DERIVATIONS WITH NILPOTENT VALUES ON Γ-RINGS

Abstract

Let M be a prime ${\Gamma}$-ring and let d be a derivation of M. If there exists a fixed integer n such that $(d(x){\alpha})^nd(x)=0$ for all $x{\in}M$ and ${\alpha}{\in}{\Gamma}$, then we prove that d(x) = 0 for all $x{\in}M$. This result can be extended to semiprime ${\Gamma}$-rings.

1. INTRODUCTION

The notion of a Г-ring was first introduced by Nobusuwa [10] as a generalization of a classical rings and then Barnes [2] generalized the same concepts in a broad sense. The concept of a derivation and a Jordan derivation of Г-rings have been first introduced by Sapanci and Nakajima [13] and they proved that every Jordan derivation in a certain prime Г-ring is a derivation. Afterwards many Mathemati- cians worked on derivations of Г-rings and developed some fruitful results. Paul and Uddin [11, 12] studied on Jordan and Lie structures in Г-rings and they proved the Levitzki’s Theorem in Г-rings. In [5], Halder and Paul proved that if d is a left derivation of a 2-torsion free semiprime Г-ring such that (d(x)α)nd(x) = 0 for all x ∈ M and α ∈ Г, then d = 0, where n is a fixed integer. Giambruno and Herstien [4] proved a classical result in rings which is stated as follows: If d is a derivation of a prime ring R, such that d(x)n = 0 for all x ∈ R, then d(x) = 0, where n is a fixed integer. He also extended this result to semiprime rings. Feng Wei [15] proved it in generalized derivations of semiprime rings. Then, Ali, Ali and Fillips [1] worked on a nilpotent and invertible values on semiprime rings with generalized derivations and they developed some remarkable results. By the same motivations as in Giambruno and Herstein [4], we develop the following result in this paper. If d is a derivation of a prime Г-ring such that (d(x)α)nd(x) = 0 for all x ∈ M and α ∈ Г, then d = 0, where n is a fixed integer. We also extend this result in semiprime Г-rings.

 

2. Г-RINGS AND DERIVATIONS

Let M and Г be additive abelian groups. If there exists a mapping (x, α, y) → xαy of M × Г × M→ M which satisfies the conditions:

(1) (x + y)αz = xαz + yαz, x(α + β)y = xαy + xβy, xα(y + z) = xαy + xαz, (2) (xαy)βz = xα(yβz),

for all x, y, z ∈ M and α, β ∈ Г, then M is called a Г-ring in the sense of Barnes [2]. A Г-ring M is prime if xГMГy = 0 implies that x = 0 or y = 0, and is semiprime if xГMГx = 0 implies x = 0. A subring A of a Г-ring M is said to be an ideal of M if AГM ⊆ A and MГA ⊆ A. Let M be a Г-ring. An additive mapping d : M → M is called a derivation if d(xαy) = d(x)αy + xαd(y) holds for all x, y ∈ M and α ∈ Г, and d is called a Jordan derivation if d(xαx) = d(x)αx + xαd(x) holds for all x ∈ M and α ∈ Г. An ideal P of a Г-ring M is said to be prime if for any ideals A and B of M, AГB ⊆ P implies A ⊆ P or B ⊆ P. A Г-ring M is said to be prime if the zero ideal is prime.

Theorem 2.1 ([10]). If M is a Г-ring, the following conditions are equivalent:

(1) M is a prime Г-ring. (2) If a, b ∈ M and aГMГb = <0>, then a = 0 or b = 0. (3) If and are principal ideals of M such that Г = <0>, then a = 0 or b = 0. (4) If A and B are right ideals of M such that AГB = <0>, then A = <0> or B = <0>. (5) If A and B are left ideals of M such that AГB = <0>, then A = <0> or B = <0>.

 

3. DERIVATIONS WITH NILPOTENT VALUES ON Г-RINGS

We begin with the following lemmas which are essential for proving our main results.

