# Some Properties of MultiplicativeHv-Rings of Polynomials over Multiplicative Hyperrings

Authors: Utpal Dasgupta
Journal: Algebra
 (2014)

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## Abstract

The set of all polynomialsRx, over a multiplicative hyperring R,+,·, form a commutative group with respect to the component-wise addition (+) of the polynomials. For polynomials f, g in Rx, f*g is a set of polynomials whose k+1th components k∈N∪0 are chosen from the set ∑i+j=kai·bj, where ai and bj are the i+1th and the j+1th components of f and g, respectively. A multiplicative hyperring is polynomially structured if the hyperstructure Rx,+,* is a multiplicative Hv-ring. The purpose of the paper is to study the properties of the multiplicative Hv-ring Rx,+,*, corresponding to those of a polynomially structured multiplicative hyperring R.


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## Body

## 1. Introduction

H
v-structures [1] are introduced by Vougiouklis, at the Fourth AHA congress in the year of 1990. Since then, the study of Hv-structure theory has been approached in several directions by many researchers (see [2–5]). The essence of the notion of Hv-structures is to generalize the well-known algebraic hyperstructures (such as hypergroup, hyperring, and hypermodule), simply by replacing some or all axioms of the respective hyperstructures by the corresponding weak axioms. The Hv-structure of our initial concern is multiplicative Hv-ring, studied in [6, 7], which is a commutative group (R,+) along with a hyperoperation ∘ such that (i) (R,∘) is an Hv-semigroup [3, 8] (i.e., a hyperstructure in which ∘ is weak associative in the sense that a∘(b∘c)∩(a∘b)∘c≠ϕ, for all a,b,c∈R) and (ii) ∘ is weak distributive with respect to + (i.e., a∘(b+c)∩(a∘b+a∘c)≠ϕ and (a+b)∘c∩(a∘c+b∘c)≠ϕ, for all a,b,c∈R). A multiplicative Hv-ring (R,+,∘) is commutative if a∘b=b∘a, for all a,b∈R. The identity element 0R of the group (R,+) is said to be absorbing in the multiplicative Hv-ring (R,+,∘) if a∘0R=0R∘a={0R}, for all a∈R. A nonempty finite subset E={e1,e2,…,en} of a multiplicative Hv-ring (R,+,∘) is called anidentity set(or i-set, in short) [9] of R if (i) ei∉〈0R〉 for at least one i=1,2,…,n and (ii) for any a∈R, a∈(∑i=1n‍ei∘a)∩(∑i=1n‍a∘ei). An element e of R is called a hyperidentity of R if the set {e} is an i-set of R.Unlike a ring, the equality of the set-expressions(-a)∘b, a∘(-b), and -(a∘b) does not hold in general on a multiplicative Hv-ring (R,+,∘) for any a,b∈R. In fact, if (Z,+,·) is the ring of integers and if ∘ is a hyperoperation on Z, defined by a∘b={a+b,-a-b,a,b}, for all a,b∈Z, then (Z,+,∘) is a commutative multiplicative Hv-ring, in which 1∘(-2)≠(-1)∘2, (-1)∘2≠-(1∘2), and -(1∘2)≠1∘(-2).A multiplicativeHv-ring (R,+,∘) is said to satisfy the condition (R) if the set equality (-a)∘b=a∘(-b)=-(a∘b) (called the condition (R) [9]) holds true for any two elements a and b of R. Let (R,+,·) be a ring and ∘ a hyperoperation on R, defined by a∘b={a,b,-a,-b}, for all a,b∈R. Then, (R,+,∘) is a multiplicative Hv-ring with condition (R).We consider now anHv-structure (R,+,∘) in which (i) (R,+) is a commutative group, (ii) (R,∘) is an Hv-semigroup, and (iii) ∘ is semidistributive across the operation + (i.e., a∘(b+c)⊆a∘b+a∘c and (a+b)∘c⊆a∘c+b∘c, for all a,b,c∈R). This Hv-structure is clearly a multiplicative Hv-ring and we thus call it a semidistributive multiplicative Hv-ring. Henceforth, throughout the paper, a multiplicative Hv-ring wherever considered will always be assumed to be a semidistributive multiplicative Hv-ring with the condition (R).In 1982, the notion of multiplicative hyperring is inducted in the study on hyperring theory by Rota, which is subsequently investigated in [10–14]. A commutative group (R,+) endowed with