# Generalized Derivations on Power Values of Lie Ideals in Prime and Semiprime Rings

Authors: Vincenzo De Filippis; Nadeem UR Rehman; Abu Zaid Ansari
Journal: International Journal of Mathematics and Mathematical Sciences
 (2014)

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## Abstract

LetR be a 2-torsion free ring and let L be a noncentral Lie ideal of R, and let F:R→R and G:R→R be two generalized derivations of R. We will analyse the structure of R in the following cases:  (a) R is prime and F(um)=G(un) for all u∈L and fixed positive integers m≠n; (b) R is prime and F((upvq)m)=G((vrus)n) for all u,v∈L and fixed integers m,n,p,q,r,s≥1; (c) R is semiprime and F((uv)n)=G((vu)n) for all u,v∈[R,R] and fixed integer n≥1; and (d) R is semiprime and F((uv)n)=G((vu)n) for all u,v∈R and fixed integer n≥1.


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## Body

## 1. Introduction

LetR be an associative ring with characteristic different from 2, Z(R) its center, U its (right) Utumi quotient ring, and C its extended centroid. The simple commutator ab-ba will be denoted by [a,b]. Recall that a derivation d:R→R is an additive map satisfying the product rule d(xy)=d(x)y+xd(y) for all x,y∈R.A left multiplierH on a ring R is an additive map satisfying the rule H(xy)=H(x)y for all x,y∈R. In case there exists an endomorphism g of R such that H(xy)=H(x)g(y) for all x,y∈R, then H is calledleftg-multiplier of R.A generalized derivationF on a ring R is an additive map satisfying F(xy)=F(x)y+xd(y) for all x,y∈R and some derivation d of R. A significative example is a map of the form F(x)=ax+xb, for some a,b∈R; such generalized derivations are called inner. Generalized derivations have been primarily studied on operator algebras. Therefore any investigation from the algebraic point of view might be interesting (see, e.g., [1]). Notice that any derivation is a generalized one and also that the generalized inner derivations include left multipliers and right multipliers. Thus the concept of generalized derivation covers both the concept of derivation and the concept of left (right) multipliers.Since the sum of two generalized derivations is a generalized derivation, of course every map of the formF(x)=cx+d(x) is a generalized derivation on R, where c is a fixed element of R and d is a derivation of R.In [1, Theorem 3] Lee proved that every generalized derivation F on a dense right ideal of R can be uniquely extended to the Utumi quotient ring U of R, and thus any generalized derivation of R can be defined on the whole U; moreover it is of the form F(x)=ax+d(x) for some a∈U and d is a derivation on U (F is said to be a generalized derivation associated with derivation d).Many results in the literature indicate that the global structure of a ringR is often tightly connected to the behaviour of additive mappings defined on R.In [2] Bergen proved that if g is an automorphism of R such that (g(x)-x)m=0, for all x∈R, where m≥1 is a fixed integer, then g=1. Daif and Bell [3] showed some results which have the same flavour, when the automorphism is replaced by a nonzero derivation d. In [3] it is proved that if R is a semiprime ring with a nonzero ideal I such that d([x,y])-[x,y]=0, or d([x,y])+[x,y]=0, for all x,y∈I, then I is central. Later Hongan [4] proved that if R is a 2-torsion free semiprime ring and I a nonzero ideal of R, then I is central if and only if d([x,y])-[x,y]∈Z(R), or d([x,y])+[x,y]∈Z(R), for all x,y∈I. Recently in [5] Ashraf and Ali obtained commutativity theorems for prime rings admitting left multipliers which satisfy similar conditions. More precisely in [5] it is showed that a prime ring R must be commutative if there exist a nonzero ideal I of R and a left multiplier H, which is not the identity map on I, such that one of the following holds: (i) H([x,y])=[x,y] for all x,y∈I [5, Theorem 2.1]; (ii) H(xy)-xy∈Z(R) for all x,y∈I [5, Theorem 3.1]; and (iii) H(xy)-yx∈Z(R) for all x,y∈I [5, Theorem 3.3]. Moreover the same results hold in case H is replaced by a generalized derivation F of R.In a more recent paper [6], Ali and Huang extended the previous cited results in the case R is a semiprime ring, I is a nonzero ideal of R, and H is aleftg-multiplier of R. They proved that I⊆Z(R) if one of the following holds: (i) H([x,y])=[x,y] for all x,y∈I [6, Theorem 2.1]; (ii) H(xy)=xy for all x,y∈I [6, Theorem 2.2]; (iii) H(xy)=yx for all x,y∈I [6, Theorem 2.3]; and (iv) H(x2)=x2, for all x∈I [6, Theorem 2.5]. Moreover, in case R is a prime ring, the same conditions force the commutativity of R.A natural question is to consider additive mapsF,G:R→R such that F(f(x,y))=G(g(x,y)), when f(x,y) and g(x,y) are both either monomials or powers of the commutator [x,y]. In this sense, in [7] it is proved that, under appropriate torsion assumptions, a prime ring R is commutative if it admits a nonzero derivation d satisfying one of the following: (i) d(xnym)=d(ymxn); (ii) d((xy)n)=d((yx)n); (iii) d((xy)n)=d(xnyn); and (iv) d((xy)n)=d(ynxn), for all x,y∈R.So it seems natural to ask about the case when the derivationd is replaced by a generalized derivation. Motivated by the previous cited results, in this paper we will introduce two different generalized derivations acting on R and satisfying some appropriate conditions on some suitable subsets of R. We will prove the following.Theorem 1.
