Aequat. Math. 82 (2011), 123–134c© Springer Basel AG 20110001-9054/11/010123-12published online February 1, 2011DOI 10.1007/s00010-010-0061-y Aequationes Mathematicae

Nilpotent and invertible values in semiprimerings with generalized derivations

Asma Ali, Shakir Ali and Vincenzo De Filippis

Abstract. Let R be a semiprime ring and F be a generalized derivation of R and n ≥ 1 afixed integer. In this paper we prove the following: (1) If (F (xy) − yx)n is either zero orinvertible for all x, y ∈ R, then there exists a division ring D such that either R = D orR = M2(D), the 2 × 2 matrix ring. (2) If R is a prime ring and I is a nonzero right idealof R such that (F (xy) − yx)n = 0 for all x, y ∈ I, then [I, I]I = 0, F (x) = ax + xb fora, b ∈ R and there exist α, β ∈ C, the extended centroid of R, such that (a − α)I = 0 and(b − β)I = 0, moreover ((a + b)x − x)I = 0 for all x ∈ I.

Mathematics Subject Classification (2000). 16N60, 16W25.

Keywords. Prime and semiprime rings, generalized derivations.

1. Introduction

Let R be a semiprime ring with center Z(R), Utumi quotient ring U andextended centroid C. For further information, definitions and basic propertieson these concepts refer to [4]. The goal of this paper is to establish that there isa relationship between the structure of the ring R and the behaviour of suitableadditive mappings defined on R that satisfy certain special identities. In par-ticular we study the case when the map is a generalized derivation of R. Recallthat an additive map d : R −→ R is a derivation of R if d(xy) = d(x)y +xd(y)for all x, y ∈ R. More specifically an additive map F : R −→ R is said tobe a generalized derivation if there is a derivation d of R such that, for allx, y ∈ R,F (xy) = F (x)y + xd(y).

A number of authors have obtained several results in this topic. For exam-ple in [7] it is proved that if R is a semiprime ring, I a non-zero ideal of R andd a derivation of R such that d([x, y]) = [x, y] for all x, y ∈ I, then I ⊆ Z(R).In [10] there is an extension of the previous result. More precisely the authorproved that I ⊆ Z(R) if d([x, y]) ± [x, y] ∈ Z(R) for all x, y ∈ I. Later in

124 A. Ali, S. Ali and V. De Filippis AEM

[8] the following result was obtained: if R is a prime ring and d([x, y]) − [x, y]is nil for all x, y ∈ R, then R is commutative. So it seems natural to askabout the case when the derivation d is replaced by a generalized derivationF : as in [19] it is proved that if R is a prime ring and I a non-zero ideal ofR such that F ([x, y]) ± [x, y] = 0 (respectively F (xy + yx) ± (xy + yx) = 0)for all x, y ∈ I, then R is commutative. Here we fix our attention on the setS = {f(xy) − yx : x, y ∈ T}, where f is a suitable additive map on R and Tis a subset of R.

In [2] it is shown that if R is a prime ring and f is a derivation of R suchthat f(xy) ± yx ∈ Z(R) (respectively f(xy) ± xy ∈ Z(R)) for all x, y ∈ I, anon-zero ideal of R, then the ring is commutative.

More recently, in [1] the authors proved that the same result holds if theadditive map f is a generalized derivation of R. In other words the results in[2] and [1] say that if T is a non-zero ideal of R and f is either a derivation ora generalized derivation of R such that any element of S is central in R, thenR is commutative.

In the present paper, we will consider the case when any element of Sis either nilpotent or invertible in R and obtain some information about thegeneralized derivation F and the structure of the semiprime (prime) ring R.

More precisely, in the first section we consider the case when S = {F (xy)−yx : x, y ∈ T} has nilpotent or invertible values, where F is a generalizedderivation of R and T = R. We prove the following result:

Theorem 1. Let R be a semiprime ring and F be a generalized derivation ofR. If for a fixed integer n ≥ 1 (F (xy) − yx)n is either zero or invertible forall x, y ∈ R, then there exists a division ring D such that either R = D orR = M2(D), the 2 × 2 matrix ring over D.

