DOI: 10.1007/s12215-008-0029-0Rendiconti del Circolo Matematico di Palermo 57, 401 – 410 (2008)

Basudeb Dhara

Derivations with power central values on multilinearpolynomials in prime rings

Received: May 31, 2008 / Accepted: October 24, 2008 – c© Springer-Verlag 2008

Abstract. Let R be a prime ring of char R �= 2 with center Z(R) and withextended centroid C, d a nonzero derivation of R and f (x1, . . .,xn) a nonzeromultilinear polynomial over C. Suppose that xsd(x)xt ∈ Z(R) for allx ∈ {d( f (x1, . . . ,xn))|x1, . . .,xn ∈ ρ}, where ρ is a nonzero right ideal of Rand s ≥ 0, t ≥ 0 are fixed integers. If d(ρ)ρ �= 0, then ρC = eRC for someidempotent e in the socle of RC and f (x1, . . . ,xn)

N is central-valued in eRCe,where N = s+ t+1.

Keywords Prime ring · Derivation · Extended centroid ·Martindale quotientring

Mathematics Subject Classification (2000) 16W25 · 16R50 · 16N60

1 Introduction

Throughout this paper R always denotes an associative prime ring with centerZ(R) and extended centroid C. The Lie commutator of x,y is denoted by [x,y]and defined by [x,y] = xy−yx for x,y ∈ R. In all that follows, we denote by da nonzero derivation of R.

In [11], Giambruno and Herstein proved that if d(x)n = 0 for all x ∈ R, thend = 0. In [12], Herstein proved that if d is nonzero and d(x)n ∈ Z(R) for allx ∈ R, then R satisfies S4, the standard identity in four variables. Bergen andCarini [3] studied the case for a noncentral Lie ideal. They proved that if charR �= 2 and d is nonzero such that d(u)n ∈ Z(R) for all u in a noncentral Lieideal of R, then R satisfies S4. Note that a noncentral Lie ideal of R containsall the commutators [x1,x2] for x1,x2 in some nonzero ideal of R except when

B. Dhara (B)Department of Mathematics, Belda College, Belda, Paschim Medinipur-721424(W.B.), IndiaE-mail: basu−dhara@yahoo.com

402 B. Dhara

char R= 2 and R satisfies S4(see [4]). So it is natural to consider the situationwhen d(x)n ∈ Z(R) for all commutators x= [x1,x2] or in the more general casewhen x = f (x1, . . . ,xn), where f (x1, . . .,xn) is a multilinear polynomial overC. In [21], Wong consider this situation and he proved that if d is nonzeroderivation such that d(x)n ∈ Z(R) for all x ∈ { f (x1, . . .,xn)|x1, . . .,xn ∈ R},then either f (x1, . . . ,xn) is central-valued on R or R satisfies S4. In [6], Changand Lin proved that if ρ is a nonzero right ideal of R, n a fixed positive integersuch that

(i) d(x)xn = 0 for all x ∈ ρ , then d(ρ)ρ = 0;(ii) xnd(x) = 0 for all x∈ ρ , then d= 0 unless R∼=M2(F), the 2×2 matrices

over a field F of two elements.In the present paper, we consider the situations when xsd(x)xt ∈ Z(R) for

all x ∈ {d( f (x1, . . . ,xn))|x1, . . .,xn ∈ R} and when xsd(x)xt ∈ Z(R) for allx ∈ {d( f (x1, . . . ,xn))|x1, . . .,xn ∈ ρ}, where ρ is a nonzero right ideal of R.

To prove the theorems we need some notations concerning quotient rings.Denote by Q the two-sided Martindale quotient ring of R and by C the cen-ter of Q, which is called the extended centroid of R. Note that Q is also aprime ring with C a field. We will make a frequent use of the following nota-tion: f (x1, . . .,xn) = x1x2 . . .xn+ ∑

I �=σ∈Sn

ασ xσ (1) . . .xσ (n) for some ασ ∈C and

we denote by f d(x1, . . .,xn) the polynomial obtained from f (x1, . . .,xn) byreplacing each coefficient ασ with d(ασ .1). Thus we write

d( f (x1, . . . ,xn)) = f d(x1, . . . ,xn)+∑i

f (x1, . . .,d(xi), . . .,xn)

and

d2( f (x1, . . . ,xn)) = d( f d(x1, . . .,xn))+d(∑i

f (x1, . . .,d(xi), . . .,xn))

= f d2(x1, . . . ,xn)+∑

if d(x1, . . . ,d(xi), . . .,xn)

+∑i

f d(x1, . . .,d(xi), . . .,xn)+ ∑i�= j

f (x1, . . . ,d(xi), . . . ,d(x j), . . . ,xn)

+∑i

f (x1, . . . ,d2(xi), . . . ,xn)

= f d2(x1, . . . ,xn)+2∑

if d(x1, . . . ,d(xi), . . .,xn)

+2 ∑i< j

f (x1, . . .,d(xi), . . .,d(x j), . . .,xn)+∑i

f (x1, . . . ,d2(xi), . . .,xn).