anassociative hyperoperation ∘ is called a multiplicative hyperring [15] if (i) ∘ is semidistributive across the operation + on R and (ii) ∘ satisfies the condition (R) for elements in R. An associative hyperoperation is eventually weakly associative and thus a multiplicative hyperring is, eventually, a semidistributive multiplicative Hv-ring with condition (R).Procesi Ciampi and Rota define in [16] polynomials over multiplicative hyperring as follows: let (R,+,·) be a multiplicative hyperring with absorbing zero 0R and let x be any symbol out of R. Then, a polynomial in x is an expression of the form f(x)=a0x0+a1x1+a2x2+⋯=∑‍akxk, (k∈N, ak∈R), in which + is a connective and only a finite number of the ak’s (called the coefficients of xk in f(x)) are different from zero (0R) of R. The degree of a polynomial f(x)=∑‍akxk (in short deg⁡.f(x)) is a nonnegative integer n such that an≠0R and ak=0R, for allk>n. A polynomial f(x)=∑‍akxk over a multiplicative hyperring R will be written as f(x)=∑k=1nakxk when and only when deg.f(x)=n. For an integer m∈N and any a,ak,b∈R(a≠0R, b≠0R,  k=1,2,…,m-1), we write the polynomials a(x)=0Rx0+⋯+0Rxm-1+axm and f(x)=ax0+a1x1+⋯+am-1xm-1+bxm simply as axm and a+a1x+⋯+am-1xm-1+bxm, respectively. Denote by R[x] the set of all polynomials in x over R and define on R[x] a binary operation + and a hyperoperation * as follows: for any two polynomials f(x)=∑‍akxk and g(x)=∑‍bkxk, from R[x], f(x)+g(x)=∑‍(ak+bk)xk and f(x)*g(x)={∑‍ckxk:ck∈∑i+j=k‍aibj}, where, for any a,b∈R, the juxtaposition ab means the set a·b. The purpose of the present paper is to study the properties of the hyperstructure (R[x],+,*), in connection to those of a particular class of multiplicative hyperrings, called polynomially structured multiplicative hyperrings which we describe formally in the following section.


## 2. Polynomially Structured Multiplicative Hyperring

It is asserted in [16] that for a multiplicative hyperring R, the hyperstructure (R[x],+,*) is always a multiplicative hyperring. But we note here that, given a multiplicative hyperring R with absorbing zero, the hyperoperation * (as is defined in Section 1) does not necessarily induce a multiplicative hyperring structure over the group of polynomials (R[x],+). In fact, we have the following example.Example 1.
Let(Z,+,·) be the ring of integers and (Z,+,∘) the multiplicative hyperring where, for any a,b∈Z, a∘b={2ab,3ab}  (denoting the product a·b of elements in the ring (Z,+,·) simply by the juxtaposition ab). Consider three polynomials f(x)=1+2x, g(x)=2+1x, and h(x)=1+3x over the multiplicative hyperring (Z,+,∘).
Then, the set of coefficients ofx in the polynomials belonging to (f(x)*g(x))*h(x) is M={4,6}∘3+{10,11,14,15}∘1. Again, the set of coefficients of x in the polynomials belonging to f(x)*(g(x)*h(x)) is N=1∘{14,15,20,21}+2∘{4,6}. By a tedious but routine calculation, one can see that {57,74,82}⊆M, whereas {57,74,82}∩N=ϕ. So, there are some polynomials in (f(x)*g(x))*h(x), the coefficient of x in each of which is an element of the set {57,74,82}. These polynomials do not belong to f(x)*(g(x)*h(x))  (since {57,74,82}∩N=ϕ). Thus, (f(x)*g(x))*h(x)⊈f(x)*(g(x)*h(x)). So, there is no question of claiming (Z[x],+,*) to be a multiplicative hyperring.However, we observe that it is possible to construct a multiplicative hyperringR corresponding to which the hyperstructure (R[x],+,*) turns out to be a multiplicative Hv-ring, if not a multiplicative hyperring at all.Let us consider a commutative group(G,+). Suppose that ϕ≠A⊆Hom(G,Hom(G)) is such that, for any a,b∈A and x,y∈G, (xa(y))b=xayb, where xa denotes the image of x∈G under a∈A and xayb is simply the mapping composition. Define a hyperoperation ∘A on G by stating that x∘Ay={xa(y):a∈A}. Then we have the following.Lemma 2.
