Γ-CENTRALIZING mappings ON Prime AND Semi-prime Γ-Rings

Transcription

1 Republic of Iraq Ministry of Higher Education and Scientific Research University of Baghdad College of Sciences Department of Mathematics Γ-CENTRALIZING mappings ON Prime AND Semi-prime Γ-Rings A Thesis Submitted to the College of Science University Of Baghdad In partial fulfillment Of the Requirements for the Degree of Master of Science in Mathematics By Sameer K. Motashar Supervisor Dr. Abdul-Rahman Hameed Majeed Professor 2011

2 ب س م الل ه الر ح م ن الر ح يم ق ل ل و ك ان ال ب ح ر م د اد ا ل ك ل م ات ر ب ي ل ن ف د ال ب ح ر ق ب ل ا ن ت نف د ك ل م ات ر ب ي و ل و ج ي ن ا ب م ث ل ه م د د ا صدق االله العظيم {الكهف/ 109 }

3 Certification I Certify that this thesis was prepared under my supervision at the university of Baghdad, College of Science Department of Mathematics as a partial requirement of the degree of Master of Science in Mathematics. Signature: Name: Dr. Abdul Rahman H. Majeed Title: Professor Date:4/10/2011 In view of the available recommendation, I forward this thesis for debate by the examining committee. Signature: Name:Dr. Raid Kamel Naji Chairman of the Department Committee of Graduate studies in Mathematics Titel: Professor Date:4/10/2011

4 We certify that we have read this thesis and examining committee examined the student in its context and that in our opinion it is adequate for the partial fulfillment of the requirements for the degree of Master of Science in Mathematics with ( ) standing. Signature: Name: Saleh M. Salih Member: Assistant Professor Date:4/10/2011 Signature: Name: Nehad Salm Member: Lecturer Date:4/10/2011 Signature: Name: Dr. Bahar Hamad Al-Bahraany Chairman: Assistant Professor Date:4/10/2011 Approved by University Committee of graduate Studies Signature: Name: Dr. Dean of the College of Science Date:4/10/2011

5 Dedication This work dedicated to : My father and mother. My darling wife and children. My brothers and sisters and sincere friends. sameer

6 Acknowledgements Thanks to God for all things Peace be upon his prophet Mohammed. I am most grateful to my supervisor Dr. Abdulrahman H. Majeed for his advice, guidance, helpful suggestion, and encouragement throughout period I have worked under his supervision. And, I wish to express my sincere thanks to the members of Mathematics Department, College of Science, and University of Baghdad. Sameer

7 Abstract Let M be a Γ-ring, the concept of centralizing in a rings was initiated by the classical result of Posner and have been extensively studied by many authors, in this work we will present and study the concept of Γ- centralizing on Γ-rings. It is known that, one of the most important problems in studding Γ-rings is, whether the Γ-ring which we consider is commutative or not? So a part of our aim here in this work is to study these algebraic structures by give their properties and looking for those conditions under which they become commutative Γ-rings or contains a central ideal. Several theorems, lemmas and corollaries are proved and many conditions are given each of which makes these algebraic structures commutative Γ-rings or contains central ideal. Also we have studied the concept of left ΓM-module over Γ-ring and the concept of left derivations on left ΓM-module.

8 CONTENTS Subject No. Introduction 1 CHAPTER ONE: The Fundamentals Basic Definitions Some Results on Prime and Semi-Prime Γ-rings Derivations of Γ-rings 25 CHAPTER TWO: ΓΓ-centralizing Derivations of Prime and 28 Semi-Prime ΓΓ-rings 2.1.Γ-centralizing Derivations on Prime Γ-rings Extend Γ- centralizing Derivations on Prime Γ-rings Γ-centralizing Derivations on Semi-Prime Γ-rings Extend Γ- centralizing Derivations on Semi-Prime Γ-rings 50 CHAPTER THREE: Γ-centralizing Homomorphism of Prime and Semi-Prime ΓΓ-rings 3.1.Γ-centralizing Homomorphism on Prime Γ-ring Γ-centralizing Homomorphism on Semi-Prime Γ-ring 63 CHAPTER FOUR :Left Derivations and Generalized Left Derivations of Left ΓM-modules 4.1.Left Derivations of Left ΓM-modules Generalized Left Derivations of Left ΓM-modules 78 References

9 Introduction Let M and Γ be two additive abelian groups. If there exists a mapping (a,α,b) aαb of M Γ M M satisfying the following for all a,b,c M and α,β Γ: 1) (a+b)αc= aαc+ bαc, a(α+β)b= aαb+aβb, aα(b+c)=aαb+aαc 2) (aαb)βc=aα(bβc) then M is called a Γ-ring. This definition is due to Barnes[2]. If,in addition to the above, there exists a mapping (α,a,β) αaβ of Γ M Γ Γ satisfying the following for all a,b,c M and α,β Γ: 3) (α+β)aγ=αaγ+βaγ,α(a+b)β=αaβ+αbβ,αa(β+γ)=αaβ+αaγ 4) (aαb)βc=a(αbβ)c=aα(bβc) 5) aαb=0 for all a,b M implies α=0, then M is called a Γ-ring in the sense of Nobusawa[16],or simply a Nobusawa Γ-ring and we say that M is Γ N -ring. It is clear that M is a Γ N -ring implies that Γ is an M-ring[7].The properties of Γ-ring were obtained by many researchers[1,4,8,9,10,12,14,20 and 27] In this thesis M denote a Γ-ring in the sense of Barnes[2].A subring of a Γ-ring M is an additive subgroup S of M such that SΓS S.Recall that M is commutative if aαb=bαa,for all a,b M and α Γ,the center of M written as Z(M) is the set of those elements in M that commutative with every element in M so Z(M)={m M mαx=xαm, x M and α Γ}. A right(left)ideal of a Γ-ring M is an additive subgroup U of M such that UΓM U (MΓU U).If U is both a right and left ideal,then we say that U is an ideal or two sided ideal of M. Now we will define a prime and semi-prime Γ-ring as follows: 1. A Г-ring M is a semi-prime if and only if aгmгa=0,with a M implies a=0. 2. M is a prime Г-ring. if a,b M and aгmгb=(0),then a=0 or b=0. 1

10 Introduction The notion of a prime and semi-prime Γ-ring was introduced by S.Kyun[11] also W.E.Barnes. [2],J.Luh[14], G. L. Booth [4],they studied the structure of Γ-rings and obtained various generalizations of corresponding parts in ring theory. In[11] S.Kyun gave definition of homomorphism on Γ-rings as follows: If M i is a Г i -ring,for i=1,2, then an ordered pair (θθ, ) of mappings is called a homomorphism of M 1 on to M 2 if it satisfies the following properties : (1) θθ is a group homomorphism from M 1 onto M 2. (2) is a group isomorphism from Γ 1 onto Γ 2. (3) for every x,y M 1, γγ Г 1, θθ(x γγy)= θθ(x) (γγ) θθ(y). In [9] F.J.Jing defined a derivation on Γ-ring as follows,an additive map d:m M is said to be a derivation of M if d(aαb)= d(a)αb+aαd(b) for all a,b M and α Γ.In[22] M.Sapanci and A.Nakajima they gave definition of Jordan derivation on Γ-ring as follows, an additive map d:m M is said to be a Jordan derivation of M if d(aαa)= d(a)αa+aαd(a),for all a M and α Γ. As what in the rings theory we will ask what the conditions under which the Γ-rings becomes commutative,therefore in this thesis we will present and study the concepts of Γ-centralizing and left ΓM-module on Γ-rings as follows: 1. If M is a Γ-ring, a mapping d of M to itself is called Γ-centralizing on a subset S of M if [x,d(x)] α Z(M) for every x S and α Γ,in the special case when [x,d(x)] α =0 hold for all x S and α Γ,the mapping d is said to be Γ-commuting on S. 2. Let M be a Γ-ring and (X,+) be an abelian groups. X is a left ΓM-module if there exists a mapping from M Γ X to X (sending (m,α,x) in to mαx) such that (m 1 +m 2 )αx=m 1 αx+m 2 αx mα(x 1 +x 2 )=mαx 1 +mαx 2 2