2 The case for ρ = R

We begin with a matrix ring case.

Derivations with power central values on multilinear polynomials in prime rings 403

Lemma 1 Let R=Mk(F), the k×k matrix algebra over the field F of charac-teristic different from 2. Suppose that a ∈ R and f (x1, . . . ,xn) is a multilinearpolynomial over F such that

[a, f (x1, . . . ,xn)]s[a, [a, f (x1, . . . ,xn)]][a, f (x1, . . .,xn)]

t ∈ Z(R)

for all x1, . . .,xn ∈ R, where s≥ 0, t ≥ 0 are fixed integers. Then either a is ascalar matrix or f (x1, . . .,xn) is central-valued on R.

Proof If k = 1, the result holds trivially. So assume that k ≥ 2. We assumefurther that char F �= 2 and then proceed to show that a ∈ F if f (x1, . . . ,xn) isnot central-valued on R. Thus let f (x1, . . .,xn) be noncentral valued in R.

Since f (x1, . . . ,xn) is not central on R, by [18, Lemma 2, Proof of Lemma 3]there exists a sequence of matrices r=(r1, . . ., rn) in R such that f (r1, . . ., rn) =γei j with 0 �= γ ∈ F and i �= j. Since the set f (R) = { f (x1, . . .,xn),xi ∈ R} isinvariant under the action of all inner automorphisms of R, for all i �= j thereexists a sequence of matrices r= (r1, . . ., rn) such that f (r) = γei j. Thus

[a, f (r1, . . ., rn)]s[a, [a, f (r1, . . . , rn)]][a, f (r1, . . ., rn)]

t

= [a,γei j]s[a, [a,γei j]][a,γei j]

t ∈ Z(R).

Commuting both sides with ei j we get

[[a,γei j]s[a, [a,γei j]][a,γei j]

t,ei j] = 0.

Left multiplying by ei j we get

0= ei j[[a,γei j]s[a, [a,γei j]][a,γei j]

t,ei j] = (−1)t+12γ s+t+1(a ji)s+t+2ei j.

This implies that a ji = 0 for any i �= j. Thus a is a diagonal matrix. Now forany F-automorphism θ of R, aθ satisfies the same property as a does, namely,

[aθ , f (x1, . . .,xn)]s[aθ , [aθ , f (x1, . . .,xn)]][a

θ, f (x1, . . . ,xn)]t = 0

for all x1, . . .,xn ∈ R. Hence, aθ must be diagonal. Write, a =k∑

i=0aiieii; then

for s �= t, we have

(1+ets)a(1−ets) =k

∑i=0

aiieii+(ass−att )ets

diagonal. Hence, ass = att and so a is a scalar matrix. ��Theorem 1 Let R be a prime ring of characteristic different from 2 andf (x1, . . .,xn) a multilinear polynomial over C. Suppose that d is a nonzeroderivation on R such that

d( f (x1, . . . ,xn))sd2( f (x1, . . .,xn))d( f (x1, . . . ,xn))

t ∈ Z(R)

for all x1, . . .,xn ∈ R, where s≥ 0, t ≥ 0 are fixed integers. Then f (x1, . . . ,xn)N

is central-valued on R, where N = s+ t+1.

404 B. Dhara

Proof On contrary, assume that f (x1, . . .,xn)N is not central-valued on R. As-

sume next that d is Q-inner derivation, i.e., d(x) = [a,x] for all x ∈ R and forsome a ∈ Q. Since d is nonzero, a /∈C. Let

g(x1, . . . ,xn+1) = [[a, f (x1, . . . ,xn)]s[a, [a, f (x1, . . . ,xn)]][a, f (x1, . . . ,xn)]

t,xn+1].

Thus R satisfies the generalized polynomial identity

g(x1, . . . ,xn+1) = [[a, f (x1, . . . ,xn)]s[a, [a, f (x1, . . . ,xn)]][a, f (x1, . . . ,xn)]

t,xn+1].

