# Products of derivations which act as Lie derivations on commutators of right ideals

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### DERIVATIONS ON COMMUTATORS OF RIGHT IDEALS

V. DE FILIPPIS

Received 5 April 2006; Revised 22 October 2006; Accepted 30 October 2006

LetRbe a prime ring of characteristic diﬀerent from 2,I a nonzero right ideal ofR,d andδnonzero derivations ofR, ands4(x1,x2,x3,x4) the standard identity of degree 4. If the composition () is a Lie derivation of [I,I] into R, then eithers4(I,I,I,I)I=0 or δ(I)I=d(I)I=0.

Throughout this note,R will be always a prime ring of characteristic diﬀerent from 2 with centerZ(R), extended centroidC, and two-sided Martindale quotient ringQ. Let f :R→Rbe additive mapping ofRinto itself. It is said to be a derivation ofRiff(xy)= f(x)y+x f(y), for allx,y∈R. LetS⊆Rbe any subset ofR. If for anyx,y∈S,f([x,y])= [f(x),y] + [x,f(y)], then the mapping f is called a Lie derivation onS. Obviously any derivation ofRis a Lie derivation on any arbitrary subsetSofR.

A typical example of a Lie derivation is an additive mapping which is the sum of a derivation and a central map sending commutators to zero.

The well-known Posner first theorem states that ifδanddare two nonzero derivations ofR, then the composition () cannot be a nonzero derivation ofR[12, Theorem 1]. An analog of Posner’s result for Lie derivations was proved by Lanski . More precisely, Lanski showed that ifδanddare two nonzero derivations ofRandLis a Lie ideal ofR, then () cannot be a Lie derivation ofLintoRunless char(R)=2 and eitherRsatisfies s4(x1,. . .,x4), the standard identity of degree 4, ord=αδ, forα∈C.

This note is motived by the previous cited results. Our main theorem gives a general-ization of Lanski’s result to the case when () is a Lie derivation of the subset [I,I] into R, whereIis a nonzero right ideal ofRand the characteristic ofRis diﬀerent from 2. The statement of our result is the following.

Theorem 1. LetRbe a prime ring of characteristic diﬀerent from 2,Ia nonzero right ideal

ofR,dandδnonzero derivations ofR, ands4(x1,. . .,x4) the standard identity of degree 4. If the composition (dδ) is a Lie derivation of [I,I] intoR, then eithers4(I,I,I,I)I=0 or δ(I)I=d(I)I=0.

Hindawi Publishing Corporation

International Journal of Mathematics and Mathematical Sciences Volume 2006, Article ID 53145, Pages1–12

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Remark 2. Notice that for allu,v∈[I,I], we obviously have that

()[u,v]=()(u),v+u, ()(v)+δ(u),d(v)+d(u),δ(v). (1)

Hence, since we suppose that () is a Lie derivation on [I,I], we will always assume as a main hypothesis that [δ(u),d(v)] + [d(u),δ(v)]=0, for anyu,v∈[I,I].

Remark 3. The assumptionS4(I,I,I,I)I=0 is essential to the main result. For example, considerR=M3(F), forFa field of characteristic 3, and letei j be the usual matrix unit inR. LetI=(e11+e22)R,δthe inner derivation induced by the elemente13,dthe inner derivation induced by the elemente12, that is,δ(x)=[e13,x]=e13x−xe13, andd(x)= [e12,x]=e12x−xe12, for allx∈R. In this case, notice thatS4(x1,x2,x3,x4)x5is an identity forI, moreover

δe11+e22x1,e11+e22x2,de11+e22y1,e11+e22y2

+de11+e22x1,e11+e22x2,δe11+e22y1,e11+e22y2

=de11+e22x1,e11+e22x2e11+e22y1,e11+e22y2e13

−de11+e22y1,e11+e22y2e11+e22x1,e11+e22x2e13=0 (2)

for anyx1,x2,y1,y2∈R, but clearlyd(I)I=[e12,I]I=0.

