On completely 0 simple semigroups

(1)

VOL. 19 NO. 3 (1996) 507-520

ON

COMPLETELY

O-SIMPLE

SEMIGROUPS

YUE-CHAN PHOEBE HO

Department

of Mathematicsand

Computer

Science

CentralMissouri StateUniversity

Warrensburg, MO 64093

(Received March 17, 1994 and in revised form August 31, 1995)

ABSTRACT.

Let Sbea completely 0-simple semigroup andFbe analgebraically closedfield.

Then for each0-minimalrightideal

M

of

S, M

B

UCU

{0},

where

B

isarightgroup andC

isazerosemigroup.

Also,

amatrixrepresentation for Sother thanReesmatrix isfound for the

conditionthatthesemigroup ring

R(F,

S)

issemisimpleArtinian.

KEY

WORDS AND

PHRASES.

Completely 0-simple semigroups, 0-minimalright ideals, right groups,zerosemigroups,representationof semigroups, semisimpleArtiniansemigroup rings. 1991 AMS

SUBJECT CLASSIFICATION

CODES. 20M30, 20M25

1.

INTRODUCTION.

A

semigroup

S

isasetof elementstogetherwith anassociative, binary operation definedon

S.

A

nonempty subset

A

ofasemigroupSisaleft (right)idealofSifSAC_

A(AS

_{C_}

A).

A

_is_a two-sidedidealof

S

if it isbothaleftidealandarightidealofS.

A

issaid tobeaminimal

(left,

right)idealofS if,forany

(left,

right) ideal

B,

B

C_

A

implies

B

A.

A

(left,

right)ideal

A

of

S issaid tobe 0-minimalifwheneverthere is a

(left,

_right)ideal BofS contained in

A,

either

B

A

or

B

{0}.

Sisa0-simple semigroupifS ₇

{0}

and

{0}

istheonlyproper idealofS.

An

elementeinSiscalledanidempotentife e.

Let E

be the set of idempotents. Define e

_<

f

if

ef

e

re.

Then anonzeroidempotent is saidto be primitive ifit is minimal with

respect to

_<

and S is said to be completely 0-simpleif it is 0-simple and contains aprimitive idempotent.

Let Fbe afield.

A

semigroup ring

R(F,

S)

is anassociative F-algebrawith the semigroup

S as its basisandwithmultiplicationdefineddistributively using the semigroup multiplication

inS. IfIis a

(left,

right) idealofS then the semigroup ring

R(F,

I)

is a

(left,

right) idealof

R(F, S).

Foreach fiin

R(F,

S),

fi

,eS,,,,eFO.z

such that only afinitenumber of

c’s

are nonzero. The set

zES, az 6F

(2)

Supp(5) anddenoteitby

An

n x n matrix

A

(a,j)

is calleda nono-row natrixif at mostone row of

A

contains nonzeroentries; i.e. a_u 0forall i,3 except *0 forsomes0. Let

T

be a semigroup andlet

3,4(n,

r

)

be theset ofall thenxn mono-rowmatricesover

r

.

Then

M(n, T

)

is asemigroup

with matrixmultiplicationasits operation.

Throughout this paper, S denotes a completely 0-simple semigroup, F denotes an

alge-braically closed field,

R(F,

T)

means the senigroup algebra generated by a semigroup

T,

and

R=

R(F,S).

2. O-MINIMAL RIGHT IDEALS.

SinceSiscompletely0-simple,it isshown in

[1]

that S isregularand contains at leastone

0-minimalrightideal. Let

M

be sucha0-minimalrightideal. Then

AI

eSforsomeprimitive idempotente whichserves as aleft identityinM.

Suppose

thereexists a nonzeroelement ain

Ssuch thataS O. Thena ash

{0}

which contradictsthe regularity ofS. Therefore for all nonzeroain

S,

aS7 O.

Hence, M

B

UCU

0}

where

B

{b e

MlbS

M

bM}

(2.1)

and

C

{c

6

MIcS

M

andcM

0}.

(2.2)

PROPOSITION

2.1.

