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Southeast Asian
Bulletin of
Mathematics
c
SEAMS. 2011
Southeast Asian Bulletin of Mathematics (2011) 35: 29–34
Jordan k-Derivations of Completely Prime ΓN-Rings
Sujoy Chakraborty
Department of Mathematics, Shahjalal University of Science and Technology, Sylhet-
3114, Bangladesh
Email: sujoy chbty@yahoo.com
Akhil Chandra Paul
Department of Mathematics, Rajshahi University, Rajshahi-6205, Bangladesh
E-mail: acpaulru math@yahoo.com
Received 7 May 2007
Accepted 29 April 2009
Communicated by Nguyen Van Sanh
AMS Mathematics Subject Classification (2000): 16N60, 16W25, 16U80
Abstract. This article is based on some derivations of certain gamma rings. Giving the
definitions of k-derivation and Jordan k-derivation of a gamma ring as well as that of
certain gamma rings, some important results relating to these concepts are developed.
Clearly, every k-derivation of a Γ-ring Mis also a Jordan k-derivation of M. But, the
converse is not true in general. We are to show that every Jordan k-derivation of a
2-torsion free completely prime ΓN-ring is also a k-derivation of the same.
Keywords: Derivation; k-derivation; Jordan k-derivation; Gamma ring.
1. Introduction
Let Mand Γ be additive abelian groups. If there is a mapping (a, α, b)7→ aαb
of M×Γ×M→Msuch that (a) (a+b)αc =aαc +bαc,a(α+β)b=aαb +aβb,
aα(b+c) = aαb +aαc and (b) (aαb)βc =aα(bβ c) hold for all a, b, c ∈Mand
α, β ∈Γ, then Mis called a Γ-ring (in the sense of Barnes [1]). [4]
In addition to the above, if there exists another map (α, a, β)7→ αaβ of
Γ×M×Γ→Γ such that (a*) (α+β)aγ =αaγ +βaγ,α(a+b)β=αaβ +αbβ,
αa(β+γ) = αaβ +αaγ, (b*) (aαb)βc =a(αbβ)c=aα(bβc) and (c*) aαb = 0
30 S. Chakraborty and A.C. Paul
implies α= 0 for all a, b ∈Mand α, β, γ ∈Γ, then Mis called a Γ-ring in the
sense of Nobusawa [5], or simply, a ΓN-ring [7]. Clearly, Mis a ΓN-ring implies
that Γ is an M-ring.
Let Mbe a Γ-ring. Then Mis said to be 2-torsion free if and only if 2a= 0
implies a= 0 for all a∈M. And, Mis said to be commutative if and only
if aαb =bαa holds for all a, b ∈Mand α∈Γ. Besides, Mis called a prime
gamma ring if and only if aΓMΓb= 0 (with a, b ∈M) implies a= 0 or b= 0.
And, Mis called semiprime if and only if aΓMΓa= 0 (with a∈M) implies
a= 0. Moreover, Mis said to be completely prime if and only if aΓb= 0 (with
a, b ∈M) implies a= 0 or b= 0. And, Mis called completely semiprime if and
only if aΓa= 0 (with a∈M) implies a= 0. Note that every prime Γ-ring is
semiprime and every completely prime Γ-ring is prime.
Now, let Mbe a Γ-ring and let d:M→Mbe an additive mapping. If the
condition d(aαb) = d(a)αb +aαd(b) holds for all a, b ∈Mand α∈Γ, then dis
said to be a derivation of M.
If Mis a Γ-ring, and d:M→Mand k: Γ →Γ are additive mappings such
that d(aαb) = d(a)αb +ak(α)b+aαd(b) is satisfied for all a, b ∈Mand α∈Γ,
then dis called a k-derivation of M.
Let Mbe a Γ-ring, and let d:M→Mand k: Γ →Γ be additive mappings.
If d(aαa) = d(a)αa +ak(α)a+aαd(a) holds for all a∈Mand α∈Γ, then dis
called a Jordan k-derivation of M.
Note that M. Sapanci and A. Nakajima [6] has introduced the notions of
derivation and Jordan derivation of a Γ-ring, and that H. Kandamar has in-
troduced the concept of k-derivation of a Γ-ring in a considerable detail in [3].
Later, Y. Ceven and M.A. Ozturk [2] has developed the concept of Jordan gen-
eralized derivation of a Γ-ring. Following those earlier developments successively
we introduce here the concept of Jordan k-derivation of a Γ-ring as above.
2. Main Results
From the very definitions it is obvious that every k-derivation of a Γ-ring Mis
also a Jordan k-derivation of M. Of course, its converse statement is not true
in general. This article aims to show that the converse result is also true if we
choose Mas a 2-torsion free completely prime ΓN-ring. To achieve this goal we
need to develop some important results in the following way.
