# On Prime-Gamma-Near-Rings with Generalized Derivations

Authors: Kalyan Kumar Dey; Akhil Chandra Paul; Isamiddin S. Rakhimov
Journal: International Journal of Mathematics and Mathematical Sciences
 (2012)

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## Abstract

LetN be a 2-torsion free prime Γ-near-ring with center Z(N). Let (f,d) and (g,h) be two generalized derivations on N. We prove the following results: (i) if f([x,y]α)=0 or f([x,y]α)=±[x,y]α or f2(x)∈Z(N) for all x,y∈N, α∈Γ, then N is a commutative Γ-ring. (ii) If a∈N and [f(x),a]α=0 for all x∈N, α∈Γ, then d(a)∈Z(N). (iii) If (fg,dh) acts as a generalized derivation on N, then f=0 or g=0.


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## Body

## 1. Introduction

The derivations inΓ-near-rings have been introduced by Bell and Mason [1]. They studied basic properties of derivations in Γ-near-rings. Then Aşci [2] obtained commutativity conditions for a Γ-near-ring with derivations. Some characterizations of Γ-near-rings and regularity conditions were obtained by Cho [3]. Kazaz and Alkan [4] introduced the notion of two-sided Γ-α-derivation of a Γ-near-ring and investigated the commutativity of a prime and semiprime Γ-near-rings. Uçkun et al. [5] worked on prime Γ-near-rings with derivations and they found conditions for a Γ-near-ring to be commutative. In [6] Dey et al. studied commutativity of prime Γ-near-ring with generalized derivations.In this paper, we obtain the conditions of a primeΓ-near-ring to be a commutative Γ-ring. If a∈N, and [f(x),a]α=0 for all x∈N,α∈Γ, then d is central. Also we prove that if (fg,dh) is the generalized derivation on N, then f and g are trivial.


## 2. Preliminaries

AΓ-near-ring is a triple (N,+,Γ), where(i)
(N,+) is a group (not necessarily abelian);(ii)
Γ is a nonempty set of binary operations on N such that for each α∈Γ,(N,+,α) is a left near-ring;(iii)
xα(yβz)=(xαy)βz, for all x,y,z∈N and α,β∈Γ.We will use the wordΓ-near-ring to mean left Γ-near-ring. For a near-ring N, the set N0={x∈N:0αx=0,α∈Γ} is called the zero-symmetric part of N. A Γ-near-ring N is said to be zero-symmetric if N=N0. Throughout this paper, N will denote a zero symmetric left Γ-near-ring with multiplicative centre Z(N). Recall that a Γ-near-ring N is prime if xΓNΓy=0 implies x=0 or y=0. An additive mapping d:N→N is said to be a derivation on N if d(xαy)=xαd(y)+d(x)αy for all x,y∈N,α∈Γ, or equivalently, as noted in [1], that d(xαy)=d(x)αy+xαd(y) for all x,y∈N,α∈Γ. Further, an element x∈N for which d(x)=0 is called a constant. For x,y∈N,α∈Γ, the symbol [x,y]α will denote the commutator xαy-yαx, while the symbol (x,y) will denote the additive-group commutator x+y-x-y. An additive mapping f:N→N is called a generalized derivation if there exits a derivation d of N such that f(xαy)=f(x)αy+xαd(y) for all x,y∈N,α∈Γ. The concept of generalized derivation covers also the concept of a derivation.


## 3. Derivations on Γ-Near-Rings

In this section we prove that a few subsidiary results (Lemmas3.1, 3.2, 3.4, 3.8, 3.9, 3.10 and 3.11) to use them for proving of our main results (Theorems 3.3, 3.5, 3.6, 3.12 and 3.13).Lemma 3.1.
Letd be an arbitrary derivation on a Γ-near-ring N. Then N satisfies the following partial distributive law: (aαd(b)+d(a)αb)βc=aαd(b)βc+d(a)αbβc and (d(a)αb+aαd(b))βc=d(a)αbβc+aαd(b)βc for all a,b,c∈N,α,β∈Γ.Proof.
For alla,b,c∈N,α,β∈Γ, we get d((aαb)βc)=aαbβd(c)+(aαd(b)+d(a)αb)βc and d(aα(bβc))=aαd(bβc)+d(a)αbβc=aα(bβd(c)+d(b)βc)+d(a)αbβc=aαbβd(c)+aαd(b)βc+d(a)αbβc. Equating these two relations for d(aαbβc) now yields the required partial distributive law.Lemma 3.2.