11 Introduction (m 1 αm 2 )βx=m 1 α(m 2 βx) for all x, x 1,x 2 X, m, m 1,m 2 M and α,β Γ. Throughout this thesis,the condition aβcαb=aαcβb, for all a,b,c M and α,β Γ will represent by (*). This thesis consists of four chapters; In chapter one we have reviewed some known definitions, some necessary lemmas and theorems which will be used in the next chapters,some basic definitions are presented which can be found in the indicated reference, we start our study with definition of a Γ-ring,several examples on Γ-rings and the definitions of prime and semi-prime ideal in Γ-rings are given. In section two we give some results on prime and semi-prime Γ-rings. In section three we give the definitions of derivations on Γ-rings, some examples and some lemmas about derivations of Γ-rings. chapter two consists of four sections: in section one we define Γ- centralizing on Γ-rings, Γ-centralizing derivation on prime Γ-rings and extend some lemmas to Γ-rings, which will be use to extend J.H. Mayne Theorem(1)[15] to Γ-rings. In section two we generalize the result of section one by extend J. Vukman Theorem (1,2) in [24] to Γ-rings. In section three we extend Theorem(3) of H.E. Bell and W.S. Martindale[3] to Γ-rings,we will prove that a semi-prime Γ-ring M must be contained a nonzero central ideal, if M satisfying (*) and 0 U left ideal in M.Where M admits a derivation which is nonzero on U and Γ-centralizing on U. Finally in section four we generalize the result of section three which it extend to Theorem (1,3) of J.Vukman[25] to Γ-rings. Chapter three consists of two section : in section one we define a homomorphism in Γ-ring and extend J.H. Mayne Theorem(1)[15] to Γ-rings as ;Let R be a prime ring and U be a nonzero ideal in R,if T is a nontrivial automorphism of R such that [x,t(x)] is in the center of R, for every x in U,then the ring R is commutative. In section two we will extend of H.E.Bell and W.S.Martindale Theorem(1)[3];Let R be 3

12 Introduction a semi-prime ring and U be a non zero left ideal of R, suppose that R admits an endomorphism T which is one-to-one on U, centralizing on U and not the identity on U,if T(U) U.Then R contains a non zero central ideal. Chapter four consists of two section : first we introduce the notion of a ΓM- module, in section one, we study the concepts of a left ΓMmodule[13], left derivation of left ΓM-module in Γ-ring[17], some lemmas explain some properties of Jordan left derivation of left ΓMmodule[17] and prove that; Let M be a Γ-ring satisfying(*) and X be a left ΓM-module.Suppose that aαx=0 with a M,x X and α Γ implies that either a=0 or x=0.if there exists a nonzero left derivation d:m X, such that 1. If d:m X is a nonzero left derivation and X be a left ΓM-module.Then M is commutative. 2. If d:m X is a nonzero Jordan left derivation and X be a 2- torsionfree. Then M is commutative. In section two we will define a generalized left derivation,generalized Jordan left derivation and we will prove that; If M be a Γ-ring satisfying (*),X be a 2-torsionfree left ΓM-module. Suppose aαx=0 with a M, x X and α Γ implies that either a=0 or x=0.if D:M X be a generalized left derivation with associated a nonzero Jordan left derivation d: M X. Then M is commutative. 4

13 CHAPTER ONE The Fundamentals

14 CHAPTER ONE Introduction: CHAPTER ONE The Fundamentals This chapter consists of three sections. In section one some basic definitions are presented which can be found in the indicated references, we start our study with definition of a Γ-ring[2], examples of a Γ-rings and we give the definitions of prime and semi-prime ideal in Γ-rings. In section two we give some results on prime and semi-prime Γ-rings. In section three we give the definitions of derivation [9],Jordan derivation [22], Jordan left derivation[6] on Γ-rings,some examples and some lemmas about derivation of Γ-rings. 1.1.Basic Definitions: The notion of Γ-ring, a concept more general than a ring, was defined by Nobusawa [16], W. E. Barnes[2], J. Luh, G. L. Booth [4] and many authors studied the structure of Γ-rings and obtained various generalizations analogous of corresponding parts in ring theory. In this section we present the definitions,examples and some lemmas about Γ- rings as follows. Definition1.1.1:[2] Let M and Γ be two additive abelian groups. If there exists a mapping (a,α,b) aαb of M Γ M M satisfying the following, for all a,b,c M and α,β Γ: 1) (a+b)αc= aαc+ bαc, a(α+β)b= aαb+aβb, aα(b+c)=aαb+aαc 2) (aαb)βc=aα(bβc) then M is called a Γ-ring. If A and B are subsets of a Γ-ring M and Ψ Γ we denote AΨB,the subset of M consisting of all finite sums of the form Σa i α i b i where a i A, 5

15 CHAPTER ONE α i Ψ and b i B,for singleton subsets we abbreviate this notation for example {a}ψb=aψb,it is well known that every ring is Γ-ring. In the follows some examples about Γ-rings Example 1.1.2:[19] Let R be any ring, the additive abelian groups M=M 2x3 (R) and Г=M 3x2 (R), denotes the sets of all 2 3 matrices and 3 2 matrices over R respectively, then M is Г-ring. Example 1.1.3: Let (R, + R,. R ) be a ring and let (Z,+) be a group of integer numbers. Let M=R and Г=Z we will define the mapping M Γ M M sending (r,n,s) to r. R ns,for all r,s R and n Γ, then M is Γ-ring. Example 1.1.4: Let V n (F) be a vector space of dimension n over a filed F and let M=V n (F), Г=F,then we can define a mapping M Γ M M (sending (x,α,y) to x.αy), for all x,y V n (F),α Г where. is the inner product of two vectors on V n (F), then M is a Г-ring. Example 1.1.5: Suppose we have a right R-module M with a ring R. Take a submodule Γ of Hom R (M,R). If a and b are elements of M and if γγ is an element of Γ,then we define aγb=a.γ(b) where γ(b) is an image of b by γγ and is an element of R. Now for all x,y,z M and α,β Г, the conditions 6

16 CHAPTER ONE 1) xαy=x.α(y) M since M is a right R-module 2) ii) (x+y)αz=(x+y).α(z)=x.α(z)+y.α(z)=xαz+yαz iiii) x(α+β)y=x.((α+β)(y) )=x.(α(y)+β(y) ) =x.α(y)+x.β(y)=xαy+xβy iiiiii) xα(y+z)=x.α(y+z)=x.(α(y)+α(z))=x.α(y)+ x.α(z)=xαy+xαz 3)(xαy)βz=(xαy).β(z)=(x.α(y)).β(z)=x.(α(y)β(z)) = x.(α(yβ(z)))=x.α(yβz)=xα(yβz) are satisfied, then M be a Γ-ring. Definition :[26] A subring of a Г-ring M is an additive subgroup S of M such that SГS S. Example 1.1.7: and aa bb cc Let R be any ring,m={ a,b,c RR} αα Г={ αα RR } then M and Г are both abelian group under matrix addition.now it is easy to show that M is a Г-ring under matrix multiplication,also we can prove that Г is subring of M. Definition1.1.8:[1] A Γ-ring M is commutative if aααb=bααa,for all a,b M and αα Γ, the center of M written as Z(M) is the set of those elements in M that commutative with every element in M, so Z(M)={m M mααx=xααm, x M and αα Γ}. 7

17 CHAPTER ONE Remark 1.1.9: If M is a Γ-ring then Z(M) is subring of M. Proof: Z(M) be a non-empty set since 0 Z(M). Let a,b Z(M),then for all m M and αα Γ, we get (a-b)ααm=aααm-bααm=mααa-mααb=mαα(a-b), which means that Z(M) is additive subgroup. Now we will prove that Z(M)ΓZ(M) Z(M). Let c,d ZZ(M), then for all ββ Γ we have cββd Z(M)ΓZ(M) Now for all m M and γγ Γ, then (cββd) γγm=cββ(dγγm)= (dγγm)ββc=(mγγd) ββc=mγγ(dββc)=mγγ(cββd) therefore cββd Z(M), thus Z(M) is subring of M. Definition :[11] A right (left ) ideal of a Г-ring M is an additive subgroup U of M such that UГM U (MГU U).If U is both a right and left ideal,then we say that U is an ideal or two sided ideal of M. Example : Let R be any ring and let M=R R, then M is also be a ring,if Г=Z then by the same way in Example1.1.2, we can get M a Г-ring. Let U be a right ideal in R then U {0} also a right ideal in M. Example :[19] Let R be a ring and (Z,+) be a group of integer numbers, we put M=M 2x2 (R) and Γ=M 2x2 (Z), then M is a Γ-ring.We use the usual addition and multiplication on matrices of M Γ M. 8

