An Investigation on Ordered Algebraic Hyperstructures

Acta Mathematica Sinica, English Series Aug., 2017, Vol. 33, No. 8, pp. 1107 1124 Published online: April 24, 2017 DOI: 10.1007/s10114-017-6093-7 Http://www.ActaMath.com Acta Mathematica Sinica, English Series Springer-Verlag Berlin Heidelberg & The Editorial Office of AMS 2017 An Investigation on Ordered Algebraic Hyperstructures Saber OMIDI Bijan DAVVAZ Department of Mathematics, Yazd University, 89195-741 Yazd, Iran E-mail : omidi.saber@yahoo.com davvaz@yazd.ac.ir Jian Ming ZHAN 1) Department of Mathematics, Hubei University for Nationalities, Enshi 445000, P.R.China E-mail : zhanjianming@hotmail.com Abstract In this paper, we present some basic notions of simple ordered semihypergroups and regular ordered Krasner hyperrings and prove some results in this respect. In addition, we describe pure hyperideals of ordered Krasner hyperrings and investigate some properties of them. Finally, some results concerning purely prime hyperideals are proved. Keywords Algebraic hyperstructure, simple ordered semihypergroup, regular ordered Krasner hyperring, right pure hyperideal, purely prime hyperideal MR(2010) Subject Classification 16Y99, 20N20 1 Introduction An ideal I of a semigroup S is called right pure if for each x I, thereexistsy I such that x = xy. In [9], the notion of pure ideals of ordered semigroups have been introduced and studied. Many authors studied different aspects of pure ideals, for instance, Ahsan and Takahashi [1], Bashir and Shabir [5], De Marco [20], Shabir and Naz [33], and many others. The notion of pure ideal on an ordered ternary semigroup was introduced and studied by Sanborisoot and Changphas [32]. By an ordered semigroup [6, 21], we mean an algebraic structure (S,, ), which satisfies the following conditions: (1) (S, ) isasemigroup;(2)(s, ) is a partially ordered set; (3) If a and b are elements of S such that a b, thenax bx and xa xb for all x S. Many authors, especially Alimov [3], Changphas [7], Conrad [11], Hion [24] and Kehayopulu [27] studied such semigroups with some restrictions. A partially ordered ring [4] is a ring (R, +, ), together with a compatible partial order, i.e., a partial order on R that is compatible with the ring operations in the sense that it satisfies: (1) For all a, b, c R, a b implies that a + c b + c; (2)Ifa, b, c R with a b and 0 c, then a c b c and c a c b. An ideal I of a ring R is called pure ideal if for every a I, Received February 17, 2016, accepted January 17, 2017 The third author is supported by a Grant of National Natural Science Foundation of China (Grant Nos. 11461025 and 11561023) 1) Corresponding author

1108 Omidi S. et al. there exists b I such that a = ab. The notion of pure ideals in commutative reduced Gelfand rings with unity has been studied in [2]. In [23], pure hyperradical in semihypergroups are introduced and some properties and related results are given. In [22], Heidari and Davvaz studied a semihypergroup (S, ) besides a binary relation, where is a partial order relation such that satisfies the monotone condition. Chvalina [10] have started the concept of ordering hypergroups in 1994 as a special class of hypergroups. Ordered semihypergroups have been studied by several authors, for example, Changphas and Davvaz [8], Davvaz et al. [19], Heidari and Davvaz [22], and many researchers. In [31], Omidi et al. studied some properties of hyperideals and interior hyperideals in ordered Krasner hyperrings. Recently, Davvaz and Omidi [17] introduced and studied some properties of hyperideals in ordered semihyperrings. The plan of this paper is the following: After an introduction, in Section 2 we present some basic notions of semihypergroups and Krasner hyperrings. In Section 3, we present some basic notions of simple ordered semihypergroups and regular ordered Krasner hyperrings and prove some results in this respect. In Section 4, we define pure hyperideals of ordered Krasner hyperrings and investigate some properties of them. In Section 5, some results concerning pure hyperideals are proved. 2 Preliminaries Hyperstructure theory was born in 1934 when Marty [29] defined hypergroups based on the notion of hyperoperation. This theory has been studied in the following decades and nowadays by many mathematicians, for example, Corsini [12], Davvaz [13 15], Davvaz and Leoreanu- Fotea [16], Davvaz and Salasi [18], Vougiouklis [34], Zhan [35, 36] and many others. As a reference for more definitions on semihypergroups we refer to [12]. Let H be a non-empty set. A mapping : H H P (H), where P (H) denotes the family of all non-empty subsets of H, is called a hyperoperation on H. The couple (H, ) is called a hyperstructure. In the above definition, if A and B are two non-empty subsets of H and x H, thenwedenote A B = a b, A x = A {x} and x B = {x} B. a A b B A hyperstructure (H, ) is called a semihypergroup if for all x, y, z H, (x y) z = x (y z), which means that u z = x v. u x y v y z A non-empty subset K of a semihypergroup (H, ) is called a subsemihypergroup of H if K K K. Let (H, ) be a semihypergroup. Then, H is called a hypergroup if it satisfies the reproduction axiom, for all x H, H x = x H = H. A non-empty subset K of H is a subhypergroup of H if K a = a K = K for every a K. In the following, we give some definitions and results in preparation of the following section. For more details the reader is referred to [16].

