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Raf. J. of Comp. & Math

’

s. , Vol. 5, No.1, 2008

21

Maximal Generalization of Pure Ideals

Raida D. Mahmood Awreng B. Mahmood

College of Computer science College of Computer science

and Mathematics and Mathematics

University of Mosul University of Mosul

Received on:6/4/2006 Accepted on:25/6/2006

ﺺﺨﻠﻤﻟا

ﻤﻟﺍ لﻜ ﺎﻬﻴﻓ ﻥﻭﻜﺘ ﻲﺘﻟﺍ ﺕﺎﻘﻠﺤﻟﺍ ﺔﺴﺍﺭﺩ ﻭﻫ ﺙﺤﺒﻟﺍ ﺍﺫﻫ ﻥﻤ ﺽﺭﻐﻟﺍ ﻰﻤﻅﻌﻟﺍ ﻰﻨﻤﻴﻟﺍ ﺕﺎﻴﻟﺎﺜ

ﻯﺭﺴﻴ ﺔﻤﻤﻌﻤ ، ﻰﻠﻋ ﺕﺎﻘﻠﺤﻟﺍ ﻩﺫﻫ ﻑﻴﺭﻌﺘ ﻡﺘ ﺎﻤﻜﺎﻬﻨﺃ ﻁﻤﻨﻟﺍ ﻥﻤ MGP ﺽﻌﺒ ﺽﺭﻋ ﻡﺘﻭ

ﺔﻴﺴﺎﺴﻷﺍ ﺎﻬﺼﺍﻭﺨ ﺵﺎﻜ ﺔﻘﻠﺤﻭ ﻑﻌﻀﺒ ﺔﻤﻅﺘﻨﻤ ﺔﻘﻠﺤ ، ﺓﻭﻘﺒ ﺔﻤﻅﺘﻨﻤﻟﺍ ﺔﻘﻠﺤﻟﺍ ﻊﻤ ﺎﻬﺘﻗﻼﻋﻭ.

ABSTRACT

The purpose of this paper is to study the class of the rings for which

every maximal right ideal is left GP-ideal. Such rings are called MGP-rings

and give some of their basic properties as well as the relation between

MGP-rings, strongly regular ring, weakly regular ring and kasch ring.

1- Introduction :

Throughout this paper, R denotes as associative ring with identity. An

ideal I of a ring R is said to be right(left) pure if for every a∈I, there exists

b∈I such that a=ab (a=ba). This concept was introduced by Fieldhouse [6],

[ 7 ], Al-Ezeh [ 2 ],[ 3 ] and Mahmood [ 9 ].

Recall that:-

1- A ring R is regular if for every a∈R there exists b∈R such that a=aba, if

a=a2 b, R is called strongly regular.

2-A ring without non-zero nilpotent elements is called reduced.

3-For any element a∈R, r(a) and l(a) denote the right annihilator and the

left annihilator of a, respectively.

4-A ring R is said to be a left(right) uniform ring if and only if every non-

zero left(right) ideals is essential .

5-Following [10], a ring R is said to be semi commutative if xy=0 implies

that xRy=0, x,y ∈R. Clearly every reduced ring is semi commutative. It

is easy to see that R is semi commutative if and only if every left(right)

annihilator in R is a two-sided ideal.

6-Y(R), J(R) are respectively the right singular ideal and the Jacobson

radical of R.

Raida D. Mahmood and Awreng B. Mahmood

22

2- MGP-rings

In this section, the concept of maximal GP-ideals is introduced and we

use it to define MGP-rings .We study such rings and give some of their

basic properties.

Following [8], an ideal I of a ring R is said to be right (left) GP-ideal

(generalized pure ideal), if for every a in I, there exists b in I and a positive

integer n such that an =an b (an =b an).

Definition 2.1 :

A ring R is called a right (left) MGP-ring if and only if every

maximal right (left) ideal is left (right) GP-ideal.

Example:

Let Z12 be the ring of the integers module 12.

Then the maximal ideals , I={0,3,6,9} , J={0,2,4,6,8,10} are GP-ideals.

The following theorem gives some interesting characteristic

properties of right MGP-rings. Before that we need the next lemma in our

proof.

Lemma 2.2:

Let a be a non zero element of a ring R and let l(a) = 0. Then

for every positive integer n, l (an) = 0.

Proof: obvious #

Theorem 2.3 :

If R is a right MGP-ring and every ideal is principal, then any left

regular element is right invertible .