18 CHAPTER ONE Let U={ aa bb a,b R} clearly U is a right ideal of M but not a left 0 0 ideal of M. Definition1.1.13:[11] For each a of a Г-ring M,the smallest right ideal containing a is called the principal right ideal generated by a and denoted by a>. Similarly we define <a and <a>,the principal left and two-sided (respectively) ideals generated by a. Now we will define prime ideals in Γ-rings and prime Г-rings. Definition1.1.14:[11] An ideal P of a Г-ring M is prime ideal if for any ideals A,B M,AГB P implies A P or B P. Definition1.1.15:[11] A Г-ring M is said to be prime Г-ring if the zero ideal is prime ideal. In the next theorem we will give some equivalent definitions of prime Г-ring as follows. Theorem :[11] If M is a Г-ring,the following conditions are equivalent 1) M is a prime Г-ring. 2) if a,b M and aгmгb=(0),then a=0 or b=0. 3) if <a> and <b> are principal ideals in M such that <a>г<b>=(0),then a=0 or b=0. 9

19 CHAPTER ONE 4) if A and B are right ideals in M such that AГB=(0),then A=(0) or B=(0) 5) if A and B are left ideals in M such that AГB=(0),then A=(0) or B=(0). Proof : 1 2 Assume that a,b are non-zero elements in M therefore the ideals generated by a and b are non-zero ideals thus <a>г<b> (0) But aгmгb <a>гmг<b> <a>г<b> (0). Therefore aгmгb (0),contradiction with assumption of (2). Thus either a=0 or b= Since <a>г<b>=(0). But aгmгb <a>гmг<b> <a>г<b>=(0). Then aгmгb=(0), by (2) either a=0 or b= Let a A and b B then aгb AГB=(0). Therefore aгb=(0). (i) Now since A is right ideal then AГMГB AГB=(0) then aгmгb=(0) We claim that <a>г<b>=(0). Assume not <a>г<b> (0). But aгb <a>г<b> (0). 10

20 CHAPTER ONE Therefore aгb (0), contradiction with (i). Then <a>г<b>=(0) and by assumption of (3) either a=0,a=(0) or b=0,b=(0). 4 5 Let a A and b B then by (5) we have aгb=0. (i) Suppose that <a>г<b>=(0). If not then <a>г<b> (0) but aгb <a>г<b> (0), we get aгb (0) contradiction with (i) but <a>, <b> are right ideals, then by (4), we get either <b>=(0), then b=0, B=(0) or <a>=(0), then a=0, A=(0). 5 1 We want prove that (0) is prime ideal. Let A and B any ideal in M with AГB=(0),but A and B are left ideal,then by(5) either A=(0) or B=(0). Now we will define semi-prime ideal in Γ-ring and semi-prime Г-rings as follows. Definition1.1.17:[11] An ideal Q of a Г-ring M is semi-prime ideal if for any ideal A of M,AГA Q implies A Q. Definition1.1.18:[11] A Г-ring M is called semi-prime Γ-ring if the zero ideal is semi-prime ideal, i.e. for any ideal U of a Г-ring M with UГU (0) implies U=(0). 11

21 CHAPTER ONE The following theorem characterize Г-rings. semi-primness for ideal in Theorem :[11] If Q is an ideal in a Г-ring M,all the following conditions are equivalent. 1) Q is a semi-prime ideal. 2) If a M such that aгmгa Q,then a Q. 3) If <a> is a principal ideal in M such that <a>г<b> Q,then a Q. 4) If U is a right ideal in M such that UГU Q then U Q. 5) If V is a left ideal in M such that VГV Q,then V Q. Proof: 1 2 Let a M and <a> is principal ideal in M, suppose that <a> Г<a> Q. (i) But aгmгa <a>гmг <a>. Now since <a> is an ideal of M, therefore by (i),we get <a>гmг <a> <a> Г<a> Q, then <a> ГMMГ <a> Q contradiction. Thus <a> Г<a> Q, but Q is a semi-prime ideal. Therefore <a> Q, a Q. 2 3 Since <a> is a principal ideal in M such that <a> Г<a> Q. Therefore aгmгa <a>гmг<a> <a> Г<a> Q. 12

22 CHAPTER ONE Thus, aгmгa Q, then by (2) we get, a Q. 3 4 We want prove that U Q, let a U and a Q. (i) Now let <a> be the principal ideal generated by a, then <a> Г<a> Q. If <a> Г<a> Q, then by (i) we get to a Q contradiction with (i). Therefore aгa <a> Г<a> Q,then aгa Q (ii) but aгa UГU Q, according to(4) we have aгa Q contradiction with (ii), therefore a Q. 4 5 If a V, then aгa V. (i) Therefore <a> is a principal ideal in M generated by a. Suppose that <a> Г<a> Q, then aгa Q contradiction with (i) so <a> Г<a> Q, but <a> is a right ideal then by assumption of (4) we have, a Q. 5 1 Let A be an ideal of a Г-ring M such that AГA Q. Since A is a left ideal, then by (5), we have A Q thus Q is semi-prime ideal. 13

23 CHAPTER ONE Corollary1.1.20:[11] A Г-ring M is a semi-prime if and only if aгmгa=0,with a M implies a=0. Definition1.1.21:[11] A subset S of a Г-ring M is called strongly nilpotent if there exists a positive integer n such that (SГ) n S=(0). Remark :[11] It is follows easily prove that if Q is a semi-prime ideal and A is an ideal such that (AГ) n A Q for an arbitrary positive integer n, then A Q. Remark :[11] (0) is a semi-prime ideal in a Г-ring M if and only if M contains no non-zero strongly nilpotent right (left) ideal. Therefore we can define three equivalent definitions of a semi-prime Г-ring as follows. Definition1.1.24: A Г-ring M is a semi-prime if the zero ideal is a semi-prime ideal. Definition1.1.25: A Г-ring M is a semi-prime if and only if has no non-zero strongly nilpotent ideal. Definition1.1.26: A Г-ring M is a semi-prime if and only if has no non-zero strongly nilpotent right(left) ideal. 14

24 CHAPTER ONE In fact,every prime Г-ring is semi-prime Г-ring,but the converse is not true in general, we will give an example about semi-prime Г-ring but not prime Г-ring, as follows : Example :[19] Let R be a prime Г 1 -ring,we put M=R R and Г=Г 1 Г 1, then M is a Г-ring,clearly M is a semi-prime Г-ring but not prime Г-ring since if a,b be a non-zero elements in R then for all (α,β),(σ,τ) Г and for all (x,y) M,then (a,0)(α,β)(x,y) (σ,τ)(0,b)=(0,0). But (a,0) and (0,b) are non-zero elements in M. Now we will define Г-residue class ring and homomorphism of a Г-rings. Definition1.1.28:[11] Let U be an ideal of a Г-ring M.If for each a+u,b+u in the factor group M/U,and each γγ Г, we define (a+u) γγ(b+u)=a γγb+u,then M/U is a Г-ring which we shall call the Г-residue class ring of M with respect to U. Definition1.1.29:[11] If M i is a Г i -ring,for i=1,2, then an ordered pair (θθ, ) of mappings is called a homomorphism of M 1 onto M 2 if it satisfies the following properties : (1) θθ is a group homomorphism from M 1 onto M 2. (2) is a group isomorphism from Γ 1 onto Γ 2. (3) for every x,y M 1, γγ Г 1, θθ(x γγy)= θθ(x) (γγ) θθ(y). The kernel of the homomorphism (θθ, ) is defined to be K={x M i θθ(x)=0}. Clearly K is an ideal of M i.if θθ is a group 15

25 CHAPTER ONE isomorphism,that is,k=(0), then (θθ, ) is called an isomorphism from the Γ 1 -ring M 1 onto the Γ 2 -ring M 2. Example :[11] Let M be a Γ-ring and let U be an ideal in M.Then the ordered pair (ρρ, ii) of the mappings where ρρ:m M/U is defined by ρρ(xx) = xx + UU for all x M,and ii is the identity mapping of Γ, is a homomorphism called natural homomorphism from M onto M/U. Theorem :[13] (the fundamental theorem of homomorphism for Γ-ring) If (θθ, ) is a homomorphism from the Γ 1 -ring M 1 onto the Γ 2 -ring M 2 with kernel K,then M 1 /K and M 2 are isomorphic. Poof: Define an ordered pair (ff, ) where ff: M 1 /K M 2 by ff(x+k)= θθ(x) for all x M 1.Then ff is will define and 1-ff is group homomorphism since ff((x+k)+(y+k))=ff(x+y+k)= θθ(x+y)= θθ(x)+ θθ(y)=ff(x+k)+ff(y+k). 2-ff is on to since let z M 2 but θθ is on to therefore there exists x M 1 such that θθ(x)=z then there exists x+k M 1 /K such that ff(x+k)= θθ(x)=z. 3-ff is one to one since Let x+k ker(f) where x M 1 therefore 0=ff(x+K)= θθ(x),therefore x ker(θθ)=k, i.e. x+k=k thus f is one to one. 4- is group isomomorphism by assumption of Theorem