Letd be a derivation on a Γ-near-ring N and suppose u∈N is not a left zero divisor. If [u,d(u)]α=0,α∈Γ, then (x,u) is a constant for every x∈N.Proof.
Fromuα(u+x)=uαu+uαx, for all x∈N, α∈Γ, we obtain uαd(u+x)+d(u)α(u+x)=uαd(u)+d(u)αu+uαd(x)+d(u)αx, which reduces uαd(x)+d(u)αu=d(u)αu+uαd(x), for all α∈Γ.
Sinced(u)αu=uαd(u), α∈Γ, this equation is expressible as uα(d(x)+d(u)-d(x)-d(u))=0=uαd((x,u)). Thus d((x,u))=0.Theorem 3.3.
LetN be a Γ-near-ring having no nonzero divisors of zero. If N admits a nontrivial commuting derivation d, then (N,+) is abelian.Proof.
Letc be any additive commutator. Then c is a constant by Lemma 3.2. Moreover, for any w∈N, α∈Γ, wαc is an additive commutator, hence also a constant. Thus, 0=d(wαc)=wαd(c)+d(w)αc and d(w)αc=0, for all α∈Γ. Since d(w)≠0 for all w∈N, we conclude that c=0.Lemma 3.4.
LetN be a prime Γ-near-ring.(i)
Ifz∈Z(N)-{0}, then z is not a zero divisor in N.(ii)
IfZ(N)-{0} contains an element z for which z+z∈Z(N), then (N,+) is abelian.(iii)
Letd be a nonzero derivation on N. Then xΓd(N)={0} implies x=0, and d(N)Γx={0} implies x=0.(iv)
IfN is 2-torsion free and d is a derivation on N such that d2=0, then d=0.Proof.
(i) Ifz∈Z(N)-{0} and zαx=0, x∈N, α∈Γ, then zαrβx=0, x, r∈N, α∈Γ. Thus we get zΓNΓx=0, by primeness of N, x=0.
(ii) Letz∈Z(N)-{0} be an element such that z+z∈Z(N), and let x,y∈N,α∈Γ. Since z+z is distributive, we get (x+y)α(z+z)=xα(z+z)+yα(z+z)=xαz+xαz+yαz+yαz=zα(x+x+y+y).
On the other hand,(x+y)α(z+z)=(x+y)αz+(x+y)αz=zα(x+y+x+y). Thus, x+x+y+y=x+y+x+y and therefore x+y=y+x. Hence (N,+) is abelian.
(iii) LetxΓd(N)=0, and let r,s be arbitrary elements of N and α,β∈Γ. Then 0=xαd(rβs)=xαrβd(s)+xαd(r)βs=xαrβd(s). Thus xΓNΓd(N)={0}, and since d(N)≠{0},x=0.
A similar argument works ifd(N)Γx={0}, since Lemma 3.1 provides enough distributivity to carry it through.
(iv) For arbitraryx,y∈N,α∈Γ, we have 0=d2(xαy)=d(xαd(y)+d(x)αy)=xαd2(y)+d(x)αd(y)+d(x)αd(y)+d2(x)αy=2d(x)αd(y). Since N is 2-torsion free, d(x)αd(y)=0,x,y∈N,α∈Γ. Thus d(x)Γd(N)={0} for each x∈N, and (iii) yields d(a)=0. Thus d=0.Theorem 3.5.
If a primeΓ-near-ring N admits a nontrivial derivation d for which d(N)∈Z(N), then (N,+) is abelian. Moreover, if N is 2-torsion free, then N is a commutative Γ-ring.Proof.
Letc be an arbitrary constant, and let x be anon-constant. Then d(xαc)=xαd(c)+d(x)αc=d(x)αc∈Z(N), α∈Γ. Since d(x)∈Z(N)-{0}, it follows easily that c∈Z(N). Since c+c is a constant for all constants c, it follows from Lemma 3.4(ii) that (N,+) is abelian, provided that there exists a nonzero constant.
Assume, then, that 0 is the only constant. Sinced is obviously commuting, it follows from Lemma 3.2 that all u which are not zero divisors belong to the center of (N,+), denoted by Z(N). In particular, if x≠0, d(x)∈Z(N). But then for all y∈N, d(y)+d(x)-d(y)-d(x)=d((y,x))=0, hence (y,x)=0.