Proof :

Let 0≠c ∈ R, such that l(c) = 0. If c R

≠

R, then there exists a

maximal right ideal M containing cR . Since R is right MGP-ring, then M is

a left GP-ideal, there exists d ∈ M and a positive integer n, such that

cn= dcn and d = cx , for some x ∈ R .

So ( 1-cx) ∈ l(cn) ,Since l( c ) = 0 , then by Lemma (2.2) we have

l(cn) = 0 , thus cx = 1 ∈ M , this contradicts cR

≠

R. Therefore

cR = R , and hence c is a right invertible. #

Lemma 2.4 :

Let R be a reduced ring. Then for every a ∈ R , and every positive

integer n , an R

∩

r (an) = 0 .

Proof: See [8]

Maximal Generalization …

23

Proposition 2.5 :

Let R be a reduced, MGP-ring. Then for every a in R and

a positive integer n, r (an ) is a direct summand of R .

Proof :

To prove r (an) is a direct summand, we claim that an R + r (an) = R.

If this is not true, let M be a maximal right ideal containing an

R + r (an).

Since R is MGP-ring , so (an)m = b (an)m for some b ∈M and a positive

integer m, this implies (1-b) ∈ l( anm ) = r (an)

⊆

M (R is reduced) , and so

1∈ M,a contradiction. Hence anR+r(an) =R .

Now, since an

R

∩

r (an) =0 ,Lemma( 2.4) ,then r (an) is a direct

summand. #

Recall that, a ring R is called a right (left) MP-ring if every maximal

right (left) ideal is a left (right) pure.

We consider the condition (*): R satisfies l( b

n

)

⊆

r(b) for any

b ∈ R and a positive integer n .

Theorem 2.6 :

Let R be a ring satisfying (*). Then R is a right MGP-ring if and

only if R is strongly regular .

Proof: If this is not true let R be a right MGP-ring and let b be any element

in R .We shall prove that bR + r (b) =R .

If this is not true let M be a maximal right ideal containing bR+r(b) . Since

R is an MGP-ring , then there exists a ∈ M and a positive integer n such

that bn = abn which implies that (1-a) ∈ l( bn)

⊆

r(b)

⊂

M , thus

1 ∈ M ,a contradiction . Therefore bR+ r(b) = R .

In particular, b u + v = 1 , for some u ∈ R , v∈r(b) .

So b = b2 u , therefore R is strongly regular .

Conversely; assume that R is strongly regular, then by [1], R is

regular and reduced. Also by [9] , R is an MP-ring and semi commutative,

then R is an MGP –ring . #

Proposition 2.7 :

Let R be a right MGP-ring satisfying (*). Then Y(R) = 0.

Proof:

If Y(R)

≠

0 ,then by a Lemma (7) of [10]; there exists

0

≠

a ∈Y (R) with a2 = 0 .From Theorem (2.6) R is strongly regular , that

is a = a2 b , for some b ∈ R . Hence a = 0, contradiction. Therefore Y(R) =

0. #

Raida D. Mahmood and Awreng B. Mahmood

24

Proposition 2.8 :

If R is a right MGP – ring, then any reduced principal right ideal of

R is a direct summand .

Proof : Let I = aR be a reduced principal right ideal of R .If aR+r(a )

≠

R،

then there exists a maximal right ideal M of R containing aR+r(a).

Now , since R is a right MGP-ring and a ∈ M , then there exists

b ∈ M and a positive integer n such that an = b an ,and hence (1-b)an= 0 .

Since I is reduced then we have (1-b) ∈ l( a

n

) = r(an) = r(a)

⊂

M, this

implies that 1 ∈ M , which contradicts M

≠

R. Therefore, aR+r(a) = R ,

thus a =a2 c for some c∈R. If we set d=a2∈I, then a=a2 d . implies that

a = ada and hence aR = e R , where e = ad is an idempotent element . Then

by [6], aR is a direct summand. #

Proposition 2.9 :

Let R be a right MGP-ring satisfying (*). If a

n

b = 0, for any

a,b ∈ R and a positive integer n, then r (an) + r(b) = R .

Proof: Assume that r (an) +r(b)

≠

R . Let M be a maximal right ideal

containing r(an) +r(b) . Since R is a right MGP-ring and an b = 0 implies

that b ∈ r ( an)

⊆

M , there exists c ∈ M and a positive integer m such that

bm = cbm , so( 1- c) ∈ l( bm)

⊆

r(b)

⊂

M,which implies that 1∈M ,which is

a contradiction . Therefore r (an) + r(b) = R .

Theorem 2.10 :

Let R be a uniform semi commutative, MGP-ring and every ideal is

principal. Then R is a division ring .

