semilattice composition of ordered semigroups

3 downloads 0 Views 188KB Size Report
[5] M. Petrich, Introduction to Semigroups, Charles E. Merrill Publ. Comp., A Bell &. Howell Comp., Columbus, Ohio, 1973. University of Athens, Department of ...
SEMILATTICE COMPOSITION OF ORDERED SEMIGROUPS NIOVI KEHAYOPULU, MICHAEL TSINGELIS ABSTRACT. We study the semilattice composition of ordered semigroups (a concept opposite to that of the semilattice decomposition), using the ideal extensions.

Semilattice decompositions of ordered semigroups have been often considered by the rst author (cf., for example, [1,2]). The opposite problem of the semilattice decomposition is the problem of semilattice composition, and the purpose of this paper is to study that problem, that is the problem of semilattice composition of ordered semigroups. We approach the problem using the concept of the ideal extension. If (S; ·; ≤) is an ordered semigroup, Y a semilattice and {S } ∈Y a family of pairwise disjoint ordered subsemigroups of S indexed by Y , then we say that S is a semilattice Y of the ordered [ semigroups {S } ∈Y if S = S and, for each ; ∈ Y , we have 1) S · S ⊆ S ∈Y

and 2) S · (S ]≤ 6= ∅ implies ¹ . We rst give the main theorem of semilattice composition of ordered semigroups. Given a semilattice Y and a family {S } ∈Y of disjoint ordered semigroups indexed by Y , we construct an ordered semigroup S which is a semilattice Y of the ordered semigroups {S } ∈Y . Conversely, we prove that each ordered semigroup S which is a semilattice Y of the ordered semigroups {S } ∈Y can be so constructed. Then we prove that each ordered semigroup S which is a semilattice Y of the ordered semigroups {S } ∈Y , is a subdirect product of some {B } ∈Y , where B is the extension of S with a zero possibly adjoined to the extension. For the semilattice composition of semigroups -without order- we refer to [5]. We restrict ourselves to the new results concerning the order of the ordered semigroups. 2000 Mathematics Subject Classi cation. 06F05 Key words and phrases. Ideal, (ideal) extension, semilattice Y of the ordered semigroups {S } ∈Y , subdirect product of ordered semigroups.

1

1. In this paragraph we give the main theorem of semilattice composition of ordered semigroups using the ideal extensions. Let (S; ·; ≤) be an ordered semigroup, ∅ 6= I ⊆ S . I is called a left (resp. right) ideal of S if 1) SI ⊆ I (resp. IS ⊆ I ) and 2) a ∈ I and S 3 b ≤ a implies b ∈ I . I is called an ideal of S if it is both a left and a right ideal of S [1]. Let (S; ·; ≤) be an ordered semigroup, ∅ 6= T ⊆ S . T is called a subsemigroup of S if a; b ∈ T implies a:b ∈ T . Every subsemigroup T of S with the relation \≤T " on T de ned by ≤T := {(a; b) ∈ T × T | a ≤ b} is an ordered semigroup (called an ordered subsemigroup of S ). Clearly, ≤T =≤ ∩(T × T ). Thus, if (S; ·; ≤S ) and (T; ∗; ≤T ) are ordered semigroups, T is an ordered subsemigroup of S if, by de nition, a; b ∈ T implies a:b ∈ T and ≤T =≤S ∩(T × T ). Let (S; ·; ≤S ), (T; ◦; ≤T ) be ordered semigroups, f : S → T a mapping of S into T . f is called isotone if x; y ∈ S , x ≤S y implies f (x) ≤T f (y). f is called reverse isotone if x; y ∈ S , f (x) ≤T f (y) implies x ≤S y. [Each reverse isotone mapping is (1 − 1): Let x; y ∈ S , f (x) = f (y). Since f (x) ≤T f (y), we have x ≤S y. Since f (y) ≤T f (x), we have y ≤S x. Then x = y]. f is called a homomorphism if it is isotone and satis es f (xy) = f (x) ◦ f (y) for all x; y ∈ S . f is called an isomorphism if it is onto, homomorphism and reverse isotone. S and T are called isomorphic, in symbol S ≈ T if there exists an isomorphism between them. S is embedded in T if S is isomorphic to a subset of T i.e. if there exists a mapping f : S → T which is homomorphism and reverse isotone [3]. A semigroup (Y; ·) is called a semilattice if x2 = x and xy = yx ∀ x; y ∈ Y (it is whatever we call commutative idempotent semigroup or commutative band [5]). If (Y; ·) is a semilattice and \¹" a relation on Y de ned by ¹ if and only if = (= );