Since f (x1, . . .,xn) is not central-valued on R and a /∈ C, this is a nontrivialGPI. By Chuang [7], this GPI is also satisfied by Q. In case C is infinite, wehave g(r1, . . ., rn+1) = 0 for all r1, . . . , rn+1 ∈Q⊗C C, where C is the algebraicclosure of C. Since both Q and Q⊗C C are centrally closed [9, theorem 2.5and 3.5] we may replace R by Q or Q⊗C C according as C is finite or infinite.Thus we may assume that R is centrally closed over C which is either finite oralgebraically closed and g(r1, . . ., rn+1) = 0 for all r1, . . . , rn+1 ∈ R. By Mar-tindale’s theorem [19], R is a primitive ring having nonzero socle H with Cas the associated division ring. In light of Jacobson’s theorem [13, p. 75], Ris isomorphic to a dense ring of linear transformations on a vector space Vover C. Now, if V is finite dimensional over C, then the density of R on Vimplies that R∼=Mk(C) with k =dimCV . By Lemma 1, we have f (x1, . . .,xn)is central-valued on R.

If V is infinite dimensional over C, then as in lemma 2 in [21], the set f (R)is dense on R and so from

[a, f (r1, . . . , rn)]s[a, [a, f (r1, . . . , rn)]][a, f (r1, . . ., rn)]

t ∈ Z(R)

for all r1, . . ., rn ∈ R, we have

[a, r]s[a, [a, r]][a, r]t ∈ Z(R)

for all r ∈ R. Let e and f be two orthogonal idempotent element of H that ise f = 0. Then for all r ∈ H, we have

0= [[a,er f ]s[a, [a,er f ]][a,er f ]t,er f ].

Left multiplying by f , we get

0 = f [[a,er f ]s[a, [a,er f ]][a,er f ]t,er f ]

= f [a,er f ]s[a, [a,er f ]][a,er f ]ter f

= f (aer f )s(−2aer f a)(er f a)ter f .

Since char R �= 2, this implies ( f aer)s+t+3= 0 for all r ∈H. By [10], it followsthat f aer = 0 for any r ∈ H, implying f ae = 0. In particular, for any idem-potent e ∈ H, we have (1− e)ae= 0= ea(1− e) that is [a,e] = 0. Therefore,[a,E] = 0, where E is the additive subgroup generated by all idempotents ofH. Since E is a non central Lie ideal of H, this implies a ∈C, a contradiction.

Derivations with power central values on multilinear polynomials in prime rings 405

Assume next that d is not Q-inner. Then R satisfies the differential identity(

f d(r1, . . . , rn)+∑i

f (r1, . . . ,d(ri), . . ., rn)

)s(f d2(r1, . . ., rn)

+2∑i

f d(r1, . . . ,d(ri), . . . , rn)+2 ∑i< j

f (r1, . . . ,d(ri), . . .,d(r j), . . . , rn)

+∑i

f (r1, . . .,d2(ri), . . ., rn)

)(f d(r1, . . . , rn)+∑

if (r1, . . .,d(ri), . . ., rn)

)t

∈ Z(R).

By Kharchenko’s [14] theorem, R satisfies the polynomial identity(

f d(r1, . . . , rn)+∑i

f (r1, . . . ,xi, . . ., rn)

)s(f d2(r1, . . ., rn)

+2∑i

f d(r1, . . . ,xi, . . ., rn)+2 ∑i< j

f (r1, . . . ,xi, . . .,x j, . . . , rn)

+∑i

f (r1, . . . ,yi, . . . , rn)

)(f d(r1, . . ., rn)+∑

if (r1, . . . ,xi, . . . , rn)

)t

∈ Z(R).

In particular, for r1 = x2 = x3 = · · ·= xn = 0, we have, R satisfies

f (x1, . . ., rn)s(2 f d(x1, . . . , rn)+ f (y1, . . . , rn)) f (x1, . . . , rn)

t ∈ Z(R)

which implies R satisfies the blended component

f (x1, . . . , rn)s f (y1, . . ., rn) f (x1, . . ., rn)

t ∈ Z(R).

In particular, we have that R satisfies

f (r1, . . ., rn)s+t+1 ∈ Z(R)

that is f (r1, . . ., rn)N is central-valued in R. ��

3 The case for ρ

We need the following lemmas:

Lemma 2 Let ρ be a nonzero right ideal of R and d a derivation of R. Thenthe following conditions are equivalent:

(i) d is an inner derivation induced by some b ∈ Q such that bρ = 0;(ii) d(ρ)ρ = 0.

