Mediterr. j. math. 3 (2006), 15-29 l660-.5446/01001.5-15, DOl 10.1007/800009-006-0061-2 Mediterranean Journal © 2006 Birkhiiuser Verlag Easel/Switzerland of Mathematics

Engel Type Identities with Derivations for Right Ideals in Prime Rings

Rajendra K. Sharma and Basudeb Dhara

Abstract. Let R be a prime ring of char R oF 2, d a non-zero derivation of R and p a non-zero right idea! of R such that ,d(Y)]n, [Y, = 0 for all x,y E P or [[d(x),d(y)j",d([y, = 0 for all x,y E p, n,m? 0 are fixed integers. If LP, pip oF 0, then d(p)p = O.

Mathematics Subject Classification (2000). Primary 16W25, 16R50: Secondary 16N60.

Keywords. Prime ring, semiprime ring, derivation, GPI.

1. Introduction

Let R be an associative ring and Z(R) be its center. Let n be a positive integer. Given x, y E R, we set yja = x, yh = y] = xy - yx. then an Engel condition is a polynomial y] k-l, y]. k = 1. 2, ... in noncommuting indeterminates. A commutative ring satisfies any such polynomial and any Lie nilpotent ring satisfies one if k is sufficiently large.

The Engel type identity with derivation appeared in the well known paper of E. C. Posner [21] who showed that for a nOll-zero derivation d of a prime ring R, if [d(x), J:] E Z(R) for all x E R, then R is commutative. Since then several authors have studied this kind of Engel type identities with derivation acting on one-sided, two-sided and Lie ideals of a prime and semiprime ring. For details we refer to ,[6],[14],[16],[18],[19], where further references can be found.

In [11], Herstein proved that if R is a prime ring with char R l' 2 and R admits a non-zero derivation d such that [d(::1 d(y)] = 0 for all x, y E R, then R is commutative. Daif and Bell [9) proved that if R is a semiprime ring with a derivation d and U is a two-sided ideal of R such that d[x, y] = ±[x, y] fClr all x, yE U, then U is central ideal. They also proved that if R is a semiprime ring with a derivation satisfying [d(l:), d(y)] = [x, y] for all x, y E U, \\There 0 l' U is a right ideal of R, then U is central. The same authors [3] showed that if R is a prime

16 R.K. Sharma and Basudeb Dhara Mediterr. j. math.

ring with a derivation d such that [d(x), = ±d[x,y] for all ;r;,y E U, where U is a non-zero right ideal of R, then either R is commutative or d2 (U) = 0 as well as

(d(U))2 = O. So it is natural to ask what happens in case ,d(y)]n' = 0 for all x,y E p or [[d(x),d(Y)]n,d( = 0 for all :1.;,y E p, with rn,n:2 0 fixed integers and p a non-zero right ideal of R.

Throughout this paper, unless specially stated, R always denotes a prime ring with center Z(R), with extended centroid C, and with two-sided Martindale quotient ring Q.

It is well known that any derivation of R can be uniquely extended to a derivation of Q, and so any derivation of R can be defined on the whole of Q. Moreover Q is a prime ring as well as R and the extended centroid C of R coincides with the center of Q. We refer to [2] and [17] for more details.

Denote by Q *c C {X, Y} the free product of the C-algebra Q and C {X, Y}, the free C-algebra in noncommuting indeterminates X, Y.

2. The Case: R is a Prime Ring

We need the following lemma.

Lemma 2.1. Let p be a non-zero -right -ideal of Rand d a de-rivation of R. Then the following condit-ions are equivalent:

d is an inner derivat'ion i-ndw:ed by some bE Q such that bp = 0; d(pJp = O.

