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 is right semi π-regular element in. Therefore is right semi π-regular local ring.

Let R be a ring. Then R is right semi π-regular local ring if and only if either or is direct summand for all and.

Proof:

Let and is direct summand. Then there exists an ideal, such that. Thus, there is and, such that and hence and therefore. Now, to prove, let. Then, that is and. But and. Then and, hence

and by the same way we can prove

from (5) and (6) we obtain. Therefore a is right semi π-regular element. Now, if and is direct summand.

Then, there exists an ideal such that, and there exists and, such that. Thus

.

Therefore. Now, to prove.

Let. Then and hence

Thus,. But and, then and therefore, hence

Now, let. Then and hence, Thus, therefore and we have

form (7) and (8) we obtain. Therefore is right semi π-regular element. That is R is right semi π-regular local ring.

Now, let R be aright semi π-regular local ring. Then either a or is right semi π-regular element in R. If a is right semi π-regular element, then there exists and such that and.

Hence, , that is, then and thus. Therefore.

Now, to prove, suppose that, then and. Hence for some and, since, then and, that is [proposition 2.2]. Thus and therefore, that is is direct summand of R.

Now, if is right semi π-regular element, then there exists such that and. Since, we have, and since.

Hence,. Thus,.

Now, to prove. Let.

Then and, hence and for some. Since then and.

Hence [proposition 2.2] and thus and then.

That is. Therefore is direct summand of R.

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

If R is local ring with for all and, then R is right semi π-regular local ring.

Proof:

Let R be local ring. Then either a or is invertible element in R [6] .

If a is invertible, then there exists an element b in R such that, hence and then. Let. Then. To prove. Let. Then, it follows that and then, that is. Hence

Now, let. Then and hence that is, thus. Therefore

from (9) and (10) we obtain. Hence a is right semi π-regular element in R. Now, if is invertible element in R, then there exists an element c in R such that. That is, it follows that. let. Then. To prove, let, then that is and hence, and then. Thus

Now, let, that is and hence, it follows that, that is. Hence

form (11) and (12) we have. Thus is right semi π-regular element. Therefore R is right semi π-regular ring.