Left Derivations and Strong Commutativity Preserving Maps on Semiprime $\Gamma$-Rings

Abstract

In this paper, firstly as a short note, we prove that a left derivation of a
semiprime Γ\Gamma-ring M must map M into its center, which improves a
result by Paul and Halder and some results by Asci and Ceran. Also we prove
that a semiprime Γ\Gamma-ring with a strong commutativity preserving
derivation on itself must be commutative and that a strong commutativity
preserving endomorphism on a semiprime Γ\Gamma-ring M must have the form
σ(x)=x+ζ(x)\sigma(x)=x+\zeta(x) where ζ\zeta is a map from M into its center, which
extends some results by Bell and Daif to semiprime Γ\Gamma-rings.

Full text

arXiv:1206.4177v1 [math.RA] 19 Jun 2012
LEFT DERIVATIONS AND STRONG COMMUTATIVITY
PRESERVING MAPS ON SEMIPRIME Γ-RINGS
XIAOWEI XU, JING M A, AND YUAN ZHOU
Abstract. In this paper, firstly as a short note, we prove that a left derivation
of a semiprime Γ-ring M must map M into its center, which improves a result
by Paul and Halder and some results by Asci and Ceran. Also we prove
that a semiprime Γ-ring with a strong commutativity preserving derivation
on itself must be commutative and that a strong commutativity preserving
endomorphism on a semiprime Γ-ring M must have the form σ(x) = x + ζ(x)
where ζ is a map from M into its center, which extends some results by Bell
and Daif to semiprime Γ-rings.
1. Introduction
In 1964, Nobusawa [11] had introduce d the notion of a Γ-ring, which was ex-
tended by Barnes [2] in 1966 so that the quotient Γ-ring of a Γ-ring c an be defined
reasonably. At present the notion by Nobusawa is called a Γ
N
-ring, and the notion
by Barnes is ca lled a Γ-ring. A Γ
N
-ring is a Γ-ring, and there exists a Γ-ring being
not a Γ
N
-ring. Barnes [2] has defined a Γ-ring as following: Let M and Γ be two
additive abelian groups. If there exists a map (a, α, b) 7→ aαb of M × Γ × M → Γ
satisfying the conditions
• (a + b)αc = aαc + bαc, a(α + β)b = aαb + aβb, aα(b + c) = aαb + aαc,
• (aαb)βc = aα(bβc)
for all a, b, c ∈ M and α, β ∈ Γ, then M is called a Γ-ring.
An associative ring R can be seen as a Γ-ring. For example it is pointed out
explicitly in [7] that an associative ring R is a Γ-ring with Γ = U or Z where Z
is the ring of integers and U is an ideal (even an additive subgroup) of R. Some
properties of Γ-rings as contrast to those of ge ne ral rings have also been obtained
by Barnes [2 ], K yuno [9] and Luh [10].
An additive subgroup U of a Γ-ring M is called a left (resp. right) ideal of M if
MΓU ⊆ U (resp. U ΓM ⊆ U). A left idea l U of a Γ-ring M is called an ideal of M if
it is also a right ideal of M . The set Z(M) = {a ∈ M |aαb = bαa, ∀b ∈ M, ∀α ∈ Γ}
is called the center of M .
A Γ-ring M is ca lled prime if aΓM Γb = 0 with a, b ∈ M implies a = 0 or b = 0.
A Γ-ring M is called semiprime if aΓMΓa = 0 with a ∈ M implies a = 0. The
notion of a (resp. semi-)prime Γ-ring is an extensio n for the notion of a (resp.
semi-)prime ring.
1991 Mathematics Subject Classification. 16W25, 16N60, 16Y99.
Key words and phrases. prime gamma ring; semiprime gamma ring; left derivation; str ong
commutativity preserving map.
The paper is supported by the NNSF of China (No. 10871023 and No.11071097), 211 Project,
985 Project and the Basic Foundation for Science R esearch from Jilin University.
1