For its proof we refer to [7, Lemma]. \Ve mention a very important result which will be used quite frequently as follows:

Theorem 2.2 (Kharchenko [13]). Let R be a prime ring, d a den:uation on Rand I a non-zero ideal of R. If I satisfies the differ-ential identity

= 0 for any rt, 1"2,···. r n E I.

then either or are sat'isfied, where

I sat2sfies the generalized polynomial identity

(ii) d is Q-inneri.e., fOT some q E (2, = [q, xl and I satisfie8 the geneml-ized polynomial 'identity

!'(", r' r fq. '1. 2"""nq, , [q, , ... , [q, TnD = O.

Theorem 2.3. Let R be a pr-ime ring of char R le 2 and d a non-zero derivation of R such that [[d(:1.; J' = 0 for all x, y E R, rn, n :2 0 are fixed integers, then R is commutative.

Vol. 3 (2006) Engel Type Identities ]7

Proof. If R is commuta.tive, we have nothing to prove. So. let R be noncommuta­tive. If d is not Q-inner. then by Kharchenko's Theorem (13).

g(x, y, 11, v) = [y, = 0,

for all x. y, u. v E R. This is a polynomial identity and hence there exists a field F such that R;; i'vh(F) with k> 1 and Rand 111i;;(F) satisfy the same polynomial identity [15. Lemma 1]. But by choosing u = e12. v = ell, x = ell and y = e21.

we get

0=

which is a contradiction. Now let d be Q-inner derivation, say d = ad(a) for some a E Q i.e .. d(x)

fa. x] for all x E R. then we have

x]. [a. = O.

for all x. y E R. Since d =1= 0, a ti C and hence R satisfies a nontrivial generalized polynomial identity (GPI). By [20]. it follows that RC is a primitive ring with H = Soc(RC) i 0 and eH e is finite dimensional over C for any minimal idempotent e E RC. Moreover we may assume that H is noncorumutative. otherwise R must be commutative which is a contradiction.

Notice that H satisfies xl. la. [y, = 0 (sce [15, proof of Theorem 1]). For any idempotent e E Hand ;1' E H we have

0= (! ­

)=0

Left multiplying by e and right multiplying by a( 1 - we get

o [a. (1 - 1 - e)xe. 1 ­(1 - ,,(1 - e)xea(1 ­

" =eI:(-1FG)[a,(I- (ae-ea)[a.(1 (l-e)xea(l-e)x )=0

= (-l )'l+t t G)e(a(l - e):r:e)lea((l - e)xea)" )(1 - e)xea(1 - e)x )=0

n = (-1 (ea(l - e)x)"+2 I: (7)

=(-1 12n (ea(l-

Since char Ri 2. (ea(1 - e)x)n+2 = O. By Levitzki's Lernma [12, Lemma 1.1]. ea(l - e)::r: = 0 for all x E H. Since H is prime ring. ea(] - e) = 0 i.e .. eae = ea for any idempotent e EH. Now replacing e with 1 - e. wc get that (1- e )ae = 0 that is eoe = ue. Therefore for any idempotent e EH. we have fa. e] C-~ O. So a commutes with all idempotents in H. Since H is a simple ring. either H is generated by its idempotents or H does not contain any nontrivial idempotent. The first case gives a E C contradicting d f O. In the last case. H is a finite dimensional division algebra over C. This implies that H. = RC = Q and a E: H. By [1 . Lemma 2]. there exists a field F such that H; l'vh(F) and Af",(F) satisfies

x, x]mJ. Then by the same argument as earlier. a commutes with

18 R.K- Sharma and Basudeb Dhara Mediterr. j. math.

all idempotents in l'vh(F), again giving the contradiction Cl E C i.e .. d = O. This completes the proof of the theorem. 0

Theorem 2.4. Let R be a prime ring of char R cl- 2, d a non-zero derivation of Rand p a nOTi-zem right ideal of R such that ,d(y)Jn. [y. = 0 for all x, yE p,m,n :::>: 0 fixed integers. [f pJp cl- 0, then d(p)p = O.

We begin the proof by first proving the following lemma.

Lemma 2.5. Ifd(p)p cl- 0 and .d(y)Jn, = 0 faT all x,y E p, m,n:::>: 0 fixed integers, then R sati5:fies nontrivial generalized 'Polynomial identity (GP[).

