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Two-sided annihilator condition with generalized derivations on multilinear polynomials.

1. Introduction

Let R be a prime ring with center Z(R). We denote by [a,b] = ab - ba the simple commutator of the elements a,b [member of] R and by [[a,b].sub.k] = [[[a,b].sub.k-1],b], for k > 1, the kth commutator of a, b. Throught this paper we will use the following notation: U will be the (two-sided) Utumi quotient ring of a ring R (sometimes, as in [1], U is called the symmetric ring of quotients). The definition, the axiomatic formulation, and the properties of this quotient ring U can be found in [1-3].

In any case, when R is a prime ring, all that we need here about this object is that

(1) R [subset or equal to] U;

(2) U is a prime ring;

(3) the center of U, denoted by C is a field which is called the extended centroid of R.

A well known result of Posner [4] says that if d is a derivation of R such that [d(x),x] [member of] Z(R), for all x [member of] R, then R is commutative. In [5] Lanski generalizes the result of Posner, by replacing the element x [member of] R with an element of a noncentral Lie ideal L of R. More precisely he proves that if [[d(x), x].sub.k] = 0 for all x [member of] L and k [greater than or equal to] 1 a fixed integer; then char(.R) = 2 and R satisfies [s.sub.4], the standard identity of degree 4.

Let f([x.sub.1], ..., [x.sub.n]) be a multilinear polynomial over C in n noncommuting variables and denote by f(X) the set of all evaluations of f([x.sub.1], ..., [x.sub.n]) in X [subset or equal to] R. In case f([x.sub.1], ..., [x.sub.n]) is not central valued on R, it is well known that the additive subgroup generated by f(R) contains a noncentral Lie ideal of R. Moreover any noncentral Lie ideal of R contains all the commutators [x, y] for x, y in some nonzero ideal of R, unless char(R) = 2 and [dim.sub.C]RC = 4.

In light of this and following the line of investigation of the previous cited papers, in [6] P. H. Lee and T. K. Lee consider the Engel-condition [[d(x), x].sub.k] = 0, in case x [member of] f(I), where I is a two-sided ideal of R. They show that either f([x.sub.1], ..., [x.sub.n]) is central valued in R or char(R) = 2 and R satisfies [s.sub.4].

These results indicate that the global structure of a prime ring R is often tightly connected to the behaviour of additive mappings defined on R, which act on suitable subsets of the whole ring. In [7] de Filippis and di Vincenzo study the left annihilator of the set [d(u)u - ud(u), u [member of] f(R)}, where d is a derivation. In case the annihilator is not zero, the conclusion is that f([x.sub.1], ..., [x.sub.n]) is central valued on R. These facts in a prime ring are natural tests which evidence that the set [d(u)u - ud(u), u [member of] f(R)} is rather large in R.

More recently, Liu [8] and Wang [9] have examined the identity a[[d(u), u].sub.k] = 0, where d is a derivation of R and u [member of] f(I), where I is a one-sided ideal of R. In particular, for I = R, if a [not equal to] 0 and f([x.sub.1], ..., [x.sub.n]) is not central valued on R, then char(R) = 2 and R satisfies [s.sub.4].

In [10] de Filippis considers a similar situation, in the case the derivation d is replaced by a generalized derivation F. An additive map F: R [right arrow] R is said to be a generalized derivation if there is a derivation d of R such that, for all x,y [member of] R, F(xy) = F(x)y + xd(y). A significative example is a map of the form F(x) = ax + xb, for some a,b [member of] R; such generalized derivations are called inner. Generalized derivations have been primarily studied on operator algebras. Therefore any investigation from the algebraic point of view might be interesting (see, e.g., [11]).

The main result in [10] is the following.

Theorem A. Let R be a prime ring of characteristic different from 2, with extended centroid C, U its two-sided Utumi quotient ring, F [not equal to] 0 a nonzero generalized derivation of R, f([x.sub.1], ..., [x.sub.n]) a noncentral multilinear polynomial over C inn noncommuting variables, and a [member of] R such that

a[F(f([r.sub.1], ..., [r.sub.n])), f([r.sub.1], ..., [r.sub.n])] = 0 (1)

for any [r.sub.1], ..., [r.sub.n] [member of] R. Then either a = 0 or one of the following holds:

(1) there exists [lambda] [member of] C such that F(x) = [lambda]x, for all x [member of] R;

(2) there exist q [member of] U and [lambda] [member of] C such that F(x) = (q + [lambda])x + xq, for all x [member of] R, and f[([x.sub.1], ..., [x.sub.n]).sup.2] is central valued on R.

We would like to remark that the same conclusions hold in case we consider the right annihilator, more precisely.

Theorem B. Let R be a prime ring of characteristic different from 2, with extended centroid C, U its two-sided Utumi quotient ring, F [not equal to] 0 a nonzero generalized derivation of R, f([x.sub.1], ..., [x.sub.n]) a noncentral multilinear polynomial over C inn noncommuting variables, and a [member of] R such that

[F(f([r.sub.1], ..., [r.sub.n])), f([r.sub.1], ..., [r.sub.n])]a = 0 (2)

for any [r.sub.1], ..., [r.sub.n] [member of] R. Then either a = 0 or one of the following holds:

(1) there exists [lambda] [member of] C such that F(x) = [lambda]x, for all x [member of] R;

(2) there exist q [member of] U and [lambda] [member of] C such that F(x) = (q + [lambda])x + xq, for all x [member of] R, and f[([x.sub.1], ..., [x.sub.n]).sup.2] is central valued on R.

Here we will consider a more general situation, involving a two-sided annihilating condition. More specifically, we study simultaneously left and right annihilators of the set {[F(x), x]: x [member of] f(R)} and prove the following.

Theorem 1. Let R be a prime ring of characteristic different from 2, with extended centroid C, U its two-sided Utumi quotient ring, F a nonzero generalized derivation of R, f([x.sub.1], ..., [x.sub.n]) a noncentral multilinear polynomial over C in n noncommuting variables, and a,b [member of] R such that

a[F(f([r.sub.1], ..., [r.sub.n])), f([r.sub.1], ..., [r.sub.n])] b = 0 (3)

for any [r.sub.1], ..., [r.sub.n] [member of] R. Then one of the following holds:

(1) a = 0;

(2) b = 0;

(3) there exists [lambda] [member of] C such that F(x) = [lambda]x, for all x [member of] R;

(4) thereexist q [member of] U and [lambda] [member of] C such that F(x) = (q + [lambda])x + xq, for all x [member of] R, and f[([x.sub.1], ..., [x.sub.n]).sup.2] is central valued on R;

(5) there exist q [member of] U and [lambda],[mu] [member of] C such that F(x) = (q + [lambda])x + xq, for all x [member of] R, and aq = [mu]a, qb = [mu]b.