In the particular caseI=Rand bothd,δare inner derivations, induced, respectively, by some elementsa,b∈R, our theorem has the following flavor.

Lemma 4. LetRbe a prime ring of characteristic diﬀerent from 2,a,b∈Rsuch that [[a,v], [b,u]] + [[b,v], [a,u]]=0, for allv,u∈[R,R]. Then eitherais a central element ofRorb is a central one.

The proof is a clear special case of [8, Theorem 6]. We first fix some notations and recall some useful facts.

Remark 5. Denote byT=Q∗CC{X}the free product overCof theC-algebraQand the freeC-algebraC{X}, withXa countable set consisting of noncommuting indeterminates {x1,. . .,xn}. The elements ofTare called generalized polynomials with coeﬃcients inQ. I,IR, andIQsatisfy the same generalized polynomial identities with coeﬃcients inQ. For more details about these objects, we refer the reader to [1,2,4].

Remark 6. Any derivation ofRcan be uniquely extended to a derivation ofQ, and so any derivation ofRcan be defined on the whole ofQ[2, Proposition 2.5.1]. MoreoverQis a prime ring as well asRand the extended centroidCofRcoincides with the center ofQ [2, Proposition 2.1.7, Remark 2.3.1].

Remark 7. Let f(x1,. . .,xn,d(x1),. . .,d(xn)) be a diﬀerential identity ofR. One of the fol-lowing holds (see ):

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(2) orRsatisfies the generalized polynomial identity

fx1,. . .,xn,y1,. . .,yn. (3)

Moreover I, IR, and IQ satisfy the same diﬀerential identities with coeﬃcients in Q (see ).

Finally, as a consequence of [11, Theorem 2], we have the following.

Remark 8. LetRbe a prime ring andmi=1aiXbi+nj=1cjXdj=0, for allX∈R, where ai,bi,cj,dj∈RC. If{a1,. . .,am}are linearlyC-independent, then eachbiisC-dependent ond1,. . .,dn. Analogously, if{b1,. . .,bm}are linearlyC-independent, then eachaiisC -dependent onc1,. . .,cn.

For the remainder of the note we will assume that the hypothesis of the theorem holds but that the conclusion is false.

Thus, we will always suppose that there existc1,c2,c3,c4,c5,c6,c7,c8,c9∈I such that s4(c1,c2,c3,c4)c5=0, and eitherδ(c6)c7=0 ord(c8)c9=0.

We begin with the following.

Lemma 9. Letδanddboth be Q-inner derivations such that eitherδ(I)I=0 ord(I)I=0. ThenRis a ring satisfying a nontrivial generalized polynomial identity.

Proof. ByRemark 2, we assume that [δ(u),d(v)] + [d(u),δ(v)]=0, for anyu,v∈[I,I]. Leta,b∈Qsuch thatδ(x)=[a,x] andd(x)=[b,x], for allx∈R.

Without loss of generality, we may assume in this context thatδ(I)I=0. Notice that if {y,ay} are linearly C-dependent for all y∈I, then there exists α∈C, such that (a−α)I=0 (see [10, Lemma 3]). If we replaceabya−α, since they induce the same inner derivation, it follows thatδ(I)I=[a−α,I]I=0, a contradiction. Thus there exists x∈Isuch that{x,ax}are linearlyC-independent.

Letx∈Isuch that{x,ax}are linearlyC-independent andr1,r2,r3,r4are any elements ofR. Then

a,xr1,xr2,b,xr3,xr4+b,xr1,xr2,a,xr3,xr4=0. (4)

Denote

F1=r1xr2−r2xr1bxr3,xr4−r1xr2−r2xr1xr3,xr4b −r3xr4−r4xr3bxr1,xr2+r3xr4−r4xr3xr1,xr2b,

F2= −r3xr4−r4xr3axr1,xr2+r3xr4−r4xr3xr1,xr2a +r1xr2−r2xr1axr3,xr4−r1xr2−r2xr1xr3,xr4a, F3=r3xr4−r4xr3baxr1,xr2−r1xr2−r2xr1baxr3,xr4

+r1xr2−r2xr1axr3,xr4b−r3xr4−r4xr3bxr1,xr2a −r1xr2−r2xr1baxr3,xr4+r1xr2−r2xr1bxr3,xr4a +r3xr4−r4xr3abxr1,xr2−r3xr4−r4xr3axr1,xr2b.