Then we will study the case when the set S = {F (xy) − yx : x, y ∈ T} hasnilpotent values where T is a right ideal of R, and we prove that:

Theorem 2. Let R be a prime ring and I be a nonzero right ideal of R. If fora fixed integer n ≥ 1 (F (xy) − yx)n = 0 for all x, y ∈ I, then [I, I]I = 0 andF (x) = ax+xb and also there exist α, β ∈ C, the extended centroid of R, suchthat (a−α)I = 0 and (b−β)I = 0. Moreover ((a+ b)x−x)I = 0 for all x ∈ I.

2. Set with invertible values

We begin with the following easy consequence of Theorem 4 in [15]:

Proposition 1. Let R be a semiprime ring and F be a generalized derivationof R. If for a fixed integer n ≥ 1 (F (xy) − yx)n = 0 for all x, y ∈ R, then Fis an identity map and R is commutative.

Vol. 82 (2011) Nilpotent and invertible values in semiprime rings 125

Proof. In view of Lee’s result [15], we consider F (x) = ax + d(x) for allx ∈ U and d a derivation of U . Thus R satisfies the differential identity(axy + d(xy) − yx)n. Since R and U satisfy the same differential identities(see [16]), U satisfies the differential identity (axy + d(xy) − yx)n. Set y = 1,so (F (r) − r)n = 0 for all r ∈ U . Moreover, the map x �−→ G(x) = F (x) − xis a generalized derivation of U . Thus G(r)n = 0, for all r ∈ U and in view ofTheorem 5 in [15] we have that G is the identity map on U ,that is F (r) = r, forall r ∈ U . Hence, by our main assumption it follows that (xy−yx)n = 0 for allx, y ∈ U . In this case it is known that U is commutative (see for example [6])as well as R. �

The following examples illustrate that the above results do not hold forarbitrary rings.

Example 1. Let S be any ring and let R =

⎧⎨

⎩

⎛

⎝0 a b0 0 c0 0 0

⎞

⎠ | a, b, c ∈ S

⎫⎬

⎭.

Define F : R −→ R by F

⎛

⎝0 a b0 0 c0 0 0

⎞

⎠ =

⎛

⎝0 0 −b0 0 c0 0 0

⎞

⎠. Then F is a gen-

eralized derivation with associated derivation d given by d

⎛

⎝0 a b0 0 c0 0 0

⎞

⎠ =

⎛

⎝0 a 00 0 −c0 0 0

⎞

⎠. It can be verified that for n ≥ 2 (F (xy) − yx)n = 0 for all

x, y ∈ R. However, F is neither an identity map on R nor R is commutative.Hence, Proposition 1 is not true for arbitrary rings.

Example 2. Let R ={(

a b0 c

)

| a, b, c ∈ Z2

}

and I ={(

0 b0 0

)

| b ∈ Z2

}

.

Define a map F : R −→ R by F

(a b0 c

)

=(

a 00 0

)

. Then F is a generalized

derivation on R with associated derivation d given by d

(a b0 c

)

=(

0 b0 0

)

satisfying (F (xy) − yx)n = 0 for all n ≥ 1 and x, y ∈ I, an ideal of R.

For a1 =(

0 b0 c

)

and b1 =(

a 00 c

)

, F (x) �= a1x + xb1 for all x ∈ I,

showing that Theorem 2 is not true for arbitrary rings.

In all that follows we consider the case when (F (xy) − yx)n is either zeroor invertible for any x, y in the semiprime ring R. We assume there exist x0,y0 ∈ R such that (F (x0y0) − y0x0)

n �= 0.

Lemma 1. R is a simple ring.

