JACODESMATH / ISSN 2148-838X

J. Algebra Comb. Discrete Appl.
2(1) • 25-37

Received: 9 July 2014; Accepted: 25 November 2014
DOI 10.13069/jacodesmath.87481

Journal of Algebra Combinatorics Discrete Structures and Applications

γ-Lie structures in γ-prime gamma rings with
derivations

Research Article

Okan Arslan1∗, Hatice Kandamar1∗∗

1. Adnan Menderes University, Faculty of Arts and Sciences, Department of Mathematics, Aydın, Turkey

Abstract: Let M be a γ-prime weak Nobusawa Γ-ring and d 6= 0 be a k-derivation of M such that k (γ) = 0 and
U be a γ-Lie ideal of M. In this paper, we introduce definitions of γ-subring, γ-ideal, γ-prime Γ-ring
and γ-Lie ideal of M and prove that if U * Cγ, charM 6= 2 and d3 6= 0, then the γ-subring generated
by d(U) contains a nonzero ideal of M. We also prove that if [u,d(u)]γ ∈ Cγ for all u ∈ U, then U is
contained in the γ-center of M when charM 6= 2 or 3. And if [u,d(u)]γ ∈ Cγ for all u ∈ U and U is
also a γ-subring, then U is γ-commutative when charM = 2.

2010 MSC: 16N60, 16W25, 16Y99

Keywords: Gamma ring, γ-Prime gamma ring, γ-Lie ideal, k-Derivation, γ-Commutativity

1. Preliminaries

Let M and Γ be additive Abelian groups. M is said to be a Γ-ring in the sense of Barnes[2] if there
exists a mapping M × Γ ×M → M satisfying these two conditions for all a,b,c ∈ M, α,β ∈ Γ:

(1) (a + b) αc = aαc + bαc
a(α + β)c = aαc + aβc
aα (b + c) = aαb + aαc

(2) (aαb) βc = aα (bβc)

In addition, if there exists a mapping Γ × M × Γ → Γ such that the following axioms hold for all
a,b,c ∈ M, α,β ∈ Γ:

(3) (aαb) βc = a (αbβ) c

∗ E-mail: oarslan@outlook.com.tr
∗∗ E-mail: hkandamar@adu.edu.tr

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γ-Lie structures with derivations

(4) aαb = 0 for all a,b ∈ M implies α = 0, where α ∈ Γ

then M is called a Γ-ring in the sense of Nobusawa[10]. If a Γ-ring M in the sense of Barnes satisfies
only the condition (3), then it is called weak Nobusawa Γ-ring[9].

Let M be a Γ-ring in the sense of Barnes. Then M is said to be a prime gamma ring if aΓMΓb = 0
with a,b ∈ M implies either a = 0 or b = 0[2]. It is also defined in [2] that M is a completely prime
gamma ring if aΓb = 0 with a,b ∈ M implies either a = 0 or b = 0.

For a subset U of M and γ ∈ Γ, the set Cγ (U) = {a ∈ M | aγu = uγa, ∀u ∈ U} and the set
Cγ = {a ∈ M | aγm = mγa, ∀m ∈ M} are called γ-center of the subset U and γ-center of M respectively.

In 2000, Kandamar[7] firstly introduced the notion of a k-derivation for a gamma ring in the sense
of Barnes and proved some of its properties and commutativity conditions for Nobusawa gamma rings.

Commutativity conditions with derivations for a gamma ring has been investigated by a number of
authors. In [8], Khan, Chaudhry and Javaid proved that if M is a prime gamma ring (in the sense of
Barnes) of characteristic not 2, I is a nonzero ideal of M and f is a generalized derivation on M, then M
is a commutative gamma ring. In [12], Suliman and Majeed showed a nonzero Lie ideal of a 2-torsion-free
prime Γ-ring M with a nonzero derivation d is central if d(U) is contained in the center of M.

In this paper, we define γ-Lie ideal for a weak Nobusawa gamma ring and show that if U * Cγ,
charM 6= 2 and d3 6= 0, then the γ-subring generated by d(U) contains a nonzero ideal of M. We also
prove that if [u,d(u)]γ ∈ Cγ for all u ∈ U, then U ⊆ Cγ when charM 6= 2 or 3. And if [u,d(u)]γ ∈ Cγ for
all u ∈ U and U is also a γ-subring, then U is γ-commutative when charM = 2.

2. γ-Lie ideals and derivations

Now we give some new definitions and make some preliminary remarks we need later.

Let M be a weak Nobusawa Γ-ring and 0 6= γ ∈ Γ. A subgroup I of M is said to be a γ-subring
if xγy ∈ I for all x,y ∈ I. A subgroup A of M is said to be a γ-left ideal(resp. γ-right ideal) if
mγa ∈ A(resp. aγm ∈ A) for all m ∈ M, a ∈ A. If A is both γ-left and γ-right ideal then A is called a
γ-ideal of M.

M is called γ-commutative gamma ring if xγy = yγx for all x,y ∈ M.
We say that the additive subgroup U of M is said to be a γ-Lie ideal of M if [U,M]γ ⊆ U. We also

say that if there exists a γ ∈ Γ such that aγMγb = 0 with a,b ∈ M implies either a = 0 or b = 0 then M
is called a γ-prime gamma ring.

An element a of M is called γ-nilpotent if there exists a positive integer n such that anγ := (aγ)
na = 0.

In what follows, let M be a γ-prime weak Nobusawa Γ-ring of characteristic not 2, d 6= 0 be a
k-derivation of M such that k (γ) = 0 and U be a γ-Lie ideal of M unless otherwise specified.

Lemma 2.1. If a ∈ M γ-commutes with [a,x]γ for all x ∈ M, then a is in the γ-center of M.

Proof. Let x,y ∈ M. Therefore, we get [a,x]γ γ [a,y]γ = 0 by hypothesis. Replacing y by mγx with
m ∈ M, we obtain [a,x]γ γMγ [a,x]γ = 0. Hence, a is in the γ-center of M since M is γ-prime.

Lemma 2.2. Suppose that U 6= (0) is both a γ-subring and a γ-Lie ideal of M. Then either U ⊆ Cγ or
U contains a nonzero ideal of M.