Lemma 3.1 ([14, Lemma 3]). If d ≠ 0 is a derivation of M, then d does not vanish on a non-zero one-sided ideal of M.

Proof. Let L ≠ 0 be the left ideal of M. Suppose that d(L) = 0. For all x ∈ L, m ∈ M and α ∈ Г, we have mαx ∈ L. Therefore, 0 = d(mαx) = d(m)αx + mαd(x) = d(m)αx. Since d ≠ 0, we have x = 0, a contradiction to the fact that L ≠ 0.           

Lemma 3.2. If L ≠ 0 is a left ideal of M and T = {x ∈ L | LГx = xГL = 0}. Then, L/T is a prime Г-ring.

Proof. It is suffcient to prove that T is a prime ideal of L. Let U and V be ideals of L such that UГV ⊆ T. Then, LГUГV ГL ⊆ LГTГL = 0. But LГU and V ГL are ideals of M. Since M is prime, either LГU = 0 or V ГL = 0. If LГU = 0, then U ⊆ T. If V ГL = 0, then V ⊆ T. Therefore, we have either U ⊆ T and V ⊆ T.           

In [11, Theorem 3.1], Paul and Uddin proved the Levitzki Theorem in Г-rings. In this paper we will frequent used of its special case.

Lemma 3.3 ([11, Theorem 3.1]). If L is a left ideal of M and (xα)nx = 0 for all x ∈ L and α ∈ Г, where n is a fixed integer, then L = 0.

We shall also use an easy variant of Lemma 3.3.

Lemma 3.4. If x, y ∈ M and ((xαmβy))n(xαmβy) = 0 for all m ∈ M and α, β ∈ Г, where n is a fixed integer, then yαx = 0 for all α ∈ Г.

Definition 3.5. Let M be a Г-ring and let R be a subset of M. Define L(R) = {x ∈ M | xαr = 0, for all r ∈ R and α ∈ Г}, and T(R) = {x ∈ M | rαx = 0, for all r ∈ R and α ∈ Г}.

It is clear that L(R) is a left ideal and T(R) is a right ideal of M.

Let M be a prime Г-ring and d be a derivation of M such that (d(x)α)nd(x) = 0 for all x ∈ M and α ∈ Г. We have to show that d = 0.

We begin with assuming that d = 0. Our first result is:

Lemma 3.6. For x ∈ M, d(L(x)) ⊆ L(x) and d(T(x)) ⊆ T(x).

Proof. If y ∈ L(x), then yαx = 0 for all α ∈ Г. Therefore,

0 = yαd(xαy)αd(xαy)    = yα(d(x)αy + xαd(y)αd(xαy)    = yαd(x)αyαd(xαy) + yαxαd(y))αd(xαy)    = yαd(x)αyα(d(x)αy + xαd(y))    = yαd(x)αyαd(x)αy.

Now, we have

0 = yα(d(xαy)α)2d(xαy)    = yαd(xαy)αd(xαy)αd(xαy)    = yαd(x)αyαd(x)αyα(d(x)αy + xαd(y))    = yα((d(x)αyα)2d(x)αy

Therefore, we have

0 = yα(d(xαy)α)nd(xαy) = yα((d(x)αy)α)nd(x)αy.

Thus, d(x)αyα(d(xαy)α)nd(x)αy = 0. This implies that ((d(x)αy)α)n+1d(x)αy = 0 for all y ∈ L(x) and α ∈ Г. But then L(x)Гd(x) is a left ideal of M in which every element is nilpotent. Therefore, by Lemma 3.3, L(x)Гd(x) = 0.

For y ∈ L(x), we have 0 = d(yαx) = d(y)αx + yαd(x) = d(y)αx. Now, we have d(L(x)) ⊆ L(x). On the other hand, the analogous argument yields d(T(x)) ⊆ T(x).           

Lemma 3.7. If x ∈ M, then either d(xГM)Гx = 0 or L(x)Гd(L(x)). Similarly, either xГd(MГx) = 0 or d(T(x))ГT(x) = 0.