B

isarightgroup;i.e.

B

Gx

E

whereGisagroup and

E

is a_right zerosemigroup.

PROOF. B

isasemigroupbecause,forallbl,

b2

6

B,

(blb:)S

bl(b2S)

bM

M

(2.3)

and

(b)M

b(b2M)

bM

M.

(2.4)

In

order to be a right group,

B

has to be right simple and contain a primitive idempotent. Obviously, thegenerator e of MisinBfor eS

M

eM. So B q). Given anyb

B,

if

bin1

brn:,forrnl,rn2

M,

then

ebern

ebem2,sincee is aleft identity ofM. Butfrom

[1]

weknow thateSeisagroupwith0 and identitye. Soebemusthaveaninverseb ineSe. Thus

rrt em

bt(b)ml

bt(lb)m2

era2 m2.

(2.5)

Therefore bm bm2ifandonly if

rn

rn2. Nowgiven a, b6

B,

a6

M

bM impliesa=bm forsomern M. rnmust bein

B;

otherwiseaM

b(rnM)

0 contradictsthe assumption that a B. HencebB

B

for all b B. Therefore,

B

isaright group and

B

Gx

E

whereGisa group and

E

isarightzerosemigroup.

Let q0 be the identity ofG. Then(g0,

e),

forany e

E,

is a left identity of

B

and ofM.

(3)

C bC.

In

particular, (g0,

e)c

c. Conversely, if(g,

e)c

cfor someg

G,

then cs (go,

e)

forsomes SbecausccS M. Hence

(, )

(,

)(g0,

)

(, )

c

(go,

);

(2.6)

i.e. g=g0.

So(g,e)c=cforanycC

g=g0. Using this result anddenotingd =(g,e)d

forg Gandd

C,

weget

d9

dh

d

(g-lh,

e)d

:=,

g-lh

go

==

g h.

(2.7)

PROPOSITION

2.2. Fixan element e E. Then there existsa subset

D

inC such that everyc Ccanbe uniquely expressed by

d

forsomeg Gandd D.

PROOF.

For the fixede,considerthecollection

‘4

{A

C

Cl(g,e)A

C_ Cand

(g,e)a

7/= (h,e)as

for all g,h G andal 7 as

A}.

(2.8)

Suppose

B

is achain in ‘4. Then forany distinct al and a2 in

tAB

thereexist A1,

As

/ such that al

A1

and a2

As.

Without loss of generosity, assume

A1

C_

A2.

Then hi,as

As.

Then

(g,

e)al

7

(h, e)a2

for allg, h

G;

i.e.

tAB

C,4.

By

Zorn’s

Lemma,

,4 containsamaximal

element

D

and soevery c C canbeuniquely expressedasc

da

forsomeg G and d D.

Otherwise

D

tA

{c}

C_ ,4whichis contradictary to the nature ofD.

Withthe result of Proposition2.2,letusdenote

(g,

d)

(g,

e)d

for each

(g,

e)d

C. Then

(h,

f)(g,

d)

(h,

f)(g, e)(go,

e)d

(hg,

d)

for allg,h

G,

f

E,

andd D. Weconclude that

(g,y)(h,x)

(gh,

x)

for allg,h

G,f

_

E,

andx

E

t.JD.

Accordingto the

Rees

Theoremin

[1],

a completely0-simple semigroupcanbe represented

by a regular

Rees

m xn matrixsemigroup,

M(H;

m, n;

P)

over the group

H,

withan n xrn sandwich matrixP. Whilethe groupHmeansthe -class ofanidempotent e, we can seethat

Gx{e}-eSe=H.

Foreachs

S,

sMiseither

{0}

or a0-minimalright ideal. So S

Us,M,

whereqX

(go,

e)

and

sM

₇ sjMfor all

#

j.

Furthermore, s,M

B,

tAC,tA

{0}

for each with

B,

{b

s,M[bS

s,M

b(s,M)}

and

C,

{c

6-

s,M[cS-

s,Mand

c(s,M)=

{0}}.