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Hence (4) isaxF1+bxF2+xF3=0. If{ax,bx,x}are linearlyC-independent, then (4) is a nontrivial generalized polynomial identity forR, sinceF1=0 inT, usingb /∈C. On the other hand, if there existα1,α2∈Csuch thatbx=α1x+α2ax, it follows thatRsatisfies

axF1+α1xF2+α2axF2+xF3=0, (6)

that is, again a nontrivial GPI, because{x,ax}are linearlyC-independent, by the choice ofxand sinceF1+F2=0 inT, usinga,b /∈C.

The same argument shows that ifd(I)I=0, then there existsx∈Isuch that{x,bx} are linearlyC-independent andRsatisfies in any case a nontrivial GPI.

At this point, we need a result that will be useful in the continuation of the note.

Remark 10. LetR=Mn(F) be the ring ofn×nmatrices over the fieldF, denote byei j the usual matrix unit with 1 in the (i,j)-entry and zero elsewhere. Since there exists a set of matrix units that contains the idempotent generator of a given minimal right ideal, we observe that any minimal right ideal is part of a direct sum of minimal right ideals adding toR. In light of this and applying [6, Proposition 5, page 52], we may assume that any minimal right ideal ofRis a direct sum of minimal right ideals, each of the formeiiR.

Lemma 11. LetR=Mn(F) be the ring ofn×nmatrices over the fieldF of characteristic

diﬀerent from 2 andn≥2. Letdbe a nonzero inner derivation ofR, andIa nonzero right ideal ofR. Ifais a nonzero element ofIsuch that (d([x1,x2])[x3,x4]−d([x3,x4])[x1,x2])a= 0, for allx1,x2,x3,x4∈I, then eithers4(I,I,I,I)I=0 ordis induced by an elementb∈R such that (b−β)I=0, for a suitableβ∈Z(R).

Proof. Letbbe an element ofRwhich induces the derivationd, that is,d(x)=[b,x], for all x∈R. As above, letei j be the usual matrix unit with 1 in the (i,j)-entry and zero elsewhere and writea=ai jei j,b=bi jei j, withai jandbi jelements ofF.

We know thatIhas a number of uniquely determinated simple components: they are minimal right ideals ofRandI is their direct sum. In light ofRemark 10, we may write I=eRfor somee=t

i=1eiiandt∈ {1, 2,. . .,n}. Sinces4(I,I,I,I)I=0 in caset≤2, we may suppose thatt≥3.

First of all, we want to prove thatbrs=0 for alls≤tandr=s. To do this, suppose by contradiction that there existi=jsuch thatbi j=0 (j≤t). Without loss of generality, we replacebbybi j−1(b−bj jIn), whereInis the identity matrix inMn(F) so that we assume bi j=1 andbj j=0. Moreovera=exfor a suitablex∈R.

Let nowk≤t,k=i,j, [x1,x2]=eki, [x3,x4]=eji. In this case, we have

0=b,ekiejia−b,ejiekia (7)

and left multiplying byekk,

ekibejia=0, (8)

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On the other hand, if we choose [x1,x2]=ekiand [x3,x4]=ejk, we have

0=b,ekiejk−b,ejkekia=b,ekiejka= −bi jekka. (9)

Thereforeerra=0 for allr=j, that is,a=ej ja. Finally, consider [x1,x2]=ekiand [x3,x4]= ekk−ej j. Then

0=b,ekiekk−ej j−b,ekk−ej jekia=ekibej ja, (10)

that is,ej ja=0. This implies thatea=0, so thata=0, a contradiction. This argument says that ifa=0, thenbi j=0 for alli=j,j≤t.