2 XIAOWEI XU, JING MA, AND YUAN ZHOU
Recall that an additive map δ from a ring R into itself is called a left derivation
if δ(xy) = xδ(y) + yδ(x) holds for all x, y ∈ R. In 1990, Breˇsar and Vukman [6]
firstly intr oduced the notion of a left derivation in a ring and pr oved that a left
derivation of a semiprime ring R must map R into its center.
Similarly an additive map δ from a Γ-ring M into itself is called a left (resp.
right) derivation if δ(xαy) = xαδ(y) + yαδ(x) (resp. δ(xαy) = δ(y)αx + δ(x)αy)
holds fo r all x, y ∈ M and α ∈ Γ.
Naturally the notions of derivations and endomorphisms for rings have extensions
for Γ-rings. An additive map µ from a Γ-ring M into itself is called a derivation
(resp. endomorphism) if µ(xαy) = µ(x)αy + xαµ(y) (r esp. µ(xαy) = µ(x)αµ(y))
holds fo r all x, y ∈ M and α ∈ Γ.
Some conclusions conce rning maps and identities in (semi-)prime Γ-rings o r Γ
N
-
rings were obtaine d as a complement of those in (semi-)prime r ings. For example
we c an search some of them in [14, 15, 16, 17, 18, 13, 12, 8, 7].
In 2007, Asci and Ceran [1] discussed the left derivation of a prime Γ -ring and
obtained some conclusions on commutativity of a prime Γ-ring. For example it
is proved that for a prime Γ-ring M with a nonzero ideal U and a nonzero right
derivation d, if charM 6= 2, 3, d
2
(U) ⊆ Z(M) and d(U) ⊆ U , then M is commuta-
tive.
In 2009, Paul and Halder [13] proved that a left derivation of a semiprime Γ-ring
M must map M into its center under the as sumption that aαbβc = aβbαc holds
for all a, b, c ∈ M and α, β ∈ Γ. In 2010, Paul and Halder [14] also considered
this problem. In this paper, we get rid of the ass umptions in Paul and Halder’s
Theorems. Particularly we obtain that a prime Γ-ring with a nonzero left or right
derivation must be commutative, which extends the conclusions app e aring in [1].
For a ring R w ith a, b ∈ R, the symbol [a, b] denotes ab − ba. Similarly in a
Γ-ring M with a, b ∈ M and α ∈ Γ, the symbol [a, b]
α
denotes a αb − bαa. For
a ring R with a, b, c ∈ R the commutator formulas [a, bc] = [a, b]c + b[a, c] and
[ab, c] = a[b , c] + [a, c]b are very useful in the course of dealing with identities in
semiprime rings. But in a Γ-ring M with a, b, c ∈ M and α, β ∈ Γ, the commutator
formulas become [a , bαc]
β
= [a, b]
β
αc + bα[a, c]
β
+ bβaαc − bαaβc and [aαb, c]
β
=
[a, c]
β
αb + aα[b, c]
β
+ aαcβb − aβcαb. The assumption that aαbβc = aβbαc holds
for all a, b, c ∈ M and α, β ∈ Γ can make the commutator formulas appe ar as
[a, bαc]
β
= [a, b]
β
αc + bα[a, c]
β
and [aαb, c]
β
= [a, c]
β
αb + aα[b, c]
β
which is same
as those in general rings. However in this paper, all results being proved need not
this assumption, which shows that some results ca n be kept from semiprime rings
to semiprime Γ-rings although the basic commutator formulas have to be changed.
For convenience aαcβb − aβcαb is usually denoted by the symbol a[α, β]
c
b although
the form [α, β]
c
= αc β − βcα has not any sense.
A map f from a ring R into itself is called strong commutativity pre serving (scp)
on a subset S of R if [f(x), f (y)] = [x, y] holds for all x, y ∈ S. The notion of a
strong commutativity preserving map was first introduced by Bell and Mas on [4].
Bell and Daif [3] gave characterizatio n of sc p derivations and endomorphisms on
onesided ideals of semiprime rings. Breˇsar and Miers [5] also described scp additive
maps on a semiprime ring.
Naturally a map f from a Γ -ring M into itself is called strong commutativity
preserving (scp) on a subset S of M if [f (x), f (y)]
α
= [x, y]
α
holds for all x, y ∈ S
and α ∈ Γ. In this paper, we also obtain that a semiprime Γ-ring with a strong

