# Derivations Acting as Homomorphisms and as Anti homomorphisms in σ Lie Ideals of σ Prime Gamma Rings

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## Gamma Rings

1

### S. Chakraborty

2,∗

1Department of Mathematics, Rajshahi University, Rajshahi-6205, Bangladesh

2Department of Mathematics, Shahjalal University of Science and Technology, Sylhet-3114, Bangladesh

### Abstract

LetU be a non-zero σ-square closed Lie ideal of a 2-torsion freeσ-primeΓ-ringM satisfying the condition aαbβc=aβbαcfor alla, b, c∈ M andα, β Γ, and letdbe a derivation ofM such that = σd. We prove here that (i) ifdacts as a homomorphism onU, thend= 0orU ⊆Z(M), whereZ(M)is the centre ofM; and (ii) ifdacts as an anti-homomorphism onU, thend= 0orU ⊆Z(M).

### Keywords

σ-Prime Gamma Ring, Lie Ideal, Derivation, Involution

### Mathematics Subject Classification 2010

16W10, 16W25, 16U80

### Introduction

SupposeM andΓare additive abelian groups. If there exists a mapping(a, α, b)7→aαbofΓ×M →M satisfying (a)

(a+b)αc=aαc+bαc,a(α+β)b=aαb+aβb,aα(b+c) =aαb+aαc, and (b)(aαb)βc=(bβc)for alla, b, c∈M andα, β∈Γ, thenM is said to be aΓ-ring in the sense of Barnes [3]. The setZ(M)={a∈M :aαm=mαafor allα∈Γ

andm∈M}is called the center of theΓ-ringM.In this article,M will represent aΓ-ring with centreZ(M).

Recall thatM is said to be 2-torsion free if2a= 0witha ∈M, thena= 0. M is called prime if, for anya, b ∈M, aΓMΓb= 0impliesa= 0orb= 0. A mappingσ:M →Mis called an involution ifσ2(a) =a,σ(a+b) =σ(a) +σ(b) andσ(aαb) =σ(b)ασ(a)for alla, b∈M andα∈Γ.

AΓ-ringM equipped with an involutionσis said to be aσ-primeΓ-ring if for alla, b∈M,aΓMΓb= 0 =aΓMΓσ(b)

impliesa= 0orb= 0. It is noted that every primeΓ-ring having an involutionσisσ-prime, but the converse is in general not true. LetSaσ(M) ={a∈M :σ(a) =±a}, which represents the set of symmetric and skew-symmetric elements ofM. For anya, b∈M andα∈Γ, the symbol[a, b]αstands for the commutatoraαb−bαa. The basic commutator identities are

[aβb, c]α=[b, c]α+a[β, α]cb+ [a, c]αβband

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where[α, β]a=αaβ−βaα, for alla, b, c∈M andα, β∈Γ.Throughout the article, we shall consider the condition (*)aαbβc=aβbαc

for alla, b, c∈M andα, β∈Γ. Using this condition (*), the above identities reduce to

[aβb, c]α=[b, c]α+ [a, c]αβband[a, bβc]α=[a, c]α+ [a, b]αβc, which are extensively used in our results.

An additive subgroupU ofM is called a left (or, right) ideal ofM ifMΓU ⊂U (or,UΓM ⊂U), whereas U is called a (two-sided) ideal ofM ifU is a left as well as a right ideal ofM.

An additive subgroupU ofM is called a Lie ideal if[U, M⊂U. IfU is a Lie ideal ofM, thenU is called aσ-Lie ideal ifσ(U) =U, andU is called a square closed Lie ideal ifuαu∈U for allu∈U andα∈Γ. A Lie idealU ofM is said to be aσ-square closed Lie ideal if it is square closed andσ(U) =U.

An additive mappingd:M →M is called a derivation ifd(aαb) =d(a)αb+aαd(b)for alla, b∈M andα∈Γ. An additive mappingϕ:M →M is said to be a homomorphism ifϕ(aαb) =ϕ(a)αϕ(b)for alla, b∈M andα∈Γ. And, an additive mappingψ:M →M is called an anti-homomorphism ifψ(aαb) =ψ(b)αψ(a)for alla, b∈M andα∈Γ.

A derivationdofM is said to act as a homomorphism [resp. as an anti-homomorphism] on a subsetSofM ifd(aαb) =

d(a)αd(b)[resp.d(aαb) =d(b)αd(a)] for alla, b∈Sandα∈Γ.

