Annihilator condition of a pair of derivations in prime and semiprime rings

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ANNIHILATOR CONDITION OF A PAIR OF DERIVATIONS IN PRIME AND SEMIPRIME RINGS1

Basudeb Dhara, Nurcan Argac¸∗∗and Krishna Gopal Pradhan

Department of Mathematics, Belda College, Belda, Paschim Medinipur 721 424,

West Bengal, India

∗∗Department of Mathematics, Science Faculty, Ege University, 35100,

Bornova, Izmir, Turkey

e-mails: basu dhara@yahoo.com; nurcan.argac@ege.edu.tr; kgp.math@gmail.com

(Received 10 December 2013; after final revision 29 October 2015;

accepted 2 November 2015)

Letn be a fixed positive integer, R be a prime ring,D andGtwo derivations of R andLa

noncentral Lie ideal ofR. Suppose that there exists06=a∈Rsuch thata(D(u)ununG(u)) = 0for allu∈L, wheren≥1is a fixed integer. Then one of the following holds:

1. D=G= 0, unlessRsatisfiess4;

2. char(R)6= 2,Rsatisfiess4,nis even andD=G;

3. char(R)6= 2,Rsatisfiess4,nis odd andDandGare two inner derivations induced by

b, crespectively such thatb+c∈C;

4. char(R) = 2andRsatisfiess4.

We also investigate the case whenRis a semiprime ring.

Key words : Prime ring; derivation; extended centroid; Martindale quotient ring.

1. INTRODUCTION

Throughout this paper,R always denotes a prime ring with centerZ(R), extended centroidC and Q is its Martindale quotient ring. Forx, y R, the Lie commutator of x, y is denoted by [x, y]

1

This work is supported by a grant from National Board for Higher Mathematics (NBHM), India. Grant No. is

NBHM/R.P. 26/ 2012/Fresh/1745 dated 15.11.12 and second author is supported by The Scientific and Technological

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and defined by[x, y] = xy −yx. An additive subgroupL ofR is said to be a Lie ideal ofR, if [L, R] L. ByDwe mean a derivation ofR, that is an additive mappingD : R R satisfying D(xy) = D(x)y+xD(y)for allx, y ∈R. A derivationDis calledQ-inner if it is inner induced

by an element, sayq ∈Qas an adjoint, that is,D(x) = [q, x]for allx R. A derivation which is

notQ-inner is called a Q-outer derivation. The standard polynomial identitys4 in four variables is

defined ass4(x1, x2, x3, x4) =

P

σ∈S4

(1)σx

σ(1)(2)(3)(4)where(1)σ is+1or1according

toσbeing an even or odd permutation in symmetric groupS4.

A well known result proved by Posner [23], states that if the commutators[D(x), x] Z(R)

for all x R, then either D = 0 or R is commutative. Then many related generalizations of

Posners result have been obtained by a number of authors in literature. Breˇsar proved in [5] that if D(x)x−xG(x) Z(R)for allx R, then eitherD = G = 0orRis commutative. Later Lee

and Wong [21] studied the same situation of Breˇsar for allxin some noncentral Lie idealLofRand

obtained that eitherD=G= 0orRsatisfiess4.

Recently, Argac¸ and De Filippis [1] obtained the following result:

LetRbe a prime ring with char(R)6= 2,La non-central Lie ideal ofR,D, Gtwo derivations of Randn≥1a fixed integer. If(D(x)x−xG(x))n= 0for allxL, then eitherD=G= 0orR

satisfies the standard identitys4andD, Gare inner derivations, induced respectively by the elements

aandbsuch thata+b∈Z(R).

Recently, first author of this paper [10] studied the case with left annihilator condition. More

precisely, he obtained the following result:

LetRbe a prime ring,DandGtwo derivations ofRandLa noncentral Lie ideal ofR. Suppose

that there exists06=a∈Rsuch thata(D(u)u−uG(u))n= 0for allu∈L, wheren≥1is a fixed

integer. Then one of the following holds:

(i)D=G= 0unlessRsatisfiess4;

(ii) char(R) = 2andRsatisfiess4;

(iii) char(R)6= 2andRsatisfiess4,DandGare two inner derivations induced byp, qrespectively

such thatp+q∈C.