Suppose that (b−β)I=0, forβ∈F. In this case, there exist 1≤r,s≤t, withr=s, such thatbrr=bss.

Let f be theF-automorphism ofRdefined by f(x)=(1−ers)x(1 +ers). Thus we have that f(x)∈I, for allx∈Iand

f(b),x1,x2x3,x4−f(b),x3,x4x1,x2f(a)=0 (11)

for allx1,x2,x3,x4∈I. Ifa=0, then f(a)=0, and as above, the (r,s)-entry of f(b) is zero. On the other hand,

f(b)=1−ers

b1 +ers

=b+brrers−bssers, (12)

that is,brr=bss, a contradiction. This means that there existsβ∈Fsuch that (b−β)I= 0. Denote b−β=p. Since b and p induce the same inner derivationd, we have that ([p, [x1,x2]][x3,x4][p, [x3,x4]][x1,x2])a=0 withpI=0.

Lemma 12. LetRbe a prime ring of characteristic diﬀerent from 2,da nonzero inner

deriva-tion ofR,Ia nonzero right ideal ofR. Ifais a nonzero element ofIsuch that (d([x1,x2])[x3, x4]−d([x3,x4])[x1,x2])a=0, for allx1,x2,x3,x4∈I, then eithers4(I,I,I,I)I=0 ordis induced by an elementb∈Rsuch that (b−β)I=0, for a suitableβ∈Z(R).

Proof. As a reduction ofLemma 9, we have that ifRis not a GPI ring, then we are done. Thus consider the only case whenRsatisfies a nontrivial generalized polynomial identity. Thus the Martindale quotient ringQofRis a primitive ring with nonzero socleH= Soc(Q). H is a simple ring with minimal right ideals. LetD be the associated division ring ofH, by Dis a simple central algebra finite-dimensional overC=Z(Q). Thus H⊗CFis a simple ring with minimal right ideals, withF an algebraic closure ofC. Let b be an element of Rwhich induces the derivationd. Moreover ([b, [x1,x2]][x3,x4] [b, [x3,x4]][x1,x2])a=0, for allx1,x2,x3,x4∈IH⊗CF(see, e.g., [4, Theorem 2]). Notice that ifCis finite, we chooseF=C.

Now we claim that for anyc∈IH, there existsβ∈Cwith (b−β)c=0. If not, then for somec∈IH, (b−β)c=0 for allβ∈C, so in particularbc=0. SinceHis regular , there existsg2=gIH, such thatcgIH, ande2=eH

CF, such that g,bg,gb,a,c,bc,cb∈eH⊗CF

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Letx1,x2,x3,x4∈ge(H⊗CF)eanda=eae=0, then

0=eb,x1,x2x3,x4−b,x3,x4x1,x2eae. (14)

ApplyingLemma 11, we have thate(b−λ)ec=0 forλ∈C, so (b−β)c=0, contradicting the choice ofc.

As in the proof ofLemma 9, by [10, Lemma 3], we conclude that there existsβ∈C

such that (b−β)I=0.

Lemma 13. Ifδanddare both inner derivations, then the theorem holds.

Proof. ByRemark 2, we assume that [δ(u),d(v)] + [d(u),δ(v)]=0, for anyu,v∈[I,I]. Let a,b∈Q such that δ(x)=[a,x] and d(x)=[b,x], for all x∈R. Since in light of Lemma 9,Rsatisfies a nontrivial GPI, then without loss of generality,R is simple and equal to its own socle andIR=I. In fact,Qhas nonzero socleHwith nonzero right ideal J=IH. Note thatHis simple,J=JH, andJsatisfies the same basic conditions asI. Now just replaceRbyH,IbyJ, and we are done.