Pmof. Suppose, on the contrary, that R does not satisfy any nontrivial GPI. We may assume that R is noncommutative, otherwise R satisfies trivially a nontrivial GPI. \Ve consider two cases.

Case I. Suppose that d is a Q-inner derivation induced by an element a E Q. Then for any :r E p

xX]. :rY]]n. [:rY, xX] m]

is a GPT for R, so it is the zero element in Q "c C{X, Y}. Expanding this we get,

[:rY, xX]m t( -IF (n) [a, :ry]j[a. xX] .rYln-j j=O J

-~(-lF(~) .rY]J[a . .rX] :rYjn-)[xY,:rX]rn=()

that is "­

(n). .[:rV. xX]rn L(- J (a1,y - 1;Ya)) (a;7;X - xXa)(axY - .rYaYc-)

.]=0

- t(-1)) (~I)(a;TY - xY (axX - xXa)(a:rY - xYa)n-i[xY.xX]·m = O. )=0 .7

If a:T and :r are linearly C-independent for some .r E p. then n

- '\' 1,-1)l (n)(L;1~Y(axY - 1,ya))-1 (axX - xXa)(axl' - :.rYa)n-) [xY. xX~mL \1/ \ /\ - " " J=(J .

= O.

Repeating the same process yields

- t(-l)J (~) (axY)) a.rX(axy),,-j[:rY, xXJm = 0 )=0 J

in Q *c C{X. Y}. This implies

-axX(axYY'[xY. xX]rn = 0

Va!. 3 (2006) Enge! Type Identities 19

in Q *c C{ X, Y}, which gives a:1: = O. a contradiction. Thus for any x E p. ax and x are C-dependent. Then (a - alp = 0 for some a E C. Replacing a with a - a, we may assume that ap = O. Then by Lemma 2.1, d(p)p = 0, a contradiction.

Case 11. Suppose that d is not a Q-inner derivation. If for all x E p, d(x) E ;r;C. then [d(x), xl = 0 which implies that R is commutative (see . Therefore there exists x E p such that d(x) ~ xC i.e., x and d(x) are linearly C-independcm. By our assumption we have that R satisfies

[[d(xX). d(xY)]n, [xY•.xX]mJ = O.

By Kharchenko's Theorem [13].

[ld(x)X + xr], d(x)Y + xrz]n.lxY. xX]m] = 0.

for all X, Y, r1. rz ER. In particular for Tt = "z = O.

[[d(x)X. d(x)Y]'n.lxY. xX]m = O.

which is a non-trivial GPI for R. because x and d(x) are linearly C-independem, a contradiction. [J

Pmoj of Theorem 2.4. Suppose dIPJp cJ 0 and then we derive a contradiction. By Lemma 2.5, R is a GPI-ring. so is also Q (see [8]). By [20J, Q is a primitive ring with H = 80c(Q) fe O. Moreover, we may assume that [pH. pHjpH cJ O. otherwise by . [pQ. pQ]pQ = O. which is a contradiction. By Lemma 2.1. we may also assume that d(pH)pH fe O. We consider the two following cases.

Case 1. Let d be a Q-inner derivation induced by an element a E Q. According to

our assumption and by [17]. we may assume that

. [y. =0 (2.1 )

is satisfied by pQ and hence by pH. Let e = eZ E pH and y EH. Replacing x with e and y with cy(l - c) in (2.1) and then left multiplying it by (1 e) and right multiplying it by aey. yield

(1 - c) eL cy(l-

After a simple manipulation it reduces to

1 -

1 - e)aey)rt-+-2 = O.

Since char RcJ 2. we have ((1 -- c)aey)n+z = O. By Levitzki's Lemma [J2. Lemma 1.1]. (1 - e)aey = 0 for all y E H. By primeness of H. (1 - e)ae = Cl i.e .. ae = we.