Now we assume thatN is 2-torsion free. By Lemma 3.1, (aαd(b)+d(a)αb)βc=aαd(b)βc+d(a)αbβc for all a,b,c∈N,α,β∈Γ, and using the fact that d(aαb)∈Z(N), α∈Γ, we get cαaβd(b)+cαd(a)βb=aαd(b)βc+aαd(b)βc,α,β∈Γ. Since (N,+) is abelian and d(N)⊆Z(N), this equation can be rearranged to yield d(b)α[c,a]β=d(a)α[b,c]β for all a,b,c∈N,α,β∈Γ.
Suppose now thatN is not commutative. Choosing b,c∈N, with [b,c]β≠0, β∈Γ, and letting a=d(x), we get d2(x)α[b,c]β=0, for all x∈N,α,β∈Γ, and since the central elements d2(x) cannot be a nonzero divisor of zero, we conclude that d2(x)=0 for all x∈N. But by Lemma 3.4(iv), this cannot happen for nontrivial d.Theorem 3.6.
LetN be a prime Γ-near-ring admitting a nontrivial derivation d such that [d(x),d(y)]α=0 for all x,y∈N,α∈Γ. Then (N,+) is abelian. Moreover, if N is 2-torsion free, then N is a commutative Γ-ring.Proof.
By Lemma3.4(ii), if both z and z+z commute element-wise with d(N), then zαd(c)=0,α∈Γ, for all additive commutators c. Thus, taking z=d(x), we get d(x)αd(c)=0 for all x∈N,α∈Γ, so d(c)=0 by Lemma 3.4(iii). Since wαc is also an additive commutator for any w∈N, α∈Γ, we have d(wαc)=0d(w)αc, and another application of Lemma 3.4(iii) gives c=0.
Now we assume thatN is 2-torsion free. By the partial distributive law, d(d(x)αy)βd(z)=d(x)αd(y)βd(z)+d2(x)αyβd(z) for all x,y,z∈N,α,β∈Γ, hence, d2(x)αyβd(z)=d(d(x)αy)βd(z)-d(x)αd(y)βd(z)=d(x)αd(y)βd(z)=d(z)α(d(d(x)βy)-d(x)βd(y))=d(z)αd2(x)βy=d2(x)αd(z)βy,α,β∈Γ. Thus d2(x)α(yβd(z)-d(z)βy)=0 for all x,yz∈N, α,β∈Γ.
Replacingyδt,δ∈Γ, we obtain d2(x)αyδtβd(z)=d2(x)αd(z)βyδt=d2(x)αyβd(z)δt, for all x,y,z,t∈N,α,β,δ∈Γ, so that d2(x)αyβ[t,d(z)]δ=0 for all x,y,z,t∈N, α,β,δ∈Γ. The primeness of N shows that either d2=0 or d(N)⊆Z(N), and since the first of these conditions is impossible by Lemma 3.4(iv), the second must hold and N is a commutative Γ-ring by Theorem 3.5.Definition 3.7.
LetN be a Γ-near-ring and d a derivation of N. An additive mapping f:N→N is said to be a right generalized derivation of N associated with d if
(3.1)f(xαy)=f(x)αy+xαd(y)∀x,y∈R,α∈Γ,
and f is said to be a left generalized derivation of N associated with d if
(3.2)f(xαy)=d(x)αy+xαf(y)∀x,y∈R,α∈Γ.
Finally, f is said to be a generalized derivation of N associated with d if it is both a left and right generalized derivation of N associated with d.Lemma 3.8.
Letf be a right generalized derivation of a Γ-near ring N associated with d. Then(i)
f(xαy)=xαd(y)+f(x)αy for all x,y∈N,α∈Γ;(ii)
f(xαy)=xαf(y)+d(x)αy for all x,y∈N,α∈Γ.Proof.
(i) For anyx,y∈N,α∈Γ, we get
(3.3)f(xα(y+y))=f(x)α(y+y)+xαd(y+y)=f(x)αy+f(x)αy+xαd(y)+xαd(y),f(xαy+xαy)=f(x)αy+xαd(y)+f(x)αy+xαd(y).
Comparing these two expressions, we obtain
(3.4)f(x)αy+xαd(y)=xαd(y)+f(x)αy∀x,y∈N,α∈Γ,
and so,
(3.5)f(xαy)=xαd(y)+f(x)αy∀x,y∈N,α∈Γ.
(ii) In a similar way.Lemma 3.9.
Letf be a right generalized derivation of a Γ-near ring N associated with d. Then(i)
(f(x)αy+xαd(y))βz=f(x)αyβz+xαd(y)βz, for all x,y∈N, α,β∈Γ.(ii)
(d(x)αy+xαf(y))βz=d(x)αyβz+xαf(y)βz, for all x,y∈N, α,β∈Γ.Proof.
