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Oct 30, 2010 - Abstract A semiring S whose additive reduct is a semilattice is called a k-regular semiring if for every a ∈ S there is x ∈ S such that a + axa ...
Semigroup Forum (2011) 82: 131–140 DOI 10.1007/s00233-010-9271-9 R E S E A R C H A RT I C L E

On semirings whose additive reduct is a semilattice M.K. Sen · A.K. Bhuniya

Received: 20 April 2007 / Accepted: 22 September 2010 / Published online: 30 October 2010 © Springer Science+Business Media, LLC 2010

Abstract A semiring S whose additive reduct is a semilattice is called a k-regular semiring if for every a ∈ S there is x ∈ S such that a + axa = axa. For a semigroup F , the power semiring P (F ) is a k-regular semiring if and only if F is a regular semigroup. An element e ∈ S is a k-idempotent if e + e2 = e2 . Basic properties of k-regular semirings whose k-idempotents are commutative have been studied. Keywords k-regular semiring · k-idempotent · k-semifield · k-Clifford semiring · Least distributive lattice congruence

1 Introduction There are many concepts of universal algebras generalizing an associative ring (R, +, ·). Semiring is one of them which is a generalization not only of associative rings but also of distributive lattices and has been proven very useful. The intriguing feature about the study of semirings is the investigation of how the distributive laws force the structural characteristics of the two reducts to interact. The present paper is a continuation of our investigation to this line on the semirings whose additive reduct is a semilattice. In our previous contributions we characterized k-regular semirings, completely k-regular semirings and k-Clifford semirings. Introducing the

Communicated by Mikhail Volkov. M.K. Sen () Department of Pure Mathematics, University of Calcutta, Kolkata, India e-mail: [email protected] A.K. Bhuniya Department of Mathematics, Visva-Bharati, Santiniketan 731235, India e-mail: [email protected]

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notion of k-semifields we proved that a semiring whose additive reduct is a semilattice is completely k-regular (k-Clifford) if and only if it is a union (distributive lattice) of k-semifields [10–12]. There are plenty of examples of semirings whose additive reduct is a semilattice. In the category of natural examples we mention the semirings of endomorphisms of semilattices, power semirings of semigroups, semirings of matrices over distributive lattices etc. The syntactic semiring of a given formal language introduced by Polak [9] is another example of such semirings. He proved a modification of the Eilenberg theorem relating certain classes of rational languages and pseudovarieties of semirings whose additive reduct is a semilattice. The preliminaries and prerequisites we need for this article are discussed in Sect. 2. In Sect. 3, we discuss a natural link between regular semigroups and k-regular semirings. For a semigroup F , the semiring P (F ) of all subsets of F is a k-regular semiring if and only if F is a regular semigroup. Moreover the k-idempotents of P (F ) commute if and only if P (F ) is a commutative semiring. This is not true for a semiring S, in general. We give an example of such a semiring in Sect. 5. Being motivated from how the idea of inverses of an element is related with the regularity in semigroups, we define k-inverses of an element analogously from kregularity. In Sect. 4, we investigate the sum of k-inverses and the k-inverses of kidempotents. We characterize the k-regular semirings S such that the k-idempotents of S commute in Sect. 5 and also give a characterization of the least distributive lattice congruence on such semirings. 2 Preliminaries A semiring (S, +, ·) is an algebra with two binary operations + and · such that both the additive reduct (S, +) and the multiplicative reduct (S, ·) are semigroups and the following distributive laws hold: x(y + z) = xy + xz

and (x + y)z = xz + yz.

The set of all multiplicative (additive) idempotents of a semiring (S, +, ·) is denoted by E · (S) (respectively E + (S)). Thus E · (S) = {e ∈ S | e2 = e} and E + (S) = {a ∈ S | a + a = a}. By SL+ we denote the variety of all semirings (S, +, ·) such that (S, +) is a semilattice, i.e. a commutative and idempotent semigroup. Let (S, +, ·) be a semiring in SL+ and let A be a nonempty subset of S. The k-closure of A in S is defined by A = {x ∈ S | x + a1 = a2 for some a1 , a2 ∈ A}. Then we have A ⊆ A, and if (A, +) is a subsemigroup of (S, +), then A = {x ∈ S | x + a = a for some a ∈ A} and A = A, since (S, +) is a semilattice. A is called a k-set if A ⊆ A. An ideal K of S is called a k-ideal if it is a k-set, i.e. K = K [5]. Similarly we define left k-ideals and right k-ideals of S.