Studies on Ordered Algebraic Hyperstructures 1109 Definition 2.1 ([30]) A canonical hypergroup is a non-empty set H endowed with an additive hyperoperation +:H H P (H), satisfying the following properties : (1) for any x, y, z H, x +(y + z) =(x + y)+z; (2) for any x, y H, x + y = y + x; (3) there exists 0 H such that 0+x = x, for any x H; (4) for every x H, there exists a unique element x H such that 0 x + x ;(we shall write x for x and we call it the opposite of x.) (5) z x + y implies that y x + z and x z y, that is (H, +) is reversible. In the above definition, for A, B H, wedenotea + B = a A a + b. The following b B equalities follow easily from the axioms: ( a) =a and (a + b) = a b. Definition 2.2 ([28]) A Krasner hyperring is an algebraic hypersructure (R, +, ) which satisfies the following axioms : (1) (R, +) is a canonical hypergroup ; (2) (R, ) is a semigroup having zero as a bilaterally absorbing element, i.e., x 0=0 x =0; (3) the multiplication is distributive with respect to the hyperoperation +. We call 0 the zero of the Krasner hyperring (R, +, ). For x R, let x denote the unique inverse of x in (R, +). Then ( x) =x for all x R. Inaddition,wehave(x + y) (z + w) x z + x w + y z + y w, ( x) y = x ( y) = (x y) for all x, y, z, w R. A Krasner hyperring R is called commutative (with unit element) if (R, ) is a commutative semigroup (with unit element). A subhyperring of a Krasner hyperring (R, +, ) is a non-empty subset A of R which forms a Krasner hyperring containing 0 under the hyperoperation + and the operation on R, that is, A is a canonical subhypergroup of (R, +) and A A A. Then a non-empty subset A of R is a subhyperring of (R, +, ) if and only if, for all x, y A, x + y A, x A and x y A. A non-empty subset I of a Krasner hyperring (R, +, ) is called a left (resp. right) hyperideal of R if (I,+) is a canonical subhypergroup of (R, +) and for every a I and r R, r a I (resp. a r I). I is called a hyperideal if I is both left and right hyperideal of R. That is, x + y I and x I for all x, y I and x r, r x I for all x I and r R. Every hyperideal (whether left, right, two-sided) is a subhyperring of R but the converse is not true in general. A hyperideal I of R is said to be proper if I R. A non-empty subset I of a Krasner hyperring (R, +, ) is a left (resp. right) hyperideal if and only if (1) a, b I implies a b I; (2) a I, r R imply r a I (resp. a r I). 3 Ordered Algebraic Hyperstructures First, we give certain definitions needed for our purpose. An ordered semihypergroup [22] (S,, ) is a semihypergroup (S, ) together with a partial order that is compatible with the hyperoperation, meaning that for any x, y, z in S, x y z x z y and x z y z. Here, z x z y means for any a z x there exists b z y such that a b. The case x z y z is defined similarly. An ordered semihypergroup (S,, ) is called an ordered hypergroup if (S, ) is a hypergroup. For a non-empty subset K of an ordered semihypergroup