126 A. Ali, S. Ali and V. De Filippis AEM

Proof. Let M be any non-zero two-sided ideal of R. If M �= R, then M doesnot contain invertible elements. Thus by our assumption (F (xy) − yx)n is anidentity in M . By Theorem 4 in [16], (F (xy) − yx)n

z = 0 for all x, y ∈ R andz ∈ M . In particular (F (x0y0) − y0x0)

nz = 0. Since 0 �= (F (x0y0) − y0x0)

n isan invertible element, we obtain z = 0, that is, M = 0 and R is simple. �Remark 1. Since R is simple with 1, it is a dense ring of linear transformationson a (right) vector space V over a division ring D. Our next purpose is to showthat R is a simple artinian ring, that is, R contains a non-zero transformationof finite rank (see p.75 in [11], Lemma 1.2.2 in [9]).

Lemma 2. R is an artinian ring.

Proof. We prove the statement by contradiction. Suppose in all that followsthat R is not artinian.

We divide the proof into a number of steps:

Step 1. Let v, w ∈ V such that rv = rw = 0 for some r ∈ R. Then d(r)v andd(r)w are linearly dependent over D.

Suppose that d(r)v and d(r)w are linearly independent over D. By thedensity of R, there exist s1, s2 ∈ R such that

s1d(r)v = 0, s2d(r)v = 0, s1d(r)w = w, s2d(r)w = w

Therefore we have

(a (s1d(r)s2r) + d (s1d(r)s2r) − s2rs1d(r))nv = 0

and also

(a (s1d(r)s2r) + d (s1d(r)s2r) − s2rs1d(r))nw = w.

This implies that

(a (s1d(r)s2r) + d (s1d(r)s2r) − s2rs1d(r))n

is neither zero nor invertible, a contradiction.

Step 2. Let v, w ∈ V be linearly independent such that rv = rw = 0 for somer ∈ R. Then d(r)v = d(r)w = 0.

At first suppose d(r)v �= 0. By the previous Lemma, there exist a, b ∈D, (a, b) �= (0, 0), such that

d(r)va + d(r)wb = 0 that is d(r)(va + wb) = 0

moreover (va+wb) �= 0, because v, w are linearly independent. By the densityof R there exist s1, s2 ∈ R such that

s1d(r)v = v, s2d(r)v = v.

Then

(a (s1d(r)s2r) + d (s1d(r)s2r) − s2rs1d(r))nv = v �= 0.

Vol. 82 (2011) Nilpotent and invertible values in semiprime rings 127

On the other hand

(a (s1d(r)s2r) + d (s1d(r)s2r) − s2rs1d(r))n (va + wb) = 0

that is,

(a (s1d(r)s2r) + d (s1d(r)s2r) − s2rs1d(r))n

is neither invertible nor zero, a contradiction.In the same way, we obtain a contradiction if we suppose d(r)w �= 0.

Step 3. Suppose v ∈ V such that rv = 0 for some r ∈ R. Then d(r)v = 0.

Suppose by contradiction d(r)v �= 0 and obviously v �= 0. Since R isnot artinian then R does not contain transformations of finite rank and sodimD(rV ) = ∞. Therefore, there exists w ∈ rV such that w, d(r)v are linearlyindependent over D.

Moreover, since rv = 0 and rw �= 0, v and w are linearly independent. Bythe density of R, there exists s ∈ R such that

s(rw) = 0, s(d(r)v) = v.

Since (sr)v = (sr)w = 0, by the previous Lemma we have 0 = d(sr)v =sd(r)v = v �= 0, a contradiction.

Step 4. If v ∈ V and r ∈ R then d(r)v = (vr)γr, where γ ∈ D is independentof the choice of v ∈ V .

We can assume rv �= 0, or else by the previous Lemma also d(r)v = 0.Suppose that d(r)v and rv are linearly independent. By the density of R,

there exists s ∈ R such that

s(rv) = 0, s(d(r)v) = v

and by the previous Lemma and (sr)v = 0 we have both d(s)rv = 0 and0 = d(sr)v = d(s)rv + sd(r)v = v �= 0, a contradiction.

Hence d(r)v and rv are linearly dependent over D: for all v ∈ V there existsγv ∈ D such that d(r)v = (rv)γv. Therefore, for any v, w ∈ V, r ∈ R

d(r)v = (rv)γv, d(r)w = (rw)γw, d(r)(v + w) = (rv + rw)γv+w.