In [4], Bell and Kappe proved that ifdis a derivation of a semiprime ringR which is either an endomorphism or an anti-endomorphism onR, thend = 0; whereas, the behavior ofdis somewhat restricted in case of prime rings in the way that ifdis a derivation of a prime ringRacting as a homomorphism or an anti-homomorphism on a non-zero right idealU ofR, thend= 0. Asma et. al. [1] extended this result of prime rings on square closed Lie ideals. Afterwards, the said result was extended toσ-prime rings by Oukhtite et. al. in [11].

InΓ-rings, Dey and Paul [9] proved that ifDis a generalized derivation of a primeΓ-ringM with an associated derivation d of M which acts as a homomorphism and an anti-homomorphism on a non-zero idealI of M, thend = 0or M is commutative. Afterwards, Chakraborty and Paul [6] worked onk-derivation of a semiprimeΓ-ring in the sense of Nobusawa [10] and proved thatd= 0wheredis ak-derivation acting as ak-endomorphism and as an anti-k-endomorphism.

In this article, the above mentioned results following [11] in classical rings are extended to those in gamma rings with derivation acting as a homomorphism and as an anti-homomorphism onσ-primeΓ-rings. Our objective is to prove that

(i) ifdis a derivation of a 2-torsion freeσ-primeΓ-ringM such that = σdand ifdacts as a homomorphism on a non-zeroσ-square closed Lie idealU ofM, thend= 0orU ⊆Z(M); and

(ii) ifdis a derivation of a 2-torsion freeσ-primeΓ-ringM satisfying the condition (*) such that =σdand ifdacts as an anti-homomorphism on a non-zeroσ-square closed Lie idealUofM, thend= 0orU ⊆Z(M).

### -rings

We start this section with an example which ensures the existence of an involution in aΓ-ring. We also give an example of a σ-primeΓ-ring which is not a primeΓ-ring along with an example of a Lie ideal in aΓ-ring.

Example 2.1LetM be aΓ-ring. DefineM1={(a, b) :a, b∈M}andΓ1={(α, α) :α∈Γ}. Addition and multiplication on M1 are defined as: (a, b) + (c, d) = (a+c, b+d) and(a, b)(α, α)(c, d) = (aαc, dαb). Under these addition and multiplication,M1is aΓ1-ring. Define a mappingσ:M1→M1byσ((a, b)) = (b, a). Thenσis an involution onM1([13], Example 3.2).

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of the primeness ofM, these yield:a= 0orc= 0;d= 0orb= 0;a= 0ord= 0andc= 0orb= 0. In all the cases, we obtain(a, b) = 0or(c, d) = 0, which establishes our claim.

But,M1is not a primeΓ-ring. For,(a,0)(α, α)(x, y)(β, β)(0, b) = (0,0)but(a,0)or(0, b)are not zero.

Example 2.2LetR be a commutative ring of characteristic 2 having unity element 1. ConsiderM = M1,2(R)andΓ = {

(

n.1

n.1

)

: n Z, 2 - n}. ThenM is aΓ-ring. SupposeN = {(x, x) : x R} ⊆ M. Then for each (x, x) N,

(a, b)∈Mand

(

n

n

)

Γ, we have

(x, x)

(

n

n

)

(a, b)(a, b)

(

n

n

)

(x, x)

= (xna−bnx, xnb−anx)

= (xna−2bnx+bnx, bnx−2anx+xna)

= (xna+bnx, bnx+xna)∈N. Therefore,N is a Lie ideal ofM.

IfU is a square closed Lie ideal (i.e. for allu∈U andα∈Γ), then for eachv∈U,uαv+vαu=(u+v)α(u+v)

uαu−vαv. Therefore,uαv+vαu∈U. On the other hand,uαv−vαu∈Ufor allu, v ∈Uandα∈Γ. Hence,2uαv∈U for allu, v∈Uandα∈Γ. We need to use this result frequently.

We proceed with the following lemmas.

Lemma 2.1 ([8], Lemma 3.1)LetU ̸= 0be aσ-ideal of a 2-torsion freeσ-primeΓ-ringM satisfying the condition (*). If

[U, U]Γ= 0, thenU ⊆Z(M).

Lemma 2.2 ([7], Lemma 2.2)LetU *Z(M)be aσ-ideal of a 2-torsion freeσ-primeΓ-ringM satisfying the condition (*) anda, b∈M such thataαU βb=aαU βσ(b) = 0for allα, β∈Γ. Thena= 0orb= 0.