(2.11)

Chooses S sothat sMis 0-minimal. Notethat s

sM;

otherwises =tm forsomeother

0-minimalright idealtM. Ifm

B

thensM traM

tM;

whilem CimpliessM traM O. SosS sM siMforsomei. Givenrn

M,

wehave

(4)

whilesm=O

(sm)S=O

<:==

s(mS)=O

m=O.

(2.13)

Now if ms 6 C then

M

(ms)S

(ms)M

0 causes acontradiction. So when srn 6

C,,

ms O.

From

here,wcget

(2.14)

forsome m E

M

such that

XI’S

(g,

)

(g--l,

)(m,)/21

(g-l,

)(m.s)m2

(go, )’r/1

(go,

)m2

1 m2.

(2.15)

3. SEMISIMPLE

ARTINIAN

SEMIGROUP ALGEBRAS.

Nowconsiderthe semigroupalgebra

R

R(F,

S)

whereSisacompletely0-simple semigroup.

Welearned from

[2]

that asimpleideal in ascmisimpleArtinianringisisomorphicto a matrix

ring. With this in mind, we would like to see if this matrix ring can help us find a matrix

semigroup representingS.

First,letuslookat twoimportant properties.

PROPOSITION

3.1.

(see

[3])

If

R(F, S)

isrightArtinian, then Sis finite.

PROPOSITION

3.2.

(see

[1])

R(F, G)

issemisimple Artinian if andonlyifchafF does not

divide

[G[.

WhenSisfinite,

w

gea(x)

is anelement of

R.

Let

J,

{siw

Z

s,(g,x)[x

_

E

tA

D}

(3.1)

for each and J U

J,

in R. For

e S,

if

(go,x)t

0 then

(w)t

aea(g,x)t

0 and if

(g0,z)t

(h,y),

forsomey6

EtA D

andh

e

G,

then

g.G

because

G

Gh.

In

addition, wew 7w with

[G[

and e E. Consequently, each

,

R(F,

Ji)

isarightidealand

R(F,

J)

isan idealofR.

LEMMA

3.3. If

R

issemisimple Artinian, then

,

is a minimalright idealof such that

i

j

forall/andj and

PROOF.

Suppose/]

is anonzeroright idealof contained in

,.

Find a nonzeroelement E

.i

so that

t()

in withrespect to thebasis

J,

isminimal. Suppose

>

1 andwrite

(5)

must be 0 otherwise

fis.w

s,w

haslength 1 in

i.

contradicting t

>

1. So f0g 0. But

since

R

issemisimple, sois

J.

Thenfi 0, whichis against the choiceofft. So g 1 andthen

fi as,w, forsome a

F\{0}

andx EOD. Sincethere exists Ssatisfying

(g0,z)t

B;

i.e.

(g0,x)t

(h,e)

forsomeh

G,

ande

E,

weobtain

a(c-17-1tw)

(as,wx)(

-1,’/-ltwy)

[--1StWxtWy

"/--1S,(WeWy)

"/--18,(Wy)

S,Wy.

(3.4)

But

t(go,y)

tM sjMforsomej implies

a-17-1twu

.

So

s,wu

,

for all y EU

D,

and

.

That is,

L

isaminimalright idealof

Notethat

J,

f3Jj for all j implies

L

Cl 0.

By

mappings,w, to

s,w,

from

L

to

07j

weobtainanisomorphism, hence

L

.

Also J

I,.Jtq=l

Js,

hence j

(L"

PROPOSITION3.4. If

R

issemisimple Artinian,

07

isasimpleideal ofR.

PROOF Let be a nonzero ideal of

R

contained in o

7.

For each i, if Cl

.

0 then f3

L

L.

Givenany0 fi

’,,,,

as,w in

,

if

(g0,

y)fi 0for all y

E

U

D

then

]fi

0 andsofi 0, contradicting

fi

0. Sothereexistsy

E

U

D

such that

0

#

(go,

a,s,z

=

EF,zEEUD

Itfollows that

sj(go,

y)fi

4

fq for eachj andso

@]i

C_

..