(i) For anyx,y,z∈N, α,β∈Γ, we get f((xαy)βz)=f(xαy)βz+xαyβd(z).
On the other hand,(3.6)f(xα(yβz))=f(x)αyβz+xαd(yβz)=f(x)αyβz+xαd(y)βz+xαyβd(z).
From these two expressions of f(xαyβz), we obtain that, for all x,y,z∈N, α,β∈Γ,
(3.7)(f(x)αy+xαd(y))βz=f(x)αyβz+xαd(y)βz.
(ii) The proof is similar.Lemma 3.10.
LetN be a prime Γ-near-ring, f a nonzero generalized derivation of N associated with the nonzero derivation d and a∈N. (i) If aΓf(N)=0, then a=0. (ii) If f(N)Γa=0, then a=0.Proof.
(i) For anyx,y∈N,α,β∈Γ, we get 0=aβf(xαy)=aβf(x)αy+aβxαd(y)=aβxαd(y). Hence aΓNΓd(N)=0. Since N is a prime Γ-near-ring and d≠0, we obtain a=0.
(ii) A similar argument works iff(N)Γa=0.Lemma 3.11.
LetN be a prime Γ-near-ring. Let f be a generalized derivation of N associated with the nonzero derivation d. If N is a 2-torsion free Γ-near-ring and f2=0, then f=0.Proof.
(i) For anyx,y∈N, α∈Γ, we get
(3.8)0=f2(xαy)=f(f(xαy))=f(f(x)αy+xαd(y))=f2(x)αy+2f(x)αd(y)+xαd2(y).
By the hypothesis,
(3.9)2f(x)αd(y)+xαd2(y)=0∀x,y∈N,α∈Γ.
Writing f(x) by x in (3.9), we get f(x)αd2(y)=0, for all x,y∈N,α∈Γ.
By Lemma3.9(ii), we obtain that d2(N)=0 or f=0. If d2(N)=0 then d=0 from Lemma 3.4(iv), a contradiction. So we find f=0.Theorem 3.12.
LetN be a prime Γ-near-ring with a nonzero generalized derivation f associated with d. If f(N)⊆Z(N), then (N,+) is abelian. Moreover, if N is 2-torsion free, then N is commutative Γ-ring.Proof.
Suppose thata∈N, such that f(a)≠0. So, f(a)∈Z(N)-{0} and f(a)+f(a)∈Z(N)-{0}. For all x,y∈N, α∈Γ, we have (x+y)α(f(a+f(a))=(f(a+f(a))α(x+y).
That is,xαf(a)+xαf(a)+yαf(a)+yαf(a)=f(a)αx+f(a)αx+f(a)αy+f(a)αy, for all x,y∈N, α∈Γ.
Sincef(a)∈Z(N), we get f(a)αx+f(a)αy=f(a)αy+f(a)αx, and so, f(a)α(x,y)=0 for all x,y∈N,α∈Γ.
Sincef(a)∈Z(N)-{0} and N is a prime Γ-near-ring, it follows that (x,y)=0, for all x,y∈N. Thus (N,+) is abelian.
Using the hypothesis, for anyx,y,z∈N,α,β∈Γ, zαf(xβy)=f(xβy)αz. By Lemma 3.4(ii), we have zαd(x)βy+zαxf(y)=d(x)αyβz+xαf(y)βz. Using f(N)⊂Z(N) and (N,+) being abelian, we obtain that
(3.10)zαd(x)βy-d(x)αyβz=[x,z]αβf(y),∀x,y∈N,α,β∈Γ.
Substituting f(z) for z in (3.10), we get f(z)β[d(x),y]α=0 for all x,y∈N,α,β∈Γ.
Sincef(z)∈Z(N) and f a nonzero generalized derivation with associated with d, we get d(N)⊂Z(N). So, N is a commutative Γ-ring by Theorem 3.3.Theorem 3.13.
LetN be a prime Γ-near-ring with a nonzero generalized derivation f associated with d. If [f(N),f(N)]α=0, α∈Γ, then (N,+) is abelian. Moreover, if N is 2-torsion free, then N is commutative Γ-ring.Proof.
By the same argument as in Theorem3.12, it is shown that if both z and z+z commute elementwise with f(N), then we have
(3.11)zαf(x,y)=0∀x,y∈N,α∈Γ.