Remark 2. By the primeness of R and in light of Theorems A and B, we may assume that R is not a domain. Moreover, since the center of a prime ring cannot contain nonzero zero-divisor, then neither a [member of] Z(R) nor b [member of] Z(R). Finally in all that follows we always suppose char(R) [not equal to] 2.

In the sequel we will make a frequent use of the following.

Remark 3. If B is a basis of U over C then any element of T = U[*.sub.C]C{[x.sub.1], ..., [x.sub.n]}, the free product over C of the C-algebra U and the free C-algebra C{[x.sub.1], ..., [x.sub.n]}, is called generalized polynomial and can be written in the form g = [[summation].sub.i][a.sub.i][m.sub.i]. In this decomposition the coefficients [[alpha].sub.i] are in C and the elements [m.sub.i] are B-monomials; that is, [m.sub.i] = [q.sub.0][y.sub.1][q.sub.1], ..., [y.sub.h][q.sub.h], with [q.sub.i] [member of] B and [y.sub.i] [member of] {[x.sub.1], ..., [x.sub.n]}. In [12] it is shown that a generalized polynomial g = [[summation].sub.i][a.sub.i][m.sub.i] is the zero element of T if and only if all [[alpha].sub.i] are zero. Let [a.sub.1], ..., [a.sub.k] [member of] U be linearly independent over C and [a.sub.1][g.sub.1]([x.sub.1], ..., [x.sub.n]) + ... + [a.sub.k][g.sub.k]([x.sub.1], ..., [x.sub.n]) = 0 [member of] T, for some [g.sub.1], ..., [g.sub.k] [member of] T. If, for any i, [g.sub.i]([x.sub.1], ..., [x.sub.n]) = [[summation].sup.n.sub.j=1][x.sub.j][h.sub.j]([x.sub.1], ..., [x.sub.n]) and [h.sub.j]([x.sub.1], ..., [x.sub.n]) [member of] T, then [g.sub.1]([x.sub.1], ..., [x.sub.n]), ..., [g.sub.k]([x.sub.1], ..., [x.sub.n]) are the zero element of T. The same conclusion holds if [g.sub.1]([x.sub.1], ..., [x.sub.n])[a.sub.1] + ... + [g.sub.k]([x.sub.1], ..., [x.sub.n])[a.sub.k] = 0 [member of] T, and [g.sub.i]([x.sub.1], ..., [x.sub.n]) = [[summation].sup.n.sub.j=1][h.sub.j]([x.sub.1], ..., [x.sub.n])[x.sub.j] for some [h.sub.j]([x.sub.1], ..., [x.sub.n]) [member of] T.

We refer the reader to [1, 12] for more details on generalized polynomial identities.

2. An Independent Result

We will dedicate this section to the proof of the following proposition on linear identities with commutators in matrix rings. This result will be useful in the sequel.

Proposition 4. Let C be a field and R = [M.sub.t](C) the algebra of t x t matrices over C and S = [R, R]. Let a,b,c [member of] R, such that c [not member of] Z(R) and a[c, x]b = 0 for all x [member of] S. Then there exists [lambda] [member of] Z(R) such that ac = [lambda]a and cb = [lambda]b.

In order to prove Proposition 4, we need several lemmas.

Lemma 5. Let K be an infinite field and t [greater than or equal to] 2. If [A.sub.1], ..., [A.sub.k] are not scalar matrices in [M.sub.t](K), then there exists some invertible matrix B [member of] [M.sub.t](K) such that each matrix B[A.sub.1][B.sup.-1], ..., B[A.sub.k][B.sup.-1] has all nonzero entries.

Proof. See Lemma 1.5 in [13]. []

Lemma 6. Let R be a prime ring with extended centroid C. Suppose [[summation].sup.n.sub.i=1][a.sub.i]x[b.sub.i] + [[summation].sup.m.sub.j=1][c.sub.j]x[d.sub.j] = 0, for all x [member of] R, where [a.sub.i], [b.sub.i], [c.sub.j], [d.sub.j] [member of] R, for i = 1, ..., n and j = 1, ..., m. If [a.sub.1], ..., [a.sub.n] are C-independent then each [b.sub.i] is C-dependent on [d.sub.1], ..., [d.sub.m]. Analogously if [b.sub.1], ..., [b.sub.n] are C-independent then each [a.sub.i] is C-dependent on [c.sub.1], ..., [c.sub.m].

Proof. It is Martindale's result contained in [14]. []

Lemma 7. Let R be a prime ring with extended centroid C. Suppose a[x, y] + [x, y]b = 0,for all x,y [member of] R, where a,b [member of] R. Then a = -b [member of] C.

Proof. It is an easy consequence of Lemma 6. []

Lemma 8. Let K be an infinite field, R = [M.sub.m](K) the algebra of m x m matrices over K, Z(R) the center of R, and S = [R, R]. Assume that there exist a, b, c, q nonzero elements of R such that axq + cxb = 0 for all x [member of] S. If q [member of] Z(R) then one of the following holds:

(1) a, b, c are central matrices and aq + bc = 0;

(2) b is a central matrix and aq + bc = 0.

Proof. Since q [member of] Z(R), by the assumption, we have that aqx + cxb = 0 for all x [member of] S. Clearly if c [member of] Z(R) then aqx + xbc = 0 for all x [member of] S, and by Lemma 7 we get aq = -bc [member of] Z(R); that is, a,b,c [member of] Z(R). On the other hand, if be Z(R), then (aq + bc)x = 0 for all x [member of] S and it follows easily that aq + bc = 0.

In light of this, we consider c and b both nonscalar matrices. We will prove that in this case we get a contradiction.

Here we denote by [e.sub.ij] the usual matrix unit with 1 in the (i, j)-entry and zero elsewhere.

By Lemma 5, we can assume that c and b have all nonzero say c = [[summation].sub.kl][c.sub.kl][e.sub.kl] and b = [[summation].sub.kl][b.sub.kl][e.sub.kl] for 0 [not equal to] [c.sub.kl], 0 [not equal to] [b.sub.kl] [member of] K.