By [10. Lemma 1]. H is a regular ring. Thus for each r E pH. there exists an idempotent e E pH such that r = er and e E rH. So. for I' E pH, ar = aCT =

eaer = ear E pH. Therefore. apH C;;; pH. Denote the left annihilator of pH in H by lH(pH). Then pH = pHn~~(pH) • a prime C-algebra with the derivation Cl such

that d(x) = d(x). for all x E pH. By assumption we have that

[y, = O.

20 R..K. Sharma and Ba:mdeb Dhara Mediterr. j. math.

for all x, y E pH. By Theorem 2.3, either d = °or pH is commutative. In the first case we have d(pH)pH = 0 and in the later case [pH, pH]pH = 0. In both cases, we have a contradiction.

Case n. Now we assume that the derivation d is not Q-inner. Let rE pH. Since H is regular ring, there exists an element e2 = e E pH such that r = er and e E rH. According to our assumption, ,ve have for all b E pH and :r. yE H,

[[d(bx), d(bY)]n, [by, = 0,

that is

[[d(b)x + bd(.T) , d(b)y ~ bd(y)jn, [by, = O.

By Kharchenko's Theorem [131,

[fd(b)::r + br1, d(b)y + bT2Jn, [by, bX!rn] = 0,

for all X,y,r1,T2 E H. In particular for r1 = r2 = 0,

[fd(b).T, d(b)Yln' [by. b:r]m] = 0,

for all X:,y E H. For b = e,.r = e and y = ey(1- c) we have

0= [[d(e)e, d(e)ey(1 - ,[ey(1 - e).

Now we have the fact that for any idempotent e, 0 = ed(c2) - cd(e) = ed(e) + ed(c)e - ed(e) = i.e .. ed(e)c = O. Using this fact we can write from above that

0= [(d(e)ey(l- e)Y'd(e)e,ey(1­

Now right multiplying by ey and left multiplying by (1 - e)d(e), this reduces after a simple manipulation to ((1 - e)d(c)ey)n-r2 = O. By Levitzki's Lemma [12, Lemma 1.1 . (1- c)d(e)eH = O. By primeness of H, (1- e)d(e)e = O. This implies (1 - e)d(e) C~ (1- c)d(e2) = (1- e)d(c)e = O. Thus d(e) = E eH c;: pH. Now d(r) = d(er) c= d(e)er + E pH. Hence d(pH) c;: pH.

Set pH = W,p,H j W ,with lH(pH) the left annihilator of pH in H. Then d p . i : I_H\P , } \

induces a canonical derivation d on pH satisfying

for all x, y E pH. By Theorem 2.3, we have either d = 0 or pH is commutative. Therefore we have that either d(pH)pH = 0 or [pH, pH]pH = O. In both cases, we have contradiction. This completes the proof of the theorem.

Theorem 2.6. Let R be a prime ring of char R i- 2 and d a non-zero derivatwn of R ,9Uch that [fd(x), d(y)Jn, d([y, :r]m)] = 0 for all x, yE R, whcre m. n 2: 0 are fixed integers. Then R is commutat'lve.

21 Vo!. 3 (2006) EngeJ Type Identities

Pmof. If R is commutative, we have nothing to prove. So, let R be noncommuta­tive. By assumption, we see that

m-I

[[d(x), d(y)]", [dry), :rlm + ~ my· Xll' d(x)L X]m-l- = Cl, 2=0

for all x. yE R. If d is not Q-inner. then by Kharchenko's Theorem [13].

m-1

g(x, y. u, v) = -+- ~ [try, .uj, X]m-l-l = 0. ;=0

for all x. y. U. v E R. In particular for y = Cl, we have v]", xlmJ = Cl. for all x. u, v E R. This is a polynomial identity and hence there exists a field F such that R c;;: Mk(F) with k > 1 and Rand MdF) satisfy the same polynomial identity [15, Lemma 1]. By choosing u = el2, v = ell.X = e12 + e2] we see that

0= m is odd. m ]s even,

which is a contradiction. Now let d be Q-inner derivation. say d = ad(a) for some a E Q i.e.. d( x)

[a, xl for all x E R~ then \ve have

xL [a. y]]", [y, = 0,

for all :1:, Y E R. Since d i= 0, a rt C and hence R satisfies a non-trivial GPI. By [201, it follows that RC is a primitive ring with H = Soc(RC) i- Cl and eH e is finite dimensional over C for any minimal idempotent e E RC. Moreover we may assume that H is noncommutative. otherwise R must be commutative which is a contradiction.

