EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 9, No. 4, 2016, 402-418 ISSN 1307-5543 – www.ejpam.com On Regular Multiplicative Hyperrings Reza Ameri∗, Ali Kordi School of Mathematics, Statistics and Computer Science, College of Sciences, University of Tehran, P.O. Box: 14155-6455, Tehran, Iran Abstract. We introduce and study regular multiplicative hyperrings, as a generalization of classical rings. Also, we use the fundamental relation γ∗ on a given regular multiplicative hyperring R and prove that the fundamental ring R/γ∗ of R is a regular ring. Finally, we investigate the algebraic properties of M(R), the regular hyperideal of R, generated by all elements of R such that its generated hyperideal is regular. 2010 Mathematics Subject Classifications: 20N20 Key Words and Phrases: Multiplicative hyperring, Regular hyperring, Fundamental relation, Hyper- ideal, Regular element, Idempotent 1. Introduction and Primary The theory of hyperstructures has been introduced by Marty in 1934 during the 8th Congress of the Scandinavian Mathematicians [19]. Marty introduced hypergroups as a gener- alization of groups. He published some notes on hypergroups, using them in different contexts as algebraic functions, rational fractions, non-commutative groups and then many researchers have been worked on this new field of modern algebra and developed it. It was later observed that the theory of hyperstructures has many applications in both pure and applied sciences; for example, semi-hypergroups are the simplest algebraic hyperstructures that possess the prop- erties of closure and associativity. The theory of hyperstructures has been widely reviewed [8, 9, 12, 19, 31]. In [9] Corsini and Leoreanu-Fotea have collected numerous applications of algebraic hy- perstructures, especially those from the last fifteen years to the following subjects: geometry, hypergraphs, binary relations, lattices, fuzzy sets and rough sets, automata, cryptography, codes, median algebras, relation algebras, artificial intelligence, and probabilities. The hy- perrings were introduced and studied by Krasner [18], Nakasis [21], Massouros [19] and especially studied by Davvaz and Leoreanu-Fotea [13], Zahedi and Ameri [33], Ameri and ∗Corresponding author. Email addresses: rameri@ut.ac.ir (R. Ameri), ali.kordi@gmail.com (A. Kordi) http://www.ejpam.com 402 c© 2016 EJPAM All rights reserved. R. Ameri, A. Kordi / Eur. J. Pure Appl. Math, 9 (2016), 402-418 403 Norouzi [1, 2]. The study on hyperrings in [31] ends with an outline of applications in chem- istry and physics, analyzing several special kinds of hyperstructures: e-hyperstructures and transposition hypergroups. The theory of suitable modified hyperstructures can serve as a mathematical background in the field of quantum communication systems. A well-known type of a hyperring, called the Krasner hyperring [18]. Krasner hyperrings are essentially hy- perrings, with approximately modified axioms in which addition is a hyperoperation, while the multiplication is an operation. Then, this concept has been studied by a variety of authors. Some principal notions of hyperring theory can be found in [12, 13, 20, 30, 32]. The another type of hyperrings was introduced by Rota in 1982 which the multiplication is a hyperopera- tion, while the addition is an operation, and it is called it a multiplicative hyperring (for more details see [26–29]) which was subsequently investigated by Olson and Ward [22] and many others. De Salvo [14] introduced hyperrings in which the additions and the multiplications are hyperoperations. Moreover, there exists other types of hyperrings that both the addition and multiplication are hyperoperations and instead associativity, commutativity and distribu- tivity satisfy in weak associativity, weak commutativity and weak distributivity, which is called Hv-hyperrings, this type of hyperrings can be seen in [31, 32]. Also, there are other types of hyperrings