Proof. First, suppose that the γ-subring U is not γ-commutative. Then, there exists x,y ∈ U such
that [x,y]γ 6= 0. Since U is a γ-Lie ideal, [x,y]γ γM ⊆ U. Hence,

[
[x,y]γ γa,b

]
γ
∈ U for all a,b ∈ M.

Expanding this, we get bγ [x,y]γ γa ∈ U leading to Mγ [x,y]γ γM ⊆ U. Moreover, Mγ [x,y]γ γM 6= 0.
We have shown that the result is correct if the γ-subring U is not commutative.

26



O. Arslan, H. Kandamar

Now suppose that U is γ-commutative. Then
[
a, [a,x]γ

]
γ

= 0 for a ∈ U and x ∈ M. Therefore, we
have U ⊆ Cγ by Lemma 2.1.

Lemma 2.3. Let U be a γ-Lie ideal of M and U * Cγ. Then there exists a nonzero ideal K of M such
that [K,M]γ ⊂ U but [K,M]γ * Cγ.

Proof. Since U * Cγ, it follows from Lemma 2.1 that [U,U]γ 6= 0. Let K = Mγ [U,U]γ γM. Then it
is clear that K is a nonzero ideal of M.

Let T (U) =
{
x ∈ M : [x,M]γ ⊆ U

}
. Then, it can be shown that U ⊆ T (U) and T (U) is both a

γ-subring and a γ-Lie ideal of M. Let u,v ∈ U such that [u,v]γ 6= 0. Replacing v by vγm with m ∈ M,
we obtain [u,v]γ γM ⊆ T (U). Hence,

[
[u,v]γ γm,n

]
γ
∈ T (U) for all m,n ∈ M. Expanding this, we get

K ⊆ T (U). Therefore, we have shown that [K,M]γ ⊆ U.

Suppose that [K,M]γ ⊆ Cγ. Then,
[
x, [x,m]γ

]
γ

= 0 for all x ∈ K, m ∈ M. Let y ∈ M. Since
K ⊆ Cγ by Lemma 2.1, we have [y,nγkγm]γ = 0 for all m,n ∈ M,k ∈ K which leads to y ∈ Cγ. But
this contradicts with U * Cγ.

Lemma 2.4. Let u ∈ M. If a ∈ Cγ and aγu ∈ Cγ, then a = 0 or u ∈ Cγ.

Proof. Suppose that a 6= 0. Since [aγu,m]γ = 0 for all m ∈ M, we get aγ [u,m]γ = 0. Replacing m
by mγn with n ∈ M, we obtain [u,n]γ = 0 for all n ∈ M. This gives that u ∈ Cγ.

Lemma 2.5. If U is a γ-Lie ideal of M and U * Cγ, then Cγ(U) = Cγ.

Proof. It is clear that Cγ (U) is both a γ-subring and γ-Lie ideal of M. We claim that Cγ (U) cannot
contain a nonzero ideal of M. Suppose K is a nonzero ideal of M which is contained in Cγ (U). Then,
it is clear that [u,kγm]γ = 0 for all u ∈ U, k ∈ K and m ∈ M. Expanding this, we get kγ [u,m]γ = 0.
Replacing k by kγm with m ∈ M, we obtain u ∈ Cγ which leads to a contradiction. Hence, Cγ (U) ⊆ Cγ
by Lemma 2.2.

Lemma 2.6. If U is a γ-Lie ideal of M, then Cγ
(

[U,U]γ

)
= Cγ (U).

Proof. First, suppose that [U,U]γ * Cγ. Since [U,U]γ is a γ-Lie ideal of M, we have Cγ
(

[U,U]γ

)
= Cγ

by Lemma 2.5. Now, suppose that [U,U]γ ⊆ Cγ. Let a =
[
u, [u,x]γ

]
γ
for u ∈ U and x ∈ M. Since

a ∈ Cγ and aγu ∈ Cγ, we write a = 0 or u ∈ Cγ by Lemma 2.4. If a = 0, we have u ∈ Cγ by Lemma 2.1.
Hence, we get U ⊆ Cγ. Thus, Cγ

(
[U,U]γ

)
= Cγ (U).

Lemma 2.7. Let U be a γ-Lie ideal of M and U * Cγ. If aγUγb = 0 for a,b ∈ M, then a = 0 or b = 0.

Proof. There exists a nonzero ideal K of M such that [K,M]γ ⊂ U but [K,M]γ * Cγ by Lemma
2.3. Thus, aγ [kγaγu,m]γ γb = 0 for u ∈ U, k ∈ K and m ∈ M by hypothesis. Expanding this, we get
aγKγaγMγUγb = 0. Since M is γ-prime, we obtain aγKγa = 0 or Uγb = 0. Let aγKγa = 0. If a 6= 0,
then we have K = 0 which is a contradiction. Now, let Uγb = 0. Therefore, [u,m]γ γb = 0 for all u ∈ U
and m ∈ M. Hence, we get UγMγb = 0 which means b = 0.

Lemma 2.8. If d is a k-derivation of M such that k (γ) = 0 and d2 = 0, then d = 0.

Proof. Since d2 (xγy) = 0 for all x,y ∈ M, we have d (x) γd (y) = 0. Replacing y by mγx with m ∈ M,
we get d (x) γMγd (x) = 0 for all x ∈ M. Thus, d = 0 since M is γ-prime.

27



γ-Lie structures with derivations

Lemma 2.9. If d 6= 0 is a k-derivation of M such that k (γ) = 0, then Cγ (d (M)) = Cγ.

Proof. Let a ∈ Cγ (d (M)) and suppose a /∈ Cγ. Thus, [a,d (xγy)]γ = 0 for all x,y ∈ M. Expanding
this, we get d (x) γ [a,y]γ + [a,x]γ γd (y) = 0. If y ∈ M γ-commutes with a, then [a,y]γ = 0. So the last
equation reduces to [a,x]γ γd (y) = 0 for all x ∈ M. Then, d (y) = 0 since a /∈ Cγ. Indeed, if d (y) 6= 0, we
get a ∈ Cγ. But this is a contradiction. Therefore, d (y) = 0 for all y ∈ Cγ (a). Thus, d = 0 by Lemma
2.8 which contradicts with the assumption.

