Annihilator Conditions with Derivations in Prime Rings of Characteristic 2

Indian J. pure appl. Math., 39(6): 459-465, December 2008 °c Printed in India.

ANNIHILATOR CONDITIONS WITH DERIVATIONS IN PRIME RINGS OF CHARACTERISTIC 2

YU WANG

College of Mathematics, Jilin Normal University, Siping, Jilin 136000, Peoples’ Republic of China e-mail: [email protected]

(Received 1 August 2006; after ﬁnal revision 11 August 2008; accepted 22 August 2008)

Let R be a prime ring of characteristic 2 with d a nonzero derivation of R,L a noncentral Lie ideal of R, a ∈ R such that a[d(u), u]n is central, for all u ∈ L, where n is a ﬁxed positive

integer. Then a = 0 unless R satisﬁes S4, the standard identity in 4 variables.

Keywords:Primering;derivation;generalizedpolynomialidentity;centraldifferential identity

Let R be a prime ring with center Z. For x, y ∈ R, we set [x, y] = xy − yx and [x, y]2 = [[x, y], y]. 2 2 Note that [x, y]2 = xy + y x when char R = 2. Let n be a ﬁxed positive integer. We denote by d a nonzero derivation of R. As usual, we denote by S4 the standard identity in 4 variables. A well known result of Posner [15] states that if [d(x), x] ∈ Z, for all x ∈ R, then R is commutative. Posner’s theorem was extended to the case of Lie ideals of prime rings by Lee and Lee [13] and Lanski [10], respectively. Carini and De Filippis [2] studied a more generalized situation when [d(u), u]n ∈ Z for all u in a noncentral Lie ideal of R of characteristic 6= 2. Furthermore, De Filippis [6] investigated the left annihilator of power values of commutators in prime rings of characteristic 6= 2 and obtained the following:

Theorem A [6] — Let R be a prime ring of characteristic different from 2, d a nonzero derivation of R, L a noncentral Lie ideal of R, n ≥ 1 a ﬁxed integer and a ∈ R. If a[d(u), u]n ∈ Z for all u ∈ L, then either a = 0 or R satisﬁes S4, the standard identity in 4 variables. 460 YU WANG

The main purpose of this paper is to remove the restriction of characteristic in Theorem A. Explicitly we prove the following:

Theorem 1 — Let R be a prime ring of characteristic 2, Z the center of R, d a nonzero derivation of R, L a noncentral Lie ideal of R and a ∈ R. If a[d(u), u]n ∈ Z for all u ∈ L, then either a = 0 or R satisﬁes S4, the standard identity in 4 variables.

As a direct consequence of the above two results we have the following general result.

Theorem 2 — Let R be a prime ring, Z the center of R, d a nonzero derivation of R, L a noncentral Lie ideal of R and a ∈ R. If a[d(u), u]n ∈ Z for all u ∈ L, then either a = 0 or R satisﬁes S4, the standard identity in 4 variables.

In all that follows, R will be a prime ring with its center Z, its extended centroid C and its two-sided Martindale quotient ring Q. All these notions are explained in detail in a book of Beidar et al. [1]. It is well known that every derivation d of R can be uniquely extended to a derivation of Q, which will be also denoted by d. Recall that a derivation d of R is called Q-inner induced by some element b ∈ Q if d(x) = [b, x] for all x ∈ R. Otherwise, d is called Q-outer. Let

T = Q ∗C C{X,Y } be the free product over C of C-algebra Q and C{X,Y }, the free C-algebra in two noncommuting indeterminates X and Y .

In order to prove our main result we need some lemmas as follows.

Lemma 1 — Let R be a prime ring of characteristic 2 and a, b ∈ Q with b ∈ C such that n a([b, [x, y]]2) = 0 for all x, y ∈ R, where n ≥ 0 is a ﬁxed integer. If R does not satisfy any nontrivial generalized polynomial identity, then a = 0.

n PROOF : By a theorem due to Chuang [4, Theorem 2] we have a([b, [x, y]]2) = 0 for all x, y ∈ Q. Since R does not satisfy any nontrivial generalized polynomial identity, we see that n a([b, [X,Y ]]2) is the zero element in the free product T = Q ∗C C{X,Y }, that is

n−1 2 n−1 2 a([b, [X,Y ]]2) b[X,Y ] + a([b, [X,Y ]]2) [X,Y ] b = 0.

It follows from b∈ / C that

n−1 2 n−1 2 a([b, [X,Y ]]2) b[X,Y ] = 0 = a([b, [X,Y ]]2) [X,Y ] b

n−1 and so a([b, [X,Y ]]2) = 0 ∈ T . Continuing this process, we obtain that a[b, [X,Y ]]2 = 0, that is ab[X,Y ]2 + a[X,Y ]2b = 0. Since b∈ / C, we get that a[X,Y ]2b = 0 and so a = 0 as desired.