26 CHAPTER ONE 5- ff((x+k)γγ(y+k))=ff(x γγy+k) = θθ(x γγy) = θθ(x) (γγ) θθ(y) =ff(x+k) (γγ)ff(y+k). Theorem :[11] An ideal Q in a Γ-ring M is a semi-prime ideal if and only if the Γ-residue class ring M/Q contains no non-zero strongly nilpotent ideal. Proof: Let A+Q={a+Q a A} be strongly nilpotent ideal in Γ-residue class ring M/Q then ((A+Q)Γ) n (A+Q)=Q, for some positive integer n. (1) Therefore we get from(1) (AΓ) n A+Q=Q. (2) We have from (2) (AΓ) n A Q. (3) It, s clear that A is an ideal of M, therefore by Remark we get A Q, thus A+Q=Q which lead to M/Q contains no non-zero strongly nilpotent ideal. Conversely let A be any ideal in M with AΓA Q,then AΓA+Q=Q. Therefore, by Definition1.1.28,we get Q=(A+Q)Γ(A+Q). But M/Q contains no non-zero strongly nilpotent ideal therefore 17

27 CHAPTER ONE A+Q=Q which means that A Q, than Q is a semi-prime ideal. Theorem :[11] If P is an ideal in the Γ-ring M,then the Γ-residue class ring M/P is a prime Γ-ring if and only if P is a prime ideal in M. Proof: For any ideals A,B in M with AΓB P, then AΓB+P=P. (1) Therefore by (1) we get P=(A+P)Γ(B+P). (2) But A+P, B+P be an ideals in M/P and M/P is a prime which means that the zero ideal is prime ideal therefore either A+P P thus A P or B+P P thus B P. Therefore P is prime ideal in a Γ-ring. Conversely let A+P,B+P are ideals of M/P with (A+P)Γ(B+P)=P. It is clear that A,B are ideals in M,therefore we have AΓB+P=P thus AΓB P but P is a prime ideal in M, then either A P or B P. D.OZden, M.A. Ozturk and Y.B.Jun in [26],they gave the definitions of right and left annihilator of subset and some results about annihilator in a Γ-rings. Definition1.1.34:[26] Let M be a Γ-ring.For a subset U of M, Ann L (U)={a M aγu=0} is called the left annihilator of U.A right annihilator Ann r (U) can be defined similarly. 18

28 CHAPTER ONE Lemma1.1.35:[26] Let M be a semi-prime Γ-ring and U be a non-zero ideal of M.Then Ann r (U)= Ann L (U)=Ann(U). Lemma1.1.36:[26] Let M be a semi-prime Γ-ring and U be a non-zero ideal of M.Then 1. Ann(U) is an ideal of M 2. U Ann(U)=(0). Definition1.1.37:[7] Let M be a Γ-ring.Then M is called n-torsion free if na=0 implies a=0, for every a M where n is positive integer. Definition : Let M be a Γ-ring.if there exists a positive integer n such that na=0 for all a M,then the smallest positive integer with this property is called the characteristic of Γ-ring,if no such positive integer exists then M is said to be of characteristic zero. Proposition1.1.39: If M be a prime Γ-ring,then M is n-torsion free if and only if characteristic of M not equal n. Proof: Obvious. 19

29 CHAPTER ONE Definition1.1.40:[6] A Γ-ring M is called a completely prime if aγb =(0) implies a=0 or b=0 where a,b M. It is clear that aγbγaγb aγmγb,therefore every completely prime is prime Γ-ring. Definition :[21] Let M be a Γ-ring,then for all a,b M and αα Γ,we define [a,b] α =aααb-bααa. In the following lemma, S. Chakraborty and A. C. Paul properties of [a,b] α. gave the Lemma :[7] If M is a Γ-ring,then for all a,b,c M and α, ββ Γ,then 1) [a,b] α +[b,a] α =0 2) [a+b,c] α =[a,c] α +[b,c] α 3) [a,b+c] α =[a,b] α +[a,c] α 4) [a,b] α+β =[a,b] α +[a,b] β 5) [aβb,c] α =a ββ[b,c] α +[a,c] α ββb+ a ββcαb-aαcββb. Now if aβcαb=aαcβb,then from Lemma (5),we get [aβb,c] α =aβ[b,c] α +[a,c] α βb. Throughout this thesis,the condition aβcαb=aαcβb, for all a,b,c M and α,β Γ will represent by (*). 20

30 CHAPTER ONE 1.2 Some Results on Prime and Semi-Prime Γ-rings: In this section we present some lemmas and theorem which will be used in our work. Lemma : Let M be a semi-prime Γ-ring satisfying (*) and let a M such that aβ[a,m] α =0 (or [a,m] α βa=0), for all m M and α, β Γ. Then a Z(M). Proof: For all m 1 M and δ Γ,then 0=aβ[a,mδm 1 ] α =aβ(mδ[a,m 1 ] α +[a,m] α δm 1 ). aβmδ[a,m 1 ] α =0. (1) By assumption and (*),we get 0=mδaβ[a,m 1 ] α =mβaδ[a,m 1 ] α. (2) Now from (1) and (2), we have [a,m] β δ[a,m 1 ] α =0. (3) In (3) replace m 1 by m 1 γm,for all γ Γ, we have 0=[a,m] β δ[a,m 1 γm] α =[a,m] β δ(m 1 γ[a,m] α +[a,m 1 ] α γm) =[a,m] β δm 1 γ[a,m] α. Now for all β ΓΓ take β = α and since M is a semi-prime Γ-ring therefore [a,m] α =0 for all m M and α Γ, thus a Z(M). Similarly we can prove the lemma,when [a,m] α βa=0. 21

31 CHAPTER ONE Lemma1.2.2: Let M be a semi-prime Г-ring satisfying (*) and U be a left ideal of M. Then Z(U) Z(M). Proof: Let a Z(U), for all α Г and x M,then xαa U and for all β Γ, we get 0=[a,xαa] β =xα[a,a] β +[x,a] β αa=[x,a] β αa,therefore by Lemma a Z(M). Theorem1.2.3.: Let M be a semi-prime Γ-ring satisfying (*) and let U be a non zero left ideal of M.If U be a commutative as a Γ-ring, then U Z(M),if in addition M is a prime Γ-ring.Then M must be commutative. Proof: By Lemma1.2.2, we get our first desired U Z(U) Z(M). (1) Now suppose that M is a prime Γ-ring. Let x,y M and a U,since U is a left ideal then xγa U, therefore by (1),we get xγa Z(M). Now for all α Γ and y M, then (0)=[y,xΓa] α =xγ[y,a] α +[y,x] α Γa=[y,x] α Γa. Therefore [y,x] α ΓU=(0). (2) But U is a left ideal then MΓU U. (3) 22

32 CHAPTER ONE Now from (2) and (3),we have, [y,x] α ΓMΓU=(0). Since M is a prime Γ-ring and U be a nonzero left ideal,then [y,x] α =0 for all x,y M and α Γ, which means that M is commutative. Lemma1.2.4.: If M is a prime Γ-ring and U is a non-zero ideal of M. Then Ann r (U)=(0). Proof: Since UΓAnn r (U)=(0) and UΓM U,then UΓMΓAnn r (U) UΓ Ann r (U)=(0). Therefore, UΓMΓAnn r (U) =(0). But M be a prime Γ-ring and U be a non-zero ideal of M then Ann r (U)=(0). Lemma 1.2.5: Let M be a prime Γ-ring and U be a non-zero right(left) ideal of M, a M and UΓa=(0) ( aγu=(0)). Then a=0. Proof: Since U is a non-zero right ideal, then UΓM U. (1) Then from (1), we have UΓMΓa UΓa=(0). But M is Prime Γ-ring and U be a non-zero right ideal, then a=0. 23