Lemma 2.10. Let d 6= 0 be a k-derivation of M such that k (γ) = 0 and U be a γ-Lie ideal of M.

(i) If d (U) = 0, then U ⊆ Cγ.

(ii) If d (U) ⊆ Cγ then U ⊆ Cγ.

Proof. (i) Let u ∈ U and x ∈ M. Since d (u) = d
(

[u,x]γ

)
= 0 by hypothesis, we get [u,d (x)]γ = 0

for all x ∈ M. Therefore, u centralizes d (M). Then, we get U ⊆ Cγ by Lemma 2.9.

(ii) Suppose that U * Cγ. Then, V = [U,U]γ * Cγ by proof of Lemma 2.6. Since d
(

[u,v]γ

)
= 0

for all u,v ∈ U, we get d (V ) = 0. It follows that V ⊆ Cγ by (i). But this is a contradiction.

Lemma 2.11. Let d be a k-derivation of M such that k (γ) = 0 and U be a γ-Lie ideal of M such that
U * Cγ. If tγd (U) = 0 (or d (U) γt = 0) for t ∈ M, then t = 0.

Proof. Let u ∈ U and x ∈ M. Using the fact [u,x]γ γu = [u,xγu]γ ∈ U, we have tγd
(

[u,x]γ γu
)

= 0.
Expanding this, we get tγ [u,x]γ γd (u) = 0 for all x ∈ M andu ∈ U. Replacing x by d (v) γy with v ∈ U,
y ∈ M, we obtain tγuγd (v) = 0 for all v,u ∈ U since tγd (U) = 0 and M is γ-prime. Hence, t = 0 by
Lemma 2.7.

Theorem 2.12. Let d 6= 0 be a k-derivation of M such that k (γ) = 0. If U is a γ-Lie ideal of M such
that d2 (U) = 0, then U ⊆ Cγ.

Proof. Suppose that U * Cγ. By proof of Lemma 2.6, we have V = [U,U]γ * Cγ. There exists a
nonzero ideal K of M such that [K,M]γ ⊂ U but [K,M]γ * Cγ by Lemma 2.3. Let y ∈ M, t ∈ [K,M]γ
and u ∈ V . If w := d (u), then d (w) = 0. By hypothesis, d2

(
[tγw,y]γ

)
= 0. Expanding this, we have

d (t) γd
(

[w,y]γ

)
= 0 for all t ∈ [K,M]γ , y ∈ M, w ∈ d (V ). Since [K,M]γ is a γ-Lie ideal of M and

[K,M]γ * Cγ, we have d
(

[d (V ) ,M]γ

)
= 0 by Lemma 2.11. Expanding last equation, we conclude that

[d (u) ,d (x)]γ = 0 for all x ∈ M,u ∈ V which means d (V ) ⊆ Cγ (d (M)). Therefore, we have V ⊆ Cγ by
Lemma 2.9 and by Lemma 2.10. But this is a contradiction.

Theorem 2.13. Let d 6= 0 be a k-derivation of M such that k (γ) = 0. If U is a γ-Lie ideal of M such
that U * Cγ, then Cγ (d (U)) = Cγ.

Proof. Let a ∈ Cγ (d (U)) and suppose that a /∈ Cγ. We have V = [U,U]γ * Cγ by proof of Lemma
2.6. Since d (V ) ⊆ U and a ∈ Cγ (d (U)) we get aγd2 (u) = d2 (u) γa and aγd (u) = d (u) γa. Now,
applying given derivation d to last equation gives d (a) ∈ Cγ (d (V )). Since a ∈ Cγ (d (U)), u ∈ V and
V is a γ-Lie ideal, we have [d (a) ,u]γ = d

(
[a,u]γ

)
∈ d (V ). It follows that [d (a) ,V ]γ = 0 which means

d (a) ∈ Cγ (V ). Therefore, d (a) ∈ Cγ (V ) = Cγ by Lemma 2.5.
Using same process for the element aγa gives d (aγa) = 2aγd (a) ∈ Cγ since aγa ∈ Cγ (d (U)). Thus,

d (a) = 0 by Lemma 2.4. Therefore, if d (b) 6= 0 for any b ∈ Cγ (d (U)) we have b ∈ Cγ. So we get

28



O. Arslan, H. Kandamar

a + b ∈ Cγ since d (a + b) = d (b) 6= 0. Then we have a ∈ Cγ. But this is a contradiction. Consequently,
when we suppose Cγ (d (U)) * Cγ, we are forced to d (a) = 0 for all a ∈ Cγ (d (U)).

Let W = {x ∈ M | d (x) = 0}. Then we have Cγ (d (U)) ⊆ W . Moreover, d
(

[a,u]γ

)
= 0 for any

a ∈ Cγ (d (U)) and u ∈ U.
There exists a nonzero ideal K of M such that [K,M]γ ⊂ U but [K,M]γ * Cγ by Lemma 2.3.

If t ∈ [K,M]γ ⊂ U ∩ K, then tγa ∈ K. Thus,
[
a,d

(
[tγa,u]γ

)]
γ

= 0. Expanding this, we get

d (t) γ
[
a, [a,u]γ

]
γ

= 0 for all t ∈ [K,M]γ , u ∈ U. Hence,
[
a, [a,U]γ

]
γ

= 0 by Lemma 2.11. Since

U * Cγ, we have a ∈ Cγ (U). Therefore, a ∈ Cγ by Lemma 2.5. But this is a contradiction.

Lemma 2.14. If d3 6= 0 and U * Cγ, then d (M) the γ-subring generated by d (M) contains a nonzero
γ-ideal of M.