LetR be a 2-torsion free prime ring and let F:R→R and G:R→R be two generalized derivations associated with derivations d:R→R and δ:R→R, respectively. Suppose that there exist m,n≥1 integers such that F([x,y]m)=G([x,y]n) for all x,y∈R. Then either R satisfies the standard identity s4 or one of the following holds:(a)
F=G is an inner ordinary derivation of R and for all x1,x2∈R there exists α∈C such that [x1,x2]m=[x1,x2]n+α;(b)
F=G and [x1,x2]m=[x1,x2]n, for all x1,x2∈R.Theorem 2.
LetR be a 2-torsion free prime ring and let L be a noncentral Lie ideal of R, and let F:R→R and G:R→R be two nonzero generalized derivations associated with derivations d:R→R and δ:R→R, respectively. Suppose that there exist m,n,p,q,r,s≥1 integers such that F((upvq)m)=G((vrus)n) for all u,v∈L. Then R satisfies the standard identity s4.In the last section we study some commutativity conditions for a semiprime ring with a generalized derivation satisfying suitable algebraic conditions. More precisely, we will prove the following.Theorem 3.
LetR be a 2-torsion free semiprime ring and let F:R→R and G:R→R be generalized derivations associated, respectively, with derivations d:R→R and δ:R→R. Set L=[R,R] and n≥1 such that F((uv)n)=G((vu)n) for all u,v∈L. Then there exists a central idempotent e of  U such that, on the direct sum decomposition U=eU⊕(1-e)U, the generalized derivations F and G vanish identically on eU and the ring (1-e)U satisfies s4.Theorem 4.
LetR be a 2-torsion free semiprime ring and let F:R→R and G:R→R be generalized derivations associated with derivations d:R→R and δ:R→R, respectively. Set n≥1 such that F((uv)n)=G((vu)n) for all u,v∈R. Then R contains a nonzero central ideal, d(R)⊆Z(R), δ(R)⊆Z(R), and there exist λ,μ∈C such that F(x)=λx+d(x), G(x)=μx+δ(x), for all x∈R.
Moreover there exists a central idempotente of  U such that, on the direct sum decomposition U=eU⊕(1-e)U, the generalized derivations F and G vanish identically on eU and the ring (1-e)U is commutative.


## 2. Action of Generalized Derivations on Prime Rings

In order to prove the main result in this section, we first fix some intermediate lemmas and theorems. We begin with the following.Remark 5 (see [8], Lemma 3).
LetR be a prime ring with extended centroid C and let f(x1,…,xn) be any polynomial, not necessarily multilinear. Then the right (left) annihilator of f(x1,…,xn) is zero, unless when f(x1,…,xn) is an identity for R.Lemma 6.
LetR be a primitive ring which is isomorphic to a dense ring of linear transformations of a vector space V over the field C such that dim⁡CV≥2 and let a,b∈R, m,n≥1, different integers. If for any x,y∈R, [x,y]ma=[x,y]nb, then a=b and [x1,x2]m=[x1,x2]n for all x1,x2∈R, unless when a=b=0.Proof.
Let0≠v∈V. Since dim⁡CV≥2, there exists w∈V such that {v,w} are linearly C-independent. By the density of R there exist x,y∈R such that
(1)vx=w,vy=0,wy=v
so that v[x,y]=v and by the hypothesis
(2)0=v([x,y]ma-[x,y]nb)=v(a-b)
that is V(a-b)=(0), which implies a=b. Hence ([x1,x2]m-[x1,x2]n)a=0 for all x1,x2∈R and by Remark 5 we have that either a=b=0 or [x1,x2]m=[x1,x2]n for any x1,x2∈R.Lemma 7.
