Pseudo-Complemented Semigroups

A theory of pseudo-complements in lattices, and particularly in distributive lattices was first developed by V. Glivenko [14], M. H. Stone [8] and Garrett Birkhoff [3]. Later, O. Frink [9] extended the concept of pseudo-complements to semilattices as a generalization of the theory of lattices. A semilattice (L, ∧) with least element 0 is said to be pseudo- complemented if, for any a ∈ L, there exists an element а∗ satisfying the condition that a ∧ x = 0 if and only if x ≤ а∗. Further, it is proved that L∗ = {а∗: a ∈ L} becomes a Boolean algebra. Lee [7] proved that any pseudo-complemented semilattice is equationally definable. A semigroup with zero is an algebra of type (2, 0) satisfying a(bc) = (ab)c and a0 = 0 = 0a for all a,b,c∈S. Let E(S) denote the central idempotent e in a semigroup S, i.e., e2 (= e.e) =e and ex = for allX ∈S . Then E(S) is a subsemigroup of S . In this paper, we shall introduce the notion of a pseudo- complemented semigroup (S, . ) with 0 and give certain important properties of these. Mainly, we prove that the set of all pseudo-complements of elements of S is a Boolean algebra.

The concept of Baer-Stone semigroup was first introduced by U.M.Swamy [13]. Here, we prove that a Baer-Stone semigroup is a common abstraction of the multiplicative semi group of Baer- ring [13] and the meet semi-lattice structure of a Stone lattice [1]. In this paper we mainly pay attention to pseudo-complemented semigroup.

2.    Pseudo-Complemented Semigroup

In this section, we give the definition of a pseudo-complemented semigroup and prove some basic properties of a pseudo-complemented semigroup.

Definition 2.1. Let S be a commutative semigroup with zero. For any x ∈ S and X ⊆ s , define the sets, as given below:

xS = {xa :   ∈S}

• ∗ = {a∈S:    = 0 for all x ∈S }.

X ^∗ is called the annihilator of X in s . If X = { x }, then simply we write ( x )^∗ for { x }^∗.

Definition 2.2. A commutative semigroup with zero is said to be a pseudo-complemented semigroup if, for each ∈  , there exists an idempotent in S such that ( )^∗ =   .

Observe that for any idempotents and in ,    =    implies = (for, ∈    =     ⇒

=     ⇒     =     =     = and similarly     =   ). This shows that in a pseudo complemented semigroup S there can be atmost one idempotent, which will be denoted by ∗ such that ∗ = ∗ . In this case, the map ′→  ∗ is called a pseudo-complementation on . Note that these can be at most one pseudo- complementation on any semigroup.

The following are some examples of pseudo-complemented semigroup.

(x)∗ ={

{0} if x ǂ 0

R if x =0

and hence ( ,.) is a pseudo-complemented semigroup.

∗      0 if x ǂ 0

• ^{x ={} 1 if x=0

Example 2.5. Let be the set of positive divisors of 50 i.e.,   = {1,2,5,10,25,50}. If we define for any

,     ∈   ,      =        { ,  }.

Then ( ,.) is a pseudo-complemented semigroup with 1 as the zero element, in which

1∗ = 50, 2∗ = 25, 10∗ = 0 = 50∗     5∗ = 2 = 25∗.

In the following lemma we give some properties of a pseudo-complemented semigroup.

Lemma 2.6. Let be a pseudo-complemented semigroup. Then for any ,   ∈  , we have the following.

• (1)    0^∗ is the unity in .

• (2)   ∗  ∗ =

• (4)    = 0 ⇔  ∗

• (5)   ∗∗   =   =

• (6)   ∗∗∗ =

• (7)    (    )∗∗ =   ∗∗∗∗

Proof.

• (1)    It follows by the fact that = (0)∗ = 0∗

• (2)    and (3) are direct implications of the definition of pseudo-complemented semigroup.

• (4) xy — 0 ^ у — x* f for some s E S ^ x * у — y.

• (5)    It follows by (3) and (4).