1110 Omidi S. et al. (S,, ), we define (K] = {x S : x a for some a K}. Let (S,, ) be an ordered semihypergroup. An element a S is said to be regular if there exists an element x S such that a a x a. An ordered semihypergroup S is called regular if for any a S, a (a S a]. Note that every ordered hypergroup is a regular ordered semihypergroup. Definition 3.1 ([22]) A non-empty subset I of an ordered semihypergroup (S,, ) is called a hyperideal of S if it satisfies the following conditions : (1) S I I and I S I; (2) If a I and b S such that b a, thenb I. Condition (2) is equivalent to the condition (I] =I. Inthefollowing,weintroducesimpleorderedsemihypergroupsandgivesomeexamples.In particular, we find the condition which is necessary and sufficient for (S,, ) tobeasimple ordered semihypergroup. Definition 3.2 An ordered semihypergroup (S,, ) is called simple if S is the only hyperideal of S, that is, for any hyperideal A of S, we have A = S. A subsemihypergroup K of S is called simple if K with the hyperoperation and the order of S is a simple ordered semihypergroup. Example 3.3 We have (S,, ) is a simple ordered semihypergroup where the hyperoperation and the order relation are defined by a b c a a {a, b} {a, c} b a {a, b} {a, b} c a {a, b} c := {(a, a), (b, b), (c, c), (b, a), (c, a)}. The covering relation and the figure of S are given by = {(b, a), (c, a)}. a b Example 3.4 We have (S,, ) is an ordered semihypergroup where the hyperoperation and the order relation are defined by a b c d a a {a, b} {a, c} {a, d} b a {a, b} {a, c} {a, d} c a b c d d a b c d c

Studies on Ordered Algebraic Hyperstructures 1111 := {(a, a), (b, b), (c, c), (d, d), (a, b), (a, c), (a, d), (b, c), (b, d), (c, d)}. The covering relation and the figure of S are given by = {(a, b), (b, c), (c, d)}. d c b a Now, it is easy to see that (S,, ) is a simple ordered semihypergroup. Example 3.5 Let S = {a, b, c, d} be a set with the hyperoperation defined as follows: a b c d a a {a, b} {a, c} a b a {a, b} {a, c} a c a {a, b} {a, c} a d a {a, b} {a, c} a Then, (S, ) is a semihypergroup. We have (S,, ) is an ordered semihypergroup where the order relation is defined by := {(a, a), (b, b), (c, c), (d, d), (a, b), (a, c), (a, d), (b, c), (d, b), (d, c)}. The covering relation and the figure of S are given by = {(a, d), (b, c), (d, b)}. c b d a Now, it is easy to see that (S,, ) is a simple ordered semihypergroup. Example 3.6 Suppose that S = {a, b, c, d, e, f, g, h, i, j, k, l}, A = {d, e, j, k}, B = {d, f, j, l}, C = {a, b, g, h}, D = {a, c, g, i}, E = {d, e}, F = {d, f}, G = {a, b}, H = {a, c}, I = {d, j, }, J = {a, g}, K = {e, k}, L = {b, h}, M = {f,l}, N = {c, i}, O = {j, k}, P = {j, l}, Q = {g, h} and T = {g, i}. We consider the ordered semihypergroup (S,, ), where the hyperoperation

1112 Omidi S. et al. is defined by the following table: a b c d e f g h i j k l a d E F a G H I A B J C D b d e F a b H I K B J L D c d E f a G c I A M J C N d a G H d E F J C D I A B e a b H d e F J L D I K B f a G c d E f J C N I A M g j O P g Q T j O P g Q T h j k P g h T j k P g h T i j O l g Q i j O l g Q i j g Q T j O P g Q T j O P k g h T j k P g h T j k P l g Q i j O l g Q i j O l and the order is defined by := {(a, a), (b, b), (c, c), (d, d), (e, e), (f,f), (g, g), (h, h), (i, i), (j, j), (k, k), (l, l), (a, g), (b, a), (b, g), (b, h), (c, a), (c, g), (c, i), (d, j), (e, d), (e, j), (e, k), (f,d), (f,j), (f,l), (h, g), (i, g), (k, j), (l, j)}. The covering relation and the figure of S are given by = {(a, g), (b, a), (b, h), (c, a), (c, i), (d, j), (e, d) (e, k), (f,d), (f,l), (h, g), (i, g), (k, j), (l, j)}. h g i a b c k j l d e f Now, it is easy to see that (S,, ) is a simple ordered semihypergroup. Let us first make the following observation. Theorem 3.7 An ordered semihypergroup (S,, ) is simple if and only if (S a S] =S for all a S. It means that for any a, b S, thereexistx, y S such that b x a y.