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We refer to [6] for the information we need concerning semigroup theory and to [4] for notions concerning semiring theory. Lemma 2.1 Let (S, +, ·) be a semiring in SL+ and a, b, c, d ∈ S. 1. If there exist x1 , x2 , y1 , y2 ∈ S such that a + x1 by1 = x2 by2 , then a + (x1 + x2 + y1 + y2 )b(x1 + x2 + y1 + y2 ) = (x1 + x2 + y1 + y2 )b(x1 + x2 + y1 + y2 ). 2. If there exist x, y ∈ S such that a + xby = xcy, then a + (x + y)(b + c)(x + y) = (x + y)(b + c)(x + y). 3. If a + xcx = xcx and b + ydy = ydy for some x, y ∈ S, then a + (x + y) × (c + d)(x + y) = (x + y)(c + d)(x + y) and b + (x + y)(c + d)(x + y) = (x + y) × (c + d)(x + y). 4. If a + b = b and c + d = d, then ac + bd = bd. Proof (1) For u = x1 + x2 + y1 + y2 we have x1 by1 + ubu = x2 by2 + ubu = ubu, since (S, +) is a semilattice. Thus a + x1 by1 + ubu = x2 by2 + ubu implies a + ubu = ubu. (2) This follows similarly, since xby + (x + y)(b + c)(x + y) = xcy + (x + y) × (b + c)(x + y) = (x + y)(b + c)(x + y). (3) This follows similarly, since xcy + (x + y)(c + d)(x + y) = xdy + (x + y) × (c + d)(x + y) = (x + y)(c + d)(x + y). (4) In fact, ac + bc = bc implies that ac + b(c + d) = b(c + d) and so ac + bd = bd.  A semigroup (S, ·) is regular if for all a ∈ S there exists x ∈ S such that a = axa. A semiring (S, +, ·) is called regular (in the sense of von Neumann) if the multiplicative reduct (S, ·) is a regular semigroup. Bourne [2] defined an additively commutative semiring S to be regular if for all a ∈ S there exist x, y ∈ S such that a + axa = aya. Clearly, in a ring this concept coincides with the von Neumann regularity. In [1], the Bourne regularity was renamed as k-regularity to distinguish it from the von Neumann regularity. If S is a semiring in SL+ , then S is k-regular if and only if for every a ∈ S there exists x ∈ S such that a + axa = axa, by Lemma 2.1(2). Every additively commutative regular semiring is k-regular. Special cases are idempotent semirings with commutative addition, which include distributive lattices, but also semifields with idempotent addition [7]. Thus every von Neumann regular semiring in SL+ is k-regular. The converse is not true. For examples of such semirings we refer to [10]. Throughout the rest of this article, unless otherwise stated, (S, +, ·) is always a semiring in SL+ . Let (L, +) be a semilattice and S = End(L) be the set of all endomorphisms of L. Define addition on S by: for α, β ∈ S, (α + β)(x) = α(x) + β(x) for all x ∈ L. Then (S, +, ◦) is a semiring in SL+ , where ◦ is the usual composition of mappings. If L has a greatest element then (S, +, ·) is a k-regular semiring [10]. In the following

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theorem we give an embedding theorem for the semirings in SL+ into the k-regular semiring of endomorphisms of a semilattice. Theorem 2.2 Every semiring whose additive reduct is a semilattice can be embedded into a k-regular semiring of endomorphisms of a semilattice. Proof Let (S, +, ·) be a semiring in SL+ . Adjoin two new elements 0, ∞ to S and denote S  = S ∪ {0, ∞}. Extend the binary operations to S  as follows: (i) 0 + 0 = 0, a + 0 = a and 0 · 0 = a · 0 = 0 · a = 0 for all a ∈ S (ii) ∞ + ∞ = a + ∞ = ∞ and ∞ · ∞ = a · ∞ = ∞ · a = ∞ for all a ∈ S (iii) 0 + ∞ = ∞ + 0 = ∞ and 0 · ∞ = ∞ · 0 = 0. Then (S  , +, ·) is also a semiring in SL+ . Let {0, 1} be the semilattice of two elements. Denote L = S  × {0, 1}. Then (L, +) is a semilattice with the greatest element (∞, 1). Hence (End(L), +, ◦) is a k-regular semiring. For any a ∈ S, define Fa : L −→ L by  Fa (x, i) =

(ax, 0) if i = 0, (ax + a, 0) if i = 1.