33 CHAPTER ONE Lemma : The center of a semi-prime Γ-ring M contains no non-zero strongly nilpotent elements. Proof: Let a Z(M) be a strongly nilpotent element then there exits smallest positive integer n such that (aγ) n a=(0). (1) Then from (1) we have (aγ) n-1 aγa=(0). (2) Since M is a Γ-ring,we get (aγ) n-1 aγaγm=(0). (3) Now from (3) and since (aγ) n-2 a M, therefore (0)= (aγ) n-1 aγmγaγ(aγ) n-2 a=(aγ) n-1 aγmγ(aγ) n-1 a. But M is a semi-prime Γ-ring we have from above relation (0)= (aγ) n-1 a But n is smallest positive integer such that (aγ) n a=(0), then a=0. 24

34 CHAPTER ONE 1.3 Derivations of Γ-rings: In this section we investigate definitions of derivations,jordan derivations, Jordan left derivation and some lemmas about derivations used in our study. Definition1.3.1.:[9] Let M be a Γ-ring and d:m M be an additive map, d is called a derivation on M if d(aαb)= d(a)αb+aαd(b) for all a,b M and α Γ, d is called a Jordan derivation on Γ-ring M if: Definition1.3.2:[6] d(aαa)= d(a)αa+aαd(a). Let M be a Γ-ring and d:m M be an additive map,d is called a Jordan left derivation on M if d(aαa)=2aαd(a),for any a M and α Γ. In [19] S.M. Salih gave an example of derivation and Jordan derivation on a Γ-rings.And In [6] Y.Ceven give an example of Jordan left derivation on Γ-rings as follows. Example1.3.3.[19] Let R be a ring,m=m 1x2 (R) and Γ={ mm mm is an integer 0 number},then M is a Γ-ring,we define d:m M by : d((a b))=(0 -b) for all a,b R. We use the usual addition and multiplication on matrices of M Γ M.Clearly d is a derivation on M. Example1.3.4.[19] Let M be a Γ-ring and let a M such that aγa=(0) and xαaβx=0,for all x M,α,β Γ but xαaβy 0, for some x,y M, x y. Let d be a map on M in to itself defined by: 25

35 CHAPTER ONE d(x)=xαa+aαx, for all x M, α Γ. It is clear that d is a Jordan derivation but not derivation on M. Example1.3.5.[6] Let R be a ring,m=m 1x2 (R) and Γ={ m m is an integer 0 number},then M is a Γ-ring if D:R R is a Jordan left derivation and N={(a a) a R} is the subset of M,then N is a Γ-ring and the map d:n N defined by d((a a))=(d(a) D(a)) is a Jordan left derivation on N. Now we will prove two lemmas which we need they in our study. Lemma1.3.6: If d is a non-zero derivation of a prime Γ-ring M then Ann r (d(m))=ann L (d(m))=(0). Proof: For all β Γ,we get d(m) β Ann r (d(m))=(0). (1) Now for all a,b M and α Γ then aαb M. Therefore from (1) we get (0)=d(aαb)β Ann r (d(m)) (0)=(d(a)αb+aαd(b))β Ann r (d(m)) (0)= d(a)αbβ Ann r (d(m)). Therefore from above relation,we get d(a)γmγann r (d(m))=(0). But M is a prime Γ-ring and d be a non-zero derivation on M, then Ann r (d(m))=(0). 26

36 CHAPTER ONE Lemma 1.3.7: Let U be a non zero right ideal in a prime Γ-ring M if M has a derivation d which is zero on U. Then d is zero on M. Proof: We have d(u)=0,then (0)=d(UΓM)=d(U)ΓM+UΓd(M)= UΓd(M) UΓd(M)=(0). (1) But U is right ideal then by Lemma and (1), d(m)=0. 27

37 CHAPTER TWO ΓΓ-centralizing Derivations of Prime and Semi-Prime ΓΓ-rings

38 CHAPTER TWO CHAPTER TWO ΓΓ-Centralizing Derivations of Prime and Semi-Prime ΓΓ-rings Introduction Let R be a ring with center Z and S a nonempty subset of R. A mapping F from R to R is called centralizing on S if [x,f(x)]=xf(x)-f(x)x Z, for all x S;in the special case where [x,f(x)]=0 for all x in S, the mapping F is said to be commuting on S[2]. The study of centralizing and commuting mappings in a rings was initiated by the classical result of Posner[18],which states that the existence of a nonzero centralizing derivation on a prime ring forces the ring to be commutative (Posner, s second theorem ).In this chapter we define Γ- centralizing mapping on Γ-ring and we extend the results of J.H.Mayne [15],J.Vukman[24],H.E.Bell and W.S.Martindale[3]and J.Vukman[25]in Γ-rings. This chapter consists of three sections: in section one we present the definitions of Γ-centralizing on Γ-rings, Γ-centralizing derivation on prime Γ-rings, and we will extend J.H. Mayne theorem. In section two we generalize the results of section one by extend the theorem of J. Vukman[24]. In section three we extend theorem of H.E. Bell and W.S. Martindale,we will prove that a semi-prime Γ-ring must be contained a nonzero central ideal, if M satisfying (*) and 0 U left ideal in M.Where M admits a derivation which is nonzero on U and centralizing on U. Finally in section four we generalize the result of section three which is extend to theorem of J.Vukman[25]. 28

39 CHAPTER TWO 2.1.Γ-centralizing Derivations on Prime Γ-rings: In this section we define Γ-centralizing mapping on Γ-rings, Jordan subring in Γ-rings and we will prove that if M is a prime Γ-ring satisfying(*),u be a non-zero ideal and d be a non-zero derivation of M such that [x,d(x)] α =xαd(x)-d(x)αx Z(M) for every x in U, then the Γ- ring M is commutative.for this aim we need to prove some lemmas in this section. Definition 2.1.1: Let M be a Γ-ring, a mapping d of M to itself is called Γ-centralizing on a subset S of M if [x,d(x)] α Z(M) for every x S and α Γ,in the special case when [x,d(x)] α =0 hold for all x S and α Γ,the mapping d is said to be Γ-commuting on S. Example 2.1.2: Let M 1 be Γ 1 -ring,put M=M 1 M 1 and Γ= Γ 1 Γ 1 then M is a Γ-ring. Define a mapping d:m M by d((x,y))=(y,x) for all x,y M 1, and let S={(x,0) x M 1 }be a subset of M. Then [(x,0),d((x,0)] α =(x,0)α(0,x)-(0,x)α(x,0)=(xα0,0αx)-(0αx,xα0)=(0,0) that is mean d is Γ-centralizing on S. Definition 2.1.3: An additive subgroup U of a Γ-ring M is called Jordan subring of M if a,b U implies aαb+bαa U for all α Γ. Remark 2.1.4: Every one sided ideal of Γ-ring is a Jordan subring. 29

40 CHAPTER TWO Lemma 2.1.4: Let M be a prime Γ-ring, if a Z(M) and aγb Z(M).Then either a=0 or b Z(M). Proof: By assumption we have, [aγb,m] α =(0), for all α Γ. Therefore (0)=aΓ[b,M] α +[a,m] α Γb+aΓ(Mαb)-aα(MΓb). (1) But a Z(M), then aα(mγb)= (MΓb)αa=MΓ(aαb)=(aΓM)αb=aΓ(Mαb). From (1) we have (0)=aΓ[b,M] α. Now by definition of Γ-ring we get (0)=MΓaΓ[b,M] α =aγmγ[b,m] α. But M is a prime Γ-ring then either a=0 or b Z(M). Lemma 2.1.5: Let M be a prime G-ring of characteristic not equal two satisfying (*) and U be a Jordan subring of M. If d is a Jordan derivation of U, such that [x, d(x)] a Z(M) for all x U and a G.Then [x, d(x)] a = 0,for all x U and a G. Proof: By assumption we have [x+y,d(x+y)] α Z(M) for all x,y U,α Γ. (1) 30