Lemma 2 — Let R be a prime ring of characteristic 2, I a nonzero ideal of R, d a nonzero derivation of R, a ∈ R and n ≥ 1 a ﬁxed integer. If a[d([x, y]), [x, y]]n = 0 for all x, y ∈ I, then a = 0 unless R satisﬁes S4, the standard identity in 4 variables. DERIVATIONS IN PRIME RINGS OF CHARACTERISTIC 2 461

PROOF : Suppose on the contrary that both a 6= 0 and R does not satisfy S4. By assumption we have

a[[d(x), y] + [x, d(y)], [x, y]]n = 0 for all x, y ∈ I. If d is not Q-inner, by Kharchenko’s theorem in [9], we get

a[[z, y] + [x, t], [x, y]]n = 0 for all x, y, t, z ∈ I. In particular, for z = 0, we obtain that a[[x, t], [x, y]]n = 0 for all x, y, t ∈ I. By a theorem due to Chuang [4, Theorem 2], a[[x, t], [x, y]]n = 0 for all x, y, t ∈ Q and so in R. Let v = [[x, t], [x, y]]n. Then av = 0. Now we can write a[[w, v], [ua, v]]n = 0 for all w, u ∈ R. Since av = 0, it reduces to a(wv2ua)n = 0. This can be written as (awv2u)n+1 = 0 for all w, u ∈ R. By Levitzki’s lemma [7, Lemma 1.1] awv2u = 0 for all w, u ∈ R. Since R is prime, v2 = 0 that is [[x, t], [x, y]]2n = 0 for all x, y, t ∈ R. This is a polynomial identity and hence there exists a ﬁeld

F such that R ⊆ Mm(F ) with m > 1 and R and Mm(F ) satisfy the same polynomial identities. 2n But choosing x = e11, y = e21 and t = e12 we get 0 = [[x, t], [x, y]] = e11 + e22 which is a contradiction.

Thus we assume that d is Q-inner. That is, there exists a noncentral element b ∈ Q such that n d(x) = [b, x] for all x ∈ R. So a([b, [x, y]]2) = 0 for all x, y ∈ I. Since I and Q satisfy the same generalized polynomial identities (or, GPIs in brief), we have

n a([b, [x, y]]2) = 0 for all x, y ∈ Q. (1)

In case, C is inﬁnite, the GPI (1) is also satisﬁed by Q ⊗C C where C is the algebraic closure of

C. Since both Q and Q⊗C C are prime and centrally closed [5], we may replace R by Q or Q⊗C C according as C is finite or infinite. Thus we may assume that R is centrally closed over Z which is n either finite or algebraically closed such that a([b, [x, y]]2) = 0 for all x, y ∈ R. By Martindale’s theorem in [14] R is a primitive ring having nonzero socle and the commuting division D is a finite dimensional central division algebra over Z. Since Z is either finite or algebraically closed, D must coincide with Z. Thus R is isomorphic to a dense subring of EndZ V for some vector space V over

Z. Since R does not satisfy S4, it is obvious that dimZ V ≥ 3. We will show that, for any given v ∈ V, v and bv are Z-dependent. Assume on the contrary that v and bv are Z-independent and set

W = Zv + Zbv. Since dimZ V ≥ 3, there exists u ∈ V such that v, bv, u are also Z-independent.

If av 6= 0, by the density of R in End Z V there exist two elements r1 and r2 in R such that

r1v = 0, r1bv = u, r1u = v; r2v = 0, r2bv = 0, r2u = u.

n It follows from (1) that 0 = a([b, [r1, r2]]2) v = av, a contradiction. 462 YU WANG

Suppose that av = 0. Since a 6= 0, there exists w ∈ V such that aw 6= 0 and so a(v − w) 6= 0. As above we get that there exist β, γ ∈ Z such that

bw = βw and b(w − v) = γ(w − v).

This yields that (β − γ)w ∈ W . Now β = γ implies the contradiction that bv = βv. Thus β 6= γ and so w ∈ W . But if u ∈ V with au = 0, then a(w + u) 6= 0. So w + u ∈ W forcing u ∈ W . Thus V = W and so dimZ V = 2, a contradiction.

Hence, in any case, for all v ∈ V , v and bv are linearly Z-dependent. Thus, standard arguments show that b ∈ Z which contradicts our hypothesis. With this the proof is complete.

The following lemma is due to Chang and Lee [3, Theorem 1].

Lemma A [3] — Let R be a prime ring and ρ a nonzero right ideal of R. If ρ possesses a central differential identity, then R is a PI-ring.

The following lemma is due to Lanski [11, Lemma 2].

Lemma B [11] — Let R be a noncommutative simple algebra, ﬁnite-dimensional over its center

Z. If g(x1, . . . , xt) ∈ R∗Z Z{xj}, the free product over Z, is an identity for R that is homogeneous in {x1, . . . , xt} of degree d, then for some ﬁeld F and n > 1, R ⊆ Mn(F ) and g(x1, . . . , xt) is an identity for Mn(F ).

Lemma 3 — Let R = Mm(F ), the ring of m × m matrices over a ﬁeld F of characteristic 2 n n and a, b ∈ R with b∈ / F . If a([b, [x, y]]2) ∈ F for all x, y ∈ R and a([b, [r1, r2]]2) 6= 0 for some r1, r2 ∈ R, then m ≤ 2.