Similarly as in Theorem 2.4 [8], it can be shown that Fa ∈ End(L) and F : S −→ End(L) defined by F (a) = Fa is an injective homomorphism.  In [10] we defined two binary relations L and R on a k-regular semiring (S, +, ·) analogous to the Green’s relations [3] on a regular semigroup as follows: for a, b ∈ S aLb

if Sa = Sb

and a R b

if aS = bS.

Let (S, +, ·) be a k-regular semiring and a, b ∈ S. Then a L b if and only if there is x ∈ S such that a + xb = xb and b + xa = xa and a R b if and only if there is x ∈ S such that a + bx = bx and b + ax = ax [10, Proposition 3.3]. In [12], we defined a new type of elements called k-idempotents to characterize the k-regular semirings which are distributive lattices of k-semifields. Definition 2.3 [12] An element e of a semiring S in SL+ is called k-idempotent if e + e2 = e2 . The set of all k-idempotents in a semiring S is denoted by Ek (S). If a, x ∈ S are such that a + axa = axa, then ax, xa ∈ Ek (S). If (S, +, ·) is a semiring, then (Ek (S), +) is a subsemilattice of (S, +). For e, f ∈ Ek (S) we have e + e2 = e2 and f + f 2 = f 2 . Then e + f + e2 + f 2 = e2 + f 2 which implies that e + f + (e + f )2 = (e + f )2 and so e + f ∈ Ek (S). The product of two k-idempotents may not be a k-idempotent.

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3 Construction of some k-regular semirings Let (F, ·) be a semigroup and S = P (F ) be the set of all subsets of F . Define A+B =A∪B

and AB = {ab | a ∈ A, b ∈ B} for all A, B ∈ S.

Then (S, +, ·) is a semiring in SL+ . In the following theorem we characterize the semigroups (F, ·) such that (P (F ), +, ·) is k-regular. Theorem 3.1 Let (F, ·) be a semigroup. Then the semiring (P (F ), +, ·) is k-regular if and only if (F, ·) is a regular semigroup. Proof Let (P (F ), +, ·) be a k-regular semiring and a ∈ F . Then A = {a} ∈ P (F ) and so there is X ∈ P (F ) such that A + AXA = AXA i.e. A ⊆ AXA. This implies that there is x ∈ X such that a = axa. Thus (F, ·) is a regular semigroup. Conversely let (F, ·) be a regular semigroup and A ∈ P (F ). Then for each a ∈ A, there is x ∈ F such that a = axa. We choose one such x and denote it by xa . Then X = {xa | a ∈ A} ∈ P (F ) is such that A ⊆ AXA which implies that A + AXA = AXA. Thus (P (F ), +, ·) is a k-regular semiring.  Let (F, ·) be a regular semigroup. Then for every A ∈ P (F ) the left (right) ideal of F generated by A is FA (respectively AF ). Theorem 3.2 Let (F, ·) be a regular semigroup. Then for A, B ∈ P (F ), 1. A L B if and only if FA = FB. 2. A R B if and only if AF = BF . Proof 1. If A, B ∈ P (F ) are such that A L B, then there exists X ∈ P (F ) such that A+XB = XB and B +XA = XA, i.e. A ⊆ XB and B ⊆ XA. Then FA ⊆ (F X)B ⊆ FB and similarly FB ⊆ FA implies that FA = FB. Let A, B ∈ P (F ) be such that FA = FB. Since (F, ·) is a regular semigroup, A ⊆ FA = FB. Then for every a ∈ A, there exist xa ∈ F and b ∈ B such that a = xa b. Let X be the set of all such xa . Then X ∈ P (F ) and A ⊆ XB, i.e. A + XB = XB. Similarly there exists Y ∈ P (F ) such that B + Y A = Y A. Thus A L B. The second statement follows dually.  Lemma 3.3 Let (S, +, ·) be a k-regular semiring such that ef = f e for all e, f ∈ Ek (S). Then for every a, b ∈ S there is x ∈ S such that ab + bxa = bxa. Proof Since (S, +, ·) is k-regular, there exist x, y, z ∈ S such that a + axa = axa, b + byb = byb and ab + abzab = abzab. Now zab + zabzab = zabzab implies that abzab + abzabzab = abzabzab. Hence ab + abzabzab = abzabzab. Then we have a + ata = ata, b + btb = btb and ab + tabt = tabt, by Lemma 2.1, where t = x + y + abz + zab. Then ta, bt ∈ Ek (S) implies that tabt = bt 2 a and so  ab + bt 2 a = bt 2 a. In the following theorem we characterize the regular semigroups (F, ·) such that the k-idempotents of P (F ) commute.