Proof. Since d2 (d (M)) 6= 0, we have y ∈ d (M) ⊆ d (M) such that d2 (y) 6= 0. Thus, we get Mγd (y) ⊆
d (M) since d (xγy) and d (x) γy in d (M) for all x ∈ M. Similarly, d (y) γM ⊆ d (M). If we act d to the
element aγd (y) γb for a,b ∈ M, we get aγd2 (y) γb ∈ d (M) by above, that is to say Mγd2 (y) γM ⊆ d (M).
We also have Mγd2 (y) ⊆ d (M) and d2 (y) γM ⊆ d (M) by above. Therefore, the γ-ideal of M generated
by d2 (y) 6= 0 contained in d (M).

Lemma 2.15. Let d3 6= 0, U * Cγ and V = [U,U]γ. If d (V ) contains a nonzero left ideal λ of M and
a nonzero right ideal ρ of M, then d (U) contains a nonzero ideal of M.

Proof. Since d (V ) ⊆ U, we have d (d (V )) ⊆ d (U). Let a ∈ λ ⊆ d (V ) and x ∈ M. Thus, d (xγa) ∈
d (U). Expanding this, we get xγd (a) ∈ d (U) for all x ∈ M and a ∈ λ. Therefore, we have Mγd (λ) ⊆
d (U). Similarly, d (ρ) γM ⊆ d (U). Let a ∈ λ and u ∈ V . If we act d to the element [a,u]γ, we get
d (a) γu ∈ d (U) by above, that is to say d (λ) γV ⊆ d (U). Similarly, V γd (ρ) ⊆ d (U).

Let I = λγV γρ. Then by Lemma 2.7, I is a nonzero ideal of M. Moreover, d (I) ⊆ d (U). By Lemma
2.14, d (I) contains a nonzero γ-ideal K of I since d3 6= 0 and I is γ-prime. Let S := λγKγρ. Then S is
an ideal of M which is contained in d (U).

Lemma 2.16. Let 0 6= I < M and U * Cγ. If d (U) does not contain a nonzero right ideal(or left ideal)
of M and [c,I]γ ⊆ d (U) then c ∈ Cγ.

Proof. Let t ∈ d (U) and i ∈ I. Then [c,tγi]γ ∈ d (U) by hypothesis. Expanding this, we get
[c,d (U)]γ γI ⊆ d (U). But, since d (U) does not contain a nonzero right ideal of M, we get [c,d (U)]γ γI =
0. Thus, [c,d (U)]γ = 0 since 0 6= I < M and M is a γ-prime gamma ring. Then by Theorem 2.13, we
get c ∈ Cγ (d (U)) = Cγ.

Lemma 2.17. Let U * Cγ, V = [U,U]γ and W = [V,V ]γ. If d
2 (U) γd2 (U) = 0 then d3 (W) = 0.

Proof. By proof of Lemma 2.6, V and W are not contained in Cγ since U * Cγ. Also, we have d (W) ⊆
V and d2 (W) ⊆ d (V ) ⊆ U. If u ∈ U, v ∈ V and w ∈ W , then we have d2 (u) γd2

([
d (v) ,d2 (w) γt

]
γ

)
= 0

for any t ∈ U. Expanding this, we get

d2 (u) γd (v) γ
(
d4 (w) γt + 2d3 (w) γd (t)

)
= 0 (1)

by hypothesis. In (1), if we choose t ∈ d (V ) ⊆ U, it follows d2 (u) γd (v) γd4 (w) γt = 0 for such t. Thus,
we have d2 (u) γd (v) γd4 (w) = 0 by Lemma 2.11. Then we get from (1) that d2 (u) γd (v) γd3 (w) γd (t) =

29



γ-Lie structures with derivations

0 for all t ∈ U. By Lemma 2.11, we conclude that d2 (u) γd (v) γd3 (w) = 0 for all u ∈ U, v ∈ V and
w ∈ W . Similarly, we have d3 (w) γd (v) γd2 (u) = 0 by reversing sides.

By hypothesis, d2 (d (t)) γd2
(

[v,d (w)]γ

)
= 0 for w,t ∈ W and v ∈ V . Expanding this, we get

d3 (t) γvγd3 (w) = 0 that is to say d3 (t) γV γd3 (w) = 0 for all w,t ∈ W . It follows that d3 (W) = 0 by
Lemma 2.7.

Lemma 2.18. If U * Cγ and d3 (U) = 0 then d3 = 0.

Proof. Let u ∈ U and m ∈ M. Then we have d3
(

[u,m]γ

)
= 0. Expanding this, we get

3
[
d2 (u) ,d (m)

]
γ

+ 3
[
d (u) ,d2 (m)

]
γ

+
[
u,d3 (m)

]
γ

= 0. (2)

Let V = [U,U]γ and W = [V,V ]γ. In (2), replacing u by d
2 (w) with w ∈ W, we get

[
d2 (w) ,d3 (m)

]
γ

= 0

by hypothesis. Again replacing u by d (w) and m by d (m) in (2), we obtain
[
d (w) ,d4 (m)

]
γ

= 0.

By proof of Lemma 2.6, W * Cγ. Thus, by Theorem 2.13, Cγ (d (W)) = Cγ. Since
[
d (w) ,d4 (m)

]
γ

=

0 for all w ∈ W and m ∈ M, it follows d4 (M) ⊆ Cγ.

By hypothesis, d4
(

[u,m]γ

)
= 0 for u ∈ U and m ∈ M. Expanding this, we get

6
[
d2 (u) ,d2 (m)

]
γ

+ 4
[
d (u) ,d3 (m)

]
γ

= 0 (3)

since d4 (m) ∈ Cγ. Similarly, expanding the equation d3
(

[u,d (m)]γ

)
= 0, we get

3
[
d2 (u) ,d2 (m)

]
γ

+ 3
[
d (u) ,d3 (m)

]
γ

= 0. (4)

Combining the equation (4) and the equation (3) we get
[
d (u) ,d3 (m)

]
γ

= 0 for all u ∈ U and m ∈ M.
Therefore, by Theorem 2.13, d3 (M) ⊆ Cγ (d (U)) = Cγ. Hence, we get d3 (m) γd2 (u) ∈ Cγ that is to say
d3 (M) γd2 (U) ⊆ Cγ.

