On Semi π Regular Local Ring

(1)

ISSN Online: 2333-9721 ISSN Print: 2333-9705

DOI: 10.4236/oalib.1104788 Oct. 12, 2018 1 Open Access Library Journal

On Semi

π

-Regular Local Ring

Zubayda M. Ibraheem, Raghad A. Mustafa, Maha F. Khalf

Department of Mathematics, College of Computer and Mathematical Sciences, University of Mosul, Mosul, Iraq

Abstract

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

( )

n

r a 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.

Subject Areas

Algebra

Keywords

Local, Ring, Semi π-Regular

1. Introduction

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 aba= [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 a ab= and r a

( ) ( )

=r b .

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

How to cite 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

Received: July 21, 2018 Accepted: October 9, 2018 Published: October 12, 2018

This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).

http://creativecommons.org/licenses/by/4.0/

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DOI: 10.4236/oalib.1104788 2 Open Access Library Journal

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 _an₌_{a b}n _and _{r a}

( )

n ₌_{r b}

_{( )}

_[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 _an ₌_{a ba}n n_{. 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

(

1−a

)

_{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

(

1−a

)

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

(

1−a

)

_{is right (left) semi}_π_{-regular element for every}_a_in_R_.

Examples:

Let

(

Z2, ,+ ⋅

)

be a ring and let G=

{

g g: 2 =1

}

is cyclic group, then

{

}

2 0,1, ,1

Z G= g +g _is_π_{-regular ring. Thus}_R_{is semi}_π_{-regular local ring.}

Let R be the set of all matrix in Z2 which is defined as:

2

: , , 0

a b

R a b d Z

d

  

 

=  ∈ 

  

 .

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 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 _an ₌_{a b}n _and

( )

n

( )

r a =r b _{, so} _an

(

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 c

)

n 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_.

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2.3. Proposition

Let R be a right semi π-regular local reduced ring. Then the idempotent asso-ciated 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}_an₌_{a b}n _and _{r a}

( )

n ₌_{r b}

_{( )}

_{. Assume that, there}

is an element b in R such that _an₌_an_b_and _{r a}

( )

n ₌_r

( )

_b _{, which implies}

that _{a b b}n

(

₋

)

₌0_{, hence}

(

_b₋_b

)

_∈_{r a}

( )

n ₌_{r b}

( )

₌_r

( )

_b _and _{b b b}

(

₋

)

₌₀ _,

that is b b b

(

−

)

=0_{and then}_{bb b}₌ 2_, _b2₌_b_b _{, which implies}_{bb b}₌ _, b=bb _.

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

2

b =bb _and _{bb b}₌ 2 _{. Hence}_b₌_b_b _and _{bb b}₌ _{, and therefore}

b b b

b= b b= = _{. Now, if}

(

1−a

)

_{is right semi}_π_{-regular element, then there}

ex-ists an element c R∈ _{such that}

(

1−a

) (

n= −1 a c

)

n _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 c

)

n _,

then

(

1₋_{a c}

)

n _{= −}

(

1 _{a c}

)

n _{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_, im-plies that _c2 ₌_c_c_and _{cc c}₌ 2_{, that is} _c₌_c_c_and _{cc 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}

2

c =cc_and _{cc c}₌ 2_{. Thus} _c₌_c_c_and _{cc 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.

2.4. Theorem

Let R be a right semi π-regular local ring. Then any element a R∈ _is_π_-regular

if _Ran₌_Rb_{for any associated element}_b_in_R_.

Proof:

Let a R∈ _and_R_{be a 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}

in R, then there exists b R∈ _{such that} _an ₌_{a b}n _and _{r a}

( )

n ₌_{r b}

( )

_.

Now, assume that _Ran₌_Rb_{. Then}_ran ₌_b_and _ran_∈_Ran_, _{b Rb}_∈ _{. Since}

b is idempotent element, then b+ −

(

1 b

)

=1_and _ran_{+ −}

(

1 _b

)

₌1_{, it follows} that _{a r a}n n n₊_an

(

1₋_b

)

₌_an_.

