Generalized derivations and commutators with nilpotent values on Lie ideals.

Abstract

Let R be a prime ring of characteristic [not equal to] 2 with right quotient ring U and extended centroid C, g [not equal to] 0 a generalized derivation of R, L a non-central Lie ideal of R and n [greater than or equal to] 1 such that [g(u), u][.sup.n] = 0, for all u [member of] L. We prove that there exists an element a [member of] C such that g(x) = ax, for all x [member of] R, unless when R satisfies [s.sub.4] and there exists an element b [member of] U such that g(x) = bx + xb, for all x [member of] R.

Keywords and Phrases: Lie ideals, Prime rings, Generalized derivations, Generalized polynomial identities.

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Let R be a prime ring and g a non-zero derivation of R. A well known result of Posner [13] states that if the commutator [g(x), x] [member of] Z(R), the center of R, for any x [member of] R, then R is commutative.

Several authors generalized the result of Posner to a Lie ideal.

In [8] C. Lanski proved that if R is a prime ring, L a non-commutative Lie ideal of R and g [not equal to] 0 a derivation of R, such that [g(x), x] [member of] Z(R), for all x [member of] L, then either R is commutative, or char(R) = 2 and R satisfies [s.sub.4], the standard identity in 4 variables.

More recently in [2] we studied what happens in case [g(x), x][.sup.n] [member of] Z(R), for any x [member of] L, a non-commutative Lie ideal of R and n [greater than or equal to] 1 a fixed integer. Our conclusion was that R satisfies [s.sub.4]. In particular in [2] it is showed that if [g(x), x][.sup.n] = 0, for any x [member of] L, then R is commutative.

In this paper we will continue this line of investigation, in the case g is not an usual derivation. More precisely we will consider g an additive mapping on R such that g(xy) = g(x)y + xd(y), for all x, y [member of] R and d a derivation of R. In literature, such a mapping is called generalized derivation of R.

In light of this definition, we will prove that

Theorem. Let R be a prime ring of characteristic [not equal to] 2 with right quotient ring U and extended centroid C, g [not equal to] 0 a generalized derivation of R, L a non-central Lie ideal of R and n [greater than or equal to] 1. If [g(u), u][.sup.n] = 0, for all u [member of] L then there exists an element a [member of] C such that g(x) = ax, for all x [member of] R, unless when R satisfies [s.sub.4] and there exists an element b [member of] U such that g(x) = bx + xb, for all x [member of] R.

We cannot expect the conclusion that R is commutative, as the following example shows:

Example. Let R = [M.sub.2](F), the ring of all 2 x 2 matrices over the field F and a [member of] R such that g(x) = ax + xa, for all x [member of] R. If L = [R, R], then we get [g(u), u] = 0, for all u [member of] [R, R], because [u.sup.2] [member of] F, but obviously R is not commutative.

1. Preliminaries

In all that follows, unless stated otherwise, R will be a prime ring of characteristic [not equal to] 2, L a Lie ideal of R, g [not equal to] 0 a generalized derivation of R and n [greater than or equal to] 1 a fixed integer such that [g(x), x][.sup.n] = 0, for all x [member of] L.

For any ring S, Z(S) will denote its center, and [a, b] = ab - ba. In addition [s.sub.4] will denote the standard identity in 4 variables.

The related object we need to mention is the right Utumi quotient ring U of a ring R (sometimes, as in [1], U is called the maximal right ring of quotients).

The definitions, the axiomatic formulations and the properties of this quotient ring U can be found in [1].

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

1) R [??] 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.

We make also a frequent use of the theory of generalized polynomial identities and differential identities (see [1], [3], [7], [10], [12]). In particular we need to recall that, when R is prime and I a two-sided ideal of R, then I, R and U satisfy the same generalized polynomial identities [3] and also the same differential identities [10].

In [9] T.K. Lee extended the definition of a generalized derivation as follows: by a generalized derivation we mean an additive mapping g : I [right arrow] U such that g(xy) = g(x)y + xd(y), for all x, y [member of] I, where I is a dense right ideal of R and d is a derivation from I into U.

In particular, in all that follows we will make use of the previous cited result in the case I is a two-sided ideal of R.

Moreover Lee also 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:

Proposition. Every generalized derivation g on a dense right ideal of R can be uniquely extended to U and assumes the form g(x) = ax + d(x), for some a [member of] U and a derivation d on U.

More details about generalized derivations can be found in [6], [9], [11].

2. Proof of the Theorem

We begin with the following:

Lemma. Let R be a prime ring of characteristic different from 2, I a two-sided ideal of R, a, b elements of R and n [greater than or equal to] 1. If ([a[x, y] - [x, y]b, [x, y]])[.sup.n] = 0, for any x, y [member of] I, then a, b [member of] Z(R), unless when R satisfies [s.sub.4] and a = -b.