which were completely studied in [12]. These hyperrings are studied by Rahnamai Barghi [25]. Procesi and Rota in [23] have studied ring of fractions in Krasner hyperrings and also they conceptualized in [24] the notion of primeness of hyperideal in a multiplicative hy- perring, and in [10], Dasgupta extended the prime and primary hyperideals in multiplicative hyperrings. Asokkumar and Velrajan [4, 5] have studied Von Neumann regularity in Krasner hyperrings. A special equivalence relations which is called fundamental relations play important roles in the the theory of algebraic hyperstructures. The fundamental relations are one of the most important and interesting concepts in algebraic hyperstructures that ordinary algebraic struc- tures are derived from algebraic hyperstructures by them. The fundamental relation β∗ on hypergroups was defined by Koskas [17], mainly studied by Corsini [19], Freni [15, 16], Vou- giouklis [32] (for more details about hyperrings and fundamental relations on hyperrings see [1, 2, 11, 12, 30, 32]). In this paper we consider the classes of multiplicative hyperring as a hyperstructures (R,+, .), where (R,+) is an abelian group, (R,+) is a semihypergroup and the hyperoperation ”.” is distributive with respect to the operation ”+ ”, i.e. a.(b+ c) ⊆ a.b+ a.c. The purpose of this paper is the study study regular multiplicative hyperrings. In this regards we study the properties of regular multiplicative hyperring R and obtain some results. We will proceed to use the fundamental relation γ∗ on R and prove that the fundamental ring R/γ∗ of R is regular whenever R is regular. Also, we show that this process make a functor from the category of regular multiplicative hyperrings to the category of regular rings. Finally, the notion of regular hyperideal M(R), consisting of the elements of R such that the generated hyperideal by these elements are regular hyperideals, are introduced and its basic properties are investigated. R. Ameri, A. Kordi / Eur. J. Pure Appl. Math, 9 (2016), 402-418 404 2. Regular Multiplicative Hyperring Recall that a hyperoperation ”.” on nonempty set H is a mapping of H × H into the family of all nonempty subsets of H. Let ”.” be a hyperoperation on H. Then, (H, .) is called a hypergroupoid. we can extend the hyperoperation on H to subsets of H as follows. For A, B ⊆ H and h ∈ H, then AB = ∪a∈A,b∈Bab, Ah= A{h}, hB = {h}B. A semihypergroup is a hypergroupoid (H, .), which is associative, that is (a.b).c = a.(b.c) or fall a, b, c ∈ H. A hypergroup is a semihypergroup (H, .), that satisfies the reproduction axioms, that is a.H = H = H.a for all a ∈ H. A non-empty set R with two hyperoperations + and . is said to be a hyperring if (R,+) is a canonical hypergroup, (R, .) is a semihypergroup with r.0 = 0.r = 0 for all r ∈ R (0 as a bilaterally absorbing element) and the hyperoperation . is distributive with respect to +, i.e., for every a, b, c ∈ R; a(b+ c) = ab+ ac and (a+ b)c = ac + bc. A multiplicative hyperring is an additive commutative group (R,+) endowed with a hyper- operation "." which satisfies the following conditions: (1) ∀a, b, c ∈ R : a(bc) = (ab)c; (2) ∀a, b, c ∈ R : (a+ b)c ⊆ ac + bc, a(b+ c) ⊆ ab+ ac; (3) ∀a, b ∈ R : (−a)b = a(−b) = −(ab). If in (2) we have equalities instead of inclusions, then we say that the multiplicative hy- perring is strongly distributive. Definition 1. Let R be a multiplicative hyperring. Then (i) An element e ∈ R is said to be a left(resp. right)identity if a ∈ e.a (resp. a ∈ a.e) for a ∈ R. An element e is called an identity element if it is both left and right identity element. (ii) An element e ∈ R is said to be a left(resp. right) scalar identity if a = e.a(resp., a = a.e) for a ∈ R. An element e is called an scalar identity element if it is both left and right scalar identity element. (iii) An element a is called a left (right) invertible (with respect to e), if there exists x ∈ R, such that e ∈ xa(e ∈ ax) and