Hence

0 = rv(γv − γv+w) + rw(γw − γv+w).

If rv and rw are linearly independent then γv = γw = γv+w. On the otherhand, if rv = λrw, of course there exists u ∈ V such that ru, rv are linearlyindependent as well as ru, rw. Hence, as we said above, γv = γu = γw. Thusd(r)v = γrv, for all v ∈ V, r ∈ R and γ is a fixed element of D, just dependingon the choice of r.Final Step.

Let v ∈ V, r ∈ R, then by the previous Lemma, we have d(r)v = (rv)γr.Left multiplying by r gives that rd(r)v = r2vγr.

128 A. Ali, S. Ali and V. De Filippis AEM

On the other hand, again by the previous Lemma, d(r)(rv) = r(rv)γr, thatis, d(r)rv = r2vγr.

The previous equalities say that [d(r), r]V = 0. For V is a faithful irreduc-ible left R-module, [d(r), r] = 0 for all r ∈ R. By Posner’s result on derivationsin [18], R must be commutative, unless d = 0. Since in the first case we getthe contradiction that R is a field and so it is artinian (because it is simple),we now assume d = 0, that is F (x) = ax for all x ∈ R.

We divide the last part of the proof into two cases:If there exists a non-zero right ideal � of R such that (a(xy) − yx)n = 0 for

all x, y ∈ �, then by Lemma 3 we have that R satisfies a non-trivial generalizedpolynomial identity. Since R is simple with 1, by Martindale’s Theorem [13],R is a finite dimensional central simple algebra, that is, R = Mk(D), the ringof all k × k matrices over a division ring D, which is a contradiction since weassume that R is not artinian.

Suppose that, for any � right ideal of R, (a(xy) − yx)n is not an identityin �.

Of course, for any non-zero right ideal � of R and for any x, y ∈ �,(a(xy)

− yx)n ⊆ a� + �.

Then a� + � contains invertible values. This implies a� + � = R. Let �1, �2

be non-zero right ideals of R, such that �1 ⊆ �2. We know that a�1 + �1 =R = a�2a+�2. Let c ∈ �2 −�1, so c = ap+q, where p, q ∈ �1. Moreover ap �= 0because c /∈ �1, and ap ∈ �2.

Since pR is a non-zero right ideal of R, pR+a(pR) = R and also a(pR) ⊆ �2.Therefore R = pR + a(pR) ⊆ �2, that is, �2 = R. It follows that either R is adivision ring or R is the ring of all 2 × 2 matrices over a division ring, whichis again a contradiction since we suppose that R is not artinian. �

Theorem 1. Let R be a semiprime ring and F be a generalized derivation ofR. If for a fixed integer n ≥ 1 (F (xy) − yx)n is either zero or invertible forall x, y ∈ R, then there exists a division ring D such that either R = D orR = M2(D), the 2 × 2 matrix ring over D.

Proof. The previous Lemma shows that R is a simple artinian ring, that is,R = Mk(D) is a ring of k × k matrices over a division ring D. Our aim is toprove that k ≤ 2. Hence suppose by contradiction that k ≥ 3.

As above we write F (x) = ax + d(x), for suitable a ∈ R and derivation dof R. By [17] there exists a derivation δ : D −→ D and a matrix A ∈ Mk(D)such that d = dA + δ, where dA is the inner derivation induced by A, thatis, dA(x) = Ax − xA, for all x ∈ Mk(D), and δ : Mk(D) −→ Mk(D) isthe derivation induced by δ which acts as follows: if B ∈ Mk(D), with B =∑

i,j βijeij , βij ∈ D and eij the usual matrix unit, then δ(B) =∑

i,j δ(βij)eij .For x = eii and y = eij (i �= j), we have that

(a(xy) + d(xy) − yx)n = (aeij + [A, eij ] + δ(α)eij)n

Vol. 82 (2011) Nilpotent and invertible values in semiprime rings 129

has rank ≤ 2, thus it is not invertible in R. Therefore

(aeij + [A, eij ] + δ(α)eij)n = 0

and right multiplying by eij it follows that (−eijA)neij = 0, that is, A is adiagonal matrix in R, say A =

∑i αieii, αi ∈ D.