Proof. For any x, y [member of] I, ([a[x, y] - [x, y]b, [x, y]])[.sup.n] = 0. Since I and U satisfy the same generalized polynomial identities, we have ([a[x, y] - [x, y]b, [x, y]])[.sup.n] = 0, for any x, y [member of] U. In particular this one holds in R and so R is a GPI-ring. Moreover, since U remains prime by the primeness of R, replacing R by U we may assume that the extended centroid C = Z(U) is just the center of R. Note that R is a centrally closed prime C-algebra in the present situation [4], i.e. RC = R. By Martindale's theorem in [12], RC (and so R) is a primitive ring which is isomorphic to a dense ring of linear transformations of a vector space V over a division ring D.

Assume first that [dim.sub.D] V = t, a finite integer.

In this case R is a simple GPI ring with 1, and so it is a central simple algebra finite dimensional over its center. From Lemma 2 in [8] it follows that there exists a suitable field F such that R [??] [M.sub.k](F), the ring of all k x k matrices over F, and moreover [M.sub.k](F) satisfies the same generalized polynomial identity of R. If k = 1, then R is commutative. Consider now the case when k [greater than or equal to] 2. In what follows we denote [e.sub.ij] the usual matrix unit in [M.sub.k](F).

Step 1.

Suppose first that k = 2. Recall that in this case [x, y][.sup.2] [member of] F, for all x, y [member of] R. For this reason ([a[x, y] - [x, y]b, [x, y]])[.sup.2] = 0, for all x, y [member of] R.

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. If we choose x = [e.sub.12], y = [e.sub.21] then we get:

0 = [a([e.sub.11] - [e.sub.22]) - ([e.sub.11] - [e.sub.22])b, ([e.sub.11] - [e.sub.22])][.sup.2] = [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Therefore either [a.sub.12] + [b.sub.12] = 0 or [a.sub.21] + [b.sub.21] = 0. We want to prove that [a.sub.12] + [b.sub.12] = 0 and also [a.sub.21] + [b.sub.21] = 0.

Without loss of generality we can pick [a.sub.12] + [b.sub.12] = 0 and suppose by contradiction that [a.sub.21] + [b.sub.21] [not equal to] 0.

Let [x, y] = [[e.sub.11], [e.sub.12] + [e.sub.21]] = [e.sub.12] - [e.sub.21]. In this case we have:

(i) 0 = [a([e.sub.12] - [e.sub.21]) - ([e.sub.12] - [e.sub.21])b, ([e.sub.12] - [e.sub.21])][.sup.2] = [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Since we suppose [a.sub.21] + [b.sub.21] [not equal to] 0 and char(R) [not equal to] 2, then [a.sub.11] + [b.sub.11] + [a.sub.22] + [b.sub.22] = 0.

Now let [x, y] = [[e.sub.11], - [e.sub.12] - [e.sub.21]] = -[e.sub.12] + [e.sub.21]. In this case we have:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

which implies [a.sub.21] + [b.sub.21] = 0, a contradiction.

Therefore must be [a.sub.12] + [b.sub.12] = 0 and also [a.sub.21] + [b.sub.21] = 0, and thanks to (i), it follows again that [a.sub.11] + [b.sub.11] + [a.sub.22] + [b.sub.22] = 0.

Finally choose [x, y] = [[e.sub.11], [e.sub.12] - [e.sub.21]] = [e.sub.12] + [e.sub.21]. In this case we have:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

and since [a.sub.11] + [b.sub.11] = -[a.sub.22] - [b.sub.22], it follows

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

that is 0 = [a.sub.11] + [b.sub.11] = [a.sub.22] + [b.sub.22].

Thus we conclude that if k = 2, then a + b = 0.

Step 2.

Let now k [greater than or equal to] 3. For all i, j, k positive integer such that anyone is different from the others, let [x, y] = [[e.sub.ik] - [e.sub.kj], [e.sub.kk]] = [e.sub.ik] + [e.sub.kj]. Then

(ii) 0 = [a([e.sub.ik] + [e.sub.kj]) - ([e.sub.ik] + [e.sub.kj])b, ([e.sub.ik] + [e.sub.kj])][.sup.n] = (-a[e.sub.ij] + [e.sub.ij]b - [e.sub.ik]b[e.sub.ik] - [e.sub.ik]b[e.sub.kj] - [e.sub.kj]b[e.sub.ik] - [e.sub.kj]b[e.sub.kj] - [e.sub.ik]a[e.sub.ik] - [e.sub.kj]a[e.sub.ik] - [e.sub.kj]a[e.sub.kj] - [e.sub.ik]a[e.sub.kj])[.sup.n].

Right multiplying the (ii) by [e.sub.ii] we get ([e.sub.ij]b)[.sup.n][e.sub.ii] = 0, for all i [not equal to] j, that is the (j, i)-entry of the matrix b is zero, for all i [not equal to] j. This means that b is a diagonal matrix.

On the other hand, left multiplying the (ii) by [e.sub.jj], we have that [e.sub.jj](-a[e.sub.ij])[.sup.n] = 0 and, as above, a is a diagonal matrix.