In [19], Lee and Zhou studied the situation for generalized derivations as follows:

Let R be a prime ring that is not commutative and such that R 6∼= M2(GF(2)), let G, H be

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G(x) =xwandH(x) =wxfor allx∈R; (2) eitherw∈C, orxmandxnareC-dependent for all x∈R.

Carini and De Filippis [6] studied the situation in a noncentral Lie ideal of a prime ringR with

central values. More precisely, they proved the following theorem:

LetRbe a prime ring,U the Utumi quotient ring ofR,C = Z(U)the extended centroid ofR, La non-central Lie ideal ofR,HandGnon-zero generalized derivations ofR. Suppose that there

exists an integern≥ 1such thatH(un)un+unG(un) C, for allu ∈L, then either there exists a∈U such thatH(x) =xa, G(x) = −ax, orRsatisfies the standard identitys4. Moreover, in the

last case the structures of the mapsG, Hare obtained.

In [7], Chang et al. proved the following:

Letnbe a fixed positive integer. LetR be a noncommutative(n+ 1)!-torsion free prime ring.

Suppose that there exist Jordan derivationsDandG:R →Rsuch thatD(x)xn−xnG(x) = 0for

allx∈R. Then we haveD= 0andG= 0.

In the present paper, our aim is to study the same situation of derivations with left annihilator

condition in a prime ringRthat is,a(D(x)xn−xnG(x)) = 0for allx∈L, whereLis a noncentral

Lie ideal ofR.

LetQbe the Martindale quotient ring of a prime ringRandCthe center ofQ, which is called

ex-tended centroid ofR. Note thatQis also a prime ring withCa field. We denote byT =Q∗CC{X},

the free product overC of theC-algebraQand the freeC-algebraC{X}, withXthe countable set

consisting of the noncommuting indeterminatesx1, x2, . . .. The elements ofT are called generalized

polynomial with coefficients inQ. Nontrivial generalized polynomial means a nonzero element ofT.

For more details about these objects we refer to [3, 14].

2. THE RESULTS ON PRIMERINGS

First we fix the following remarks:

Remark 2.1 : Let R be a prime ring and La noncentral Lie ideal of R. If char(R) 6= 2, by

[4, Lemma 1] there exists a nonzero idealI ofR such that0 6= [I, R] L. If char(R) = 2and dimCRC > 4i.e., char(R) = 2andRdoes not satisfys4, then by [18, Theorem 13] there exists a

nonzero idealIofRsuch that06= [I, R]⊆L. Thus if either char(R)6= 2orRdoes not satisfys4,

then we may conclude that there exists a nonzero idealI ofRsuch that[I, I]⊆L.

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a derivation ofQ, and so any derivation ofRcan be defined on the whole ofQ. We refer to [3, 20]

for more details.

We begin with lemmas

Lemma 2.3 — Let nbe a fixed positive integer, R be a prime ring with extended centroid C

and a, b, c R. If a 6= 0 such that a(b[x1, x2]n+1 [x1, x2]b[x1, x2]n [x1, x2]nc[x1, x2] + [x1, x2]n+1c) = 0 for all x1, x2 R, where n 1 is a fixed integer, then either R satisfies a

nontrivial generalized polynomial identity (GPI) orb, c∈C.

PROOF : Assume that R does not satisfy any nontrivial GPI. If R is commutative, trivially R satisfies a nontrivial GPI which is a contradiction. So, R must be noncommutative. Let T = Q∗C C{X1, X2}, the free product ofQandC{X1, X2}, the freeC-algebra in noncommuting

in-determinates X1 and X2. Then, since a(b[x1, x2]n+1 [x1, x2]b[x1, x2]n [x1, x2]nc[x1, x2] + [x1, x2]n+1c) = 0is a GPI forR, we see that

a(b[X1, X2]n+1[X1, X2]b[X1, X2]n−[X1, X2]nc[X1, X2] + [X1, X2]n+1c) = 0 (1)

inT =Q∗CC{X1, X2}. Ifc /∈C, thencand1are linearly independent overC. Thus, (1) implies

a([X1, X2]n+1c) = 0

inT. This impliesc= 0, a contradiction. Hence we conclude thatc∈C. Then (1) reduces to

a(b[X1, X2][X1, X2]b)[X1, X2]n= 0 (2)

inT. Again by the same way, this implies thatb∈C. 2

Lemma 2.4 — Let n be a positive integer, R be a noncommutative prime ring with extended

centroidC anda, b, c R. Suppose that a = 06 such that a(b[x1, x2]n+1 [x1, x2]b[x1, x2]n [x1, x2]nc[x