41 CHAPTER TWO Therefore [x+y,d(x+y)] α =[x, d(x+y)] α +[y,d(x+y)] α =[x,d(x)] α +[y,d(y)] α +[x,d(y)] α +[y,d(x)] α Z(M). Now by assumption and since Z(M) is an additive subgroup of M then [x,d(y)] α +[y,d(x)] α Z(M) (2) In (2) replace y by xβx,for all β Γ,we have [x,d(xβx)] α +[xβx,d(x)] α =[x,d(x)βx+xβd(x)] α +[ xβx,d(x)] α =[x,d(x)βx] α +[x,xβd(x)] α +[xβx,d(x)] α = [x,d(x)] α βx+xβ[x,d(x)] α +xβ[x,d(x)] α +[x,d(x)] α βx But [x,d(x)] α Z(M), then from the above relation we have 4xβ[x,d(x)] α Z(M). By Lemma then either [x,d(x)] α =0 or 4x Z(M) which lead to [4x,d(x)] α =4[x,d(x)] α =0, but ChM 2,then [x,d(x)] α =0, for all x U and α Γ. Lemma 2.1.6: Let U be a right ideal in a prime Γ-ring M satisfying (*).If d is a derivation of M such that[x,d(x)] α Z(M) for all x U and for all α Γ.Then [x,d(x)] α =0, for all x U and for all α Γ. Proof: If characteristic of M is not two,lemma implies that [x,d(x)] α =0 for all x U. 31

42 CHAPTER TWO Now suppose that M has characteristic equal two.let x,y be in U and d be an additive mapping,then [[x,y] β,d(x)] α =[xβy-yβx,d(x)] α =[xβy,d(x)] α -[yβx,d(x)] α. The fact that M has characteristic two means y=-y for all y M therefore [[x,y] β,d(x)] α =[xβy,d(x)] α +[yβx,d(x)] α So relation above becomes =xβ[y,d(x)] α +[x,d(x)]βy+yβ[x,d(x)] α +[y,d(x)] α βx = xβ[y,d(x)] α + [y,d(x)] α βx+2[x,d(x)] α βy. [[x,y] β,d(x)] α =xβ[y,d(x)] α +[y,d(x)] α βx. (1) Now we want to prove that [[x,y] β,d(x)] α + [xβx,d(y)] α =0. (2) Therefore from (1), we have [[x,y] β,d(x)] α +[xβx,d(y)] α =xβ[y,d(x)] α +[y,d(x)] α βx+xβ[x,d(y)] α +[x,d(y)] α βx =xβ([y,d(x)] α +[x,d(y)] α ) +([y,d(x)] α +[x,d(y)] α )βx. Since [y,d(x)] α +[x,d(y)] α Z(M) and ChM=2 then [[x,y] β,d(x)] α +[xβx,d(y)] α =2 xβ([y,d(x)] α +[x,d(y)] α )=0. In above relation let d(x)=z,we have [[x,y] β,z] α + [xβx,d(y)] α =0 (3) As a special case of (3) when y=x,we get [xβx,z] α =0. (4) 32

43 CHAPTER TWO For all x U and δ Γ let y=xδz. Now from (3), we have 0=[[x, xδz] β,z] α +[xβx,d(xδz)] α =[xδ[x,z] β +[x,x] β δz, z] α +[xβx,d(xδz)] α =xδ[[x,z] β,z] α +[x,z] α δ[x,z] β +[xβx,d(xδz)] α. But [x,z] β Z(M), thus by above relation we have [x,z] α δ[x,z] β =-[xβx,d(xδz)] α =[xβx,d(xδz)] α. (5) So by (5) then [x,z] α δ[x,z] β =[xβx,d(x)δz+xδd(z)] α =[xβx,zδz]+[ xβx,xδd(z)] α Now from (4) we get [xβx,zδz] α =0, therefore from above relation we get [x,z] α δ[x,z] β = [xβx,d(xδz)] α =xδ[xβx,d(z)] α +[xβx,x] α δd(z)=xδ[xβx,d(z)] α [x,z] α δ[x,z] β =xδ([xβx,d(z)] α +2d(xβx)αz+2zαd(xβx)) =xδ(xβxαd(z)-d(z)αxβx+2d(xβx)αz+2zαd(xβx)) =xδ(d(xβx)αz+xβxαd(x)+ d(z)αxβx+zαd(xβx) + d(xβx)αz-zαd(xβx)) = xδ(d(xβxαz)-d(z α xβx)+[d(xβx),z] α ) = xδ(d([xβx,z] α )+ [d(xβx),z] α ) Now from (4), [xβx,z] α =0 and since d(xβx)=d(x)βx+xβd(x) =xβd(x) - d(x)βx =[x,d(x)] β Z(M). Therefore [d(xβx),z] α =0. So [x,z] α δ[x,z] β =0. 33

44 CHAPTER TWO Therefore (0)=MΓ[x,z] α Γ[x,z] β =[x,z] α ΓMΓ[x,z] β But M is a prime Γ-ring then [x,z] α =[x,d(x)] α =0. So d is Γ-commuting on U. Theorem 2.1.7: Let M be a prime Γ-ring satisfying(*) and U be a nonzero ideal in M. If d is a nonzero derivation of M such that [x,d(x)] α Z(M) for every x U and α Γ. Then M is commutative. Proof: By Lemma , we have [x,d(x)] α =0 for all x U and α Γ. Therefore 0=[x+y,d(x+y)] α =[x,d(y)] α +[y,d(x)] α x,y U and α Γ (1) Since U is an ideal, in(1) replace y by xβy U for all β Γ, we have 0=[x,d(xβy)] α +[xβy,d(x)] α =[x,d(x)βy] α +[x,xβd(y)] α +[ xβy, d(x)] α =d(x)β[x,y] α +[x,d(x)] α βy+xβ[x,d(y)] α +xβ[y,d(x)] α +[x,d(x)] α βy = d(x)β[x,y] α +xβ([x,d(y)] α +[y,d(x)] α )= d(x)β[x,y] α. Thus 0= d(x)β[x,y] α for all x,y U and α,β Γ. (2) In (2) replace y by yδa for all a M,δ Γ,we have 0=d(x)β[x,yδa] = d(x)β(yδ[x,a] α +[x,y] α δa) 34

45 CHAPTER TWO = d(x)βyδ[x,a] α + d(x)β[x,y] α δa = d(x)βyδ[x,a] α. Therefore from the above relation,we have d(x)γ UΓ[x,a] α =(0), but MΓU U, which means that d(x)γmγu[x,a] α =(0) but M is a prime Γ-ring,then either d(x)=0 or [x,a] α =0 for all a M So for any element x U, d(x)=0 or x Z(M). By assumption, d is not zero on M then by Lemma1.3.7 d is not zero on U. Thus there is an element x 0 in U such that d(x) 0 and x Z(M). Let y be any other element in U. If y Z(M) then y+x Z(M) and d(y)=0. But,0=d(x+y)=d(x)+d(y)=d(x) and so d(x)=0, a contradiction. Hence y Z(M) for every y in U. Therefore U Z(M) and by Theorem1.2.3 then M must be commutative. 2.2.Extend Γ- centralizing Derivations on Prime Γ-rings: Our aim in this section is to present two theorems; Let M be a prime Γ-ring of characteristic not two satisfying(*). Suppose there exists a nonzero derivation d:m M,such that (i) The mapping x [d(x),x] β is Γ-commuting on M, in this case M is commutative. (ii) The mapping x [d(x),x] β is Γ- centralizing on M and characteristic of M not three.then M is commutative. 35

46 CHAPTER TWO Theorem 2.2.1: Let M be a prime Γ-ring of characteristic not two satisfying(*). Suppose there exists a nonzero derivation d:m M,such that the mapping x [d(x),x] β is Γ-commuting on M. In this case M is commutative. Proof: By assumption, we have [[d(x),x] β, x] α =0 for all x M and α,β Γ (1) Let us introduce a mapping B(.,.):M M M by the relation B(x,y)=[d(x),y] α +[d(y),x] α, x,y M,α Γ. It is obvious that bi-additive. B(.,.) is symmetric (i.e. B(x,y)=B(y,x)) and A simple calculation shows that the relation B(xβy,z)=B(x,z)βy+xβB(y,z)+d(x)β[y,z] α +[x,z] α βd(y) and α,β Γ for all x,y,z M (2) We introduce also a mapping f: M M by f(x)=b(x,x). We have f(x)=2[d(x),x] α for all x M and αα Γ. (3) Obviously the mapping f satisfies the relation f(x+y)=f(x)+f(y)+2b(x,y) for all x,y M and α Γ. (4) Throughout the proof we shall use the mapping B(.,.) and the relations (2),(3) and (4) without specific reference. The relation (1) can now be written in the form [f(x),x] α =0 (5) The linearization of (5) gives 36