• (6)    We have    = 0 ⇔  ∗∗   = 0. It follows that ∗∗∗ =

• (7)    Let ,    ∈  . We have

• (   )^∗ = 0 ⇒  ^∗ (   )^∗ =  (   )^∗ (by(4))

⇒  ∗∗ (  )∗ =0

⇒    ∗∗(  )∗ =0

⇒   ^{∗ ∗∗}(   )^∗ =   ^∗∗(   )^∗ (by(4))

⇒  ∗∗ ∗∗(  )∗ =0

⇒ (   )∗ ∗∗  ∗∗ =   ∗∗∗∗ and ∗    = 0 ⇒ (  )∗ ∗ =

⇒(  )∗∗ ∗ =0

⇒  ∗∗(   )∗∗ = ()

and similarly  ∗∗(   )∗∗ = ()

⇒  ∗∗ ∗∗(   )∗∗ =   ∗∗(   )∗∗ = ()

From the above two arguments, (   )∗∗ =  ∗∗∗∗

For any idempotent e in a commutative semigroup , the set    = {   ∶    ∈  } is a subsemigroup of , since (   )(   ) =.

Theorem 2.7. Let S be a pseudo-complemented semigroup and e an idempotent in S, then eS is a pseudocomplemented semigroup.

Proof. Let x ^ x * be a pseudo-complementation on S. We have eS is a commutative semigroup with zero. For any e x and e у G eS with x and у € S, we have

( e x)( e у) = 0 ^ x e у = 0 (since e2 = e ) « x*ey = ey « ex*ey = ey and hence ( ex)* = ex* in eS. Therefore ex ^ ex* is a pseudo-complementation on eS. Thus eS is pseudo-complemented semigroup.

Definition 2.8. Let S be a pseudo-complemented semigroup. For any subset X of S, define

X * = {x * : x G X}.

For any pseudo-complemented semigroup S, we have

S * = F(S) * = { e G T(S): e ** = e} = {x G S: x ** = x} = {x G S: x = у * / о r sоme у G S}

Definition 2.9. An element x in a pseudo-complemented semigroup S is called closed if x** = x; The set of all closed elements in S is called the centre of S and is denoted by B(S).

We observe that (S) = S * = T(S) * . Now, we prove that B(S) is a Boolean algebra of all closed elements of .

Theorem 2.10. Let S be a pseudo-complemented semigroup. For any x, у G B(S), define x Л у = xy and x V у = (x*y*)*.

Then (B(S),Л,V,*) is a Boolean algebra.

Proof. Clearly (B(S),Л) is a semilattice and we have a partial order < on B(S) defined by a < b «а Л b = a for any a,b G B(S) and а Л b = glb {a,b}. Let a,b G B(S) . Since a*b*a = 0 = a*b *b , (a*b*)*a = a and (a*b*)*b = b, which implies that a < ( a*b *)* = a V b and b < a V b and hence a V b is an upper bound of {a,b} in B(S). Now, for any c G B(S), a < c and b < c ^ ac — a and bc — b

^ c*a =0 =

^ a*c* = c* =b

^ a*b*c*

^ (a*b *)*c* =0

^ (a * b * ) * c ** = (a * b * ) *

^ (a V b)c = a V b ( since c**

^ a V b < c

Therefore a V b is the lub of { a, b} in B(S). Now а Л ( a V b) = a( a*b *)* = a (since ( a*b*)a = 0)

Also, a * (ab ) = 0 and hence (ab) * a * = a * which implies that a V (a Л b) = (a * (ab ) * ) * = a ** = a. Thus (B(S)^,V) is a lattice with smallest element 0 and greatest element 1(= 0 * ). Further, for any a G B (S), а Л a * = aa * = 0 and a V a * = (a * a ** ) * = 0 * = 1. Thus a * is a complement of a in B(S). Finally, let a, b,c G B(S). Put x = а Л (b V c ) and у = (a Л b ) V (a Л c ). In any lattice, it is true that у < x. We have а Л b < у ^ у * пЬ = 0 ^ у * пЬ * = у*a and а Л c < у ^ y*ac =

• 0 ^ y*ac* = у * п.

