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Raf. J. of Comp. & Math
’
s. , Vol. 4, No. 1, 2007
57
On Rings whose Maximal Essential Ideals are Pure
Raida D. Mahmood Awreng B. Mahmood
College of Computer sciences and Mathematics
University of Mosul
Received on: 6/4/2006 Accepted on: 25/6/2006
ﺹﺨﻠﻤﻟﺍ ﻉﻭﻨ ﻥﻤ ﺕﺎﻘﻠﺤﻟﺍ ﻡﻭﻬﻔﻤ ﺙﺤﺒﻟﺍ ﺍﺫﻫ ﻡﺩﻘﻴ MEP) ﻟﺍ ﻲﻟﺎـﺜﻤ ﺀﺯﺠ لﻜ ﺎﻬﻴﻓ ﻲﺘﻟﺍ ﺕﺎﻘﻠﺤ
ﻥﻤﻴﺍﻲﻤﻅﻋﺃ ﻲﺴﺎﺴﺃ ﻲﻘﻨ ﻭﻫ ﺭﺴﻴﺃ (ﺀﺎﻁﻋﺇﻭ ﺎـﻬﻟ ﺔﻴﺴﺎﺴﻷﺍ ﺹﺍﻭﺨﻟﺍ . ﻁﻭﺭﺸـﻟﺍ ﺀﺎـﻁﻋﺇ ﻙﻟﺫـﻜ
ﺔﻘﻠﺤﻠﻟ ﺔﻴﻓﺎﻜﻟﺍﻭ ﺔﻴﺭﻭﺭﻀﻟﺍMEPﻑﻌﻀﺒ ﺔﻤﻅﺘﻨﻤﻭ ﺓﻭﻘﺒ ﺔﻤﻅﺘﻨﻤ ﺔﻘﻠﺤ ﻥﻭﻜﺘ ﻲﻜﻟ .
ABSTRACT
This paper introduces the notion of a right MEP-ring (a ring in
which every maximal essential right ideal is left pure) with some of their
basic properties; we also give necessary and sufficient conditions for MEP –
rings to be strongly regular rings and weakly regular rings.
1- Introduction
An ideal I of a ring R is said to be right (left)pure if for every Ia
∈
,
there exists Ib
∈
such that a=ab (a=ba),[1],[2].
Throughout this paper, R is an associative ring with unity.
Recall that:
1) R is called reduced if R has no non _zero nilpotent elements.
2) For any element a in R we define the right annihilator of a by
r(a)={ 0:
=
∈
axRx } , and likewise the left annihilator l(a).
3) R is strongly regular [4], if for every Ra
∈
,there exists Rb
∈
such that
baa 2
=.
4) Z,Y,J(R) are respectively the left singular ideal right singular ideal and
the Jacobson radical of R .
5) A ring R is said to be semi-commutative if xy=0 implies that xRy=0,for
all x,y
∈
R .It is easy to see that R is semi-commutative if and only if
every right (left) annihilator in R is a two-sided ideal [8]
Raida D. Mahmood and Awreng B. Mahmood
58
2-MEP-Rings:
In this section we introduce the notion of a right MEP-ring with
some of their basic properties;
Definition 2.1:
A ring R is said to be right MEP-ring if every maximal essential
right ideal of R is left pure.
Next we give the following theorem which plays the key role in
several of our proofs.
Theorem 2.2:
Let R be a semi commutative, right MEP–ring. Then R is a reduced
ring.
Proof: Let a be a non zero element of R, such that a2 = 0 and let M be a
maximal right ideal containing r (a). We shall prove that M is an essential
ideal. Suppose that M is not essential, then M is a direct summand, and
hence there exists 0
≠
e = e2 ∈ R such that M = r (e) (Lemma 2-3, of [8]).
Since R is semi commutative and a ∈ M , then e a = 0 and this implies
that e ∈ r (a) ⊆ M = r (e).
Therefore e=0, is a contradiction. Thus M is an essential right ideal.
Since R is a right MEP- ring, then M is left pure for every a ∈ M. Hence
there exists b∈M such that a = ba implies that (1- b) ∈ l(a) = r (a) ⊆ M,
so 1 ∈ M and this implies that M=R, which is a contradiction. Therefore
a= 0 and hence R is a reduced ring.r
Theorem 2.3:
If R is a semi commutative, right MEP-ring, then every essential
right ideal of R is an idempotent.
Proof:
Let I = bR be an essential right ideal of R . For any element b ∈ I,
RbR+ r (b) is essential in R (Proposition 3 of [5]).
If RbR + r (b)
≠
R, let M be a maximal right ideal containing
RbR + r(b) . Since R is MEP – ring , then there exists a ∈ M such that
b = ab and (1-a) ∈ l(b) = r (b)
⊂
M. So 1 ∈ M is a contradiction .
On Rings whose Maximal…
59
Thus RbR + r (b) = R , and 1 = u + d , u ∈ RbR
⊆
I , d ∈ r (b) .