1, x2] + [x1, x2]n+1c) = 0for allx1, x2 ∈R, wheren≥1is a fixed integer. Then one of the following holds:

1. b, c∈C, unlessRsatisfiess4;

2. char(R)6= 2,Rsatisfiess4,nis even andb−c∈C;

3. char(R)6= 2,Rsatisfiess4,nis odd andb+c∈C;

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PROOF: By assumption,Rsatisfies generalized polynomial identity

g(x1, x2) =a(b[x1, x2]n+1[x1, x2]b[x1, x2]n−[x1, x2]nc[x1, x2] + [x1, x2]n+1c). (3)

By Lemma 2.3, b, c C which gives the conclusion (i) unless R satisfies a nontrivial GPI.

Now, assume thatR satisfies a nontrivial GPI. SinceRandQsatisfy same generalized polynomial

identities (see [9]), Qsatisfiesg(x1, x2). Moreover, if C is infinite, we haveg(x1, x2) = 0for all

x1, x2 ∈Q⊗C C, whereCis the algebraic closure ofC. Since bothQandQ⊗C Care prime and

centrally closed [12], we may replaceRbyQorQ⊗C Caccording toC finite or infinite. Thus we

may assume thatC =Z(R)andRisC-algebra centrally closed, which satisfiesg(x1, x2) = 0. By

Martindale’s theorem [22],Ris then a primitive ring and hence is isomorphic to a dense ring of linear

transformations of a vector spaceV overC.

Suppose that dimCV 3.

We shall show that for anyv ∈V,vandcvare linearlyC-dependent. Suppose thatvandcvare

linearly independent for somev V. Since dimCV 3, there existsu V such thatv, cv, uare

linearlyC-independent set of vectors. By density, there existx1, x2∈Rsuch that

x1v= 0, x1cv = 0, x1u=cv; x2v=cv, x2cv =u, x2u= 0.

Then[x1, x2]v= 0,[x1, x2]cv =cvand so0 =a(b[x1, x2]n+1[x1, x2]b[x1, x2]n−[x1, x2]nc[x1, x2]+ [x1, x2]n+1c)v=acv.

This implies that ifacv6= 0, by contradiction, we can say thatvandcvare linearlyC-dependent.

Now choosev∈V such thatvandcvare linearlyC-independent.

Then acv = 0. SetW = SpanC{v, cv}. Let ac 6= 0. Then there existsw V such that

acw 6= 0 and thenac(v−w) = −acw 6= 0. By the previous argument we have that w, cw are

linearlyC-dependent and(v−w), c(v−w)too. Thus there existα, β∈Csuch thatcw =αwand c(v−w) =β(v−w). Thencv=β(v−w) +cw=β(v−w) +αwi.e.,(α−β)w=cv−βv ∈W.

Nowα=βimplies thatcv=βv, a contradiction. Henceα6=βand sow∈W.

Again, ifu ∈V withacu= 0thenac(w+u)6= 0. So,w+u ∈W forcingu∈W. Thus it is

observed thatw∈V withacw 6= 0impliesw∈W andu ∈V withacu= 0impliesu ∈W. This

implies thatV =W i.e., dimCV = 2, a contradiction. Hence we conclude thatac= 0.

Again sincev, cv, uare linearlyC-independent set of vectors, by density, there existx1, x2 ∈R

such that

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Then [x1, x2]v = 0, [x1, x2]cv = v + cv and so 0 = a(b[x1, x2]n+1 [x1, x2]b[x1, x2]n

[x1, x2]nc[x

1, x2] + [x1, x2]n+1c)v =av. Sincea6= 0, by above argument this leads to a contra-diction.

Hence,v andcvare linearly C-dependent for allv V. Thus for each v V, cv = αvv for

someαv C. It is very easy to prove thatαv is independent of the choice ofv V. Thus we can

writecv=αvfor allv ∈V andα∈Cfixed. Now letr∈R,v∈V. Sincecv=αv,

[c, r]v= (cr)v−(rc)v=c(rv)−r(cv) =α(rv)−r(αv) = 0.

