^{1}

^{1}

^{1}

A ring
*R*
is said to be a right (
left) semi
*π*
-regular local ring if and only if for all
*a*
in
*R*
, either a or (1-a) is a right (left) semi π-regular element. The purpose of this paper is to give some characterization and properties of semi π-regular local rings, and to study the relation between semi π-regular local rings and local rings. From the main results of this work: 1) Let R be a semi π-regular reduced ring. Then the idempotent associated element is unique. 2) Let R be a ring. Then R is a right semi π-regular local ring if and only if either *r*(*a** ^{n}*) or

*r*((1-

*a*)

^{n}) is direct summand for all

*a*∈

*R*and

*n*∈

*Z*

^{ }. If R is a local ring with

*r*(

*a*)

^{n}*r*(

*a*) for all

*a*∈

*R*and

*n*∈

*Z*

^{ }, then

*R*is a right semi

*π*-regular local ring.

Throughout this paper, R will be an associative ring with identity. For a ∈ R , r ( a ) , ( l ( a ) ) denote the right (left) annihilator of a. A ring R is reduced if R contains, no non-zero nilpotent element.

A ring R is said to be Von Neumann regular (or just regular) if and only if for each a in R, there exists b in R such that a = a b a [

By extending the notion of a right semi π-regular ring to a right semi-regular ring is defined as follows:

A ring R is said to be right semi π-regular if and only if for each a in R, there exist positive integers n and b in R such that a n = a n b and r ( a n ) = r ( b ) [

Following [

A ring R is said to be a local semi-regular ring, if for all a in R, either a or ( 1 − a ) is a semi-regular element [

We extend the notion of the local semi-regular ring to the semi π-regular local ring defined as follows:

A ring R is said to be a semi π-regular local ring, if for all a in R, either a or ( 1 − a ) is a semi π-regular element.

Clearly that every π-regular ring is a semi π-regular local ring.

In this section we give the definition of a semi π-regular local ring with some of its characterization and basic properties.

A ring R is said to be right (left) semi π-regular local ring if and only if for all a in R, either a or ( 1 − a ) is right (left) semi π-regular element for every a in R.

Examples:

Let ( Z 2 , + , ⋅ ) be a ring and let G = { g : g 2 = 1 } is cyclic group, then Z 2 G = { 0 , 1 , g , 1 + g } is π-regular ring. Thus R is semi π-regular local ring.

Let R be the set of all matrix in Z 2 which is defined as:

R = { [ a b 0 d ] : a , b , d ∈ Z 2 } .

It easy to show that R is semi π-regular local ring.

Let R be a right semi π-regular local ring. Then the associated elements are idempotents.

Proof:

Let a ∈ R , since R is right semi π-regular local ring. Then either a or ( 1 − a ) is right semi π-regular element, that there exists b in R such that a n = a n b and r ( a n ) = r ( b ) , so a n ( 1 − b ) = 0 , gives ( 1 − b ) ∈ r ( a n ) = r ( b ) . Thus b ( 1 − b ) = 0 , which implies b = b 2 . Now, if ( 1 − a ) is right semi π-regular element, then there exists c in R such that ( 1 − a ) n = ( 1 − a ) n c c and r ( ( 1 − a ) n ) = r ( c ) . So ( 1 − a ) n ( 1 − c ) = 0 , thus ( 1 − c ) ∈ r ( ( 1 − a ) n ) = r ( c ) . Hence c ( 1 − c ) = 0 and therefore c = c 2 .

In general the associated element is not unique. But the following proposition give the necessary condition to prove the associated element is unique.

Let R be a right semi π-regular local reduced ring. Then the idempotent associated element is unique.

Proof:

Let a ∈ R , since R is right semi π-regular local ring. Then either a or ( 1 − a ) is right semi π-regular element in R. If a is right semi π-regular element, then there exists b ∈ R such that a n = a n b and r ( a n ) = r ( b ) . Assume that, there is an element b ¯ in R such that a n = a n b ¯ and r ( a n ) = r ( b ¯ ) , which implies that a n ( b − b ¯ ) = 0 , hence ( b − b ¯ ) ∈ r ( a n ) = r ( b ) = r ( b ¯ ) and b ¯ ( b − b ¯ ) = 0 , that is b ( b − b ¯ ) = 0 and then b ¯ b = b ¯ 2 , b 2 = b b ¯ , which implies b ¯ b = b ¯ , b = b b ¯ .