(
G
,
+
,
∘
A
) is a multiplicative hyperring with absorbing zero.Proof.
Letx,y,z∈G. Then, p∈x∘A(y∘Az)⇒p∈x∘As  (for some s∈y∘Az)⇒p=xa(s) and s=yb(z)  (for some a,b∈A)⇒p=xa(yb(z))=(xayb)(z)=(xa(y))b(z)⇒p∈(xa(y))∘Az ⇒ p∈(x∘Ay)∘Az (since xa(y)∈x∘Ay). So, x∘A(y∘Az)⊆(x∘Ay)∘Az. The reverse inclusion can also be shown to be true by adopting similar arguments. Hence, x∘A(y∘Az)=(x∘Ay)∘Az. Now, p∈x∘A(y+z)⇒p∈xa(y+z) (for some a∈A) =xa(y)+xa(z)∈x∘Ay+xa∘Az. So, x∘A(y+z)⊆x∘Ay+xa∘Az. Again, p∈(y+z)∘Ax⇒p∈(y+z)ax (for some a∈A) =(ya+za)x (since a∈Hom(G,Hom(G))) =ya(x)+za(x)⊆y∘Ax+z∘Ax. So, (y+z)∘Ax⊆y∘Ax+z∘Ax. Moreover, x∘A(-y)={xa(-y):a∈A}={-(xa(y)):a∈A} (since xa∈Hom(G)) =-{xa(y):a∈A}=-(x∘Ay) and (-x)∘Ay={(-x)a(y):a∈A}={(-(xa))(y):a∈A} (since a∈Hom(G,Hom(G))) ={xa(-y):a∈A}=x∘A(-y). Thus, (G,+,∘A) is a multiplicative hyperring. Finally, if 0G denotes the identity element of the group (G,+), then x∘0G={xa(0G):a∈A}={0G} (since for a∈A, xa∈Hom(G)) and also 0G∘x={(0G)a(x):a∈A}={0G} (since, for a∈Hom(G,Hom(G)),(0G)a is the zero homomorphism from G to G). Thus, 0G is absorbing in the multiplicative hyperring (G,+,∘A).
The multiplicative hyperring(G,+,∘A), defined in Lemma 2, is called a multiplicative A-hyperring (of course if such a set A exists for the group (G,+)). That a multiplicative A-hyperring exists is evident in the following example.Example 3.
Let(R,+,·) be a ring. Then, as is shown in [9, 13], (R,+,P*) is a multiplicative hyperring with absorbing zero, where P∈P(R) with |P|≥2 and P* is the P-hyperoperation [8, 17] on the semigroup (R,·); that is, xP*y={x·a·y:a∈P} for all x,y∈R. Now, for each a∈P and x∈R, we define a mapping xa:R→R by stating that xa(y)=x·a·y, for all y∈R. Then xa∈Hom(R,+). Thus, corresponding to each a∈P, we have a mapping a:(R,+)→Hom(R,+), given by a(x)=xa, for any x∈R. Then, P⊆Hom(R,Hom(R)). Moreover, for any x,y,z∈R and a,b∈P, (xa(y))b(z)=(x·a·y)·b·z=x·a·(y·b·z)=x·a·(yb(z))=xa(yb(z))=(xayb)(z); that is, (xa(y))b=xayb. Thus, the hyperoperation ∘A  (for A=P) is defined on the group (R,+). Note that, for any x,y∈R, x∘Ay={xa(y):a∈A}={x·a·y:a∈P}=aP*y. Thus, (R,+,P*) is a multiplicative A-hyperring, for A=P.Proposition 4.
