Survey of Linear Transformation Semigroups Whose The Quasi-ideals are Bi-ideals

Survey of Linear Transformation Semigroups

Whose The Quasi-ideals are Bi-ideals

Karyati1,Sri Wahyuni2

1 Department of Mathematics Education, Faculty Of Mathematics and Natural Sciences, Yogyakarta State University

2 Department of Mathematics Education, Faculty Of Mathematics and Natural Sciences, Gadjah Mada University

Abstract

A sub semigroup of a semigroup is called a quasi-ideal of if . A sub semigroup of a semigroup is called a bi-ideal of if . For nonempty subset

Q S S SQ∩QS⊆Q

B S S BSB⊆B A of a

semigroup S,

( )

A_q and

( )

A_b denote respectively the quasi-ideal and bi-ideals of S generated by A . Let denote the class of all semigroup whose bi-ideals are quasi-ideals. Let be a module over a division ring and be a semigroup under composition of all homomorphisms

BQ M_D

D Hom(M_D)

D

D M

M →

:

α . The semigroup Hom(MD)is a regular semigroup.

In this paper we will survey which sub semigroups of whose the quasi-ideals are bi-ideals.

) (M_D Hom

Keywords

: Semigroup, BQ, quasi-ideal, bi-ideal

I. Introduction

The notion of quasi-ideal for semigroup was introduced by O. Steinfeld, 4), in 1956. Although bi-ideals are a generalization of quasi-ideals, the notion of bi-ideal was introduced earlier by R>A Good and D.R Hughes in 1952.

A sub semigroup of a semigroup is

called a quasi-ideal of if . A sub

semigroup of a semigroup is called a bi-ideal of if . Then Quasi-ideals are a neralization of left ideals and right ideals and bi-ideals are a generalization of quasi-bi-ideals.

Q S

S SQ∩QS⊆Q

B S

S BSB⊆B

O. Steinfeld has defined a bi-ideal and quasi-ideal as follows: For nonempty subset A of a semigroup , the quasi-ideal of S

generated by

S

( )

A_q

A is the intersection of all quasi-ideal of S containing Aand bi-ideal

( )

A_b of S

generated by A is the intersection of all bi-ideal of S containing A .1).

We use the symbol to denote a semigroup with an identity, otherwise, a semigroup with an identity 1 adjoined . 4). A.H. Clifford and G.H. Preston in 1) have proved that

1

S S

S

Proposition 1.1. ([1], page 133) For a nonempty subset A of a semigroup S,

( )

A_q =S1A∩AS1=

(

SA∩AS

)

∪A

Proposition 1.2. ([1], page 133) For a nonempty subset A of a semigroup S,

( )

_A

( )

_AS1_{A A}

(

_ASA

)

_{A A}2

b = ∪ = ∪ ∪

By these definitions,

( )

A_qand are the smallest quasi-ideal and bi-ideal, respectively, of

containing

( )

A_b

S A. Since every quasi- ideal of S is a bi-ideal, it follows that for a nonempty subset

A of S,

( )

A_b ⊆

( )

A _q. Hence, if

( )

is a

quasi-ideal of , then

b

A

S

( )

A_b=

( )

A_q, then we have:

Proposition 1.3. ([1], page 134) If A is a nonempty subset of a semigroup S such that

( )

A_b ≠

( )

A _q, then

( )

A_b is a bi-ideal of which is not a quasi-ideal.

S

S Lajos has defined a , that is the class of all semigroup whose bi-ideals are quasi-ideals. He has proved that:

BQ

Proposition 1.4. ([5], page 238) Every regular semigroup is a BQ-semigroup.

The next proposition is given by K.M Kapp in 5).

Proposition 1.5. ([5], page 238) Every left [right] simple semigroup and every left [ right] 0-simple semigroup is a BQ-semigroup.

In fact, J.Calais has characterized -semigroups in 5) as follows

Proposition 1.6. ([5], page 238) A semigroup is a -semigroup if and only if

S BQ

( ) ( )

x,y _q = x,y_b

for all x,y∈S

Let be a module over a division

ring and be a semigroup under

composition of all homomorphisms

D

M

D Hom(M_D)

D

D M

M →

:

α . The semigroup is a

regular semigroup. 6).

) (M_D Hom

II. Main Results

Let be a set of all bijective

homomorphisms of , i.e.:

(

MD

Is

)

)

(

M_D Hom

(

M_D

)

{

α Hom(M_D) αisan isomorfisma

}

Is = ∈

The is a regular semigroup, because for

all

(

MD

Is

)

)

(

MD

Is ∈

α , there is an α'=α−1∈Is

(

M_D

)

such that αoα'oα =αoα−1oα =α. By

Proposition 1.4. the semigroup is in .