Thus

ff

issimple

Under the assumption that

F

isalgebraically

closed, R

is semisimpleArtinianimplies that

]

is a matrix ring such that

]

MatnF.

As was mentioned by

Jacobson[4],

there exists a set ofmatrixunits

{e,s

}

such that

]i

ei,d

.

As we cansee, eachminimalrightideal

],

is an n-dimensionalsubspaceof with basisJ,. So

IE

U

D]

n and the numberofthe elements in

{

J,

}

isalson.

Foreach i, let

07i

be isomorphic to the ithrow-subspacein

Mat,,F

anduse

----

to denotethe twocorrespoondingelements between the two sets. Thenwehave

0 0 0

s,w,

-

a

a2 a,., ithwherea,

F

for eachx

E

UD.

(3.6)

0 0 0

Let us begin bystudying thefirst row. For any e

E,

recall that

ww

7(w)

and suppose

al a2 an

I

0 0 0

w, Then

(6)

"),we 9’

(ll 2

0 0

0 ()

0 0

=a

al a2

n

/

0 0 0

Astod E

D,

weknow that wdwd 0. So

O)

l0

0

0 0 0

(3.7)

0 0 0 0

w

"

..

=:=

0

0 0 0 0

a2

O)

IO

0

0 0 0

al a2

an

0 0

O

al

..

al 0

0 0

(3.8)

,x2

We

concludethat,forxE

E

U

D,

w=

7

..

0 0

1,

ifxE

Azl

_O,

ifxD"

In

general,sincesiwewx 7S,Wxfor each i, givens,w,

where

ithweobtain

0 0 0

"IS,W= al a2 a,, ith

L

0

0 0 0

a

7

,=

,

ith.

0 0 0

(7)

Consequently,s,wx al ,’zl ,x2

0 0

0

ith.

Suppose

thereexists

f

<5

E\{e}

and

0 0 0

s,w!

"

b

b2

b, ith. Then

0 0 0

s,wlwx 7s,w

=

a

A

A,

ith

ba

0 0 0

ith,

(3.10)

hence

a

b

#

0.

Now

let 3’, al. We get siwx

"

7i

0 0 0

ith for each

x <5

E

tOD.

In

orderto study

At,,

weneed to lookat twodifferentcasesofsi(go,

z)

for each mad eachi.

Case

1. If

s,(go,x)

6_

Ci

then

(90,z)si

0 andw,s,wy

=

0 for all!/q

E

U

D.

Thus

0 0 0

O=wswy7

"’i

7i

Az A:

A,,

ith

0 0 0

Al

A2

A

n/

0 0

7A,

7i

..

0 0 0

(3.11)

But3’and7i arenotzero So

A,

0.

(8)

Sow,s, we andw,s,wy wewy 7w for ally E

E

tOD. That is,

0

..

3’ 7,

0 0 0 0 0 0

0

7A,%

...

0 0 0

zth

-I

Hence

A,

7’7,

Let71 7, weobtainournext proposition.

PROPOSITION 3.5.

0 0 0

ith, where

A,

7(%)

-1 _if

s,(go,x)

B,;

and

A,,

0 if

s,(go,x)

C,. Thus, for all x,y

E

U

D,

either withboth

s,(go,

x)

andsi(g0,

!/)

areinB,or

A,,

A,

0.

Wit

h

thisresult,wearereadyto findarepresentation for each elementofS. Given xE

EtOD,

let

f

g,, if

(go,

x)s,

(gi,

e) e

B

h,

_0, _if

(go,x)s,

_O.

(3.13)

In

particular,

f

go, ifz E

E

h,

(3.14)

0, if

D"

Defineamapping

-

S

A4(n,

G

)

by

0 0 0

(s,(g,x))=g

hl

h2

h=,= ith.

(3.15)

0 0 0

iswell-definedfor if

s,(g,x)

sj(h,y)

then j and

(g,x)

(h,y).

(9)

aE Ssuch that

0 0 0

(a)

gl g2 gn ith with g, 0.

(3.16)

0 0 0

PROOFWefirstclaimthat isamonomorphism.