then, one can easily seen, that (Y; ·; ¹) is an ordered semigroup. In the following, when we refer to a semilattice, we consider it as an ordered semigroup de ned by the order \¹" given above.

Remark 1. Let (Y; ·) be a semilattice and ; ∈ Y . Then ¹ and ¹ .

Indeed: ( ) = ( ) = 2 = and ( ) = 2 = .

If (S; ·; ≤) is an ordered semigroup and ∅ 6= H ⊆ S , we denote (H ]≤ := {t ∈ S | t ≤ h

2

for some h ∈ H }:

De nition 1. Let (S; ·; ≤) be an ordered semigroup, Y a semilattice and {S } ∈Y

a family of ordered subsemigroups of S . S is called a semilattice Y of the ordered semigroups {S } ∈Y if conditions A)-D) below are satis ed: A) S ∩ S = ∅ ∀ ; ∈ Y , 6= . [ B) S = S . ∈ Y

C) S · S ⊆ S ∀ ; ∈ Y . D) S ∩ (S ]≤ 6= ∅ ⇒ ¹ .

Proposition. Let (S; ·; ≤) be an ordered semigroup, Y a semilattice and {S } ∈Y

a family of ordered subsemigroups of S . Then S is a semilattice Y of the ordered semigroups {S } ∈Y if and only if there is a homomorphism  : S → Y such that −1 ({ }) = S for every ∈ Y . PROOF. =⇒. Let S be a semilattice Y of the ordered semigroups {S } ∈Y . We consider the mapping  : S → Y | (x) := ; ∈ Y such that x ∈ S :

Let x ≤ y and ; ∈ Y such that x ∈ S , y ∈ S . Since S 3 x ≤ y ∈ S , we have x ∈ (S ]≤ . Since S ∩ (S ]≤ 6= ∅, we have ¹ . ⇐=. D) Let ; ∈ Y such that S ∩ (S ]≤ 6= ∅. Then ¹ . Indeed: Let x ∈ S and x ∈ (S ]≤ . Then (x) = and x ≤ y for some y ∈ S . Since  is a homomorphism, we have (x) ¹ (y). Since y ∈ S := −1 ({ }), we have (x) = . Thus we have ¹ . The rest is the proof given in semigroups without order.

De nition 2. Let (S; ·; ≤S ) be an ordered semigroup. An ordered semigroup is called an ideal extension (or just an extension) of S if S in an ideal of D. That is, S ⊆ D; S ◦ D ⊆ S; D ◦ S ⊆ S; (D; ◦; ≤D )

a ∈ S; D 3 b ≤D a implies b ∈ S:

For abbreviation, since no confusion is possible, we denote the multiplication in the di erent extensions D ; ∈ Y be the notation \·". Because of condition i), condition iii) in the Theorem below clearly holds for = .

Theorem 1. Let Y be a semilattice and {(S ; ◦ ; ≤ )} ∈Y a family of disjoint ordered semigroups indexed by Y . Let {D } ∈Y be a family of ordered semigroups such that D is an extension of S for every ∈ Y . Suppose that for each ; ∈ Y , ¹ , there is a mapping ; : S → D having the properties: 3

i) ii)

; (x) = x ∀

∈ Y ∀ x ∈ S .

