On n-Weakly Regular Rings

As a generalization of right weakly regular rings, we introduce the notion of right n-weakly regular rings, i.e. for all aN(R), aaRaR. In this paper, first give various properties of right n-weakly regular rings. Also, we study the relation between such rings and reduced rings by adding some types of rings, such as NCI, MC2 and SNF rings.

Throughout this paper a ring R denotes an associative ring with identity and all modules are unitary. For a subset X of R, the left(right) annihilator of X in R is denoted by r(X)(l(X)). If X={a}, we usually abbreviate it to r(a)(l(a)). We write J(R), and N(R), for the Jacobson radical and the set of nilpotent elements respectively.
The center of the ring R is denoted by Cent(R) and it is ra A ring R is called n-regular if for all aN(R), aaRa [7]. A right R-module M is called N flat if for any a  N(R), the mapping 1M  i:M  R Ra → M  R R is monic, where i : Ra → R is the inclusion mapping [8]. A ring R is called right (left) SNF if every simple right(left) R-module is N flat [8].
A ring R is called semiprime if it has no nilpotent ideals [6]. A ring R is called reduced if N(R) = 0 [6], or equivalently, a 2 =0 implies a=0 in R for all a  R. Recall that a ring R is MERT(resp. MELT), if every maximal essential right (resp. left) ideal of R is an ideal [9].

n -Weakly Regular Ring
This section is devoted to give the definition of n-weakly regular rings with some of its characterizations and basic properties.
A ring R is right (left) weakly regular [6], if aaRaR (RaRa) for every aR. We called R is weakly regular if it is both right and left weakly regular.

Definition 2.1
A ring R is to be right (left) n-weakly regular if aaRaR (aRaRa) for all aN(R). We say that R is n-weakly regular if it is right and left n-weakly regular ring.
Clearly every weakly regular rings are n-weakly regular.

Examples:
1-Every reduced ring is n-weakly regular. 2-Every n-regular ring is n-weakly regular ring. 3-The ring Z6 of integers modulo 6, is reduced, n-regular, weakly regular ring, so it is n-weakly regular.

4-Let
. R is nregular, weakly regular ring, so it is n-weakly regular but R not reduced ring. 5-The ring Z of integer number is reduced, n-regular, so it is n-weakly regular but Z is not weakly regular ring.

Proposition 2.2
R is a right n-weakly regular ring if and only if aR is idempotent right ideal for all aN(R). Proof: Let R is a right n-weakly regular ring and I is a principal right ideal of R generated by a nilpotent element, then there exists aN(R) such that I=aR, clearly On the other hand, since R is right n-weakly, then there exists y,zR such that a=ayaz. Now let xI, then there exists rR such that x=ar=ayazrI 2 . Therefore Conversly, Let aN(R), since aR is idempotent right ideal of R, so aaR=aRaR. Therefore R is right n-weakly regular ring. ■

Proposition 2.3
Let R be a right n-weakly regular ring. If aR  I, for aN(R) and I is right or left ideal. Then aRI=aR. Proof: It is clearly that bRI  bR for any bR. Now let aN(R) and xaR, then there exists rR such that x=ar. Since R is right n-weakly regular ring then there exists y,zR such that a=ayaz, x=ayazr, hence azraR  I. So aR  aRI. Therefore aRI=aR. ■

Corollary 2.4
Let R be a ring for all aN(R) and any I right or left ideal of R such that aR  I. Then the following condition are equivalent: 1-R is right n-weakly regular. Let R be a right n-weakly regular ring.
. since R is right n-weakly regular then there exists y,z  R such that a=ayaz=a 2 yz=0yz=0 (a  Cent(R)). Therefore a=0. This is shows that

Corollary 2.6
Let R be a commutative ring. Then R is reduced if and only if R is right nweakly regular.

Lemma 2.7 [2]
1-Every one sided or two sided nil ideal of R is contained in J(R).

2-In any ring R, aJ(R) if and only if 1-ar is invertible for all rR.
Now we have the following proposition

Theorem 2.8
Let R be a right n-weakly regular ring. Then N(R)  J(R)=0.