For its proof, we refer to [5, Lemma].

406 B. Dhara

Lemma 3 Let ρ be a nonzero right ideal of R with char R �= 2, d a nonzeroderivation of R and f (x1, . . .,xn) a nonzero multilinear polynomial over Csuch that d( f (x1, . . . ,xn))

sd2( f (x1, . . .,xn))d( f (x1, . . . ,xn))t ∈ Z(R) for all

x1, . . . ,xn ∈ ρ , where s ≥ 0, t ≥ 0 are fixed integers. Then either d(ρ)ρ = 0or R satisfies nontrivial generalized polynomial identity.

Proof Suppose, on the contrary, that R does not satisfy any nontrivial gener-alized polynomial identity. Then R is noncommutative, otherwise R satisfiestrivially a nontrivial GPI, a contradiction. Now we consider the following twocases:Case I. Suppose that d is a Q-inner derivation induced by an element a ∈ Q.Then for any x ∈ ρ ,

[[a, f (xX1, . . .,xXn)]s[a, [a, f (xX1, . . .,xXn)]][a, f (xX1, . . .,xXn)]

t,xXn+1]

is a GPI for R, so it is the zero element in Q ∗C C{X1,X2, . . .,Xn+1}. DenotelR(ρ) the left annihilator of ρ in R. Suppose first that {1,a,a2} are linearlyC-independent modulo lR(ρ), that is (αa2+β a+ γ)ρ = 0 if and only if α =β = γ = 0. Since R is not a GPI-ring, a fortiori it cannot be a PI-ring. Thus,by [16, Lemma 3] there exists x0 ∈ ρ such that {a2x0,ax0,x0} are linearlyC-independent. In this case, we have that

[[a, f (x0X1, . . . ,x0Xn)]s[a, [a, f (x0X1, . . .,x0Xn)]]

[a, f (x0X1, . . .,x0Xn)]t,x0Xn+1] = 0. (1)

In this expansion the term

(a f (x0X1, . . . ,x0Xn))s(−2a f (x0X1, . . . ,x0Xn)a)( f (x0X1, . . . ,x0Xn)a)

tx0Xn+1

appears nontrivially which is a contradiction.Therefore, {1,a,a2} are linearly C-dependent modulo lR(ρ), that is there

exist α ,β ,γ ∈ C, not all zero, such that (αa2+β a+ γ)ρ = 0. Suppose thatα = 0. Then β �= 0, otherwise γ = 0. Thus by (β a+ γ)ρ = 0, we have that(a+β−1γ)ρ = 0. Since a and a+β−1γ induce the same inner derivation, wemay replace a by a+β−1γ in the basic hypothesis. Therefore, in any case wemay suppose aρ = 0 and then by Lemma 2, d(ρ)ρ = 0.

Next suppose that α �= 0. In this case there exist λ ,μ ∈C such that a2x0 =λ ax0 + μx0 for all x0 ∈ ρ . If ax0 and x0 are linearly C-dependent for allx0 ∈ ρ , then again we get aρ = 0 and hence by Lemma 2, d(ρ)ρ = 0, end-ing the proof. Therefore, choose x0 ∈ ρ such that ax0 and x0 are linearly C-independent. Then (1) implies that

(a f (x0X1, . . .,x0Xn))s(a2 f (x0X1, . . . ,x0Xn)−2a f (x0X1, . . .,x0Xn)a

+ f (x0X1, . . . ,x0Xn)a2)( f (x0X1, . . .,x0Xn)a)

tx0Xn+1 = 0. (2)

Derivations with power central values on multilinear polynomials in prime rings 407

Replacing a2x0 with λ ax0+μx0 in (2), we get that R satisfies

(a f (x0X1, . . . ,x0Xn))s((λ a+μ) f (x0X1, . . .,x0Xn)−2a f (x0X1, . . . ,x0Xn)a

+ f (x0X1, . . .,x0Xn)(λ a+μ))( f (x0X1, . . . ,x0Xn)a)tx0Xn+1 = 0. (3)

This gives

(a f (x0X1, . . .,x0Xn))s(−2a f (x0X1, . . .,x0Xn)a)