LetR be a 2-torsion free prime ring, J a nonzero two-sided ideal of R, and f(x1,…,xn) a noncentral polynomial over C, the extended centroid of R. If F:R→R is a generalized derivation associated with a derivation d:R→R such that F(f(r1,…,rn))∈C for all r1,…,rn∈J, then F(x)=0, for all x∈R.Proof.
In light of previous remarks, we have that there existsa∈U, the Utumi quotient ring of R, such that F(x)=ax+d(x), for all x∈R. Thus J satisfies the generalized differential identity
(3)[af(x1,…,xn)+d(f(x1,…,xn)),xn+1].
Since, by [9], J and R satisfy the same differential identities, then we have that R satisfies F(f(x1,…,xn))∈C. Let S be the additive subgroup generated by the subset
(4){f(r1,…,rn)∣r1,…,rn∈R}.
S is a Lie ideal of R; indeed for any r∈R, y1,…,yn∈R one has
(5)[r,f(y1,…,yn)]=∑i‍f(y1,…,[r,yi],…,yn)∈S.
IfS is noncommutative then, by [10, pages 4-5], there exists a nonzero two-sided ideal I of R such that 0≠[I,R]⊆S. In this case it is easy to see that F([r1,r2])∈C, for all r1,r2∈I. In particular [F(u),u]=0 for all u∈[I,I]. Since [I,I] is a noncentral Lie ideal of R, it follows easily that F must be zero (see, e.g., Theorem 3.3 in [11]).
Hence we may consider it the only case whenS is commutative.
Thus[f(x1,…,xn),f(y1,…,yn)] is an identity in R. This means that there exist a field K and a positive integer m such that [f(x1,…,xn),f(y1,…,yn)] is also an identity in Mm(K). If m=1, R is commutative, thus we suppose m≥2.
Sincef(x1,…,xn) is not central valued on R, there exist r1,…,rn∈Mm(K) such that f(r1,…,rn)=a∉Z(R), so that [a,f(y1,…,yn)] is also a generalized identity in Mm(K). By a result of Lee (see [12], Theorem), we have the contradiction that f(x1,…,xn) is central valued on R.Lemma 8.
LetR be a 2-torsion free primitive ring which is isomorphic to a dense ring of linear transformations of a vector space V over the field C such that dim⁡CV≥3, and let a,b∈R, γ∈C, m,n≥1, different integers. If for any x,y∈R, γ[x,y]m+a[x,y]m+[x,y]mb=a[x,y]n+[x,y]nb, then γ=0, a+b=0 and one of the following holds:(a)
a,b∈C;(b)
for allx1,x2∈R there exists α∈C such that [x1,x2]m=[x1,x2]n+α.Proof.
Let0≠v∈V such that {v,va} are linearly C-independent. Since dim⁡CV≥3, there exists w∈V such that {v,va,w} are linearly C-independent. By the density of R there exist x,y∈R such that
(6)vax=w,vay=0,vx=w,vy=0,wy=v
so that v[x,y]=v, va[x,y]=0 and by the hypothesis
(7)0=v(γ[x,y]m+a[x,y]m+[x,y]mb-a[x,y]n-[x,y]nb)=γv
which implies γ=0. Thus a([x,y]m-[x,y]n)+([x,y]m-[x,y]n)b=0 for all x,y∈R and by Lemma 7 we get the required conclusions.
Assume now that, for anyv∈V, {v,va} are linearly C-dependent. In this case standard arguments show that a∈C; hence [x,y]m(γ+a+b)=[x,y]n(a+b) and by Lemma 6 we get γ+a+b=a+b; that is, γ=0 and a=-b∈C.Lemma 9.
LetR be a 2-torsion free primitive ring which is isomorphic to a dense ring of linear transformations of a vector space V over the field C such that dim⁡CV≥3, and let a,b,c,q∈R and m,n≥1 be such that a[x,y]m+[x,y]mb=c[x,y]n+[x,y]nq, for all x,y∈R. Then one of the following holds:(a)
a,b,c,q∈C, a+b=0 and c+q=0;(b)
a-c=q-b∈C, a+b=c+q=0 and for all x1,x2∈R there exists α∈C such that [x1,x2]m=[x1,x2]n+α;(c)
a-c=q-b∈C and [x1,x2]m=[x1,x2]n for all x1,x2∈R.Proof.