For a multiplicativeA-hyperring (R,+,∘A), the hyperoperation * induces a multiplicative Hv-ring structure on the group (R[x],+) of polynomials over R.Proof.
Letf(x)=∑‍akxk, g(x)=∑‍bkxk, and h(x)=∑‍dkxk be three polynomials in R[x]. Then, (f(x)*g(x))*h(x)={∑‍tkxk:tk∈∑u+v=k‍(∑i+j=u‍ai∘Abj)∘Adv} and f(x)*(g(x)*h(x))={∑‍skxk:sk∈∑u+v=k‍au∘A(∑i+j=v‍bi∘Adj)}. Now, we choose and fix an element α∈A. Then, for each k∈N0,
(1)pk=∑u+v=k‍(∑i+j=u‍(ai)α(bj))α(dv)∈∑u+v=k‍(∑i+j=u‍ai∘Abj)∘Adv,qk=∑u+v=k‍(au)α(∑i+j=v‍(bi)α(dj))∈∑u+v=k‍au∘A(∑i+j=v‍bi∘Adj).
Again,pk=∑u+v=k‍(∑i+j=u‍(ai)α(bj))α(dv)=∑u+v=k∑i+j=u‍((ai)α(bj))α(dv)  (since α is a homomorphism) = ∑u+v=k‍∑i+j=u‍((ai)α(bj)α)(dv) (since (xα(y))α=xαyα), qk=∑u+v=k‍(au)α(∑i+j=v‍(bi)α(dj))=∑u+v=k‍∑i+j=v‍(au)α((bi)α(dj)) (since (au)α is a homomorphism) = ∑u+v=k‍∑i+j=v‍((au)α(bi)α)(dj). Clearly then, for each k∈N0,
(2)pk=qk∈(∑u+v=k‍(∑i+j=u‍ai∘Abj)∘Adv)⋂‍(∑u+v=k‍au∘A(∑i+j=v‍bi∘Adj)).
Hence, (f(x)*g(x))*h(x)∩f(x)*(g(x)*h(x))≠ϕ. It is shown in [16] that, for any multiplicative hyperring R the hyperoperation * defined on R[x] is semidistributive over the operation + on R[x] and also satisfies the condition (R) for any two polynomials in R[x]. Thus, for the multiplicative A-hyperring (R,+,∘A), the hyperstructure (R[x],+,*) is a multiplicative Hv-ring (with absorbing zero 0R[x]=0xR0+0xR1+0xR2+⋯).
We call a multiplicative hyperring(R,+,·) with absorbing zero polynomially structured if (R[x],+,*) is a multiplicative Hv-ring. The class of multiplicative A-hyperrings is a subclass of the class of polynomially structured multiplicative hyperrings (by Proposition 4). Throughout the rest of the paper, R will stand for a polynomially structured multiplicative hyperring.


## 3. Polynomials over Integral Hyperrings

An elementa(≠0R) of a multiplicative Hv-ring (S,+,∘) is a left (resp., right) divisor of zero in S if there exists b∈S* (resp., c∈S*=S∖{0S}) such that 0S∈a∘b (resp., 0s∈c∘a) and adivisor of zero in S if it is either a left or a right divisor of zero in S. An element a(≠0S) of S is a left (resp., right) strong divisor of zero in S if there exists b∈S*=S∖{0S} (resp., c∈S*) such that a∘b={0S} (resp., c∘a={0S}) and astrong divisor of zero in S if it is either a left or a right strong divisor of zero in S.Definition 5.