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Theorem 3.4 Let (F, ·) be a regular semigroup. Then the following statements are equivalent. 1. The k-idempotents of P (F ) commute. 2. (F, ·) is a commutative semigroup. 3. (P (F ), +, ·) is a commutative semiring. Proof Since (2) ⇒ (3) and (3) ⇒ (1) are immediate, we are to prove (1) ⇒ (2) only. Let a ∈ F and e be an idempotent in F . Then {e}, {a} ∈ P (F ). Then there are X, Y ∈ P (F ) such that {e}{a} + {a}X{e} = {a}X{e} and {a}{e} + {e}Y {a} = {e}Y {a} i.e. {e}{a} ⊆ {a}X{e} and {a}{e} ⊆ {e}Y {a}, by Lemma 3.3. Hence there are x, y ∈ X such that ae = exa and ea = aye. Then eae = e2 xa = exa = ae and eae = aye2 = aye = ea shows that ae = ea. Hence (F, ·) is a Clifford semigroup and so F = [L; Gα , φα,β ] is a strong semilattice L of groups Gα with the structure homomorphisms φα,β such that ab = (φα,αβ a)(φβ,αβ b), if a ∈ Gα , b ∈ Gβ . Let eα be the identity in Gα and a, b ∈ Gα . Then both E1 = {eα , a} and E2 = {eα , b} are k-idempotents in P (F ) and so E1 E2 = E2 E1 which implies that {eα , a, b, ab} = {eα , a, b, ba}. If a = eα or b = eα then ab = ba. Otherwise ab = a, b and hence either ab = eα or ab = ba. If ab = eα then b = a −1 , the inverse of a in the group Gα and so ab = ba. Thus Gα is Abelian for all α ∈ L. Then for a ∈ Gα , b ∈ Gβ , we get ab = (φα,αβ a)(φβ,αβ b) = (φβ,αβ b)(φα,αβ a) = ba. Hence (F, ·) is a commutative semigroup. 

4 k-inverses of k-regular elements Let (S, +, ·) be a semiring and a ∈ S. Then b ∈ S is called a k-inverse of a if a + aba = aba and b + bab = bab. The set of all k-inverses of a ∈ S is denoted by Vk (a). Every k-regular element has a k-inverse. Let a, x ∈ S such that a + axa = axa. This implies that a + ax(a + axa) = ax(a + axa) and so a + a(xax)a = a(xax)a. Again this implies that xax + (xax)a(xax) = (xax)a(xax). Thus xax ∈ Vk (a). Lemma 4.1 Let (S, +, ·) be a k-regular semiring and a, b ∈ S. Then for all a ∈ Vk (a) and b ∈ Vk (b) one has a + b ∈ Vk (a + b), in particular (Vk (a), +) is a subsemilattice of (S, +). Proof From a + aa a = aa a and b + bb b = bb b we get a + b + aa a + bb b = aa a + bb b. Adding (a + b)(a + b )(a + b) on both sides yields a + b + (a + b)(a + b )(a + b) = (a + b)(a + b )(a + b). Similarly we have a + b + (a + b )(a + b)(a + b ) = (a + b )(a + b)(a + b ). Thus a + b ∈ Vk (a + b).  In the following theorem we find a necessary and sufficient condition so that every k-inverse of a k-idempotent is a k-idempotent. Theorem 4.2 Let (S, +, ·) be a k-regular semiring such that Ek (S) is a k-set. Then the following statements are equivalent:

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1. Vk (e) ⊆ Ek (S) for all e ∈ Ek (S). 2. ef ∈ Ek (S) for all e, f ∈ Ek (S). Proof (1) ⇒ (2): Let e, f ∈ Ek (S). Since (S, +, ·) is k-regular, Vk (ef ) = ∅. Let x ∈ Vk (ef ). Then ef + ef xef = ef xef and x + xef x = xef x. Let g = f xe. Then x + xef x = xef x implies that f xe + f xef xe = f xef xe and so g = f xe ∈ Ek (S). Now x + xef x = xef x implies that x + x(e + e2 )(f + f 2 )x = x(e + e2 )(f + f 2 )x and so f xe + f xe(ef )f xe = f xe(ef )f xe. Again ef + ef xef = ef xef implies that ef + e(f + f 2 )x(e + e2 )f = e(f + f 2 )x(e + e2 )f, i.e. ef + ef (f xe)ef = ef (f xe)ef . Thus ef ∈ Vk (g). Since g ∈ Ek (S), (1) implies that ef ∈ Ek (S). (2) ⇒ (1): Let e ∈ Ek (S) and x ∈ Vk (e). Then xe, ex ∈ Ek (S) and so by (2) we have xe2 x ∈ Ek (S). Now x + xex = xex implies that x + x(e + e2 )x = x(e + e2 )x  i.e. x + xe2 x = xe2 x. Since Ek (S) is a k-set, this implies that x ∈ Ek (S).

5 k-regular semirings with commutative subsemiring Ek (S) Let (S, +, ·) be a semiring in SL+ . Define the relation σ on S by: for a, b ∈ S aσ b

if a + r1 bs1 = a m

and b + r2 as2 = bn

for some r1 , r2 , s1 , s2 ∈ S 1 and m, n ∈ N. Then σ is reflexive and symmetric since (S, +) is a semilattice. So the transitive closure ρ = σ ∗ is an equivalence relation and γ = {(x + ras, x + rbs) ∈ S × S | aρb, x ∈ S 0 , r, s ∈ S 1 } is a congruence on S. Let ef ∈ Ek (S) for all e, f ∈ Ek (S). Then ef + (ef )2 = (ef )2 implies that ef + e(f e)f = (ef )2 . Similarly f e + f (ef )e = (f e)2 , which shows that ef σf e. Thus γ is a congruence on S such that ef γf e for all e, f ∈ Ek (S) when ef ∈ Ek (S) for all e, f ∈ Ek (S). Though usually the product of two k-idempotents need not be a k-idempotent, this becomes the case in the semirings (S, +, ·) such that ef = f e for all k-idempotents e, f . In fact, Lemma 2.1(4) implies that ef + e2 f 2 = e2 f 2 and so ef + (ef )2 = (ef )2 , since ef = f e. The semirings for which ef = f e for all e, f ∈ Ek (S) is one of our main interests in this paper. It is natural to ask whether such semirings must be commutative. In the following we give an example of a semiring S which is noncommutative but ef = f e for all e, f ∈ Ek (S). Example 5.1 Take N = N \ {1}. Define two binary operations on R = N × N by: for (a, b), (c, d) ∈ R, (a, b) + (c, d) = (min{a, c}, min{b, d}) and (a, b) ◦ (c, d) = (ac, b). Then (R, +, ◦) is a non-commutative semiring in SL+ without any k-idempotent. We adjoin an element 0 to R and extend the binary operations to R 0 by: x + 0 =