Notice that H satisfies :c], [a. [g. = 0 (see [15. proof of The­orem 1J). For any idempotent e E H and x E H we have

o ej, (1 - ,[a, - e)xe,

= [2..: (-IFG)[a. (1 - (ae - ea) (1- e)xe]"-J. [a, (1­J=O

Right multiplying by (1 - e )2' and left multiplying by e we get

o e{ f. (-l)"G)(a(1 - e):ce)J ea((l - e)xea)"-J(l - e)xea )=0

-+-ea(1 - e)xe j~)(" --I)"G) 1 - ea((l - e):rear'-J }(1 - e)x

n 2 2..: (-1)" Cj)(ea(1 - e)x)"~2

j=O

(-1 r'2"-rl(ea(1 - e)x)"+2

Since char R f=- 2, (ea(l - e)x)"+2 = Cl. By Levitzki's lemma [12, Lemma 1.1'. ea(l - e)x = 0 for all x EH. Since H is prime ring, ear 1 - e) = 0 i.e.. eae = ea for any idernpotent e E H. Now replacing e with 1 - e we get that (l - e)ae ~-" Cl that

22

0

R.K. Sharma and Basudeb Dhara Mediterr. j. rnath.

is eae = ae. Thus eae = ae = ea. Therefore for any idempotent e E JJ, we have [a, e] = O. So a commutes with all idempotents in H. Since H is a simple ring, either H is generated by its idempotents or H does not contain any nontrivial idernpotent. The first case implies a E C contradicting d i O. In the last case, H is a finite dimensional division algebra over C. This implies that H = RC = Q and a E H. By [1.5, Lemma 2], there exists a field F such that H ~ lvh(F) and Ah(F) satisfies :c]' , [a, x]m]]. Then by the same argument as earlier. a commutes with all idempotents in AI/<; (F), again giving the contradiction a E C i.e., d = O. This completes the proof of the theorem.

Theorem 2.7. Let R be a prime ring of char R i 2. d a non-zero derivation of Rand p a non-zero right ideal of R such that [ld(x). d(Y)]n, d([y, x]m)] = 0 for all x, yE p, rn, n 2' 0 fixed integers. If rp, pjp i 0, then d(p)p = O.

\Ve present an example.

Example. Let R = AI2(GF(2)), the 2 x 2 matrix ring over the field F = GF(2) and p = ellR. Consider the inner derivation d(x) = [e21, x] for all x E R. Then it is easily verified that p satisfies the identity [[d(x), d(Y)]n. d([y, x]m)] = 0 where fp, p]p = 0 but d(p)p i o.

To prove the theorem we need the following lemma.

Le=a 2.8. If d(p)p i 0 and [[d(x). d(y)Jn, d([y, = 0 for all x, y E p. rn, n ::: 0 fixed integers, then R satisfies nontr'ivwl generalized polynomial identity (GPI).

Proof. Suppose on the contrary that R does not satisfy any nontrivial GP1. \Ve may assume that R is noncommutative, otherwise R satisfies trivially a nontrivial GP1.