Suppose that d3 (M) 6= 0. By Lemma 2.4, we have d2 (U) ⊆ Cγ. Since d4 (mγd (u)) ∈ Cγ it follows
d4 (M) γd (U) ⊆ Cγ. Since d (U) cannot be contained in Cγ by Lemma 2.10, we get d4 (M) = 0 by
Lemma 2.4. So d4 (mγd (u)) = 4d3 (m) γd2 (u) = 0. Hence, d3 (M) γd2 (U) = 0. On the other hand, we
have d2 (U) 6= 0 by Theorem 2.12. Since d2 (U) ⊆ Cγ and M is γ-prime gamma ring, d3 (M) = 0, that is
to say d3 = 0.

Lemma 2.19. If [U,U]γ ⊆ Cγ then U ⊆ Cγ.

Proof. Let [U,U]γ = 0. Then, we get
[
u, [u,x]γ

]
γ

= 0 for all u ∈ U and x ∈ M by hypothesis.
Therefore, u ∈ Cγ by Lemma 2.1.

Now, let [U,U]γ 6= 0. Then, there exist u,v ∈ U such that [u,v]γ 6= 0. Let d(x) = [x,v]γ for
x ∈ M. Then d2(x) = [d(x),v]γ ∈ Cγ for all x ∈ M by hypothesis. Let a = d(u) and b = d

2(x).
Therefore, d2(uγx) = 2aγd(x) + bγu ∈ Cγ. Then we have [u, 2aγd(x) + bγu]γ = 0. Expanding this,
we get 2aγ [u,d(x)]γ = 0 for all x ∈ M. Replacing x by uγv in the last equation, we obtain a

3
γ = 0.

Therefore, we have a nonzero γ-nilpotent element a in the γ-center of the γ-prime gamma ring M. But
this is a contradiction.

Lemma 2.20. Let M be a gamma ring of characteristic not 2 and U be a γ-Lie ideal of M. If [u,d(u)]γ ∈
Cγ and u2 ∈ U for all u ∈ U, then [u,d(u)]γ = 0.

30



O. Arslan, H. Kandamar

Proof. We know that
[
u + u2,d(u + u2)

]
γ
∈ Cγ for all u ∈ U by hypothesis. Expanding this, we get

4 [u,d(u)]γ γu ∈ Cγ. Hence, [u,d(u)]γ γ [u,m]γ = 0 for all u ∈ U and m ∈ M. Replacing m by mγx with
x ∈ M, we obtain [u,d(u)]γ γmγ [u,x]γ = 0 that leads to [u,d(u)]γ = 0 or [u,x]γ = 0 for all x ∈ M since
M is γ-prime gamma ring.

Lemma 2.21. Let U be a γ-Lie ideal of M and [u,d(u)]γ ∈ Cγ for all u ∈ U. Then
[
[d(m),u]γ ,u

]
γ
∈ Cγ

for all u ∈ U and m ∈ M. Moreover, if [u,d(u)]γ = 0 for all u ∈ U, then
[
[d(m),u]γ ,u

]
γ

= 0 for all

u ∈ U and m ∈ M.

Proof. Let u ∈ U and m ∈ M. By hypothesis,
[
u + [u,m]γ ,d(u + [u,m]γ)

]
γ
∈ Cγ. Expanding this,

we get [
[u,m]γ ,d(u)

]
γ

+
[
u, [d(u),m]γ

]
γ

+
[
u, [u,d(m)]γ

]
γ
∈ Cγ.

But for any u ∈ U and m ∈ M one can show that[
[u,m]γ ,d(u)

]
γ

+
[
u, [d(u),m]γ

]
γ

=
[
m, [d(u),u]γ

]
γ

= 0.

Therefore, we get desired result. Similary, the other statement can easily be shown.

Lemma 2.22. Let a ∈ M. If aγd(x) = 0 for all x ∈ M, then a = 0 or d = 0.

Proof. By hypothesis, aγd(xγy) = 0 for all x,y ∈ M. Expanding this, we get aγxγd(y) = 0 for all
x,y ∈ M. Since M is γ-prime gamma ring we have the desired result.

3. Main results

Theorem 3.1. Let M be a γ-prime weak Nobusawa Γ-ring of characteristic not 2 and d be a k-derivation
of M such that k (γ) = 0 and d3 6= 0. If U is a γ-Lie ideal of M such that U * Cγ then d (U) contains a
nonzero ideal of M.

Proof. Let V = [U,U]γ and W = [V,V ]γ. According to Lemma 2.15, it is enough to show that the
γ-subring d (V ) contains a nonzero left ideal of M and a nonzero right ideal of M. Suppose that d (V )
does not contain a nonzero right ideal of M.

Let w ∈ [W,W]γ and a := d (w). Since aγ [a,x]γ ∈ W , we have d
(
aγ [a,x]γ

)
∈ d (W). Expanding

this, we get

d (a) γ [a,x]γ ∈ d (V ), ∀a ∈ d
(

[W,W]γ

)
, x ∈ M. (5)

On the other hand, since d
(

[a,u]γ

)
∈ d (V ) and [a,d (u)]γ ∈ d (V ) for u ∈ V , we have

[d (a) ,V ]γ ⊆ d (V ), ∀a ∈ d
(

[W,W]γ

)
. (6)

For m ∈ M we also have

d (a) γ [d (a) ,m]γ + d (a) γ [a,d (m)]γ ∈ d (V )

31



γ-Lie structures with derivations

since d (a) γd
(

[a,m]γ

)
∈ d (V ). Hence, by (5)

d (a) γ [d (a) ,m]γ ∈ d (V ), ∀a ∈ d
(

[W,W]γ

)
, m ∈ M. (7)

In (7) replacing a by a + b with a, b ∈ d
(

[W,W]γ

)
we obtain

s := d (a) γ [d (b) ,m]γ + d (b) γ [d (a) ,m]γ ∈ d (V ), ∀a,b ∈ d
(

[W,W]γ

)
.

If t := [d (a) γd (b) ,m]γ = d (a) γ [d (b) ,m]γ + [d (a) ,m]γ γd (b) then

s− t = d (b) γ [d (a) ,m]γ − [d (a) ,m]γ γd (b) =
[
d (b) , [d (a) ,m]γ

]
γ
.