Recall thats4(c1,c2,c3,c4)c5=0 and eitherδ(c6)c7=0 ord(c8)c9=0. By the regularity of R, there exists an element e2=eIR such that eR=c1R+c2R+c3R+c4R+c5R+ c6R+c7R+c8R+c9Randeci=ci, fori=1,. . ., 9. We note thats4(eRe,eRe,eRe,eRe)=0 (and dimC(eRe)9).

Letx,y,z∈R, so

a,e,ex(1−e),b, [ey,ez]+b,e,ex(1−e),a, [ey,ez]=0. (15)

DenoteA=(1−e)ae,B=(1−e)be. Assume thatA=0 butB=0. Consider first the case when{1−e, (1−e)a}are linearlyC-independent. Equation (15), multiplied on the left by (1−e), says that

(1−e)b[ey,ez]aex(1−e) + (1−e)b[ey,ez]ex(1−e)a=0. (16)

ByRemark 8and since{1−e, (1−e)a}are linearlyC-independent, it follows that there existsλ1∈Csuch that(1−e)b[ey,ez]ae=λ1(1−e)b[ey,ez]e.

Therefore

(1−e)b[ey,ez]exλ1(1−e) + (1−e)b[ey,ez]ex(1−e)a=0, (17)

which implies that (1−e)b[ey,ez]e=0, again since {1−e, (1−e)a} are linearly C-independent. If we denoteT=eR, (1−e)b[T,T]T=0 forces (1−e)bT[T,T]T=0, so either (1−e)bT=0 or [T,T]T=0. Thus we have that eitherB=(1−e)be=0 or [x1,x2]x3 is an identity foreR. In this last case, a fortiori s4(x1,x2,x3,x4)x5 is an iden-tity for eR. In both cases, we have a contradiction, since we suppose that B=0 and s4(c1,c2,c3,c4)c5=0.

Suppose now that (1−e)a=λ(1−e), for someλ∈C. Equation (16) says that

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and so

(1−e)b[ey,ez]ae+λ(1−e)b[ey,ez]e=0, (19)

that is, fora=λe−ae,

(1−e)b[ey,ez]a=0. (20)

DenoteU=[ey,ez]a. Since (1−e)be[Ux1,ex2]a=0, for allx1,x2∈R, it follows that (1−e)bex2Ux1a=0, and so eithera=0 orU=0. Again denoteT=eR. IfU=0, we have [T,T]a=0, so that [T,T]Ta=0, which implies eithera=0 or [T,T]T=0. Since [eR,eR]e=0, we haveae=λein any case.

All the previous arguments say that (a−λ)e=0. Replacingabya−λ=a, since they induce the same inner derivation, we may assume that for allx,y,z,t∈R,

a, [ex,ey],b, [ez,et]+b, [ex,ey],a, [ez,et]=0. (21)

Left multiplying (21) by (1−e), we have

(1−e)be[ez,et], [ex,ey]a=0, (22)

in particular

0=(1−e)be[ez,et],ex,ey(1−e)a=(1−e)be[ez,et]exey(1−e)a, (23)

and by the previous same argument, (1−e)a=0, that is,a=ea. In light of this, by (22),

eR(1−e)be[eze,ete], [exe,eye](eaRe)=0. (24)

Let Gbe the subgroup of eRegenerated by the polynomial [[eze,ete], [exe,eye]]. It is easy to see thatGis a noncentral Lie ideal ofeRe. In this condition, it is well known that [eRe,eRe]⊆G, and soeR(1−e)be[eRe,eRe]eaRe=0.

Consider now the simple Artinian ringeRe, then we have that

eR(1−e)beex1e,ex2e(eaRe)=0 ∀x1,x2∈R. (25)

LetU=[ex1e,ex2e](eaRe), soeR(1−e)beU=0. Since

eR(1−e)beUex1e,ex2e(eaRe)=0, (26)

then

eR(1−e)bex2Uex1(eaRe)=0. (27)

It follows that if (1−e)be=0, thena=0, that is,a=λ∈C, a contradiction. Thus the conclusion is that ifA=(1−e)ae=0, thenB=(1−e)be=0.

