Right Sided Ideals and Multilinear Polynomials with Derivations on Prime Rings
BASUDEBDHARA(*) - RAJENDRAK. SHARMA(**)
ABSTRACT- Le tRbean associativeprimering of charR62with centerZ(R) and extended centroidC,f(x1;. . .;xn) a nonzero multilinear polynomial overCinn noncommuting variables,da nonzero derivation ofRandra nonzero right ideal of R. Weprovethat: (i) if [d2(f(x1;. . .;xn));f(x1;. . .;xn)]0for allx1;. . .;xn2r thenrCeRCfor some idempotent elementein thesocleofRCandf(x1;. . .;xn) is central-valued ineRCeunlessdis an inner derivation induced byb2Qsuch that b20 and br0; (ii) if [d2(f(x1;. . .;xn));f(x1;. . .;xn)]2Z(R) for all
x1;. . .;xn2rthenrCeRCfor some idempotent elementein thesocleofRC
and eitherf(x1;. . .;xn) is central ineRCeoreRCesatisfies the standard identity S4(x1;x2;x3;x4) unlessdis an inner derivation induced byb2Qsuch thatb20 andbr0.
Throughout this paper,Ralways denotes a prime ring with extended centroidCandQits two-sided Martindale ring of quotient. Bydwemean a nonzero derivation ofR. Forx;y2R, thecommutator ofx;yis denoted by [x;y] and defined by [x;y]xy yx. Wedenote[x;y]2[[x;y];y]
[x;y]y y[x;y].
A well known result proved by Posner [17] states that R must be commutativeif [d(x);x]2Z(R) for all x2R. In [10] Lanski generalized the Posner's result to a Lie ideal. More precisely Lanski proved that ifL is a noncommutativeLieideal ofRsuch that [d(x);x]2Z(R) for allx2L,
(*) Indirizzo dell'A.: Department of Mathematics, Belda College, Belda, Paschim Medinipur-721424, India.
E-mail: basu dhara@yahoo.com
(**) Indirizzo dell'A.: Department of Mathematics, Indian Institute of Technol- ogy, Delhi, Hauz Khas, New Delhi-110016, India.
E-mail: rksharma@maths.iitd.ac.in
Mathematics Subject Classification: 16W25, 16R50, 16N60.
This work is supported by a grant from University Grants Commission, India.
then char R2 and R satisfies S4(x1;x2;x3;x4), thestandard identity.
Notethat a noncommutativeLieideal ofR contains all thecommutators [x1;x2] forx1;x2 in some nonzero ideal ofR( see [10, Lemma 2 (i), (ii)]).
So, it is natural to consider the situation when [d(x);x]2Z(R) for all commutators x[x1;x2] or more general case xf(x1;. . .;xn) where
f(x1;. . .;xn) is a multilinear polynomial. In [11] Lee and Lee proved that
if [d(f(x1;. . .;xn));f(x1;. . .;xn)]2Z(R) for allx1;. . .;xn in somenonzero ideal of R, then f(x1;. . .;xn) is central-valued on R, except when char R2 and Rsatisfies S4(x1;x2;x3;x4). Recently, De Filippis and Di Vin- cenzo (see [7]) consider the situationd([d(f(x1;. . .;xn));f(x1;. . .;xn)])0 for all x1;. . .;xn2R, where d and d aretwo derivations of R. The statement of De Filippis and Di Vincenzo's theorem is the following:
THEOREMA ([7, Theorem 1]). Let K be a noncommutative ring with unity, R a prime K-algebra of characteristic different from 2, d and d nonzero derivations of R and f(x1;. . .;xn) a multilinear polynomial over K. If d([d(f(x1;. . .;xn));f(x1;. . .;xn)])0 for all x1;. . .;xn2R, then f(x1;. . .;xn) is central-valued on R.
In casedand d are two same derivations, the differential identity be- comes [d2(f(x1;. . .;xn));f(x1;. . .;xn)]0 for all x1;. . .;xn2R. So, it is natural to ask, what happen in cases[d2(f(x1;. . .;xn));f(x1;. . .;xn)]2Z(R) for all x1;. . .;xn2R and [d2(f(x1;. . .;xn));f(x1;. . .;xn)]2Z(R) for all
x1;. . .;xn 2r, whereris a non-zero right ideal of R. In the present paper our
object is to study these cases.