Proof : Let 0

≠

a ∈ R and aR

≠

R ,and let M be a maximal right ideal

containing aR .Since R is an MGP-ring, then there exists b ∈ aR

⊆

M ،

and a positive integer n such that an= ban .

This implies that an =acan , for some c ∈ R. Since R is uniform so every

ideal is an essential ideal.

Let x ∈ r (ar)

∩

an R .Then acx =0 and x= an z for some z∈R, so

acan z= 0 , yields an

z=0=x .Therefore, r (ac)

∩

an

R = 0 ,since R is a

uniform ring and an R ≠0 ,then r(ac) = 0 . Since R is semi commutative ,

l(ac) = 0 , then by Theorem (2.3) ac is a right invertible element ,so there

exists v ∈R such that acv = 1 . Hence a (cv) = 1 ∈ M ,which is a

contradiction. Therefore aR = R .

Now, since ar=1 (aR=R),we have ara=a which implies that

(1-ra )∈ r(a)=l(a)⊆l(ar)=r(ar)=0.Therefore, (1-ra)=0 ,whence ra=1,so a is a

left invertible .Thus R is a division ring. #

Maximal Generalization …

25

3-The relation between MGP-rings and other rings

In this section we give further properties of the MGP-rings and link

between MGP- rings and other rings .

We shall begin this section with the following result, which gives the

connection between MGP-rings and weakly regular rings.

Following [11], a ring R is a right (left) weakly regular if I2 =I for

each right (left) ideal I of R . Equivalently, if a∈aRaR (a∈RaRa) for every a

in R . Then R is called weakly regular.

Theorem 3.1 :

Let R be a right MGP-ring and satisfying (*). Then R is a reduced

weakly regular ring .

Proof : Let a be a non zero element in R with a2 = 0. Let M be a maximal

right ideal containing r (a). Since a ∈ r (a)

⊆

M and R is an MGP-ring ,

then there exists b ∈ M and a positive integer n such that an = ban , which

implies that ( 1- b ) ∈ l(an)

⊆

r (a)

⊂

M , yielding 1 ∈ M,which is a

contradiction .

Therefore, a = 0, and hence R is a reduced ring .We show that

RxR + r(x) = R, for any x ∈ R.

Suppose that there exists y ∈ R such that RyR+ r (y)

≠

R .

Then there exists a maximal right ideal M of R containing RyR+r(y). Since

R is a right MGP-ring , there exists a in M and a positive integer n such

that yn = a yn implying that ( 1-a) ∈ l(yn)

⊆

r(y)

⊂

M , whence (1-a) ∈M

and so 1 ∈ M implies that M=R , which is a contradiction.

Therefore, RxR +r(x) =R, for any x ∈R.

Hence R is a right weakly regular ring. Since R is reduced, it also can be

easily verified that R is a weakly regular ring. #

Definition 3.2: [9]

A ring R is said to be a right (left) Kasch ring if every maximal right

(left) ideal is a right (left) annihilator .

Theorem 3.3 :

Every semi commutative right MGP-ring is a right Kasch ring .

Proof : Let M be any maximal right ideal of R and let Y(R) be the right

singular ideal of R .

If M

∩

Y(R) = 0, then for any y ∈ Y(R), y

∉

M, this implies that

r(y) is an essential right ideal of R .

Let x ∈ r (y)

∩

r (1-y), then yx= 0 and (1-y) x = 0 yields x=yx=0.

Raida D. Mahmood and Awreng B. Mahmood

26

Therefore r(y)

∩

r (I-y) = 0, whence r (1-y) = 0. Since R is semi

commutative ring, then we have l(1-y) = 0 .

By Theorem (2.3),(1-y) is an invertible element of R. Hence

y ∈ J

⊂

M ,a contradiction.

Thus M

∩

Y(R)

≠

٠ . Let 0

≠

a∈ M

∩

Y(R).

Since R is an MGP-ring, then there exists b ∈ M and a positive

integer n such that an = ban = aran .We claim that r (ar)

∩

an R = 0.

If not, let d∈r(ar)

∩

an R .Then ar d = 0 and d=an x for some x∈R , so

aran x = 0 implies that an x = 0=d. Therefore, r (ar)

∩

an R = 0.But r(ar) is

essential ,then an R = 0 and hence an x = 0 , for all x ∈ R implies that

an ∈ l(x) = r(x). Therefore, M = r(x).Thus R is a right Kasch ring. #

Corollary 3.4:

Let R be a reduced MGP-ring. Then R is a Kasch ring .

Proof: Since R is a reduced right MGP-ring .Then by Theorem (3.3) R is a

Kasch ring .#

Maximal Generalization …

27

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