LEFT DERIVATIONS AND STRONG COMMUTATIVITY PRESERVING MAPS 3
commutativity preserving derivation on itself must be commutative and that the
strong commutativity preserving endomorphism on a semiprime Γ-r ing M must be
the form σ(x) = x + ζ(x) where ζ is a map from M into its center Z(M ), which
have ever been obtained for semiprime rings by Bell and Daif ([3, Corollary 1] and
the results implied by [3, Theorem 3]).
2. Left derivations on semiprime Γ-rings
The method employed in this paper is straightforward c omputation. Firstly we
introduce some remarks.
Remark 2.1. Let M b e a Γ-ring and δ : M → M a left derivation. Then both
δ([a, b ]
α
) = 0 and [c, b]
β
αδ(a) = aαcβδ(b) − cβaαδ(b) hold for all a, b, c ∈ M and
α, β ∈ Γ.
Proof. Using the definition of a left derivation on a Γ-ring and δ((aαb)βc) =
δ(aα(bβc)) for all a, b, c ∈ M and α, β ∈ Γ, we can obtain the co nc lusion through
the straightforward computation. 
The following simple observation is important for (semi-)prime Γ-rings.
Remark 2.2. Let M be a Γ-ring with c ∈ Z(M), a
1
, . . . , a
n
∈ M and β
1
, . . . , β
n
∈
Γ.Then
cβ
1
a
1
· · · β
n
a
n
= a
1
β
σ(1)
· · · a
i
β
σ(i)
cβ
σ(i+1)
a
i+1
· · · β
σ(n)
a
n
for all i ∈ {1, . . . , n} and σ ∈ S
n
the symmetric g roup of degree n.
Proof. It is easy to see that for each i ∈ {1, 2, . . . , n}
cβ
1
a
1
· · · β
n
a
n
= a
1
β
1
· · · a
i
β
i
cβ
i+1
a
i+1
· · · β
n
a
n
since c ∈ Z(M). So it is sufficient to prove the equation
cβ
1
a
1
· · · β
n
a
n
= cβ
σ(1)
a
1
· · · β
σ(n)
a
n
for all σ ∈ S
n
in order to complete the proof. Now we prove this equation by
induction on n. For n = 1 the equation is obvious. Set n > 1. Suppose that for
every k < n the equation holds . If σ(n) = n, then there exists τ ∈ S
n−1
such that
τ(i) = σ(i) for all 1 ≤ i ≤ n − 1. By the inductive assumption we get
cβ
1
a
1
· · · β
n−1
a
n−1
= cβ
τ (1)
a
1
· · · β
τ (n−1)
a
n−1
= cβ
σ(1)
a
1
· · · β
σ(n−1)
a
n−1
.
Hence cβ
1
a
1
· · · β
n
a
n
= cβ
σ(1)
a
1
· · · β
σ(n)
a
n
. If σ(n) 6= n, then by the inductive
assumption and c ∈ Z(M ) we deduce that
cβ
1
a
1
· · · β
n
a
n
= (cβ
σ(n)
a
1
β
j
1
· · · a
n−2
β
j
n−2
a
n−1
)β
n
a
n
= cβ
σ(n)
(a
1
β
j
1
· · · a
n−2
β
j
n−2
a
n−1
β
n
a
n
)
= (a
1
β
j
1
· · · a
n−2
β
j
n−2
a
n−1
β
n
a
n
)β
σ(n)
c
= (cβ
j
1
a
1
· · · β
j
n−2
a
n−2
β
n
a
n−1
)β
σ(n)
a
n
= cβ
σ(1)
a
1
· · · β
σ(n−1)
a
n−1
β
σ(n)
a
n
,
which completes the proof. 
Remark 2.2 is important for prime or semiprime Γ-rings although it is easy to
prove. For example if a semiprime Γ-ring M has its center nonze ro then every
commutator formula has a neat form with the help of any nonzero center element,
i.e., dα[aβb, c ]
γ
= dα(aβ[b, c]
γ
+ [a, c]
γ
βb) holds for all d ∈ Z(M ), a, b, c ∈ M