Since [e.sub.ji] [member of] S for all i [not equal to] j, then, for any i [not equal to] j,

X = aq[e.sub.ji] + c[e.sub.ji]b = 0 (4)

in particular the (i, j)-entry of X is [c.sub.ij][b.sub.ij] = 0, a contradiction. []

For sake of clearness, we may write the previous lemma as follows.

Lemma 9. Let K be an infinite field, R = [M.sub.m](K) the algebra of m x m matrices over K, Z(R) the center of R, and S = [R, R]. Let [a.sub.1], [a.sub.2], [a.sub.3], [a.sub.4] be nonzero elements of R such that [a.sub.1] x [a.sub.2] + [a.sub.3] x [a.sub.4] = 0 for all x [member of] S. Assume there exists i [member of] {1, 2, 3, 4} such that [a.sub.i] [member of] Z(R). Then [a.sub.1] = [alpha][a.sub.3] and [a.sub.2] = -[alpha][a.sub.4], for a suitable [alpha] [member of] Z(R).

Lemma 10. Let K be an infinite field, R = [M.sub.m](K) the algebra of m x m matrices over K, and Z(R) the center of R. Assume that there exist a, b, c, q nonzero elements of R such that axq + cxb = 0 for all x [member of] S = [R, R]. If q [not member of] Z(R) and b - [alpha]q [member of] Z(R), for a suitable [alpha] [member of] K, then b - [alpha]q = a + [alpha]c = 0.

Proof. Assume that a + [alpha]c is not a scalar matrix. By Lemma 5, we can assume that a + [alpha]c and q have all nonzero entries, say a + [alpha]c = [[summation].sub.kl][t.sub.kl][e.sub.kl] and q = [[summation].sub.kl][q.sub.kl][e.sub.kl], for 0 [not equal to] [t.sub.kl], 0 [not equal to] [q.sub.kl] [member of] K.

Since b = [beta]I + [alpha]q, for a suitable [beta] [member of] K, by our assumption we have that

axq + cx([beta] + [alpha]q) = 0; (5)

that is,

[beta]cx + (a + [alpha]c)xq = 0, (6)

for all x [member of] S. In particular for x = [[e.sub.ii], [e.sub.ij]], with i [not equal to] j,

0 = X = [beta]c[e.sub.ij] + (a + [alpha]c)[e.sub.ij]q. (7)

By calculations one has that the (j, i)-entry of X is 0 = [t.sub.ji][q.sub.ji], a contradiction.

Therefore a + [alpha]c must be a central matrix. In light of this, there exist [beta],[gamma] [member of] K such that b = [alpha]q + [beta] and a = -[alpha]c + [gamma], so that 0 = (-[alpha]c + [gamma])xq + cx([alpha]q + [beta]) = ([beta]c)x + x([gamma]q), for all x [member of] S = [R, R]. Once again by Lemma 7 and since q [not member of] Z(R), it follows that [beta] = [gamma] = 0; that is, b = [alpha]q and a = -[alpha]c. []

Lemma 11. Let K be an infinite field, R = [M.sub.m](K) the algebra of m x m matrices over K, and S = [R, R]. Suppose there exist a, b, c, q [member of] R such that axq + cxb = 0 for all x [member of] S. Denote

a = [summation over (kl)][a.sub.kl][e.sub.kl], b = [summation over (kl)][b.sub.kl][e.sub.kl],

c = [summation over (kl)][c.sub.kl][e.sub.kl], q = [summation over (kl)][q.sub.kl][e.sub.kl] (8)

for suitable [a.sub.kl], [b.sub.kl], [c.sub.kl], and [q.sub.kl] elements of K. If there are i [not equal to] j such that [q.sub.ji] [not equal to] 0, = 0, and [b.sub.ji] = 0, then [a.sub.ri] = 0 and [b.sub.rk] = 0 for all r [not equal to] i and k [not equal to] r (i.e., the only nonzero off-diagonal elements of b fall in the ith row).

Proof. Consider the assumption

axq + cxb = 0 [for all]x [member of] [R, R]. (9)

In particular, for x = [e.sub.ij], we have

X = a[e.sub.ij]q + c[e.sub.ij]b = 0 (10)

so that, for all r [not equal to] i, the (r, i)-entry of the matrix X is 0 = [a.sub.ri][q.sub.ji] + [c.sub.ri][b.sub.ji] = [a.sub.ri][q.sub.ji]. Since [q.sub.ji] [not equal to] 0, one has [a.sub.ri] = 0 for all r [not equal to] i, in particular [a.sub.ji] = 0. Thus, in case m = 2 we are done (since [b.sub.ji] = [a.sub.ji] = 0).

Assume in what follows that m [greater than or equal to] 3, and choose x = [e.sub.it], with t [not equal to] i, j. Hence we also have

Y = a[e.sub.it]q + c[e.sub.it]b = 0. (11)

From the previous equalities it follows that

(1) for all s [not equal to] i, the (j, s)-entry of the matrix [lambda] is [a.sub.ji][q.sub.js] + [c.sub.ji][b.sub.js] = 0;

(2) for all s [not equal to] i,j, the (j, s)-entry of the matrix Y is [a.sub.ji][q.sub.ts] + [c.sub.ji][b.sub.ts] = 0;

(3) the (j, i)-entry of the matrix Y is [a.sub.ji][q.sub.ti] + [c.sub.ji][b.sub.ti] = 0;

(4) for all k [not equal to] i,t, the(j, k)-entry of the matrix Y is [a.sub.jt][q.sub.tk] + [c.sub.ji][b.sub.tk] = 0 (note that this holds also in case k = j).