Case 1. Suppose that d is an Q-inner derivation induced by an element a E Q. Then for any x E p

[[[a, xX], [a, [xY. xX]m]]

is a GPI for R, so it is the zero element in Q *c C{X, Y}. Expanding this we get,

T F• • j n . j r n- j r • 1n ( .) [a, xr] [a. xX] la, xr ]2)-1) la, [xY, xX]mJ j=O J

r i rn i .. - . rn-i1 r ". 1 _ m ()-[0.,2:)-1) i (:rX) xY(xX) do.,xXLla,xYLn -0,

t=O

Vol. 3 (2006) Engel Type Identities

that is

((axx - xXa)[a,xYj"

j+ t(-1)5(;)(aXY -xYa)j XXHa,xYjn- ) fa, rxY,xX]m]

- (axY(xx)m - xY(xx)ma + a ~(_1)i (7) (xX)'xy(xx)m-i t=1. -t( _1)i (7) (xX)iXY(xX)m-i a) xx], [a, xY]]n = o.

t=1

Ifax and x are linearly C-independent for some x E p, then

(axX[a,Xyr

+ f) -1)5 (~) axY(axY - xYa)j-1 xX][a, xYjn- j ) [a, [xv' xX]m] j=1 ]

- (axY(xx)m

+a t( _1)' (7) xX(xX)'-IXY(xx)m-i) [[a, xx], [a,xYJln = O. ,=1

This implies that

axX[a,xYJn[a. [xY,xX]m]

\-i In 1,£-1 m-i~1 1 r I m () (::r:X) xY(xX) lla,xXj,la,.TY]Jn=O.-axX~(-1) i '/,=1.

This can be re-written as

axX {(axY - xYa)[a,xy]"-1[a, [xY,xXJml

-t( -1)' (7) (xX)i-1 x y(xx)m-t xX], ra, xYj]n} = O. 1.=1

Again. since ax and x are C-independent. we have

axXaxY[a, xYln-1 [a, [xY,xX]ml = O.

Expanding this we can write

24 R.K. Sharma and Basudeb Dhara Mediterr. j. rnath.

By a similar approach, since ax and x are C-independent, this reduces to

in Q *c C {X. Y}. This implies ax: = 0, a contradiction. Thus for any x E p, ax and .r are C-dependent. Then (a - a)p = 0 for some a E C. Replacing a with a - a, we may assume that ap = O. Then by Lemma 2.1, d(p)p = 0, a contradiction.

Case II. Suppose that d is not Q-inner derivation. If for all x E p. d(x) E xC, then x] = 0 which implies that R is commutative (see [5]). Therefore there exists x E p such that d(x) fc xC Le., x and d(x) are linearly C-independent. By our assumption we see that R satisfies

[ld(xX), d(xY)]w d(lxY, xX]m)] = O.

This implies

l[d(x)X + xd(X). d(:r:)Y + xd(Y)Jn, ld(x)Y + xd(Y), xX]", rn-l

+L xX]" d(x)X + xd(X)], xX]m-t-d = O. i=O

By Kharchenko'::; Theorem [13],

m-I

,xXJi,d(:r)X + i=O

for all X, Y, rj. T2 E R. In particular for Tj = T2 = O.

m-j

[[d(:r)X. d(:r:)YJn, [d(x)Y, xX]m + L [[[:r1", :rX],. d(x)X], :.z:X]md-l] = 0, 1.=0

which is a non-trivial GPI for R. because x and d(x) are linearly C-illdependent, a contradiction. 0

Now we are ready to prove the theorem.

Proof of TheOTern 2.7. Suppose d(p)p le 0 and then we derive a contradiction. By Lemma 2.8, R is a GPI-ring, so is also Q (see . By [20], Q is a primitive ring with H = 50c(Q) le o. Moreover we may assume that [pH,pH]pH le D, otherwise by ,[pQ,pQ]pQ = 0, which is a contradiction. By Lemma 2.1, we may assume that d(pH)pH i- D. Now we consider the two cases.