By (6), s − t ∈ d (V ). Thus, we get t ∈ d (V ), that is [d (a) γd (b) ,M]γ ⊆ d (V ). Since d (V ) does not
contain a nonzero right ideal of M, d (a) γd (b) ∈ Cγ for all a,b ∈ d

(
[W,W]γ

)
by Lemma 2.16. Let

n := d (a) γd (b). By (5), d (b) γ [b,x]γ ∈ d (V ). It follows nγ [b,x]γ = d (a) γd (b) γ [b,x]γ ∈ d (V ) since
d (a) ∈ d (V ). On the other hand, since nγ [b,x]γ = [b,nγx]γ ∈ d (V ), we have [b,nγM]γ ⊆ d (V ). Let
I = nγM. If I 6= 0, then b ∈ Cγ for all b ∈ d

(
[W,W]γ

)
by Lemma 2.16. Thus, by Lemma 2.10

we get [W,W]γ ⊆ Cγ that is to say U ⊆ Cγ by Lemma 2.19. But this is a contradiction. Therefore,
I = nγM = 0. So we get n = d (a) γd (b) = 0 for all a,b ∈ d

(
[W,W]γ

)
since M is γ-prime gamma ring.

That is, d2
(

[W,W]γ

)
γd2

(
[W,W]γ

)
= 0. Hence, we conclude that the contradiction d3 = 0 by Lemma

2.17 and Lemma 2.18.

Theorem 3.2. Let M be a gamma ring of characteristic not 2 or 3 and U be a γ-Lie ideal of M. If
[u,d(u)]γ ∈ Cγ for all u ∈ U, then U ⊂ Cγ.

Proof. By Lemma 2.21 we have[
[d(m),u]γ ,u

]
γ
γu = uγ

[
[d(m),u]γ ,u

]
γ
.

Expanding this, we get

3u2γd(m)γu + d(m)γu3 = 3uγd(m)γu2 + u3γd(m). (8)

Let d(m) = m′. Replacing m by u in (8), we obtain

u3γu′ −u′γu3 = 3(uγu′ −u′γu)γu2 (9)

for all u ∈ U. Since [u,u′]γ γu = uγ [u,u
′]γ we have

2γ(uγu′ −u′γu)γu = u2γu′ −u′γu2. (10)

Again replacing m by uγm′ in (8), we obtain

3uγu′γm′γu2 + u3γu′γm′ − 3u2γu′γm′γu−u′γm′γu3 = 0 (11)

for all u ∈ U and m ∈ M. Multiplying (8) with u′ gives

3u′γuγm′γu2 + u′γu3γm′ − 3u′γu2γm′γu−u′γm′γu3 = 0.

Substracting the last equation from (11) we get

3(uγu′ −u′γu)γm′γu2 + (u3γu′ −u′γu3)γm′ − 3(u2γu′ −u′γu2)γm′γu = 0.

32



O. Arslan, H. Kandamar

Using the equations (9) and (10) we conclude

(uγu′ −u′γu)γ(m′γu2 + u2γm′ − 2uγm′γu) = 0

for all u ∈ U and m ∈ M. If uγu′ −u′γu 6= 0 for some u, then

m′γu2 + u2γm′ − 2uγm′γu = 0 (12)

for that u. Replacing m by uγm, we obtain

(u′γm + uγm′)u2 + u2γ(u′γm + uγm′) − 2uγ(u′γm + uγm′) = 0.

Expanding last equation, we have

u′γmγu2 + u2γu′γm− 2uγu′γmγu = 0 (13)

for all m ∈ M. If we replace m by u in (12) and multiply by m on the right, then we get

u′γu2γm + u2γu′γm− 2uγu′γuγm = 0. (14)

Substracting (14) from (13) gives

u′γ(mγu2 −u2γm) − 2uγu′γ(mγu−uγm) = 0. (15)

Replacing m by uγm in (15), we obtain

u′γuγ(mγu2 −u2γm) − 2uγu′γuγ(mγu−uγm) = 0. (16)

Mulyiplying (15) by u we get

uγu′γ(mγu2 −u2γm) − 2u2γu′γ(mγu−uγm) = 0. (17)

Substracting (16) from (17) gives

(uγu′ −u′γu)γ(mγu2 −u2γm− 2uγ(mγu−uγm)) = 0

for all m ∈ M. Since M is γ-prime gamma ring we have

mγu2 −u2γm− 2uγ(mγu−uγm) = 0.

Now think the inner Iγu-derivation Iuγ on M. From the last equation we write I2uγ = 0 that leads to
Iuγ = 0 by Lemma 2.8. Hence, u ∈ Cγ.

So far we proved that if [u,u′]γ 6= 0 for u ∈ U, then u ∈ Cγ. Now assume that [u,u
′]γ = 0 for all

u ∈ U. By Lemma 2.21,
[
[d(m),u]γ ,u

]
γ

= 0 for all m ∈ M and u ∈ U. Replacing u by u + w with
w ∈ U, we obtain [

[d(m),u]γ ,w
]
γ

+
[
[d(m),w]γ ,u

]
γ

= 0. (18)

Choose v,w ∈ U such that wγv ∈ U. Replacing w by wγv in (18), we obtain

[w,u]γ γ [d(m),v]γ + [d(m),w]γ γ [v,u]γ = 0.

For any t ∈ M and w ∈ U the element v = [t,w]γ ensures the condition wγv ∈ U. So by above we have

[w,u]γ γ
[
d(m), [t,w]γ

]
γ

+ [d(m),w]γ γ
[
[t,w]γ ,u

]
γ

= 0 (19)

33



γ-Lie structures with derivations

for all t,m ∈ M and u,w ∈ U. Replacing u by w, we obtain

[d(m),w]γ γ
[
[t,w]γ ,w

]
γ

= 0. (20)

Replacing t by tγd(a) with a ∈ M, we obtain

[d(m),w]γ γ [t,w]γ γ [d(a),w]γ = 0 (21)

for all m,t,a ∈ M and w ∈ U. Replacing u by [t,w]γ in (19), we get
[
[t,w]γ ,w

]
γ
γ
[
[t,w]γ ,d(m)

]
γ

= 0.