Studies on Ordered Algebraic Hyperstructures 1113 Proof Suppose that S is a simple ordered semihypergroup and a S. We show that (S a S] is a hyperideal of S. Sincea 3 (S a S], it follows that (S a S]. Wehave S (S a S] =(S] (S a S] (S 2 a S] (S a S], (S a S] S =(S a S] (S] (S a S 2 ] (S a S]. Now, suppose that x (S a S] andy S such that y x. Sincex (S a S], it follows that x b for some b S a S. Sincey x and x b, wegety b. So,wehavey (S a S]. Hence (S a S] is a hyperideal of S, as desired. Therefore, (S a S] =S for all a S. Conversely, assume that (S a S] =S for all a S. LetI be a hyperideal of S containing x. Then, (S x S] =S. If y S, theny (S x S]. So, y (u x v] forsomeu, v S. Since (u x v] I, wegety I. ThisimpliesthatI = R. Hence the proof is completed. An immediate consequence of Theorem 3.7 is the following: Corollary 3.8 Every ordered hypergroup is a simple ordered semihypergroup. Proof If (S, ) is a hypergroup, then for all a S, a S = S a = S. So,wehaveS = a S S S S. ThusS S = S, and it follows that (S a S] =(S S] =(S] =S. By Theorem 3.7, S is simple. In a recent paper [25], the authors proved the following theorem: Theorem 3.9 ([25]) Let S and T be two simple semihypergroups. Then the product S T is a simple semihypergroup. In the following, we extend the above result based on ordered semihypergroups. Theorem 3.10 Let (S,, S ) and (T,, T ) be two simple ordered semihypergroups. Then, (S T,, ) is a simple ordered semihypergroup, where for all (s 1,t 1 ) and (s 2,t 2 ) in S T, hyperoperation and partial order are defined by (1) (s 1,t 1 ) (s 2,t 2 )=(s 1 s 2 ) (t 1 t 2 ), (2) (s 1,t 1 ) (s 2,t 2 ) if and only if s 1 S s 2 and t 1 T t 2. Proof It is easy to see that S T is an ordered semihypergroup. Let (a, b) be an arbitrary element of S T. Then, S T =(S a S] (T b T]. If (u, v) S T, then there exist (c, d) S a T band (s, t) S T such that (u, v) (c, d) (s, t). Also, there exist elements s in S and t in T such that c s a and d t b.since(c, d) s a t b =(s,t ) (a, b), we conclude that ( ] (u, v) (c, d) (s, t) c s d t (c,d) S a T b =((S a T b) (s, t)] (S T (a, b) (s, t)] (S T (a, b) S T ]. Thus we have S T (S T (a, b) S T ]. It is clear that (S T (a, b) S T ] S T.So, we have (S T (a, b) S T ]=S T. Therefore, S T is a simple ordered semihypergroup. We would like to first define the notion of ordered Krasner hyperring as it appears in [31].

1114 Omidi S. et al. Definition 3.11 An ordered Krasner hyperring (R, +,, ) is a Krasner hyperring equipped with a partial order relation such that for all a, b, c R, we have (1) a b implies a + c b + c, meaning that for any x a + c, thereexistsy b + c such that x y. (2) a b and 0 c imply a c b c and c a c b. Example 3.12 Krasner hyperrings are viewed as ordered Krasner hyperrings under the equality order relation. Example 3.13 For any ordered ring (R, +,, ), define two binary hyperoperations, and order relation as x y = {x + y}, x y = {x y} and = {(x, y) x y}. Then(R,,, ) is an ordered hyperring. Example 3.14 ([31]) Let R = {0,a,b,c} be a set with the hyperaddition and the multiplication defined as follows: 0 a b c 0 0 a b c a a {0,b} {a, c} b b b {a, c} {0,b} a c c b a 0 0 a b c 0 0 0 0 0 a 0 a b c b 0 b b 0 c 0 c 0 c Then, (R,, ) is a Krasner hyperring. We have (R,,, ) is an ordered Krasner hyperring where the order relation is defined by := {(0, 0), (a, a), (b, b), (c, c), (0,b),(c, a)}. The covering relation and the figure of R are given by = {(0,b),(c, a)}. b a 0 c Definition 3.15 A homomorphism from an ordered Krasner hyperring (R, +,, ) into an ordered Krasner hyperring (T,,, ) is a mapping ϕ : R T such that (1) ϕ(a + b) ϕ(a) ϕ(b); (2) ϕ(a b) =ϕ(a) ϕ(b); (3) for all a, b R, a b implies ϕ(a) ϕ(b). Also ϕ is called a good homomorphism if in the previous condition (1), the equality is valid. The mapping ϕ is called isotone if a, b R, a b implies ϕ(a) ϕ(b) andreverse isotone if a, b R, ϕ(a) ϕ(b) implies a b. Note that each reverse isotone mapping is one-to-one. The mapping ϕ is called an isomorphism if it is a reverse isotone onto homomorphism. The ordered