Substituting f(t),t∈N for z in (3.11), we get f(t)αf(x,y)=0, α∈Γ. By Lemma 3.9(i), we obtain that f(x,y)=0 for all x,y∈N, α∈Γ. For any w∈N, β∈Γ, we have 0=f(wβx,wβy)=f(wβ(x,y))=d(w)β(x,y)+wβf(x,y) and so, we obtain d(w)β(x,y)=0, for any w∈N, β∈Γ. From Lemma 3.4(iii), we get (x,y)=0 for any x,y∈N.
Now we assume thatN is 2-torsion free. By the assumption [f(N),f(N)]α=0, α∈Γ, we have
(3.12)f(z)αf(f(x)βy)=f(f(x)βy)αf(z)∀x,y,z∈N,α,β∈Γ.
Hence we get
(3.13)f(z)αd(f(x))βy+f(z)αf(x)βf(y)=d(f(x))αyβf(z)+f(x)αf(y)βf(z),f(z)αd(f(x))βy+f(x)αf(z)βf(y)=d(f(x))αyβf(z)+f(x)αf(z)βf(y),
and so,
(3.14)f(z)αd(f(x)βy)=d(f(x))αyβf(z)∀x,y,z∈N,α,β∈Γ.
If we take yδw instead of y in (3.14), then
(3.15)d(f(x))αyδwβf(z)=f(z)αd(f(x)βyδw)=d(f(x))αyδf(z)βw∀x,y,z∈N,α,β,δ∈Γ,
and so,
(3.16)d(f(x))αyδwβf(z)-d(f(x))αyδf(z)βw=d(f(x))αyδ[f(z),w]δ=0∀x,y,z∈N,α,β,δ∈Γ.
Thus we get d(f(x))ΓNΓ[f(z),w]δ=0, for all x,y,z∈N, α,β,δ∈Γ. Since N is a prime Γ-near-ring, we have d(f(N))=0 or f(N)⊂Z(N). Let us assume that d(f(N))=0. Then
(3.17)0=d(f(xαy))=d(d(x)αy+xαf(y))
and so,
(3.18)d2(x)αy+d(x)αd(y)+d(x)αf(y)=0,∀x,y∈N,α∈Γ.
Replacing y by yβz, β∈Γ, in (3.18), we get
(3.19)0=d2(x)αyβz+d(x)αd(yβz)+d(x)αf(yβz)=d2(x)αyβz+d(x)αd(y)βz+d(x)αyβd(z)+d(x)αf(y)βz+d(x)αyβd(z)={d2(x)αy+d(x)αd(y)+d(x)αf(y)}βz+2d(x)αyβd(z)∀x,y,z∈N,α,β∈Γ.
Using (3.18) and N being 2-torsion free Γ-near-ring, we get d(N)ΓNΓd(N)=0.
Thus we obtain thatd=0. It contradicts by d≠0. The theorem is proved.


## 4. Generalized Derivations of Γ-Near-Rings

We denote a generalized derivationf:N→N determined by a derivation d of N by (f,d). We assume that d is a nonzero derivation of N unless stated otherwise.Theorem 4.1.
Let(f,d) be a generalized derivation of N. If f([x,y]α)=0  for all x,y∈N,α∈Γ, then N is a commutative Γ-ring.Proof.
Assume thatf([x,y]α)=0 for all x,y∈N, α∈Γ. Substitute xβy instead of y, obtaining
(4.1)f([x,xβy]α)=f(xβ[x,y]α)=d(x)β[x,y]α+xβf([x,y]α)=0.
Since the second term is zero, it is clear that
(4.2)d(x)αxβy=d(x)αyβx∀x,y∈N,α,β∈Γ.
Replacing y by yδz in (4.2) and using this equation, we get
(4.3)d(x)αyβ[x,z]δ=0∀x,y,z∈N,α,β,δ∈Γ.
Hence either x∈Z(N) or d(x)=0. Let L={x∈N∣d(x)=0}. Then Z(N) and L are two additive subgroups of (N,+)=Z(N)∪L. However, a group cannot be the union of proper subgroups, hence either N=Z(N) or N=L. Since d≠0, we are forced to conclude that N is a commutative Γ-ring.Theorem 4.2.
Let(f,d) be a generalized derivation of N. If f([x,y]α)=±[x,y]α  for all x,y∈N, α∈Γ, then N is a commutative Γ-ring.Proof.
Assume thatf([x,y]α)=±[x,y]α for all x,y∈N, α∈Γ. Replacing y by xβy, β∈Γ, in the hypothesis, we have
(4.4)f([x,xβy]α)=±(xαxβy-xαyβx)=±xβ[x,y]α.
On the other hand,
(4.5)f([x,xβy]α)=f(xβ[x,y]α)=d(x)β[x,y]α+xβf([x,y]α)=d(x)β[x,y]α+xβ(±[x,y]α).