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0 = 0 + x and x ◦ 0 = 0 ◦ x = 0 for all x ∈ R. Then R 0 is a non-commutative kregular semiring with a unique k-idempotent 0. Further, (N, max, ·) is a commutative k-regular semiring such that every element is a k-idempotent. Hence S = R 0 × N is a non-commutative k-regular semiring such that Ek (S) = {(0, n) ∈ S | n ∈ N} is a commutative subsemiring of S. Theorem 5.2 Let (S, +, ·) be a semiring such that ef = f e for all e, f ∈ Ek (S). Then aea ∈ Ek (S) for all a ∈ S, a ∈ Vk (a) and e ∈ Ek (S). Proof Since a ∈ Vk (a), we have a +aa a = aa a which implies that aea +aa a(e + e2 )a = aa a(e + e2 )a , i.e. aea + aa ae2 a = aa ae2 a . Then a ae = ea a implies that aea + (aea )(aea ) = (aea )(aea ), i.e. aea ∈ Ek (S).  Theorem 5.3 Let (S, +, ·) be a semiring such that ef = f e for all e, f ∈ Ek (S) and a, b ∈ S. Then yx ∈ Vk (ab) for all x ∈ Vk (a), y ∈ Vk (b). Proof Let x ∈ Vk (a) and y ∈ Vk (b). Then xa, by ∈ Ek (S) and so xaby = byxa. Now, by Lemma 2.1(4), we have yx + (yby)(xax) = (yby)(xax) which implies that yx + yxabyx = yxabyx. Similarly we have ab + abyxab = abyxab. Thus yx ∈ Vk (ab).  A semiring (S, +, ·) in SL+ is called a k-semifield if for all a ∈ S and all nonzero b ∈ S there is u ∈ S such that a + ub = ub and a + bu = bu. A semiring (S, +, ·) in SL+ is a k-Clifford semiring if S is a distributive lattice of k-semifields. Also a semiring (S, +, ·) is k-Clifford if and only if for all a, b ∈ S there is x ∈ S such that ab + bxa = bxa. For details we refer to [12]. Thus if (S, +, ·) is a k-regular semiring such that ef = f e for all e, f ∈ Ek (S), then it is a k-Clifford semiring, by Theorem 3.3. Hence L=R on every semiring (S, +, ·) such that ef = f e for all e, f ∈ Ek (S). In the following theorem we give an alternative characterization of the Green equivalences on such semirings. Theorem 5.4 Let (S, +, ·) be a k-regular semiring such that ef = f e for all e, f ∈ Ek (S). Then for a, b ∈ S, a L b if and only if a ∈ Vk (a) and b ∈ Vk (b) implies that there is x ∈ S such that a a + b xb = b xb and b b + a xa = a xa. Proof Let a, b ∈ S such that a L b. Then for a ∈ Vk (a) and b ∈ Vk (b) there are a

∈ Vk (a) and b

∈ Vk (b) such that a a + b

b = b

b and b b + a

a = a

a [11, Theorem 3.6]. Thus a a + b

(b + bb b) = b

(b + bb b) implies that a a + (b

b)(b b) = (b

b)(b b) and so a a + b (bb

)b = b (bb

)b. Similarly we have b b + a (aa

)a = a (aa

)a. Then by Lemma 2.1, it follows that a a + b xb = b xb and b b + a xa = a xa, where x = bb

+ aa

. Conversely let a, b ∈ S and a ∈ Vk (a), b ∈ Vk (b). Then there is x ∈ S such that

a a + b xb = b xb and b b + a xa = a xa. Then a + aa a = aa a implies that a + ab xb = ab xb and b + bb b = bb b implies that b + ba xa = ba xa. Thus a L b.  Now we investigate the conditions we need for a semiring (S, +, ·) with ef = f e for all e, f ∈ Ek (S) to be a distributive lattice.

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Lemma 5.5 Let (S, +, ·) be a k-regular semiring in SL+ such that ef = f e for all e, f ∈ Ek (S). If a + ab = a for all a, b ∈ S then Ek (S) = E · (S) = S, i.e. (S, ·) is a band. Proof Let e ∈ Ek (S). Then from e + e2 = e2 and a + ab = a we get e = e2 , whence Ek (S) = E · (S). For a ∈ S there is some x ∈ S such that a = a + a(xa) = axa. Then  a = a + ax implies that a 2 = a 2 + axa = a 2 + a and so a ∈ Ek (S) = E · (S). Theorem 5.6 For any k-regular semiring (S, +, ·) in SL+ the following statements are equivalent: 1. (S, +, ·) is a distributive lattice. 2. ef = f e for all e, f ∈ Ek (S) and a + ab = a for all a, b ∈ S. 3. ef = f e for all e, f ∈ E · (S) and a + ab = a for all a, b ∈ S. Proof Since (1) ⇒ (2) and (2) ⇒ (3) are clear, we have to show (3) ⇒ (1) only. From the above lemma we get that (S, ·) is a semilattice and this implies a(a + b) =  a 2 + ab = a + ab = a. Thus (S, +, ·) is a distributive lattice. Now we characterize the least distributive lattice congruence on the k-regular semirings (S, +, ·) such that ef = f e for all e, f ∈ Ek (S). Since such semirings are k-Clifford, J is the least distributive lattice congruence on S [12]. We give here an alternative proof. Define a relation η on S by: for a, b ∈ S, aηb

if and only if

a + axb = b + axb

for some x ∈ S.