4 XIAOWEI XU, JING MA, AND YUAN ZHOU
and α, β, γ ∈ Γ. Particularly for a prime Γ-ring with its center nonzero every
commutator formula has the same fo rm as one in a prime ring. That is [a βb, c]
γ
=
aβ[b, c]
γ
+ [a, c]
γ
βb always holds for all a, b, c ∈ M and β, γ ∈ Γ in a prime Γ-ring
with its c enter nonzero. But in general for most prime or semiprime Γ-rings the
center is zero. However Remark 2.2 is still useful for a semiprime Γ-ring when
proving some results on commutativity even though the center is equal to z ero.
The following characteriz ation for left derivations in prime or semiprime Γ-rings
will make use of this observation.
Theorem 2.3. A left derivation of a semiprime Γ-ring M must map M into its
center.
Proof. Let δ : M → M be a le ft derivation. By Remark 2 .1 we have
[c, b]
β
αδ(a) = aαcβδ(b) − cβaαδ(b), a, b, c ∈ M, α, β ∈ Γ.(2.1)
Putting b = [b, d]
γ
in (2.1) and applying Remark 2.1 we obtain
[c, [b, d]
γ
]
β
αδ(a) = 0 , a, b, c, d ∈ M, α, β, γ ∈ Γ.(2.2)
Then for all a, b, c, d, a
1
∈ M and α, β, γ, γ
1
∈ Γ
[a, [c, [b, d]
γ
]
β
]
α
γ
1
δ(a
1
) = aα[c, [b, d]
γ
]
β
γ
1
δ(a
1
) = 0.(2.3)
Putting a = aγ
1
a
1
in (2.2) and applying (2.3) we get that
0 = [c, [b, d]
γ
]
β
αa
1
γ
1
δ(a) + [c, [b, d]
γ
]
β
αaγ
1
δ(a
1
)
= [c, [b, d]
γ
]
β
αa
1
γ
1
δ(a) + [c, [b, d]
γ
]
β
αaγ
1
δ(a
1
) + [a, [c, [b, d]
γ
]
β
]
α
γ
1
δ(a
1
)
= [c, [b, d]
γ
]
β
αa
1
γ
1
δ(a) + aα[c, [b, d]
γ
]
β
γ
1
δ(a
1
)
= [c, [b, d]
γ
]
β
αa
1
γ
1
δ(a)
holds for all a, b, c, d, a
1
∈ M and α, β, γ, γ
1
∈ Γ . That is [c, [b, d]
γ
]
β
ΓMΓδ(a) = 0
holds for all a, b, c, d ∈ M a nd β, γ ∈ Γ . Hence [c , [b, δ(a)]
γ
]
β
ΓMΓ[c, [b, δ(a)]
γ
]
β
= 0
holds for all a, b , c ∈ M and β, γ ∈ Γ. Then [δ(a), b]
γ
∈ Z(M ) for all a, b ∈ M and
γ ∈ Γ since M is semipr ime.
Put a = dγa in (2.1) then for all a, b, c, d ∈ M and α, β, γ ∈ Γ
[c, b]
β
αdγδ(a) + [c, b]
β
αaγδ(d) = dγaαcβδ(b) − cβdγaαδ(b).(2.4)
Multiply the two sides of (2.1) with “dγ” from the left hand side then for all
a, b, c, d ∈ M and α, β, γ ∈ Γ
dγ[c, b]
β
αδ(a) = dγaαcβδ(b) − dγcβaαδ(b).(2.5)
Compute (2.4)−(2.5) then for all a, b, c, d ∈ M and α, β, γ ∈ Γ
[c, b]
β
αdγδ(a) + [c, b]
β
αaγδ(d) − dγ[c, b]
β
αδ(a) = dγcβaαδ(b) − cβdγaαδ(b).
(2.6)
Setting c = b, d = [δ(b), b ]
β
σδ(b) in (2.6), and then applying [δ(b), b]
β
∈ Z(M ) and
Remark 2.2 we have that
[δ(b), b ]
β
σ[δ(b), b]
β
γaαδ(b) = 0
holds fo r all a, b ∈ M and α, β, γ, σ ∈ Γ. That is
[δ(b), b ]
β
σ[δ(b), b]
β
ΓMΓ[δ(b), b]
β
σ[δ(b), b]
β
= 0
holds for all b ∈ M and β, σ ∈ Γ. Then [δ(b), b]
β
σ[δ(b), b]
β
= 0 holds for all b ∈ M
and β, σ ∈ Γ. Hence [δ(b), b]
β
= 0 for all b ∈ M and β ∈ Γ since [δ(b), b]
β
∈ Z(M )