Since R is reduced ring, then r ( b ) = l ( b ) = l ( b ¯ ) . Hence ( b − b ¯ ) ∈ l ( b ) = l ( b ¯ ) and then ( b − b ¯ ) b = 0 and ( b − b ¯ ) b ¯ = 0 which implies b 2 = b b ¯ and b b ¯ = b ¯ 2 . Hence b = b ¯ b and b b ¯ = b ¯ , and therefore b = b ¯ b = b b ¯ = b ¯ . Now, if ( 1 − a ) is right semi π-regular element, then there exists an element c ∈ R such that ( 1 − a ) n = ( 1 − a ) n c and r ( ( 1 − a ) n ) = r ( c ) . Now, we assume that the associated element c is not unique.

Then, there exists c ¯ ∈ R such that r ( ( 1 − a ) n ) = r ( c ¯ ) , ( 1 − a ) n = ( 1 − a ) n c ¯ , then ( 1 − a ) n c = ( 1 − a ) n c ¯ which implies that ( 1 − a ) n ( c − c ¯ ) = 0 , that is ( c − c ¯ ) ∈ r ( ( 1 − a ) n ) = r ( c ) = r ( c ¯ ) . Hence c ( c − c ¯ ) = 0 and c ¯ ( c − c ¯ ) = 0 , implies that c 2 = c c ¯ and c ¯ c = c ¯ 2 , that is c = c c ¯ and c ¯ c = c ¯ . Since R is reduced ring, then l ( c ¯ ) = r ( c ) = l ( c ) and then ( c − c ¯ ) c = 0 , ( c − c ¯ ) c ¯ = 0 , that is c 2 = c ¯ c and c c ¯ = c ¯ 2 . Thus c = c ¯ c and c c ¯ = c ¯ . Therefore c = c ¯ c = c c ¯ = c ¯ .

The following theorem give the condition to a semi π-regular local ring to be π-regular ring.

Let R be a right semi π-regular local ring. Then any element a ∈ R is π-regular if R a n = R b for any associated element b in R.

Proof:

Let a ∈ R and R be a right semi π-regular local ring. Then either or ( 1 − a ) is right semi π-regular element in R. If a is right semi π-regular element in R, then there exists b ∈ R such that a n = a n b and r ( a n ) = r ( b ) .

Now, assume that R a n = R b . Then r a n = b and r a n ∈ R a n , b ∈ R b . Since b is idempotent element, then b + ( 1 − b ) = 1 and r a n + ( 1 − b ) = 1 , it follows that a n r n a n + a n ( 1 − b ) = a n .

Thus a n r a n = a n . Therefore a is π-regular element in R.

Now, if ( 1 − a ) is right semi π-regular element, then there exists an element c ∈ R such that : ( 1 − a ) n = ( 1 − a ) n c and r ( ( 1 − a ) n ) = r ( c ) .

If R ( 1 − a ) n = R c , assume that s ( 1 − a n ) = c , where s ( 1 − a ) ∈ R ( 1 − a ) , c ∈ R . Since c is idempotent element, then c + ( 1 − c ) = 1 and S ( 1 − a ) n + ( 1 − c ) = 1 , it follows that ( 1 − a ) n S ( 1 − a ) n + ( 1 − a ) n ( 1 − c ) = ( 1 − a ) n , that is ( 1 − a ) n S ( 1 − a ) n + ( 1 − a ) n − ( 1 − a ) n c = ( 1 − a ) n .

Thus ( 1 − a ) n S ( 1 − a ) n = ( 1 − a ) n . Therefore ( 1 − a ) is π-regular element in R.

The epimorphism image of right semi π-regular local ring is right semi π-regular local ring.

Proof:

Let f : R → R ¯ be epimorphism homomorphism function from the ring π in to the ring R ¯ , where R is right semi π-regular local ring and let e ¯ , y , 1 ¯ be element s in R ¯ . Then there exists elements e , x , 1 in R such that

f ( e ) = e ¯ , f ( x ) = y , f ( 1 ) = 1 ¯ .

Now, since R is right semi π-regular local ring, then either x or ( 1 − x ) is right semi π-regular element, that is x n = x n e and r ( x n ) = r ( e ) . Then

y n = ( f ( x ) ) n = f ( x n ) = f ( x n e ) = f ( x n ) f ( e ) = y n e ¯ .