It follows from the two expressions for f([x,xβy]α) that
(4.6)d(x)αxβy=d(x)αyβx∀x,y∈N,α,β∈Γ.
Using the same argument as in the proof of Theorem 4.1, we get that N is a commutative Γ-ring.Theorem 4.3.
Let(f,d) be a nonzero generalized derivation of N. If f acts as a homomorphism on N, then f is the identity map.Proof.
Assume thatf acts as a homomorphism on N. Then one obtains
(4.7)f(xαy)=f(x)αf(y)=d(x)αy+xαf(y)∀x,y∈N,α∈Γ.
Replacing y by yβz in (4.7), we arrive at
(4.8)f(x)αf(yβz)=d(x)αyβz+xαf(yβz).
Since (f,d) is a generalized derivation and f acts as a homomorphism on N, we deduce that
(4.9)f(xαy)βf(z)=d(x)αyβz+xαd(y)βz+xαyβf(z).
By Lemma 3.9(ii), we get
(4.10)d(x)αyβf(z)+xαf(y)βf(z)=d(x)αyβz+xαd(y)βz+xαyβf(z),
and so
(4.11)d(x)αyβf(z)+xαf(yβz)=d(x)αyβz+xαd(y)βz+xαyβf(z).
That is,
(4.12)d(x)αyβf(z)+xαd(y)βz+xαyβf(z)=d(x)αyβz+xαd(y)βz+xαyβf(z).
Hence, we deduce that
(4.13)d(x)αyβ(f(z)-z)=0∀x,y,z∈N,α,β∈Γ.
Because N is prime and d≠0, we have f(z)=z for all z∈N. Thus, f is the identity map.Theorem 4.4.
Let(f,d) be a nonzero generalized derivation of N. If f acts as an antihomomorphism on N, then f is the identity map.Proof.
By the hypothesis, we have(4.14)f(xαy)=f(y)αf(x)=d(x)αy+xαf(y)∀x,y∈N,α∈Γ.
Replacing y by xβy in the last equation, we obtain
(4.15)f(xβy)αf(x)=d(x)βxαy+xβf(xαy).
Since (f,d) is a generalized derivation and f acts as an antihomomorphism on N, we get
(4.16)(d(x)βy+xβf(y))αf(x)=d(x)αxβy+xαf(y)βf(x).
By Lemma 3.9(ii), we conclude that
(4.17)d(x)αyβf(x)+xαf(y)βf(x)=d(x)αxβy+xαf(y)βf(x),
and so
(4.18)d(x)αyβf(x)=d(x)αxβy∀x,y∈N,α,β∈Γ.
Replacing y by yδz and using this equation, we have
(4.19)d(x)αyβ[f(x),z]α=0∀x,z∈N,α,β∈Γ.
Hence we obtain the following alternatives: d(x)=0 or f(x)∈Z(N), for all x∈N. By a standard argument, one of these must hold for all x∈N. Since d≠0, the second possibility gives that N is commutative Γ-ring by Theorem 3.12, and so we deduce that f is the identity map by Theorem 4.3.Theorem 4.5.
Let(f,d) be a generalized derivation of N such that d(Z(N))≠0, and a∈N. If [f(x),a]α=0  for all x∈N, α,β∈Γ, then a∈Z(N).Proof.
Sinced(Z(N))≠0, there exists c∈Z(N) such that d(c)≠0. Furthermore, as d is a derivation, it is clear that d(c)∈Z(N). Replacing x by cβx, β∈Γ, in the hypothesis and using Lemma 3.9(ii), we have
(4.20)f(cβx)αa=aαf(cβx),d(c)αxβa+cαf(x)βa=aαd(c)βx+aαcβf(x).
Since c∈Z(N) and d(c)∈Z(N), we get
(4.21)d(c)αxβ[y,a]α=0∀y∈N,α,β,δ∈Γ.
By the primeness of N and 0≠d(c)∈Z(N), we obtain that a∈Z(N).Theorem 4.6.
Let(f,d) be a generalized derivation of N, and a∈N. If [f(x),a]α=0  for all x∈N, then d(a)∈Z(N).Proof.
Ifa=0, then there is nothing to prove. Hence, we assume that a≠0.
Replacingx by aβx in the hypothesis, we have
(4.22)f(aβx)αa=aαf(aβx),d(a)αxβa+aαf(x)βa=aαd(a)βx+aαaβf(x).
Using f(x)αa=aαf(x), we have
(4.23)d(a)αxβa=aαd(a)βx∀x∈N,α,β∈Γ.