LEFT DERIVATIONS AND STRONG COMMUTATIVITY PRESERVING MAPS 5
the center of the semiprime Γ ring M. Setting c = δ(b), d = [d, δ(b)]
β
σd in (2.6),
and then applying [d, δ(b)]
β
∈ Z(M ) and Remark 2.2 we deduce that
[d, δ(b)]
β
σ[d, δ(b )]
β
γaαδ(b) = 0
holds for all a, b, d ∈ M and α, β, γ, σ ∈ Γ . Similar to proving that [δ(b), b]
β
= 0 we
also obtain [d, δ(b)]
β
= 0 for all b, d ∈ M and β ∈ Γ, which completes the proof. 
Furthermore we will get the result for prime Γ-rings.
Corollary 2.4. A prime Γ-ring with a nonzero left derivation must be commutative.
Proof. Let δ : M → M be a nonzero left derivation of a prime Γ-ring M . By
Theorem 2.3, Remark 2.1 and 2.2
[c, b]
β
αδ(a) = [a, c]
β
αδ(b), a, b, c ∈ M, α, β ∈ Γ.(2.7)
Putting c = a and applying δ(a) ∈ Z(M ) we deduce that [a, b]
β
ΓMΓδ(a) = 0 holds
for all a, b ∈ M and β ∈ Γ. Hence for every a ∈ M we deduce that either a ∈ Z(M )
or δ(a) = 0. That is M = ker δ ∪ Z(M) is the union of its two subgroups. Thus M
is commutative since δ 6= 0. 
3. Strong commutativity preserving maps on semiprime Γ-rings
The following results (Theorem 3.1 and 3.2) on scp maps have been proved
in semiprime rings by Bell and Daif (see [3] for reference in which more general
situation were considered). Here we will indicate that some results appea ring in
[3] also hold in semiprime Γ-rings altho ugh the commutator formulas have become
complicated.
Theorem 3.1. A semiprime Γ-ring with a strong commutativity preserving deriva-
tion must be commutative.
Proof. Suppose that M is a semiprime Γ-ring with a strong commutativity
preserving derivation δ on M . That is [δ(x), δ(y)]
α
= [x, y]
α
for all x, y ∈ M and
α ∈ Γ . Then for all x, y, z ∈ M and α, β ∈ Γ
[xβz, y]
α
= [δ(xβz), δ(y)]
α
= [δ(x)βz, δ(y)]
α
+ [xβδ(z), δ(y)]
α
.
Moreover we get that
xβ[z, y]
α
+ [x, y]
α
βz + x[β, α]
y
z
= δ(x)β[z, δ(y)]
α
+ [δ(x), δ(y)]
α
βz + δ(x)[β, α]
δ(y)
z+
xβ[δ(z), δ(y)]
α
+ [x, δ(y)]
α
βδ(z) + x[β, α]
δ(y)
δ(z)
holds fo r all x, y, z ∈ M and α, β ∈ Γ. That is fo r all x, y, z ∈ M and α, β ∈ Γ
x[β, α]
y
z = δ(x)β[z, δ(y)]
α
+ δ(x)[β, α]
δ(y)
z + [x, δ(y)]
α
βδ(z) + x[β, α]
δ(y)
δ(z).
(3.1)
Putting z = zγt in (3.1) we obtain that for all x, y, z, t ∈ M and α, β, γ ∈ Γ
x[β, α]
y
zγt = δ(x)β

zγ[t, δ(y)]
α
+ [z, δ(y)]
α
γt + z[γ, α]
δ(y)
t

+ δ(x)[β, α]
δ(y)
zγt + [x, δ(y)]
α
βδ(z)γt
+ [x, δ(y)]
α
βzγδ(t) + x[β, α]
δ(y)
δ(z)γt + x[β, α]
δ(y)
zγδ(t).
(3.2)
Multiplying the two sides of (3.1) by “γt” from the right hand side, and then
comparing with (3.2) we deduce that for all x, y, z, t ∈ M and α, β, γ ∈ Γ
δ(x)β

zγ[t, δ(y)]
α
+ z[γ, α]
δ(y)
t

+

[x, δ(y)]
α
βz + x[β, α]
δ(y)
z

γδ(t) = 0.(3.3)

