Applied Mathematics E-Notes, 1(2001), 24-30 \begin{matrix}c \\ ◯ \end{matrix} Available free at mirror sites of http://math2.math.nthu.edu.tw/∼amen/

Some Results for Generalized Lie Ideals in Prime Rings with Derivation II ^{∗}

Kâzım Kaya^{†}, Öznur Gölbaşi^{‡}, Neşet Aydin^{§}

Received 18 December 2000

Abstract Let R be a prime ring of characteristic different from two, d : R → R a nonzero derivation, and M a non-zero left ideal of R. We prove the following results:

(1) if a ∈ R and [d(R), a]_{σ,τ} = 0, then σ(a) + τ(a) ∈ Z, the center of R, (2) if d([R, a]_{σ,τ}) = 0, then σ(a) + τ(a) ∈ Z, (3) if ([R, M]_{σ,τ} , a)_{σ,τ} = 0, then a ∈ Z, (4) (d(R), a) = 0 if, and only if, d((R, a)) = 0.

Let R be a ring and σ, τ be two mappings from R into itself. We write [x, y], (x, y), [x, y]_{σ,τ} , (x, y)_{σ,τ} for xy−yx, xy+yx, xσ(y)−τ(y)x and xσ(y)+τ(y)x respectively, and will make extensive use of the following basic commutator identities: (xy, z) = x[y, z]+ (x, z)y = x(y, z) − [x, z]y, [xy, z]_{σ,τ} = x[y, z]_{σ,τ} + [x, τ(z)]y = x[y, σ(z)] + [x, z]_{σ,τ} y.

An additive mapping D : R → R is called a derivation if D(xy) = D(x)y + xD(y) holds for all x, y ∈ R. A derivation D is inner if there exists an a ∈ R such that D(x) = [a, x] for all x ∈ R. For subsets A, B ⊂ R, let [A, B] ([A, B]_{σ,τ}) be the additive subgroup generated by all [a, b] ([a, b]_{σ,τ}) for all a ∈ A and b ∈ B. We recall that a Lie ideal L is an additive subgroup of R such that [R, L] ⊂ L. We first introduce the generalized Lie ideal in [4] as following. Let U be an additive subgroup of R, and let σ, τ : R → R be two mappings. Then (i) U is a (σ, τ)-right Lie ideal of R if [U, R]_{σ,τ} ⊂ U, (ii) U is a (σ, τ)-left Lie ideal of R if [R, U]_{σ,τ} ⊂ U, (iii) if U is both a (σ, τ)-right Lie ideal and (σ, τ)-left Lie ideal of R, then U is a (σ, τ)-Lie ideal of R. Every Lie ideal of R is a (1, 1)-Lie ideal of R, where 1 : R → R is the identity map. As an example, let I be the set of integers,

R = \{( \begin{matrix}x & y \\ z & t \end{matrix} ) : x, y, z, t ∈ I\} ,

U = \{( \begin{matrix}x & y \\ 0 & x \end{matrix} ) : x, y ∈ I\} ⊂ R,

and τ : R → R the mapping defined by τ(x) = bxb, where b = ( \begin{matrix}1 & −1 \\ 0 & −1 \end{matrix} ) ∈ R. Then

U is a (1, τ)-left Lie ideal but not a Lie ideal of R.

∗Mathematics Subject Classifications: 16N10, 16W25, 16U80.

†Çanakkale Onsekiz mart Univ., Math. Department, Çanakkale, Turkey

‡Cumhuriyet Univ., Math. Department, 58140 Sivas, Turkey

§Mersin Univ., Math. Department, 33342 Mersin, Turkey

Some algebraic properties of (σ, τ)-Lie ideals are considered in [1], [2] and [5], where further references can be found.

Let R be a prime ring of characteristic different from two, d : R → R a nonzero derivation, Z the center of R and a ∈ R. Lee and Lee in [6] proved that if [d(R), d(R)] ⊂ Z, then R is commutative. In the present paper, we generalize this result for generalized Lie ideal. In [3], Herstein proved that in a prime ring of characteristic different from two, [d(R), a] = 0 implies a ∈ Z. We shall extend Herstein’s theorem by proving that [d(R), a]_{σ,τ} = 0 implies σ(a) + τ(a) ∈ Z.

Throughout this note, R will be a prime ring with characteristic different from 2, Z the center of R, d a non-zero derivation of R and U is (σ, τ)-left Lie ideal of R.