Suppose first thatdim⁡CV=t is finite. In this case R≅Mt(C), the ring of all t×t matrices over C with t≥3. Let eij be the usual matrix unit and denote a=∑‍arsers, c=∑‍crsers, with ars,crs∈C. Fix i≠j and choose [x,y]=[eij,eji]=eii-ejj. By the hypothesis
(8)a(eii-ejj)m+(eii-ejj)mb=c(eii-ejj)n+(eii-ejj)nq
and left multiplying by ekk, for any k≠i,j, it follows that aki-cki=0. This means that the matrix a-c is diagonal. Let now φ be any automorpism of R and note that φ(a)[x,y]m+[x,y]mφ(b)=φ(c)[x,y]n+[x,y]nφ(q), for all x,y∈R. Therefore φ(a-c) must be a diagonal matrix. In particular, for any i≠j, (1+eij)(a-c)(1-eij) must be a diagonal matrix. By easy computation it follows that ajj-cjj=aii-cii; that is, a-c=α∈Z(R). Analogously one can prove that b-q=β∈Z(R).
Therefore we have that(9)(α+β)[x,y]m+c[x,y]m+[x,y]mq=c[x,y]n+[x,y]nq.
Suppose now thatdim⁡CV=∞. By Lemma 2 in [13], R satisfies the following generalized identity axm+xmb-cxn-xnq. Let e∈H=soc(R) be any idempotent minimal element and recall that H is generated by such idempotent elements. Since H satisfies axm+xmb-cxn-xnq, then in particular ae+eb-ce-eq=0. Thus we can easily obtain both (1-e)(a-c)e=0 and e(b-q)(1-e)=0, which imply [a-c,e]=0 and [b-q,e]=0; that is, [a-c,H]=0 and [b-q,H]=0. Also in this case there exist α,β∈C such that a-c=α and b-q=β. Therefore in any case Identity (9) holds and the conclusion follows as an application of Lemma 8.Lemma 10.
LetR be a 2-torsion free prime ring, a,b,c,q∈R, and let m,n≥1 be such that a[x,y]m+[x,y]mb=c[x,y]n+[x,y]nq, for all x,y∈R. Then either R satisfies s4, the standard identity of degree 4, or one of the following holds:(a)
a,b,c,q∈C, a+b=0 and c+q=0;(b)
a-c=q-b∈C, a+b=c+q=0 and for all x1,x2∈R there exists α∈C such that [x1,x2]m=[x1,x2]n+α;(c)
a-c=q-b∈C and [x1,x2]m=[x1,x2]n for all x1,x2∈R.Proof.
Firstly assume thatR does not satisfy any nontrivial generalized polynomial identity. In light of [14] and by our assumption, it follows that both
(10)a[x,y]m-c[x,y]n,[x,y]mb-[x,y]nq
are trivial generalized polynomial identities of R. This means that a,b,c,q∈C, so that (a+b)[x,y]m-(c+q)[x,y]n is a trivial generalized polynomial identity for R.
Hencea+b=0 and c+q=0.
Consider now the case thatR satisfies some nontrivial generalized polynomial identity. By Theorem 3 in [15] it follows that S=RC is a primitive ring with soc(R)≠0, where C=Z(U) is the extended centroid of R, and the Utumi quotient ring U is a C-algebra centrally closed. Since R and U satisfy the same generalized polynomial identities (see [14]), without loss of generality, we may replace R by U and Z(R) by C and R is a C-algebra centrally closed. Then R is a dense ring of linear transformations of a vector space V over C. In case dimCV≤2, then R satisfies the standard identity s4. In case dimCV≥3 we may apply Lemma 9. In any case we are done.We consider now the more general situation.Proof of Theorem1.
In all that follows we assume thatR does not satisfy s4; if not we are done.
Since any generalized derivationH of R can be implicitly assumed to be defined on the whole U and assumes the form H(x)=cx+h(x) for some c∈U and h a derivation on U, we may assume that there exist a,b∈U and d,δ derivations on U such that
(11)F(x)=ax+d(x),G(x)=bx+δ(x).
SinceR and U satisfy the same generalized polynomial identities [14] as well as the same differential identities [9], then, without loss of generality, to prove our results we may assume that
(12)F([x,y]m)=G([x,y]n)∀x,y∈U,
where d, δ are derivations on U. We divide the proof into 3 cases.
