0, we see that for any z∈U, f(xy)z=f(xyz)=f(yz)f(x)= f(y)zf(x)= f(y)f(zx)= f(y)f(x)f(z)= f(xy)f(z), and hence f(xy)(z-f(z))=0. Since f(xy)∈Z∖{0}, we conclude that f(z)=z for all z∈U, and it follows easily that f is the identity map on N.
We note now that if the identity map onN acts as an antihomomorphism on U, then U is commutative, so that by Lemmas 1(ii) and 4N is a commutative ring.
To complete the proof of our theorem, we need only to argue thatd=0. By our antihomomorphism hypothesis
(16)f(xy)=d(x)y+xf(y)=f(y)f(x)∀x,y∈U.
Replacing y by xy in the above relation, we get
(17)f(xy)f(x)=f(xxy)=d(x)xy+xf(xy)∀x,y∈U.
This implies that
(18)(d(x)y+xf(y))f(x)=d(x)xy+xf(y)f(x)∀x,y∈U.
Using Lemma 5(ii), we get
(19)d(x)yf(x)+xf(y)f(x)=d(x)xy+xf(y)f(x)∀x,y∈U.
Thus
(20)d(x)yf(x)=d(x)xy∀x,y∈U.
Replacing y by yr in (20) and using (20), we get
(21)d(x)yrf(x)=d(x)xyr,andsod(x)y[r,f(x)]=0∀x,y∈U,r∈N.
Application of Lemma 3(i) yields that for each x∈U either d(x)=0 or [r,f(x)]=0; that is d(x)=0 or f(x)∈Z.
Suppose that there existsw∈U such that f(w)∈Z∖{0}. Then for all v∈U such that d(v)=0, f(wv)=f(w)v=f(v)f(w)=f(w)f(v), and hence f(w)(v-f(v))=0=v-f(v). Now consider arbitrary x,y∈U. If one of f(x),f(y) is in Z, then f(xy)=f(x)f(y). If d(x)=0=d(y), then d(xy)=d(x)y+xd(y)=0, so f(xy)=xy=f(x)f(y). Therefore f(xy)=f(x)f(y) for all x,y∈U, and by Theorem 15, f is the identity map on N, and therefore d=0.
The remaining possibility is that for eachx∈U, either d(x)=0 or f(x)=0. Let u∈U∖{0}, and let U1=uN. Then U1 is a nonzero semigroup right ideal contained in U and U1 is an additive subgroup of N. The sets {x∈U1|d(x)=0} and {x∈U1|f(x)=0} are additive subgroups of U1 with union equal to U1, so d(U1)={0} or f(U1)={0}. If d(U1)={0}, then d=0 by Lemma 1(i). Suppose, then, that f(U1)={0}. Then for arbitrary x,y∈N  f(uxy)=f(ux)y+uxd(y)=0=uxd(y), so uNd(y)={0}, and again d=0. This completes the proof.



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