For the sake of completeness we recall some basic notations, de- finitions and some easy consequences of the result of Kharchenko [8]
about the differential identities on a prime ring R. First, wedenoteby Der(Q) theset of all derivations on Q. By a derivation word D of R we
mean Dd1d2d3. . .dm for somederivations di of R. For x2R, we
denote by xD theimageof xunder D, that isxD( (xd1)d2 )dm. By a differential polynomial, we mean a generalized polynomial, with coefficients in Q, of the form F(xDij) involving noncommutativein- determinates xi on which thederivations words Dj act as unary op- erations. F(xDij)0 is said to be a differential identity on a subset Tof Q if it vanishes for any assignment of values from T to its in- determinates xi.
Now letDint betheC-subspaceofDer(Q) consisting of all inner deri- vations onQ. By Kharchenko's theorem [8, Theorem 2], we have the fol- lowing result:
LetRbe a prime ring of characteristic different from 2. If two nonzero derivationsdanddareC-linearly independent moduloDintandF(xDij) is a differential identity onR, whereDjarederivations words of thefollowing formd;d;d2;dd;d2, thenF(yji) is a generalized polynomial identity onR, whereyjiare distinct indeterminates.
As a particular case, we have:
If d is a nonzero derivation on R and F(x1;. . .;xn;xd1;. . .;xdn; xd12;. . .;xdn2) is a differential identity onR, then one of the following holds:
(i) eitherd2Dint
or
(ii) R satisfies the generalized polynomial identity F(x1;. . .;xn; y1;. . .;yn;z1;. . .;zn)
Denote byQCCfX1;. . .;Xngthefreeproduct of theC-algebraQand
CfX1;. . .;Xng, thefreeC-algebra in noncommuting indeterminates
X1;. . .;Xn.
Sincef(x1;. . .;xn) is a multilinear polynomial, we can write f(x1;. . .;xn)x1x2. . .xn X
I6s2Sn
asxs(1). . .xs(n)
where Sn is the permutation group over n elements and any as2C.
Wedenoteby fd(x1;. . .;xn) thepolynomial obtained from f(x1;. . .;xn) by replacing each coefficient as with d(as:1). In this way wehave
d(f(x1;. . .;xn))fd(x1;. . .;xn)X
i
f(x1;. . .;d(xi);. . .;xn) and
d2(f(x1;. . .;xn))d(fd(x1;. . .;xn))dX
i
f(x1;. . .;d(xi);. . .;xn)
fd2(x1;. . .;xn)X
i
fd(x1;. . .;d(xi);. . .;xn)
X
i
fd(x1;. . .;d(xi);. . .;xn)X
i6j
f(x1;. . .;d(xi);. . .;d(xj);. . .;xn)
X
i
f(x1;. . .;d2(xi);. . .;xn)
fd2(x1;. . .;xn)2X
i
fd(x1;. . .;d(xi);. . .;xn)
2X
i5j
f(x1;. . .;d(xi);. . .;d(xj);. . .;xn)X
i
f(x1;. . .;d2(xi);. . .;xn):
1. The case forrR.
LEMMA1.1. Let RMk(F)be the ring of all kk matrices over a field F of characteristic 62, b2R and f(x1;. . .;xn) is a multilinear poly- nomial over F. If k2and[[b;[b;f(x1;. . .;xn)]];f(x1;. . .;xn)]0for all x1;. . .;xn2R or if k3 and [[b;[b;f(x1;. . .;xn)]];f(x1;. . .;xn)]2Z(R) for all x1;. . .;xn2R, then either b2FIk or f(x1;. . .;xn)is central-va- lued on R.
PROOF. Le tb(bij)kk. Le teijbetheusual matrix unit with 1 in (i;j) entry and zero else where. Now we proceed to show that b2Z(R) if 2f(x1;. . .;xn) is non central valued onR.