Case
1. Let d(x)=[c,x] and δ(x)=[q,x] be both inner derivations in U, so that F(x)=ax+[c,x]=(a+c)x-xc and G(x)=bx+[q,x]=(b+q)x-xq, for suitable elements c,q∈U. Thus we have (a+c)[x,y]m+[x,y]m(-c)=(b+q)[x,y]n+[x,y]n(-q), for all x,y∈U. In this case we apply Lemma 10 and we have that either R satisfies s4 or one of the following holds:(a)
a=b=0 and d=δ=0; that is, F=G=0;(b)
a=b=0, c-q∈C, and d=δ=F=G; that is, F=G is the inner derivation induced by the element c; moreover for all x1,x2∈R there exists α∈C such that [x1,x2]m=[x1,x2]n+α;(c)
[x1,x2]m=[x1,x2]n, for all x1,x2∈R,q-c∈C (i.e., d=δ); moreover a=b+β for a suitable β∈C. In this case, our assumption implies that β[x1,x2]m=0 for all x1,x2∈R, which means β=0, since R is not commutative. Therefore a=b and F=G.
Case
2. Assume that d and δ are C-independent modulo U-inner derivations.
By the main assumption,U satisfies
(13)a[x,y]n+∑i+j=m-1‍[x,y]i([d(x),y]+[x,d(y)])[x,y]j=b[x,y]n+∑i+j=n-1‍[x,y]i([δ(x),y]+[x,δ(y)])[x,y]j.
Then by the result in [16], we have that U satisfies
(14)a[x,y]n+∑i+j=m-1‍[x,y]i([x1,y]+[x,x2])[x,y]j=b[x,y]n+∑i+j=n-1‍[x,y]i([x3,y]+[x,x4])[x,y]j.
In particularU satisfies the blended component
(15)∑i+j=m-1‍[x,y]i[x1,y,x,y]j.
ThusU is a PI-ring and there exists Mt(C) satisfying (15). For x=e12, y=e21, and x1=e32 we have the contradiction e31=0.
Case
3. Assume finally that d and δ are not both inner derivations (otherwise we are done by Case 1) such that d and δ are C-dependent modulo U-inner derivations. Without loss of generality say d=αδ+adq; that is, d(x)=αδ(x)+[q,x], for suitable α∈C and q∈U. Under this assumption, we prove that a number of contradictions follow. We may assume the following:(i)
d,δ are both nonzero derivations of R. In fact, if either d or δ is zero then both F and G are inner generalized derivations of R, which is a contradiction;(ii)
δ is not an inner derivation, if not both F and G are inner generalized derivations of R, which is again a contradiction.
Suppose first thatα=0; then d(x)=[q,x]. Thus U satisfies
(16)a[x,y]m+[q,[x,y]m]=b[x,y]n+δ([x,y]n);
that is,
(17)(a+q)[x,y]m+[x,y]m(-q)=b[x,y]n+∑i+j=n-1[x,y]i([δ(x),y]+[x,δ(y)])[x,y]j.
By [16] it follows that U satisfies
(18)(a+q)[x,y]m+[x,y]m(-q)=b[x,y]n+∑i+j=n-1‍[x,y]i([x1,y]+[x,x2])[x,y]j
and in particular U satisfies the blended component
(19)∑i+j=n-1‍[x,y]i[x1,y][x,y]j.
ThereforeU is a PI-ring, so that there exists a field K such that U and the matrix ring Mt(K) satisfy the same polynomial identities. Moreover, we may assume t≥3, since U does not satisfy s4. Notice that if we choose x=e12, y=e21, and x1=e32 then the contradiction e31=0 follows.
Then assumeα≠0. Thus U satisfies
(20)a[x,y]m+αδ([x,y]m)+[q,[x,y]m]=b[x,y]n+δ([x,y]n);
that is,
(21)(a+q)[x,y]m+[x,y]m(-q)+α∑i+j=m-1‍[x,y]i([δ(x),y]+[x,δ(y)])[x,y]j=b[x,y]n+∑i+j=n-1‍[x,y]i([δ(x),y]+[x,δ(y)])[x,y]j.
By [16] it follows that U satisfies
(22)(a+q)[x,y]m+[x,y]m(-q)+α∑i+j=m-1‍[x,y]i([x1,y]+[x,x2])[x,y]j=b[x,y]n+∑i+j=n-1‍[x,y]i([x1,y]+[x,x2])[x,y]j
and in particular U satisfies both
(23)(a+q)[x,y]m+[x,y]m(-q)=b[x,y]n,(24)α∑i+j=m-1‍[x,y]i([x1,y]+[x,x2])[x,y]j=∑i+j=n-1‍[x,y]i([x1,y]+[x,x2])[x,y]j.