Proof. Let a, b ∈ L(x). Then, aαx = 0 and bαx = 0 for all α ∈ Г. Now, we obtain that d(b)αxαa = 0, and so, d(b)αd(xαa) = 0. Since xαa ∈ L(x), we have d(xαa)αd(xαa) = 0. Now, we have

0 = d(xαa + b)αd(xαa + b)    = (d(xαa) + d(b))α(d(xαa) + d(b))    = d(xαa)αd(xαa) + d(b)α)d(xαa) + d(xαa)αd(b) + d(b)αd(b)    = d(xαa)αd(b).

Hence,

0 = d(xαa + b)αd(xαa + b)αd(xαa + b)    = (d(xαa)αd(b))α(d(xαa) + d(b))    = d(xαa)αd(b)αd(xαa) + d(xαa)αd(b)αd(b)    = d(xαa)αd(b)αd(b).

Using the same argument, we obtain,

(1)      0 = (d(xαa + b)α)nd(xαa + b) = d(xαa)α(d(b)α)n−1d(b).

Let m ∈ M, a, b ∈L(x), aαx = 0 and bαx = 0 for all α ∈ Г. Therefore, aαxαmαa = bαxαmαa = 0. Hence, the result of (2) gives us

0 = d(aαmαxαa)α(d(b)α)n−1d(b)    = (d(aαm)α(xαa) + aαmαd(xαa)αd(b)α)n−1d(b)    = d(aαm)α(xαa)α d(b)α)n−1d(b) + aαmαd(xαa)α(d(b)α)n−1d(b)    = d(aαm)α(xαa)αd(b)α)n−1d(b); using (2).

In other words, we write the above relation as d(aαm)αxαL(x)αd(b)α)n−1d(b) = 0 for all m ∈ M, b ∈ L(x) and α ∈ Г. If L(x)αd(b)α)n−1d(b) ≠ 0, then by the primeness of M, we obtain d(aαm)αx = 0 for all m ∈ M and α ∈ Г. Hence, we have d(a ГM) Гx = 0. On the other hand, suppose that L(x)α(d(b)α)n−1d(b) = 0 for all b ∈ L(x) and α ∈ Г. Since d(L(x)) ⊆ L(x) and d(T) ⊆ T where T = {c ∈ L(x) | L(x)αc = 0}, d induces a derivation which we write as d on B = L(x)/T .

By Lemma 3.2, B is a prime Г-ring. The fact that L(x)α(d(b)α)n−1d(b) = 0 for all b ∈ L(x) translates into (d(c)α)n−1d(c) = 0 for all c∈ B. Thus, d(c) = 0 for all c ∈ B (by induction). This yields us that d(L(x)) ⊆ T and so L(x)αd(L(x)) = 0 for all α ∈ Г. The same argument yields the right-handed version of what we have just proved. Thus, the proof is completed.           

Lemma 3.6 gives us two sets of elements which have rather particular properties, and which yield the following definition:

We set A = {x ∈ M | xГd(MГx) = 0} and B = {x ∈ M |d(xГM)Гx = 0}. These two subsets A and B play a key role in which what is to follow. Their basic algebraic characterizations are expressed in the following.

Lemma 3.8. A is a non-zero left ideal of M, B is a nonzero right ideal of M and AГB = 0. Furthermore, d(A) ⊆ A, d(B) ⊆ B and AГd(A) = d(B)ГB = 0.

Proof. The stated properties of A and B are the same, so we have to show that A ≠ 0 is a left ideal of M, d(A) ⊆ A and d(A)ГA = 0. If x, y ∈ M are such that (x)Гd(L(x)) = 0 and L(y)Гd(L(y)) = 0. Then, we shall prove that L(x)Гd(L(y)) = 0. In order to see this, let a, b ∈ L(x) and t, z ∈ L(y). By our assumption on L(x) and L(y), we obtain d(aαb) = d(a)αb and d(tαz) = d(t)αz. Therefore,

0 = bα(d(aαb + tαz)α)2nd(aαb + tαz)    = bα((d(a)αb + d(t)αz)α)2n(d(a)αb + d(t)αz)    = bαd(t)αzα((d(a)αb + d(t)αz)α)2n−1(d(a)αb + d(t)αz)    …    = (bα((d(t)αzαd(a)αb)α)nd(t)αzαd(a)αb.