; (S ) · ; (S ) ⊆ S ∀

; ∈ Y .

iii) If ; ; ∈ Y , ¹ , a ∈ S , b ∈ S , then ; ( ; (a) · ; (b)) = ; (a) · ; (b).

We assume that for each ; ∈ Y , ¹ , there is a set r ; ⊆ S × S such that iv) r ; =≤ ∀ ∈ Y . v) If (a; b) ∈ r ; and (b; c) ∈ r ; , then (a; c) ∈ r ; . vi) If (a; b) ∈ r ; and (c; d) ∈ r ; , then ( ; (a) · ; (c); ;  (b) · ;  (d)) ∈ r ;  .

We put S :=

[ ∈Y

S .

We de ned an operation \∗" and a relation \≤∗ " on S as follows: If a ∈ S and b ∈ S , then a ∗ b := ; (a) · ; (b) [ r ; . ≤∗ := ¹

Then (S; ∗; ≤∗ ) is an ordered semigroup and it is a semilattice Y of the ordered semigroups {(S ; ◦ ; ≤ )} ∈Y . Conversely, every ordered semigroup S which is a semilattice Y of ordered semigroups {S } ∈Y can be so constructed. PROOF. 1) (S; ∗; ≤∗ ) is an ordered semigroup. In fact: (S; ∗) is a semigroup (cf. [5]). The relation \≤∗ " is an order relation on S . Indeed: Let a ∈ S . If a ∈ S for some ∈ Y then, by iv), (a; a) ∈≤ = r ; ⊆≤∗ . Let a ≤∗ b and b ≤∗ a. Then, there exist ; ; ;  ∈ Y such that ¹ , ¹ , (a; b) ∈ r ; ⊆ S × S and (b; a) ∈ r ; ⊆ S × S . Since S ∩ S 6= ∅ (a ∈ S ∩ S ), we have =  . Since S ∩ S 6= ∅, we have = . Hence we have ¹ = ¹  = , then = = =  , and r ; = r ; = r ; =≤ . Since (a; b) ∈≤ and (b; a) ∈≤ , we have a = b. Let a ≤∗ b and b ≤∗ c. Then, there exist ; ; ;  ∈ Y such that ¹ , ¹ , (a; b) ∈ r ; ⊆ S × S and (b; c) ∈ r ; ⊆ S × S . Since S ∩ S 6= ∅, we have = , then ¹ = ¹  . Since ¹  , the element r ; is well de ned. Since (a; b) ∈ r ; and (b; c) ∈ r ; , by v), we get (a; c) ∈ r ; ⊆≤∗ . 4

The multiplication \∗" is compatible with the ordering. Indeed: Let a ≤∗ b and c ∈ S . Let ; ∈ Y , ¹ such that (a; b) ∈ r ; ⊆ S × S and let c ∈ S for some ∈ Y . By iv), we have (c; c) ∈≤ = r ; . Since (a; b) ∈ r ; , and (c; c) ∈ r ; , by vi), we have ( ; (a) · ; (c); ; (b) · ; (c)) ∈ r ; :

By the de nition of \∗", we have (a ∗ c; b ∗ c) ∈ r ; ⊆≤∗ (since ¹ ), i.e. a ∗ c ≤∗ b ∗ c. In a similar way we prove that a ≤∗ b and c ∈ S implies c ∗ a ≤∗ c ∗ b. 2) (S ; ◦ ; ≤ ) is an ordered subsemigroup of (S; ∗; ≤∗ ) for every ∈ Y . In fact: Let ∈ Y . If a; b ∈ S , then a ∗ b :=

; (a) · ; (b) = ; (a) · ; (b)

(where \·" is the

multiplication of D ): Since

a and ; (b) := b, we have a ∗ b = a:b ∈ S :S ⊆ S :D ⊆ S . ≤ =≤∗ ∩(S × S ). Indeed: ; (a) :=

≤∗ ∩(S × S ) =

µ[ ¹



r ; ∩ (S × S ) =

[ ¹

(r ; ∩ (S × S ))

