DERIVATIONS WITH NILPOTENT VALUES ON Γ-RINGS
DERIVATIONS WITH NILPOTENT VALUES ON Γ-RINGS
The Pure and Applied Mathematics. 2014. Nov, 21(4): 237-246
• Received : May 27, 2014
• Accepted : November 01, 2014
• Published : November 30, 2014 PDF e-PUB PubReader PPT Export by style
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KALYAN KUMAR, DEY
AKHIL CHANDRA, PAUL
BIJAN, DAVVAZ
DEPARTMENT OF MATHEMATICS, YAZD UNIVERSITY, YAZD, LRANEmail address:davvaz@yazd.ac.ir

Abstract
Let M be a prime Г-ring and let d be a derivation of M . If there exists a fixed integer n such that ( d ( x ) α ) n d ( x ) = 0 for all x M and α ∈ Г, then we prove that d ( x ) = 0 for all x M . This result can be extended to semiprime Г-rings.
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1. INTRODUCTION
The notion of a Г-ring was first introduced by Nobusuwa  as a generalization of a classical rings and then Barnes  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  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  , Halder and Paul proved that if d is a left derivation of a 2-torsion free semiprime Г-ring such that ( d ( x ) α ) n d ( x ) = 0 for all x M and α ∈ Г, then d = 0, where n is a fixed integer. Giambruno and Herstien  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  proved it in generalized derivations of semiprime rings. Then, Ali, Ali and Fillips  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  , we develop the following result in this paper. If d is a derivation of a prime Г-ring such that ( d ( x ) α ) n d ( 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  . A Г-ring M is prime if x Г M Г y = 0 implies that x = 0 or y = 0, and is semiprime if x Г M Г x = 0 implies x = 0. A subring A of a Г-ring M is said to be an ideal of M if A Г M A and M Г A 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 , A Г B 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 (  ). If M is a Г- ring, the following conditions are equivalent:
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. PPT Slide
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Lemma 3.2. If L ≠ 0 is a left ideal of M and T = { x L | L Г x = x Г L = 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 U Г V T . Then, L Г U Г V Г L L Г T Г L = 0. But L Г U and V Г L are ideals of M . Since M is prime, either L Г U = 0 or V Г L = 0. If L Г U = 0, then U T . If V Г L = 0, then V T . Therefore, we have either U T and V T . PPT Slide
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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 ( ) n x = 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 ) n d ( 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+1 d ( x ) αy = 0 for all y L ( x ) and α ∈ Г. But then L ( x d ( x ) is a left ideal of M in which every element is nilpotent. Therefore, by Lemma 3.3, L ( x d ( x ) = 0.
For y L ( x ), we have 0 = d ( y α x ) = d ( y x + y α d ( x ) = d ( y x . Now, we have d ( L ( x )) ⊆ L ( x ). On the other han d , the analogous argument y iel d s d ( T ( x )) ⊆ T ( x ). PPT Slide
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Lemma 3.7. If x M , then either d( x Г M x = 0 or L ( x d ( L ( x )). Similarly , either x Г d ( M Г x ) = 0 or d ( T ( x ))Г T ( 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 ) n d ( xαa + b ) = d ( xαa ) α ( d ( b ) α ) n−1 d ( 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−1 d ( b ) = 0 for all m M , b L ( x ) and α ∈ Г. If L ( x ) αd ( b ) α ) n−1 d ( 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 Г M ) Г x = 0. On the other hand, suppose that L ( x ) α ( d ( b ) α ) n−1 d ( 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−1 d ( b ) = 0 for all b L ( x ) translates into ( d ( c ) α ) n−1 d ( 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. PPT Slide