Studies on Ordered Algebraic Hyperstructures 1115 Krasner hyperrings R and T are called isomorphic if there exists an isomorphism between them. Let (R, +,, ) be an ordered Krasner hyperring and A R. Then, (A] is the subset of R defined as follows: (A] ={x R : x a for some a A}. ForA, B R, wehave (1) A (A], (2) (A](B] (AB], (3) A B implies (A] (B] and (4) ((A]] = (A]. The preceding discussion focused on properties of regular ordered Krasner hyperrings. An element a of an ordered Krasner hyperring (R, +,, ) issaidtoberegular if there exists an element x R such that a (a x) a, i.e., a (ara] for all a R or A (ARA] for all A R. An ordered Krasner hyperring (R, +,, ) is said to be regular if every element of R is regular. Theorem 3.16 Let ϕ : R T be a good homomophism from a regular ordered Krasner hyperring (R, +,, ) into an ordered Krasner hyperring (T,,, ). Then, Im ϕ is a regular ordered Krasner hyperring. Proof Note that every subhyperring K of an ordered Krasner hyperring (T,,, ) isan ordered Krasner hyperring under the order relation on R. Every subhyperring K of T with the relation K on K defined by K := {(x, y) K K x y} is an ordered Krasner hyperring. Let x Im ϕ be an arbitrary element. Then, there exists an element a R such that ϕ(a) =x. Since R is a regular ordered Krasner hyperring, it follows that a (ara]. So, there exists an element b R such that a a b a. Since ϕ is a good homomophism, we have x = ϕ(a) ϕ(a b a) =ϕ(a) ϕ(b) ϕ(a) =x ϕ(b) x. Hence, Im ϕ is a regular ordered Krasner hyperring. Theorem 3.17 Let (R, + 1, 1, 1 ) and (T,+ 2, 2, 2 ) be two regular ordered Krasner hyperrings. Then, the product R T is a regular ordered Krasner hyperring, where for all (a, b) and (c, d) in R T, hyperoperation, operation and a partial order relation are defined as (1) (a, b) (c, d) ={(x, y) :x a + 1 c, y b + 2 d}, (2) (a, b) (c, d) =(a 1 c, b 2 d), (3) (a, b) (c, d) if and only if a 1 c and b 2 d. Proof It is easy to see that R T is an ordered Krasner hyperring. Let (r, t) be an arbitrary element of R T. Then, there exist elements x in R and y in T such that r 1 r 1 x 1 r and t 2 t 2 y 2 t. Thuswehave (r, t) (r, t) (x, y) (r, t). This implies that R T is a regular ordered Krasner hyperring. The concept of hyperideal in ordered Krasner hyperrings is given as follows:

1116 Omidi S. et al. Definition 3.18 Let (R, +,, ) be an ordered Krasner hyperring. A non-empty subset I of R is called a left (resp. right) hyperideal of R if it satisfies the following conditions : (1) (I,+) is a canonical subhypergroup of (R, +); (2) R I I (resp. I R I); (3) If a I and b R such that b a, thenb I. By two-sided hyperideal or simply hyperideal, we mean a non-empty subset I of R which is both a left and a right hyperideal of R. Note that the condition (3) in Definition 3.18 is equivalent to I =(I]. Remark 3.19 hyperring. Every hyperideal of an ordered Krasner hyperring R is an ordered Krasner Theorem 3.20 If I is a hyperideal of a commutative ordered Krasner hyperring (R, +,, ), then (I : r) ={x R x r I}, for all r R, is also a hyperideal of R. Proof Since 0 (I : r), it follows that (I : r). Let a, b (I : r). Then a r I and b r I. So,wehave(a b) r = a r b r I. Thusa b (I : r). Now, let a (I : r) and s R. Sincea r I, it follows that (a s) r = a (s r) =a (r s) =(a r) s I s I. So, we have a s (I : r). Let a (I : r) andb R be such that b a. Then, we have b r a r. Since a r I and I is a hyperideal of R, weobtainb r I. So,b (I : r). This shows that I is a hyperideal of R, as desired. Theorem 3.21 of R, then Let (R, +,, ) be an ordered Krasner hyperring. If A and B are hyperideals B A 1 = {x R a A, a x B}, A 1 B = {x R a A, x a B} are hyperideals of R. Proof We show that B A 1 is a hyperideal of R. Since0 B A 1, it follows that B A 1. Let x, y B A 1 and r R. Ifa A, thena x, a y B. So,wehavea (x y) =a x a y B. Thus x y B A 1. Let r R and x B A 1. If a A, thena r A. So, we have (a r) x B. This implies that a (r x) B. Thusr x B A 1.Thatx r B A 1 can be proved similarly. Now, let x B A 1 and y R be such that y x. Leta A. Then, we have a y a x. Since a x B and B is a hyperideal of R, weobtaina y B. So, y B A 1. This shows that B A 1 is a hyperideal of R, as desired. Similarly, A 1 B is a hyperideal of R. 4 Some Notes on Pure Hyperideals In the remainder of the paper we will use extensively the concept of pure hyperideals in ordered Krasner hyperrings. Let us first give the following definitions. Definition 4.1 Let (R, +,, ) be an ordered Krasner hyperring. A right hyperideal I of R is called a right pure right hyperideal if for each x I, there is an element y I such that (x, x y).