Lemma 2.3 LetU ̸= 0be aσ-ideal of a 2-torsion freeσ-primeΓ-ringM satisfying the condition (*) andda derivation of

M such that=σdandd(U) = 0. Thend= 0orU ⊆Z(M).

Proof. For all u U, m M andα Γ, we have[u, m]α U. So, we get0 = d([u, m]α) = [d(u), m]α+

[u, d(m)]α= [u, d(m)]α. That is, for allu∈U,m∈M andα∈Γ,

[u, d(m)]α= 0. (1)

Puttingmβtformin (1), wheret∈M andβ∈Γ, we have

0 = [u, d(mβt)]α= [u, d(m)βt+mβd(t)]α

=d(m)β[u, t]α+ [u, d(m)]αβt+ [u, m]αβd(t) +[u, d(t)]α

=d(m)β[u, t]α+ [u, m]αβd(t), by using (1). Thus, for allu∈U,m, t∈Mandα, β∈Γ, we have

d(m)β[u, t]α+ [u, m]αβd(t) = 0. (2) Takingt=min (2), we find that

d(m)β[u, m]α+ [u, m]αβd(m) = 0 . Sinced(m)β[u, m]α= [u, m]αβd(m)[by (1)], therefore, we have

2d(m)β[u, m]α= 0 .

By the 2-torsion freeness ofM, for allu∈U,m∈M andα, β∈Γ, we obtain

d(m)β[u, m]α= 0. (3)

Replacinguby2uγvin (3), withv∈Uandγ∈Γ, we have

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= 2d(m)βuγ[v, m]α+ 2d(m)β[u, m]αγv

= 2d(m)βuγ[v, m]α, by using (3). SinceM is 2-torsion free, for allv∈U,m∈Mandα, β, γ∈Γ, we obtain

d(m)βU γ[v, m]α= 0. (4)

Letm∈Saσ(M). Then the fact thatσ(U) =U leads to

d(m)βU γ[v, m]α= 0 =d(m)βU γσ([v, m]α). (5) In view of Lemma 2.2, it gives thatd(m) = 0or[v, m]α= 0for allv∈U. Asm+σ(m)∈Saσ(M), thend(m+σ(m)) = 0 or[U, m+σ(m)]α = 0. If[U, m+σ(m)]α= 0, then[U, m]α=[U, σ(m)]α. If[U, m−σ(m)]α= 0, then[U, m]α =

[U, σ(m)]α. Adding these two relations, we obtain that2[U, m]α= 0, and hence[U, m]α= 0for allm∈M andα∈Γ, by 2-torsion freeness ofM.

Now we assume thatd(m+σ(m)) = 0. It givesd(m) +σd(m)) = 0(since=σd), and we obtaind(m)∈Saσ(M). Applying this in (4), we conclude thatd(m) = 0or[U, m]α = 0. Ifd(m−σ(m)) = 0, thend(m) ∈Saσ(M), and once again by using (4), we obtain thatd(m) = 0or[U, m]α= 0.

LetA={m ∈M : d(m) = 0}andB ={m ∈M : [U, m]α = 0}. ThenAandBare two additive subgroups ofM such thatA⊂MandB⊂M. We also haveA∪B=M. But, a group cannot be a union of two of its proper subgroups, and thusM =AorM =B. IfM =A, thend(m) = 0for allm∈M, i.e.d= 0. IfM =B, thenU ⊆Z(M). Consequently, we haved= 0orU ⊆Z(M).

Now we have the position to prove our main results.

Theorem 2.1 LetU ̸= 0be aσ-square closed Lie ideal of a 2-torsion freeσ-primeΓ-ringM satisfying the condition (*) andda derivation ofM which acts as a homomorphism onU. If=σd, thend= 0orU ⊆Z(M).

