ON REGULARITY AND FLATNESS

A ring R is called a right SF-ring if all its simple right Rmodules are flat. It is well known that Von Neumann regular rings are right and left SF-rings. In this paper we study conditions under which SF-rings are strongly regular. Finally, some new characterstic properties of right SF-rings are given.


INTRODUCTION
In this paper all rings are assumed to be associative with identity, and all modules are unital right R-modules.
Following [2], a ring R is called a right (left) SF-ring if all of its simple right (left) R-modules are flat . It is well known that a ring R is Von Neumann regular if and only if every right (left) R-module is flat [3]. Ramamurthi in [8] asked whether left and right SF-ring is Von Neumann regular. Many authors have given various conditions for SF-rings to be regular (see, e.g. Chen [1], Ming [4], Rege [9] and Xu- [10] ). In this paper, to the list of equivalent conditions, we shall add several news. We recall that:

1-
A ring R is called reduced if R contains no non-zero nilpotent elements. 2-R is said to be Von Neumann regular (or just regular) if aaRa for every aR, and R is called strongly regular if aa 2 R . Clearly, every strongly regular ring is a regular reduced ring. 3-R is said to be right duo-ring if every right ideal is a two-sided ideal. 4-r(a) and L(a) will denote right and left annihilator of a respectively. 5-Following [9], for any ideal I of R, R/I is flat if and only if for each aI , there exists bI such that a=ba. 6-Y and J will stand for the right singular ideal and Jacobson radical of R.

RINGS WHOSE SIMPLE MODULES ARE FLAT
Following [7], a ring R is called ERT-ring if every essential right ideal of R is a two-sided ideal. Ming [6] proved the following: We use a similar method of proof in Prop.2.1 to establish the following lemma.

Lemma 2.2:
If R is an ERT-ring , then R/Y is a reduced ring.
Proof. Suppose that R/Y is not reduced, then there exists an element Y  a + YR/Y , aR , such that (a+Y) 2 = Y. This implies that aY and a 2 Y . So r(a 2 ) is essential right ideal of R. Since R is ERT, then r(a 2 ) is a two-sided ideal . Let I be any subideal of r(a 2 ) Such that Iis essential in (a)I , this means that r  (a)r  Ia, then (a)r  r(a 2 ) and hence in R , this contradicts aY.
The following theorem gives the condition of being right SF-rings are strongly regular.

Theorem 2.3:
Let R be a ring. Then the following are equivalent.
(2) R is a right SF-and ERT ring.
(1) By Lemma 2.2, R/Y is a reduced ring. We claim that Y=0. Suppose that Y  0 then by [5], there exists 0  yY such that y 2 = 0.
Let M be a maximal right ideal containing r (y). Since r (y) is an essential two-sided ideal of R , then M must be an essential two-sided ideal of R. On the other hand, since R/M is flat module , and since yM , there exists cM such that y=yc, whence 1-cr(y)  M, yielding 1M which contradicts M  R.
This proves that R is a reduced ring. In order to show that R is regular we need to prove that aR+r(a) = R for any aR. Suppose that aR + r(a)  R , then there exists a maximal right ideal L containing aR + r(a). But aL and R/M is flat , there exists bL such that a=ba, whence 1-bL(a) = r(a)  M.
Yielding 1M which contradicts L  R. In particular ar+d = 1, for some rR and dr(a), whence a 2 r=a . This proves that R is a strongly regular ring.
We now consider an other condition for right SF-ring to be strongly regular. contradicts M  R. Therefore, R is a reduced ring. By a similar method of proof used in Theorem 2.3,R is strongly regular.

BASIC PROPERTIES
Recall that a ring R is a right uniform if every right ideal of R is essential .
We are now in a position to give new characteristic properties of a right SF-ring.

Theorem 3.1:
If R is a right SF-ring , then 1-If L(a) = 0 , then a is a right invertable . 2-Every reduced ideal of R is strongly regular. 3-If J is reduced , then J = 0 . 4-If R is a right uniform ring, then R is a division ring.

Proof.
(1) Let aR with L(a)=0. If a R  R , there exists a maximal right ideal M containing aR. Since aM and R/M is flat , there exists bM, such that a=ba. Whence 1-bL(a) = 0, yielding LM, which contradicts M  R. Therefore aR=R.
(3) Let aJ, then by (2) J is strongly regular, and hence there exists bJ such that a=a 2 b . But aJ gives (1-ab) u = 1 for some uR, this implies that (a-a 2 b)u=a. Thus a=0, consequently, J=0. (4) Suppose that Y  0, then there exists a maximal right ideal M containing Y. For any 0  yY, gives yM, but R/M is flat, then there exists xM such that y=xy, whence yr(1-x). On the other hand, since R is a right uniform , then r(1-x) is an essential right ideal of R. Thus 1-x Y  M, this implies that 1M, contradicting M  R. Therefore, Y=0. On the other hand, since R is uniform, then for every aR, r(a) = 0, then by (1) , R is a division ring.
Before closing this section, we present the following result. Proposition 3.2: Let R be a reduced right SF-ring, for any a, bR with a.b=0 , then r(a) +r(b) = R.
Proof. Suppose that a.b=0 and r(a) + r(b)  R. Then there exists a maximal right ideal M containing r(a) + r(b). Since ar(b)  M, and since R/M is flat, there exists cM such that a = ca, whence 1-cL(a) = r(a)  M, yielding LM, which contradicts M  R. Therefore r(a) + r(b) = R.