Therefore, (bα(d(t)αzα((d(a)α)n+1bαd(t)αzαd(a) = 0 for all a,b ∈ L(x), t, z ∈ L(y) and α ∈ Г. Making several uses of Lemma 3.3, we obtain from the above relation that L(x)αd(L(y))αL(y)αd(L(x)) = 0 for all α ∈ Г. Since M is prime, we have L(x)αd(L(y)) = 0 or L(y)αd(L(x)) = 0 for all α ∈ Г. Suppose that L(y)αd(L(x)) = 0 for all α ∈ Г. Then, for all b ∈ L(y), z, t ∈ L(x). Since L(x)αd(L(x)) = 0, 0 = zαd(bαt) = zα(d(b)αt + bαd(t)) = zαd(b)αt and bαd(t) ∈ L(y)αd(L(x)) = 0. Thus, zαd(b)αL(x) = 0 and so zαd(b) = 0. This says that L(x)αd(L(y)) = 0. Thus, our assertion has been verified. We shall now show that A ≠ 0. Suppose that A = 0. By Lemma 3.6, we get that L(x)αd(L(x)) = 0 for all x ∈ M. Take y ∈ M such that L(y) ≠ 0. By Lemma 3.1, d(L(y)) ≠ 0. Since (d(x)α)nd(x) = 0 for all x ∈ M, α ∈ Г, d(x) ∈ L(d(x)α)n−1d(x), hence d(x)αd(L(y)) = 0. Since d(L(y)) ≠ 0, d(x) = 0 which is a contradiction to the fact that d ≠ 0. Thus, indeed, A ≠ 0.

Our next goal is to show that A is a left ideal of M. From the definition of A = {x ∈ M | xГd(MГx) = 0}. It is clear that x ∈ A, t ∈ M, α ∈ Г forces tαx ∈A. So, all we need to show that if x, y ∈ A, then x + y ∈ A. If a, b, z, t ∈ M, then d(aαxαbαx) = d(aαx)αbαx + aαxαd(bαx) = d(aαx)αbαx, since x ∈ A. Similarly, d(zαyαtαy) = d(zαy)αbαy. Now, we have

0 = ((d(aαx)αbαx + zαyαtαy)α)2n(d(aαxαbαx + zαyαtαy)αd(aαx)αbαx    = ((d(aαx)αbαx + d(zαy)αtαy)α)2n(d(aαx)αbαx + d(zαy)αtαy)αd(aαx)αbαx    = d(aαx)αbαxα((d(zαy)αbαyαd(aαx)μ)n(d(zαy)αtαyαd(aαx))αbαx.

Since aαxαd(bαx). Thus, we get that

((d(aαx)αbαxαd(zαy)αtαy)α)n+1(d(aαx)αbαxαd(zαy)αtαy = 0,

for all a,b, z, t ∈ M and α ∈ Г. In view of Lemma 3.3, we obtain that xαd(zαy) = 0 for all x, y, z ∈ M and α ∈ Г. This yields that xαd(mαy) = 0 or yαd(mαx) = 0 for all m ∈ M and α ∈ Г. If xαd(mαy) = 0, then since xαd(mαx) = 0, we have

0 = xαd(mαyαmαx)    = xαd(mαy)αmαx + xαmαyαd(mαx)    = xαmαyαd(mαx)

and so yαd(mαx) = 0 by the primeness of M. Thus, we obtain that

(x + y)αd(mαx + d(mαy)) = xαd(mαx) = xαd(mαx + yαd(mαx) + yαd(mαy) = 0,

this implies that x + y ∈ A. Therefore, so far, we have seen that A ≠ 0 is a left ideal of M. Of course, we also now have by the same argument that B ≠ 0 is a right ideal of M. In view of the definition of A and Lemma 3.5, twice yields that d(A) ⊆ A. Now, we want to prove that AГd(A) = 0. Let x, y ∈ A. We have seen that yαd(mαx) = 0, hence yαd(mαxαy) = 0. This gives that 0 = yαd(mαx)αy + yαmαxαd(y) = yαMαxαd(y). The primeness of M gives that xαd(y) = 0 and so, AГd(A) = 0. Similarly, we can prove that d(B)ГB = 0. Now, we also have to show that AГB = 0. Let x ∈ A, y ∈ B, z ∈ M and α ∈ Г,