By

letting

s,(g,x),

sj(h,y) be anytwo elementsin

L,

wehave

0 0 0

(s,(g,x))=g

h

hn

ithand

(3.17)

0 0 0

(sj(h,y))=h

h

hv,

jth.

(3.18)

0 0 0

SO

(s,(g,z))(sj(h,y)

ghx,

0 0 0

h

hvi

h2

h,

0 0 0

ith.

(3.19)

If

(go,x)s

_i

e B,

then

(go, z)sj

(hx,,

e)

and

(8i(g, z)sj(h, y))

(si(g,

e)(h,

y))

(si(gh,

h,

y))

ghj

h

0 0 0

h

h,

0 0 0

ith.

(3.20)

But if(go, z)sj 0then

h=,

0.

In

both cases,weseethat

(10)

Suppose (s,(g,

x))

(s,(h,

y)).

Then

0 0

....

0

/

0 0 0

g

hl

h2

hn

ith=h

h

h2

hn

jth.

(3.22)

0 0

}

0 0 0

First, j.

Next,

gh

hhk

for all k. Then,for eachk,either

h,

h

5 Gor

h

0

h.

Consequently,

Ak

A,

forall k, and x y by Proposition 3.5. Thus g h and

s,(g,x)

sj(h,y);

i.e. is amonomorphism.

Nowwewanttoshow that

(S)

isaleftidealof

.M(n,

G).

Given anysi(g,x)

e

Swith

0 0 0

(s,(g,x))=g

h:l h:

h.

ith

(3.23)

0 0 0

and any

0

jth E

.M(n,

GO),

theproduct

0 0 0 0 0 0

bl

b2

bn

jth g

ha

h

hn

0 0 0 0 0 0

ith

isstill in

(S).

Therefore

(S)is

leftidealof

M(n,

G).

For each i, there exists x

e

Esuch that

s,(go,x)

e

Bi, hence

(go,x)s,

(h,,,e)

forsome e E. Thus

(s,(go,x))

is anelementin

(S)

whose iithentryisnonzero.

(11)

(i)

Sis finite,

(ii)

S is isomorphic to aleft idealofan n x n mono-rowmatrixsemigroupover afinitegroup

G,

denoted by

Y

.Azi(n,

GO),

such that for each i, there exists an element a E Swith

0 0 0

a 9a 92 9,., ith andg,

:/:O,

0 0 0

(iii)

the characteristic of

F

doesnot divide

IG

I.

By

assumption(iii),

R(F, G)

isasemisimpleArtinianring. Thenitis statedin

[2]

that is

thedirectsumofitsminimalleftidealswhicharegenerated bya setoforthorgonalidempotents

{fa,f2,"’f,}

andthe identity1

fa

+

f2

+’"+ fp.

Notethat

.

R(F,M)=

Mat,(().

Let

(f).

jthfor l<i<pandl<j

<n.

Then

{(f,)jj[i

1, 2, p; j 1, 2,

n}

isaset oforthorgonalidempotentsin

Q

suchthat

,,i(fi)ji

isequaltothe identitymatrixin

Q.

LEMMA

3.7

.4(f,)aj

isaminimalleftidealof

Q

for each and j.

PROOF

Foreach and j, the left ideal

all

./(f,)jj

{

anl

aln

’

"..

)

(fi)jjlak,

e [

for each k and

l}

art

Also

jth

(3.26)

all

anl

jth

aln 0

bafi

0

)

a,,,

0

b,,f,

0

(12)

ll

forall

aln

"..

)

E

Q.

Since

f,

is aminimalleft idealof

(,

either

biifi

(fi

atlrl

or

b,

f

0. But if

(b

f

0then

b,.

f, 0. So

.

jth

blf,

0

bnlfi

0

iseither 0or

](fi)

for any

jth

0 bljft 0

0

bnjfi

0

That is,

.4(fi)jj

isaminimalleftideal

Lete,,

ith

0 0 0

0 1 0

0 0 0

ithfor eachi. Then

.