Therefore y * ab * c * = y * a which implies that y * a(b * c * ) * = 0 and hence y ** x = x. Since у ** = , We have    = and hence <  ,thus   = and therefore   Л ( V ) = ( Л ) V ( Л

• ^c ).

3.    Baer-Stone Semigroups

Therefore (B(S),Л,V) is a distributive complemented lattice and hence a Boolean Algebra.

The following is a consequence of above theorem.

Corollary 2.11. Let B(S) be the centre of a pseudo-complemented semigroup. For anyx and у in B(S) define x.у = xy an d x + у = ((xy*)*(x*y)*)*.

Then ( в ( s ),+,.) is a Boolean ring; that is a ring with unity in which every element is idempotent.

The concept of Baer-Stone semigroup was first introduced by U. M. Swamy [13] in a short note published in the journal ‘Semigroup Forum’ and mentioned a characterization theorem without proof. In this section, we characterize Baer-Stone semigroups, as a common abstraction of Baer-rings and Stone lattices. First, we start with the following theorem.

Theorem 3.1. Let *^1 and *^2 be pseudo-complemented semi groups. Then the product *^1 × S2 is a pseudocomplemented semigroup under the coordinate-wise operations.

Proof. Since*^1 S₁ and S2 are commutative semigroups with zero, so is *^1 × S2 and (0,0) is the zero in ■$1 × ^2 .

For any (X-у , X2 ) and (У1 , y2 ) in *^1 × ^2 ,

(^1 , *2 ) ( У1 , У2 ) = (0, 0) ⇔ Х1У1 = 0 and Х2У2 = 0

⇔ *1 ∗ У1 =         ∗ У2 =

⇔ ( ^1 ∗, *2 ∗) (У1 , У2)=(У1 , У2 )

and hence ( X-у ∗, ■^2 ^∗) is the pseudo-complement of ( ^"1 , ■^2 ) in *^1 × S2 .

Note that, the converse of above theorem is also true; that is, if *^1 × S2 is pseudo-complement semigroup then so are S^ and ■^2 . If ( X^ ,0)^∗=( a , b ) then *1 ∗ ₌ a Also, if (0, x₂ )^∗=( s , t ), then x₂ ∗₌ t .

The above result can be extended to arbitrary products of pseudo-complemented semigroups.

Theorem 3.2. Let ( s ,.) be a pseudo-complemented semigroup and x ∈ s . Define

dx ∶   →x∗S×x ∗∗ S by dx (У) = (x∗У,x ∗∗ У).

Then d_x is a homomorphism of semigroups and preserves pseudo-complements.

Proof. By theorem 2.7, x ^∗ S and x ∗∗ S are pseudo-complemented semi groups such that

(X∗У)∗ = x∗У ∗ in x∗S and (x ∗∗ У)∗ = x ∗∗ У ∗ in X ∗∗ S.

Since X ^∗ is an idempotent and S is commutative, we have

dx (yz ) = (X∗yz,X ∗∗ yz) = (X∗yx ∗z,X ∗∗ yx ∗∗ z)

= (X∗У,x ∗∗ У)(x∗z,X ∗∗ z)

=   ( У ) d_x ( z )

and   dx (У∗) = (x∗У,x ∗∗ У ∗)

= ((x∗У)∗,(х ∗∗ У)∗)

= (х∗У,х ∗∗ У)∗ = dx (У)∗, the p.c of dx (У).

Thus d_x is a homomorphism and preserves pseudo-complements.

Definition 3.3. For any pseudo-complemented semigroup S , the map dx ∶   →х∗S×х ∗∗ S defined by dx (У)=(х∗У,х ∗∗ У ) for all У∈S is called the decomposition map corresponding to х∈S.

Definition 3.4. A pseudo-complemented semigroup S is called a Baer-Stone semigroup if for each х ∈ S , the decomposition map d_x is an isomorphism of S onto х ^∗ S × х ∗∗ S .

Theorem 3.5. Let ( S ,.) be a Baer-Stone semigroup and e an idempotent in S. Then eS is Baer-Stone semigroup.