a is called invertible if it is both a left and a right invertible. A multiplicative hyperring R is called a left (right) invertible if every element of R has a left (right) invertible and R is called invertible if it is both a left and a right invertible. Denote the set of all invertible elements in R by U(R) (with respect to the identity e by Ue(R)). Definition 2. Let R be a multiplicative hyperring. We called a ∈ R is regular if there exists x ∈ R such that a ∈ axa. So, we can define that R is regular multiplicative hyperring, if all of elements in R are regular elements. The set of all regular elements in R is denoted by V (R). R. Ameri, A. Kordi / Eur. J. Pure Appl. Math, 9 (2016), 402-418 405 Example 1. Let (R,+, .) be the regular commutative ring with an unitary element. For every sub- set A∈ P∗(R) = P(R)−{;}, |A| ≥ 2, and 1 ∈ A, define a multiplicative hyperring (RA,+,◦), where RA = R and for all x , y ∈ RA, x ◦ y = {xa y|a ∈ A}. Then (RA,+,◦) is a regular multiplicative hyperring. Since, for all a ∈ R, there exists r ∈ R such that a = ara. Now, by setting x = r we have, a ◦ x ◦ a = {asx |s ∈ A} ◦ a = {asx ta|s, t ∈ A} = {axast|s, t ∈ A} = {ast|s, t ∈ A}, since 1 ∈ A, we have a ∈ a ◦ x ◦ a. Hence (RA,+,◦) is regular. Example 2. Let (R,+, .) be a non-zero regular ring and for all a, b ∈ R, define a hyperoperation a ◦ b = {a.b, 2a.b, 3a.b, . . .}. Then (R,+,◦) is a regular multiplicative hyperring, which is not strongly distributive. Also, for all a ∈ R, there exists r ∈ R such that a = ara. Now by setting x = r we have a ◦ x ◦ a = {ar, 2ar, . . . , nar, . . .} ◦ a = {ara, 2ara, . . . , nara, . . .}. Then a ∈ a ◦ x ◦ a. 2.1. Applications of the γ∗-Relation in Regular Multiplicative Hyperrings Let (R,+, .) be a hyperring. We define the relation γ as follows: aγb if and only if {a, b} ⊆ U where U is a finite sum of finite products of elements of R, i.e., aγb⇔∃z1, . . . , zn ∈ R such that {a, b} ⊆ ∑ j∈J ∏ i∈I j zi; I j , J ⊆ {1, . . . , n}. We denote the transitive closure of γ by γ∗. The relation γ∗ as the smallest equivalence relation on a multiplicative hyperring (R,+, .) such that the quotient R/γ∗, the set of all equivalence classes, is a fundamental ring. Let U be the set of all finite sums of products of elements of R we can rewrite the definition of γ∗ on R as follows: aγ∗b⇔∃z1, . . . , zn+1 ∈ R with z1 = a, zn+1 = b and u1, . . . , un ∈ U such that {zi , zi+1} ⊆ ui for i ∈ {1, . . . , n}. Suppose that γ∗(a) is the equivalence class containing a ∈ R. Then, both the sum ⊕ and the product � in R/γ∗ are defined as follows:γ∗(a)⊕γ∗(b) = γ∗(c) for all c ∈ γ∗(a)+γ∗(b) and γ∗(a)�γ∗(b) = γ∗(d) for all d ∈ γ∗(a).γ∗(b). Then R/γ∗ is a ring, which is called fundamental ring of R (see also [31]). Theorem 1. Let R be a regular multiplicative hyperring. Then R/γ∗ is regular ring. Proof. Assume that x ∈ R/γ∗. Thus there is a r ∈ R such that x = γ∗(r). Since R is a regular hyperring, then there exists r ′ ∈ R such that r ∈ r r ′r. So γ∗(r) = γ∗(r r ′r) = γ∗(r)� γ∗(r ′)� γ∗(r). Therefore R/γ∗ is a regular ring. Remark 1. The converse of Theorem 1 is not valid. For example let (R,+, .) be a non regular ring. Consider (R,+, .) as a hyperring under operators "+" and ".". Clearly R/γ∗ ∼= R. R. Ameri, A. Kordi / Eur. J. Pure Appl. Math, 9 (2016), 402-418 406 Definition 3. A multiplicative hyperring R is said multiplicatively n-complete (n-MC) if for all x1, . . . , xn ∈ R we have γ( n ∏ i=1 x i) = n ∏ i=1 x i . Also [7], we called that a hyperring R is n-complete if for all (k1, . . . , kn) ∈ Nn and for all (x1 j , . . . , x iki ) ∈ Rki we have γ( n ∑ i=1 ( ki ∏ j=1 x i j)) = n ∑ i=1 ( ki ∏ j=1 x i j). Theorem 2. Let R be a 3-MC multiplicative hyperring. If R/γ∗ is a regular ring then R is a regular multiplicative hyperring. Proof. Assume that a ∈ R. Then there exists r ∈ R such that γ∗(a)� γ∗(r)� γ∗(a) = γ∗(a). Then γ∗(a) = γ∗(ara). Thus a ∈ γ∗(ara) = ara, since R is 3-MC and hence a ∈ ara. Proposition 