PROOF : Suppose on the contrary that m ≥ 3. By assumption we easily see that a is an P invertible matrix. We set b = bstest with bst ∈ F . For distinct integers 1 ≤ i, j, r ≤ m, if we choose [r1, r2] = [eij, ejr − eri] = eir + erj, then X X [b, [r1, r2]]2 = beij + eijb = bjieii + bjiejj + bjteit + bsiesj t6=i s6=j and by assumption we get X X n n n −1 ([b, [r1, r2]]2) = (bji) eii + (bji) ejj + αteit + βsesj ∈ F a t6=i s6=j

n with αt, βs ∈ F . Since m ≥ 3, we easily see that ([b, [r1, r2]]2) is not an invertible matrix. Thus X X n n n ([b, [r1, r2]]2) = (bji) eii + (bji) ejj + αteit + βsesj = 0. t6=i s6=j DERIVATIONS IN PRIME RINGS OF CHARACTERISTIC 2 463

n It follows that in particular, (bji) = 0 and so bji = 0. Therefore, b is a diagonal matrix, that is P b = bssess with bss ∈ F . P Next we show that b = bssess is a central matrix, that is, bii = bjj for distinct i, j. Let

ϕij ∈ AutF (R) such that ϕij(x) = (1 + eij)x(l − eij) for all x ∈ R. On the one hand, for any n x, y ∈ R, we have ϕ(a)([ϕij(b), [x, y]]2) ∈ F . Since ϕ(a) 6= 0, as above we have that ϕij(b) is also diagonal. On the other hand, ϕij(b) = b + (bjj − bii)eij, i.e., bjj = bii for any j 6= i. Hence b is a central matrix, a contradiction. The proof is complete.

We are now in a position to prove our main result.

PROOF : Suppose that R does not satisfy S4. That is dimC RC > 4. Our goal is to prove that a = 0. By a theorem of Lanski and Montgomery [12, Theorem 13], we have [L, L] 6= 0 and 0 6= [I,R] ⊆ L, where I is the ideal of R generated by [L, L]. By assumption we have a[d([x, y]), [x, y]]n ∈ Z for all x, y ∈ I. Thus

a[[d(x), y] + [x, d(y)], [x, y]]n ∈ Z for allx, y ∈ I. (2)

If a[d([x, y]), [x, y]]n = 0 for all x, y ∈ I, by Lemma 2 we get that a = 0 as desired. Otherwise, n there exist r1, r2 ∈ I such that a[d([r1, r2]), [r1, r2]] 6= 0. Thus I satisﬁes the central differential identity a[[d(X),Y ] + [X, d(Y )], [X,Y ]]n.

In view of Lemma A, R must be a PI-prime ring and so is Q. If d is not Q-inner, applying Kharchenko’s theorem [9] to (2), we have

[a([[t, y] + [x, z], [x, y]])n, r] = 0 for all x, y, t, z, r ∈ I.

In particular, for z = 0,

[a[[t, y], [x, y]]n, r] = 0 for all x, y, t, r ∈ I. (3)

Since I and Q satisfy the same generalized polynomial identities, (3) is satisfied by Q. If a 6= 0, then (3) is a nontrivial GPI on Q. By Martindale’s theorem in [14] Q is a PI-primitive ring. By the famous Kaplansky-Amitsur’s theorem (see [8, Page 17, Theorem]) Q is finite dimensional central simple algebra over C. It follows from Lemma B that there exists a suitable field F of characteristic

2 such that Q ⊆ Mm(F ), the ring of all m × m matrices over F , and moreover Mm(F ) satisﬁes 464 YU WANG

P (3). Since dimC RC > 4, it implies that m > 2. We set a = astest with ast ∈ F . For distinct integers 1 ≤ i, j, s ≤ m, if we choose x = esj, y = ess, t = ejs, r = eji in (3), then X akjeki + aijejj + aisejs = 0 k

This implies that akj = 0 for any k, j, that is a = 0, a contradiction.

We next assume that d is an Q-inner derivation induced by an element b ∈ Q. Since d 6= 0, b∈ / n C. Hence a([b, [x, y]]2) ∈ Z for all x, y ∈ I and so

n [a([b, [x, y]]2) , z] = 0 for all x, y, z ∈ Q. (4)

n Since there exist r1, r2 ∈ I such that a([b, [r1, r2]]2) 6= 0 it is obvious that (4) is a nontrivial GPI on Q. By Martindale’s theorem in [14] Q is a PI-primitive ring. As above we have that Q is finite dimensional central simple algebra over C. It follows from Lemma B that there exists a suitable field F of characteristic 2 such that Q ⊆ Mm(F ), the ring of all m × m matrices over F with m ≥ 3 and moreover Mm(F ) satisfies the same (4). But it follows from Lemma 3 that m ≤ 2, a contradiction. The proof is thereby complete.

ACKNOWLEDGEMENT

The author are thankful to the referees for some helpful comments and suggestions.

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