Theorem 5.7 Let (S, +, ·) be a k-regular semiring such that ef = f e for all e, f ∈ Ek (S). Then η is the least distributive lattice congruence on S. Proof Let a, b ∈ S such that a + axb = b + axb for some x ∈ S. Now there exists y ∈ S such that axb + byax = byax, by Theorem 3.3. Similarly there exists z ∈ S such that ax + xza = xza and so byax + byxza = byxza. This implies that axb + byax +byxza = byax +byxza, i.e. axb +byxza = byxza. Thus a +axb +byxza = b + axb + byxza implies that b + byxza = a + byxza and so η is symmetric. Now let a, b, c ∈ S such that a η b and b η c. Then, by Lemma 2.1, there is x ∈ S such that a + axb = b + axb and b + bxc = c + bxc. Then a + axb + bxc = c + axb + bxc implies that a + ax(b + bxc) + (b + axb)xc = c + ax(b + bxc) + (b + axb)xc and so a + ax(c + bxc) + (a + axb)xc = c + ax(c + bxc) + (a + axb)xc, i.e. a + a(x + xbx)c = c + a(x + xbx)c. Thus, η is an equivalence relation on S. Let a η b and c ∈ S. Then, by Lemma 2.1, there is x ∈ S such that a + axb = b + axb, b + bxb = bxb and c + cxc = cxc. Now a + c + axb + cxc = b + c + axb + cxc implies that a + c + (a + c)x(b + c) = b + c + (a + c)x(b + c). Thus (a + c) η (b + c). Again xb, cx ∈ Ek (S) implies that xbcx = cx 2 b. Thus ac + axbc = bc + axbc implies that ac + axbcxc = bc + axbcxc, i.e. ac + acx 2 bc = bc + acx 2 bc. Hence ac η bc. Similarly we have ca η cb. Thus η is a congruence on (S, +, ·).

140

M.K. Sen, A.K. Bhuniya

Let a ∈ S. Then there exists x ∈ S such that a +axaxa = axaxa and a 2 +a 2 xa 2 = Now, by Theorem 3.3, there is y ∈ S such that ax + xya = xya. Then

a 2 xa 2 .

a + a 2 xa 2 + ax(ax + xya)a = a 2 + a 2 xa 2 + ax(ax + xya)a implies that a +a 2 xa 2 +ax 2 ya 2 = a 2 +a 2 xa 2 +ax 2 ya 2 and so a +a(ax +x 2 y)a 2 = a 2 + a(ax + x 2 y)a 2 . Thus a η a 2 . Let a, b ∈ S. Then there is x ∈ S such that a + axa = axa and ab + abxab = abxab. Now, repeated applications (similarly as the proof that η is symmetric) of Theorem 3.3 imply that there is y ∈ S such that xab + xyba = xyba. Then ab + ab(xab + xyba) = ab(xab + xyba) implies that ab + abxyba = abxyba. Similarly there is z ∈ S such that ba + abzba = abzba. Thus we have ab + abxyba + abzba = ba + abxyba + abzba and so ab + ab(xy + z)ba = ba + ab(xy + z)ba. Hence ab η ba. Again a + axa = axa implies that a + abxab + axa = a + ab + abxab + axa which in turn implies that a + a(x + bx)(a + ab) = a + ab + a(x + bx)(a + ab). Thus aη η (a + ab). Hence η is a distributive lattice congruence on (S, +, ·). Let ξ be a distributive lattice congruence on (S, +, ·) and a, b ∈ S such that a + axb = b + axb for some x ∈ S. Then a ξ (a + axb) = (b + axb) ξ b and so a ξ b. Hence η is the least distributive lattice congruence on (S, +, ·).  Acknowledgement The authors express their sincere thanks to the learned referee for many useful comments, including some suggested revision of the proofs of Lemma 4.1, Theorem 5.6 and Theorem 5.7 incorporated in this version.

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