Proof. By theorem 2.7, eS is a pseudo-complemented semigroup in which the pseudo-complemented of any ex is precisely ex∗. Now, for any ex ∈ eS, consider the map f ∶    → ex ∗s× ex ∗∗ s defined by

f ( ey) = (ex ∗ ey, ex ∗∗ ey) = (X∗ ey,X ∗∗ ey).

Clearly f is a homomorphism and an injection, since the decomposition d_x is so. To prove f is surjective, let ( ex ^∗ У , ex ∗∗ z ) ∈ ex ^∗ S × ex ∗∗ S . Then there exists s ∈ S such that ^x ( s ) = ( x ^∗ ey , x ∗∗ ez ) and therefore x ^∗ s = ^∗ ey and x ∗∗ s = ∗∗ ez . Which implies that f ( es ) = ( x ^∗ ey , x ∗∗ ez ). Thus f is an isomorphism. Therefore eS is a Baer-Stone semigroup.

Next, we establish a correspondence between the closed elements in a Baer-Stone semigroup S and decompositions of S into products of two semigroups with zero and unity.

Theorem 3.6 . Let ( S ,.) be a Baer-Stone semigroup and x ∈ S . Then x ∈ В ( S ) if and only if there exist Baer-Stone semigroups ^i and ^2 and an isomorphism f ∶   → S 1 × ^2 such that f ( x ) = (1,0); in this

case (X∗) = (0,1).

Proof. Suppose that X ∈В(S). Then х∗∗ = х . Note that both X and X∗∗are idempotents. Put S-^  = and ^2 =  ∗S. Then, by theorem 3.5, ^i and S^ are Baer-Stone semigroups in which X and X ∗ are unities in *^1

and ^2respectively. Define f ∶   → “^l × ^2 by f ( У ) = ( xy , X ^∗ У ) for all У ∈ s .

Then f ( x ) = ( x ,0) and f ( X ^∗) = (0, X ^∗). Since S is Baer-Stone semigroup, it can be easily verified that f is an isomorphism. Conversely, suppose that S^ and ^2 are Baer-Stone semigroups and f ∶    → S-f ×

S₂ is an isomorphism such that /( X ) = (1,0). Then/( X ^∗) = (0, 1) and f ( X ^∗∗) = (1,0). Since/ is an injection, X ∗∗ = X . Thus X ∈ В ( S ).

Definition 3.7. [6]. A commutative ring ( R ,+,.) is called a Baer-ring if, for each a ∈ R , there exists an idempotent e in

R such that the principal ideal generated by e is the annihilator of a in R ; that is, ( a )^∗ = eR .

Note that for any element a in a commutative ring R , there can be at most one idempotent e in R such that ( a )^∗ = eR ; for , eR =     for any idempotents e and f implies e = and f = and hence e =  .

Also, note that every Baer ring R has unity, since (0)^∗(= R ) = eR for an idempotent e , which becomes the unity in R . We prove that the multiplicative semigroup of a Baer ring is a Baer-Stone semigroup.

Theorem 3.8. If ( R ,+,.) is a Baer-ring, then ( R ,.) is a Baer-Stone semigroup.

Proof. Let ( R ,+,.) be a Baer-ring. Then, for any a ∈ R , there exists unique idempotent e in R such that ( a )^∗ = eR . Define a ∗ ₌ e . Then, for any x ∈ R , ax = 0 ⇔ X ∈ a ^∗ R ⇔ a ^∗ x =  .

Therefore, the mapping a → a ^∗ is the pseudo-complementation on R . Thus ( R ,.) is a p.c.semigroup. Next, we observe that, for any a ∈ R , a is an idempotent if and only if a + a ^∗ = 1; if a is an idempotent, then 1 - a is also an idempotent and, for any x ∈ R , xa = 0 ⇔ (1 - a ) x = and hence ( a )^∗ = (1 - a ) R , so that a ^∗ = 1 - a .

To prove the decomposition map d_x is an isomorphism, let У , z ∈ R such that

X ^∗ У =  ^∗ z and X ∗∗ У =   ∗∗ z .