Thus[c, r]v = 0for allv∈V i.e.,[c, r]V = 0. Since[c, r]acts faithfully as a linear

transforma-tion on the vector spaceV,[c, r] = 0for allr ∈R. Therefore,c∈Z(R).

Therefore, from (3) we have thatRsatisfies generalized polynomial identity

g(x1, x2) =a[b,[x1, x2]][x1, x2]n= 0. (4)

Letvandbv are linearly independent overC. Since dimCV 3, there existsw ∈V such that

v, cv, ware linearlyC-independent set of vectors. By density, there existx1, x2 ∈Rsuch that

x1v= 0, x1bv=v, x1w=−v+ 2bv; x2v=bv, x2bv=w, x2w= 0.

Then[x1, x2]v =v,[x1, x2]bv=−v+bvand so0 =a[b,[x1, x2]][x1, x2]nv=av. Then again

by above arguments, sincea 6= 0, this leads a contradiction. Hence we conclude thatvandbv are

linearly dependent for allv∈V, implyingb∈C.

If dimCV = 2, thenR∼=M2(C), that is,Rsatisfiess4. If char(R) = 2, we get our conclusion

(4). So let char(R) 6= 2. Now we use the fact [x, y]2 Z(M2(C))for all x, y M2(C), and

consider the following two cases:

Letnbe an even integer. Then we have from (3) thatRsatisfies

a[x1, x2]n[b−c,[x1, x2]] = 0. (5)

By Lemma 2.1 in [11], since char(R)6= 2, we haveb−c∈C, which is our conclusion (2).

Letnbe an odd integer. Then we have from (3) thatRsatisfies

a[x1, x2]n[b+c,[x1, x2]] = 0. (6)

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Theorem 2.5 — Letnbe a fixed integer,Rbe a prime ring,DandGtwo derivations ofR,La

noncentral Lie ideal ofR. Suppose that there exists06=a∈Rsuch thata(D(u)un−unG(u)) = 0

for allu∈L, wheren≥1is a fixed integer. Then one of the following holds:

1. D=G= 0, unlessRsatisfiess4;

2. char(R)6= 2,Rsatisfiess4,nis even andD=G;

3. char(R) 6= 2,R satisfiess4,nis odd andDandGare two inner derivations induced byb, c

respectively such thatb+c∈C;

4. char(R) = 2andRsatisfiess4.

PROOF: If char (R) = 2andR satisfiess4, we obtain our conclusion (4). So, let either char (R)6= 2orRdoes not satisfys4. SinceLis a noncentral Lie ideal ofR, by Remark 1, there exists a

nonzero idealI ofRsuch that[I, I]⊆L. Hence, by our assumption we have,

a(D([x1, x2])[x1, x2]n−[x1, x2]nG([x1, x2])) = 0 (7)

for allx1, x2 ∈I.

Now we divide the proof in the two cases:

Case I : Let DandGare both Q-inner derivations ofR i.e.,D(x) = [b, x]for allx R and G(x) = [c, x]for allx∈R, whereb, c∈Q. Then from (7), we obtain that

a(b[x1, x2]n+1[x1, x2]b[x1, x2]n−[x1, x2]nc[x1, x2] + [x1, x2]n+1c) = 0 (8)

for allx1, x2 ∈I. By Chuang [9], this GPI is also satisfied byQand hence byR. By Lemma 2.4, if

Rdoes not satisfys4,b, c∈C, that is,D=G= 0, which is our conclusion (1). If char(R)6= 2and

Rsatisfiess4, then by Lemma 2.4,b+c ∈Cwhennis odd andb−c∈ Cwhennis even. Ifnis

even, thenb−c∈CimpliesD=G. Thus conclusions (2) and (3) are obtained.

Case II : Next assume that D andG are not bothQ-inner derivations of R, but they are C

-dependent modulo inner derivations ofR. SupposeD=λG+adb, that is,D(x) =λG(x) + [b, x]

for allx∈R, whereλ∈C,b∈Q. ThenGcannot be an inner derivation ofR. From (7), we obtain

thatI satisfies

a µ

λG([x1, x2])[x1, x2]n+ [b,[x1, x2]][x1, x2]n−[x1, x2]nG([x1, x2])

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SinceGis not inner derivation ofR, by Kharchenko’s theorem [17], we have that

a µ

λ([u, x2] + [x1, v])[x1, x2]n+ [b,[x1, x2]][x1, x2]n−[x1, x2]n([u, x2] + [x1, v])

= 0. (9)

for allx1, x2, u, v ∈I. By Chuang [9], this GPI is also satisfied byQand hence byR. In particular

foru=v= 0, we have thatRsatisfies

a[b,[x1, x2]][x1, x2]n= 0.