For simplicity, wewritef(x1;. . .;xn)f(x), where x(x1;. . .;xn) RnR R(ntimes). Then by assumption,
[[b;[b;f(x)]];f(x)][b2f(x) 2bf(x)bf(x)b2;f(x)]2Z(R) for all x2Rn. Since f(x1;. . .;xn) is assumed to be noncentral on R, by [15, Lemma 2, Proof of Lemma 3] there exists a sequence of matrices
r(r1;. . .;rn) in R such that f(r)f(r1;. . .;rn)aeij60 where
06a2F and i6j. Thus
[b2aeij 2baeijbaeijb2;aeij]2Z(R):
Sincetherank of [b2aeij 2baeijbaeijb2;aeij] is 2, [b2aeij 2baeijb
aeijb2;aeij]0. Left multiplying by eij, we get 0eij( 2baeijbaeij)
2a2b2jieij. SincecharF62,bji0. Fors6t, le tsbea permutation in thesymmetric group Sm such that s(i)s and s(j)t. Le t c bethe automorphism of R defined by xcP
p;qjpqepqc
P
p;qjpqes(p);s(q). Then
f(rc)f(r1c;. . .;rnc)f(r)c aest60 and wehaveas abovebts0 for
s6t. Thusbis a diagonal matrix. For anyF-automorphismuofR,buenjoys thesameproperty as b does, namely, [[bu;[bu;f(x)]];f(x)]2Z(R) for all x2Rn. Hence,bumust bediagonal. WritebPk
i1aiieii; then for eachj61, wehave
(1e1j)b(1 e1j)Xk
i1
aiieii(bjj b11)e1j
diagonal. Therefore,bjjb11and sobis a scalar matrix.
LEMMA1.2. Let R be a prime ring of characteristic different from2 and f(x1;. . .;xn) a multilinear polynomial over C. If for any i1;. . .;n,
[f(x1;. . .;zi;. . .;xn);f(x1;. . .;xn)]0
for all x1;. . .;xn;zi2R, then the polynomial f(x1;. . .;xn)is central-valued on R.
PROOF. Le t a be a noncentral element ofR. Then replacingzi with [a;xi] wehavethat for anyi1;. . .;n
[f(x1;. . .;[a;xi];. . .;xn);f(x1;. . .;xn)]0 and so
"
Xn
i0
f(x1;. . .;[a;xi];. . .;xn);f(x1;. . .;xn)
#
0
which implies, [a;f(x1;. . .;xn)]20 for allx1;. . .;xn2R. By [11, Theorem], f(x1;. . .;xn) is central-valued onR.
THEOREM 1.3. Let R be a prime ring of characteristic different from 2, d a nonzero derivation of R, f(x1;. . .;xn) a multilinear polynomial over C. If
[d2(f(x1;. . .;xn));f(x1;. . .;xn)]2Z(R) for allx1;. . .;xn2R;
then either f(x1;. . .;xn)is central-valued on R or R satisfies the standard identity S4(x1;x2;x3;x4).
PROOF. Le t I be any nonzero two-sided ideal of R. If for every r1;. . .;rn 2I, [d2(f(r1;. . .;rn));f(r1;. . .;rn)]0, then by [14], this generalized differential identity is also satisfied by Q and hence by R as well. By Theorem A, f(r1;. . .;rn) is then central-valued on R and wearedone. Now weassumethat for somer1;. . .;rn 2I, 06[d2(f(r1;. . .;rn)); f(r1;. . .;rn)]2I\Z(R). Thus I\Z(R)60. Let K be a nonzero two-sided ideal of RZ, the ring of central quotients of R.
Since K\R is a nonzero two-sided ideal of R, (K\R)\Z(R)60.
Therefore, K contains an invertible element in RZ and so RZ is a simplering with identity 1.
By assumption,Rsatisfies the differential identity g(x1;. . .;xn;d(x1);. . .;d(xn);d2(x1);. . .;d2(xn))
[[fd2(x1;. . .;xn)2X
i
fd(x1;. . .;d(xi);. . .;xn)
2X
i5j
f(x1;. . .;d(xi);. . .;d(xj);. . .;xn)
X
i
f(x1;. . .;d2(xi);. . .;xn);f(x1;. . .;xn)];xn1]:
Ifdis notQ-inner, then by Kharchenko's theorem [8],
"
fd2(x1;. . .;xn)2X
i
fd(x1;. . .;yi;. . .;xn)
2X
i5j
f(x1;. . .;yi;. . .;yj;. . .;xn)
X
i
f(x1;. . .;zi;. . .;xn);f(x1;. . .;xn)
;xn1
#
0 (1)
for all xi;yi;zi;xn12R for i1;2;. . .;n. In particular, for any i, as- sumingy1 yi 1yi1 yn0;z1 zn0, wehave
[[fd2(x1;. . .;xn)2fd(x1;. . .;yi;. . .;xn);f(x1;. . .;xn)];xn1]0 and so
"
fd2(x1;. . .;xn)2X
i
fd(x1;. . .;yi;. . .;xn);f(x1;. . .;xn)
;xn1
#
0 for allxi;yi;xn12R,i1;2;. . .;n. Thus from (1), weobtain
"
2X
i5j
f(x1;. . .;yi;. . .;yj;. . .;xn)
X
i
f(x1;. . .;zi;. . .;xn);f(x1;. . .;xn)
;xn1
#
0 (2)
for allxi;yi;zi;xn12Rfori1;2;. . .;n.