By (1) and (2) and since [a.sub.ji] = 0 and [c.sub.ji] [not equal to] 0, we have both [b.sub.js] = 0, for all s [not equal to] i, and [b.sub.ts] = 0 for all t [not equal to] i, j and s [not equal to] i, j. So by (3) [b.sub.ti] = 0 for all t [not equal to] i. Finally by (4), [b.sub.tk] = 0 for all t [not equal to] i, j and k [not equal to] t. []

Lemma 12. Let K be an infinite field, R = [M.sub.m](K) the algebra of m x m matrices over K, and S = [R, R]. Suppose there exist a, b, c, q [member of] R such that axq + cxb = 0 for all x [member of] S. Denote

b = [summation over (kl)][b.sub.kl][e.sub.kl], c = [summation over (kl)][c.sub.kl][e.sub.kl], q = [summation over (kl)][q.sub.kl][e.sub.kl] (12)

for suitable [b.sub.kl], [c.sub.kl], and [q.sub.kl] elements of K. Assume there are i [not equal to] j such that [b.sub.ji] = 0. If [q.sub.rs] [not equal to] 0, [c.sub.rs] [not equal to] 0 for all r [not equal to] s, then one of the following holds:

(1) a = b = 0;

(2) m = 2, cq = 0, and there exists 0 [not equal to] [lambda] [member of] K such that

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (13)

Proof. Firstly we consider the case m [greater than or equal to] 3. The first step is to apply twice Lemma 11: this forces b to be a diagonal matrix. In fact [b.sub.ji] = 0, [q.sub.ji] [not equal to] 0, and [c.sub.ji] [not equal to] 0 imply that [b.sub.rk] = 0 for all r [not equal to] i and k [not equal to] r; in particular, since m [greater than or equal to] 3, there exists t [not equal to] i such that [b.sub.lt] = 0, for all l [not equal to] t, i. Since [q.sub.lt] [not equal to] 0, [c.sub.lt] [not equal to] 0, we have [b.sub.rk] = 0 for all r [not equal to] t and k [not equal to] r, so [b.sub.ik] = 0 for all k [not equal to] i, as required. Say b = [[summation].sub.k][b.sub.kk][e.sub.kk].

Consider now the inner automorphism of R induced by the invertible matrix P = I + [e.sub.rj], for r [not equal to] i, j: [phi](x) = [P.sup.-1]xP. Of course [phi](a)x[phi](q) + [phi](c)x[phi](b) = 0, for all x [member of] S. Moreover the (j, i)-entries of [phi](q), [phi](c), and [phi](b) are, respectively, [q.sub.ji] [not equal to] 0, [c.sub.ji] [not equal to] 0, and [b.sub.ji] = 0. Therefore, again by Lemma 11, any (r, j)-entry of [phi](b) is zero, for all r [not equal to] i. By calculations 0 = [([phi](b)).sub.rj] = [b.sub.jj] - [b.sub.rr]; that is, [b.sub.jj] = [b.sub.rr].

On the other hand, if [chi] is the inner automorphisms induced by the invertible matrix Q = I + [e.sub.ri], as above [chi](a)x[chi](q) + [chi](c)x[chi](b) = 0, for all x [member of] R. Since the (i, j)-entries of [chi](q), [chi](c), and [chi](b) are, respectively, [q.sub.ij] [not equal to] 0, [c.sub.ij] [not equal to] 0, and [b.sub.ij] = 0, and again any (r, i)-entry of [chi](b) is zero, for all r [not equal to] j; that is, 0 = [([phi](b)).sub.ri] = [b.sub.ii] - [b.sub.rr] and [b.sub.ii] = [b.sub.rr] = [b.sub.jj] = [beta], for all r [not equal to] i, j. Thus b = [beta]I is a central matrix in R. By Lemma 9, either b = [alpha]q for some [alpha] [member of] K or b = 0. Since the first case cannot occur, we get b = 0 and also a = 0 which follows from a[R, R]q = 0 and q [not equal to] 0.

Let now m = 2; that is, R = [M.sub.2](K). In this case it is well known that for any element x [member of] [R, R] there exist [alpha], [beta], [gamma] [member of] K such that [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Without loss of generality we may assume [b.sub.21] = 0. In case [b.sub.12] = 0, then by the same above argument we show that b [member of] Z(R) and we are done again. Thus we consider the case [b.sub.12] [not equal to] 0. Moreover, by applying Lemma 11 it follows [a.sub.21] = 0. Hence we may write

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (14)

For x = [e.sub.12] [member of] [R, R] we have

X = a[e.sub.12 q + c[e.sub.12]b = 0 (15)

so that the (2,2)-entry of the matrix X is 0 = [c.sub.21][b.sub.22]; that is, [b.sub.22] = 0 and the (1,1)-entry of the matrix X is 0 = [a.sub.11][q.sub.21]; that is, [a.sub.11] = 0. On the other hand, for x = [e.sub.21] [member of] [R, R],we have

Y = a[e.sub.21]q + c[e.sub.21]b = 0. (16)

The (1,2)-entry of the matrix Y is 0 = [a.sub.12][q.sub.12] + [c.sub.12][b.sub.12]; that is, [a.sub.12] [not equal to] 0 and [b.sub.12]/[q.sub.12] = -[a.sub.12]/[c.sub.12]. Moreover the (2,2)-entry of the matrix Y is 0 = [a.sub.22][q.sub.12] + [c.sub.22][b.sub.12]. Therefore, if denoted [lambda] = -[b.sub.12]/[q.sub.12], one has [a.sub.22] = [lambda][c.sub.22] and [a.sub.12] = [lambda][c.sub.12].

Analogously, the (1,1)-entry of the matrix Y is 0 = [a.sub.12][q.sub.11] + [c.sub.12][b.sub.11]. Thus [b.sub.11] = -[lambda][q.sub.11] and [b.sub.12] = -[lambda][q.sub.12].

Finally, by our assumption and for [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], with [alpha] [not equal to] 0, we also have

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (17)

and by easy calculations it follows cq = 0. []

Lemma 13. Let K be an infinite field, R = [M.sub.2](K) the algebra of m x m matrices over K, and S = [R, R]. Let a,b,c [member of] R and denote

a = [summation over (kl)][a.sub.kl][e.sub.kl], b = [summation over (kl)][b.sub.kl][e.sub.kl],

c = [summation over (kl)][c.sub.kl][e.sub.kl], cb = [summation over (kl)][p.sub.kl][e.sub.kl] (18)

for suitable [a.sub.kl], [b.sub.kl], [c.sub.kl], and [p.sub.kl] elements of K. Suppose c [not member of] Z(R) and a[c, x]b = 0 for all x [member of] S. Assume there are i [not equal to] j such that [p.sub.ji] = 0. If [a.sub.rs] [not equal to] 0, [b.sub.rs] [not equal to] 0,for all r [not equal to] s, then ac = cb = 0.