Moreover, if χ is a Z(D)-automorphism of Mk(D), then the generalizedderivation Fχ(x) = (χaχ−1)x + (χdχ−1)(x) satisfies the same condition of F ,that is, (Fχ(xy) − yx)n is either zero or invertible for all x, y ∈ R.

For k ≥ t ≥ 3 and b ∈ D, let ϕ = ϕt,b be the automorphism of Mk(D)defined by ϕ(x) = (1 − be1t)x(1 + be1t). Then ϕ−1(x) = (1 + be1t)x(1 − be1t)and

dϕ(x) = ϕdA(ϕ−1(x)) + ϕδ(ϕ−1(x))

= ϕ(Aϕ−1(x) − ϕ−1(x)A

)

+ϕ(δ(b)e1tx(1 − be1t) + (1 + be1t)δ(x)(1 − be1t)

+(1 + be1t)x(−δ(b)e1t))

= ϕ(A)x − xϕ(A) + (1 − be1t)δ(b)e1tx + δ(x)+x(−δ(b)e1t)(1 + be1t)

= ϕ(A)x − xϕ(A) + δ(b)e1tx − xδ(b)e1t + δ(x)

= dB(x) + δ(x), where B = ϕ(A) + δ(b)e1t.

Therefore, as above, B must be a diagonal matrix. Since A =∑

i αieii, weobtain

(−bαt + α1b + δ(b)) e1t = 0 for any b ∈ D and k ≥ t ≥ 2.

In particular set b = 1, then α1 = αt for any t, that is, A is a scalar matrix,A = αIk, (α = α1). Therefore δ(b) = −(αb− bα), which implies δ = −dαIk

andd = dA + δ = dαIk

− dαIk= 0.

Hence, if k ≥ 3 then F (x) = ax. As above for x = eii and y = eij (i �= j),we have that (a(xy) − yx)n = (aeij)

n has rank ≤ 2, thus it is not invertible inR. Therefore (aeij)

n = 0. Left multiplication by eij yields that eij(aeij)n = 0,that is, a is a diagonal matrix in R, say a =

∑i γieii, γi ∈ D.

Let now x = eij and y = eji (i �= j). Therefore, we have (a(xy) − yx)n =(aeii − ejj)

n = γni eii + (−1)nejj , which is neither zero nor invertible since

k ≥ 3.This contradiction says that k ≤ 2 and we are done. �

Also here we provide an example showing that the above result does nothold for arbitrary rings:

130 A. Ali, S. Ali and V. De Filippis AEM

Example 3. Let D be a division ring and R ={(

a b0 c

)

| a, b, c ∈ D

}

. Define

F : R −→ R by

F

(a b0 c

)

=(

a 00 0

)

.

Then F is a generalized derivation with associated derivation d given by

d

(a b0 c

)

=(

0 −b0 0

)

. It can be easily seen that for x1 =(

a1 b1

0 c1

)

and y1 =(

a2 b2

0 c2

)

one has

[F (x1y1) − y1x1]n =

[

F

((a1 b1

0 c1

)(a2 b2

0 c2

))

−(

a2 b2

0 c2

) (a1 b1

0 c1

)]n

=(

(a1a2 − a2a1) −(a2b1 + b2c1)0 −c2c1

)n

�= 0.

Since a1a2 − a2a1 �= 0 and c2c1 �= 0 it follows that the matrix(

(a1a2 − a2a1) −(a2b1 + b2c1)0 −c2c1

)

is invertible in R. Hence(

(a1a2 − a2a1) −(a2b1 + b2c1)0 −c2c1

)n

is invertible in R, for n ≥ 1. Also there exist elements x0 =(

0 a0 0

)

and

y0 =(

a b0 c

)

in R such that (F (x0y0) − y0x0)n = 0 for n ≥ 1. However,

neither R is division ring nor R = M2(D).