By localizingRatZ(R), weobtain that (2) is also an identity ofRZ. Since RandRZsatisfy the same polynomial identities, in order to prove thatR satisfiesS4, wemay assumethatRis a simplering with 1. ThusRsatisfies theidentity (2). Now putting yi[b;xi]d(xi) and zi[b;[b;xi]]
d2(xi),i1;2;. . .;n for someb2=Z(R), where d is an inner derivation induced by someb2R, weobtain thatRsatisfies
[[d2(f(x1;. . .;xn));f(x1;. . .;xn)];xn1]0:
Thus by Martindale's theorem [16], Ris a primitivering with a minimal right ideal, whose commuting ringDis a division ring which is finitedi- mensional overZ(R). However, sinceRis simplewith 1,Rmust beArti- nian. HenceRDk0, thering ofk0k0 matrices overD, for somek01.
Again, by [9, Lemma 2], it follows that there exists a field F such that RMk(F), thering of all kkmatrices over the fieldF, andMk(F) sa- tisfies
[[d2(f(x1;. . .;xn));f(x1;. . .;xn)];xn1]0:
Ifk3, then by Lemma 1.1, we haveb2Z(R), a contradiction. Thusk2, that is,RsatisfiesS4(x1;x2;x3;x4).
Similarly, thesameconclusion can bedrawn in cased is an Q-inner derivation induced by someb2Q.
2. The case for one-sided ideal.
We begin with the following lemmas
LEMMA2.1. Letrbe a nonzero right ideal of R and d a derivation of R.
Then the following conditions are equivalent:
(i) d is an inner derivation induced by some b2Q such that br0;
(ii) d(r)r0.
For its proof, we refer to [2, Lemma].
LEMMA 2.2. Let R be a prime ring, r a nonzero right ideal of R,
f(x1;. . .;xt)a multilinear polynomial over C, a2R and n a fixed positive
integer. If f(x1;. . .;xt)na0 for all x1;. . .;xt2r, then either a0 or f(r)r0.
For its proof, we refer to [3, Lemma 2 (II)].
LEMMA2.3. Let R be a prime ring. If[d2(f(x1;. . .;xn));f(x1;. . .;xn)]2 Z(R) for all x1;. . .;xn2r, then R satisfies nontrivial generalized polynomial identity unless d is an inner derivation induced by b2Q such that b20and br0.
PROOF. Supposeon thecontrary thatRdoes not satisfy any nontrivial generalized polynomial identity (GPI). Thus we may assume that R is noncommutative, otherwiseRsatisfies trivially a nontrivial GPI. Now we consider the following two cases:
CASEI. Supposethatdis aQ-inner derivation induced by an element b2Qsuch thatb260. Then for anyx02r
[[b;[b; f(x0X1;. . .;x0Xn)]]; f(x0X1;. . .;x0Xn)]2Z(R) that is
[[b2f(x0X1;. . .;x0Xn) 2bf(x0X1;. . .;x0Xn)b
f(x0X1;. . .;x0Xn)b2; f(x0X1;. . .;x0Xn)];x0Xn1] (3)
is a GPI for R, so it is the zero element in QCCfX1;. . .;Xn1g.
Denote lR(r) theleft annihilator of r inR. Supposefirst that f1;b;b2g arelinearly C-independent modulo lR(r), that is (ab2bbg)r0 if and only if abg0. Since R is not a GPI-ring, a fortiori it can not be a PI-ring. Thus, by [13, Lemma 3] there exists x02r such that fb2x0;bx0;x0g arelinearly C-independent. Then we have that
[[b2f(x0X1;. . .;x0Xn) 2bf(x0X1;. . .;x0Xn)b
f(x0X1;. . .;x0Xn)b2;f(x0X1;. . .;x0Xn)];x0Xn1]0 is a nontrivial GPI forR, a contradiction.