Hence b = bu. Therefore I = I2 (Lemma 3 of [7]). r
Proposition 2.4:
Let R be a semi commutative, right MEP-ring. Then the J (R) =(0).
Proof:
Let 0≠a∈ J(R). If aR+r (a)
≠
R, then there exists a maximal right
ideal M containing aR +r (a ). Since a ∈ M and r (a)
⊂
M, then by a
similar method of proof used in Theorem (2.2) M is an essential ideal .
Since R is MEP – ring , then there exists b ∈ M, such that a = ba , but
a∈J(R)
⊂
M so 1∈M, is a contradiction. Therefore aR + r(a) = R
(Proposition 5 of [8]) and ar +d = 1 , for some r ∈ R and d ∈ r(a) , this
implies that a = a2r.
Since a ∈J, then there exists an invertible element v in R such that
( 1-ar) v = 1 , so (a-a2r) v = a , yields a = 0 . This proves that J(R) =(0). r
Recall that a ring R is said to be MERT-ring [7], if every maximal
essential right ideal of R is a two-sided ideal.
Theorem 2.5:
If R is MERT, MEP-ring, then Y(R) = (0).
Proof: If 0)(
≠
RY , by Lemma (7) of [6] , there exists )(0 RYy
∈
≠
with y2 = 0 . Let L be a maximal right ideal of R, containing r(y) .We
claim that L is an essential right ideal of R.
Suppose this is not true, then there exists a non-zero ideal T of R such that
L IT = (0) . Then yRT
⊆
LT
⊆
L I T = 0 impolies T
⊆
r(y)
⊆
L,
so L I T ≠ 0. This contradiction proves that L is an essential right ideal.
Since R is an MEP-ring, then L is a left pure.
Thus for every y ∈ L, there exists c ∈ L such that y = cy (L is a left pure).
Since R is MERT , then cy ∈L (two sided ideal)and thus 1∈L, is a
contradiction. Therefore Y (R) =(0).r
3- The connection between MEP-Rings and other rings
In this section, we study the connection between MEP-Rings and
strongly regular rings, weakly regular rings.
Raida D. Mahmood and Awreng B. Mahmood
60
Following [3],a ring R is right (left) weakly regular if I2 = I for
each right (left) ideal I of R. Equivalently, a ∈ aRaR ( a ∈ RaRa) for every
a ∈ R . R is weakly regular if it’s both right and left weakly regular.
The following result is given in [3]:
Lemma 3.1:
A reduced ring R is right weakly regular if and only if it is left
weakly regular.
Next we give the following lemma:
Lemma 3.2:
If R a semi-commutative ring then RaR+r(a) is an essential right
ideal of R for any a in R.
Proof: Given 0≠ a ∈ R, assume that [ RaR + r (a) ] II = 0 ,where I is a
right ideal of R. Then I a
⊆
I I RaR = 0, and so I
⊆
l(a) = r (a) (R is
semi commutative).Hence I = 0; whence RaR +r (a) is an essential right
ideal of R.r
Theorem 3.3:
If R is a semi commutative, right MEP-ring, then R is a reduced
weakly regular ring.
Proof: By Theorem (2.2), R is a reduced ring .We show that RaR+r(a)=R,
for any a ∈ R.
Suppose that RaR + r (a)
≠
R, then there exists a maximal right
ideal M containing RaR + r(a).By a similar method of proof used in
Theorem (2.2), M is an essential ideal.
Now R is MEP- ring, so a = ba , for some b ∈ M , hence
(1-b) ∈ l (a) = r (a) ⊆ M and so 1 ∈ M which is a contradiction. Therefore
M=R and hence RaR + r (a) = R, for any a ∈ R. In particular 1= cab + d,
for some c, b ∈ R, d ∈ r (a).
Hence a = acab and R is right weakly regular. Since R is reduced,
then by Lemma (3.1) R is a weakly regular ring. r
Before closing this section, we give the following result.
On Rings whose Maximal…
61
Theorem 3.4:
A ring R is strongly regular if and only if R is a semi-commutative,
MEP, MERT- ring.
Proof:
Assume that R is MEP, MERT-ring, let 0
≠
a ∈ R, we shall prove
that aR + r (a) = R . If aR + r(a)
≠
R , then there exists a maximal right
ideal M containing aR + r(a) . Since M is essential, then M is left pure.
Hence a= ba , for some b ∈ M , so 1 ∈ M, a contradiction . Therefore M=R
and hence aR+r(a) = R . In particular ar +d = 1, for some r ∈ R, d ∈ r(a).
So a=a²r.Therefore R is strongly regular.
Conversely: Assume that R is strongly regular, then by [3], R is
regular and reduced .Also R is MEP and semi-commutative.r
Raida D. Mahmood and Awreng B. Mahmood
62
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