( f (x0X1, . . . ,x0Xn)a)tx0Xn+1 = 0 (4)

which is a nontrivial GPI, because ax0 and x0 are linearly C-independent, acontradiction.Case II. Next suppose that d is an outer derivation. If for all x ∈ ρ , d(x) ∈xC, then [d(x),x] = 0 which implies that R is commutative (see [2]). There-fore there exists x ∈ ρ such that d(x) /∈ xC i.e., x and d(x) are linearly C-independent. By our assumption, we have that R satisfies

[( f d(xX1, . . .,xXn)+∑

if (xX1, . . .,d(x)Xi+xd(Xi), . . .,xXn))

s

(f d2(xX1, . . . ,xXn)+2∑

if d(xX1, . . . ,d(x)Xi+xd(Xi), . . .,xXn)

+2 ∑i< j

f (xX1, . . .,d(x)Xi+xd(Xi), . . . ,d(x)Xj+xd(Xj), . . . ,xXn)

+∑i

f (xX1, . . .,d2(x)Xi+2d(x)d(Xi)+xd2(Xi), . . . ,xXn)

)

( f d(xX1, . . .,xXn)+∑i

f (xX1, . . .,d(x)Xi+xd(Xi), . . .,xXn))t,xXn+1

].

By Kharchenko’s theorem [14],

[( f d(xX1, . . . ,xXn)+∑

if (xX1, . . . ,d(x)Xi+xri, . . .,xXn))

s

(f d2(xX1, . . .,xXn)+2∑

if d(xX1, . . . ,d(x)Xi+xri, . . . ,xXn)

+ ∑i< j

f (xX1, . . .,d(x)Xi+xri, . . .,d(x)Xj+xr j, . . . ,xXn)

+∑i

f (xX1, . . . ,d2(x)Xi+2d(x)ri+xsi, . . . ,xXn)

)

( f d(xX1, . . . ,xXn)+∑i

f (xX1, . . .,d(x)Xi+xri, . . .,xXn))t,xXn+1

]= 0

408 B. Dhara

for all X1, . . .,Xn, r1, . . . , rn, s1, . . . , sn ∈ R. In particular, for X1 = r2 = . . . =rn = s1 = 0,

[ f (xr1, . . .,xXn)s(

2 f d(xr1, . . . ,xXn)+2 ∑j≥2

f (xr1, . . . ,d(x)Xj, . . . ,xXn)

+ f (2d(x)r1, . . .,xXn)

)f (xr1, . . . ,xXn)

t,xXn+1] = 0

which is a nontrivial GPI for R, because x and d(x) are linearly C-independent,a contradiction. ��Theorem 2 Let R be a prime ring of char R �= 2 with center Z(R) and withextended centroid C, d a nonzero derivation of R and f (x1, . . .,xn) a nonzeromultilinear polynomial over C such that

d( f (x1, . . .,xn))sd2( f (x1, . . . ,xn))d( f (x1, . . .,xn))

t ∈ Z(R)

for all x1, . . . ,xn in some nonzero right ideal ρ of R, where s≥ 0, t≥ 0 are fixedintegers. If d(ρ)ρ �= 0, then ρC = eRC for some idempotent e in the socle ofRC and f (x1, . . .,xn)

N is central-valued in eRCe, where N = s+ t+1.

Proof If [ f (ρ),ρ ]ρ = 0, that is [ f (x1, . . .,xn),xn+1]xn+2 = 0 for allx1,x2, . . .,xn+2 ∈ ρ , then by [15, Proposition], ρC= eRC for some idempotente ∈ soc(RC) and f (x1, . . . ,xn) is central-valued on eRCe and so f (x1, . . . ,xn)

N

is central-valued on eRCe.So, assume that [ f (ρ),ρ ]ρ �= 0, that is [ f (x1, . . . ,xn),xn+1]xn+2 is not an

identity for ρ . By Lemma 3, R is a prime GPI-ring and so is Q (see [1]and [7]). Since Q is centrally closed over C, it follows from [19] that Q isa primitive ring with H = Soc(Q) �= 0. Then f (ρH)ρH �= 0. For otherwise,f (ρQ)ρQ= 0 by [1] and [7], a contradiction. Choose a1, . . .,an+2 ∈ ρH suchthat [ f (a1, . . .,an),an+1]an+2 �= 0. Let a ∈ ρH. Since H is a regular ring, thereexists e2 = e ∈ H such that eH = aH + a1H + · · ·+ an+2H. Then e ∈ ρHand a = ea, ai = eai for i= 1, . . .,n+2. Thus, we have f (eHe) = f (eH)e �=0. By our assumption and by [17, Theorem 2], we may also assume that[d( f (x1, . . . ,xn))

sd2( f (x1, . . . ,xn))d( f (x1, . . . ,xn))t,xn+1] is an identity for ρQ.