If we replace t by t + d(a) we get[
[t,w]γ ,w

]
γ
γ
[
[d(a),w]γ ,d(m)

]
γ

= 0 (22)

for all m,t,a ∈ M and w ∈ U. Replacing t by d(t)γs with s ∈ M in (22), we obtain

[d(t),w]γ γ [s,w]γ γd(m)γ [d(a),w]γ = 0

for all m,t,a,s ∈ M and w ∈ U.
Replacing t by tγd(s) in (21), we conclude

[d(m),w]γ γMγ [d(s),w]γ γ [d(a),w]γ = 0

for all m,a,s ∈ M and w ∈ U. Since M is γ-prime gamma ring we get [d(m),w]γ = 0 or
[d(s),w]γ γ [d(a),w]γ = 0 for all m,a,s ∈ M and w ∈ U. If [d(m),w]γ = 0 for all m ∈ M and w ∈ U,
then w ∈ Cγ by Lemma 2.9, so we are done. Suppose there is a pair m ∈ M and w ∈ U such that
[d(m),w]γ 6= 0. Hence, w /∈ Cγ and

[d(s),w]γ γ [d(a),w]γ = 0 (23)

for all a,s ∈ M. Replacing a by bγc with b,c ∈ M in (23), we get

[d(s),w]γ γd(b)γ [c,w]γ = 0.

If we replace b by [t,w]γ in this equation we have

[d(s),w]γ γ [t,d(w)]γ γ [w,c]γ = 0

for all c,t,s ∈ M and w ∈ U. Replacing c by cγm1 with m1 ∈ M, we obtain

[d(s),w]γ γ [t,d(w)]γ = 0.

Hence, replacing t by tγk with k ∈ M in the last equation, we get d(w) ∈ Cγ.
Now suppose u ∈ U ∩Cγ. Then d([u,a]γ) = 0 for all a ∈ M. Therefore, we have d(u) ∈ Cγ. Hence,

d(u) ∈ Cγ for all u ∈ U and then we get d([w,a]γ) ∈ Cγ for all a ∈ M. Expanding this, we obtain
[w,d(a)]γ ∈ Cγ and replacing a by aγw, we have

[w,d(a)]γ γw + [w,a]γ γd(w) ∈ Cγ. (24)

Therefore, commuting this element by w we get[
w, [w,a]γ

]
γ
γd(w) = 0

for all a ∈ M. Since M is γ-prime gamma ring we have
[
w, [w,a]γ

]
γ

= 0 or d(w) = 0 for all a ∈ M.

If
[
w, [w,a]γ

]
γ

= 0, then w ∈ Cγ by Lemma 2.1. But this is a contradiction. Hence, d(w) = 0 and
[w,d(a)]γ γw ∈ Cγ for all a ∈ M by (24). It follows that [d(a),w]γ γ [w,b]γ = 0 for a,b ∈ M. Replacing
b by bγc with c, we obtain [d(a),w]γ = 0 or [w,b]γ = 0. So in both cases we have w ∈ Cγ which is a
contradiction.

34



O. Arslan, H. Kandamar

Corollary 3.3. Let M be a gamma ring of characteristic 3 and U be a γ-Lie ideal of M. If [u,d(u)]γ ∈ Cγ
and u2 ∈ U for all u ∈ U, then U ⊂ Cγ.

Theorem 3.4. Let M be a gamma ring of characteristic 2 and U be a γ-Lie ideal and γ-subring of M.
If [u,d(u)]γ ∈ Cγ for all u ∈ U, then U is γ-commutative.

Proof. By Lemma 2.21,
[
[d(m),u]γ ,u

]
γ
∈ Cγ for all u ∈ U and m ∈ M. Hence,

d(m)γu2 + u2γd(m) ∈ Cγ (25)

for all u ∈ U and m ∈ M. Then [
d(m),d(m)γu2 + u2γd(m)

]
γ

= 0

and [
u2,d(m)γu2 + u2γd(m)

]
γ

= 0.

Expanding these equations, we get

u2γ(d(m))2 = (d(m))2γu2 (26)

and

u4γd(m) = d(m)γu4 (27)

respectively.

Since d(u2) = uγd(u) + d(u)γu ∈ Cγ for u ∈ U, replacing m by v + u2γv with v ∈ U, we obtain(
u2γd(v) + d(v)γu2

)2
= 0

for all u,v ∈ U. Using γ-primeness of M we have

u2γd(v) = d(v)γu2 (28)

for all u,v ∈ U by (25).
Replacing u by u + w with w ∈ U in (28), we get

(uγw + wγu) γd(v) = d(v)γ (uγw + wγu) .

Replacing w by wγu, we get

(uγw + wγu) γ (uγd(v) + d(v)γu) = 0

for all u,v,w ∈ U. We conclude(
u21γw + wγu

2
1

)
γ (uγd(u) + d(u)γu) = 0, ∀u,u1,w ∈ U

replacing u by u + u21 with u1 ∈ U and taking v = u.
Hence, if [d(u),u]γ 6= 0 for some u ∈ U, then u

2
1γw = wγu

2
1 for all u1,w ∈ U. Then, we have

u2γ (wγm + mγw) = (wγm + mγw) γu2 for all m ∈ M and u,w ∈ U. Expanding this, and replacing m
by mγu, we obtain (

u2γm + mγu2
)
γ (wγu + uγw) = 0

35



γ-Lie structures with derivations

for all u,w ∈ U and m ∈ M. Replacing w by [u,t]γ with t ∈ M, we get(
u2γm + mγu2

)
γ
(
u2γt + tγu2

)
= 0

for all u ∈ U and m,t ∈ M. Again replacing t by tγp with p ∈ P , we conclude u2 ∈ Cγ for all u ∈ U.

Assume that [d(u),u]γ = 0 for all u ∈ U. Then, by Lemma 2.21
[
[d(m),u]γ ,u

]
γ

= 0 for all u ∈ U

and m ∈ M. Expanding this, we get u2γd(m) = d(m)γu2. Replacing m by mγa with a ∈ M, we obtain

d(m)γ
(
u2γa + aγu2

)
+
(
u2γm + mγu2

)
γd(a) = 0.