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Now we assume that neitherA=0 norB=0, and proceed to get contradictions, prov-ing thatA=B=0.

Left multiplying (15) by (1−e) and right multiplying bye, we get

(1−e)aex(1−e)b[ey,ez]e+ (1−e)b[ey,ez]ex(1−e)ae

+ (1−e)bex(1−e)a[ey,ez]e+ (1−e)a[ey,ez]ex(1−e)be=0. (28)

If we denoteA=A[ye,ze],B=B[ye,ze], it follows that

AxB+BxA+BxA+AxB=0. (29)

Consider now the case whenA,Bare linearlyC-independent.

In light ofRemark 8and (29), it follows that there existα1,α2,α3,α4inCsuch that B=α1A+α2B,A=α3A+α4B. So we rewrite (29) as follows:

2α1AxA+ 2α4BxB+α2+α3AxB+α2+α3BxA=0, (30)

that is,

Ax2α1A+α2+α3B+Bx2α4B+α2+α3A=0. (31)

SinceA,BareC-independent, by (31) and againRemark 8, it follows that 2α1A+ (α2+ α3)B=0 and 2α4B+ (α2+α3)A=0, so the independence ofAandB forcesα1=α4= α2+α3=0.

Therefore we have that B[eRe,eRe]⊆CB. Notice that B[eRe,eRe]=0. In fact, if B[eRe,eRe]=0, since [eRe,eRe]=(0) is a noncentral Lie ideal of the simple Artinian ringeRe, the contradictionB=0 is immediate.

Letu,v∈[eRe,eRe]. Hence there existω1,ω2, 0=ω∈Csuch that

B[u,v]=ωB=0, Bu=ω1B, Bv=ω2B, (32)

and by calculation we get the contradiction

0=ωB=B[u,v]=0. (33)

Hence we may assume thatAandBare linearlyC-dependent, sayA=αB, for 0=α∈C, so alsoA=αB. Equation (29) is now 2αBxB+ 2αBxB=0, and it follows thatBandB must be linearlyC-dependent, so thatBxB=0 andB=B=0.

Therefore in any case, we have that ifs4(eR,eR,eR,eR)e=0, then (1−e)be=(1 e)ae=0.

LetJ=eR,J=JJ∩lR(J);Jis a primeC-algebra. Sinced(J)⊆Jandδ(J)⊆J,dandδ induce onJthe following two derivations:

d:J−→J such thatd(x)=d(x),

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Therefore, we have

0=δr1,r2,dr3,r4+dr1,r2,δr3,r4 (35)

for allr1,r2,r3,r4∈J. ByLemma 4, we have that one of the following holds:

δ=0, d=0, Jis commutative. (36)

Sinces4(J,J,J,J)J=0, the last case cannot occur. On the other hand, now we prove that also the other cases lead us to contradictions.

Suppose that the first case occurs, that is,δ(J)J=0. By the lemma in , there exists an elementq=a−α∈Q, withα∈C, such that (a−α)J=0. Moreoveraandqinduce the same inner derivationδ, so that we have

b,x1,x2x3,x4−b,x3,x4x1,x2q=0 ∀x1,x2,x3,x4∈J. (37)

In particular, for anyr∈R, choose [x1,x2]=[e,er(1−e)]=er(1−e). From (37), it fol-lows that

b,x3,x4eR(1−e)q=0. (38)

If (1−e)q=0, we haveq=eq∈J. Under this condition, byLemma 12, it follows from (37) that eitherq=0, which implies the contradictiona∈Candδ=0, or (b−β)J=0 for a suitableβ∈C, that is,d(eR)eR=0. So consider the case when [b, [x3,x4]]e=0 for all x3,x4∈J, and remember thatbe=ebe. This implies that [ebe, [y1,y2]]=0 for ally1,y2∈ eRe, that is, eithereReis a commutative central simple algebra orebe∈Ce. In the first case, we have the contradiction 0=s4(ec1,ec2,ec3,ec4)ec5=s4(c1,c2,c3,c4)c5=0. In the second one, we get againd(eR)eR=0. Therefore we conclude that in any case,δ(eR)eR= d(eR)eR=0, which is again a contradiction because ofδ(c6)c7=0 ord(c8)c9=0.