Therefore, f1;b;b2g arelinearly C-dependent modulo lR(r), that is there exista;b;g2C, not all zero, such that (ab2bbg)r0. Suppose that a0. Then b60, otherwise g0. Thus by (bbg)r0, wehave that (bb 1g)r0. Sincebandbb 1ginduce the same inner derivation, wemay replaceb by bb 1g in the basic hypothesis. Therefore, in any casewemay supposebr0 and then from (3), R satisfies
x0Xn1f2(x0X1;. . .;x0Xn)b20. Since R does not satisfy any nontrivial
GPI, b20, a contradiction.
Next suppose that a60. In this casethereexist l;m2C such that b2x0lbx0mx0 for all x02r. If bx0 and x0 arelinearly C- dependent for all x02r, then again we obtain br0 and so b20.
Therefore choose x02r such that bx0 and x0 arelinearly C-in- dependent. Then replacing b2x0 with lbx0mx0, weobtain from (3)
that R satisfies
h(lbm)f2(x0X1;. . .;x0Xn) 2bf(x0X1;. . .;x0Xn)bf(x0X1;. . .;x0Xn)
f(x0X1;. . .;x0Xn)(lbm)f(x0X1;. . .;x0Xn)g f(x0X1;. . .;x0Xn)(lbm)f(x0X1;. . .;x0Xn)
2f(x0X1;. . .;x0Xn)bf(x0X1;. . .;x0Xn)bf2(x0X1;. . .;x0Xn)b2 ;x0Xn1i : This is a nontrivial GPI forR, because the term
lbf2(x0X1;. . .;x0Xn) 2bf(x0X1;. . .;x0Xn)bf(x0X1;. . .;x0Xn) x0Xn1
appears nontrivially, a contradiction.
CASEII. Supposethatdis an inner derivation induced by an elementb2Q such that b20. Thus wehavethat [ 2bf(X1;. . .;Xn)b;f(X1;. . .;Xn)]2 Z(R) is satisfied by r. In casethereexists x02r such that fbx0;x0g arelinearly C-independent, we have that [[ 2bf(x0X1;. . .;x0Xn)b;
f(x0X1;. . .;x0Xn)];x0Xn1] is a non trivial GPI forR, a contradiction. Hence fbx0;x0garelinearlyC-dependent for allx02r, that is there existsa2C such that (b a)r0. Thus wehavethat [af2(X1;. . .;Xn)(a b);Xn1] is satisfied byr, in particularRsatisfies:
[af2(x0X1;. . .;x0Xn)(a b);f(x0X1;. . .;x0Xn)]af3(X1;. . .;Xn)(a b) for anyx02r. SinceRis not GPI, it follows that eitherba2C, which is a contradiction, ora0 which meansbr0, as required.
CASEIII. Supposethatdis an inner derivation induced by an element b2Q such that br0. Thus wehavethat [ f2(X1;. . .;Xn)b2;Xn1] is satisfied byr, in particularRsatisfies:
[ f2(x0X1;. . .;x0Xn)b2;f(x0X1;. . .;x0Xn)]f3(x0X1;. . .;x0Xn)b2 for anyx02r. Again sinceRis not GPI weconcludethatb20.