Proof. Let us suppose thatd(aαb) =d(a)αd(b)for alla, b∈U andα∈Γ. Assume thata, b, c∈Uandα, β∈Γ. As

4aαbβc= 2(2aαb)βc, it follows that4aαbβc∈U. SinceM is 2-torsion free, we obtain

d(aαbβc) =d(aαb)βc+aαbβd(c) =d(a)αd(b)βc+aαbβd(c). (6) On the other hand,

d(aαbβc) =d(a)αd(bβc) =d(a)αd(b)βc+d(a)αbβd(c). (7) Comparing (6) and (7), we obtain(d(a)−a)αbβd(c) = 0for alla, b, c∈U andα, β∈Γ. Therefore, for alla, c∈U and α, β∈Γ, we get

(d(a)−a)αU βd(c) = 0. (8)

As=σdandσ(U) =U, we conclude thatd(c) = 0for allc∈U or,d(a) =afor alla∈U. Ifd(c) = 0for allc∈U, then in view of Lemma 2.3, we conclude thatd = 0orU ⊆Z(M). Now considerd(a) =afor alla∈ U. Letm ∈M, u∈U andα∈Γ. Usingd(u) =uandd([u, m]α) = [u, m]α, we have seen that[u, d(m)]α= 0for allu∈U,m∈M and α∈Γ. By the similar argument as in the proof of Lemma 2.3, we are forced to conclude thatd= 0orU ⊆Z(M).

Theorem 2.2 LetM be a 2-torsion freeσ-primeΓ-ring satisfying the condition (*), and letU ̸= 0be aσ-square closed Lie ideal ofM. Letdbe a derivation ofM which acts as an anti-homomorphism onU. If=σd, thend= 0orU ⊆Z(M).

Proof. Suppose thatdacts as an anti-homomorphism onU. For alla, b∈Uandα∈Γ, we then get

d(aαb) =d(a)αb+aαd(b) =d(b)αd(a). (9) Substituting2aβbforain (9) withβ Γ, and using the 2-torsion freeness ofM, we get

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=⇒d(b)βd(a)αb+aβbαd(b) =d(b)αd(a)βb+d(b)αaβd(b). By using (*), for alla, b∈U andα, β∈Γ, we then obtain

aαbβd(b) =d(b)βaαd(b). (10) Replacingaby2cγain (10), wherec∈U andγ∈Γ, and using the 2-torsion freeness ofM, we find that

cγaαbβd(b) =d(b)βcγaαd(b) (11) for alla, b, c∈U andα, β, γ∈Γ. But, from (10), we get

cγaαbβd(b) =cγd(b)βaαd(b). Comparing this with (11), we obtain

cγd(b)βaαd(b) =d(b)βcγaαd(b). By using (*), we find that

[c, d(b)]γβaαd(b) = 0, and hence, for allb, c∈U andα, β, γ∈Γ, we get

[c, d(b)]γβU αd(b) = 0. (12)

Forb∈U∩Saσ(M), asσ(U) =U, we have

[c, d(b)]γβU αd(b) = 0 = [c, d(b)]γβU ασ(d(b)).

In view of Lemma 2.2, we obtain[c, d(b)]γ = 0ord(b) = 0for allc∈U. Sincedσ=σdandσ(U) =U, using the fact thatc+σ(c), c−σ(c)∈U ∩Saσ(M), by virtue of (12), we clearly find thatd(b) = 0or[U, d(b)]γ = 0for allb ∈U and γ∈Γ.

SetF ={b∈U :d(b) = 0}andG={b∈U : [U, d(b)]γ = 0}. Clearly,FandGare additive subgroups ofUsuch that U =F∪G, and henceU =F orU =G. IfU =F, thend(U) = 0, and by virtue of Lemma 2.3, we find thatd= 0or U ⊆Z(M). Now assume thatU =G. Then, for allu∈U, we obtain

[u, d(b)]γ = 0. (13)

Puttingbαbforbin (13), we get

0 = [u, d(bαb)]γ = [u, d(b)αb+bαd(b)]γ

=d(b)α[u, b]γ+ [u, d(b)]γαb+[u, d(b)]γ+ [u, b]γαd(b). By using (13) in the above relation, we obtain

d(b)α[u, b]γ+ [u, b]γαd(b) = 0. (14) From (13), we getuγd(b) = d(b)γu, which forces tod(b)α[u, b]γ = [u, b]γαd(b), since[u, b]γ ∈U. Using this in (14), by the 2-torsion freeness ofM, for allu, b∈Uandα, γ∈Γ, we find

d(b)α[u, b]γ = 0. (15)

Putting2uβvin place ofu, wherev∈Uandβ Γ, and using (15) and 2-torsion freeness ofM, we getd(b)αuβ[v, b]γ = 0 so thatd(b)αU β[v, b]γ= 0for allb, v∈U andα, β, γ∈Γ.

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### Acknowledgements

We are thankful to the reviewers for their useful suggestions and valuable comments to improve this article significantly.

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