1. ([25]) Let R be a commutative ring and a ∈ V (R). Then there is a unique x ∈ R with axa = a and xax = x. Obviously, if R is a regular commutative multiplicative hyperring, then for each a ∈ R there exists a unique element γ∗(x) in R/γ∗ such that γ∗(axa) = γ∗(a) and γ∗(xax) = γ∗(x). Theorem 3. If R is a commutative 3-MC multiplicative hyperring and a ∈ R be a regular element of R. Then there exists x ∈ R such that a ∈ axa and x ∈ xax. Proof. Since R is a commutative multiplicative hyperring then it is easy to check that R/γ∗ is commutative ring. As a ∈ R is a regular element then γ∗(a) is a regular element of R/γ∗, by Theorem 1, then, by Proposition 1, there is a unique γ∗(x) ∈ R/γ∗ for x ∈ R such that γ∗(a)�γ∗(x)� γ∗(a) = γ∗(a) γ∗(x)�γ∗(a)� γ∗(x) = γ∗(x). Then γ∗(axa) = γ∗(a) and γ∗(xax) = γ∗(x). Therefore, a ∈ axa and x ∈ xax , since R is 3-MC. Definition 4. Let R be a multiplicative hyperring. Then we called that Mn(R), as the set of all hypermatrices of R. Also we called that for allA = (Ai j)n×n,B = (Bi j)n×n ∈ P∗(Mn(R)),A ⊆B if and only if Ai j ⊆ Bi j . R. Ameri, A. Kordi / Eur. J. Pure Appl. Math, 9 (2016), 402-418 407 Remark 2. Let R be a multiplicative hyperring with a scalar identity 1 and Mn(R) denotes the set of all n× n matrices with entries in R. It is easy to verify that Mn(R) is a non-commutative multiplicative hyperring with unitary element under usual matrix operations. Let A= (ai j)n×n be a matrix, where ars = ab for some 1 ≤ r, s ≤ n and in other positions ai j = 0. Then A = BC, where B = (bi j)n×n such that brs = a and C = (ci j) such that css = b and in other entries of B, C we have bi j and ci j = 0. Recall that R has a zero absorbing property if for all a ∈ R, 0 ◦ a = a ◦ 0= {0}. Theorem 4. Let (R,+, .) be a multiplicative hyperring such that it has zero absorbing property. Then Mn(R)/γ ∗ ∼= Mn(R/γ ∗). Proof. Consider the projection homomorphism φ : Mn(R)→ Mn(R/γ∗) defined by φ((ai j)n×n) = (γ∗(ai j))n×n for all ai j ∈ R and 1≤ i, j ≤ n. We denote the equivalence relation associated with φ by ρ. That is, (ai j)n×nρ(bi j)n×n⇐⇒(γ∗(ai j))n×n =(γ∗(bi j))n×n, ∀ai j , bi j ∈ R, 1≤ i, j ≤ n. In fact, ρ = ker(φ). Since φ is an epimorphism, we have Mn(R)/ρ = Mn(R)/ker(φ)∼= Mn(R/γ ∗). We know that Mn(R/γ∗) is a ring, and so Mn(R)/ρ is a ring. Thus γ∗ ⊆ ρ, since γ∗ is the smallest equivalence relation on Mn(R) such that Mn(R)/γ∗ is a ring. Let (a′i j)n×nρ(ai j)n×n for all ai j ∈ R and 1≤ i, j ≤ n. Hence (γ∗(a′i j))n×n = (γ ∗(ai j))n×n⇐⇒ γ∗(a′i j) = γ ∗(ai j), ∀ai j ∈ R. Then we conclude that ai j , a′i j ∈ m ∑ s=1 ks ∏ t=1 xst for some (xs1, . . . , xsks ) ∈ Rks and 1≤ i, j ≤ n. Then we have (ai j)n×n, (a′i j)n×n ∈ ( m ∑ s=1 ks ∏ t=1 x i j st)n×n = m ∑ s=1 ( ks ∏ t=1 x i j st)n×n = m ∑ s=1 n ∑ i=1, j=1 Ai j , where Ai j = (bpq)n×n, such that bpq = ¨ ∏ks t=1 x i j st if p = i, q = j, 0 otherwise R. Ameri, A. Kordi / Eur. J. Pure Appl. Math, 9 (2016), 402-418 408 now by the Remark 2, we have Ai j = Bi j(B j j)ks−1 for s = 1, . . . m, where Bi j = (cuv)n×n, where cuv = ¨ x i j st if u= s, v = t, 0 otherwise so we have {(ai j)n×n, (a′i j)n×n} ∈ m ∑ s=1 n ∑ i=1, j=1 Ai j = m ∑ s=1 n ∑ i=1, j=1 Bi j(B j j)ks−1, i.e., (ai j)n×nγ(a′i j)n×n. Hence (ai j)n×nγ ∗(a′i j)n×n. Consequently, (a′i j)n×n ∈ γ∗((ai j)n×n) and therefore ρ ⊆ γ∗. Then γ∗ = ρ and so Mn(R)/γ∗ ∼= Mn(R/γ∗). Theorem 5. Let R be a multiplicative hyperring. R is n-complete (n-MC) if and only if Mn(R) is n-complete (n-MC). Proof. (⇒) Assume that R is n-complete and T = (ai j)n×n ∈ γ( ∑n s=1( ∏ks t=1(xst)n×n)), then {(ai j)n×n, n ∑ s=1 ( ks ∏ t=1 (xst)n×n)} ⊆ n ∑ z=1 ( wz ∏ `=1 (yuv) z` n×n). Now, for convenience, let A= (Ai j)n×n = n ∑ s=1 ( ks ∏ t=1 (xst)n×n) and B = (Bi j)n×n = n ∑ z=1 ( wz ∏ `=1 (yuv) z` n×n), then ai j ∈ Bi j , Ai j ⊆ Bi j , so γ(ai j) = γ(Bi j),γ(Ai j) = γ(Bi j). Since R is n-complete, then ai j ∈ γ(Bi j) = γ(Ai j) = Ai j , i.e., ai j ∈ Ai j for all 1 ≤ i, j ≤ n. Hence (ai j)n×n ∈ A, and so γ( ∑n