Taking xδy instead of x in the last equation and using this, we conclude that
(4.24)d(a)αNβ[a,y]α=0∀y∈N,α,β∈Γ.
Since N is a prime Γ-near-ring, we have either d(a)=0 or a∈Z(N). If 0≠a∈Z(N), then (N,+) is abelian by Lemma 3.2(ii). Thus
(4.25)f(xαa)=f(aαx)f(x)αa+xαd(a)=d(a)αx+aαf(x)
and so
(4.26)[d(a),x]α=0∀x∈N,α∈Γ.
That is,d(a)∈Z(N). Hence in either case we have d(a)∈Z(N). This completes the proof.Theorem 4.7.
Let(f,d) be a generalized derivation of N. If N is a 2-torsion free Γ-near-ring and f2(N)⊂Z(N), then N is a commutative Γ-ring.Proof.
Suppose thatf2(N)⊂Z(N). Then we get
(4.27)f2(xαy)=f2(x)αy+2f(x)αd(y)+xαd2(y)∈Z(N)∀x,y∈N,α∈Γ.
In particular, f2(x)αc+2f(x)αd(c)+xαd2(c)∈Z(N) for all x∈N, c∈Z(N), α∈Γ. Since the first summand is an element of Z(N), we have
(4.28)2f(x)αd(c)+xαd2(c)∈Z(N)∀x∈N,c∈Z(N),α∈Γ.
Taking f(x) instead of x in (4.28), we obtain that
(4.29)2f2(x)αd(c)+f(x)αd2(c)∈Z(N)∀x∈N,c∈Z(N),α∈Γ.
Since d(c)∈Z(N), f2(x)∈Z(N), and so f2(x)αd(c)∈Z(N) for all x∈N, c∈Z(N), α∈Γ, we conclude f(x)αd2(c)∈Z(N) for all x∈N, c∈Z(N), α∈Γ.
SinceN is prime, we get d2(Z(N))=0 or f(N)⊆Z(N). If f(N)⊆Z(N), then N is a commutative Γ-ring by Lemma 3.8. Hence, we assume d2(Z)=0. By (4.28), we get 2f(x)αd(c)∈Z(N) for all x∈N, c∈Z(N), α∈Γ.
SinceN is a 2-torsion free near-ring and d(c)∈Z(N), we obtain that either f(N)⊂Z(N) or d(Z(N))=0. If f(N)⊂Z(N), then we are already done. So, we may assume that d(Z(N))=0. Then
(4.30)f(cαx)=f(xαc),f(c)αx+cαd(x)=f(x)αc+xαd(c),
and so
(4.31)f(c)αx+cαd(x)=f(x)αc∀x∈N,c∈Z(N).
Now replacing x by f(x) in (4.31), and using the fact that f2(N)⊂Z(N), we get
(4.32)f(c)αf(x)+cαd(f(x))=f^2(x)αc∀x∈N,c∈Z(N).
That is,
(4.33)f(c)αf(x)+cαd(f(x))∈Z∀x∈N,c∈Z(N),α∈Γ.
Again taking f(x) instead of x in this equation, one can obtain
(4.34)f(c)αf2(x)+cαd(f2(x))∈Z∀x∈N,c∈Z(N),α∈Γ.
The second term is equal to zero because of d(Z)=0. Hence we have f(c)αf2(x)∈Z(N) for all x∈N, c∈Z(N), α∈Γ.
Sincef2(N)⊂Z(N) by the hypothesis, we get either f2(N)=0 or f(Z(N))⊂Z(N). If f2(N)=0, then the theorem holds by Definition 3.7. If f(Z)⊂Z(N), then f(xαf(c))=f(f(c)αx) for all x∈N, c∈Z(N), and so
(4.35)d(x)αf(c)=f(c)αf(x)∀x∈N,c∈Z(N).
Using f(c)∈Z(N), we now have f(c)α(d(x)-f(x))=0 for all x∈N, c∈Z(N), α∈Γ. Since f(Z(N))⊂Z(N), we have either f(Z(N))=0 or d=f. If d=f, then f is a derivation of N and so N is commutative Γ-ring by Lemma 3.11.
Now assume thatf(Z(N))=0. Returning to the equation (4.31), we have
(4.36)cα(d(x)-f(x))=0∀x∈N,c∈Z(N),α∈Γ.
Since c∈Z(N), we have either d=f or Z(N)=0. Clearly, d=f implies the theorem holds. If Z(N)=0, then f2(N)=0 by the hypothesis, and so N is a commutative Γ-ring by Lemma 3.4(iv). Hence, the proof is completed.Corollary 4.8.