47 CHAPTER TWO [f(x+y),x+y] α =[f(x),y] α +[f(y),x] α +2[B(x,y),x] α +2[B(x,y),y] α =0. (6) The substitution x for x in the above relation leads to [f(x),y] α -[f(y),x] α +2[B(x,y), x] α -2[B(x,y), y] α =0. (7) From (6) and (7) we obtain 2[f(x),y] α +4[B(x,y),x] α =0,but ChM 2, then we have [f(x),y] α +2[B(x,y),x] α =0 (8) Let us replace in (8) y by xβy.then 0=[f(x),xβy] α +2[B(xβy,x), x] α =[f(x),xβy] α +2[f(x)βy+xβB(y,x)+d(x)β[y,x] α,x] α =[f(x),x] α βy+xβ[f(x),y] α +2[f(x),x] α βy+2f(x)β[y,x] α +2xβ[B(x,y),x] α +2[d(x),x] α β[y,x] α + 2d(x)β[[y,x] α,x] α. Using the above calculation, (5) and (8) we arrive at 3f(x)β[y,x] α +2d(x)β[[y,x] α,x] α =0 for all x,y M and α,β Γ. (9) Now from (8) replace y by yβx, for all β Γ we have 0=[f(x),yβx] α +2[B(yβx,x),x] α =3[y,x] α βf(x)+2[[y,x],x] α βd(x). (10) We intend to prove that 0=(3f(x)βd(x)-d(x)βf(x))δ[y,x] for all x,y M and α,δ,β ΓΓ (11) holds for this purpose we write yδz instead of y in (9) we have 0=3f(x)βyδ[z,x] α +4d(x)β[y,x] α δ[z,x] α +2d(x)βyδ[[z,x] α,x] α, x,y,z M. Putting in the above relation y=d(x) and z=y, we obtain 0=3f(x)βd(x)δ[y,x] α +4d(x)β[d(x),x] α δ[y,x] α +2d(x)βd(x)δ[[y,x] α,x] α 0=3f(x)βd(x)δ[y,x] α + 2d(x)βf(x)δ[y,x] α +2d(x)βd(x)δ[[y,x] α,x] α. Now from (9) and since M be a Γ-ring then 37

48 CHAPTER TWO 0=3d(x)βf(x)δ[y,x] α +2d(x)βd(x)δ[[y,x] α,x] α relation we have replace it in the above 0=3f(x)βd(x)δ[y,x] α - d(x)βf(x)δ[y,x] α 0=(3f(x)βd(x) - d(x)βf(x)) δ[y,x] α. (12) Now from (12) replace y by yμμz for all z M and μμ Γ, we get 0=(3f(x)βd(x) - d(x)βf(x)) δ[yμμz,x] α 0=(3f(x)βd(x) - d(x)βf(x))δyμμ[z,x] α. In the above relation replace z by 2d(x), get to 0=(3f(x)βd(x) - d(x)βf(x))δyμμ(2[d(x),x] α ) 0=(3f(x)βd(x) - d(x)βf(x))δyμμf(x). (13) Now since M be a Γ-ring then by (13) we have 0=(3f(x)βd(x) - d(x)βf(x))δyμ(3f(x)βd(x)). (14) In (13) also we can get to 0=(3f(x)βd(x) - d(x)βf(x))δyβf(x). In the above relation replace y by yμμd(x) then 0=(3f(x)βd(x) - d(x)βf(x))δyμμd(x)βf(x). (15) Now from (15) and (14) we have 0=(3f(x)βd(x) - d(x)βf(x))δyμμ(3f(x)βd(x) - d(x)βf(x)). (16) But M is a prime Γ-ring then 3f(x)βd(x) - d(x)βf(x)=0. (17) By the same way we can prove that 3d(x)βf(x)- f(x)βd(x)=0. (18) Therefore from (18), we have f(x)βd(x) =3d(x)βf(x), which replace in (16) we get 38

49 CHAPTER TWO 0=(9 d(x)βf(x)- d(x)βf(x)) δyμμ(9 d(x)βf(x)- d(x)βf(x)) =(8 d(x)βf(x)) δyμμ(8 d(x)βf(x)) But M is prime Γ-ring and ChM 2, then d(x)βf(x)=0. Now in the above relation replace x by x+y,we have 0=d(x+y)βf(x+y)=( d(x)+ d(y))β(f(x)+ f(y)+2b(x,y)) = d(x)βf(x)+d(y)βf(x)+d(x)βf(y)+d(y)βf(x)+2d(x)βb(x,y) +2d(y)βB(x,y) d(x)βf(y)+ d(y)βf(x)+2d(x)βb(x,y) +2d(y)βB(x,y)=0 (19) Now in (19) replace x by x which gives -d(x)βf(y)+ d(y)βf(x)+2d(x)βb(x,y) -2d(y)βB(x,y)=0 (20) Adding (19) and (20) we arrive to 0=2d(y)βf(x)+4d(x)βB(x,y)=2(d(y)βf(x)+2d(x)βB(x,y)), therefore d(y)βf(x)+2d(x)βb(x,y)=0. (21) In (21) replace y by yδx,then 0= d(yδx)βf(x)+2d(x)βb(x, yδx) = d(y)δxβf(x)+yδd(x)βf(x)+2 d(x)βb(xδy,x) = d(y)δxβf(x)+ 2d(x)βB(y,x)δx+2d(x)βyδf(x)+2d(x)β[y,x] α δd(x) = d(y)βxδf(x)-d(y)βf(x)δx +2d(x)βyδf(x) +2d(x)β[y,x] α δd(x) =d(y)β[x,f(x)] δ +2d(x)βyδf(x) +2d(x)β[y,x] α δd(x) Now from (5) and ChM 2, then d(x)βyδf(x) +d(x)β[y,x] α δd(x)=0. (22) In (22) replace xβy by y we get d(x)βxβyδf(x) +d(x)β[xβy,x] α δd(x)=0, d(x)βxβyδf(x) + d(x)βxβ[y,x] α δd(x)=0. (23) 39

50 CHAPTER TWO Since M be a Γ-ring, then for all x M and by (22) we have xβd(x)βyδf(x) +xβd(x)β[y,x] α δd(x)=0. (24) Therefore from (23) and (24) we arrive to [d(x),x] β βyδf(x) +[d(x),x] β β[y,x] α δd(x)=0. Therefore 2[d(x),x] β βyδf(x) +2[d(x),x] β β[y,x] α δd(x)=0 f(x)βyδf(x)+ f(x)β[y,x] α δd(x)=0. (25) In (10) replace yδz instead of y we have 0=3[yδz,x] α βf(x)+2[[yδz,x], x] α βd(x) =3(yδ[z,x] α +[y,x] α δz )βf(x)+ 2[yδ[z,x] α +[y,x] α δz, x] α βd(x) = 3yδ[z,x] α βf(x)+3[y,x] α δz βf(x)+ 2[yδ[z,x] α,x] α βd(x) +2[[y,x] α δz,x] α βd(x) =3yδ[z,x] α βf(x)+3[y,x] α δz βf(x)+ 2yδ[[z,x] α,x] α βd(x) +2[y,x] α δ[z,x] α βd(x)+2[y,x] α δ[z,x] α βd(x)+2[[y,x] α,x] δz β d(x) 3[y,x] α δzβf(x)+ 2[[y,x] α,x] δz β d(x)+4[y,x] α δ[z,x] α βd(x)=0 (26) Putting in (26) y=2d(x) and making use of (1) 0=3[2d(x),x] α δzβf(x)+ 2[[2d(x),x] α,x] δz β d(x)+4[2d(x),x] α δ[z,x] α βd(x) 0=3f(x)δzβf(x)+ 4f(x)δ[z,x] α βd(x) for all x,z M and δ,β Γ. In above replace z by y we have 0=3f(x)δyβf(x)+ 4f(x)δ[y,x] α βd(x), but M satisfy (*) then 3f(x)βyδf(x)+ 4f(x)β[y,x] α δd(x)=0 (27) Now from (25) and (27) we have 40

51 CHAPTER TWO -f(x)βyδf(x)=0 thus f(x)βyδf(x)=0,since M is a prime Γ-ring and ChM 2, f(x)=2[d(x), x] α =0, then [d(x), x] α =0 for all x M and αα Γ. Now by Theorem ,M is commutative. Theorem 2.2.2: Let M be a prime Γ-ring of characteristic deferent from two and three satisfying(*). Suppose that, there exists a nonzero derivation d:m M,such that the mapping x [d(x),x] α is Γ-centralizing on M. Then M is commutative. Proof: Let B(x,y)=[d(x),y] α +[d(y),x] α,for all x,y M and αα Γ. And f(x)= B(x,x)=2[d(x),x] α. Since the map x [d(x),x] α is Γ-centralizing on M then [f(x),x] α Z(M). (1) By the same steps of Theorem we can prove that [f(x),y] α +2[B(x,y),x] α Z(M). (2) Put in (2) xβx for y,we obtain [f(x), xβx] α +2[B(x, xβx),x] α Z(M). Now from Theorem (2),we have xβ[f(x),x] α +[f(x),x] α βx+2([b(x,x)βx+xβb(x,x),x] α ) =2[f(x),x] α βx+[f(x)βx,x] α +2[xβf(x),x] α =2[f(x),x] α βx +2[f(x),x] α βx+2[f(x),x] α βx =6[f(x),x] α βx Z(M). Now for all y M and δ Γ,we have 41