Then

У = ( x ∗ ₊ x ∗∗₎ У =  ^∗ У + x ∗∗ У

=  ^∗ z + X ∗∗ z = ( X ^∗ + X ∗∗₎ z =

Therefore dx ∶   →x∗R×x ∗∗ R is a monomorphism. Further, let (X∗У,x ∗∗ z) ∈X∗R×x ∗∗ R with У,z∈R.

Put s = ( X ^∗ У + x ∗∗ z ). Then since X ^∗ X ^∗∗ = 0, we get

X ^∗ s =   ∗₍ X ^∗ У + x ∗∗ z ) = X ^∗ У ;

X ∗∗ s =   ∗∗₍ X ^∗ У + x ∗∗ z ) = X ∗∗ z .

Therefore d_x is a surjection also. Hence d_x ∶    → x ^∗ R × x ∗∗ R is an isomorphism and thus ( R ,.) is a

Baer-Stone semigroup.

Next, we prove that meet-semi lattice of a Stone lattice is a Baer-Stone semigroup. Let us recall from [1, 2, 5] that, a lattice L= (L,∧,∨) with 0 is called a pseudo-complemented lattice if (L,∧) is a pseudo-complemented semi lattice that is, there exists a mapping x→ X∗of L into itself such that, for any x and У in L,x∧У= 0 ⇔x∗ ∧ У= ; equivalently (x)∗ = {X∗ ∧У∶   ∈L}, where (x)∗ = {У∈L∶   ∧У = 0}, the annihilator of x . In this case x ∗ is the largest element in L such that x∧X∗ = 0.

Definition 3.9.  [1]. A bounded distributive lattice (L,∧,∨) is called a Stone lattice if it is pseudo complemented and

x∗ ∨x∗∗ = 1 for all x ∈L.

Now, we prove that the decomposition map on a Stone lattice is an isomorphism.

Lemma 3.10. Let (L,∧,∨) be a Stone lattice. Then, for any x∈L , the mapping dx ∶   →x∗L×x ∗∗ L defined by dx (У) = (x∗ ∧У,x ∗∗ ∧ У) is an isomorphism.

Proof. By the distributivity in L , it is clear that dx (У∧z) = dx (У) ∧ dx (z)

and dx (У∨z) = dx (У) ∨ dx (z) for all У,z∈L.

Note that X ∗ and X∗∗ are the greatest elements in X∗L and x ∗∗ L respectively. Also, dx(0) = (0,0) and dx (1) = (x,X∗∗). Further, for any У,z∈L, dx (У) = dx (z) ⇒X∗ ∧У= ∗ ∧ z and x∗∗ ∧ У=  ∗∗ ∧ z

⇒ У = ( x ∗ ∨ x ^∗∗) ∧ У = ( X ∗ _∧ У ) ∨ ( x ∗∗ У )

= ( х ∗ _∧ z ) ∨ ( X ^∗∗ ∧ z )

= ( X ^∗ ∨ X ^∗∗) ∧ z = 1 ∧ z =  .

Therefore f is a monomorphism. Next, for any (X∗ ∧У,X ∗∗ ∧ z) ∈X∗L× X ∗∗ L, Put s= (X∗ ∧У) ∨ (X∗∗ ∧ z). Then x∗ ∧s=  ∗ ∧У (since ∗ ∧ x∗∗ ∧ z = 0)

and ^∗∗ ∧ s =   ∗∗ _∧ z (since X ∗∗ _∧ X ∗ _∧ У = 0)

and therefore ^X ( S ) = ( X ∗ ∧ У , X ∗∗ _∧ z ). Therefore ^x is an epimorphism. Thus &x is an isomorphism of L onto x ^∗ L × x ∗∗ L .

Theorem 3.11. If ( L , ∧ , ∨ ) is a Stone lattice, then ( L , ∧ ) is a Baer-Stone semigroup.

Proof. It follows the above lemma.

4.    Conclusion

This is the initial work on Pseudo-complement semigroup and the authors are working on more details which will be adding in subsequent publications.