Proof. By our hypothesis, we have acxb - axcb = 0 for all x [member of] S. By Lemma 12 it follows that either ac = cb = 0 or ab = 0 and there exists 0 [not equal to] [lambda] [member of] K such that

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (19)

Notice that ab = 0 implies that the following holds:

[a.sub.11][b.sub.11] + [a.sub.12][b.sub.21] = 0, (20)

[a.sub.21][b.sub.11] + [a.sub.22][b.sub.21] = 0. (21)

Moreover, by computing the product ac we get

[a.sub.11][c.sub.11] + [a.sub.12][c.sub.21] = 0, (22)

[a.sub.21][c.sub.11] + [a.sub.22][c.sub.21] = 0, (23)

[a.sub.11][c.sub.12] + [a.sub.12][c.sub.22] = [lambda][a.sub.12]. (24)

Finally, by computing the product cb we also have

[c.sub.21][b.sub.11] + [c.sub.22][b.sub.21] = 0, (25)

[c.sub.21][b.sub.12] + [c.sub.22][b.sub.22] = 0. (26)

Notice that, in case [a.sub.11] = 0, by (20) it follows the contradiction [a.sub.12][b.sub.21] = 0. Thus [a.sub.11] [not equal to] 0 and multiply (25) by [a.sub.11], so that [c.sub.21][b.sub.11][a.sub.11] + [c.sub.22][b.sub.21][a.sub.11] = 0. Again by (20) we have [b.sub.21]([c.sub.22][a.sub.11] - [c.sub.21][a.sub.12]) = 0 and using (22) it follows [b.sub.21]([c.sub.22][a.sub.11] + [c.sub.11][a.sub.11]) = 0. Since [b.sub.21] [not equal to] 0 and [a.sub.11] [not equal to] 0, then [c.sub.11] = -[c.sub.22] = [mu].

Assume [mu] [not equal to] 0, denoted by I the identity matrix in R, and let [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Since c and c' induce the same inner derivation, then by our assumptions we have that a[c',x]b = 0 for all x [member of] S. By applying again Lemma 12, it follows that either ac' = c'b = 0 or ab = 0 and there exists 0[not equal to] v [member of] K such that

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (27)

In the latter case, by using the same above argument, the matrix c' satisfies the equalities (22) and (25); that is, respectively,

[a.sub.12][c.sub.21] = 0 (28)

implying [c.sub.21] = 0, and

-2[mu][b.sub.21] = 0 (29)

which is a contradiction.

Therefore

ac' = c'b = 0, i.e., ac = [mu]a, cb = [mu]b. (30)

In this case, by using both (22) and (30), the (1,1)-entry of the matrix ac should be

0 = [a.sub.11][c.sub.11] + [a.sub.12][c.sub.21] = [mu][a.sub.11] [not equal to] 0. (31)

The previous contradiction implies [mu] = 0; that is, [c.sub.22] = 0 and by (26) also [c.sub.21] = 0. Hence [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Now consider the following elements in S:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (32)

Thus

Z = a[c, [x.sub.0]] b = 0 T = a[c, [y.sub.0]]b = 0 (33)

and in particular the (1,1)-entry of Z is

[c.sub.12](-[a.sub.11][b.sub.11] - 2[a.sub.11][b.sub.21] + [a.sub.21][b.sub.21]) = 0, (34)

and the (1,1)-entry of T is

[c.sub.12]([a.sub.11][b.sub.11] - 2[a.sub.11][b.sub.21] - [a.sub.21][b.sub.21]) = 0. (35)

Since c [not equal to] 0, then [c.sub.12] [not equal to] 0. Therefore the sum of (34) and 35) forces the contradiction -4[a.sub.11][b.sub.21] = 0. []

Lemma 14. Let K be an infinite field, R = [M.sub.t](K) the algebra of t x t matrices over K, and S = [R, R]. Let a, b, c [member of] R and denote

a = [summation over (kl)][a.sub.kl][e.sub.kl], b = [summation over (kl)][b.sub.kl][e.sub.kl],

cb = [summation over (kl)][p.sub.kl][e.sub.kl], ac = [summation over (kl)][q.sub.kl][e.sub.kl] (18)

for suitable [a.sub.kl], [b.sub.kl], [p.sub.kl], and [q.sub.kl] elements of K. Suppose c [not member of] Z(R) and a[c, x]b = 0 for all x [member of] S. Then there exists [lambda] [member of] Z(R) such that ac = [lambda]a and cb = [lambda]b.

Proof. Clearly if one of a, b, ac, or cb is a scalar matrix we are done by Lemma 9. In order to prove this lemma, we may assume that a, b, ac, and cb are noncentral matrices.

By Lemma 5, there exists some invertible matrix Q [member of] [M.sub.t](K) such that Qa[Q.sup.-1] = a', Qb[Q.sup.-1] = b', Q(ac)[Q.sup.-1] = (ac)', and Q(cb)[Q.sup.-1] = (cb)' have all nonzero entries.

Notice that {ac, a} are linearly Z(R)-dependent if and only if {(ac)', a'} are linearly Z(R)-dependent; analogously {cb, b} are linearly Z(R)-dependent if and only if {(cb)', b'} are linearly Z(R)-dependent. Moreover ac = cb = 0 if and only if (ac)' = (cb)' = 0. Therefore, in order to prove our result, we may replace a, b, ac, cb, respectively, by a', b', (ac)', (cb)', so that a, b, ac, and cb have all nonzero entries.

For x = [e.sub.ij] [member of] S we have

X = ac[e.sub.ij]b - a[e.sub.ij]cb = 0; (37)

in particular the (j, i)-entry of X is [q.sub.ji][b.sub.ji] - [a.sub.ji][p.sub.ji] = 0. Denote 0 [not equal to] [eta] = [q.sub.ji]/[a.sub.ji], so that [p.sub.ji] = [eta][b.sub.ji]. Let I be the identity matrix in R and u = c - [eta]I. Since u and c induce the same inner derivation in R,then a[u, x]b = 0;thatis, a(c - [eta]I)xb - ax(c - [eta]I)b = 0, for all x [member of] S. Moreover a and b have all nonzero entries, and the (j, i)-entry of (c - [eta]I)b is zero. Thus we may apply Lemmas 12 and 13 and obtain ac = [eta]a and cb = [eta]b, as required. []

Proof of Proposition 4. If one assumes that C is infinite, the conclusion follows from Lemma 14.

