1. Introduction
Throughout this paper, R will be an associative ring with identity. For , , 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 [1] . Following [2] , a ring R is said to be right semi-regular if and only if for each a in R, there exists b in R such that and .
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 and [3] .
Following [4] , a ring R is said to be π-regular if and only if for each a in R, there exist positive integers n and b in R such that . A ring R is called a local ring, if it has exactly one maximal ideal [5] .
A ring R is said to be a local semi-regular ring, if for all a in R, either a or is a semi-regular element [6] .
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 is a semi π-regular element.
Clearly that every π-regular ring is a semi π-regular local ring.
2. A Study of Some Characterization of 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.
2.1. Definition
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 is right (left) semi π-regular element for every a in R.
Examples:
Let be a ring and let is cyclic group, then is π-regular ring. Thus R is semi π-regular local ring.
Let R be the set of all matrix in which is defined as:
.
It easy to show that R is semi π-regular local ring.
2.2. Proposition
Let R be a right semi π-regular local ring. Then the associated elements are idempotents.
Proof:
Let , since R is right semi π-regular local ring. Then either a or is right semi π-regular element, that there exists b in R such that and , so , gives . Thus , which implies . Now, if is right semi π-regular element, then there exists c in R such that and . So , thus . Hence and therefore .
In general the associated element is not unique. But the following proposition give the necessary condition to prove the associated element is unique.
2.3. Proposition
Let R be a right semi π-regular local reduced ring. Then the idempotent associated element is unique.
Proof:
Let , since R is right 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 such that and . Assume that, there is an element in R such that and , which implies that , hence and , that is and then , , which implies , .
Since R is reduced ring, then . Hence and then and which implies and . Hence and , and therefore . Now, if is right semi π-regular element, then there exists an element such that and . Now, we assume that the associated element c is not unique.
Then, there exists such that , , then which implies that , that is . Hence and , implies that and , that is and . Since R is reduced ring, then and then , , that is and . Thus and . Therefore .
The following theorem give the condition to a semi π-regular local ring to be π-regular ring.
2.4. Theorem
Let R be a right semi π-regular local ring. Then any element is π-regular if for any associated element b in R.
Proof:
Let
and R be a right semi π-regular local ring. Then either
Now, assume that . Then and , . Since b is idempotent element, then and , it follows that .
Thus . Therefore a is π-regular element in R.
Now, if is right semi π-regular element, then there exists an element such that : and .
If , assume that , where , . Since c is idempotent element, then and , it follows that , that is .
Thus . Therefore is π-regular element in R.
2.5. Proposition
The epimorphism image of right semi π-regular local ring is right semi π-regular local ring.
Proof:
Let be epimorphism homomorphism function from the ring π in to the ring , where R is right semi π-regular local ring and let be element s in . Then there exists elements in R such that
.
Now, since R is right semi π-regular local ring, then either x or is right semi π-regular element, that is and . Then
.
Now, to prove . If , then , that is , then , and , hence .
Thus , that is . Then , thus . Hence . Therefore,
(1)
Now, let . Then , it follows that and then .
Thus and hence . Therefore
(2)
from (1) and (2), we obtain .
Now, if is right semi π-regular element in R, then and .
Now, . Thus .
Now, to prove .
Let . Then . That is , then and . Then and hence , that is , it follows that .
Hence , thus . Therefore
(3)
Now, let , implies to , hence , thus . Hence . Therefore
(4)
from (3) and (4) we obtain
, that is either y or is right semi π-regular element in
. Therefore
is right semi π-regular local ring.
2.6. Theorem
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
(5)
and by the same way we can prove
(6)
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
(7)
Now, let. Then
and hence
, Thus
, therefore
and we have
(8)
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.
2.7. Theorem
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
(9)
Now, let. Then
and hence
that is
, thus
. Therefore
(10)
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
(11)
Now, let, that is
and hence
, it follows that
, that is
. Hence
(12)
form (11) and (12) we have. Thus
is right semi π-regular element. Therefore R is right semi π-regular ring.
3. The Conclusion
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 is π-regular if
for any associated element b in R.
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 or
is direct summand for all
and
.
If R is a local ring with for all
and
, then R is a right semi π-regular local ring.
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https://doi.org/10.1080/00927872.2010.524184
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