Obviously by a similar argument and (36), we are also finished whend(J)J=0. For the proof of the main theorem, we need the following results.

Lemma 14. LetRbe a prime ring of characteristic diﬀerent from 2 andI a nonzero right

ideal ofR. If for anyx1,x2,x3,x4∈I, [[x1,x2], [x3,x4]]=0, then [I,I]I=0.

Proof. First note that if [y, [I,I]]=0, for somey∈R, then, for anys,t∈I, we have 0= [y, [st,t]]=[s,t][y,t]. In particular, for anyx∈IR, 0=[sx,t][y,t]=[s,t]x[y,t], that is [s,t]IR[y,t]=0. By the primeness ofR, we have that either [s,t]I=0, that is, [I,I]I=0, or [y,I]=0. In this last case, 0=[y,IR]=I[y,R] forcingy∈Z(R).

Therefore, if we assume that [I,I]I=0, the assumption [[I,I], [I,I]]=0 forces 0= [I,I]⊆Z(R). Lets,t∈Ibe such that [s,t]I=0 and [s,t]∈Z(R). Then 2s[s,t]=[s2,t] Z(R), so [s,t]=0 forcess∈Z(R) and we have the contradiction [s,I]=0.

Lemma 15. LetR be a noncommutative prime ring of characteristic diﬀerent from 2, q

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Proof. Suppose that [I,I]I=0. As inLemma 14, first we recall that the condition [y, [I, I]]=0 forces y∈Z(R). This means that [q, [I,I]]⊆Z(R), since [[q, [I,I]], [I,I]]=0. Moreover we may assume that [q, [I,I]]=0, if not, then we are finished byLemma 14.

Note that from [q, [I,I]]⊆Z(R), it follows that [q, [[I,I],I]]⊆Z(R). Expanding this yields [[I,I], [q,I]]⊆Z(R). Since for allx∈I, we have [[I,I], [q,xq]]⊆Z(R), then [[I,I], [q,I]q]⊆Z(R). Hence

0=[I,I], [q,I]q,q=q, [q,I]q, [I,I]. (39)

Since the second factor is nonzero and central, we have [q, [q,I]]=0, which implies that for allx,y∈I,

0=q, [q,xy]=q, [q,x]y+x[q,y]=2[q,x][q,y]. (40)

This means that [q,I][q,I]=0 and a fortiori [q, [I,I]][q, [I,I]]=0 giving the

We are ready to prove the following main result.

Theorem 16. LetR be a prime ring of characteristic diﬀerent from 2,I a nonzero right

ideal ofR,dandδ nonzero derivations ofR,s4(x1,. . .,x4) the standard identity of degree 4. If the composition (dδ) is a Lie derivation of [I,I] intoR, then eithers4(I,I,I,I)I=0 or δ(I)I=d(I)I=0.

Proof. ByRemark 2, we assume that [δ(u),d(v)] + [d(u),δ(v)]=0, for anyu,v∈[I,I]. Suppose by contradiction that there existc1,c2,c3,c4,c5, c6, c7, c8, c9 in I such that s4(c1,c2,c3,c4)c5=0 and eitherδ(c6)c7=0 ord(c8)c9=0.

First suppose thatδanddareC-independent moduloDint. Lett1,t2,t3,t4∈I, by assumption,Rsatisfies

δt1x1,t2x2+t1x1,δt2x2,dt3x3,t4x4+t3x3,dt4x4

+dt1x1,t2x2+t1x1,dt2x2,δt3x3,t4x4+t3x3,δt4x4 =δt1x1+t1δx1,t2x2+t1x1,δt2x2+t2δx2,

dt3x3+t3dx3,t4x4+t3x3,dt4x4+t4dx4

+dt1x1+t1dx1,t2x2+t1x1,dt2x2+t2dx2,

δt3x3+t3δx3,t4x4+t3x3,δt4x4+t4δx4=0.