Now let K be an infinite field which is an extension of the field C and let [bar.R] = [M.sub.t](K) [congruent to] R[[cross product].sub.C]K. Consider the generalized polynomial

P ([x.sub.1],[x.sub.2]) = a(c[[x.sub.1],[x.sub.2]] - [[x.sub.1],[x.sub.2]]c)b (38)

which is a generalized polynomial identity for R. Since P([x.sub.1],[x.sub.2]) is a multilinear generalized polynomial in the indeterminates [x.sub.1], [x.sub.2], then it is a generalized polynomial identity for [bar.R] and the conclusion follows again from Lemma 14. []

3. The Inner-Case in Prime Rings

In this section we consider f(R), the set of all evaluations of the noncentral multilinear polynomial f([x.sub.1], ..., [x.sub.n]) over C, and assume that F is an inner generalized derivation, so that there exist c,q [member of] U such that F(x) = cx + xq, for all x [member of] R, and f(R) satisfies

a[cx + xq,x]b, (39)

where a, b are nonzero elements of R.

In order to prove the first result we premit the following.

Fact 1. Let R = [M.sub.t](C) be the algebra of t x t matrices over C of characteristic different from 2. Notice that the set f(R) = {f([r.sub.1], ..., [r.sub.n]): [r.sub.1], ..., [r.sub.n] [member of] R} is invariant under the action of all inner automorphisms of R. Hence if denoted by r = ([r.sub.1], ..., [r.sub.n]) [member of] R x R x R x ... x R = [R.sup.n], then for any inner automorphism [phi] of [M.sub.t](C), we have that [r.bar] = ([phi]([r.sub.1]), ..., [phi]([r.sub.n])) [member of] [R.sup.n] and [phi](f(r)) = f([r.bar]) [member of] f(R).

Since f([x.sub.1], ..., [x.sub.n]) is not central then, by [15] (see also [16]), there exist [u.sub.1], ..., [u.sub.n] [member of] [M.sub.t](C) and [alpha] [member of] C - {0}, such that f([u.sub.1], ..., [u.sub.n]) = [alpha][e.sub.kl], with k [not equal to] l. Moreover, since the set {f([v.sub.1], ..., [v.sub.n]): [v.sub.1], ..., [v.sub.n] [member of] [M.sub.t](C)} is invariant under the action of all C-automorphisms of [M.sub.t](C), then for any i [not equal to] j there exist [r.sub.1], ..., [r.sub.n] [member of] [M.sub.t](C) such that f([r.sub.1], ..., [r.sub.n]) = [alpha][e.sub.ij].

Now we may start with the following.

Proposition 15. Let C be a field, R = [M.sub.t](C) the algebra of txt matrices over C, and f([x.sub.1], ..., [x.sub.n]) a noncentral multilinear polynomial over C. Let 0 [not equal to] a, 0 [not equal to] b, c, q [member of] R and denote

a = [summation over (kl)][a.sub.kl][e.sub.kl], b = [summation over (kl)][b.sub.kl][e.sub.kl],

c = [summation over (kl)][c.sub.kl][e.sub.kl], q = [summation over (kl)][q.sub.kl][e.sub.kl] (40)

for suitable [a.sub.kl], [b.sub.kl], [c.sub.kl], and [q.sub.kl] elements of C. Suppose that

a[cf([r.sub.1], ..., [r.sub.n]) + f([r.sub.1], ..., [r.sub.n])q, f {[r.sub.1], ..., [r.sub.n]}]b = 0 (41)

for all [r.sub.1], ..., [r.sub.n] [member of] R. Then one of the following holds:

(1) c,q [member of] Z(R);

(2) there exists [lambda] [member of] Z(R) such that c - q = [lambda], and f[([x.sub.1], ..., [x.sub.n]).sup.2] is central valued on R;

(3) there exist [lambda],[mu] [member of] Z(R) such that c - q = [lambda], ac = [mu]a and cb = [mu]b.

Proof. By our assumption, R satisfies the following generalized polynomial identity:

a[cf{[x.sub.1], ..., [x.sub.n]) + f{[x.sub.1], ..., [x.sub.n])q, f{[x.sub.1], ..., [x.sub.n])]b. (42)

As in the previous section [e.sub.ij] denotes the matrix unit with 1 in (i, j)-entry and zero elsewhere.

Firstly we assume C is an infinite field.

Since f([x.sub.1], ..., [x.sub.n]) is not central then, by Fact 1, for any i [not equal to] j, there exist [r.sub.1], ..., [r.sub.n] [member of] [M.sub.t](C) such that f([r.sub.1], ..., [r.sub.n]) = [e.sub.ij].

Then we obtain

0 = a [c[e.sub.ji] + [e.sub.ji]q, [e.sub.ji]]b = a[e.sub.ji][q.sub.eji]b - a[e.sub.ji]c[e.sub.ji]b. (43)

In particular,

[a.sub.ij]([q.sub.ij] - [c.sub.ij])[b.sub.ij] = 0. (44)

In light of Remark 2, we assume that a and b are not central matrices. Denote w = q - c and suppose that w is not scalar. By Lemma 5 there exists an C-automorphism [phi] of [M.sub.t](C) such that w' = [phi](w), a' = [phi](a), and b' = [phi](b) have all nonzero entries. Clearly w', a', and b' must satisfy the condition (44) and this is a contradiction.

This means that q - c = [gamma]I, for some y [member of] C, and the main condition is now

a[cf([r.sub.1], ..., [r.sub.n]) + f([r.sub.1], ..., [r.sub.n])c, f ([r.sub.1], ..., [r.sub.n])] b = 0, (45)

for all [r.sub.1], ..., [r.sub.n] [member of] R; that is, a[c, f[([r.sub.1],..., [r.sub.n]).sup.2]]b = 0, for all [r.sub.1], ..., [r.sub.n] [member of] R.

Consider the additive subgroup of R, generated by the set S = {[x.sup.2]: x [member of] f(R)}. By [17], either S [subset or equal to] Z(R) or the noncentral Lie ideal [R, R] of R is contained in S. In the first case we conclude that f[([x.sub.1], ..., [x.sub.n]).sup.2] is central valued in R and we are done. In either case we have a[c, [[r.sub.1], [r.sub.2]]]b = 0, for all [r.sub.1], [r.sub.2] [member of] R, and by Proposition 4 we get the required conclusions.