Proof:
If a N(R)  J(R), then there exists y,z R such that a= ayaz. Hence a(1-yaz)=0, since a  J, yaz  J, then by Lemma 2.7(2), there exists invertible element v  R such that (1-yaz)v=1. So (a-ayaz)v=a, yield a=0. Therefore N(R)  J(R)=0. ■ Let R be a ring we denoted to the upper nil radical for a ring R by Nil * (R) and it is the sum of nil ideal in the ring R.

Corollary 2.9
Let R be a right n-weakly regular ring .Then Nil * (R)=0.

Proof:
Let I be a left or right or two sided nil ideal, by Lemma 2.7(1), we have that I  J(R), I  N(R)  J(R)=0, Theorem 2.8, I=0, which is a contradiction. So R not contain any nil ideal. Therefore N * (R)=0. ■

Corollary 2.10
Let R be a right n-weakly regular ring. Then R is semiprime ring.

Proof:
Let I be a nilpotent right ideal then I  N(R)  J(R)=0 (Theorem 2.8) I=0. Therefore R is semiprime ring. ■ In this section we gives the connection between n-weakly regular rings and reduced rings, SNF rings.

Proposition 3.1
The following conditions are equivalent for a ring R. 1-R is reduced. 2-R is right n-weakly regular and N(R) forms a right ideal of R. 3-R is right n-weakly regular and N(R) forms a left ideal of R. 4-R is right n-weakly regular and NI ring. 5-R is right n-weakly regular and N(R)  J(R). Proof: Suppose that R is right n-weakly regular ring. So N(R)  J(R)=0, (Theorem 2.8). Since N(R)  J(R), then N(R)  J(R)= N(R) =0. Therefore R is reduced. ■

Theorem 3.2
Let R be a ring with aR=Ra, for all aN(R). Then the following conditions are equivalent: 1-R is right n-weakly regular.

Proof: 1 →2
Let 0  aR , such that a 2 =0. Since R is a right n-weakly regular, then aaR=aRaR (Proposition 2.1) = aRRa (Ra=aR) =aRa. so aaRa, hence R is n-regular.

→3
Let 0  aR , such that a 2 =0. Since R is a right n-regular, then there exists bR such that a=aba since aR=Ra there exists xR such that ab=xa, so a=aba=xa 2 =x0=0. Therefore R is reduced. 3 →1 It is trivial. ■ A ring R is called NCI provided that N(R) contains a non zero ideal of R whenever N(R)  0 [1].

Lemma 3.3 [1]
Let R be a ring with N(R)  0. Then R is NCI if and only if N*(R)  0.

Proportion 3.4
Let R be a NCI ring , then R is right n-weakly regular if and only if R is reduced.

Proof:
Let N(R)  0, sinc R is NCI ring from Lemma 3.3, we get that N*(R)  0 but R is right n-weakly regular, N*(R)=0 (Corollary 2.9), which is contradiction. So N(R)=0. Therefore R is reduced. ■ A ring R is called weakly reversible if and only if for all a,b,rR such that ab=0, Rbra is a nil left ideal of R (equivalently braR is nil right ideal of R). Clearly ZI ring is weakly reversible [4].

Proposition 3.5
A ring R be right n-weakly regular ring and weakly reversible if and only if R is reduced.

Proof:
Let aR with a 2 =0. Then a=ayaz for some y,zR because R is right nweakly regular ring. Since R is weakly reversible then ayaR is nil right ideal of R so ayaR 0 (Lemma 2.7(1) & Theorem 2.8) we get ayaR=0, in particular a=ayaz=0. Therefore R is reduced.
Converse, it is trivial. ■ Recall that a ring R is right MC2 if eR K  is simple, e 2 =e, , then K=gR for some g 2 =g [5].

Lemma 3.6 [9]
Let R be a left MC2,right SNF ring and MELT ring. Then R is a semiprime ring and right non singular.