LEMMA 1. If a ∈ R and [d(R), a]_{σ,τ} = 0, then σ(a) + τ(a) ∈ Z.

PROOF. If a ∈ Z then the proof of the theorem is obvious. So we assume that a ∉ Z. By hypothesis, we have for all x ∈ R,

0 = [d(xσ(a)), a]_{σ,τ} = [d(x)σ(a) + xd(σ(a)), a]_{σ,τ}
= d(x)[σ(a), σ(a)] + [d(x), a]_{σ,τ} σ(a) + x[d(σ(a)), a]_{σ,τ} + [x, τ(a)]d(σ(a)).

Hence we obtain

[x, τ(a)]d(σ(a)) = 0, x ∈ R  (1)

Replacing x by xy, y ∈ R in (1) and using (1), we get

[R, τ(a)]Rd(σ(a)) = 0.

Since R is prime ring and a ∉ Z, we obtain d(σ(a)) = 0. Now let us consider the following mappings on R : D(x) = [x, σ(a)] and H(x) = [x, a]_{σ,τ} , where D is a non-zero derivation of R such that Hd(x) = 0. For any x, y ∈ R, we have H(xy) = [xy, a]_{σ,τ} = x[y, σ(a)] + [x, a]_{σ,τ} y. Hence we get

H(xy) = H(x)y + xD(y)  (2)

But this can also be calculated in a different way. Indeed, H(xy) = [xy, a]_{σ,τ} = x[y, a]_{σ,τ} + [x, τ(a)]y and so one obtains

H(xy) = [x, τ(a)]y + xH(y)  (3)

For any r ∈ R, 0 = [d(r), a]_{σ,τ} = d(r)σ(a) − τ(a)d(r), and so,

0 = d(0) = d(d(r)σ(a) − τ(a)d(r))
= d^{2}(r)σ(a) + d(r)d(σ(a)) − d(τ(a))d(r) − τ(a)d^{2}(r)
[d^{2}(r), a]_{σ,τ} − d(τ(a))d(r).

This implies that d(τ(a))d(r) = 0 for all r ∈ R. Using [7, Lemma 1], we obtain d(τ(a)) = 0. On the other hand, for any x ∈ R,

dH(x) = d([x, a]_{σ,τ}) = d(xσ(a) − τ(a)x)
= d(x)σ(a) + xd(σ(a)) − d(τ(a))x − τ(a)d(x) = [d(x), a]_{σ,τ} = 0.

26 Generalized Ideals in Prime Rings Thus we get,

dH(R) = 0.  (4)

In view of (2), (3) and (4), one obtains, for any x, y ∈ R,

0 = Hd(xH(y)) = H(d(x)H(y) + xdH(y)) = H(d(x)H(y))
= Hd(x)H(y) + d(x)DH(y) = d(x)DH(y),

and so, d(R)DH(R) = 0. By using [7, Lemma 1] we arrive at

DH(x) = 0.

That is,

[[x, a]_{σ,τ} , σ(a)] = 0, x ∈ R.  (5)

Taking τ(a)x, instead of x in (5), we get

0 = [[τ(a)x, a]_{σ,τ} , σ(a)] = [τ(a)[x, a]_{σ,τ} + [τ(a), τ(a)]x, σ(a)]
= [τ(a)[x, a]_{σ,τ} , σ(a)] = τ(a)[x, a]_{σ,τ} , σ(a)] + [τ(a), σ(a)][x, a]_{σ,τ} ,

and so,

[τ(a), σ(a)][x, a]_{σ,τ} = 0, x ∈ R.  (6)

Replacing x by xy, y ∈ R in (6) and using (6), we obtain

[τ(a), σ(a)]R[R, σ(a)] = 0.

Since R is a prime ring and a ∉ Z, we get

[τ(a), σ(a)] = 0.  (7)

Now, expanding (5) and using (7) one obtains

0 = [[x, a]_{σ,τ} , σ(a)] = [xσ(a) − τ(a)x, σ(a)]
= [x, σ(a)]σ(a) − τ(a)[x, σ(a)] = [[x, σ(a)], a]_{σ,τ} ,

that is,

HD(x) = 0, x ∈ R.  (8)

Linearizing (8), we get

0 = HD(xy) = H(D(x)y + xD(y))
= HD(x)y + D(x)D(y) + xHD(y) + [x, τ(a)]D(y)
= D(x)D(y) + [x, τ(a)]D(y) = [x, σ(a)]D(y) + [x, τ(a)]D(y),

that is,

[x, σ(a) + τ(a)]D(y) = 0, x, y ∈ R.