3. Set with nilpotent values on right ideals

Finally we consider the case when the generalized derivation F acts on theelement of a right ideal of a prime ring R. In all that follows we denote by Qthe Martindale quotient ring of R. We begin with:

Lemma 3. Let R be a prime ring and � a non-zero right ideal of R. If(F (xy)

− yx))n = 0 for any x, y ∈ �, then R is a GPI-ring.

Proof. By Theorem 3 in [15] every generalized derivation g on a dense rightideal of R can be uniquely extended to the Utumi quotient ring U of R, andthus we can think of any generalized derivation of R to be defined on the wholeof U and to be of the form F (x) = ax+d(x) for some a ∈ U and d a derivationon U . Assume that R is not commutative, otherwise we conclude trivially that

Vol. 82 (2011) Nilpotent and invertible values in semiprime rings 131

R is a GPI-ring. Moreover, since R is not commutative, there exists x0 ∈ �−C.Suppose now that R does not satisfy any non-trivial generalized polynomialidentity.

Notice that in the case d = 0, we have the contradiction that R satisfies thenon-trivial GPI (ax0Xx0Y − x0Y x0X)n. Then assume d �= 0. Suppose that dis an inner derivation, d = ad(b), for some b ∈ Q, d(x) = [b, x], for all x ∈ Q.Since d �= 0, then is b /∈ C. Thus

(a(x0X1x0X2) + [b, x0X1x0X2] − x0X2x0X1)n

= (a(x0X1x0X2) + bx0X1x0X2 − x0X1x0X2b − x0X2x0X1)n

.

Since R is not a GPI-ring, then R satisfies

(a(x0X1x0X2) + bx0X1x0X2 − x0X1x0X2b − x0X2x0X1)n−1

x0X1x0X2b

and again since we suppose that R does not satisfy any non-trivial generalizedpolynomial identity, it follows that

(a(x0X1x0X2) + bx0X1x0X2 − x0X1x0X2b − x0X2x0X1)n−1

is trivially satisfied by R. By repeating the same process a number of timeswe’ll obtain that

((a + b)(x0X1x0X2) − x0X1x0X2b − x0X2x0X1)

is a generalized polynomial identity for R, moreover it is non-trivial sinceb /∈ C, a contradiction.

Suppose now that d is not an inner derivation of R. If for all r ∈ �, d(r) ∈rC, then [d(r), r] = 0, that is either d = 0 or R is commutative (see [18]),in any case we get a contradiction. Therefore, there exists x0 ∈ � such thatd(x0) /∈ x0C. Thus

(a(x0X1x0X2) + d(x0)X1x0X2 + x0d(X1)x0X2

+x0X1d(x0)X2 + x0X1x0d(X2) − x0X2x0X1)n

is a generalized differential identity for R, and by Kharchenko’s theorem in[12], we have that

(a(x0X1x0X2) + d(x0)X1x0X2 + x0X3x0X2

+x0X1d(x0)X2 + x0X1x0X4 − x0X2x0X1)n

is satisfied by R. In particular for X2 = 0,

(x0X1x0X4)n

is satisfied by R, a contradiction again. �

We conclude the paper with the following:

132 A. Ali, S. Ali and V. De Filippis AEM

Theorem 2. Let R be a prime ring admitting a generalized derivation F andI be a nonzero right ideal of R. If for a fixed integer n ≥ 1 (F (xy) − yx)n = 0for all x, y ∈ I, then [I, I]I = 0 and F (x) = ax + xb and also there existα, β ∈ C, the extended centroid of R, such that (a−α)I = 0 and (b−β)I = 0.Moreover ((a + b)x − x)I = 0 for all x ∈ I.

Proof. Suppose first that [x1, x2]x3 is not an identity for I. We proceed toderive a contradiction. Since by Lemma 3 R is a GPI ring, so is also U (see[3] and [5]). By [13] Q is a primitive ring with H = Soc(RC) �= 0, moreoverwe may assume that [x1, x2]x3 is not an identity for IH, otherwise it shouldbe an identity also for �Q, which is a contradiction. Let a1, a2, a3 ∈ IH suchthat [a1, a2]a3 �= 0. Since H is a regular ring, there exists e2 = e ∈ H suchthat eH = a1H + a2H + a3H, e ∈ eH, and ai = eai for all i = 1, 2, 3.