Now let K be an infinite field which is an extension of the field C and let [bar.R] = [M.sub.m](K) [congruent to] R[[cross product].sub.C]K. Notice that the multilinear polynomial f([x.sub.1], ..., [x.sub.n]) is central-valued on R if and only if it is central-valued on [bar.R]. Consider the generalized polynomial

P([x.sub.1], ..., [x.sub.n]) = a(cf[([x.sub.1], ..., [x.sub.n].sup.2]b - f[([x.sub.1], ..., [x.sub.n]).sup.2]qb

+ af([x.sub.1], ..., [x.sub.n])(q - c)f([x.sub.1], ..., [x.sub.n]))b (46)

which is a generalized polynomial identity for R. Moreover it is multihomogeneous of multidegree (2, ..., 2) in the indeterminates [x.sub.1], ..., [x.sub.n].

Hence the complete linearization of P([x.sub.1], ..., [x.sub.n]) is a multilinear generalized polynomial [THETA]([x.sub.1], ..., [x.sub.n], [y.sub.1], ..., [y.sub.n]) in 2n indeterminates; moreover

[THETA]([x.sub.1], ..., [x.sub.n], [x.sub.1], ..., [x.sub.n]) = [2.sup.n]P([x.sub.1], ..., [x.sub.n]). (47)

Clearly the multilinear polynomial [THETA]([x.sub.1], ..., [x.sub.n], [y.sub.1], ..., [y.sub.n]) is a generalized polynomial identity for R and [bar.R] too. Since char(C) [not equal to] 2 we obtain P([r.sub.1], ..., [r.sub.n]) = 0, for all [r.sub.1], ..., [r.sub.n] [member of] [bar.R], and the conclusion follows from the argument contained in the first part of this proposition. []

Lemma 16. If there exist 0 [not equal to] a [member of] R, 0 [not equal to] b [member of] R, c,q [member of] U such that a[cx + xq, x]b = 0, for all x [member of] f(R), then R satisfies a nontrivial generalized polynomial identity, unless when one of the following holds:

(1) c,q [member of] C;

(2) c - q [member of] C and there exists [lambda] [member of] C such that ac = [lambda]a, qc = [lambda]b.

Proof. Assume that R does not satisfy any nontrivial generalized polynomial identity with coefficients in U. Therefore,

[PHI]([x.sub.1], ..., [x.sub.n])

= a[cf([x.sub.1], ..., [x.sub.n]) + f([x.sub.1], ..., [x.sub.n])q, f([x.sub.1], ..., [x.sub.n])]b (48)

is a trivial generalized polynomial identity for R. By calculations

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (49)

for all [x.sub.1], ..., [x.sub.n] [member of] R. If c [member of] C and q [member of] C, the proof is completed; hence we suppose that c and q are not simultaneously central. By Remark 3 and by (49), if {b, qb} are linearly C-independent then R satisfies the trivial generalized polynomial identity af[([x.sub.1], ..., [x.sub.n]).sup.2]qb = 0. It means, since a [not equal to] 0, qb = 0, a contradiction. Analogously, if we suppose {a, ac} linearly C-independent, we get ac = 0, a contradiction.

Therefore there exist [alpha],[beta] [member of] C such that qb = [beta]b and ac = [alpha]a; now (49) becomes

a(f([x.sub.1], ..., [x.sub.n])(q - c)f([x.sub.1], ..., [x.sub.n])

+ ([alpha] - [beta])f[([x.sub.1], ..., [x.sub.n]).sup.2])b = 0 (50)

for all [x.sub.1], ..., [x.sub.n] [member of] R. Since it is a trivial generalized polynomial identity, then c - q = [alpha] - [beta]. Moreover, [beta]b = qb = cb + [beta]b - [alpha]b; that is, cb = [alpha]b. []

Proposition 17. Let 0 [not equal to] a, 0 [not equal to] b, c,q [member of] R such that

a[cf([r.sub.1], ..., [r.sub.n]) + f([r.sub.1], ..., [r.sub.n])q, f([r.sub.1], ..., [r.sub.n])]b = 0 (51)

for all [r.sub.1], ..., [r.sub.n] [member of] R. Then one of the following holds:

(1) c,q [member of] Z(R);

(2) there exists A [member of] C such that c - q = [lambda], and f[([x.sub.1], ..., [x.sub.n]).sup.2] is central valued on R;

(3) there exist [lambda],[mu] [member of] C such that c - q = [lambda], ac = [mu]a, and cb = [mu]b.

Proof. By Remark 2 we assume that R is not a domain.

Moreover, by Lemma 16, R satisfies the nontrivial generalized polynomial identity:

P([x.sub.1], ..., [x.sub.n]) = a[cf([x.sub.1], ..., [x.sub.n]) + f([x.sub.1], ..., [x.sub.n])q,

f([x.sub.1], ..., [x.sub.n])]b. (52)

By a theorem due to Beidar (Theorem 2 in [18]) this generalized polynomial identity is also satisfied by U. In case C is infinite, we have P([r.sub.1], ..., [r.sub.n]) = 0 for all [r.sub.1], ..., [r.sub.n] [member of] U[[cross product].sub.C][bar.C], where [bar.C] is the algebraic closure of C. Since both U and U[[cross product].sub.C]C are centrally closed [19, Theorems 2.5 and 3.5], we may replace R by U or U[[cross product].sub.C][bar.C] according to C being finite or infinite. Thus we may assume that R is centrally closed over C which is either finite or algebraically closed. By Martindale's theorem [14], R is a primitive ring having a nonzero socle H with C as the associated division ring, and eHe is a simple central algebra finite dimensional over C, for any minimal idempotent element [member of] e RC.

In light of Jacobson's theorem [20, page 75] R is isomorphic to a dense ring of linear transformations on some vector space V over C.

Assume first that V is finite-dimensional over C. Then the density of R on V implies that R [congruent to] [M.sub.k](C), the ring of all k x k matrices over C. Since R is not commutative we assume k [greater than or equal to] 2. In this case the conclusion follows by Proposition 15.