(41)

By Kharchenko’s theorem ,Rsatisfies the generalized polynomial identity

δt1x1+t1y1,t2x2+t1x1,δt2x2+t2y2,dt3x3+t3z3,t4x4+t3x3,dt4x4+t4z4

(11)

in particularRsatisfies [[t1y1,t2x2], [t3x3,t4z4]], soQsatisfies this as well, and for ally1= x2=x3=z4=1∈Q, it follows that [[I,I], [I,I]]=0. Thus byLemma 14, we conclude that [I,I]I=0, that is,s4(I,I,I,I)I=0, which contradictss4(c1,c2,c3,c4)c5=0.

Now letδanddbeC-dependent moduloDint. There existγ1,γ2∈C, such thatγ1δ+ γ2d∈Dint, and byLemma 13, it is clear that at most one of the two derivations can be inner.

Without loss of generality, we may assume thatγ1=0, so thatδ=αd+ad(q), for α∈Candad(q) the inner derivation induced by the elementq∈Q.

Ifdis inner, then alsoδ is inner, and we have thatdis an outer derivation ofR. Let t1,t2,t3,t4∈I,Rsatisfies

αdt1x1,t2x2+t1x1,dt2x2,dt3x3,t4x4+t3x3,dt4x4

+q,t1x1,t2x2,dt3x3,t4x4+t3x3,dt4x4

+αdt1x1,t2x2+t1x1,dt2x2,dt3x3,t4x4+t3x3,dt4x4

+dt1x1,t2x2+t1x1,dt2x2,q,t3x3,t4x4 =αdt1x1+t1dx1,t2x2+t1x1,dt2x2+t2dx2,

dt3x3+t3dx3,t4x4+t3x3,dt4x4+t4dx4

+q,t1x1,t2x2,dt3x3+t3dx3,t4x4+t3x3,dt4x4+t4dx4

+αdt1x1+t1dx1,t2x2+t1x1,dt2x2+t2dx2,

dt3x3+t3dx3,t4x4+t3x3,dt4x4+t4dx4

+dt1x1+t1dx1,t2x2+t1x1,dt2x2+t2dx2,q,t3x3,t4x4,

(43)

and so the Kharchenko theorem shows thatRsatisfies

αdt1x1+t1y1,t2x2+t1x1,dt2x2+t2y2,dt3x3+t3y3,t4x4+t3x3,dt4x4+t4y4

+q,t1x1,t2x2,dt3x3+t3y3,t4x4+t3x3,dt4x4+t4y4

+αdt1x1+t1y1,t2x2+t1x1,dt2x2+t2y2,dt3x3+t3y3,t4x4+t3x3,dt4x4+t4y4

+dt1x1+t1y1,t2x2+t1x1,dt2x2+t2y2,q,t3x3,t4x4.

(44)

In caseα=0, forx1=x4=0 in (44), we have thatRsatisfies

2αt1y1,t2x2,t3x3,t4y4, (45)

(12)

Now letα=0. In this case forx4=0 in (44), we have thatRsatisfies

q,t1x1,t2x2,t3x3,t4y4. (46)

As aboveQsatisifes this and, takingx1,x2,x3,y4=1, it follows that

q, [I,I], [I,I]=0. (47)

Then, byLemma 15, we conclude again with the contradiction [I,I]I=0.

Similarly, whenγ2=0, thend=βδ+ad(q), for someβ∈C, and mimicking the

Acknowledgment

The author wishes to thank the referee for her/his valuable suggestions on how to improve the paper.

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V. De Filippis: Dipartimento di Matematica, Universit´a di Messina, Salita Sperone 31, Messina 98166, Italy

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