6 XIAOWEI XU, JING MA, AND YUAN ZHOU
Setting t = δ(y) and γ = α in (3.3) we get that for all x, y, z ∈ M and α, β ∈ Γ
(xβδ(y)αz − δ(y)αxβz)αδ
2
(y) = 0.(3.4)
Putting α = α + γ into (3.4) and applying (3.4) we deduce that for all x, y, z ∈ M
and α, β, γ ∈ Γ
(xβδ(y)αz − δ(y)αxβz)γδ
2
(y) = −(xβδ(y)γz − δ(y)γxβz)αδ
2
(y).
Then by (3.4) for all x, y, z ∈ M and α, β, γ ∈ Γ
(xβδ(y)αz − δ(y)αxβz)γδ
2
(y)ΓM Γ(xβδ(y)αz − δ(y)αxβz)γδ
2
(y)
= − (xβδ(y)αz − δ(y)αxβz)γδ
2
(y)ΓM Γ(xβδ(y)γz − δ(y)γxβz)αδ
2
(y)
= 0,
which implies (xβδ(y)αz − δ(y)αxβz)γδ
2
(y) = 0 for all x, y, z ∈ M and α, β, γ ∈ Γ
since M is semiprime. Set β = α and x = δ(x) in (xβδ(y)αz−δ(y)αxβz)γδ
2
(y) = 0.
Then for all x, y, z ∈ M and α ∈ Γ
[x, y]
α
αzΓMΓ[x, y]
α
αz
= [δ(x), δ(y)]
α
αzΓMΓ[δ(x), δ(y)]
α
αz
= [δ(x), δ(y)]
α
αzΓMΓ[δ
2
(x), δ
2
(y)]
α
αz
= 0,
which s hows [x, y]
α
αz = 0 for all x, y, z ∈ M and α ∈ Γ . Then for all t, x, y, z ∈ M
and α, γ ∈ Γ
0 = [tγx, y]
α
αz = tγ[x, y]
α
αz + (tγyαx − yαtγx)αz = (tγyαx − yαtγx)αz.(3.5)
Putting α = α + β into (3.5) and applying (3.5) we deduce that for all t, x, y, z ∈ M
and α, β, γ ∈ Γ
(tγyαx − yαtγx)βz = −(tγyβx − yβtγx)αz.
Then by (3.5) for all t, x, y, z ∈ M and α, β, γ ∈ Γ
(tγyαx − yαtγx)βzΓM Γ (tγyαx − yαtγx)βz
= − (tγyαx − yαtγx)βzΓM Γ(tγyβx − yβtγx)αz
= 0,
which implies (tγyαx−yαtγx)βz = 0 for all t, x, y, z ∈ M and α, β, γ ∈ Γ. Moreover
tγyαx − yαtγx = 0 for all t, x, y ∈ M and α, γ ∈ Γ. Hence for all t, x, y ∈ M and
α, γ ∈ Γ
[tγx, y]
α
= tγxαy − yαtγx = tγxαy − tγyαx = tγ[x, y]
α
.
Then for all x, y, z ∈ M and α, β ∈ Γ
xβ[z, y]
α
= [xβz, y]
α
= [δ(xβz), δ(y)]
α
= [δ(x)βz, δ(y)]
α
+ [xβδ(z), δ(y)]
α
= δ(x)β[z, δ(y)]
α
+ xβ[δ(z), δ(y)]
α
.
So δ(x)β[z, δ(y)]
α
= 0 for all x, y, z ∈ M and α, β ∈ Γ. Hence for all t, x, y, z ∈ M
and α, β, γ ∈ Γ
0 = δ(x)β[tγz, δ(y)]
α
= δ(x)βtγ[z, δ(y)]
α
,
which implies [δ(x), δ(y)]
α
ΓMΓ[δ(x), δ(y)]
α
= 0 fo r all x, y ∈ M and α ∈ Γ. Thus
[x, y]
α
= [δ(x), δ(y)]
α
= 0 for all x, y ∈ M and α ∈ Γ completes the proof. 
It is implied by [3, Theorem 3] that for a semiprime ring R with an endomorphism
T , then T is scp o n R if and only if T (x) = x + ζ(x) for all x ∈ R where ζ is a map
from R into its center. We will show this also holds for s emiprime Γ-rings.

