On ERT and MERT-Rings

The main purpose of this paper is to study ERT and MERT rings, in order to study the connection between such rings and IIregular rings. Keyword: MERT-Rings, ERT-Rings and π-Regular Rings لوح عونلا نم تاقلحلا ERT و MERT ميها ربإ دمحم ةديبز تايضايرلاو بوساحلا مولع ةيلك لصوملا ةعماج :ثحبلا ملاتسا خيرات 26 / 06 / 2002 :ثحبلا لوبق خيرات 01 / 09 / 2002


1-Introduction:
Throughout this paper, R denotes an associative ring with identity, and all modules are unitary right R-module. Recall that; 1-An ideal I o f the ring R is essential if I has a non-zero intersection with every non-zero ideal of R; 2-A ring R is said to be -regular if for every a in R there exist a positive integer n and b in R such that a n = a n b a n 3-A right R-module M is said to be GP-injective if, for any 0  a R , there exists a positive integer n such that a n 0 and any right R-homomorphism of a n R into M extends to one of R into M. 4-For any element a in R, r(a), I (a) denote the right annihilator of a and the left annihilator of a, respectively.

2-ERT-R1NGS:
Following [3J, a ring R is said to be ERT-ring if every essential right ideal of R is a two-sided ideal.

Definition 2-1:
A ring R is said to be right weakly regular if for all a in R , there exists b in RaR such that a = ab, or equivalently every right ideal of R is idempotent.
We begin this section with the following main result:

Theorem 2.2:
If R is ERT-ring with every essential right ideal is idempotent, then R is weakly regular.

Proof:
For any a  R, if RaR not essential, then there exists an ideal I, such that K = RaR  I is essential then K = K 2 . In order to prove that R is weakly regular, we need to prove RaR = (RaR) 2 . For a K , we have aK 2 ,that is a (RaR I) 2 Thus a = (rar ' This implies that a = (rar' + i)sas' + (rar' + i) i' = rar'sas' + isas' + (rar 1 2 , this proves that R is weakly regular ring. Following [2], the singular submodule of R is

Theorem 2.3:
Let R be a semi-prime ERT right GP-injective ring. Then R is a right non singular.

Proof:
Let E be an essential right ideal of R. Then E is a two-sided ideal, and hence l(E) is a two-sided ideal ofR.
Since R is semi-prime, then l(E) E = 0 , whence l(E) = 0. This proves that R is right non singular.

3-MERT-R1NGS:
Following [3], a ring R is said to be MERT-ring if every maximal essential right ideal of/? is a two-sided ideal.

Theorem 3.1:
Let R be an MERT-ring, if for any maximal right ideal A/of R, and for any b e M, bR/bM is GP-injective, then R is strongly Pi-regular ring.

Proof:
Let b be a non-zero element in R, we claim that b n r+r(b")=R.
If b n r + r(b n )  R, let M be a maximal right ideal containing b n r + r(b n ). Then M is essential right ideal of R.
If bR = bM, then b = bc, for some c in M, this implies Since R/M is GP-injective, then there exists c  R, such that: 1+M=f(b")=cb"+M and so (1-c b n )  M, since b n M, and R is MERT-ring, this implies that M is a two-sided ideal, and hence c b"  M. Thus I M, a contradiction. Therefore b n R + r (b n ) = R.
In particular l=b n u+v;v r(b n ), u  R. Thus b n = b 2n u and therefore R is strongly -regular ring.

Theorem 3.2:
If R is MERT-ring with every simple singular right ideal is GP-injective, then Y(R)=0.

Proof:
If Y(R) 0, by Lemma (7) of [6], there exists 0  y  Y(R) with y 2 =0. Let L be a maximal right ideal of R, set L = y R+ r(y) , we claim that L is essential right ideal o f 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 implies T  r(y) L, so L  T=(0). This contradiction proves that L is an essential right ideal, that is R/L is simple singular and hence R/L is GP-injective. Now; Let f;yR -> R/L be defined by f(yr)=r+L, then f is a well-defined Rhomomorphism.
Since R/L is GP-injective, so  c R, such that l+L=f(y)=cy+L.
Since R is MERT, then eye L and thus 1  L, a contradiction.

Theorem 3.3:
Let R be a ZT ring. If every simple singular rightsmodules is GP-injective which is left self-injective, then R is strongly H-regular ring.
Thus for any left ideal I, L(I)  l = 0.
Since R is simple singular GP-injective and ZI, then R is reduced and hence r(a)=l(a) for any element a in R.
Since R is left self-injective ring, then aR is a right annihilator, by Proposition (4)of [4].
Since r(a)  r(a n ), then a n R = r(a fl ). Now, since R= r(l(r(a)))+ r(l(a)) 9 then we have R = r(l(r(a n )) + r(l((a n )) = r(a n ) +a n R In particular, for some b in R, and d in r(a n ). Thus a" = a n2 b. Therefore R is strongly -regular.