On Rings whose Maximal Essential Ideals are Pure

Raida D. Mahmood Awreng B. Mahmood raida.1961@uomosul.edu.iq awring2002@yahoo.com College of Computer sciences and Mathematics University of Mosul, Iraq Received on: 06/04/2006 Accepted on: 25/06/2006 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 I a  , there exists I b  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 : =  ax R x } , and likewise the left annihilator l(a). 3) R is strongly regular [4], if for every 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] 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 a 2 = 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 = e 2  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.

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 Since a J, then there exists an invertible element v in R such that ( 1-ar) v = 1 , so (a-a 2 r) v = a , yields a = 0 . This proves that J(R) =(0).  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: , by Lemma (7) of [6] , there exists ) ( 0 R Y y   with y 2 = 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  T = (0) . Then yRT  LT  L  T = 0 impolies T  r(y)  L, so L  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).

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.
Following [3],a ring R is right (left) weakly regular if I 2 = 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 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.  Before closing this section, we give the following result.

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.