CASE IV. Next suppose that d is not Q-inner derivation. By our as- sumption wehavethatRsatisfies
0h
fd2(xX1;. . .;xXn)2P
i fd(xX1;. . .;d(x)Xixd(Xi);. . .;xXn)
2P
i5j f(xX1;. . .;d(x)Xixd(Xi);. . .;d(x)Xjxd(Xj);. . .;xXn)
P
i f(xX1;. . .;d2(x)Xi2d(x)d(Xi)xd2(Xi);. . .;xXn);f(xX1;. . .;xXn)
;Xn1
i:
By Kharchenko's theorem [8], hfd2(xX1;. . .;xXn)2P
i fd(xX1;. . .;d(x)Xixri;. . .;xXn)
2P
i5j f(xX1;. . .;d(x)Xixri;. . .;d(x)Xjxrj;. . .;xXn)
P
i f(xX1;. . .;d2(x)Xi2d(x)rixsi;. . .;xXn);f(xX1;. . .;xXn)
;Xn1i
0 for all X1;. . .;Xn;r1;. . .;rn;s1;. . .;sn2R. In particular, for r1
r2 rn0, wehave hfd2(xX1;. . .;xXn)2P
i fd(xX1;. . .;d(x)Xi;. . .;xXn) 2P
i5jf(xX1;. . .;d(x)Xi;. . .;d(x)Xj;. . .;xXn)P
i f(xX1;. . .;d2(x)Xi;. . .;xXn)
P
i f(xX1;. . .;xsi;. . .;xXn);f(xX1;. . .;xXn)
;Xn1
i0:
HenceRsatisfies the blended component
[[f(xs1;. . .;xXn);f(xX1;. . .;xXn)];Xn1]0 which is a nontrivial GPI forR, a contradiction.
THEOREM2.4. Let R be an associative prime ring of char R62with center Z(R)and extended centroid C, f(x1;. . .;xn)a nonzero multilinear polynomial over C in n noncommuting variables, d a nonzero derivation of R andra nonzero right ideal of R. If[d2(f(x1;. . .;xn));f(x1;. . .;xn)]0 for all x1;. . .;xn 2rthenrCeRC for some idempotent e in the socle of RC and f(x1;. . .;xn)is central-valued on eRCe unless d is an inner deri- vation induced by b2Q such that b20and br0.
PROOF. Supposedis not aQ-inner derivation induced by an element b2Qsuch thatb20 andbr0.
Now assumefirst that f(r)r0, that is f(x1;. . .;xn)xn10 for all
x1;x2;. . .;xn12r. Then by [12, Proposition], rCeRC for someidem-
potent e2soc(RC). Since f(r)r0, wehavef(rR)rR0 and hence f(rQ)rQ0 by [4, Theorem 2]. In particular,f(rC)rC0, or equivalently, f(eRC)e0. Thenf(eRCe)0, that is,f(x1;. . .;xn) is a PI foreRCeand, a fortiori, central valued oneRCe.
Next assume thatf(r)r60, that isf(x1;. . .;xn)xn1is not an identity forrand then we derive a contradiction. By Lemma 2.3,Ris a GPI-ring
and so is also Q (see [1] and [4]). By [16], Q is a primitivering with Hsoc(Q)60. Moreover, we may assumef(rH)rH60, otherwise by [1]
and [4], f(rQ)rQ0, which is a contradiction. Choosea0;a1;. . .;an2rH such thatf(a1;. . .;an)a060. Leta2rH. SinceHis a regular ring, there existse2e2Hsuch that
eHaHa0Ha1H anH:
Then e2rH and aea;aieai for i0;1;. . .;n. Thus wehave
f(eHe)f(eH)e60. By our assumption and by [14, Theorem 2], we also assumethat
[d2(f(x1;. . .;xn));f(x1;. . .;xn)]
is an identity for rQ. In particular [d2(f(x1;. . .;xn));f(x1;. . .;xn)] is an identity forrHand so foreH. It follows that, for allr1;. . .;rn2H,
0[d2(f(er1;. . .;ern));f(er1;. . .;ern)]:
Wemay writef(x1;. . .;xn)t(x1;. . .;xn 1)xnh(x1;. . .;xn), where xn never appears as last variable in any monomials ofh. Le tr2H. Then re- placingrnwithr(1 e), wehave
0[d2(t(er1;. . .;ern 1)er(1 e));t(er1;. . .;ern 1)er(1 e)]:
(4)
Now, weknow thefact thatd(x(1 e))e x(1 e)d(e) and (1 e)d(ex)
(1 e)d(e)exand so
(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 using this facts, wehavefrom (4),
0(1 e)[d2(t(er1;. . .;ern 1)er(1 e));t(er1;. . .;ern 1)er(1 e)]
(1 e)d2(t(er1;. . .;ern 1)er(1 e))t(er1;. . .;ern 1)er(1 e)
2(1 e)d(e)t(er1;. . .;ern 1)er(1 e)d(e)t(er1;. . .;ern 1)er(1 e)
2((1 e)d(e)t(er1;. . .;ern 1)er)2(1 e):
This implies
0 2
(1 e)d(e)t(er1;. . .;ern 1)er 3 that is
0 2
(1 e)d(e)t(er1;. . .;ern 1)eH 3:
By [6], (1 e)d(e)t(er1;. . .;ern 1)eH0 which implies (1 e)d(e)t(er1e;. . .;ern 1e)0:
SinceeHeis a simpleArtinian ring and t(eHe)60 is invariant under the action of all inner automorphisms ofeHe, by [5, Lemma 2],(1 e)d(e)0 and so d(e)ed(e)2eH. Thus d(eH)d(e)Hed(H)eHrH and d(a)d(ea)2d(eH)rH. Therefore,d(rH)rH. Denote the left anni- hilator ofrHinHbylH(rH). ThenrH rH
rH\lH(rH), a primeC-algebra with thederivationdsuch thatd(x)d(x), for allx2rH. By assumption, wehavethat
[d2(f(x1;. . .;xn));f(x1;. . .;xn)]0
for allx1;. . .;xn2rH. By TheoremA, eitherd0 orf(x1;. . .;xn) is cen- tral-valued onrH.