Thus _{a ra}n n ₌_an_{. 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 c

)

n _and r

(

1−a

)

n

)

=r c

( )

_.

If _R

(

1₋_a

)

n ₌_Rc_{, assume that}_s

(

₁₋_an

)

₌_c _{, where} _s

(

₁₋_a

)

_∈_R

(

₁₋_a

)

_, c R∈ _{. Since} _c_{is idempotent element, then} c+ −

(

1 c

)

=1 _and

(

1

) (

n 1

)

1

S −a + −c = _{, it follows that}

(

1₋_{a S}

) (

n 1₋_a

) (

n_{+ −}1 _a

) (

n 1₋_c

) (

_{= −}1 _a

)

n_,

(4)

Thus

(

1−a S

) (

n 1−a

) (

n = −1 a

)

n_{. Therefore}

(

1−a

)

_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 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

( )

,

( )

, 1

( )

1

f e =e f x =y f = .

Now, since R is right semi π-regular local ring, then either x or

(

1−x

)

is right semi π-regular element, that is _xn₌_{x e}n _and _{r x}

( )

n ₌_{r e}

_{( )}

_{. Then}

( )

(

)

n

( ) ( ) ( )

( )

n n n n n

y = f x = f x = f x e = f x f e = ye.

Now, to prove _{r y}

( )

n ₌_r

( )

_e _{. If} _{a r y}_∈

( )

n _{, then} _{y a}n ₌₀ _{, that is}

( )

(

f x

)

na=0 _{, then} _{f x a}

( )

n ₌0 _{, and} _{f f x f}−1

( )

n −1

_{( )}

_a ₌₀ _{, hence}

( )

1 ₀

n

x f− a ₌ _.

Thus _f−1

( )

_a _∈_{r x}

( )

n ₌_{r e}

( )

_{, that is}_ef−1

( )

_a ₌₀_{. Then} _{f e a}

( )

₌₀_{, thus}

0

ea= _{. Hence} a∈r

( )

e _{. Therefore,}

( )

n

( )

r y ⊆r e ₍₁₎

Now, let b∈r

( )

e . Then eb=0, it follows that yeb=0_{and then} _{y eb}n ₌0_. Thus _{y b}n ₌0_{and hence} _{b r y}_∈

( )

n _{. Therefore}

( )

n

r e ⊆r y ₍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 e

)

n

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 e

)

n

)

= f

(

1−x

) ( )

n f e =

(

1−y e

)

n .

Now, to prove r

(

1−y

)

n=r e

( )

_.

Let c r∈

(

1−y

)

n_{. Then}

(

1−y c

)

n =0_{. That is}

(

f

( )

1 − f x

( )

)

nc=0_{, then}

(

)

(

f 1−x

)

nc=0_and f

(

1−x c

)

n =0 _{. Then}

(

₁

)

n 1

( )

₀

x f− c

− = _{and hence}

( ) (

)

( )

1 ₁ n

f− c _∈r ₋x ₌r e _{, that is} _ef−1

_{( )}

_c ₌₀_{, it follows that} _{f e c}

_{( )}

₌₀_.

Hence ec=0_{, thus} c∈r

( )

e _{. Therefore}

(

1

)

n

( )

r −y ⊆r e (3)

Now, let d∈r

( )

e , implies to ed =0 _{, hence}

(

1−y ed

)

n =0 , thus

(

1−y d

)

n =0. Hence d r∈

(

1−y

)

n. Therefore

( )

(

1

)

n

r e ⊆r −y ₍₄₎

(5)

semi π-regular element in R. Therefore R 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

( )

n

r a or r

(

1−a

)

n

)

_{is direct summand for all} a R∈ _and n z_∈ +_.