Theorem 3.7 [9]
Let I be a right ideal of a ring R. Then R/I is N flat if and only if Ia=I  Ma for all aN(R).

Theorem 3.8
Let R be right SNF, left MC2 and MELT ring. Then R is left n-weakly regular ring.

Proof:
From Lemma 3.6, we get that R is a semiprime ring and a N(R). If RaR+l(a)  R, then there exists a maximal left ideal M of R containing RaR+l(a), if M is not essential then we can write M=l(e), where e 2 =e R and e  0, since RaRe=0 because RaR  M, (ReRa) 2 =0 implies ReRa=0 (since R is semiprime) ReRa=0 in particular ea=0 and el(a)  M=l(e), then e 2 =0, which is a contradiction. Therefore M is an essential, since R is MELT ring, then M is a two sided ideal then there exists a maximal right ideal L in R containing M, since R is right SNF ring then R/ L is N flat right R-module, a=ma for some mM (Theorem 3.7), (1-m)a=0, 1-ml(a)  M  L therefore 1-m L implies 1 L which is a contradiction, therefore RaR+l(a) =R for all aN(R). Thus R is left n-weakly regular ring. ■

Theorem 3.9
Let R be MELT and right SNF ring with RaR is essential for all aN(R), Then R is left n-weakly regular ring.

Proof:
Let a  N(R). If RaR+l(a)  R then there exists a maximal left ideal M of R containing RaR+l(a), since RaR is left annihilator of a nilpotent element by the hypothesis RaR is essential left ideal in R, M is an essential ideal of R (MELT ring), there exists a maximal right ideal L in R such that M  L. Since R is right SNF ring, then R/ L is N flat right R-module, a=ma for some mM (Theorem 3.7), 1-ml(a)  M  L implies 1 L, which is a contradiction. Therefore RaR+l(a) =R for all aN(R), and so R is left n-weakly regular ring. ■

Definition 3.10 [8]
A right R-module M is said to be nil-injective, if for any aN(R), any right Rhomomorphism f:aR →M can be extended to R →M, or equivalently f=m., where mM.
The ring R is called right nil-injective if RR is right nil-injective. Clearly a reduced ring is a right nil-injective and n-regular ring is a right nil-injective [8].

Theorem 3.11
Let R be a semiprime ring whose simple singular right R-module are nilinjective. Then R is right n-weakly regular ring.

Proof:
Let aN(R). We claim that RaR+r(a)=R if not, there exists a maximal right ideal M of R containing RaR+r(a). If M is not essential in R then M=r(e), e 2 =eR. Since Rae  RaR  M=r(e), eRae=0, (aeR) 2 =0 but R is semiprime, aeR=0, so ae=0. Thus e r(a)  M=r(e), which is a contradiction. Hence M is essential right ideal in R and so R/M is nil-injective. Define a mapping M R aR f / : → such that f(ar)=r+M, let x,yR such that ax=ay, a(x-y)=0, (x-y)+M=f(a(x-y))=f(0)=M, then x-yM, x+M=y+M, f(ax)=x+M=y+M=f(ay), f is well define. 1+M=f(a)=(b+M)(a+M)=ba+M, 1-baM, since baRaR  M then 1M which is a contradiction, so RaR+r(a)=R, in particular there is y,zR and vr(a) such that yaz+v=1, ayaz+av=a, a=ayaz. Therefore R is right n-weakly regular ring. ■

Proposition 3.12
Let R be a ring whose simple right R-module are nil-injective. Then R is right nweakly regular ring.

Proof:
Assume that aR such that aRa=0. Then RaR  r(a). If a  0 then there exists a maximal right ideal M containing r(a). By hypothesis R/M is nil-injective. We define a mapping M R aR f / : → such that f(ar)=r+M, f is well define similar to Theorem 3.11, so there exists bR such that 1+M=f(a)=ba+M, 1-baM because baRaR  M then 1M which is a contradiction, so a=0. Therefore R is a semiprime ring, by Theorem 3.11 we get that R is right n-weakly regular ring. ■