A multiplicativeHv-ring is called an integral Hv-ring if there is no strong divisor of zero in it. A commutative integral Hv-ring is an Hv-domain. A strong integral Hv-ring is a multiplicative Hv-ring in which there is no divisor of zero. A strong Hv-domain is a commutative strong integral Hv-ring. We call an integral Hv-ring (resp., an Hv-domain)  S simply an integral hyperring (resp., a hyperdomain) [13], when the Hv-ring S is a multiplicative hyperring.Before entering into the study of the multiplicativeHv-ring R[x] of polynomials over integral hyperring and hyperdomain, let us go through the following useful observations.Remark 6.
(a) For any polynomially structured multiplicative hyperringR, the identity element 0R[x]=0Rx0+0Rx1+0Rx2+⋯ of the group (R[x],+) is absorbing in the multiplicative Hv-ring (R[x],+,*).
(b)R[x] is commutative if R is a commutative multiplicative hyperring.
(c) The mappingψ:R→R[x], defined by, for all r∈R*, ψ(r)=rx0 and ψ(0R)=0R[x], is a strong monomorphism. In fact, for any r,s∈R, ψ(rs)={ψ(c):c∈rs}={cx0:c∈rs}=(rx0)*(sx0)=ψ(r)*ψ(s). Thus R can be identified with its isomorphic image in R[x] and for any r∈R*, we can write the polynomial rx0 simply as r and the zero polynomial 0R[x] as 0R.
(d) IfE={e1,e2,…,en} is an i-set of R, then the set Ex⊆R[x] is an i-set in the multiplicative Hv-ring R[x], where, for any A∈P*(R), Ax denotes the set {ax0:a∈A}. In fact, for any f(x)=∑‍akxk∈R[x], we have that ∑i=1n‍ei*f(x)=∑i=1n‍(ei*∑‍akxk)=∑i=1n‍{∑‍aikxk:aik∈eiak}={∑‍(∑i=1n‍aik)xk:aik∈eiak} = {∑‍bkxk:bk∈∑i=1n‍eiak}. Then, since ak∈∑i=1n‍eiak, we have that f(x)∈∑i=1n‍ei*f(x), identifying eix0 with ei. Similarly, one can see that f(x)∈∑i=1n‍f(x)*ei.
On the other hand, if, for somen∈N, ER[x]={ϵi(x)=∑‍ei,kxk∈R[x]:i=1,2,…,n} is an i-set in R[x], then E={ei,0:i=1,2,…,n} is an i-set in R. In fact, for any a∈R, ax0∈∑i=1n‍(ax0)*ϵi(x)=∑i=1n‍{∑‍ai,kxk∈R[x]:ai,k∈aei,k}={∑‍(∑i=1n‍ai,k)xk∈R[x]:ai,k∈aei,k} = {∑‍akxk∈R[x]:ak∈∑i=1n‍aei,k}⇒a∈∑i=1n‍aei,0. Similarly, from ax0∈∑i=1n‍ϵi(x)*(ax0), one may arrive at a∈∑i=1n‍ei,0a.This is clear from Remark6 (d) that, for any hyperidentity e of R, the polynomial ex0∈R[x] is a hyperidentity in the multiplicative Hv-ring R[x]. Is every hyperidentity of R[x] of the form ex0 for some hyperidentity e of R? Following is an example of a multiplicative hyperring R such that R[x] has a hyperidentity ϵ(x)≠ex0 for any hyperidentity e of R.Example 7.
Let(Z,+,·) be the ring of integers and P={0,1}. Then (Z,+,P*) is a commutative polynomially structured multiplicative hyperring (as is shown in Example 3). Denote the multiplicative hyperring (Z,+,P*) by ZP. Consider a polynomial 1+1x∈ZP[x]. Then, for any f(x)=∑‍akxk∈ZP[x], we see that (1+1x)*f(x)={∑‍bkxk:b0∈1P*a0 and bk∈1P*ak+1P*ak-1 (for k∈N)}. Clearly, a0∈1P*a0 and for any k∈N, ak∈1P*ak+1P*ak-1  (since, 0∈1P*ak-1). Thus, f(x)∈(1+1x)*f(x) and so 1+1x is a hyperidentity in ZP[x], which is not in the form ex0 for any hyperidentity e of ZP.Remark 8.