Proof:

Let a R∈ _and _{r a}

( )

n _{is direct summand. Then there exists an ideal} _{I R}_⊂ _,

such that _{R r a}₌

( )

n _⊕_I _{. Thus, there is}_{d r a}_∈

( )

n _and _{b I}_∈ _{, such that} 1

d b+ = _{and hence}_{a d a b a}n ₊ n ₌ n_{and therefore}_{a b a}n ₌ n_{. Now, to prove}

( )

n

( )

r a =r b _{, let}_{x r a}_∈

( )

n _{. Then}_{a x}n ₌₀_{, that is}_{a bx}n ₌₀_and _{bx r a}_∈

( )

n _.

But bx I∈ _and _{r a}

( )

n __I₌0_{. Then} _bx₌₀_and _{x r b}_∈

( )

_{, hence}

( )

n

( )

r a ⊆r b ₍₅₎

and by the same way we can prove

( )

n

r b ⊆r a (6)

from (5) and (6) we obtain _{r a}

( )

n ₌_{r b}

( )

_{. Therefore}_a_{is right semi}_π_-regular

element. Now, if

(

1−a

)

n

)

∈R and r

(

1−a

)

n

)

is direct summand.

Then, there exists an ideal I R⊂ such that, R r=

(

1−a

)

n

)

⊕J and there exists c J∈ _and f r∈

(

1−a

)

n

)

, such that 1= +f c_{. Thus}

(

1−a

) (

n= −1 a f

)

n + −

(

1 a c

)

n _.

Therefore

(

1−a

) (

n = −1 a c

)

n . Now, to prove r

(

1−a

)

n

)

=r c

( )

. Let _{w r}_∈

(

1₋_a

)

n

)

_{. Then}

₍

₁

₎

n ₀

a w

− = _{and hence}

(

1−a cw

)

n =0

Thus, cw r∈

(

1−a

)

n

)

. But cw J∈ and I_r

(

1−a

)

n

)

=0, then cw=0 and therefore w r c∈

( )

_{, hence}

(

)

(

1 n

)

( )

r −a ⊆r c (7)

Now, let z r c∈

( )

_{. Then} cz=0 _{and hence}

(

1−a cz

)

n =0 _{, Thus}

(

1−a z

)

n =0, therefore z r∈

(

1−a

)

n

)

and we have

( )

(

1

)

n

)

r c ⊆r −a (8)

form (7) and (8) we obtain r c

( )

=r

(

1−a

)

n

)

_{. Therefore}

(

1−a

)

n_{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

(

1−a

)

is right semi π-regular element in R. If a is right semi π-regular element, then there exists b R∈ and n Z_∈ +_{such that} _an ₌_{a b}n _and _{r a}

( )

n ₌_{r b}

( )

_.

Hence, _an

(

1₋_b

)

₌0_{, that is}

(

₁_{− ∈}_b

)

_{r a}

( )

n _{, then}₁_{= + −}_b

(

₁ _b

)

_{and thus}

(

1

)

R bR= + −b R. Therefore _{R bR r a}₌ ₊

( )

n _.

Now, to prove _{bR r a}_

( )

n ₌0_{, suppose that}_{x bR r a}_∈ _

( )

n _{, then} _{x bR}_∈

and _{x r a}_∈

( )

n _{. Hence}_{x br}₌ _{for some}_{r R}_∈ _and _ax₌₀ _{, since}

( )

n

( )

x r a∈ =r b _{, then} bx=0 and b br⋅ =0, that is br=0 [proposition 2.2]. Thus x=0_{and therefore}_{bR r a}_

( )

n ₌0_{, that is} _{r a}

( )

n _{is direct summand}

(6)

Now, if

(

1−a

)

n_{is right semi}_π_{-regular element, then there exists}c R∈

such that

(

1−a

) (

n = −1 a c

)

n _and r

(

1−a

)

n

)

=r c

( )

. Since

(

1−a

) (

n 1−c

)

=0_, we have

(

1− ∈c

)

r

(

1−a

)

n

)

_{, and since}1= + −c

(

1 c

)

_.

Hence, R cR= + −

(

1 c R

)

_{. Thus,} R cR r= +

(

1−a

)

n

)

. Now, to prove r

(

1−a

)

n

)

_cR=0_{. Let} y r∈

(

1−a

)

n

)

_cR_.