Now, to prove r ( y n ) = r ( e ¯ ) . If a ∈ r ( y n ) , then y n a = 0 , that is ( f ( x ) ) n a = 0 , then f ( x n ) a = 0 , and f − 1 f ( x n ) f − 1 ( a ) = 0 , hence x n f − 1 ( a ) = 0 .

Thus f − 1 ( a ) ∈ r ( x n ) = r ( e ) , that is e f − 1 ( a ) = 0 . Then f ( e ) a = 0 , thus e ¯ a = 0 . Hence a ∈ r ( e ¯ ) . Therefore,

r ( y n ) ⊆ r ( e ¯ ) (1)

Now, let b ∈ r ( e ¯ ) . Then e ¯ b = 0 , it follows that y e ¯ b = 0 and then y n e ¯ b = 0 .

Thus y n b = 0 and hence b ∈ r ( y n ) . Therefore

r ( e ¯ ) ⊆ r ( y n ) (2)

from (1) and (2), we obtain r ( e ¯ ) = r ( y n ) .

Now, if ( 1 − x ) is right semi π-regular element in R, then ( 1 − x ) n = ( 1 − x ) n e and r ( 1 − x ) n = r ( e ) .

Now, f ( 1 − x ) n = ( f ( 1 − x ) ) n = ( f ( 1 ) + f ( − x ) ) n = ( f ( 1 ) − f ( x ) ) n = ( 1 ¯ − y ) n . Thus ( 1 ¯ − y ) n = f ( 1 − x ) n = f ( ( 1 − x ) n e ) = f ( 1 − x ) n f ( e ) = ( 1 ¯ − y ) n e ¯ .

Now, to prove r ( 1 ¯ − y ) n = r ( e ¯ ) .

Let c ∈ r ( 1 ¯ − y ) n . Then ( 1 ¯ − y ) n c = 0 . That is ( f ( 1 ) − f ( x ) ) n c = 0 , then ( f ( 1 − x ) ) n c = 0 and f ( 1 − x ) n c = 0 . Then ( 1 − x ) n f − 1 ( c ) = 0 and hence f − 1 ( c ) ∈ r ( 1 − x ) n = r ( e ) , that is e f − 1 ( c ) = 0 , it follows that f ( e ) c = 0 .

Hence e ¯ c = 0 , thus c ∈ r ( e ¯ ) . Therefore

r ( 1 ¯ − y ) n ⊆ r ( e ¯ ) (3)

Now, let d ∈ r ( e ¯ ) , implies to e ¯ d = 0 , hence ( 1 ¯ − y ) n e ¯ d = 0 , thus ( 1 ¯ − y ) n d = 0 . Hence d ∈ r ( 1 ¯ − y ) n . Therefore

r ( e ¯ ) ⊆ r ( 1 ¯ − y ) n (4)

from (3) and (4) we obtain r ( e ¯ ) = r ( 1 ¯ − y ) n , that is either y or

Let R be a ring. Then R is right semi π-regular local ring if and only if either

Proof:

Let

and by the same way we can prove

from (5) and (6) we obtain

Then, there exists an ideal

Therefore

Let

Thus,

Now, let

form (7) and (8) we obtain

Now, let R be aright semi π-regular local ring. Then either a or

Hence,

Now, to prove

Now, if

Hence,

Now, to prove

Then

Hence

That is

Now, to give the relation between semi π-regular local ring and local ring.

If R is local ring with

Proof:

Let R be local ring. Then either a or

If a is invertible, then there exists an element b in R such that

Now, let

from (9) and (10) we obtain

Now, let

form (11) and (12) we have

From the study on characterization and properties of semi π-regular local rings, we obtain the following results:

1) Let R be a right semi π-regular local ring. Then the associated elements are idempotents.

2) Let R be a right semi π-regular local ring. Then the idempotent associated element is unique.

3) Let R be a right semi π-regular local ring. Then any element

4) The epimorphism image of right semi π-regular local ring is right semi π-regular local ring.

5) Let R be a ring. Then R is a right semi π-regular local ring if and only if either

If R is a local ring with

The authors declare no conflicts of interest regarding the publication of this paper.

Ibraheem, Z.M., Mustafa, R.A. and Khalf, M.F. (2018) On Semi π-Regular Local Ring. Open Access Library Journal, 5: e4788. https://doi.org/10.4236/oalib.1104788