(

MD

Is

)

)

BQ

The next, we construct some subsets of as follow:

(

MD

Hom

(

M_D

)

{

α Hom

(

M_D

)

αis injective

}

In = ∈

(

M

)

{

Hom

(

M

)

Ran V

}

Sur _D = α∈ _D α =

(

M

)

{

α Hom

(

M

)

dim

(

V\Ranα

)

isinfinite

}

OSur _D = ∈ _D

(

M_D

)

{

α Hom

(

M_D

)

dimKerαisinfinite

}

OIn = ∈

( )

( )

(

)

⎪⎭ ⎪ ⎬ ⎫ ⎪⎩ ⎪ ⎨ ⎧ ∈ = infinite is Ran \ dim injective, is α α α V M Hom M

BHom _{D D}

By these definitions, so we get:

, . The

and are sub semigroups of

:

(

MD

)

Is ⊂In

(

MD

)

Is

(

MD

)

⊂Sur

(

MD

)

)

)

(

MD

In Sur

(

M_D

(

MD

)

Hom

If α,β∈In

(

M_D

)

, then Kerα={0}, Kerβ={0}. From this condition, we have , such that In

(

MD

)

is a sub semigroup of Hom

(

M_D

)

If α,β ∈Sur

(

M_D

)

,then Ranα =V , Ranβ =V . From these conditions, we have V=Ran

(

αoβ

)

,

such that the is a sub semigroup of

.

(

MD

Sur

)

)

)

(

MD

)

Hom

The is a semigroup of

, it is caused by:

(

MD

OSur

(

MD

Hom

If α,β ∈OSur

(

M_D

)

, so dim

(

V \Ran

α

)

and

(

V \Ran

β

)

dim are infinite. For x∈Ran

(

αoβ

)

, there is y∈V such that (y)

(

αoβ

)

=x or

(

(y)α

)

β =x. So,x∈Ranβ and we conclude that

(

α β

)

Ranβ

Ran o ⊆ , so dim

(

V\Ran(αoβ

)

is infinite.

The BHom

(

MD

)

is a sub semigroup of

(

MD

)

Hom :

Ifα,β∈BHom

(

M_D

)

, then α,β are injective and the dim

(

V \Ranα

)

and dim

(

V \Ranβ

)

are infinite. By the previous proofing, so

(

\ ( )

)

dimV Ranαoβ are infinite and

(

αoβ

)

is bijective

In order to prove that . In

(

MD

)

and

(

MD

)

Sur are in if and only if is

finite, we need this Lemma:

BQ dimV

Lemma 2.1. ([2], page 407) If is a basis of

, and

B

D

M A⊆B α∈Hom

(

M_D

)

is one-to-one, then

(

Ranα/ Aα

)

dim = B\A

From this Lemma, so we can prove that:

Theorem 2.2. ([2], page 408) if and only if is finite.

(

M

)

BQ In _D ∈

D

M dim

Proof:

If dimM_D is finite, then In

(

M_D

)

=Is

(

M_D

)

. So,

(

MD

)

In is a regular semigroup. By Proposition 1.4. In

(

M_D

)

∈BQ.

The other side, assume that is infinite. Let

D

M dim

B be a basis of M_D, so B is infinite. Let

{

u n N

}

A= _n ∈ is a subset of B, where for any distinct i, j∈N, u_i ≠u_j.

Let α,β,γ∈Hom

(

M_D

)

be defined as follow:

⎩ ⎨ ⎧ ∈ ∈ = = A B v v N n u v u

v n n

\ if some for if ) ( α 2

⎩ ⎨ ⎧ ∈ ∈ = = + A B v v N n u v u

v n n

\ if some for if )

( β 1

⎩ ⎨ ⎧ ∈ ∈ = = + A B v v N n u v u

v n n

\ if some for if )

( γ 2

By this definition, so kerα=kerβ =kerγ =

{ }

0 , such that α,β,γ∈In

(

M_D

)

. Next, we have

( )(

u_n βoα

) ( )(

= u_n α•γ

)

, for all n∈N and for all

A B

and

( )(

v αoγ

) (

= (v)α

) ( )

γ = vγ =v. So we conclude that α≠βoα=αoγ. By this conditions we have βoα∈In

(

MD

)

α, because

(

MD

In

∈

)

β and αoγ∈α In

(

M_D

)

. We know that β oα=αoγ , so by the Proposition 1.1 we have βoα∈In

(

M_D

)

α∩αIn

(

M_D

)

=

( )

α _q.

Suppose that

β

o

α

∈

( )

α

_q, because α≠βoα, by Proposition 1.1 we get βoα∈αIn

(

M_D

)

α. Let λ∈In

(

MD

)

such that βoα=αoλoα.

Since

α

is injective, so β=λoα. Then we have:

{ }

1

\ u

B =Bβ=B

(

αoλ

) ( )

= Bαλ=

(

B\

{

u2n−1n∈N

}

)

λ

By Lemma 2.1, we have:

{

}

(

Ran / (B\ u_2n−1 n∈N)

)

=

{

u_2n−1 n∈N

}

dim λ λ

From these conditions, so this condition is hold:

{ }

(

Ran / B\ u₁

) {

= u_2n−1 n∈N

}

dim λ , but in the

other hand :

{ }

(

/ \

)

dim

(

/ \

{ }

)

1

dimRanλ B u₁ ≤ V B u₁ = .

So, there is a contradiction. By Proposition 1.3, we have βoα∉

( )

α _b, then In

(

MD

)

∉BQ.