Then y r∈

(

1−a

)

n

)

and

y cR

∈

, hence

(

1−a y

)

n =0_and y cr= for some r R∈ _{. Since} y r∈

(

1−a

)

n

)

=r c

( )

_then

cy

=

0

_and c cr⋅ =0_.

Hence cr=0_{[proposition 2.2] and thus} y cr= and then

y

=

0

.

That is r

(

1−a

)

n_cR=0_{. Therefore} r

(

1−a

)

n

)

_{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 _{r a}

( )

n _⊆_{r a}

( )

_{for all} _{a R}_∈ _and _{n z}_∈ +_{, then}_R_{is right}

semi π-regular local ring. Proof:

Let R be local ring. Then either a or

(

1−a

)

is invertible element in R [6]. If a is invertible, then there exists an element b in R such that ab=1, hence

aba a= and then _{a ba a}n ₌ n _{. Let} _{e ba}₌ _{. Then} _{a e a}n ₌ n _{. To prove}

( )

_an ₌_{r e}

( )

_{. Let}_{x r a}_∈

( )

n _⊆_{r a}

_{( )}

_{. Then} _x₌₀_{, it follows that}_bax₌₀_and

then ex=0, that is x r e∈

( )

. Hence

( )

n

( )

r a ⊆r e ₍₉₎

Now, let y r e∈

( )

_{. Then}

ey

=

0

_{and hence} _{a ey}n ₌0_{that is} _{a y}n ₌₀_{, thus}

( )

n

y r a∈ . Therefore

( )

n

r e ⊂r a ₍₁₀₎

from (9) and (10) we obtain _{r a}

( )

n ₌_{r e}

( )

_{. Hence}_a_{is right semi}_π_-regular

ele-ment in R. Now, if

(

1−a

)

_{is invertible element in}_R_{, then there exists an}

ele-ment c in R such that

(

1−a c

)

=1_{. That is}

(

1−a c

) (

1−a

) (

= −1 a

)

_{, it follows}

that

(

1₋_{a c}

) (

n 1₋_a

) (

_{= −}1 _a

)

n_{. let}_{d c}₌

₍

₁₋_a

₎

_{. Then}

₍

₁₋_{a d}

₎

n _{= −}

₍

₁ _a

₎

n_{. To}

prove r

(

1−a

)

n

)

=r d

( )

_{, let} x r∈

(

1−a

)

n⊆r

(

1−a

)

_{, then}

(

1−a x

)

=0_that is c

(

1−a x

)

=0_{and hence} x=0_{, and then} x r a∈

( )

_{. Thus}

(

)

(

1 n

)

( )

r −a ⊆r a (11)

Now, let y r d∈

( )

_{, that is}dy=0_{and hence}

(

1−a dy

)

n =0_{, it follows that}

(

1−a y

)

n =0, that is y r∈

(

1−a

)

n. Hence

( )

(

1

)

n

r a ⊆r −a (12)

form (11) and (12) we have r

(

1−a

)

n=r d

( )

_{. Thus}

(

1−a

)

_{is right semi} π-regular element. Therefore R is right semi π-regular ring.

3. The Conclusion

(7)

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 a R∈ is

π-regular if _Ran₌_Rb_{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 ei-ther _{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.

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this pa-per.

References

[1] Von Neumann, J. (1936) On Regular Rings. Proceedings of the National Academy of Sciences of the United States of America, 22, 707-713.

https://doi.org/10.1073/pnas.22.12.707

[2] Shuker, N.H. (1994) On Semi-Regular Ring. Journal of Education and Science, 21, 183-187.

[3] Al-Kouri, M.R.M. (1996) On π-Regular Rings. M.Sc. Thesis, Mosul University, Mosul.

[4] Kim, N.K. and Lee, Y. (2011) On Strongly π-Regularity and π-Regularity. Commu-nications in Algebra, 39, 4470-4485. https://doi.org/10.1080/00927872.2010.524184

[5] Burton, D.M. (1970) A First Course in Rings and Ideals. Addison Wesley Publishing Company, Boston.