Rings in which Every Simple Right R-Module is Flat

The objective of this paper is to initiate the study of rings in which each simple right R-module is flat, such rings will be called right SF-rings. Some important properties of right SFrings are obtained. Among other results we prove that: If R is a semi prime ERT right SF-ring with zero socle, then R is a strongly regular ring. Raf. J. Of Comp. Sc. And Math’s.,Vol. .1, No.1.2004


Introduction:
Throughout this paper, R denotes associative ring with identity and all modules are unitary.J( R ) and Y( R ) denote the Jacobson radical and the singular right ideal of R, respectively.For any nonempty subset X of a ring R, the right (resp.left) annihilater of X will hdbenoted by r(X) (resp.1(X)).Recall that: (1) A ring R is ERT if every essential right ideal of R is asided, [6] (2) R is said to be von Nuumann regular (or just regular) if, a∈aRa,for every a∈ R, and R is called strongly regular if a∈a 2 R.
In [3] Ming asked the following question: Is a semi prime right SF-ring, all of whose essential right ideals are two-sided von Neumann regular ?
In this paper, we give conditions for a semi prime right SF-ring all of whose essential right ideals are two sided to be von Neumann regular.

Basic Properties:
Following [5], a ring R is called a right (left) SF-ring, if every simple right (left) R-module is flat.
The following lemma which is due to [7], plays a central role in several of our proofs: Lemma 2.1: Let I be a right ideal of R. Then R/I is a flat R-module if and only if for each ae I, there exists be I such that a=ba.
We shall begin with the following result: If R is a right SF-ring.Then (1) Any reduced principal right ideal of R is a direct summand.
(2) Every left or right R-module is divisible.

Proof (1):
Let I=aR be a reduced principal right ideal of R and let aR+r(a) # R. Then there exists a maximal right ideal M of R containing aR+r(a).Now, since R/M is flat, then a=ba, for some b in R.
Whence l-b∈l(a)=r(a)⊆M.Yielding leM which contradicts M*R.In particular ar+c=l, for some reR and c∈r(a), whence a 2 r=a.If we set d=ar 2 ∈l, then a=a 2 d.Clearly, (a-ada) 2 =0 implies a=ada and hence I=eR, where e=ad, is idempotent element.Thus I is a direct summand.

Proof (2):
It is sufficient to prove that any non-zero divisor c of R is invertible.For then, if dc=cd=l, any right R-module M satisfies M=Mdc⊆Mc⊆M, whence M=Mc (similarly, any left right ideal containing cR.Since R/K is flat, there exist ueK, such that c=uc.Now, r(c) =1 (c )=0 implies u=l, contradicting K≠R.This proves that cd=l for some deR and hence dc=l.

Proposition 2.3:
Let R be a right SF-ring.Then either r(M)=0 or M is a direct summand.

Proof:
Suppose that r(M)≠0 and let b∈M∩r(M).Then b∈M and Mb=0.Since R/M is flat then there exists a∈M such that b=ab.Now b=ab∈Mb=0, so b=0.Thus M∩r(M)=0, this means that M not can be essential and hence M is a direct summand.

The Connection Between SF-Rings and Other Rings:
In this section we study the connection between SF-rings, biregular rings and strongly regular rings.
Recall that the right (left) socle of a ring R is defined to be the sum of all minimal right (left) ideals of R. It is well know that in a semi prime ring R, the right and left socles of R coincide, which will be denoted by socR [8].
Following [4], a ring R is biregular if RaR is generated by a central idempotent for each a∈R.this implies that b=bcb.Therefore RaR=bR=eR where e=bc is idempotent.R is therefore semi-prime and hence e is central in R. Thus R is biregular.

Theorem 3.2:
If R is a reduced ring and every maximal right ideal of R is either a right annihilator or flat, then R is strongly regular.

Theorem 3. 1 :
Let R be an ERT SF-ring with right zero socle and for every a∈R, RaR is a principal right of R. Then R is biregular.Proof:For any a∈R, set M=RaR+l(RaR).Since M is a maximal right ideal, then M is a direct summand or essential.If M is a direct summand of R, then its complement is a minimal right ideal.This implies that R has a no-zero socle, which is a contradiction.So every maximal right ideal is essential and hence two-sided.By hypothesis R/M is flat.Also RaR=bR for some b ∈ R. Since b ∈ M, b=db for some d ∈ M. Then l-d∈l(b)cM which yields l∈M.Therefore l=bc+v, c∈R, v∈l(b)