LEFT DERIVATIONS AND STRONG COMMUTATIVITY PRESERVING MAPS 7
Theorem 3.2. Let M be a semiprime Γ-ring with an endomorphism σ. Then σ is
strong commutativity preserving on M if and only if there exists a map ζ : M →
Z(M ) such that σ(x) = x + ζ(x) for all x ∈ M .
Proof. We will only consider the necessity since the sufficiency is obvious.
From [σ(xαz), σ(x)]
α
= [xαz, x]
α
for all x, z ∈ M and α ∈ Γ we obtain (σ(x) −
x)α[z, x]
α
= 0. Then for all x, y, z ∈ M and α, β ∈ Γ
0 = (σ(x) − x)α[yβz, x]
α
= (σ(x) − x)α[y, x]
α
βz + (σ(x) − x)α(yβzαx − yαxβz)
= (σ(x) − x)α(yβzαx − yαxβz).
(3.6)
Linearizing α in (3.6) we have that for all x, y, z ∈ M and α, β, γ ∈ Γ
(σ(x) − x)α(yβzγx − yγxβz) = −(σ(x) − x)γ(yβzαx − yαxβz).
Then by (3.6) for all x, y, z ∈ M and α, β, γ ∈ Γ
(σ(x) − x)α(yβzγx − yγxβz)ΓM Γ(σ(x) − x)α(yβzγx − yγxβz)
= −(σ(x) − x)α(yβzγx − yγxβz)ΓMΓ(σ(x) − x)γ(yβzαx − yαxβz)
= 0
which implies (σ(x) − x)α(yβzγx − yγxβz) = 0 fo r all x, y, z ∈ M and α, β, γ ∈ Γ.
Linearizing x in (σ(x) − x)α(yβzγx − yγxβz) = 0 we get for all x, y, z, t ∈ M and
α, β, γ ∈ Γ.
(σ(x) − x)α(yβzγt − yγtβz) = −(σ(t) − t)α(yβzγx − yγxβz).
Then for all x, y, z, t ∈ M and α, β, γ ∈ Γ
(σ(x) − x)α(yβzγt − yγtβz)ΓM Γ(σ(x) − x)α(y βzγt − yγtβz)
= −(σ(x) − x)α(yβzγt − yγtβz)ΓM Γ(σ(t) − t)α(yβzγx − yγxβz)
= 0,
which shows (σ (x) − x)α(yβzγt − yγtβz) = 0 for all x, y, z, t ∈ M and α, β, γ ∈ Γ.
Then for all x, y, z ∈ M and α, β, γ ∈ Γ we obtain both
((σ(y) − y)βzγx − (σ(y) − y)γxβz)ΓM Γ((σ(y) − y)βzγx − (σ(y) − y)γxβz) = 0
and
(yβ(σ(x) − x)γz − yγzβ(σ(x) − x))ΓMΓ(yβ(σ(x) − x)γz − yγzβ(σ(x) − x)) = 0.
So both (σ(y)−y)βzγx−(σ (y)−y)γxβz = 0 and yβ(σ(x)−x)γz−yγzβ(σ(x)−x) = 0
hold for all x, y, z ∈ M and β, γ ∈ Γ. Hence for all x, y, z ∈ M and α, β ∈ Γ we get
[(σ(x) − x)αy, z]
β
= [σ(x) − x, z]
β
αy, [yα(σ(x) − x), z]
β
= [y, z]
β
α(σ(x) − x)
and [yαz, σ(x) − x]
β
= [y, σ(x) − x]
β
αz. Then for all x, y, z ∈ M and α, β ∈ Γ
0 = [σ(x)ασ(y), σ(z)]
β
− [xαy, z]
β
=