We start as above from the expression (a(xy) + d(xy) − yx)n = 0 for allx, y ∈ I.

Suppose first that d is an inner derivation of Q induced by q ∈ Q, thatis, F (x) = (a + q)x + x(−q), and denote a + q = a′,−q = q′. Thus we have(a′(xy) + (xy)q′ − yx)n = 0 for all x, y ∈ I. Let x = ex1 and y = ey1(1 − e)for x, y ∈ R. By calculations we have that

0=(1−e) (a′(ex1ey1(1−e))+(ex1ey1(1 − e))q′)n =(1−e) (a′(ex1ey1(1−e)))n

which implies (1 − e)ae = 0, that is, ae = eae, and ax ∈ eH for all x ∈eH. Therefore F (x) ∈ eH for all x ∈ eH, that is, the generalized derivationF induces another F , which is defined in the prime ring IH = IH

IH∩lH(IH) ,

where lh(IH) is the left annihilator in H of IH, and F (x) = F (x) for allx ∈ IH. Moreover, we obviously have that

(F (xy) − yx

)n= 0 is a differen-

tial identity for IH. So, by Proposition 1, we have that IH is commutative,that is, [IH, IH]IH = 0. But this is a contradiction since 0 �= [a1, a2]a3 =[ea1, ea2]ea3 = [ea1e, ea2e]ea3.

Therefore [I, I]I = 0 and in this case, the Proposition and the proof ofTheorem 6 of [14] show that there exists an idempotent element h ∈ Soc(RC)such that CI = hRC. By repeating the argument above, we may prove that(1 − h)ah = 0, that is ah = hah, and ax ∈ hR for all x ∈ hR. ThereforeF (x) ∈ hR for all x ∈ hR, that is, the generalized derivation F induces anotherF , which is defined in the prime ring hR = hR

hR∩lR(hR) , and F (x) = F (x) forall x ∈ hR. By applying Proposition 1 again to the ring hR = hR

hR∩lR(hR) , wehave that F is the identity map on hR, that is, ((a + b)x − x) = 0 in hR forall x ∈ hR, that is, ((a + b)x − x)y = 0 for all x, y ∈ hR; and also [a, x] = 0 inhR and [b, x] = 0 in hR, that is, [a, I]I = 0 and [b, I]I = 0.

Let x, y ∈ I, r ∈ R. Then [a, xr]y = 0, whence axry − xray = 0, that is,ax = αxx, with αx ∈ C, and analogously ay = αyy, a(x + y) = αx+y(x + y).From this, it is easy to see that αx is independent of the choice of x ∈ I,

Vol. 82 (2011) Nilpotent and invertible values in semiprime rings 133

therefore there exists β ∈ C such that (a − α)I = 0 (and analogously one canprove (b − β)I = 0, for β ∈ C).

Suppose now that d is not an inner derivation of R. Let x0 ∈ I − C. Since

(a(x0X1x0X2) + d(x0)X1x0X2 + x0d(X1)x0X2 + x0X1d(x0)X2

+x0X1x0d(X2) − x0X2x0X1)n

is a generalized differential identity for R, and by Kharchenko’s theorem in[12], we have that

(a(x0X1x0X2) + d(x0)X1x0X2 + x0X3x0X2 + x0X1d(x0)X2

+x0X1x0X4 − x0X2x0X1)n

is satisfied by R. In particular for X2 = 0,

(x0X1x0X4)n

is satisfied by R, a contradiction. �

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Asma Ali and Shakir AliDepartment of MathematicsAligarh Muslim UniversityAligarh 202002Indiae-mail: asma ali2@rediffmail.come-mail: shakir50@rediffmail.com

Vincenzo de FilippisDI.S.I.A.Faculty of EngineeringUniversity of MessinaContrada Di Dio98166MessinaItalye-mail: defilippis@unime.it

Received: October 7, 2010

Revised: November 24, 2010