Ca..se 1. Let d be Q-inner derivation induced by an element a E Q. According to our assumption and by [17], we assume that

[y, =0 (2.2)

Vo!. 3 (2006) Engel Type Identities

is satisfied by pQ and hence by pH. Let c = c2 E pH and yE H. Replacing x with c and y with cy(I - c) in (2.2) and then left multiplying it by (1 - e) and right multiplying it by cy we obtain

0= (1 - e) eL [a, cy(I - e)]]". [a, [cy(1 - c),

= (-1)"ffi(1 - e) cl, [a, cy(I - , aey(1 - c) - ey(1 - c)a]ey

= (_I),n+1(I_ c) cl, [a.ey(l- c)]]ney(l- e)aey

+( _1)m+1(1 - e)aey(I - e) eL [a, ey(1 - e)]]"ey

(1 - c) ~(-I)J G) [a. cy(I - eHa, ey(I - e)]"- J ey(I - e)aey

+(- (I-e)aeY(I-e)i)-I)J(n]la,eY(I-eWa.c][a,CY(I- ey. J=() JJ

This implies n

o (-1 L (-1)" - e)acy)"+2 J=O

" +(_I)1TL+1 L(-I)"G)((l - e)aey)n-c2 j=()

(_I)"+m+l2n+ 1 ((I _ e)aey)"+2.

Since char R cl 2, we have ((1 - e)aey)"+2 = O. By Levitzki's Lemma [12, Lemma 1.1], (I - e)acy = 0 for all y E H. By primeness of H, (1 - = 0 i.e.. ae = eae. By [10. Lemma 1], H is a regular ring. Thus for each r E pH. there exists an idempotent e E pH such that r = er and c E rH. So, for r E pH. ar = aer =

eaCT = ear E pH. Therefore, apH c:; pH. Denote the left annihilator of pH in H by iJ;(pH). Then pH = PHn~~(iJH) , a prime C-algebra with the derivation d such

that d(x) = d(x:). for all x E pH. By assumption ,ve have that

[[d(x), dCY)]n. d([y, = 0

for all x, yE pH. By Theorem 2.6, either d = 0 or pH is commutative. In the first case we have d(pH)pH = 0 and in the later case [pH. pH]pH = O. In both cases. we have a contradiction.

Case n. Now we assume that the derivation d is not Q-inner. Let I' E pH. Since 2H is regular ring, there exists an element e = e E pH such that I' = eT and

c ErH. According to our assumption. we have for all b E pH and x. yE H,

that is

[[d(b)x + bd(x} d(b)y + bd(y)Jn, [d(b)y + bd(y). b:r;]m rn---l

+ L [[[by, bx];, d(b)x + bd(T)], = o. i=O

26 RK. Sharrna and Basudeb Dhara Mediterr. j. math.

By Kharchenko's Theorem [13].

rn-~ 1

i=O

for all x, y, 1'1, 1'2 EH. In particular for 1'1 = 1'2 = 0,

rn-1

[[d(b)x, d(b)y]" , [d(b)y. bX]m + L bX]i. d(b)x], bX]m-I-1] = 0, 1,=0

for an x, yE H. For b = e, .r ~~ e and y = ey(l - e) we have

d(e)ey(l - e)]", [d(e)ey(l - e), m-1

- cl, d(e)e], e]rn-i-d = Cl. i=O

Since ed(e)e = () for any idernpotent e, this gives

rn-l

L l)i[ey(l - e)d(e)e - d(e)ey(l - e), d = Cl. ;=0

Now again using the fact that = Cl for any idempotent e and that ehe commutes with e, we see that the sum term contains a nonzero value for i = rn - 1. Thus

[( -l)"(d(e)ey(l - e)f'a(e)e. (_1)m-1(ey(l - e)d(e)e - d(e)ey(l - e))] = Cl.

Right multiplying it by cy and left multiplying it by (1 - e) we get

r (-1 t(1 - eHd(e)ey(l -- e))"d(e)ey(l - e)d(e)ey

-t-(1- e)d(e)eyO - e)(d(e)ey(l - e))"d(e)ey} = Cl

This implies (-~ 1),,+m-l 2((1 - = Cl. for all y EH. By Levitzki's Lemma [12], (1- e)d(e)eH = Cl. By primeness of H. (1- e)d(e)e = O. This implies (1 - c)d(e) = (1- e)d(c2 ) = (1 - e)d(e)e = Cl. Thus die) = ed(e) E cH c:;; pH. Now d(T) = deer) = d(e)er + ed(er) E pH. Hence d(pH) ~ pH.