[σ(x)ασ(y), σ(z)]
β
− [σ(x)αy, σ(z)]
β

+

[σ(x)αy, σ(z)]
β
− [xαy, σ(z)]
β

+

[xαy, σ(z)]
β
− [xαy, z]
β

= [σ(x)α(σ(y) − y), σ(z)]
β
+ [(σ(x) − x)αy, σ(z)]
β
+ [xαy, σ(z) − z]
β
= [σ(x), σ(z)]
β
α(σ(y) − y) + [σ(x) − x, σ(z)]
β
αy + [x, σ(z) − z]
β
αy
= [x, z]
β
α(σ(y) − y) + [x, z]
β
αy − [x, σ(z)]
β
αy + [x, σ(z) − z]
β
αy
= [x, z]
β
α(σ(y) − y).
Hence for all x, y, z ∈ M and α, β, γ ∈ Γ
0 = [(σ(y) − y)γx, z]
β
α(σ(y) − y) = [σ(y) − y, z]
β
γxα(σ(y) − y).

8 XIAOWEI XU, JING MA, AND YUAN ZHOU
Thus [σ(y) − y, z]
β
ΓMΓ[σ(y) − y, z]
β
= 0 holds for all y, z ∈ M and β ∈ Γ. Then
[σ(y) − y, z]
β
= 0 for all y, z ∈ M and β ∈ Γ completes the proof. 
For prime Γ-ring s we get a further result.
Corollary 3.3. In a noncommutative prime Γ-ring M the identity map is the
unique strong commutativity preserving endomorphism on M .
Proof. Let σ : M → M be a str ong commutativity preserving endomorphism
on M. Then by Theorem 3.2 there exists a map ζ : M → Z(M) such that σ(x) =
x + ζ(x) for all x ∈ M . For all x, y ∈ M and α ∈ Γ
xαy + ζ(xαy) = σ(xαy) = σ(x)ασ(y) = xαy + ζ(x)αy + ζ(y)αx + ζ(x)αζ(y).
(3.7)
Then for x ∈ M such that ζ(x) 6= 0 we get ζ(x)α[x, y]
β
= 0 for all y ∈ M and
α, β ∈ Γ from (3.7) and Remark 2.2. So [x, y]
β
= 0 for all y ∈ M and β ∈ Γ since M
is prime and 0 6= ζ(x) ∈ Z(M ). That is fo r every x ∈ M once ζ(x) 6= 0 we always
have x ∈ Z(M). Now we assume that there exists x
0
∈ M such that ζ(x
0
) 6= 0 and
proceed to obtain a contradiction so that the proof could be completed. We may
choose an element y
0
∈ M \ Z(M ) since M is noncommutative. Then ζ(y
0
) = 0,
i.e. σ(y
0
) = y
0
. Putting x = x
0
and y = y
0
in (3.7) we obtain ζ(x
0
)αy
0
∈ Z(M ).
So ζ(x
0
)α[y
0
, z]
β
= 0 holds for all z ∈ M and α, β ∈ Γ which means y
0
∈ Z(M ) a
contradiction. 
Next we will give an example showing that there exists a non-identity strong
commutativity preserving endomorphism on certain noncommutative semiprime Γ-
ring.
Example Let R = M
2
(C) × C where C is the field of complex numbers. Then
R is a noncommutative semiprime ring and σ : R → R such that σ(A, a) = (A,
a)
for all (A, a) ∈ R is a non-ide ntity strong co mmutativity preserving automorphism
on R. 
Acknowledgement We would like to thank Professor Paul for sending the paper
required by us during prepar ing this manusc ript.
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LEFT DERIVATIONS AND STRONG COMMUTATIVITY PRESERVING MAPS 9
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Xiaowei Xu: College of Mathematics, Jilin University, Changchun 130012, PR China
E-mail address: xuxw@jlu.edu.cn
Jing Ma: College of Mathematics, Jilin University, Changchun 130012, PR China
E-mail address: jma@jlu.edu.cn
Yuan Zhou: College of Mathematics, Jilin University, Changchun 130012, PR China
E-mail address: zhouyuan150630@126.com