LetN be a 2-torsion free near-ring, and (f,d) a nonzero generalized derivation of N. If [f(N),f(N)]α=0, α∈Γ, then N is a commutative Γ-ring.Lemma 4.9.
Let(f,d) and (g,h) be two generalized derivations of N. If h is a nonzero derivation on N and f(x)αh(y)=-g(x)αd(y) for all x,y∈N, then (N,+) is abelian.Proof.
Suppose thatf(x)αh(y)+g(x)αd(y)=0 for all x,y∈N, α∈Γ.
We substitutey+z for y, thereby obtaining
(4.37)f(x)αh(y)+f(x)αh(z)+g(x)αd(y)+g(x)αd(z)=0.
Using the hypothesis, we get
(4.38)f(x)αh(y,z)=0∀x,y,z∈N,α∈Γ.
It follows by Lemma 3.10(ii) that h(y,z)=0 for all y,z∈N. For any w∈N, we have h(wαy,wαz)=h(wα(y,z))=h(w)α(y,z)+wαh(y,z)=0 and so h(w)α(y,z)=0 for all w,y,z∈N, α∈Γ.
An appeal to Lemma3.4(iii) yields that (N,+) is abelian.Theorem 4.10.
Let(f,d) and (g,h) be two generalized derivations of N. If N is 2-torsion free and f(x)αh(y)=-g(x)αd(y) for all x,y∈N,α∈Γ, then f=0 or g=0.Proof.
Ifh=0 or d=0, then the proof of the theorem is obvious. So, we may assume that h≠0 and d≠0. Therefore, we know that (N,+) is abelian by Lemma 4.9.
Now suppose that(4.39)f(x)αh(y)+g(x)αd(y)=0∀x,y∈N,α∈Γ.
Replacing x by uβv in this equation and using the hypothesis, we get
(4.40)f(uβv)αh(y)+g(uβv)αd(y)=uαf(v)βh(y)+d(u)αvβh(y)+uαg(v)βd(y)+h(u)αvβd(y)=0,
and so
(4.41)d(u)αvβh(y)=-h(u)αvβd(y)∀u,v,y∈N,α∈Γ.
Taking yδt instead of y in the above relation, we obtain
(4.42)d(u)αvβh(y)δt+d(u)αvyβδh(t)=-h(u)αvβd(y)δt-h(u)αvβyδd(t).
That is,
(4.43)d(u)αvβyδh(t)=-h(u)αvβyδd(t)∀u,v,y,t∈N,α,β,δ∈Γ.
Replacing y by h(y) in (4.43) and using this relation, we have
(4.44)h(u)αNβ(d(y)δh(t)-h(y)αd(t))=0∀u,y,t∈N.
Since N is a prime Γ-near-ring and h≠0, we obtain that
(4.45)d(y)αh(t)=h(y)αd(t),∀y,t∈N.
Now again taking uλv instead of x in the initial hypothesis, we get
(4.46)f(u)αvβh(y)+uαd(v)βh(y)+g(u)αvβd(y)+uαh(v)βd(y)=0.
Using (4.45) yields that
(4.47)f(u)αvβh(y)+2uαh(v)βd(y)+g(u)αvβd(y)=0∀u,v,y∈N,
Taking h(v) instead of v in this equation, we arrive at
(4.48)f(u)αh(v)βh(y)+2uαh2(v)βd(y)+g(u)αh(v)βd(y)=0.
By the hypothesis and (4.45), we have
(4.49)0=-g(u)αd(v)βh(y)+2uαh2(v)βd(y)+g(u)αh(v)βd(y)=-g(u)αh(v)βd(y)+2uαh2(v)βd(y)+g(u)αh(v)βd(y),
and so
(4.50)2uαh2(v)βd(y)=0∀u,v,y∈N,α,β∈Γ.
Since N is a 2-torsion free prime Γ-near-ring, we obtain that h2(N)Γd(N)=0. An appeal to Lemma 3.4(iii) and (iv) gives that h=0. This contradicts by our assumption. Thus the proof is completed.Theorem 4.11.
Let(f,d) and (g,h) be two generalized derivations of N. If (fg,dh) acts as a generalized derivation on N, then f=0 or g=0.Proof.
By calculatingfg(xαy) in two different ways, we see that g(x)αd(y)+f(x)αh(y)=0 for all x,y∈N, α∈Γ. The proof is completed by using Theorem 4.10.



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*Source: 625968-2012-06-03.xml*