As above,U is a PI-ring, so that there exists a field K such that U and the matrix ring Mt(K) satisfy (24). Since we may assume that t≥3, then for x=e12, y=e21, x1=e32, and x2=0 in (24) we get αe31=e31; that is, α=1. Moreover, by (23) and using Lemma 10 we have in any case q∈C, that is, d(x)=δ(x), and one of the following holds. (a)
um=un for all u∈[U,U] and there exists β∈C such that b=a+β. In this case we have F(x)=ax+d(x) and G(x)=(a+β)x+d(x) for all x∈U. By the main assumption we also get aum+d(um)=aum+βum+d(um), for all u∈[U,U]. Hence βum=0, which implies β=0 and F=G.(b)
a=b=0 and F=d=δ=G. This implies d(um-un)=0, for all u∈[U,U]. Since d≠0, by Lemma 7 we have that for all u∈[U,U] there exists βu∈C such that um-un=βu. Without loss of generality we consider n>m and let α be any nonzero element of C. Denote v=[x,y]∈[U,U] and u=[α-mx,y]=α-mv∈[U,U]. Thus there exist βu,βv∈C such that vm=vn+βv and um=un+βu. Hence αvm=αn-mvn+βu implying
(25)αvn+αβv=αn-mvn+βu.
In other words, for allα∈C and for all v∈[U,U], we have (α-αn-m)vn∈C. In case α≠αn-m then vn∈C for all v∈[U,U] and as above there exists a field K such that U and the matrix ring Mt(K) satisfy [[x,y]n,z]=0. Since we may assume that t≥3, a contradiction follows easily. Thus α=αn-m for any α∈C. In this last case and by (25), it follows that αβv=βu, for any α≠0. Hence, for 0≠α1≠α2∈C, we get α1βv=α2βv; that is, βv=0 for any choice of v∈[U,U], which means vm=vn for all v∈[U,U].Corollary 11.
LetR be a 2-torsion free prime ring and let L be a noncentral Lie ideal of R, and let F:R→R and G:R→R be two generalized derivations associated with derivations d:R→R and δ:R→R, respectively. Suppose that there exist m,n≥1 integers such that F(um)=G(un) for all u∈L. Then either R satisfies the standard identity s4 or one of the following holds:(a)
F=G is an inner ordinary derivation of R and for all x1,x2∈R there exists α∈C such that [x1,x2]m=[x1,x2]n+α;(b)
F=G and [x1,x2]m=[x1,x2]n, for all x1,x2∈R.Proof.
SinceL is not central in R, then by [10, pages 4-5] there exists a nonzero two-sided ideal I of R such that [I,I]⊆L. Therefore F([x,y]m)=G([x,y]n) for all x,y∈I. As above we write
(26)F(x)=ax+d(x),G(x)=bx+δ(x)
for suitable a,b∈U and d,δ derivations of R. Thus I satisfies the differential identity
(27)a[x,y]m+d([x,y]m)-b[x,y]n-δ([x,y]n).
SinceI, R, and U satisfy the same differential identities (see [9]), then (27) is satisfied by U. Hence we conclude by Theorem 1.Proof of Theorem2.
Firstly we notice that foru=v we have F(um(p+q))=G(un(r+s)), for all u∈L. By Corollary 11 we have that R is a PI-ring and, if assumed that R does not satisfy s4, then one has F=G. As remarked above, there exists an ideal I of R such that [I,I]⊆L; therefore
(28)F((upvq)m-(vrus)n)=0
for all u,v∈[I,I]. Since U and I satisfy the same differential identities, we also have that (28) is satisfied by U. By Lemma 7, and since F≠0, (upvq)m-(vrus)n is central for all u,v∈[U,U], moreover there exists a field K such that Mt(K) and U satisfy the same polynomial identities. Of course we assume that t≥3, since U does not satisfy s4. Fix u=[e12,e21+e31]=e11-e22-e32 and v=[e12,e21]=e11-e22.
Thus(29)(upvq)m=e11+αe22+βe32,(vrus)n=e11+γe22,
where α,β,γ∈{+1,-1}. Hence
(30)(upvq)m-(vrus)n=βe32∉Z(Mt(K)),
which is a contradiction.Corollary 12.