0 = (d(xαzαx + y)α)2n(d(xαzαx + y)    = (d(xαzαx) + d(y))α)2n(d(xαzαx) + d(y))    = ((d(x)αzαx + xαd(zαx) + d(y))α)2n(d(x)αzαx + xαd(zαx) + d(y))    = ((d(x)αzαx + d(y))α)2n(d(x)αzαx + d(y)); since xαd(zαx) = 0.

Therefore,

((d(x)αzαx + d(y))α)2n(d(x)αzαx + d(y)) = 0.

This gives that

0 = (d(x)αzαx + xαd(zαx) + d(y))2n(d(x)αzαx + xαd(zαx) + d(y))

and since d(y)αd(y) = 0 and xαd(x) = 0. We obtain (d(x)αzαx+d(y))2n(d(x)αzαx+ d(y)) = 0. Therefore, (d(x)αzαx + d(y))2n(d(x)αzαx + d(y)) = 0. By Lemma 3.4, we obtain that xαd(y)αd(x) = 0 for all x ∈ A, y ∈ B and α ∈ Г. So, d(xαy)αd(x) = d(x)αyαd(x) + xαd(x)αd(x) = d(x)αyαd(x). But,

0 = (d(xαy)α)nd(xαy)αd(x)    = (d(xαy)α)n(d(x)αy + xαd(y))αd(x)    = (d(xαy)α)nd(x)αyαd(x)    …    = (d(x)αy)α)n+1d(x).

Therefore, (d(x)αyα)n+1d(x)αy = 0. This shows that d(x)αy is a nilpotent element of a nil right ideal d(A)ГB of bounded index of nilpotent n + 1. By Lemma 3.3, d(x)αy = 0. This gives that d(A)ГB = 0 since A is a left ideal of M, 0 = d(MГA)ГB = d(M)ГAГB + MГd(A)ГB = d(M)ГAГd(B). We conclude that AГB = 0.           

Lemma 3.9. If x ∈ M and xαx = 0 for all α ∈ Г, then x ∈ A ∪ B.

Proof. Suppose that x ∉ B, by Lemma 3.6, L(x)Гd(L(x)) = 0. Since xαx = 0, x ∈ L(x) and MГd(L(x)) ⊆ MГL(x) ⊆ L(x). Now, we have d(MГx) ⊆ d(L(x)). Therefore, xГd(MГx) ⊆ L(x)Гd(L(x)) = 0. Hence, xГd(MГx) = 0 and by the definition of A, we obtain x ∈ A.           

Now, we are in a position to prove our main result.

Theorem 3.10. If M is a prime Г-ring and d : M → M a derivation such that (d(x)α)nd(x) = 0 for all x ∈ M and α ∈ Г, where n ≥ 1 is a fixed integer, then d = 0.