In particular, [d( f (x1, . . . ,xn))sd2( f (x1, . . . ,xn))d( f (x1, . . . ,xn))

t,xn+1] is anidentity for ρH and so for eH. It follows that, for all r1, . . . , rn+1 ∈ H,

0= [d( f (er1, . . . ,ern))sd2( f (er1, . . . ,ern))d( f (er1, . . . ,ern))

t,ern+1(1−e)].

We may write f (x1, . . . ,xn) = t(x1, . . . ,xn−1)xn+h(x1, . . .,xn), where xn neverappears as last variable in any monomials of h and t(x1, . . . ,xn−1) is a suitablemultilinear polynomial in n−1 variables. Let r ∈ H. Then replacing rn withr(1−e) we have

0= [d(t(er1, . . . ,ern−1)er(1−e))sd2(t(er1, . . . ,ern−1)er(1−e))

.d(t(er1, . . . ,ern−1)er(1−e))t,ern+1(1−e)]. (5)

Derivations with power central values on multilinear polynomials in prime rings 409

Now we know the fact that d(x(1− e))e = −x(1− e)d(e), (1− e)d(ex) =(1−e)d(e)ex and thus

(1−e)d2(ex(1−e))e = (1−e)d

{d(e)ex(1−e)+ed(ex(1−e))

}e

= (1−e)d(e)d(ex(1−e))e+(1−e)d(e)d(ex(1−e))e

= −2(1−e)d(e)ex(1−e)d(e).

Thus left multiplying by (1−e), (5) gives

0 = (1−e)d(t(er1, . . . ,ern−1)er(1−e))sd2(t(er1, . . .,ern−1)er(1−e))

d(t(er1, . . . ,ern−1)er(1−e))tern+1(1−e)

= (1−e){d(e)t(er1, . . .,ern−1)er(1−e)}sd2(t(er1, . . . ,ern−1)er(1−e))

{−t(er1, . . . ,ern−1)er(1−e)d(e)}trn+1(1−e)

= (1−e){d(e)t(er1, . . .,ern−1)er(1−e)}s

(−2d(e)t(er1, . . .,ern−1)er(1−e)d(e))

{−t(er1, . . . ,ern−1)er(1−e)d(e)}trn+1(1−e)

= (−1)t+12{(1−e)d(e)t(er1, . . .,ern−1)er}N(1−e)d(e)rn+1(1−e).

Since char R �= 2, this gives replacing rn+1 with t(er1, . . . ,ern−1)er that

0= {(1−e)d(e)t(er1, . . .,ern−1)er}N+1(1−e)

which implies

0= {(1−e)d(e)t(er1e, . . .,ern−1e)H}N+2.

By [10], (1−e)d(e)t(er1e, . . .,ern−1e) = 0 for all r1, . . . , rn−1 ∈H. Since eHeis a simple Artinian ring and t(eHe) �= 0 is invariant under the action of allinner automorphisms of eHe, by [8, Lemma 2], (1−e)d(e) = 0 and so d(e) =ed(e) ∈ eH. Thus d(eH) ⊆ d(e)H + ed(H) ⊆ eH ⊆ ρH andd(a) = d(ea) ∈ d(eH)⊆ ρH. This means that d(ρH)⊆ ρH. It is easily seenthat d(lH(ρH))⊆ lH(ρH) holds and so d naturally induces a derivation δ onthe prime ring ρH = ρH

ρH∩lH (ρH) defined by δ (x) = d(x) for x ∈ ρH, wherelH(ρH) denotes the left annihilator of ρH in H. Thus by assumption, wehave δ ( f (x1, . . . ,xn))

sδ 2( f (x1, . . .,xn))δ ( f (x1, . . . ,xn))t is a central differen-

tial identity for ρH. By theorem 1, either δ (ρH) = 0 or f (x1, . . .,xn)N is

central-valued on ρH. If δ (ρH) = 0 that is d(ρH)ρH = 0, then0 = d(ρρH)ρH = d(ρ)ρHρH implying d(ρ)ρ = 0, a contradiction. Iff (x1, . . .,xn)

N is central-valued on ρH, then [ f (x1, . . . ,xn)N ,xn+1]xn+2 is an

identity for ρH as well as ρ . By [15], ρC = eRC for some e2 = e ∈ soc(RC)and f (x1, . . .,xn)

N is central-valued on eRCe. ��

410 B. Dhara

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