(∗)

On the other hand, r ; ∩ (S × S ) ⊆ (S × S ) ∩ (S × S ) ⊆ (S ∩ S ) × (S ∩ S ):

If 6= or 6= , then r ; ∩ (S × S ) = ∅. Thus, by (∗), we have ≤∗ ∩(S × S ) = r ; ∩ (S × S ):

Since r ; =≤ ⊆ S × S , we have ≤∗ ∩(S × S ) = r ; =≤ . 3) (S; ∗; ≤∗ ) is a semilattice Y of the ordered semigroups {(S ; ◦ ; ≤ )} ∈Y . In fact: (S; ∗; ≤∗ ) is an ordered semigroup, Y is a semilattice, and {(S ; ◦ ; ≤ )} ∈Y a family of ordered subsemigroups of S . By hypothesis, A) S ∩ S = ∅ ∀ ; ∈ Y , 6= . [ B) S = S . ∈Y

C) Let ; ∈ Y . Then S ∗ S ⊆ S . Indeed: Let x ∈ S , y ∈ S . By the de nition of \∗" and condition ii), we have x ∗ y :=

; (x) · ; (y ) ∈ ; (S ) · ; (S ) ⊆ S :

5

D) Let ; ∈ Y , S ∩ (S ]≤∗ 6= ∅. Then ¹ . Indeed: Let b ∈ S ∩ (S ]≤∗ . Then b ∈ S and b ≤∗ a for some a ∈ S . Since b ≤∗ a, there exist

;  ∈ Y such that ¹  and (b; a) ∈ r ; ⊆ S × S . Since S ∩ S 6= ∅, we have = . Since S ∩ S 6= ∅, we have =  . Then we have = ¹  = . The converse statement: Let (S; ·; ≤S ) be a semilattice Y of the ordered semigroups {(S ; ◦ ; ≤ )} ∈Y . This means that (S ; ◦ ; ≤ ) is an ordered subsemigroup of S for every ∈ Y and conditions A)-D) mentioned in De nition 1 are satis ed. Then, we [ have S ∩ S = ∅ ∀ ; ∈ Y , 6= and S = S . For each ∈ Y , we put D :=

[ ¹

∈Y

S .

D is a subsemigroup of S ( ∈ Y ). Indeed: ∅ 6= D ⊆ S (since ∅ 6= S ⊆ D , and S ⊆ S ∀ ∈ Y ). Let a; b ∈ D . Then, there exist ; ∈ Y , ¹ , ¹ such that a ∈ S , ∈ S . Then a:b ∈ S :S ⊆ S (by C)). Since ¹ and ¹ , we have = 2 ¹ , so S ⊆ D . Then we have a:b ∈ D . Hence (D ; ·; ≤S ∩(D × D )) is an ordered semigroup. We put ≤D :=≤S ∩(D × D ).

is an extension of S ( ∈ Y ). In fact: First of all, ∅ 6= S ⊆ D (since ¹ ). Let a ∈ S , b ∈ D . Since b ∈ D , there exists ∈ Y such that ¹ and b ∈ S . Then a:b ∈ S : S ⊆ S . Since ¹ , we have = . Then S = S , and a:b ∈ S . Similarly we get D : S ⊆ S . Let b ∈ S and D 3 a ≤D b. Then a ∈ S . Indeed: Since a ∈ D , there exists ∈ Y such that ¹ and a ∈ S . Since a ≤D b, we have a ≤S b. (S; ·; ≤S ) is an ordered semigroup, S 3 a ≤S b ∈ S . By the de nition of (S ]≤S , we have a ∈ (S ]≤S . Then we have a ∈ S ∩ (S ]≤S . By condition D) of De nition 1, we have ¹ . Since ¹ and ¹ , we have = . Then S = S , and a ∈ S .

(D ; ·; ≤D )

For ; ∈ Y , ¹ , we de ne a mapping ; : S → D :=

[ ¹

S | x →

Since ¹ , we have S ⊆ D . Thus the mapping 6

;

; (x) := x:

is well de ned.