Studies on Ordered Algebraic Hyperstructures 1117 Left pure left hyperideals are defined similarly. Definition 4.2 Let (R, +,, ) be an ordered Krasner hyperring. A hyperideal I of R is called a left (resp. right) pure hyperideal if for every x I, thereexistsy I such that (x, y x) (resp. (x, x y) ). Equivalent Definition x (I x](resp.x (x I]). If I is both left pure and right pure hyperideal, then I is called a pure hyperideal of R. Example 4.3 {0} is a right pure hyperideal of R. Example 4.4 Let R = {0,a,b,c} be a set with the hyperoperation and the multiplication defined as follows: 0 a b c 0 0 a b c a a {0, a} c {b, c} b b c {0,b} {a, c} c c {b, c} {a, c} R 0 a b c 0 0 0 0 0 a 0 0 0 0 b 0 a b c c 0 a b c Then, (R,, ) is a Krasner hyperring [26]. We have (R,,, ) is an ordered Krasner hyperring where the order relation is defined by := {(0, 0), (a, a), (b, b), (c, c), (0,a),(0,b),(0,c),(a, c), (b, c)}. The covering relation and the figure of R are given by = {(0,a),(0,b),(a, c), (b, c)}. a c b 0 It is easy to check that {0}, {0,a} and R are hyperideals of R. Wehave 0 0 0, a b a, a c a, b b b, b c b, c b c, c c c, b b c and there is no x in R such that a a x. We conclude that (1) {0} is a left and right pure hyperideal of R. (2) {0,a} is not a left pure hyperideal of R as well as a right pure hyperideal of R. (3) {0,b} is a left pure left hyperideal of R, but {0,a} is not a left pure left hyperideal of R.

1118 Omidi S. et al. Example 4.5 LetR = {0,a,b,c,d,e} be a set with the hyperaddition and the multiplication defined as follows: 0 a b c d e 0 0 a b c d e a a a {0, a, b} d d {c, d, e} b b {0, a, b} b e {c, d, e} e c c d e 0 a b d d d {c, d, e} a a {0, a, b} e e {c, d, e} e b {0, a, b} b and 0 a b c d e 0 0 0 0 0 0 0 a 0 a b 0 a b b 0 a b 0 a b c 0 0 0 c c c d 0 a b c d e e 0 a b c d e Then, (R,, ) is a Krasner hyperring [26]. We have (R,,, ) is an ordered Krasner hyperring where the order relation is defined by := {(0, 0), (a, a), (b, b), (c, c), (d, d), (e, e), (0,a), (0,b),(a, b), (c, d), (c, e), (d, e)}. The covering relation and the figure of R are given by = {(0,a),(a, b), (c, d), (d, e)}. b e a d 0 c {0}, {0,a,b} and R are hyperideals of R. It is easy to see that {0,a,b} is a left and right pure hyperideal of R. We now prove the following lemma which is the crucial lemma in the establishment of our main theorems. Lemma 4.6 Let (R, +,, ) be an ordered Krasner hyperring. Then, (1) If {I k : k Ω} is a family of left pure hyperideals of R such that I i I j or I j I i for all i, j Ω, then k Ω I k is a left pure hyperideal of R.