Set pH = pH"q~(pH) ,with IH(pH) the left annihilator of pH in H. Then d

induces a canonical derivation don pH satisfying

[[d(x). d(y)]n, defy, = D.

for all X, y E pH. By Theorem 2.6, we have either d = 0 or pH is commutative. Therefore we have that either d(pH)pH = 0 or [pH, pH]pH = 0. In both cases, we have contradiction. This completes the proof of the theorem. C

27 Vol. 3 (2006) EllgeJ Type Identities

3. The Case: R is a Semiprime Ring

In this section we extended Theorem 2.3 and Theorem 2.6 to the semiprime case. Let R be a semiprime ring and U be its right Utumi quotient ring. It is well known that any derivation of a semiprime ring R can be uniquely extended to a derivation of its right Utumi quotient ring U and so any derivation of R can be defined on the whole of U i17, Lemma 2].

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

Lemma 3.1. ([1. Lemma 1 and Theorem 1] or [17, pages 31-32]) Let R be a 2-torsion free semiprime ring and P a maximal ideal of C. Then PU is a pnme ideal of U invanant under all derivations of U. Moreover, n{PU I P is a maximal ideal of Cwith UI PU 2-torsion free} = o.

Theorem 3.2. Let R be a 2-to'f"8ion free 8emipr-ime ring, d a non-zem denvation of R such that d(Y)]n. = 0 for all x, y E R, n, m 2" 0 fixed integer8. Then d maps R into its centre.

Proof. Since any derivation d can be uniquely extended to a derivation in U and Rand U satisfy the same differential identities [17, Theorem 3] we have

for all x. y E U. Let P be any maximal ideal of C such that UI PU is 2-torsioIl free. Then, by Lemma 3.1, PU is a prime ideal of U invariant under d. Set U = UIPG'. Then derivation d canonically induces a derivation !l on U defined by !lex) = d(x) for all x E U. Therefore,

for all x. y E U. By Theorem 2.3, either !l ~~ 0 or [U, U] = 0 i.e ... d(U) t::: PU or [U, Ul t::: PU. In any case d(U)[U, <;;: PU for any maximal ideal P of C. By Lemma 3.1, n{PU I P is a maximal ideal of C with UIPU 2-torsion free} = O. Thus d(U)[U. U] = O. Without loss of generality we have d(R)[R, R] = O. This implies

0= d(R)[R2, R] = d(R)R[R, R] + d(R)[R,R]R = d(R)R[R, R].

Therefore [R, d(R)]R[R, d(R)] = O. By semiprimeness of R, we have [R, d(R)] = 0 that is d(R) <;;: Z(R). This completes the proof of the theorem. 0

Similarly. we can proof the following result.

Theorem 3.3. Let R be a 2-torsion free semip'r'ime ring, d a non-zem derivation of R such that d(y)Jr,.d([y,x]m)] = 0 for all X.y E R, n,m 2" 0 fixed intege'f"8. Then d map8 R into its centre.

28 R.K Sharma and Basudeb Dhara Mediterr. j. math.

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29 VoL 3 (2006) Bngel Type Identities

Acknowledgment

The authors would like to thank the referee for providing very helpful comments and suggestions.

Rajendra K. Sharma Department of Mathematics Indian Institute of Technology. Delhi Hauz Khas, New Delhi-l100l6 India e-mail: rksharma@maths. iitd. ernet . in

Basudeb Dhara Department of Mathematics Indian Institute of Technology, Kharagpur Kharagpur-721302 India e-mail: basu_dhara@yahoo.com

Received: September 18. 2005

Revised: January 5, 2006

Accepted: January 23. 2006

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