Theorem 2.3. ([2], page 408) Sur

(

M_D

)

∈BQ if and only if dimHom

(

MD

)

is finite

)

)

)

)

)

)

)

)

Proof:

The proof of this theorem is similar with the previous theorem.

The other semigroups e.i. a always belongs to but it is not regular and is neither right 0-simple nor left 0-simple, if is finite. These condition is guarantied by these propositions:

(

MD

OSur BQ

D

M dim

Propositions 2.4. ([2], page 409) The semigroup isn’t regular.

(

MD

OSur

Propositions 2.5. ([2], page 410) The semigroup is neither right simple nor left 0-simple.

(

MD

OSur

Although has properties

likes above, is a left ideal of

and is always in .

(

MD

OSur

(

MD

OSur

(

MD

Hom BQ

For the other semigroup, i.e. is in if and only if is countably infinite. It is caused by this lemma:

(

MD

BHom BQ dimM_D

Lemma 2.6. ([2], page 411) If is

countably infinite , then is right

simple.

D

M dim

(

MD

BHom

The next, we construct the other subsets of Hom

(

M_D

)

:

(

MD

)

OInSur

(

)

(

)

{

α∈HomM_D dimKerα,dimV\Ranα areinfinite

}

=

(

MD

)

OBHom

(

)

{

α∈HomM_D Ranα=M_D,dimKerαisinfinite

}

=

From the definition, we get:

(

MD

)

OInSur =OSur

(

MD

)

∩OIn

(

MD

)

, this set

is not empty set, because

∈

0 OSur

(

M_D

)

∩OIn

(

M_D

)

= . This

set is a sub semigroup of .

(

MD OInSur

)

)

(

MD

)

Hom

Lemma 2.7. ([5], page 240) For every infinite

dimention of , is a regular sub

semigroup of

D

M OInSur

(

MD

(

MD

)

Hom

The following theorem is the corollary of the previous lemma:

Theorem 2.8. ([5], page 240) For every infinite dimention of MD, OInSur

(

MD

)

is in BQ

The set OBHom

(

M_D

)

is an intersection of In

(

MD

)

and OSur

(

MD

)

. Let be a basis

of , since is infinite, there is a subset

B

D

M B Aof

such that

B A = B\A = B. Then there exist a

bijection ϕ:A→B. Define a homomorphisms in

(

MD

)

Hom as follow:

( )

( )

⎩ ⎨ ⎧

∈ ∈ =

A B v if

A v if v v

\ 0

ϕ α

Hence α∈OBHom

(

MD

)

Lemma 2.9. ([5], page 241). If is

countably infinite, then is left

simple.

D

M dim

(

MD

OBHom

)

As the corollary, we get:

Theorem 2.10. ([5], page 242). The smeigroup

(

MD

)

OBHom is in if and only if is countably infinite.

BQ dimM_D

III. Conclusion

From this survey, we can conclude that there is a different conditions such that the sub semigroup of Hom

(

MD

)

is in BQ, i.e:

1. In

(

M_D

)

∈BQ if and only if is finite.

D

M dim

2. Sur

(

M_D

)

∈BQ if and only if

(

MD

)

Hom

dim is finite

4. The semigroup is neither right 0-simple nor left 0-0-simple.

(

MD

OSur

)

)

)

)

)

)

5. If is countably infinite , then is right simple.

D

M dim

(

MD

BHom

6. For every infinite dimension of ,

is a regular sub semigroup of

D

M

(

MD OInSur

(

MD

)

Hom

7. For every infinite dimention of ,

is in

D

M

(

MD

OInSur BQ

8. If is countably infinite, then

is left simple.

D

M dim

(

MD

OBHom

9. The smeigroup is in if

and only if is countably infinite.

(

MD

OBHom BQ

D

M dim

10. Finally, we can conclude that not every semigroup in is a regular semigroup and not every semigroup in is either right 0-simple or left 0-simple.

BQ

BQ

V. References

1. Baupradist, S, Kemprasit, Y,” Existence of Bi-ideals which are not Quasi-ideals in Semigroups of Continuous Functions and Semigroups of Differentiable Functions”, Proceeding of the International Conference on Algebra and Its Applications, 133,143 (2002) 2. Namnak, C, Kemprasit, Y,”On Semigroups of

Linear Transformations whose Bi-ideals are Quasi-ideals”, PU.M.A Vol 12 No4, 405,413 (2001)

3. Namnak, C, Kemprasit, Y, “Some

Semigroups of Linear Transformations Whose Sets of Bi-Ideals and Quasi-ideals Coincide”, Proceeding of the International Conference on Algebra and Its Applications, 215,224 (2002) 4. Namnak, C, Kemprasit, Y, “ Generalized

Transformatioan Semigroups Whose Bi-ideals and Quasi-ideals Coinside”, Southeast Asian Bulletin of Mathematics Vol 27, SEAMS, 623-630, (2003)

5. Namnak, C, Kemprasit, Y, “Some BQ-

Semigroups of Linear Transformations”, Kyungpook Mathematics Journal 43, 237, 246 (2003)