Conditions i)-iii) mentioned on the rst part of the Theorem are satis ed: i) If ∈ Y and x ∈ S then ; (x) := x. ii) Let ; ∈ Y . Then ; (S ) · ; (S ) ⊆ S . (The operation \·" is the operation of D which is the same with the operation of S -since D is a subsemigroup of S ). We have ; (S ) = S and ; (S ) = S . Indeed: Let x ∈ S . Since ¹ (cf. Remark 1), we have ; (S ) 3 ; (x) := x. Let x ∈ ; (S ). Then there exists t ∈ S such that x = ; (t). Since ; (t) = t, we have x = t ∈ S . Similarly, ; (S ) = S . Hence we have ; (S ) · ; (S ) = S · S ⊆ S : iii) Let ; ; ∈ Y , ¹ , a ∈ S , b ∈ S . Then ; ( ; (a): ; (b)) = ; (a): ; (b):

Indeed: We have

; (a) := a

a:b =

and

; (b) := b

(since ¹ , ¹ ). Then, by ii),

; (a): ; (b) ∈ ; (S ): ; (S ) ⊆ S :

Since ¹ , we have ; ( ; (a): ; (b)) = ; (a:b) := ab:

Since ¹ and ¹ , we have ; (a) := a and ; (b) := b. Hence, for each ; ∈ Y , ¹ , there is a mapping ; : S → D having the properties i)-iii) mentioned on the rst part of the Theorem. For ; ∈ Y , ¹ , we de ne r ; :=≤S ∩(S × S ). Then, conditions iv)-vi) mentioned on the rst part of the Theorem are satis ed. In fact: iv) If ∈ Y , then ≤ :=≤S ∩(S × S ) := r ; (since ¹ 2 ). v) Let (a; b) ∈ r ; and (b; c) ∈ r ; . Then (a; c) ∈ r ; . Indeed: (a; b) ∈ r ; :=≤S ∩(S × S ) ⇒ a ≤S b; a ∈ S ; b ∈ S . (b; c) ∈ r ; :=≤S ∩(S × S ) ⇒ b ≤S c; b ∈ S ; c ∈ S . Since a ≤S b and b ≤S c, we have a ≤S c. Then we have (a; c) ∈≤S ∩(S × S ) := r ; :

7

vi) Let (a; b) ∈ r ; and (c; d) ∈ r ; . Then ( ; (a): ; (c); ;  (b): ;  (d)) ∈ r ;  :

Indeed: Since ¹ and ¹  , we have ¹  , so the set r ;  is well de ned. (a; b) ∈ r ; ⇒ a ≤S b; a ∈ S ; b ∈ S . (c; d) ∈ r ; ⇒ c ≤S d; c ∈ S ; d ∈ S . Since ¹ and ¹ ,  ¹ and  ¹ , by C), we have ; (a): ; (c) = a:c ∈ S :S ⊆ S . ;  (b): ;  (d) = b:d ∈ S :S ⊆ S  . Since a ≤S b and c ≤S d, we have ac ≤S bd. Then we have ( ; (a): ; (c); ;  (b): ;  (d)) = (ac; bd) ∈≤S ∩(S × S  ) := r ;  : [ According to the rst part of the Theorem, the set S = S with the operation \∗" ∈Y

and the order \≤∗ " on S de ned by: a ∗ b := ; (a): ; (b) if a ∈ S and b ∈ S [ r ; ≤∗ := a¹

is a semilattice Y of the ordered semigroups {(S ; ◦ ; ≤ )} ∈Y . Then (S ; ◦ ; ≤ ) is a subsemigroup of (S; ∗; ≤∗ ) for every ∈ Y . We have ∗ = · and ≤∗ =≤S . In fact: Let a; b ∈ S , a ∈ S and b ∈ S for some ; ∈ Y . Then we have a ∗ b :=