Letϵ(x)=∑‍ekxk∈R[x] be a hyperidentity in the multiplicative Hv-ring R[x]. Then, from Remark 6 (d), e0 is a hyperidentity in the multiplicative hyperring R.Proposition 9.
LetR be a strong integral hyperring. Then every hyperidentity in the multiplicative Hv-ring R[x] is of the form ex0, for some hyperidentity e of R.Proof.
Suppose thatϵ(x)=∑‍ekxk∈R[x] is a hyperidentity in the multiplicative Hv-ring R[x]. Then, by the Remark 8, e=e0 is a hyperidentity in R. Now, let a∈R∖{0R} be arbitrary and f(x)=∑‍akxk∈R[x], where a0=a and ak=0R, for all k∈N. Then, f(x)∈ϵ(x)*f(x)∩f(x)*ϵ(x), whereby 0R∈eka∩aek, for all k∈N, whence ek=0R for all k∈N (since R is a strong integral hyperring and a≠0R). Thus, ϵ(x)=ex0+0Rx1+0Rx2+⋯, where e=e0 is a hyperidentity in R.Definition 10.
If0R[x]≠f(x)=∑‍akxk∈R[x], then the smallest integer n such that an≠0R is called the order of f(x) and is denoted by ordf(x). The order of 0R[x] is defined to be zero. For a nonempty set A∈P(R[x]), the smallest element in the set {ordf(x):f(x)∈A} does exist and is called the order of A, being denoted by ordA; that is, ordA=min.{ordf(x):f(x)∈A}.In the next proposition we will find some properties ofordA, for some A∈P*(R[x])(=P(R[x])∖{ϕ)}. For that, it is necessary, at this point, to frame some suitable notations corresponding to different types of hyperproducts of elements in the multiplicative Hv-ring R[x]. Indeed, in any multiplicative Hv-ring (S,+,∘) (which is not a multiplicative hyperring), the expression like s1∘s2∘s3∘⋯∘sn (si∈S,n≥3) bears no connotation in S, unless the parentheses “(” and “)” are meaningfully inserted in. Note that the following two expressions(3)s1∘(1s2∘(2s3∘(3s4∘⋯∘(n-3sn-2∘(n-2sn-1∘sn)n-2)n-3⋯)3)2)1,(1(2(3⋯(n-3(n-2s1∘s2)n-2∘s3)n-3∘⋯∘sn-3)3∘sn-2)2∘sn-1)1∘snare meaningful, called the finite hyperproducts of type l and type r and written in notations, respectively, as [s1∘s2∘⋯∘sn]l and [s1∘s2∘⋯∘sn]r.Proposition 11.
(i) For anyf(x),g(x)∈R[x]∖{0R[x]},
(4)either0R[x]∈f(x)*g(x)or,ord(f(x)*g(x))≥ordf(x)+ordg(x).
(ii) IfR is a strong hyperdomain, then the implication that
(5)fi(x)∈R[x]∖{0R[x]}⟹∑i=1n‍ordfi(x)=ord([f1(x)*f2(x)*⋯*fn(x)]r)=ordg(x),
for all g(x)∈[f1(x)*f2(x)*⋯*fn(x)]r holds true for any n∈N with n≥2.Proof.
(i) Letordf(x)=n and ordg(x)=m. Then, f(x)=∑‍akxk, where an≠0R,  ak=0R for k<n and g(x)=∑‍bkxk, where bm≠0R,  bk=0R for k<m. So, for any ∑‍ckxk∈f(x)*g(x), we see that, whenever k<n+m,  ck∈∑i+j=k‍aibj={0R} (since 0R is absorbing in R and ak=0R(k<n), bk=0R(k<m)). Thus, if 0R[x]∉f(x)*g(x), then for any h(x)=∑‍ckxk∈f(x)*g(x), we have that ordh(x)≥n+m. So, ord(f(x)*g(x))≥n+m=ordf(x)+ordg(x).