Assume next that V is infinite-dimensional over C. As in Lemma 2 in [21], the set f(R) is dense on R and so from P([r.sub.1], ..., [r.sub.n]) = 0, for all [r.sub.1], ..., [r.sub.n] [member of] R, we have that R satisfies the generalized identity P(x) = a[cx + xq, x]b. We remark that H satisfies P(x) = a(c[x.sup.2] - [x.sup.2]q + x(q - c)x)b = 0 (see, e.g., [5, proof of Theorem 1]); that is, for all r [member of] H,

a(c[r.sup.2] - [r.sup.2]q + r(q - c)r)b = 0. (53)

In this equality we substitute r with ex(1 - e), for any nontrivial idempotent element [member of] = [e.sup.2] [member of] H, and obtain

aex(1 - e)(q - c)ex(1 - e)b = 0. (54)

By the primeness of R, it follows that either ae = 0 or (1 - e)b = 0 or (1 - e)(q - c)e = 0. Here our aim is to prove that in any case (1 - e)(q - c)e = 0. To do this, we firstly assume that ae = 0. In (53) replace r by ex, so that ac[(ex).sup.2]b = 0, which implies ace = 0.

Moreover we substitute in (53) r with ex + y(1 - e) and by easy computation it follows ay(1 - e)(q - c)exb = 0; that is, (1 - e)(q - c)e = 0.

On the other hand, if one supposes (1 - e)b = 0 and replacing in (53) r by x(1 - e), one has a[(x(1 - e)).sup.2]qb = 0, which implies (1 - e)qb = 0. Finally, if substituted in (53) r with x(1 - e) + ey, as above we have ax(1 - e)(q - c)eyb = 0. Thus in any case it follows (1 - e)(q - c)e = 0.

Similarly one can prove also that e(q - c)(1 - e) = 0.

Hence [q - c, e] = 0, for any idempotent element [member of] e H. Since H is not a domain, then H is generated by its minimal idempotent elements; therefore [q - c, H] = (0); that is, q - c [member of] C. Let [lambda] [member of] C such that q = c + [lambda]. By our assumption it follows that H satisfies a[cx + xc, x]b that is H satisfies a[c, [x.sup.2]]b. In this last replace x by x + 1 and obtain that H satisfies a[c, 2x]b. Since char(H) [not equal to] 2, then acrb - arcb = 0, for all r [member of] H. By 14, Lemma 1] it follows that there exists [mu] [member of] C such that ac = [mu]a and cb = [mu]b, unless ac = cb = 0. []

Corollary 18. Let a,b,c [member of] R such that c [not member of] C and f([x.sub.1], ..., [x.sub.n]) be a noncentral multilinear polynomial over C. If a[[c, f([r.sub.1], ..., [r.sub.n])].sub.2]b = 0, for all [r.sub.1], ..., [r.sub.n] [member of] R, then either a = 0 or b = 0.

4. The Main Result

In [11] Lee proved that every generalized derivation can be uniquely extended to a generalized derivation of U and thus all generalized derivations of R will be implicitly assumed to be defined on the whole U and obtained the following result.

Theorem 19 (Theorem 3 in [11]). Every generalized derivation F on a dense right ideal of R can be uniquely extended to U and assumes the form F(x) = cx + d(x), for some c [member of] U and a derivation d on U.

In this section we denote by [f.sup.d]([x.sub.1], ..., [x.sub.n]) the polynomial obtained from f([x.sub.1], ..., [x.sub.n]) by replacing each coefficient [[alpha].sub.[sigma]] with d([[alpha].sub.[sigma]]). Thus we write d(f([r.sub.1], ..., [r.sub.n])) = [f.sup.d]([r.sub.1], ..., [r.sub.n]) + [[summation].sub.i]f([r.sub.1], ..., d([r.sub.i]), ..., [r.sub.n]), for all [r.sub.1], ..., [r.sub.n] in R.

In light of this, we finally prove our main result.

Proof of Theorem 1. Suppose both a [not equal to] 0 and b [not equal to] 0. Since R satisfies the generalized differential identity

a[F(f([x.sub.1], ..., [x.sub.n])), f([x.sub.1], ..., [x.sub.n])]b, (55)

the above cited Lee's result says that R satisfies

a[cf([x.sub.1], ..., [x.sub.n]) + d(f([x.sub.1], ..., [x.sub.n])), f([x.sub.1], ..., [x.sub.n])]b. (56)

If d is an inner derivation induced by an element q [member of] U, then R satisfies the generalized polynomial identity:

a [cf([x.sub.1], ..., [x.sub.n]) + qf([x.sub.1], ..., [x.sub.n]) - f([x.sub.1], ..., [x.sub.n])q,

f([x.sub.1], ..., [x.sub.n])]b (57)

which is

a[(c + d)f([x.sub.1], ..., [x.sub.n]) - f([x.sub.1], ..., [x.sub.n])d, f([x.sub.1], ..., [x.sub.n])]b. (58)

In this case we are done by Proposition 17.

Hence let d be an outer derivation of R. In this case R satisfies the differential identity:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (59)

By Kharchenko's theorem (see [16, 22]), R satisfies the generalized polynomial identity:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (60)

and in particular, for all i = 1, ..., n, R satisfies the blended component

a[f([x.sub.1], ..., [y.sub.i], ..., [x.sub.n]), f([x.sub.1], ..., [x.sub.n])]b. (61)

Let q [member of] R - Z(R) and replace any [y.sub.i] by [q, [x.sub.i]]. Thus R satisfies

a[summation over (i)]f([x.sub.1], ..., [q, [x.sub.i]], ..., [x.sub.n]), f([x.sub.1], ..., [x.sub.n]) b. (62)

that is,

a[[q,f([x.sub.1], ..., [x.sub.n])].sub.2]b. (63)

By Corollary 18, we get the contradiction q [member of] Z(R). []

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

http://dx.doi.org/10.1155/2014/563284

References

[1] K. I. Beidar, W. S. Martindale III, and A. V. Mikhalev, Rings with Generalized Identities, Pure and Applied Mathematics, Dekker, New York, NY, USA, 1996.

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V. De Filippis, G. Scudo, and L. Sorrenti

Department of Mathematics, Faculty of Sciences, University of Messina, Via F. Stagno D'Alcontres 31, 98166 Messina, Italy

Correspondence should be addressed to V De Filippis; defilippis@unime.it

Received 30 April 2014; Accepted 15 July 2014; Published 28 October 2014

Academic Editor: Ali Jaballah
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Title Annotation:Research Article
Author:De Filippis, V.; Scudo, G.; Sorrenti, L.
Publication:International Scholarly Research Notices
Article Type:Report
Date:Jan 1, 2014
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