Since ≤∗ :=

[ ¹

; (a) · ; (b) = a · b:

r ; and, moreover, for each ; ∈ Y , ¹ , r ; :=≤S ∩(S ×

S ) ⊆≤S , we have ≤∗ ⊆≤S . Let now a; b ∈ S , a ≤S b. Then a ≤∗ b. Indeed: Let a ∈ S , b ∈ S for some ; ∈ Y . Since b ∈ S and S 3 a ≤S b, we have a ∈ (S ]≤S . Since S ∩ (S ]≤S 6= ∅, we have ¹ . Then the set r ; is well de ned, and (a; b) ∈≤S ∩(S × S ) := r ; ⊆

[

¹

r ; :=≤∗ :

Remark 2. If S is a semilattice of the ordered semigroups {S } ∈Y then, for each

∈ Y , we have D = { ; (b) | b ∈ S ; ≤ }. In fact: Let ∈ Y . If ¹ then, from the de nition of ; , we have ; (S ) ⊆ D . Thus ; (b) ∈ D ∀

b ∈ S ∀ ¹ :

8

Let b ∈ D := other hand,

[

S . Then, there exists ∈ Y such that ¹ and b ∈ S . On the

¹ ; (b) := b.

2. In this paragraph we prove that the semilattices Y of the ordered semigroups {S } ∈Y , are subdirect products of extensions with a zero possibly adjoined to the extensions. Let (S; ·; ≤) be an ordered semigroup without zero and let 0 be an element which does not belong to S (0 ∈= S ). We consider the set S 0 := S ∪ {0}. We de ne a multiplication \∗" and an order \≤0 " on S 0 as follows: ( ab if a; b ∈ S a ∗ b := 0 otherwise ≤0 :=≤ ∪{(0; x) | x ∈ S 0 }.

Then (S 0 ; ∗; ≤0 ) is an ordered semigroup and 0 is the zero element of S 0 . (The proof is easy). This is the problem of the adjunction of a zero to S . The set S is a subsemigroup of S 0 . Clearly, if a; b ∈ S , then a ∗ b := ab ∈ S and ≤=≤0 ∩(S × S ), so S is an ordered subsemigroup of S 0 (cf. also [4]). Y Let {(B ; ◦ ; ≤ )} ∈Y be a family of ordered semigroups. The set B := {(x ) ∈Y | x ∈ B } with the operation \·" and the order \¹" on

Y ∈Y

∈Y

B de ned by

(x ) ∈Y · (y ) ∈Y := (x ◦ y ) ∈Y (x ) ∈Y ¹ (y ) ∈Y ⇔ x ≤ y ∀

∈Y

is an ordered semigroup. For each ∈ Y , the projection  :

Y ∈Y

B → B | (x ) ∈Y → x

is a homomorphism and onto (Fix a x ∈ B . For each ∈ Y , take an arbitrary x ∈ B (B 6= ∅). Then  (x ) ∈Y = x ).

De nition 3. Let S , {B } ∈Y be ordered semigroups. S isYcalled a subdirect product

of {B } ∈Y if there exists an (ordered) subsemigroup T of

∈Y

B which is isomorphic

to S and  (T ) = B ∀ ∈ Y . Y Equivalent De nition: There exists a reverse isotone homomorphism F : S → B such that

∈Y

 (F (S )) = B ∀ ∈ Y (cf. also [3]).

9

In the following, we use the following notation: Let S be a semilattice Y of the ordered semigroups {S } ∈Y and, for each ∈ Y , let D be the (ideal) extension of S constructed in the second part of the Theorem above. Then: If is the least element of Y (i.e. if ¹ ∀ ∈ Y ), then we put B = D . If is not the least element of Y (i.e. if ∃ ∈ Y such that 6¹ ), then we put B = D 0 where D 0 is the ordered semigroup arising from D by the adjunction of a zero element.

Theorem 2. Let (S; ·; ≤S ) be a semilattice Y of the ordered semigroups {S } ∈Y .