s=1( ∏ks t=1(xst)n×n)) ⊆ A. (⇐) Suppose that Mn(R) is n-complete. Let x ∈ γ( ∑n i=1( ∏ki j=1 x i j)). Thus {x , n ∑ i=1 ( ki ∏ j=1 x i j)} ⊆ n ∑ s=1 ( ks ∏ t=1 yst). Since Mn(R) is n-complete then x ∈ ∑n i=1( ∏ki j=1 x i j), i.e., R is n-complete. Proposition 2. ([6]) Let R be a strongly distributive multiplicative hyperring and a ∈ R. If there exists x ∈ R and c ∈ axa− a such that c is regular, then a is regular. R. Ameri, A. Kordi / Eur. J. Pure Appl. Math, 9 (2016), 402-418 409 In 1950, Brown and McCoy [3] proved that, R is a regular ring if and only if Mn(R) is regular. Now we can extend it by the following Theorem: Theorem 6. Let R be a strongly distributive multiplicative hyperring. R is regular if and only if Mn(R) is regular multiplicative hyperring. Proof. (⇒) First of all we’ll prove it for n = 2, and then we will extend it to arbitrary n. Assume that A= � a b c d � ∈ M2(R), since b is regular there exists b′ ∈ R such that b ∈ bb′b. If we set, X = � 0 0 b′ 0 � ∈ M2(R) and denote B = AXA− A then by an easy calculation we can see that B′ = � bb′a− a bb′b− b d b′a− c d b′b− d � ⊆ B. Now, we can consider B∗ = � s 0 t u � ∈ B′, where s ∈ bb′a − a, t ∈ d b′a − c, u ∈ d b′b − d. Since s, u are regular there exist s′, u′ ∈ R such that s ∈ ss′s, u ∈ uu′u. If L = � s′ 0 0 u′ � , then by simple calculation we can show that C = B∗LB∗ − B∗ ⊇ � ss′s− s 0 −t + ts′s+ uu′ t uu′u− u � = C ′. Let C∗ = � 0 0 m 0 � ∈ C ′, where m ∈ −t + ts′s+ uu′ t. Since m is regular there exists m′ ∈ R, such that m ∈ mm′m. Finally, if K = � 0 m′ 0 0 � we can see that 0 ∈ C∗KC∗ − C∗. So, C∗ is regular, then by the Proposition 2, B∗ is regular and hence A is regular. Therefore, for n = 2, if R is regular then M2(R) is regular. Since M2(M2(R))∼= M4(R), then M4(R) is regular. Thus by continuing this process we can show that for any positive integer k, M2k(R) is regular. Now, assume that n is an arbitrary positive integer, choose k such that 2k ≥ n. If A∈ Mn(R), let A1 be the matrix of M2k(R) with A in the upper left-hand corner and zeros elsewhere. Assume that A1 ∈ M2k(R), is regular, then there exists an element T = � B C D E � of M2k(R) such that A1 ∈ A1TA1. However, this implies that A∈ ABA and hence A is regular. (⇐) Assume that Mn(R) is regular and a is an arbitrary element of R, then A=      a 0 · · · 0 0 0 · · · 0 ... ... . . . ... 0 0 · · · 0      R. Ameri, A. Kordi / Eur. J. Pure Appl. Math, 9 (2016), 402-418 410 is in Mn(R). Since Mn(R) is regular, then there exists B =    b11 · · · b1n ... . . . ... bn1 · · · bnn    ∈ Mn(R) such that A∈ ABA, this means that a ∈ ab11a, i.e., a is regular. Theorem 7. Let (R,+, .) be a commutative multiplicative hyperring with zero absorbing property. Then R[x]/γ∗ ∼= (R/γ∗)[x]. Proof. Consider the map φ : R[x]→ (R/γ∗)[x] defined by φ( ∑n i=1 ai x i) = ∑n i=1 γ ∗(ai)x i . By [13, Theorem 5.6.5], φ is a projection homomorphism. We denote the equivalence relation associated with φ by ρ. That is, ( n ∑ i=1 ai x i)ρ( n ∑ i=1 bi x i)⇐⇒ n ∑ i=1 γ∗(ai)x i = n ∑ i=1 γ∗(bi)x i , ∀ai , bi ∈ R. Since φ is epimorphism we have R[x]/ρ = R[x]/ker(φ)∼= (R/γ∗)[x]. We know that (R/γ∗)[x] is a ring, then R[x]/ρ is a ring. Since γ∗ is the smallest equivalence relation on R[x] such that R[x]/γ∗ is a ring, then γ∗ ⊆ ρ. Now, let ( ∑n i=1 ai x i)ρ( ∑n i=1 bi x i) for all bi ∈ R. Hence n ∑ i=1 γ∗(ai)x i = n ∑ i=1 γ∗(bi)x i ⇐⇒ γ∗(ai) = γ ∗(bi), ∀ai ∈ R. Thus, {ai , bi} ⊆ ∏m s=1 ts. Therefore, { n ∑ i=1 ai x i , n ∑ i=1 bi x i} ⊆ n ∑ i=1 ( m ∏ s=1 ts)i x i . This means that ( ∑n i=1 ai x i)γ( ∑n i=1 bi x i). Therefore ( ∑n i=1 ai x i)γ∗( ∑n i=1 bi x i). Henceρ ⊆ γ∗, i.e., ρ = γ∗, and so R[x]/γ∗ ∼= (R/γ∗)[x]. Remark 3. If R is a commutative multiplicative hyperring with zero absorbing property and R[x] is regular, then R is regular. But the converse is not true. R. Ameri, A. Kordi / Eur. J. Pure Appl. Math, 9 (2016), 402-418 411 2.2. Some Properties of Regular Multiplicative Hyperrings Definition 5. Let R