Ifd0, then d(rH)rH0 and so d(r)r0. By Lemma 2.1,d is an inner derivation induced by an elementb2Qsuch thatbr0. Then for all
x1;. . .;xn2r, wehaveby assumption that
0[[b;[b;f(x1;. . .;xn)]];f(x1;. . .;xn)] f2(x1;. . .;xn)b2: By [3, Lemma 4], either b20 orf(r)r0. In both cases we have con- tradiction.
Iff(x1;. . .;xn) is central-valued onrH, thenrH, as well asr, satisfies
[f(x1;. . .;xn);xn1]xn20. ThenrCeRCfor some idempotent element
e2soc(RC) by [12, Proposition] andf(x1;. . .;xn) is central-valued oneRCe and wearedone.
THEOREM2.5. Let R be an associative prime ring of char R62with center Z(R)and extended centroid C, f(x1;. . .;xn)a nonzero multilinear polynomial over C in n noncommuting variables, d a nonzero derivation of R andra nonzero right ideal of R. If[d2(f(x1;. . .;xn));f(x1;. . .;xn)]2Z(R) for all x1;. . .;xn2rthenrCeRC for some idempotent e in the socle of RC and either f(x1;. . .;xn) is central-valued on eRCe or eRCe satisfies S4(x1;x2;x3;x4)unless d is an inner derivation induced by b2Q such that b20and br0.
PROOF. Supposedis not aQ-inner derivation induced by an element b2Qsuch thatb20 andbr0.
If [f(r);r]r0, that is [f(x1;. . .;xn);xn1]xn20 for all
x1;x2;. . .;xn22r, then by [12, Proposition], rCeRC for someidem- potente2soc(RC) andf(x1;. . .;xn) is central-valued oneRCe.
So, assumethat [f(r);r]r60, that is [f(x1;. . .;xn);xn1]xn2is not an identity forrand then we derive thateRCesatisfiesS4. By Lemma 2.3,Ris a GPI-ring and so is alsoQ(see [1] and [4]). By [16],Qis a primitivering with Hsoc(Q)60. Moreover, we may assume [f(rH);rH]rH60, otherwise by [1] and [4], [f(rQ);rQ]rQ0, which is a contradiction.