Then S is a subdirect product of {B } ∈Y where, for each ∈ Y , we have B = D or B = D 0 . PROOF. Let ∈ Y . If is the least element of Y , then we put B := D . Then { ∈ Y | ¹ } = Y;

D :=

[ ¹

S =

[ ∈Y

S = S:

In that case, we consider the identity mapping 1S : (S; ·; ≤S ) → B := D = S | x → x

which is (clearly) homomorphism and onto, and de ne the multiplication \∗ " and the order \≤∗ " on B as ∗ = · and ≤∗ =≤S (the same as the multiplication and the order in S ). If is not the least element of Y , then we put B := D 0 , where 0 is an arbitrary element not included in D and D 0 := D ∪ {0 } is the ordered semigroup -with the multiplication \∗ " and the order \≤∗ " below, arising from (D ; ·|D ×D ; ≤D ) (where ≤D :=≤S ∩(D × D )) by the adjunction of the zero {0 }. ( ab if a; b ∈ D For a; b ∈ D 0 we have a ∗ b := 0 otherwise. ≤∗ :=≤D ∪{(0 ; x) | x ∈ D 0 }:

We consider the mapping:

(

 : (S; ·; ≤S ) → B := D 0 | x →  (x) :=

x if x ∈ D 0 if x ∈ = D :

The mapping  is a homomorphism. In fact: Let b; c ∈ S . Then  (bc) =  (b) ∗  (c). Indeed: Let b; c ∈ D . Then  (b) := b,  (c) := c, bc ∈ D (since D is a subsemigroup of S ),  (bc) := bc, b ∗ c := bc, and  (bc) = bc = b ∗ c =  (b) ∗  (c):

10

Let b ∈ D , c ∈= D . Then  (b) := b,  (c) := 0 ,  (b) ∗  (c) = b ∗ 0 := 0 . Since [ [ b ∈ D := S , there exists ∈ Y , ¹ such that b ∈ S . Since c ∈ S = S and c ∈= D :=

[ ¹ ¹

∈Y

S , there exists ∈ Y , 6¹ such that c ∈ S . Then bc ∈ S · S ⊆ S .

If ¹ then, since ¹ (cf. Remark 1), we have ¹ . Impossible. Thus 6¹ . Then S ∩ D = ∅. Indeed: Let x ∈ S ∩ D . Then x ∈ S and there exists  ∈ Y , ¹  such that x ∈ S . Since S ∩ S 6= ∅; ;  ∈ Y , we have =  . Then ¹ . Impossible. Since bc ∈ S and S ∩ D = ∅, we have bc ∈= D . Then  (bc) := 0 . Hence  (bc) =  (b) ∗  (c). The case b ∈= D , c ∈ D is proved similarly. Let b ∈= D , c ∈= D . Then  (b) := 0 ,  (c) := 0 ,  (b) ∗  (c) = 0 . Since [ [ S , there exists ∈ Y , 6¹ such that b ∈ S . Since d ∈= D := S and b ∈ S =

∈Y

¹

c ∈= D and c ∈ S , there exists ∈ Y , 6¹ such that c ∈ S . Then bc ∈ S · S ⊆ S . As in the previous case, we prove that 6¹ , then S ∩ D = ∅. Then bc ∈= D and  (bc) := 0 .

Let b ≤S c. Then  (b) ≤∗  (c). Indeed: If b ∈= D , then  (b) := 0 . Since c ∈ S , we have  (c) ∈ D 0 . Since 0 is the zero of D 0 , we have  (b) ≤∗  (c). [ Let b ∈ D := S . Then  (b) := b and there exists ∈ Y , ¹ such that b ∈ S . Since c ∈ S =

[ ¹

∈Y

S , there exists ∈ Y such that c ∈ S . Since c ∈ S and S 3 b ≤S c,

we have b ∈ (S ]≤S . Since S ∩ (S ]≤S 6= ∅, we have ¹ . Then ¹ , S ⊆ D , c ∈ D , and  (c) := c. Then ( (b);  (c)) = (b; c) ∈≤S ∩(D × D ) :=≤D ⊆≤∗