Replacing a by v2, we get d(m)γ
(
u2γv2 + v2γu2

)
= 0 for all u,v ∈ U,m ∈ M since d(v2) =

vγd(v) + d(v)γv = 0 for v ∈ U. Therefore, u2γv2 = v2γu2 for all u,v ∈ U by Lemma 2.22. Hence,
u2γ (vγw + wγv) = (vγw + wγv) γu2 for all u,v,w ∈ U. Replacing v by vγw in the last equation,
we have (vγw + wγv) γ

(
u2γw + wγu2

)
= 0. Again replacing v by [w,m]γ with m ∈ M, we obtain(

w2γm + mγw2
)
γ
(
u2γw + wγu2

)
= 0. That is, Iw2γ(m)γ

(
u2γw + wγu2

)
= 0 for the inner Iγw2-

derivation Iw2γ on M. So by Lemma 2.22, if w2 /∈ Cγ for some w ∈ U, then u2γw = wγu2 for that
w. Therefore,

[
[u,v]γ ,w

]
γ

= 0 for all u,v ∈ U. Expanding last equation and replacing v by vγw, we
obtain [v,w]γ γ [w,u]γ = 0 for all u,v ∈ U. Replacing v by [w,r]γ with m,t ∈ M and u by [w,t]γ, we
get

(
w2γm + mγw2

)(
w2γt + tγw2

)
= 0 for all m,t ∈ M. Again replacing t by tγp with p ∈ M, we

conclude
(
w2γm + mγw2

)
γtγ

(
w2γp + pγw2

)
= 0 for all p,t ∈ M. Since M is γ-prime gamma ring we

get w2 ∈ Cγ from the last equation. But this is a contradiction.
So far we conclude u2 ∈ Cγ for all u ∈ U. Hence, uγv + vγu ∈ Cγ and (uγv + vγu) γu ∈ Cγ for all

u,v ∈ U. Therefore, we have u ∈ Cγ or uγv + vγu = 0. Then, U is γ-commutative.

If we assume that U is only γ-Lie ideal of M or only γ-subring of M, then U may not be γ-
commutative. Moreover, according to the assumptions of the theorem, the result U ⊆ Cγ cannot be
obtained.

Example 3.5. Let R be a noncommutative prime ring with identity. If M is the set of all matrices over

R of the form
(
a b a
c d c

)
, Γ = M3×2(R) and γ =


 1 00 1

0 0


 ∈ Γ, then M is a γ-prime gamma ring. It

can be shown that the subset U =
{(

a 0 a
0 b 0

)
| a,b ∈ R

}
of M is a γ-subring but it is not a γ-Lie ideal

of M. Define the inner Iγn-derivation Inγ on M with n =
(

1 0 1
0 1 0

)
∈ M. Then it is easily verified

that [u,Inγ(u)]γ ∈ Cγ for all u ∈ U but U is not γ-commutative.

Example 3.6. Let M =
{(

a b a
c d c

)
| a,b,c,d ∈ Z2

}
, Γ = M3×2(Z2) and γ =


 1 00 1

0 0


 ∈ Γ. Then M

is a γ-prime gamma ring. Let U =
{(

a b a
c a c

)
| a,b,c ∈ Z2

}
. It is easily seen that U is a γ-Lie ideal

but it is not a γ-subring of M. Define the inner Iγn-derivation Inγ on M with n =
(

1 0 1
0 1 0

)
∈ M.

Then it is easily verified that [u,Inγ(u)]γ ∈ Cγ for all u ∈ U but U is not γ-commutative.

Example 3.7. Let M =
{(

a b a
c d c

)
| a,b,c,d ∈ Z2

}
, Γ = M3×2(Z2) and γ =


 1 00 1

0 0


 ∈ Γ. Then

M is a γ-prime gamma ring. Let U =
{(

a b a
b a b

)
| a,b ∈ Z2

}
. It is easily seen that U is a γ-Lie

36



O. Arslan, H. Kandamar

ideal and a γ-subring of M but it is not a γ-ideal. Define the inner Iγn-derivation Inγ on M with

n =

(
1 0 1
0 1 0

)
∈ M. Then it is easily verified that [u,Inγ(u)]γ ∈ Cγ for all u ∈ U. Hence, U is

γ-commutative but cannot be contained in the γ-center of M.

Acknowledgment: The authors would like to thank the referees for their valuable suggestions and
comments.

References

[1] R. Awtar, Lie and Jordan structure in prime rings with derivations, Proc. Amer. Math. Soc., 41,
67-74, 1973.

[2] W. E. Barnes, On the Γ-rings of Nobusawa, Pacific J. Math., 18, 411-422, 1966.
[3] J. Bergen, J.W. Kerr, I.N. Herstein, Lie ideals and derivations of prime rings, J. Algebra, 71, 259-267,

1981.
[4] I. N. Herstein, A note on derivations, Canad. Math. Bull., 21(3), 369-370, 1978.
[5] I. N. Herstein, A note on derivations II, Canad. Math. Bull., 22(4), 509-511, 1979.
[6] I. N. Herstein, Topics in Ring Theory, The Univ. of Chicago Press, 1969.
[7] H. Kandamar, The k-Derivation of a Gamma-Ring, Turk. J. Math., 23(3), 221-229, 2000.
[8] A. R. Khan, M. A. Chaudhry, I. Javaid, Generalized Derivations on Prime Γ-Rings, World Appl.

Sci. J., 23(12), 59-64, 2013.
[9] S. Kyuno, Gamma Rings, Hadronic Press, 1991.
[10] N. Nobusawa, On a generalization of the ring theory, Osaka J. Math., 1, 81-89, 1964.
[11] E. C. Posner, Derivations in prime rings, Proc. Amer. Math. Soc., 8, 1093-1100, 1957.
[12] N. N. Suliman, A. H. Majeed, Lie Ideals in Prime Γ-Rings with Derivations, Discussiones Mathe-

maticae, 33, 49-56, 2013.

37


	Preliminaries
	-Lie ideals and derivations
	Main results
	References