Proof. Let C = A ∪ B ⊇BГA ≠ 0, where A ≠ 0 and B ≠ 0 are respectively, left and right ideals of the prime Г-ring M. Let p ∈ C, since d(x) is nilpotent, we have 0 = pαd(x)αp = pαd(x)αd(x)αp. Because pαp ∈ CГC ⊆ AГB = 0, we get that (pαd(x) − d(x)αp)α(pαd(x) − d(x)αp) = 0. But then, by Lemma 3.8, pαd(x) − d(x)αp ∈ A ∪ B for all x ∈ M. Suppose that pαd(x) − d(x)αp ∈ A, say, since p ∈ C ⊆ A, d(x)αp ∈ A. Hence, pαd(x) ∈A. If pαd(x) − d(x)αp ∈ B, then the similar argument end up with d(x)αp ∈ B, since pαd(x) ∈ B. So, for every x ∈ M, either pαd(x) ∈ A or d(x)αp ∈ B. This implies that pГd(M) ⊆ A or d(M)Гp ⊆ B. If pГd(M) ⊆ A, then since pГC ⊆ B, pГd(M) ⊆ B, hence pГd(M) ⊆ C. Similarly, if d(M)Гp ⊆ B, then we get d(M)Гp ⊆ C. So, for every p ∈ C, pГd(M) ⊆ C or d(M)Гp ⊆ C. This implies that CГd(M) ⊆ C or d(M)ГC ⊆ C. Suppose that CГd(M) ⊆ C. Hence, CГd(M)Гd(A) ⊆ CГd(A) ⊆ AГd(A) = 0. Now, BГA ⊆ C, thus BГA ⊆ d(M)Гd(A) ⊆ CГd(M)Гd(A) = 0. Since B is a right ideal of M and B ≠ 0, the primeness of M forces that AГd(M)Гd(A) = 0.

Consider the left ideal AГd(M) of M. Let t = ∑aiαid(mi), ai ∈ A, αi ∈ Г, mi ∈ M, be any element of AГd(M). Thus, if v = ∑aiαimi, then

d(v) = ∑d(ai)αimi + ∑aiαid(mi) = t + w,

where w = ∑d(ai)αimi. Furthermore,

tГw = ∑aiαid(mi)Гd(ai)αimi ∈ AГd(M)Гd(A)ГM = 0,

so tГw = 0. Now, we have

0 = (d(v)α)nd(v) = ((t + w)α)n(t + w) = (tα)nt + (wα)nw + (wα)n−1wαt + … + wα(tα)n−1t,

since tαw = 0 for every α ∈ Г. Therefore, 0 = tα(tα)nt + (wα)nw + … + wα(tα)n−1t) = (tα)n+1t. In other words, every element in AГd(M) is nilpotent of degree at most n + 1. Therefore, by Lemma 3.3, AГd(M) = 0. Since A ≠ 0, we have d(M) = 0. Similarly, if d(M)ГC ⊆ C, then we have d(M)ГB = 0. Since B ≠ 0, d(M) = 0. This proves the theorem.           

Now we prove the more general result.

Theorem 3.11. Let I ≠ 0 be an ideal of a prime Г-ring M and d be a derivation of M. If (d(x)α)nd(x) = 0 for all x ∈ I, where n ≥ 1 is a fixed integer, then d = 0.

Proof. If d(I) ⊆ I, the result is obvious. But even if d(I) ⊈ I, our proof is easily adjusted to prove the result.           

Theorem 3.11 can be extended to semiprime Г-rings which is given in the following theorem.

Theorem 3.12. Let M be a semiprime Г-ring and d be a derivation of M such that (d(x)α)nd(x) = 0 for all x ∈ M. Then, d = 0.

Proof. Since M is semiprime, ∩J = 0 where the intersection runs over all prime ideals J of M. Now, we claim that d(J) ⊆ J for every prime ideal J. Let a ∈ J, x ∈ M and α ∈ Г. Then, 0 = (d(aαx))nd(aαx) = ((d(a)αx + aαd(x)))n(d(a)αx + aαd(x)) = (d(aαx)α)nd(a)αx mod J. Thus, in the prime for all , α ∈ Г. Hence, the right ideal is a nil of bounded index n. Therefore, by Theorem 3.1 of [11] has a nilpotent ideal-which it can not, since it is prime unless . But d(a) ∈ J. So, d(J) ⊆ J. Hence, d(J) ⊆ J for all prime ideals J of M and so d induces a derivation on the prime such that for all , α ∈ Г. By Theorem 3.11, Hence, that is d(M) ⊆ J for all prime ideals J of M. Since ∩ J = 0, we get d(M) = 0. Hence, d = 0.