By Lemma 2.4, we get thatb∈C. Hence, we get from (9) thatRsatisfies

a µ

λ([u, x2] + [x1, v])[x1, x2]n−[x1, x2]n([u, x2] + [x1, v])

= 0. (10)

Assumingv = 0andu=x1, we have thatRsatisfies

a(λ−1)[x1, x2]n+1= 0 (11)

that is

a0[x1, x2]n+1 = 0, (12)

wherea0 = a(λ−1). For any two fixedx, y R, setw = [x, y]n+1. Thena0w = 0. From (12),

we can writea0[u, wva0]n+1 = 0for allu, v∈ R. Sincea0w = 0, it reduces toa0(uwva0)n+1 = 0.

This can be written as(wva0u)n+2 = 0for allu, v R. By Levitzki’s Lemma [15, Lemma 1.1],

wva0u= 0for allu, vR. SinceRis prime, eithera0 = 0orw= 0. Ifw= 0, then[x, y]n+1 = 0

for allx, y R. In this case by Herstein [13, Theorem 2],Ris commutative, contradicting the fact

that0 6= L is noncentral. Thusa0 = a(λ−1) = 0. Sincea 6= 0, this impliesλ = 1and hence D=G. Now (10) gives

a µ

([u, x2] + [x1, v])[x1, x2]n−[x1, x2]n([u, x2] + [x1, v])

= 0 (13)

for allu, v, x1, x2 R. SinceRis noncommutative, we may choosep ∈R such thatp /∈C. Then

replacinguwith[p, x1]andvwith[p, x2]in (13), we get

a µ

[p,[x1, x2]][x1, x2]n−[x1, x2]n[p,[x1, x2]]

= 0 (14)

for allx1, x2 ∈R. Then by Lemma 2.4, we conclude that sincep /∈C, char(R)6= 2,Rsatisfiess4

andnis even. Thus conclusion (2) is obtained.

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Next assume that DandGareC-independent modulo inner derivations ofR. Since neitherD

norGis inner, by Kharchenko’s theorem [17], we have from (7) thatIsatisfies

a(([u, x2] + [x1, v])[x1, x2]n−[x1, x2]n([x, x2] + [x1, y])) = 0. (15)

In particular,I satisfies the blended component

a[u, x2][x1, x2]n= 0. (16)

By [9], this GPI is also satisfied byQand hence byR. In particular, assumingu =x1, we getR

satisfiesa[x1, x2]n+1 = 0. SinceRis noncommutative, as above this leadsa= 0, a contradiction.2

Theorem 2.6 — Letnbe a positive integer,Rbe a noncommutative prime ring with char(R)6= 2, DandGtwo derivations ofR. Suppose that there exists06=a∈Rsuch thata(D(x)xn−xnG(x)) = 0for allx∈R, wheren≥1is a fixed integer. ThenD=G= 0.

PROOF: By Theorem 2.5, we have only to consider the case whenRsatisfiess4. IfD=G= 0,

we are done. So, let one of them must be nonzero. By Theorem 2.5, we have the following two cases:

Case I :nis even andD=G.

LetDbeQ-inner, that isD(x) = [b, x]for allx∈Randb∈Q. ThenRsatisfies

a(bxn+1−xbxn−xnbx+xn+1b) = 0. (17)

SinceRsatisfiess4, there exists a fieldKsuch thatR⊆M2(K)andM2(K)satisfies (l7). Since

M2(F)is a dense ring ofK-linear transformations over a vector spaceV, we assume that there exists v6= 0, such that{v, bv}is linearK-independent. By the density ofM2(K), there existr ∈M2(K)

such thatrv= 0, rbv=bv. Then

0 =a(brn+1−rbrn−rnbr+rn+1b)v=abv.