LetR be a 2-torsion free prime ring, U its Utumi quotient ring, C its extended centroid, L noncentral Lie ideal of R, and F:R→R and G:R→R two generalized derivations associated with derivations d:R→R and δ:R→R, respectively. Suppose that there exists n≥1 integer such that F((uv)n)=G((vu)n) for all u,v∈L. Then either F=G=0 or R satisfies the standard identity s4 and F-G is an ordinary derivation of R.Proof.
Foru=v∈L in our main assumption we get F(u2n)=G(u2n); moreover by applying Theorem 2 it follows that R must satisfy the standard identity s4.
ThereforeU≅M2(C) and we may assume that R⊆M2(C), the 2×2 matrix ring over C. Since L is not central and R is 2-torsion free, as remarked above, it follows that [R,R]⊆L. Hence we have that F([x1,x2]2n)=G([x1,x2]2n) is satisfied by R. Let a,b∈U such that F(x)=ax+d(x) and G(x)=bx+δ(x), for all x∈R; then R satisfies the generalized differential identity
(31)(a-b)[x1,x2]2n+(d-δ)([x1,x2]2n).
Firstly notice that in caseF=0 then R satisfies G([x1,x2][y1,y2]n) and by Lemma 7 it follows that G=0 unless [x1,x2][y1,y2] is central valued on R. In the latter case for [x1,x2]=e12 and [y1,y2]=e21 a contradiction follows.
Thus we assume bothF≠0 and G≠0 and denote H(x)=d(x)-δ(x) (of course H is a derivation of R). Here we apply again Kharchenko’s theory, using the fact that [x1,x2]2n is central valued on R, and we prove that a=b. To do this, we divide the proof into two cases.
Case  1. Let H(x)=[c,x] be the inner derivations in R. Thus we have that (a-b)[x,y]2n=0, for all x,y∈R, which implies a=b.
Case  2. Assume that H is not inner. By Kharchenko’s theory in [16], and since R satisfies
(32)(a-b)[x1,x2]2n+∑i+j=2n-1‍[x1,x2]i([H(x1),x2]+[x1,H(x2)])[x1,x2]j,
then R satisfies
(33)(a-b)[x1,x2]2n+∑i+j=2n-1‍[x1,x2]i([y1,x2]+[x1,y2])[x1,x2]j;
in particular (a-b)[x1,x2]2n is a generalized identity for R, which implies again that a=b.Corollary 13.
LetR be a 2-torsion free prime ring, U its Utumi quotient ring, C its extended centroid, and F:R→R and G:R→R two generalized derivations associated with derivations d:R→R and δ:R→R, respectively. Suppose that there exists n≥1 integer such that F((uv)n)=G((vu)n) for all u,v∈R. Then either F=G=0 or R is commutative.Proof.
By Corollary12, if we assume that F≠0 and G≠0, it follows that R⊆M2(C) and F-G=H is ordinary derivation of R. In particular H(x2)=0 for all x∈R and by Lemma 7 it follows that either H=0 or x2∈Z(R), for all x∈R.
In caseH=0, then F=G and F((uv)n-(vu)n)=0 for all u,v∈R. Again by Lemma 7, since F≠0, one has (uv)n-(vu)n∈Z(R); in particular for u=e11 and v=e11+e12, it follows the contradiction e12∈Z(R). Therefore R must be commutative.
LetH≠0, so that x2∈Z(R), for all x∈R follows by the previous argument. Thus R is commutative and by the main assumption it follows H((uv)n)=0. Once again by Lemma 7, since H≠0, it follows (uv)n∈Z(R), which leads to a contradiction, as remarked above.An easy consequence of Corollary13 is the following.Corollary 14.
LetR be a 2-torsion free prime ring, U its Utumi quotient ring, C its extended centroid, and F:R→R a generalized derivation of R. Suppose that there exists n≥1 integer such that F((uv)n)=0 for all u,v∈R. Then either F=0 or R is commutative.


## 3. Results in Semiprime Rings

In order to prove the main result of this section we will make use of the following facts.Remark 15.
LetR be a semiprime ring and let F be a generalized derivation of R associated with derivation d. If F(R)=0, then d(R)=0.Claim 1 (see [17, Proposition 2.5.1]).
Any derivation of a semiprime ringR can be uniquely extended to a derivation of its left Utumi quotient ring U, and so any derivation of R can be defined on the whole U.Claim 2 (see [18, page 38]).
IfR is semiprime then so is its left Utumi quotient ring.
The extended centroidC of a semiprime ring coincides with the center of its left Utumi quotient ring.Claim 3 (see [18, p. 42]).