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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 | x Г d ( M Г x ) = 0} and B = { x M | d ( x Г M x = 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 A Г B = 0. Furthermore , d ( A ) ⊆ A , d ( B ) ⊆ B and A Г d ( A ) = d ( B B = 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 A = 0. If x , y M are such that ( x d ( L ( x )) = 0 and L ( y d ( L ( y )) = 0. Then, we shall prove that L ( x d ( 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, ( ( d ( t ) αzα (( d ( a ) α ) n+1 bαd ( 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 ) = ( d ( b ) αt + bαd ( t )) = d ( 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 ) n d ( x ) = 0 for all x M , α ∈ Г, d ( x ) ∈ L ( d ( x ) α ) n−1 d ( 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 | x Г d ( M Г x ) = 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 A Г d ( 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, A Г d ( A ) = 0. Similarly, we can prove that d ( B B = 0. Now, we also have to show that A Г B = 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)); sincexαd(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+1 d ( x ) αy = 0. This shows that d ( x ) αy is a nilpotent element of a nil right ideal d ( A B of bounded index of nilpotent n + 1. By Lemma 3.3, d ( x ) αy = 0. This gives that d ( A B = 0 since A is a left ideal of M, 0 = d ( M Г A B = d ( M A Г B + M Г d ( A B = d ( M A Г d ( B ). We conclude that A Г B = 0. PPT Slide
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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 d ( L ( x )) = 0. Since xαx = 0, x L ( x ) and M Г d ( L ( x )) ⊆ M Г L ( x ) ⊆ L ( x ). Now, we have d ( M Г x ) ⊆ d ( L ( x )). Therefore, x Г d ( M Г x ) ⊆ L ( x d ( L ( x )) = 0. Hence, x Г d ( M Г x ) = 0 and by the definition of A , we obtain x A . PPT Slide
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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 ) n d ( x ) = 0 for all x M and α ∈ Г, where n ≥ 1 is a fixed integer , then d = 0.
Proof . Let C = A B B Г A ≠ 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 C Г C A Г B = 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 p Г d ( M ) ⊆ A or d ( M p B . If p Г d ( M ) ⊆ A , then since p Г C B , p Г d ( M ) ⊆ B , hence p Г d ( M ) ⊆ C . Similarly, if d ( M p B , then we get d ( M p C . So, for every p C , p Г d ( M ) ⊆ C or d ( M p C . This implies that C Г d ( M ) ⊆ C or d ( M C C . Suppose that C Г d ( M ) ⊆ C . Hence, C Г d ( M d ( A ) ⊆ C Г d ( A ) ⊆ A Г d ( A ) = 0. Now, B Г A C , thus B Г A d ( M d ( A ) ⊆ C Г d ( M d ( A ) = 0. Since B is a right ideal of M and B ≠ 0, the primeness of M forces that A Г d ( M d ( A ) = 0.
Consider the left ideal A Г d ( M ) of M . Let t = ∑ aiαid ( mi ), ai A , αi ∈ Г, mi M , be any element of A Г d ( M ). Thus, if v = ∑ aiαimi , then
• d(v) = ∑d(ai)αimi+ ∑aiαid(mi) =t+w,
where w = ∑ d ( a i ) αimi . Furthermore,
• tГw= ∑aiαid(mi)Гd(ai)αimi∈AГd(M)Гd(A)ГM= 0,
so t Г w = 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 = ( ) n t + ( ) n w + … + ( ) n−1 t ) = ( ) n+1 t . In other words, every element in A Г d ( M ) is nilpotent of degree at most n + 1. Therefore, by Lemma 3.3, A Г d ( M ) = 0. Since A ≠ 0, we have d ( M ) = 0. Similarly, if d ( M C C , then we have d ( M B = 0. Since B ≠ 0, d ( M ) = 0. This proves the theorem. PPT Slide
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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 ) n d ( 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. PPT Slide
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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 ) n d ( 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 )) n d ( aαx ) = (( d ( a ) αx + aαd ( x ))) n ( d ( a ) αx + aαd ( x )) = ( d ( aαx ) α ) n d ( a ) αx mod J . Thus, in the prime PPT Slide
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for all PPT Slide
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, α ∈ Г. Hence, the right ideal PPT Slide
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is a nil of bounded index n . Therefore, by Theorem 3.1 of  has a nilpotent ideal-which it can not, since it is prime unless PPT Slide
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. 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 PPT Slide
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such that PPT Slide
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for all PPT Slide
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, α ∈ Г. By Theorem 3.11, PPT Slide
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Hence, PPT Slide
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that is d ( M ) ⊆ J for all prime ideals J of M . Since ∩ J = 0, we get d ( M ) = 0. Hence, d = 0. PPT Slide
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Acknowledgements
The authors are highly grateful to referees for their valuable comments and suggestions for improving the paper.
References