Studies on Ordered Algebraic Hyperstructures 1119 (2) The intersection of finite number of left pure hyperideals of R is a left pure hyperideal of R. Proof (1) Since 0 k Ω I k, it follows that k Ω I k. Letx, y k Ω I k.thenx, y I k for some k Ω. Since I k is a hyperideal of R, weobtainx y I k for some k Ω. Thus x y k Ω I k.also, ( ) I k R = I k R k Ω k Ω k Ω I k and ( R k Ω ) I k = R I k I k. k Ω k Ω So, for each x k Ω I k and s R, x s k Ω I k. Similarly, s x k Ω I k. Now, let x k Ω I k, y R and y x. Then, x I k for some k Ω. Since I k is a hyperideal of R, it follows that y I k k Ω I k. Therefore, k Ω I k is a hyperideal of R. Let a k Ω I k. Then, a I i for some i Ω. Since I i is a left pure hyperideal of R, thereexistsb I i such that (a, b a). So,wehaveb I i k Ω I k. This implies that k Ω I k is a left pure hyperideal of R. (2) Let {I 1,I 2,...,I n } be a finite indexed family of left pure hyperideals of R. We have n i=1 I i is a hyperideal of R. Leta n i=1 I i.sincea I 1 and I 1 is a left pure hyperideal, there exists b 1 I 1 such that (a, b 1 a). Similarly, for j {2,...,n}, thereexistsb j I j such that (a, b j a). Thuswehave(a, b n b 2 b 1 a). Sinceb n b 2 b 1 I n I 2 I 1 n i=1 I i, it follows that n i=1 I i is left pure. Theorem 4.7 Let I be a hyperideal of an ordered Krasner hyperring (R, +,, ). If I is a right pure hyperideal of R, then(i I] =I. Proof Suppose that I is a right pure hyperideal of R. SinceI is a hyperideal of R, it follows that I I I R I. So, we have (I I] (I] =I. Now, let a I. Since I is a right pure hyperideal, there exists b I such that (a, a b). Asa I, a b I I. So,wehavea (I I]. This implies that I (I I]. Thus I =(I I]. Recently, pure ideals in ordered semigroups has been studied by Changphas and Sanborisoot [9]. Changphas and Sanborisoot proved that the set of all pure prime ideals is topologized. Changphas and Sanborisoot [9] defined right pure ideals of ordered semigroups as follows: Definition 4.8 Let (S,, ) be an ordered semigroup. An ideal I of S is called a right pure ideal if for every x I, thereexistsy I such that x x y. Equivalent Definition x (x I]. Let S be an ordered semigroup such that SS = S. The set of all right pure ideals of S and the set of all proper pure prime ideals of S will be denoted by P (S) andp (S), respectively. For A P (S), let I A = {J P (S) A J} and τ(s) ={I A A P (S)}. Changphas and Sanborisoot in [9, Theorem 4.1] proved the next theorem: Theorem 4.9 ([9]) τ(s) forms a topology on P (S).

1120 Omidi S. et al. The following examples show that S is not a right pure ideal of S. Example 4.10 Let (S,, ) be an ordered semigroup such that S = {0,a,b,c} and 0 a b c 0 0 0 0 0 a 0 0 0 0 b 0 a b c c 0 a b c := {(0, 0), (a, a), (b, b), (c, c), (0,a),(0,b),(0,c),(a, b), (a, c), (b, c)}. The covering relation and the figure of S are given by = {(0,a),(a, b), (b, c)}. c b a 0 Since there is no x in S such that a a x, we conclude that S is not a right pure ideal of S. Example 4.11 ([37]) Suppose that S = {a, b, c, d, e, f}. We consider the ordered semigroup (S,, ), where the multiplication is defined by the following table: a b c d e f a d e d d e d b d e d d e d c d e d d e d d d e d d e d e d e d d e d f a b c d e f and the order is defined by := {(a, a), (b, b), (c, c), (d, d), (e, e), (f,f), (d, a), (d, c), (d, f), (e, a), (e, b), (e, c), (e, d), (e, f)}. The covering relation and the figure of S are given by = {(d, a), (d, c), (d, f), (e, b), (e, d)}.