be a multiplicative hyperring. A subset A of R is idempotent if A ⊆ A2. The set of all idempotent elements of R is denoted by Idem(R). Definition 6. We say that I is a hyperideal of multiplicative hyperring (R,+, .) if it satisfies the following conditions: (1) I − I ⊆ I , (2) ∀x ∈ I , r ∈ R, x r ∪ r x ⊆ I . Definition 7. Let R be a multiplicative hyperring. The element a ∈ R is nilpotent, if there exists an n such that an = {0}. Denote the set of all nilpotent elements of R by nil(R). Definition 8. Let R be a multiplicative hyperring and x ∈ R. Then a left(right) annihilator of x is Ann(x) = {r ∈ R|r x = 0} (Ann(x) = {r ∈ R|x r = 0}). For a non-empty subset B of a multiplicative hyperring R, the annihilator of B is Ann(B) = ∩{Ann(x)|x ∈ B}. Theorem 8. Let R be a commutative multiplicative hyperring with a scalar identity 1, and a ∈ R. Then we have the following statements: (1) If a ∈ ara for r ∈ R, then ar ∈ Idem(R). (2) V (R)∩ nil(R) = {0}. Proof. (1) This is clear. (2) Suppose a ∈ V (R) ∩ nil(R). Then there is a r ∈ R and n ∈ N such that a ∈ ara and an = {0}. Thus we have, a ∈ ara ⊆ (ara)r(ara) = a4r3 = (ara)a2r2 ⊆ . . . ⊆ anr ′ = {0}r ′ = {0}, because 0.r ′ = (1− 1).r ′ ⊆ 1.r ′ − 1.r ′ = {r ′} − {r ′}= {0}. Therefore, a = 0. Theorem 9. Suppose that R is a commutative multiplicative hyperring with a scalar identity 1. Then we have the following statements: (1) If for some u ∈ U(R) and a ∈ R, a ∈ aua, then a ∈ ve for some v ∈ U(R) and e ∈ Idem(R). (2) If ζ= ue for some u ∈ U(R) and e ∈ Idem(R), then for some v ∈ U(R), ζ ⊆ ζvζ. (3) If for a ∈ R, there exists b ∈ R such that ab = 0, with a+ b ∈ U(R), then a is regular. (4) If η = ue for some u ∈ U(R) and e ∈ Idem(R) and |η2| = 1, then there exists ` ∈ η such that `b = 0 for b ∈ R with `+ b ∈ U(R). R. Ameri, A. Kordi / Eur. J. Pure Appl. Math, 9 (2016), 402-418 412 Proof. (1) Assume that a ∈ aua for some u ∈ U(R). Since u is invertible then there is a v ∈ U(R) such that 1 ∈ uv. Let e = au, thus we have e2 = (au)2 = (au)(au) = (aua)u ⊇ au= e, then e ∈ Idem(R). Hence ve = v(au) = a(vu) ⊇ a.1 3 a. (2) Since u ∈ U(R), then there exists v ∈ U(R) such that 1 ∈ vu. Thus we have: e = 1.e ⊆ (vu)e = v(ue) = vζ⇒ e ⊆ e2 ⊆ evζ⇒ ζ= ue ⊆ uevζ= ζvζ, i.e., ζ ⊆ ζvζ for some v ∈ U(R). (3) Let u = a + b. Then au = a(a + b) ⊆ a2 + ab = a2. Since u is invertible, so there is a v ∈ U(R) such that 1 ∈ uv. Therefore, we have: a ∈ a.1 ⊆ a(uv) = (au)v ⊆ a2v = ava. Hence a is regular. (4) Assume that η = ue for some u ∈ U(R) and e ∈ Idem(R). Let τ = u(1 − e). Then ητ = ue(u(1 − e)) ⊆ ueu1 − ueue = ueu − ueue ⊆ ueue − ueue = η2 − η2 = 0, i.e., for all ` ∈ η, b ∈ τ, `b = 0. Now, we need to show that η+ b ∩ U(R) 6= ;. We have 1 ∈ u.1.u ⊆ u(e+ (1− e))u ⊆ (ue+ u(1− e))u= (η+ b)u, i.e., 1 ∈ (η+ b)u, so there exist b ∈ τ,` ∈ η such that 1 ∈ b+ `. Therefore `+ b ∈ U(R). Definition 9. Let R be a multiplicative hyperring. Then R is said to be π-regular if for all a ∈ R there are r ∈ R and an integer n≥ 1, such that an ⊆ anran. Clearly, a regular multiplicative hyperring is π-regular multiplicative hyperring. Also, if a ∈ R is π-regular multiplicative hyperring then for an n≥ 1, an is regular. Theorem 10. Let R be a commutative multiplicative hyperring with a scalar identity 1. Then for a ∈ R, the following hold: (1) a is a π-regular if and only if for some n≥ 1, an is regular. (2) If an ⊆ anran for r ∈ R and n≥ 1, then anr ⊆ Idem(R). (3) If an = ue for some u ∈ U(R) and e ∈ Idem(R) and |a2n|= 1 then there exists ` ∈ an such that ` is π-regular. Proof. It’s straightforward by Theorem 9. For each hyperideal I of multiplicative hyperring R, letting OI = {a ∈ I : a ∈ aI}. Then OI is called a pure part of I . An hyperideal I is called a pure hyperideal if I = OI . Theorem 11. Let R be a commutative multiplicative hyperring with a scalar identity 1 and a ∈ R and also for r ∈ R there exists n ∈ N such that |arn| = 1. If hyperideal < a > is pure hyperideal, then R=< a > +Ann(a). R. Ameri, A. Kordi / Eur. J. Pure Appl. Math, 9 (2016), 402-418 413 Proof. Suppose that< a > is pure hyperideal, then there exists x = sa+ ∑n i=1 a · x i ⊆< a >, such that a ∈ ax , where s ∈ N, r, x i ∈ R. So, there exists ` ∈ x such that a ∈ a`. Thus we have a(1− `) ⊆ a − a` ⊆ a − a`2 ⊆ . . . ⊆ a − a`n = 0, i.e., a(1− `) = 0, which implies that 1− ` ∈ Ann(a). Therefore 1= `+ (1− `) ∈< a > +Ann(a). Hence R=< a > +Ann(a). Theorem 12. Let R be a multiplicative hyperring with a scalar identity 1 and M be a maximal hyperideal of R. Also, for a ∈ M and r ∈ R there exists s ∈ N, such that |ars| = 1. Then we have the following statements: (1) if a ∈ V (R) then for a ∈ M, a ∈ OM , (2) a ∈ OM for a ∈ M if and only if Ann(a) is not contained in M. Proof. (1) Since a ∈ V (R), then for some r ∈ R, a ∈ ara. Now, for a maximal hyperideal M such that a ∈ M , we have a ∈ ara = a(ra) ⊆ aM . Hence a ∈ OM . (2): (⇒) Suppose that a ∈ OM and Ann(a) ⊆ M . Then there is a m ∈ M such that a ∈ am. Thus a(1 − m) ⊆ a − am ⊆ a − am2 ⊆ . . . ⊆ a − ams = 0, i.e., a(1 − m) = 0. Therefore, 1−m ∈ Ann(a) ⊆ M , i.e., 1−m ∈ M and it’s contradiction. Hence Ann(a) 6⊆ M . (⇐) Assume that Ann(a) is not contained in M . Then R = M + Ann(a). So, there exist m ∈ M , x ∈ Ann(a) such that 1 = m + x , then a ∈ a.1 = a(m + x) ⊆ am + ax = am, i.e., a ∈ am. Hence a ∈ OM . 3. Properties of M(R) Definition 10. Let R be a multiplicative hyperring. Denote by M(R) the set of all elements in R such that the generated hyperideals by each of these elements are regular. Clearly, M(R) is a regular hyperring. Lemma 1. If a and b are two regular elements in a commutative multiplicative hyperring R. Then the following statements hold: (1) γ∗(ab) is regular in R/γ∗, (2) Moreover, if |ab|= 1, then so is ab. Proof. (1) Since a, b are regular in R, then there exist r1, r2 ∈ R, such that a ∈ ar1a, b ∈ br2 b. So γ∗(ab) = γ∗(ab)� γ∗(r1r2)� γ∗(ab). Hence γ∗(ab) is regular in R/γ∗. (2) By definition of regular element, there exist r1, r2 ∈ R, such that a ∈ ar1a and b ∈ br2 b. Since |ab|= 1, we have ab ∈ (ar1a)(br2 b) = (ab)r1r2(ab). Theorem 13. Let R be a strongly distributive multiplicative hyperring and a ∈ R. Thus we have the following statements: R. Ameri, A. Kordi / Eur. J. Pure Appl. Math, 9 (2016), 402-418 414 (1) If there is a regular element c in a2 − a, then a and 1− a are regular. (2) If a and 1− a are regular in R such that |a(1− a)|= 1, then so is a(1− a). Proof. (1) It immediately follows by Proposition 2. (2) By Lemma 1(2) it is clear. Theorem 14. If R be a strongly distributive multiplicative hyperring, then a right hyperideal I in the hyperring M(R) is a right hyperideal in R. Proof. Suppose a ∈ I , r ∈ R, then ar ⊆ M(R), hence for some element r ′ ∈ R, ar ⊆ arr ′ar. But r r ′ar ⊆ M(R), so ar ⊆ I . Thus I is a right hyperideal in R. Theorem 15. Let R be a strongly distributive multiplicative hyperring. Then M(R) is a hyperideal of R. Proof. Let z ∈ M(R) and r ∈ R. Since < zr >⊆< z > and < rz >⊆< z >, we have zr ⊆ M(R) and rz ⊆ M(R), thus zr ∪ rz ⊆ M(R). Now, assume that t1, t2 ∈ M(R). We need to prove that all of elements in < t1 − t2 > is regular. For achieving to it, let a ∈< t1 − t2 >. Then for some u ∈< t1 > and v ∈< t2 > we have a = u− v. As < t1 > is regular then there exists r ∈ R such that u ∈ uru. Thus by distributive property of R we have ara =(u− v)r(u− v)− u+ v =uru− urv − vru+ vrv − u+ v ⊆uru− uru+ v − urv − vru+ vrv =u(r − r)u+ v − urv − vru+ vrv =u0u+ v − urv − vru+ vrv =u(v − v)u+ v − urv − vru+ vrv =uvu− uvu+ v − urv − vru+ vrv, since the right side is in < t2 >, then ara − a ⊆< t2 >, i.e., ara − a is regular and by Propo- sition 2, a is regular. Hence < t1 − t2 > is a regular hyperideal and t1 − t2 ∈ M(R), i.e., M(R)−M(R) ⊆ M(R). Let (R,+, .) be a multiplicative hyperring and I be a hyperideal of it. We consider the usual addition of cosets and the multiplication defined as follows: (a+ I) ∗ (b+ I) = {c + I |c ∈ a.b}. On the set R/I = {r + I |r ∈ R} of all cosets of I . Then (R/I ,+,∗) is a multiplicative hyperring. here here here Proposition 3. ([19]) If (R,+, .) is a strongly distributive multiplicative hyperring and I is a hyperideal of R, then (R/I ,+,∗) is ring. R. Ameri, A. Kordi / Eur. J. Pure Appl. Math, 9 (2016), 402-418 415 Corollary 1. If (R,+, .) is a strongly distributive multiplicative hyperring, then (R/M(R),+,∗) is ring. Theorem 16. If (R,+, .) is a strongly distributive multiplicative hyperring, then M(R/M(R)) = {0}. Proof. Assume that a + M(R) denote the residue class modulo M(R) which contains the element a of R. If b + M(R) ∈ M(R/M(R)) and a ∈< b >, then a + M(R) ∈< b + M(R) >. Since < b + M(R) > is a regular ideal in R/M(R), then a + M(R) is regular. Thus, for some x+M(R) ∈ R/M(R), a+M(R) = (a+M(R))∗(x+M(R))∗(a+M(R)) or a+M(R) = axa+M(R), i.e., axa−a ⊆ M(R). Therefore axa−a is regular element of strongly distributive multiplicative hyperring R and by Proposition 2, a is regular. Thus < b > is regular hyperideal and hence b ∈ M(R). So, b+M(R) = 0R/M(R). Theorem 17. Let B be a hyperideal of a strongly distributive multiplicative hyperring R. Then M(B) = B ∩M(R). Proof. Assume that B is a hyperideal of R and let b be a element of B such that generated a regular hyperideal < b >∗ in the strongly distributive multiplicative hyperring B. Suppose < b > be the hyperideal in R generated by the element b, and let c = nb+ r b+ bs+Σri bsi , where n ∈ Z, r, s, ri , si ∈ R be any element of < b >. Since b is regular in B we have for some x ∈ B, b ∈ bx b. Hence c ⊆nb+ r(bx b) + (bx b)s+Σri(bx b)si =nb+ r(bx b) + (bx b)s+Σri(bx bx b)si =nb+ r(bx b) + (bx b)s+Σ(ri bx)b(x bsi), and thus c ⊆< b >∗. As < b > coincides with < b >∗, then < b > is regular. Hence, for b ∈ M(B) we have b ∈ B ∩M(R), i.e., M(B) ⊆ B ∩M(R). Conversely, let b ∈ B ∩M(R), then b ∈ B, b ∈ M(R). As b is regular in R, then there exists r ∈ R such that b ∈ br b and since B is a hyperideal and b ∈ B we have r b ⊆ B then b ∈ br b ⊆ b(r br)b, where r br ⊆ B. It shows that b is regular in B. Therefore B ∩ M(R) is regular hyperideal in the strongly distributive multiplicative hyperring B. Thus B ∩M(R) ⊆ M(B). This completes the proof. Theorem 18. If R is a multiplicative hyperring and it has zero absorbing property, then M(R)∩ Ann(M(R)) = {0}. Proof. For all a ∈ M(R)∩Ann(M(R)) we have a ∈ M(R), a ∈ Ann(M(R)). As a ∈ M(R) thus there exists r ∈ R such that a ∈ ara. Since ra ⊆ M(R), we have a ∈ a(ra) = {0}. Therefore a = 0. Proposition 4. ([19]) Every strongly distributive hyperring (R,+, .) with a scalar identity 1 is a ring. REFERENCES 416 Theorem 19. Let R be a multiplicative hyperring and I be a strongly distributive hyperideal of R such that it has a scalar identity e and for all x ∈ R, |xe|= 1, then R= I + Ann(I). Proof. Let x be an arbitrary element of R, then xe∪ex ⊆ I . So (xe)e = e(xe), i.e., xe = exe and similarly ex = exe. Thus xe = ex and e is in center of R. As, by Proposition 4, ex ∈ I and for all y ∈ I , y(x−ex) = ye(x−ex) = y(ex−e2 x) = y(ex−ex) = y0= 0, i.e., x−ex ∈ Ann(I). Thus we have, x = ex + (x − ex). Hence for all x ∈ R as a sum of elements ex ∈ I and x − ex ∈ I of Ann(I). This completes the proof. The following example show that under Theorem 19, R is not a ring: Example 3. Let R= Z ⊕ Z and define (a, b) ◦ (c, d) = ¨ (ac,Z) bd 6= 0 (ac, 0) bd = 0 Then (R,+,◦) is a multiplicative hyperring such that is not strongly distributive, because by con- sidering a = (1,1), b = (0,2), c = (0,−2), we have a ◦ (b + c) = (1, 1) ◦ (0, 0) = (0,0) but a◦ b+a◦ c = (0,Z)+(0,Z) = (0,Z+Z). Now, let I = (Z, 0) be a strongly distributive hyperideal of R and e = (1, 0) is scalar identity element of I such that for all r ∈ R we have |xe| = 1. It is clear that Ann(I) = (0,Z) and R= I + Ann(I) but (R,+,◦) is not a ring. Corollary 2. ([7]) If an ideal I in a ring R has a unit element e, then R= I + Ann(I). 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