Lemma 2.1. Let Mbe a ΓN-ring and let dbe a Jordan k-derivation of M. Then
for all a, b, c ∈Mand α, β ∈Γ, the following statements hold:
(i) d(aαb +bαa) = d(a)αb +d(b)αa +ak(α)b+bk(α)a+aαd(b) + bαd(a);
(ii) d(aαbβa +aβbαa) = d(a)αbβa +d(a)βbαa +ak(α)bβa +ak(β)bαa +
aαd(b)βa +aβd(b)αa +aαbk(β)a+aβbk(α)a+aαbβd(a) + aβbαd(a).
In particular, if Mis 2-torsion free, then
(iii) d(aαbαa) = d(a)αbαa +ak(α)bαa +aαd(b)αa +aαbk(α)a+aαbαd(a);
Jordan k-Derivations of Completely Prime ΓN-Rings 31
(iv) d(aαbαc +cαbαa) = d(a)αbαc +d(c)αbαa +ak(α)bαc +ck(α)bαa +
aαd(b)αc +cαd(b)αa +aαbk(α)c+cαbk(α)a+aαbαd(c) + cαbαd(a).
If Mis 2-torsion free and aαbβc =aβbαc for all a, b, c ∈Mand α, β ∈Γ,
then
(v) d(aαbβa) = d(a)αbβa +ak(α)bβa +aαd(b)βa +aαbk(β)a+aαbβd(a);
(vi) d(aαbβc +cαbβa) = d(a)αbβc +d(c)αbβa +ak(α)bβc +ck(α)bβa +
aαd(b)βc +cαd(b)βa +aαbk(β)c+cαbk(β)a+aαbβd(c) + cαbβd(a).
Proof. Compute d((a+b)α(a+b)) and cancel the like terms from both sides
to obtain (i). Then replace aβb +bβa for bin (i) to get (ii). Since Mis 2-
torsion free, (iii) is easily obtained by replacing αfor βin (ii), and then (iv) is
obtained by replacing a+cfor ain (iii). Again, since Mis 2-torsion free and
aαbβc =aβbαc for all a, b, c ∈Mand α, β ∈Γ, (v) follows from (ii) and then
finally, (vi) is obtained by replacing a+cfor ain (v).
Lemma 2.2. Let dbe a Jordan k-derivation of a 2-torsion free ΓN-ring M. Then
for all b∈Mand β∈Γ,k(βbβ) = k(β)bβ +βd(b)β+βbk(β).
Proof. For all a∈Mand α∈Γ, we have d(aαa) = d(a)αa +ak(α)a+aαd(a).
Let b∈Mand β∈Γ. Then putting βbβ for α, we get
d(aβbβa) = d(a)βbβa +ak(βbβ)a+aβbβd(a).
Expanding the LHS by Lemma 2.1(iii), we obtain
a(k(βbβ)−k(β)bβ −βd(b)β−βbk(β))a= 0.
So, the Nobusawa condition (c*) proves the lemma.
Lemma 2.3. If dis a Jordan k1-derivation as well as a Jordan k2-derivation of
a 2-torsion free ΓN-ring M, then k1=k2.
Proof. It is obvious.
Hence, it follows that if dis a Jordan k-derivation of a 2-torsion free ΓN-ring
M, then kis uniquely determined.
Definition 2.4. Let Mbe a Γ-ring, a, b ∈Mand α∈Γ. Then [a, b]α=aαb −bαa
is called the commutator of aand bwith respect to α.
Lemma 2.5. If Mis a Γ-ring, then for all a, b, c ∈Mand α, β ∈Γ:
(i) [a, b]α+ [b, a]α= 0;
(ii) [a+b, c]α= [a, c]α+ [b, c]α;
(iii) [a, b +c]α= [a, b]α+ [a, c]α;
32 S. Chakraborty and A.C. Paul
(iv) [a, b]α+β= [a, b]α+ [a, b]β.
Proof. The proofs are clear.
Thus we conclude that a Γ-ring Mis commutative if and only if [a, b]α= 0
holds for all a, b ∈Mand α∈Γ.
Definition 2.6. Let dbe a Jordan k-derivation of a ΓN-ring M. For a, b ∈M
and α∈Γ, we define Gα(a, b) = d(aαb)−d(a)αb −ak(α)b−aαd(b).
Lemma 2.7. If dis a Jordan k-derivation of a ΓN-ring M, then the following
are true for all a, b, c ∈Mand α, β ∈Γ:
(i) Gα(a, b) + Gα(b, a) = 0;
(ii) Gα(a+b, c) = Gα(a, c) + Gα(b, c);
(iii) Gα(a, b +c) = Gα(a, b) + Gα(a, c);
(iv) Gα+β(a, b) = Gα(a, b) + Gβ(a, b).
Proof. Obvious.
As a result, we can say that dis a k-derivation of a ΓN-ring Mif and only if
Gα(a, b) = 0 holds for all a, b ∈Mand α∈Γ.