We can write a = [[SIGMA].sub.i] [a.sub.i][e.sub.ii], b = [[SIGMA].sub.i] [b.sub.i][e.sub.ii]. Let [phi] be any automorphisms of [M.sub.k](F). Since ([[phi](a)[[phi](x), [phi](y)] - [[phi](x), [phi](y)][phi](b), [[phi](x), [phi](y)]])[.sup.n] = 0, then [phi](a) and [phi](b) are both diagonal matrices. In particular, for any r [not equal to] s, let [[phi].sub.rs](x) = (1 - [e.sub.rs])x(1 + [e.sub.rs]), for all x [member of] [M.sub.k](F). Therefore [[phi].sub.rs](a) and [[phi].sub.rs](b) must be diagonal, moreover

[[phi].sub.rs](a) = a - ([a.sub.ss] - [a.sub.rr])[e.sub.ij]

[[phi].sub.rs](b) = b - ([b.sub.ss] - [b.sub.rr])[e.sub.ij].

This implies that [a.sub.ss] = [a.sub.rr] and [b.sub.ss] = [b.sub.rr], for all r [not equal to] s, i.e. a and b are central in [M.sub.k](F).

Consider now the case when [dim.sub.D] V = [infinity]. By lemma 2 in [14], in this case R satisfies the generalized identity [ax - xb, x][.sup.n].

Suppose that there exists [member of] 2 V such that v, va are lineraly D-independent. By the densisy of R, there exists w [member of] V such that v, va, w are linearly D-independent and [x.sub.0] [member of] R such that

v[x.sub.0] = 0, (va)[x.sub.0] = w, w[x.sub.0] = v.

From this we get the contradiction

0 = v[a[x.sub.0] - [x.sub.0]b, [x.sub.0]][.sup.n] = v [not equal to] 0.

Hence v, va are linearly D-dependent, for all v [member of] V and standard arguments show that a [member of] Z(R). So 0 = [ax - xb, x][.sup.n] = [-xb, x][.sup.n], for all x [member of] R.

Now suppose that there exists v [member of] V such that v, vb are linearly D-independent By the density of R we get that there exists [x.sub.0] [member of] R such that

v[x.sub.0] = v, (vb)[x.sub.0] = vb - v.

Once again it follows the contradiction

0 = v[-[x.sub.0]b, [x.sub.0]][.sup.n] = v [not equal to] 0.

As above we conclude that b is central in R.

As a consequence of the previous lemma, we state here the following:

Corollary. Let R be a prime ring of characteristic different from 2, I a two-sided ideal of R, a an element of R and n [greater than or equal to] 1. If ([a[x, y], [x, y]])[.sup.n] = 0, for any x, y [member of] I, then a [member of] Z(R).

We are ready to prove the main result of this note:

Theorem. Let R be a prime ring of characteristic [not equal to] 2 with right quotient ring U and extended centroid C, g [not equal to] 0 a generalized derivation of R, L a non-central Lie ideal of R and n [greater than or equal to] 1. If [g(u), u][.sup.n] = 0, for all u [member of] L, then there exists an element a [member of] C such that g(x) = ax, for all x [member of] R, unless when R satisfies [s.sub.4] and there exists an element b [member of] U such that g(x) = bx + xb, for all x [member of] R.

Proof. Since we assume that char(R) [not equal to] 2, by a result of Herstein [5], L [??] [I, R], for some I [not equal to] 0, an ideal of R, and also L is not commutative. Therefore we will assume throughout that L [??] [I, R]. Without loss of generality we can assume L = [I, I]. In section 1 we remarked that every generalized derivation g on a dense right ideal of R can be uniquely extended to U and assumes the form g(x) = ax + d(x), for some a [member of] U and a derivation d on U. Therefore U satisfies the following differential identity

[a[x, y] + d([x, y]), [x, y]][.sup.n].

In light of Kharchenko's theory ([7], [10]), we divide the proof into two cases:

Case 1. Let d be the inner derivation induced by the element q [member of] U, that is d(x) = [q, x], for all x [member of] U. Thus U satisfies the generalized polynomial identity

[a[x, y] + q[x, y] - [x, y]q, [x, y]][.sup.n] = [(a + q)[x, y] - [x, y]q, [x, y]][.sup.n].

By our lemma we get that either (a + q) and q are central in U, or a + q = -q and U satisfies [s.sub.4].

In the first case d = 0 and for all x [member of] U, g(x) = ax, for a [member of] C.

In the either one we conclude that, for all x [member of] U,

g(x) = ax + [q, x] = ax + qx - xq = (a + q)x - xq = (-q)x - xq = (-q)x + x(-q).

In both cases we are done.

Case 2. Let now d be an outer derivation of U. Since U satisfies

[a[x, y] + [d(x), y] + [x, d(y)], [x, y]][.sup.n]

then, by Kharchenko's result in [7], U also satisfies

[a[x, y] + [t, y] + [x, z], [x, y]][.sup.n]

and in particular it satisfies the blended component

[a[x, y], [x, y]][.sup.n].

Thanks to corollary, we conclude that a 2 [member of] = Z(U).

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Vincenzo De Filippis ([dagger])

Dipartimento di Matematica, Universita di Messina

Salita Sperone 31, 98166 Messina, Italia

Received April 8, 2005, Accepted June 11, 2005.

*2000 Mathematics Subject Classification. 16N60 (primary), 16W25 (secondary)

([dagger]) E-mail:[e-mail]