$./4(fj)ii

because

.,Q

e

fzl

f

@ (9

./,4

h

and

(3.28)

(3.29)

Therefore

.,Q

issemisimpleArtinian.

PROPOSITION

3.8

R

PROOF.

Foreach i, thereexistsanelementa Ssuchthat

0 0 0

a

-

ga g2 g ith

0 0 0

(3.30)

(13)

ith

{th aE

R(F,

(S)),

(3.31)

hence 0

R(F,(S))(f),,

4(f),,.

Itfollows that

R

-

af,4.

Withall the properties foundhere, weobtainthe majorresult,

THEOREM3.9. Let

F

beanalgebraically closedfieldandSbea0-simple semigroup. Then

R

is semisimpleArtinianifandonlyifthefollowinghold:

(1)

Sisfinite;

(2)

5’is isomorphicto a left idealofa mono-rowmatrixsemigroup

.h4(n,

G

)

where G eh’e ande isanidempotent of

S,

such thatfor each i, thereis anelementaE5’with

(3)

ThecharacteristicofFdoes notdivide

[G[.

AccordingtoTheorem 5.20 in

[1],

thesandwichmatrix

P

inthe regular

Rees

matrix

semi-group,

)t4(G;

h,k;

P),

isnonsingular

(in

particular, h

k).

Here

h k meansthe number of

distinct0-minimalright

(and left)

idealsofS. Sothesizeof

P

isthesame asthat of themono-row matricesfoundin thispart.

Somereadersmay findthatthe representation described inTheorem 3.9isaspecialcaseof the dualSchutazenbergerrepresentationmentioned in

[1].

Buttheapproachesaredifferent. 4.

RELATION

TO REES

MATRIX

REPRESENTATION.

Hereafter,

we would like to check ifwe can find the

Rees

matrix representation from the

MainTheorem directly. Firstlet uslook atthe left idealsgenerated by elementsfrom

E

tO

D.

LEMMA

4.1.

S(9o, z) =/:

S(go, y)

for alldistinct x,y E

E

tO

D.’

PROOF. Supposethereexist x,y E

E

tOD suchthat

5"(go,

x)

5"(go, y).

Then

(go,

)

(po,

)(go,

)

S(go,

)

(go,

:)

(go,

u)

for some s

e

S. s

e

M

for ifsM

=

M

then

(go,x)

M

is a contradiction. Ifs

e

C then

(go,

x)

s(go,

y)

0 whichcausesanothercontradiction. Sos must bein

B

ands

(g,

f)

for someg Gand

f

EE. Hence

(14)

Since theset

E

t.J

D

is finite, we canlist the elementsin anorderwithone of the element

eE

E

tobe thefirst Using thesamenotations in

[1],

let

(g0,zx)

qx,

s,(go,e)

r,, and

qxr,, if qxr, E

Hll

px,

0, otherwise

(4.3)

The lemma abovehelpsusobtaining the nonsingular sandwichmatrix

P

(px,)

over

HI

(which

isthesameas

G).

Notethatfor eachi,px,

(go,xx)s,(go,e)

h,x,.

So

hel

he2

hen

P=

h: h:2

h,

(4.4)

h:.a h:.2

h’..

and for each s S

0 0 0

s=g

h

h.

0 0 0

ith

(4.5)

si(g,

xx)= s,(go,e)(g,e)(go,xx)

r,(g, e)qx

(g)ix; theReesmatrix.

Thisshowstherelation betweenthe

Rees

matrixandthematrixdescribed in thisarticle.

REFERENCES

1. Clifford

A.

H.and

G.

B.

Preston, Thee

AlgebraicTheory ofSemigroups

I,

Amer.

Math.

Soc.,

Providence,

R.I.,

1961.

2. Hungerford T.

W.,

Algebra, Springer-Verlag, New York, 1974. 3. Oknifiski

J.,

SemigroupAlgebras, Dekker, New York, 1991

4. Jacobson

N.,

StructureofRings, Amer. Math.

Soc.,

Providence,

R.I.,

1964.

5. StenstrSm

B.,

Rings of Quotients, Springer-Verlag, New York,1975.