Studies on Ordered Algebraic Hyperstructures 1121 a c f d b e Since there is no x in S such that a a x, b b x and c c x, we conclude that S is not a right pure ideal of S. Therefore, Theorem 4.1 in [9] is not true in a general case. 5 Purely Prime Hyperideals In this section, we define purely prime hyperideals of ordered Krasner hyperrings and investigate some related results. Definition 5.1 A right pure hyperideal I of an ordered Krasner hyperring (R, +,, ) is said to be purely maximal if for any proper right pure hyperideal J of R, I J implies I = J. Example 5.2 In Example 3.14, I 1 = {0}, I 2 = {0,b}, I 3 = {0,c} and I 4 = {0,b,c} are right pure hyperideals of R. It is easy to see that I 4 is purely maximal hyperideal. Example 5.3 Let (R,,, ) be the ordered Krasner hyperring defined as in Example 4.5. It is clear that {0,a,b} is purely maximal hyperideal. Definition 5.4 Let I be a proper right pure hyperideal of an ordered Krasner hyperring (R, +,, ). Then, I is called purely prime if for any right pure hyperideals I 1 and I 2 of R, I 1 I 2 I implies I 1 I or I 2 I. Example 5.5 In Example 3.14, I 1 = {0}, I 2 = {0,b}, I 3 = {0,c} and I 4 = {0,b,c} are right pure hyperideals of R. It is not difficult to verify that I 2, I 3 and I 4 are purely prime hyperideals of R, but I 1 is not purely prime hyperideal. Indeed, {0,b} {0,c} {0}, but {0,b} {0} and {0,c} {0}. Theorem 5.6 Let (R, +,, ) be an ordered Krasner hyperring and I ahyperidealofr. Then, I contains a largest right pure hyperideal of R (It is called the pure part of I and denoted by R(I)). Proof Let R(I) be the union of all right pure hyperideals contained in I. By Lemma 4.6, R(I) is a right pure hyperideal of R. Clearly, {0} is a right pure hyperideal of R contained in each hyperideal I of R. Then, R(I) exists and it is the largest right pure hyperideal of R contained in I. Theorem 5.7 Let (R, +,, ) be an ordered Krasner hyperring. Let I, J and {A k } k Ω be hyperideals of R. Then, (1) R(I J) =R(I) R(J). (2) k Ω R(A k) R( k Ω A k). Proof (1) Since R(I) I and R(J) J, weobtainr(i) R(J) I J. By Lemma 4.6, R(I) R(J) is a right pure hyperideal of R. So, R(I) R(J) R(I J). Since R(I J) I J

1122 Omidi S. et al. I, wehaver(i J) R(I). Similarly, R(I J) R(J). Thus we have R(I J) R(I) R(J). So, R(I J) =R(I) R(J). (2) Since R(A k ) A k for all k Ω, we have k Ω R(A k) k Ω A k.sincer(a k )isright pure, by Lemma 4.6, k Ω R(A k) is right pure. Then, k Ω R(A k) R( k Ω A k). Lemma 5.8 Let (R, +,, ) be an ordered Krasner hyperring. Then, (1) If I is a purely maximal hyperideal of R, theni is purely prime. (2) If I is a maximal hyperideal of R, thenr(i) is purely prime. Proof (1) Let I be a purely maximal hyperideal of R. LetA and B be right pure hyperideals of R such that A B I and A I. Then, A I is a right pure hyperideal such that I A I. Since I is purely maximal, it follows that R = A I. Wehave B = B R = B (A I) =(B A) (B I) I I = I. Hence, I is purely prime. (2) Let I be a maximal hyperideal of R. LetA and B be right pure hyperideals of R such that A B R(I). If A I, thena R(I). Suppose A I. It is easy to see that A I is a hyperideal of R. SinceI is a maximal hyperideal, it follows that R = A I. Wehave B = B R = B (A I) =(B A) (B I) R(I) I I I = I. Since R(I) is the largest right pure hyperideal contained in I, wegetb R(I). Hence, R(I) is purely prime. Theorem 5.9 Let A be a right pure hyperideal of an ordered Krasner hyperring (R, +,, ) and a R such that a/ A. Then, there exists a purely prime hyperideal I of R such that A I and a/ I. Proof Let C = {I : I is a right pure hyperideal of R, A I and a / I}. Since A C,it follows that C. Also,C is a partially ordered set under the usual inclusion. Let {I k : k Ω} be a chain in C. Consider B = k Ω I k. By Lemma 4.6, B is a right pure hyperideal of R. Since each I k Ccontains A and a/ I k,wehavea k Ω I k = B and a/ B. Hence B C is an upper bound for chain {I k : k Ω}. By Zorn s lemma, C has a maximal element, say K, such that K is a right pure hyperideal of R, A K and a/ K. We show that K is a purely prime hyperideal of R. Suppose A 1 and A 2 are right pure hyperideals of R such that A 1 K and A 2 K. SinceA i (i =1, 2) and K are right pure, A i K is a right pure hyperideal of R such that K A i K. By the maximality of K, wehavea A 1 K and a A 2 K. Since a A i K and a/ K, wehavea A 1 A 2.Sincea/ K, it follows that A 1 A 2 K. This shows that K is purely prime. Hence the proof is completed. Corollary 5.10 A proper right pure hyperideal A of an ordered Krasner hyperring (R, +,, ) is the intersection of all the purely prime hyperideals of R containing A. Proof By Theorem 5.9, there exists purely prime hyperideals containing A. LetI = {I k : k Ω} be the set of all purely prime hyperideals of R containing A. Since A I k for all k Ω, we have A k Ω I k. Suppose that B = k Ω I k. We show that B A. Let a / A. By Theorem 5.9, there exists a purely prime hyperideal I of R such that A I and a/ I. Hence

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