Lemma 2.8. Let Mbe a ΓN-ring, a, b ∈Mand α, γ ∈Γ. If dis a Jordan
k-derivation of M, then Gα(a, b)γ[a, b]α+ [a, b]αγGα(a, b) = 0.
Proof. Compute d((aαb)γ(bαa)+(bαa)γ(aαb)), using Lemma 2.1(i) and also,
d(a(αbγbα)a+b(αaγ aα)b) = d(a(αbγbα)a) + d(b(αaγ aα)b), using Lemma 2.2.
Since these two are equal, cancelling the similar terms from both sides of this
equality and then rearranging them using Lemma 2.7(i), we get the result.
Lemma 2.9. Let Mbe a 2-torsion free completely semiprime ΓN-ring. Suppose
a, b ∈Msuch that aΓb+bΓa= 0. Then aΓb=bΓa= 0.
Proof. By using the hypothesis, we obtain
(aΓb)Γ(aΓb) = −(bΓa)Γ(aΓb) = −(b(ΓaΓ)a)Γb
= (a(ΓaΓ)b)Γb=aΓ(aΓb)Γb=−aΓ(bΓa)Γb
=−(aΓb)Γ(aΓb).
Thus, 2(aΓb)Γ(aΓb) = 0. Since Mis 2-torsion free, (aΓb)Γ(aΓb) = 0. So, by
the completely semiprimeness of M, we get aΓb= 0. This completes the proof.
Jordan k-Derivations of Completely Prime ΓN-Rings 33
Corollary 2.10. Let Mbe a 2-torsion free completely semiprime ΓN-ring. Then
for all a, b ∈Mand α, γ ∈Γ,Gα(a, b)γ[a, b]α= [a, b]αγGα(a, b) = 0.
Proof. Application of Lemma 2.9 into Lemma 2.8 gives this corollary.
Theorem 2.11. Let Mbe a 2-torsion free completely semiprime ΓN-ring. Then
for all a, b, c, d ∈Mand α, β, γ ∈Γ,Gα(a, b)γ[c, d]β= 0.
Proof. By putting a+cfor ain the first part of Corollary 2.10, we have
Gα(a, b)γ[c, b]α+Gα(c, b)γ[a, b]α= 0.
This yields,
Gα(a, b)γ[c, b]αγGα(a, b)γ[c, b]α=−Gα(a, b)γ[c, b]αγGα(c, b)γ[a, b]α= 0.
By the completely semiprimeness of M, we obtain Gα(a, b)γ[c, b]α= 0. Likewise,
by replacing b+dfor bin this equality, we get Gα(a, b)γ[c, d]α= 0. Then replace
α+βfor αhere to obtain Gα(a, b)γ[c, d]β+Gβ(a, b)γ[c, d]α= 0. So, it gives
Gα(a, b)γ[c, d]βγGα(a, b)γ[c, d]β=−Gα(a, b)γ[c, d]βγGβ(a, b)γ[c, d]α= 0.
Hence, by the completely semiprimeness of M, we obtain Gα(a, b)γ[c, d]β= 0.
Corollary 2.12. Let Mbe a 2-torsion free completely semiprime ΓN-ring. Then
for all a, b, c, d ∈Mand α, β, γ ∈Γ,[c, d]βγGα(a, b) = 0.
Proof. Proceeding in the same way as in the proof of Theorem 2.11, by the
similar replacements in the second part of Corollary 2.10, we get this result.
Theorem 2.13. Let Mbe a 2-torsion free completely prime ΓN-ring. Then every
Jordan k-derivation of Mis a k-derivation of M.
Proof. Let Mbe a 2-torsion free completely prime ΓN-ring. Suppose dis a
Jordan k-derivation of M. Then, from Theorem 2.11, the completely primeness
of Mimplies Gα(a, b) = 0 or [c, d]β= 0 for all a, b, c, d ∈Mand α, β ∈Γ.
Consider first [c, d]β= 0 for all c, d ∈Mand β∈Γ; then Gα(a, b) = 0 for
all a, b ∈Mand α∈Γ, which implies that dis a k-derivation of M. Also, if
[c, d]β= 0 for all c, d ∈Mand β∈Γ, then Mis commutative [by definition];
accordingly, 2d(aαb)=2d(a)αb + 2ak(α)b+ 2aαd(b) [from Lemma 2.1(i)], and
hence, dis then a k-derivation of M[since Mis 2-torsion free].
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Hacettepe J. Math. and Stat. 33 (2004) 11–14.
[3] H. Kandamar, The k-derivation of a Gamma ring, Turk. J. Math. 24 (2000) 221–
231.
34 S. Chakraborty and A.C. Paul
[4] Md. Mahbubur Rashid and A.C. Paul, Regular Gamma rings, Southeast Asian
Bull. Math. 31 (5) (2007) 933–947.
[5] N. Nobusawa, On the generalization of the ring theory, Osaka J. Math. 1(1964)
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