Maximal Generalization of Pure Ideals

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.

Full text

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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