Choosea1;. . .;an2;b1;. . .;b52rHsuch that [f(a1;. . .;an);an1]an260 and S4(b1;b2;b3;b4)b560. Let a2rH. Since H is a regular ring, there existse2e2Hsuch that
eHaHa1H an2Hb1H b5H:
Then e2rH and aea;aieai for i1;. . .;n2, biebi for
i1;. . .;5. Thus wehavef(eHe)f(eH)e60. Moreover, by [14, Theo-
rem 2], we may also assume that
[[d2(f(x1;. . .;xn));f(x1;. . .;xn)];xn1]
is an identity forrQ. In particular, [[d2(f(x1;. . .;xn));f(x1;. . .;xn)];xn1] is an identity forrHand so foreH. It follows that, for allr1;. . .;rn12H,
0[[d2(f(er1;. . .;ern));f(er1;. . .;ern)];ern1]:
Wemay writef(x1;. . .;xn)t(x1;. . .;xn 1)xnh(x1;. . .;xn), where xn
never appears as last variable in any monomials ofh. Le tr2H. Then re- placingrnwithr(1 e) andrn1withrn1(1 e), wehave
(5) 0[[d2(t(er1;. . .;ern 1)er(1 e));t(er1;. . .;ern 1)er(1 e)];ern1(1 e)]:
Now, weknow thefact that d(x(1 e))e x(1 e)d(e), (1 e)d(ex)
(1 e)d(e)exand (1 e)d2(ex(1 e))e 2(1 e)d(e)ex(1 e)d(e). Thus using these facts, we have from (5),
0[[d2(t(er1;. . .;ern 1)er(1 e));t(er1;. . .;ern 1)er(1 e)];ern1(1 e)]
[d2(t(er1;. . .;ern 1)er(1 e));t(er1;. . .;ern 1)er(1 e)]ern1(1 e) ern1(1 e)[d2(t(er1;. . .;ern 1)er(1 e));t(er1;. . .;ern 1)er(1 e)]
t(er1;. . .;ern 1)er(1 e)d2(t(er1;. . .;ern 1)er(1 e))ern1(1 e) ern1(1 e)d2(t(er1;. . .;ern 1)er(1 e))t(er1;. . .;ern 1)er(1 e)
t(er1;. . .;ern 1)er(1 e)d(e)t(er1;. . .;ern 1)er(1 e)d(e)rn1(1 e)
ern1(1 e)d(e)t(er1;. . .;ern 1)er(1 e)d(e)t(er1;. . .;ern 1)er(1 e):
Replacingrn1witht(er1;. . .;ern 1)erin theaboverelation, weget 2t(er1;. . .;ern 1)er((1 e)d(e)t(er1;. . .;ern 1)er)2(1 e)0:
This implies
2((1 e)d(e)t(er1;. . .;ern 1)er)40 that is
2
(1 e)d(e)t(er1;. . .;ern 1)eH 40:
By [6], (1 e)d(e)t(er1;. . .;ern 1)eH0 which implies (1 e)d(e)t(er1e;. . .;ern 1e)0:
SinceeHeis a simpleArtinian ring and t(eHe)60 is invariant under the action of all inner automorphisms ofeHe, by [5, Lemma 2],(1 e)d(e)0 and so d(e)ed(e)2eH. Thus d(eH)d(e)Hed(H)eHrH and d(a)d(ea)2d(eH)rH. Therefore,d(rH)rH. Denote the left anni- hilator ofrHinHbylH(rH). ThenrH rH
rH\lH(rH), a primeC-algebra with thederivationdsuch thatd(x)d(x), for allx2rH. By assumption, wehavethat
[[d2f(x1;. . .;xn);f(x1;. . .;xn)];xn1]]0
for all x1;. . .;xn2rH. By Theorem 1.3, either d0 or f(x1;. . .;xn) is central-valued onrHorrHsatisfies the standard identityS4(x1;. . .;x4).
Ifd0, then as in the proof of Theorem 2.4, we have d(r)r0 and hence by Lemma 2.1,dis an inner derivation induced by an elementb2Q such thatbr0. Thus for allr1;. . .;rn2rH,
[d2(f(r1;. . .;rn));f(r1;. . .;rn)] f(r1;. . .;rn)2b22C:
Commuting both sides withf(r1;. . .;rn), weobtainf(r1;. . .;rn)3b20. In this caseby Lemma 2.2, sinceb260,f(rH)rH0. If f(rH)rH0, then [f(rH);rH]rH0, a contradiction.
Iff(x1;. . .;xn) is central-valued onrH, then we obtain that
[f(x1;. . .;xn);xn1]xn2 is an identity forr, a contradiction.
Finally, if S4(x1;. . .;x4) is an identity for rH, S4(x1;. . .;x4)x5 is an identity forrHand so forrCand this contradicts the choices of the ele- mentsb1;. . .;b5 2rH. Therefore, we conclude that in any caserCsatisfies a polynomial identity, hence by [12, Proposition], there exists an idempo- tente2Soc(RC) such thatrCeRC, as desired.
Acknowledgments. The authors would like to thank the referee for his/
her valuable comments and suggestions to modify some arguments of this paper.
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Manoscritto pervenuto in redazione il 16 marzo 2008.