52 CHAPTER TWO 0=[6[f(x),x] α βx,y] δ =6 [ [f(x),x] α βx,y] δ. But ChM 2,3,then 0=[[f(x),x] α βx,y] δ which leads to [f(x),x] α β[x,y] δ =0. (3) Now when x Z(M) then for any fixed x Z(M) a mapping y [ x,y] δ is a nonzero derivation which means that with(3) and Lemma , [f(x),x] α =0. Therefore But ChM 2, then 0=[2[d(x),x] α,x] δ =2[ [d(x),x] α,x] δ. [[d(x),x] α,x] δ =0 i.e. the map x [d(x),x] α is Γ-commuting on M then by Theorem2.2.1., M is commutative Γ-centralizing Derivations on Semi-Prime Γ-rings: In this section we will prove that, a semi-prime Γ-ring M contains a non zero central ideal,when M admits a nonzero Γ-centralizing derivation d on a nonzero left ideal U of M. For this aim we need to prove some lemmas as follows. Lemma 2.3.1:[8] Let U be an ideal of a Γ-ring M. Then the following conditions are equivalent : 1. U is a prime ideal of M. 2. If a, b M and aγmγb U,then either a U or b U. 42

53 CHAPTER TWO Lemma 2.3.2: Let M be a semi-prime Γ-ring satisfying(*) and U be a nonzero left ideal. If d is a derivation of M which is Γ- centralizing on U.Then d is Γ-commuting on U. Proof: For all x U and β Γ, xβx U. Therefore by assumption,we have That is [xβx,d(xβx)] α Z(M) for all α Γ. [xβx,d(xβx)] α =[ xβx,2xβd(x)-[x,d(x)] β ] α =2[ xβx, xβd(x) ] α =2(xβ[xβx,d(x)] α +[ xβx,x] α βd(x)) =2(xβxβ[x,d(x)] α +xβ[x,d(x)] α βx) Since [x,d(x)] α Z(M) for all x U and α Γ, then from above relation we have 4xβxβ[x,d(x)] α Z(M). (1) Now for all α Γ with (1) we get 0=4[xβxβ[x,d(x)] α,d(x)] α =4(xβxβ[[x,d(x)] α,d(x)] α +[xβx,d(x) ] α β[x,d(x)] α ) =4([xβx,d(x) ] α β[x,d(x)] α ) =4((xβ[x,d(x)] α +[x,d(x)] α βx) β[x,d(x)] α ) =4(2xβ[x,d(x)] α β[x,d(x)] α ) 8xβ[x,d(x)] α β[x,d(x)] α =0. (2) We intend to prove that 43

54 CHAPTER TWO (2[x,d(x)] α β) 2 2[x,d(x)] α =0. (3) Now since 8[x,d(x)] α β[x,d(x)] α β[x,d(x)] α =8[x,d(x)] α β[x,d(x)] α βxαd(x) -8[x,d(x)] α β[x,d(x)] α βd(x)αx =0. By (2), we get 0=8[x,d(x)] α β[x,d(x)] α β[x,d(x)] α =2[x,d(x)] α β(2[x,d(x)] α )β(2[x,d(x)] α ). Thus (2[x,d(x)] α β) 2 2[x,d(x)] α =0. Now by (3) and Lemma we have 2[x,d(x)] α =0,for all x U, α Γ. (4) But [xβx,d(x)] α =xβ[x,d(x)] α +[x,d(x)] α βx=2xβ[x,d(x)] α =0. Therefore [xβx,d(x)] α =0. (5) By assumption,we have [x,d(x)] α Z(M), for all x U,α Γ then it is easly prove to that for all x,y U,α Γ [x,d(y)] α +[y,d(x)] α Z(M) and 2([x,d(y)] α +[y,d(x)] α )=0. (6) Now we intend to prove that [xβy+yβx,d(x)] α +[xβx,d(y)] α =0 (7) [xβy+yβx,d(x)] α +[xβx,d(y)] α =[xβy,d(x)] α +[yβx, d(x)] α +xβ[x,d(y)] α +[x,d(y)] α βx =xβ[y,d(x)] α +[x,d(x)] α βy+yβ[x,d(x)] α +[y,d(x)] α βx+xβ[x,d(y)] α +[x,d(y)] α βx = xβ([y,d(x)] α + [x,d(y)] α ) 44

55 CHAPTER TWO +( [y,d(x)] α +[x,d(y)] α ) βx+2yβ[x,d(x)] α [xβy+yβx,d(x)] α +[xβx,d(y)] α =2xβ([y,d(x)] α + [x,d(y)] α ) +2yβ[x,d(x)] α. Therefore from the above relation,(4) and (6) we have [xβy+yβx,d(x)] α +[xβx,d(y)] α = 0. In (6) replace y by yβx we have 0=[xβyβx+yβxβx,d(x)] α +[xβx,d(yβx)] α =[xβyβx,d(x)] α +[yβxβx,d(x)] α +[xβx,d(yβx)] α =xβyβ [x,d(x)] α +[xβy,d(x)] α βx+yβxβ[x,d(x)] α +[yβx,d(x)] α βx +[xβx,d(y)βx] α +[xβx,yβd(x)] α =(xβy+yβx)β[x,d(x)] α +[ xβy+yβx,d(x)] α βx+d(y)β[xβx,x] α +[xβx,d(y)] α βx+yβ[xβx,d(x)] α +[xβx,y] α βd(x) =(xβy+yβx)β[x,d(x)] α +([ xβy+yβx, d(x)] α +[xβx,d(y)] α )βx +[xβx,y] α βd(x) 0=(xβy+yβx)β[x,d(x)] α +[xβx,y] α βd(x) =([x,y] β +2yβx)β[x,d(x)] α +[xβx,y] α βd(x) =[x,y] β β[x,d(x)] α +2yβxβ[x,d(x)] α +[xβx,y] α βd(x) 0 =[x,y] β β[x,d(x)] α +[xβx,y] α βd(x) In the above relation let y=d(x)βx 0 =[x,d(x)βx] β β[x,d(x)] α +[xβx,d(x)βx] α βd(x) =[x,d(x)βx] β β[x,d(x)] α =[x,d(x)] β βxβ[x,d(x)] α =xβ[x,d(x)] β β[x,d(x)] α. Replace α by β,we have 45

56 CHAPTER TWO 0=xβ[x,d(x)] β β[x,d(x)] β. Therefore ([x,d(x)] β β) 2 [x,d(x)] β =0. Since [x,d(x)] β Z(M), then by Lemma1.2.6, we have [x,d(x)] β =0,for all x U. Theorem 2.3.3: Let M be a semi-prime Γ-ring satisfying(*) and U a nonzero left ideal,if M admits a Γ-centralizing derivation d which is nonzero on U. Then M contains a nonzero central ideal. Proof: By Lemma 2.3.2, d is a Γ-commuting on U. Now since [x,d(x)] δ =0, for all x U,δ Γ then [x,d(y)] δ +[y,d(x)] δ =0 for all x,y U,α Γ. (1) In (1) replace y by yβx we have 0=[x,d(yβx)] δ +[yβx,d(x)] δ =[x,d(y)βx+yβd(x)] δ +[yβx,d(x)] δ =[x,d(y)] δ βx+d(y)β[x,x] δ +yβ[x,d(x)] δ +[x,y] δ βd(x)+yβ[x,d(x)] δ +[y,d(x)] δ βx =([x,d(y)] δ +[y,d(x)] δ )βx+[x,y] δ βd(x) [x,y] δ βd(x)=0. (2) In (2) replace y by wαy for all w,y U, α Γ we have 0=[x, wαy] δ βd(x)=(wα[x,y] δ +[x,w] δ αy)βd(x) = wα[x,y] δ βd(x)+[x,w] δ αyβd(x). 46