LetB be the set of all the idempotents in C, the extended centroid of R. Assume that R is B-algebra orthogonal complete. For any maximal ideal P of B, PR forms a maximal prime ideal of R, which is invariant under any derivation of R.We are now ready to prove Theorems3 and 4.Proof of Theorem3.
Leta,b∈U and d,δ derivations of U be such that F(x)=ax+d(x) and G(x)=bx+δ(x), for all x∈R. By Claim 2, Z(U)=C, and by Claim 1  d and δ can be uniquely defined on the whole U. Since U and R satisfy the same generalized differential identities, then F((xy)n)=G((yx)n), for all x,y∈[U,U]. Let B be the complete Boolean algebra of idempotents in C and M any maximal ideal of B.
SinceU is B-algebra orthogonal complete (see [18, page 42, (2) of Fact 1]), by Claim 3, MU is a prime ideal of U, which is both d-invariant and δ-invariant. Let d¯ and δ¯ be the derivations induced, respectively, by d and δ on U¯=U/MU and denote F¯(x¯)=a¯x¯+d¯(x¯), G¯(x¯)=b¯x¯+δ¯(x¯). For any x¯,y¯∈[U¯,U¯], F¯((x¯y¯)n)=G¯((y¯x¯)n). In particular U¯ is a prime ring and so, by Corollary 12, either F¯=0 and G¯=0 in U¯ or U¯ satisfies s4. This implies that, for any maximal ideal M of B, F(U)⊆MU and G(U)⊆MU or s4(x1,x2,x3,x4)⊆MU, for all x1,x2,x3,x4∈U. In any case F(U)s4(x1,x2,x3,x4)⊆⋂M‍MU=0 and also G(U)s4(x1,x2,x3,x4)⊆⋂M‍MU=0. From [17, Chapter 3] there exists a central idempotent element e of U, such that F(eU)=0, G(eU)=0 and (1-e)U satisfies s4.Proof of Theorem4.
Here we repeat the same argument above. Leta,b∈U such that F(x)=ax+d(x) and G(x)=bx+δ(x). Let B be the complete Boolean algebra of idempotents in C and M any maximal ideal of B.
SinceU is B-algebra orthogonal complete (see [18, p. 42, (2) of Fact 1]), by Claim 3, MU is a prime ideal of U, which is both d-invariant and δ-invariant. Let d¯ and δ¯ be two derivations induced, respectively, by d and δ on U¯=U/MU and denote F¯(x¯)=a¯x¯+d¯(x¯), G¯(x¯)=b¯x¯+δ¯(x¯). For any x¯,y¯∈[U¯,U¯], F¯((x¯y¯)n)=G¯((y¯x¯)n). Since U¯ is a prime ring so, by Corollary 13, either F¯=0 and G¯=0 in U¯ or U¯ is commutative. Moreover, if F¯=0¯ and G¯=0¯ in U¯ then, by Remark 15, it follows that d¯=0¯ and δ¯=0¯ in U¯.
This implies that, for any maximal idealM of B, one of the following holds:(a)
eitherF(U)⊆MU, d(U)⊆MU, G(U)⊆MU and δ(U)⊆MU;(b)
or[U,U]⊆MU.
In any case[F(U),U]⊆⋂MU=0 and [d(U),U]⊆⋂MU=0 imply that [aU,U]=0. Analogously [G(U),U]⊆⋂MU=0 and [δ(U),U]⊆⋂MU=0 implying that [bU,U]=0.
As mentioned above it follows that there exists a central idempotent elemente in U such that, on the direct sum decomposition eU⊕(1-e)U, F and G vanish identically on eU and the ring (1-e)U is commutative.
Moreover,[aU2,U]=(0) implies aU[U,U]=(0); that is, both UaU2[U,U]=(0) and aUU2[U,U]=(0). Therefore [a,U]U2[a,U]=(0) and, by the semiprimeness of U, it follows [a,U]=(0); that is, a∈C. Analogously one can prove that b∈C.
On the other hand, since by Theorem 3 in [9] U and R satisfy the same differential identities, then [d(R),R]=0 and [δ(R),R]=0, which imply that d(R)⊆Z(R), δ(R)⊆Z(R), and R contains some nonzero central ideals, unless when d(R)=0 and δ(R)=0.
In the last case,λ(uv)n=μ(vu)n for all u,v∈R, and in particular λ(u)2n=μ(u)2n for all u∈R. The semiprimeness of R forces λ=μ≠0 (since F≠0), so that (uv)n=(vu)n for all u,v∈R. Finally by Theorem 1 in [19], the commutativity of R follows which is a contradiction.



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