Since D is a non-zero derivation of the prime ring R, using [7, Lemma 1], we obtain σ(a) + τ(a) ∈ Z.

COROLLARY 2. Let R be a prime ring of characteristic different from two, d be a nonzero derivation of R and U is (σ, τ)-left Lie ideal of R. If [d(R), U]_{σ,τ} = 0, then σ(u) + τ(u) ∈ Z for all u ∈ U.

THEOREM 3. Let R be a prime ring of characteristic different from two, d a nonzero derivation of R and a ∈ R. If d([R, a]_{σ,τ}) = 0, then σ(a) + τ(a) ∈ Z.

PROOF. For all r ∈ R,

0 = d([τ(a)r, a]_{σ,τ}) = d(τ(a)[r, a]_{σ,τ}) = d(τ(a))[r, a]_{σ,τ} + τ(a)d([r, a]_{σ,τ})

and so,

d(τ(a))[r, a]_{σ,τ} = 0, r ∈ R.  (9)

Taking rs, s ∈ R instead of r in (9), we have

0 = d(τ(a))[rs, a]_{σ,τ} = d(τ(a))r[s, σ(a)] + d(τ(a))[r, a]_{σ,τ} s.

Using (9), we get

d(τ(a))R[R, σ(a)] = 0.

Since R is a prime ring, we see that d(τ(a)) = 0 or a ∈ Z. If a ∈ Z, then σ(a)+τ(a) ∈ Z. The proof is then complete. Therefore, let us assume that a ∉ Z. For r ∈ R,

0 = d([rσ(a), a]_{σ,τ}) = d([r, a]_{σ,τ} σ(a))
= d([r, a]_{σ,τ})σ(a) + [r, a]_{σ,τ} d(σ(a))

and so,

[r, a]_{σ,τ} d(σ(a)) = 0, r ∈ R.  (10)

Replacing r by rs, s ∈ R in (10), we get

0 = [rs, a]_{σ,τ} d(σ(a)) = r[s, a]_{σ,τ} d(σ(a)) + [r, τ(a)]sd(σ(a))
= [r, τ(a)]sd(σ(a)).

That is,

[r, τ(a)]Rd(σ(a)) = 0, r ∈ R.

Since R is a prime ring, and since we have assumed that a ∉ Z, we obtain

d(σ(a)) = 0.

Now, from our hypothesis, we have, for any r ∈ R,

0 = d([r, a]_{σ,τ}) = d(rσ(a) − τ(a)r)
= d(r)σ(a) + rd(σ(a)) − d(τ(a))r − τ(a)d(r).

Since d(σ(a)) = 0 and d(τ(a)) = 0, we get

[d(r), a]_{σ,τ} = 0, r ∈ R.

By Lemma 1, we have σ(a) + τ(a) ∈ Z.

COROLLARY 4. Let R be a prime ring of characteristic different from two, d a nonzero derivation of R and U is (σ, τ)-left Lie ideal of R. If d([R, U]_{σ,τ}) = 0, then σ(u) + τ(u) ∈ Z for all u ∈ U.

EXAMPLE. Let

R = \{( \begin{matrix}a & b \\ c & d \end{matrix} ) : a, b, c, d ∈ I\}

and

σ ( \begin{matrix}a & b \\ c & d \end{matrix} ) = ( \begin{matrix}d & −c \\ −b & a \end{matrix} ) , τ ( \begin{matrix}a & b \\ c & d \end{matrix} ) = ( \begin{matrix}a & −b \\ −c & d \end{matrix} )

be two autumorphisms of R and a = ( \begin{matrix}1 & 0 \\ 0 & 2 \end{matrix} ) ∉ Z. If we define d : R → R by

d ( \begin{matrix}a & b \\ c & d \end{matrix} ) = ( \begin{matrix}0 & −b \\ c & 0 \end{matrix} ) , then d is a derivation of R such that d([R, a]_{σ,τ}) = 0, but

σ(a) + τ(a) ∈ Z.