Of course for any u V, {u, v}linearly K-dependent impliesabu = 0. Letab 6= 0. Then

there existsw∈V such thatabw 6= 0and so{w, v}are linearlyK-independent. Alsoab(w+v) = abw6= 0andab(w−v) =abw6= 0. By the above argument, it follows thatwandbware linearlyK

-dependent, as are{w+v, b(w+v)}and{w−v, b(w−v)}. Therefore there existαw, αw+v, αw−v ∈K

such that

bw=αww, b(w+v) =αw+v(w+v), b(w−v) =αw−v(w−v).

In other words we have

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and

αww−bv=αwvw−αwvv. (19)

By comparing (18) with (19) we get both

(2αw−αw+v−αw−v)w+ (αw−v−αw+v)v= 0 (20)

and

2bv= (αw+v−αw−v)w+ (αw+v+αw−v)v. (21)

By (20), and since{w, v}areK-independent andchar(K) 6= 2, we haveαw =αw+v =αw−v.

Thus by (21) it follows2bv = 2αwv. This leads a contradiction with the fact that{v, bv}is linear

K-independent.

Therefore, we conclude thatab = 0. Again, by the density ofM2(K), there existr M2(K)

such thatrv= 0, rbv=v+bv. Then

0 =a(brn+1−rbrn−rnbr+rn+1b)v =av. By above argument, this impliesa= 0, a contradiction.

LetDbe notQ-inner, then by Kharchenko’s Theorem [17],Rsatisfies

a(yxn−xny) = 0. (22)

ThusR ⊆M2(K)andM2(K)satisfies (22). Assumingy =e12, x= e11, we get0 = −ae12,

implyinga11=a21 = 0. Again, assumingy =e21, x=e11, we havea22 =a12= 0, that isa= 0,

a contradiction.

Case II :nis odd andD(x) = [b, x], G(x) = [c, x]for allx∈Rwithb+c∈C.

In this case by hypothesis,Rsatisfies

a(bxn+1−xbxn+xnbx−xn+1b) = 0. (23)

By the above argument of case-I, this impliesa= 0, a contradiction. 2

3. THERESULTS ONSEMIPRIMERINGS ANDBANACHALGEBRAS

In this section we extended Theorem 2.6 to the semiprime ring. LetR be a semiprime ring andU

be its left Utumi quotient ring. ThenC =Z(U)is the extended centroid ofR([8, p-38]). It is well

known that any derivation of a semiprime ringRcan be uniquely extended to a derivation of its left

Utumi quotient ringU and so any derivation ofRcan be defined on the whole ofU [20, Lemma 2].

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By the standard theory of orthogonal completions for semiprime rings, we have the following

lemma.

Lemma 3.1 — ([2], Lemma 1 and Theorem 1). LetRbe a 2-torsion free semiprime ring andP a

maximal ideal ofC. ThenP U is a prime ideal ofU invariant under all derivations ofU. Moreover,

T

{P U :P ∈M(C) with U/P U 2−torsion free}= 0.

Theorem 3.2 — Let R be a 2-torsion free noncommutative semiprime ring, U the left Utumi

quotient ring ofR,da nonzero derivation ofRand0 6=a∈ R. Ifad(R) ⊆Z(R), then there exist

orthogonal central idempotentse1,e2,e3 ∈U withe1+e2+e3= 1such thatd(e1U) = 0,e2a= 0,

ande3U is commutative.

PROOF: Since any derivation dcan be uniquely extended to a derivation in U, andU andR

satisfy the same differential identities ( see [20]), thenad(x)∈Z(U), for allx∈U. LetP ∈M(C)

such thatU/P U is 2-torsion free. Note thatU is also 2-torsion free semiprime ring. P U is a prime

ideal ofU, which is d−invariant by Lemma 3.1 denote U = U/P U anddthe derivation induced

bydon U. For anyx U,ad(x) Z(U). SinceZ(U) = (C+P U)/P U = C/P U,ad(x) (C+P U)/P U = C/P U, for allx U. MoreoverU is a prime ring. Then it is clear that either d = 0; ora = 0inU; or[U , U] = 0. This implies that, for anyP M(C), eitherd(U) P U;

ora∈P U; or[U, U]⊆P U. In any casead(U)[U, U] P U for anyP M(C). By Lemma 3.1,

T

{P U :P ∈M(C) with U/P U 2torsion free}= 0. Thusad(U)[U, U] = 0.