:

The mapping  is onto. In fact: Let b ∈ D 0 := D ∪ {0 }. If b ∈ D , then b ∈ S (since D ⊆ S ), and  (b) = b. Let b = 0 . Since is not the least element of Y , there exists ∈ Y such that 6¹ . Then S ∩ D = ∅. Indeed: If x ∈ S ∩ D , then x ∈ S and there exists ∈ Y , ¹ such that x ∈ S . Since S ∩ S 6= ∅, we have = . We get 6¹ , ¹ , = . This is impossible. For each ∈ Y , we put I) f := 1S if is the least element of Y . 11

II) f :=  if is not the least element of Y . Then f is a homomorphism and onto for every ∈ Y . We consider now the mapping: F : (S; ·; ≤S ) →

Let us denote by \¹" the order on

Y ∈Y

Y

∈Y

B | x → (f (x)) ∈Y :

B .

The ordered semigroup (S; ·; ≤S ) is a subdirect product of the ordered semigroups {B } ∈Y . In fact: a) The mapping F is well de ned and it is a homomorphism (this is clear). b) F is reverse isotone. In fact: Let a; b ∈ S , F (a) ¹ F (b). Then a ≤S b. Indeed: We have (f (a)) ∈Y ¹ (f (a)) ∈Y , and f (a) ≤∗ f (b) ∀ ∈ Y

(∗)

If is the least element of Y , then f (a) := a, f (b) := b, ≤∗ =≤S , and a ≤S b. Suppose that there is no least element in Y . Then f =  ∀ ∈ Y (∗∗) [ Since a ∈ S = S , there exists ∈ Y such that a ∈ S . By (∗), we have ∈Y

f (a) ≤∗ f (b). Then, by (∗∗),  (a) ≤∗  (b). [ S := D , we have a ∈ D , then  (a) := a. Hence we have Since a ∈ S ⊆ ¹

a ≤∗  (b) and a ∈ D . Since (a;  (b)) ∈≤∗ :=≤D ∪{(0 ; x) | x ∈ D ∪ {0 }} and a 6= 0 (since a ∈ D ), we have (a;  (b)) ∈≤D ⊆ D × D . Since  (b) ∈ D , we have  (b) 6= 0 , b ∈ D ,  (b) = b. Then (a; b) ∈≤D :=≤S ∩(D × D ) ⊆≤S i.e. a ≤S b.

c) We have  (F (S )) = f (S ) = B for every ∈ Y . In fact: If x ∈ S , then F (x) := (f (x)) ∈Y , then  (F (x)) =  (f (x)) ∈Y := f (x) ∈ f (S ). If x ∈ S , then f (x) :=  (f (x)) ∈Y ∈  (F (S )), thus f (S ) ⊆  (F (S )). Since f is an onto mapping ( ∈ Y ), we have f (S ) = B for every ∈ Y .

References [1] N. Kehayopulu, On left regular and left duo poe-semigroups, Semigroup Forum 44, No. 3 (1992), 306{313. [2] N. Kehayopulu, On intra-regular ordered semigroups, Semigroup Forum 46, No. 3 (1993), 271{278. 12

[3] N. Kehayopulu and M. Tsingelis, On subdirectly irreducible ordered semigroups, Semigroup Forum 50, No. 2 (1995), 161{177. [4] N. Kehayopulu and M. Tsingelis, A note on ordered groupoids-semigroups, Scientiae Mathematicae 3, No. 2 (2000), 251{255. [5] M. Petrich, Introduction to Semigroups, Charles E. Merrill Publ. Comp., A Bell & Howell Comp., Columbus, Ohio, 1973. University of Athens, Department of Mathematics, Panepistimiopolis, Athens 157 84, Greece e-mail address: [email protected]

13