THEOREM 6. Let R be a prime ring of characteristic different from two, M a non-zero ideal of R and a ∈ R. If ([R, M]_{σ,τ} , a)_{σ,τ} = 0, then a ∈ Z.

PROOF. Let m be a noncentral element of M. Consider the following mappings on R : d_{1}(r) = [r, m]_{σ,τ} and d_{2}(r) = (r, a)_{σ,τ} . Then for any r ∈ R, we have d_{2}d_{1}(r) = d_{2}([r, m]_{σ,τ}) = ([r, m]_{σ,τ} , a)_{σ,τ} = 0. That is,

d_{2}d_{1}(r) = 0, r ∈ R.  (11)

For r ∈ R,

d_{1}(rσ(m)) = [rσ(m), m]_{σ,τ}
= r[σ(m), σ(m)] + [r, m]_{σ,τ} σ(m) = d_{1}(r)σ(m),

and so we get,

d_{1}(rσ(m)) = d_{1}(r)σ(m), r ∈ R.  (12)

Let us consider (11) together with (12), we obtain

0 = d_{2}d_{1}(rσ(m)) = d_{2}(d_{1}(r)σ(m)) = (d_{1}(r)σ(m), a)_{σ,τ}
= d_{1}(r)[σ(m), σ(a)] + (d_{1}(r), a)_{σ,τ} σ(m),

and so we have

d_{1}(R)[σ(m), σ(a)] = 0.  (13)

Putting in (13) rs, s ∈ R for r we obtain

0 = d_{1}(rs)[σ(m), σ(a)] = rd_{1}(s)[σ(m), σ(a)] + [r, τ(m)]s[σ(m), σ(a)],

and using (13) we get

[r, τ(m)]R[σ(m), σ(a)] = 0, r ∈ R.

Since R is a prime ring, we have m ∈ Z or σ([m, a]) = 0. If m ∈ Z, then σ([m, a]) = 0. That is, [M, a] = 0 since σ is automorphism of R. For all x ∈ R and m ∈ M, 0 = [xm, a] = x[m, a] + [x, a]m = [x, a]m. That is, [x, a]RM = (0), the zero ideal of R. By the primeness of R we have a ∈ Z.

THEOREM 7. Let R be a prime ring of characteristic different from two, d is a nonzero derivation of R and a ∈ R. Then (d(R), a) = 0 if, and only if, d((R, a)) = 0.

PROOF. Suppose (d(R), a) = 0. In this case we claim that d(a) = 0. If a = 0 then d(a) = 0. So we assume that a ≠ 0. For any x ∈ R, the relation (d(x), a) = 0 gives

0 = (d(xa), a) = (d(x)a+xd(a), a) = d(x)[a, a]+(d(x), a)a+x(d(a), a)−[x, a]d(a) (14)

and so,

[x, a]d(a) = 0, x ∈ R.  (15)

If we take xy, y ∈ R instead of x in (15), we obtain

[R, a]Rd(a) = (0).

Since R is a prime ring we have a ∈ Z or d(a) = 0. Now, if a ∈ Z then 0 = (d(a), a) = d(a)a + ad(a) = 2d(a)a. Then d(a)a = 0. Since we assumed that 0 ≠ a and R is a prime ring, we get d(a) = 0. Thus, we conclude that d(a) = 0. Hence for any r ∈ R we have d((r, a)) = (d(r), a) + (r, d(a)), and so, d((R, a)) = 0. Conversely, for x ∈ R,

0 = d((ax, a)) = d(a(x, a) + [a, a]x)
= d(a(x, a)) = d(a)(x, a) + ad((x, a)).

Hence we have

d(a)(x, a) = 0, x ∈ R.  (16)

In (16) replace x by xy and use (16), we get

0 = d(a)(xy, a) = d(a)x[y, a] + d(a)(x, a)y = d(a)x[y, a],

that is, d(a)R[R, a] = 0. Since R is a prime ring, we have d(a) = 0 or a ∈ Z. If d(a) = 0, then for any r ∈ R we have 0 = d((r, a)) = (d(r), a) + (r, d(a)) = (d(r), a), and so (d(R), a) = 0. If a ∈ Z, then 0 = d((a, a)) = 4d(a)a. Since the characteristic of R is different from 2, d(a)a = 0. Since a ∈ Z, we get d(a)Ra = 0, and so, d(a) = 0 or a = 0. Thus, d(a) = 0 is obtained. Finally, (d(R), a) = 0 as required.

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