By using the theory of orthogonal completion for semiprime rings ( see, [3, Chapter 3]), it follows

that there exist orthogonal central idempotents e1, e2, e3 U with e1 +e2 +e3 = 1 such that

d(e1U) = 0,e2a= 0, ande3U is commutative. 2

Theorem 3.3 — Letnbe a positive integer,Rbe noncommutative2-torsion free semiprime ring, U the left Utumi quotient ring ofRand0 6= a∈ R. Letd, gbe two nonzero derivations ofRsuch

thata(d(x)xn−xng(x)) = 0for allx∈R. Then there exist orthogonal central idempotentse1,e2,

e3 ∈U withe1+e2+e3 = 1such thatd(e1U) =g(e1U),e2a= 0, ande3U is commutative.

PROOF: Since any derivation dcan be uniquely extended to a derivation in U, andU andR

satisfy the same differential identities (see [20]), thena(d(x)xn−xng(x)) = 0, for allx ∈U. Let P ∈M(C)such thatU/P U is 2-torsion free. Note thatU is also 2-torsion free semiprime ring. By

Lemma 3.1P Uis a prime ideal ofU, which isd−invariant. DenoteU =U/P U anddthe derivation

induced bydonU. For anyx U, we geta(d(x)xn−xng(x)) = 0. MoreoverU is a prime ring

so by Theorem 2.6 we get eitherd = 0andg = 0 or[U , U] = 0, ora = 0. In any case we both

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P M(C) with U/P U 2−torsion free} = 0. Thenad(U)[U, U] = 0 andag(U)[U, U] = 0.

In particular ad(R)[R, R] = 0 and ag(R)[R, R] = 0. These imply that 0 = ad(R)[R2, R] = ad(R)R[R, R] +ad(R)[R, R]R=ad(R)R[R, R]. In particularad(R)R[R, ad(R)] = 0. Therefore, [ad(R), R]R[ad(R), R] = 0and similarly[ag(R), R]R[ag(R), R] = 0. Since Ris semiprime, we

obtain thatad(R) ⊆Z(R)andag(R) ⊆Z(R). Hence, we geta(d−g)(R)⊆Z(R). By Theorem

3.2, we have the conclusion. 2

Corollary 3.4 — LetR be a noncommutative 2-torsion free semiprime ring, U the left Utumi

quotient ring ofRand06=a∈R. Letdbe a nonzero derivation ofRsuch thata(d(x)xn+xnd(x)) = 0for allx∈R. Then there exist orthogonal central idempotentse1,e2,e3 ∈U withe1+e2+e3= 1

such thatd(e1U) = 0,e2a= 0, ande3U is commutative.

By a Banach algebra, we shall mean a complex normed algebraAwhose underlying vector space

is Banach space. The Jacobson radical ofA is the intersection of all primitive ideals ofA and is

denoted byrad(A).

Theorem 3.5 — LetAbe a Banach algebra,d, ga pair of continuous Jordan derivations ofA,

0 6= a A andna fixed positive integer. Ifa(d(x)xn−xng(x)) rad(A) for allx A, then d(A)⊆rad(A)andg(A)⊆rad(A).

PROOF : Let P be any primitive ideal of A. Sinced, g are both continuous, d(P) P and g(P) P by Lemma 3.2 in [24]. Then dandgcan be induced to a pair of Jordan derivations on

Banach algebraA/P as follows :

d(x) =d(x) +P and g(x) =g(x) +P

for all x A/P and x A. Since P is a primitive ideal ofA, quotient algebra A/P is prime

and semisimple. On the other hand, we should remark that the pair of Jordan derivationsdandg

onA/P are also a pair of derivations onA/P by Breˇsar [5]. Moreover, by a result of Johnson and

Sinclair [16] every derivation on a semisimple Banach algebra is continuous. Hence there are no

nonzero continuous derivations on commutative semisimple Banach algebras. Hence, we getd= 0

andg= 0whenA/P is commutative. It remains to show thatd= 0andg= 0in case of whenA/P

is noncommutative. By the hypothesis we have

a(d(x)xnxng(x)) = 0

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d(A)⊆rad(A)andg(A)⊆rad(A). 2

Corollary 3.6 — LetAa semisimple Banach algebra, d, gbe a pair of Jordan derivations onA

and06=a∈ A. Ifa(d(x)xn−xng(x)) = 0for allx ∈A, wherenis a fixed positive integer, then d= 0andg= 0.

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