(ii) Now suppose thatR is a strong hyperdomain with absorbing zero. Consider two polynomials f(x),g(x)∈R[x]∖{0R[x]}. Let ordf(x)=p and ordg(x)=m. Then, f(x)=∑‍akxk, where ap≠0R, ak=0R for k<p and g(x)=∑‍bkxk, where bm≠0R, bk=0R for k<m. So, 0R∉apbm (since R is a strong hyperdomain) and also apbm=∑i+j=p+m‍aibj (since ak=0R for k<p and bk=0R for k<m). Thus, for any h(x)=∑‍ckxk∈f(x)*g(x), cp+m≠0R (since cp+m∈∑i+j=p+maibj‍=apbm) and ck=0R, for all k with 0≤k<p+m. Thus, ordh(x)=p+m for any h(x)∈f(x)*g(x) and so ord(f(x)*g(x))=p+m. Hence, ordf(x)+ordg(x)=ord(f(x)*g(x))=ordh(x), for all h(x)∈f(x)*g(x). So, the implication is true for n=2 (noting that [f(x)*g(x)]r=f(x)*g(x)). Suppose that, for some integer k≥2, the implication holds true for each value of n ranging from 2 to k-1 and take any fi(x)∈R[x]∖{0R[x]} for i=1,2,…,k. Then, ∑i=1k-1‍ordfi(x)=ord([f1(x)*f2(x)*⋯*fk-1(x)]r)=ordh(x), for all h(x)∈[f1(x)*f2(x)*⋯*fk-1(x)]r. Now, let g(x)∈[f1(x)*f2(x)*⋯*fk(x)]r be arbitrary. Then, g(x)∈h(x)*fk(x) for some h(x)∈[f1(x)*f2(x)*⋯*fk-1(x)]r. So, ordg(x)=ord(h(x)*fk(x))=ordh(x)+ordfk(x)=∑i=1k-1‍ordfi(x)+ordfk(x)(since h(x)∈[f1(x)*f2(x)*⋯*fk-1(x)]r)=∑i=1k‍ordfi(x). Thus, ∑i=1k‍ordfi(x)=ord([f1(x)*f2(x)*⋯*fk(x)]r)=ordg(x), for all g(x)∈[f1(x)*f2(x)*⋯*fk(x)]r.
Hence, by strong induction the implication follows for anyn∈N with n≥2.Corollary 12.
IfR is a strong hyperdomain then the implication
(6)fi(x)∈R[x]∖{0R[x]}⟹∑i=1n‍ordfi(x)=ord([f1(x)*f2(x)*⋯*fn(x)]l)=ordg(x),
for all g(x)∈[f1(x)*f2(x)*⋯*fn(x)]l holds true for any n∈N with n≥2.Proof.
Since the multiplicative hyperringR is a hyperdomain, it is commutative and so R[x] is also a commutative multiplicative Hv-ring. Hence, for any fi(x)∈R[x]∖{0R[x]} (i=1,2,…,k), we have that [f1(x)*f2(x)*f3(x)*⋯*fk(x)]l=[fk(x)*fk-1(x)*⋯*f2(x)*f1(x)]r. Hence, the assertion follows straight from Proposition 11.Proposition 13.
If the multiplicative hyperringR is a strong hyperdomain, the multiplicative Hv-ring R[x] is a strong Hv-domain.Proof.
R being a strong hyperdomain is a commutative multiplicative hyperring. Thus R[x] is a commutative multiplicative Hv-ring. Again, since 0R is absorbing in R, 0R[x] is also absorbing in R[x]. Thus we take f(x),g(x)∈R[x]∖{0R[x]}. Then, by Proposition 11, ord(f(x)*g(x))=ordf(x)+ordg(x)>0. Hence 0R[x]∉f(x)*g(x). So, R[x] is a strong Hv-domain.


## 4.C-Ideals in R[x]

A subgroupI of the group (S,+) is called a left (resp., right) Hv-ideal of a multiplicative Hv-ring (S,+,∘) if, for any s∈S and a∈I, s∘a⊆I (resp., a∘s⊆I). I is an Hv-ideal of S if it is both a left and a right Hv-ideal of S.We call anHv-ideal of a multiplicative Hv-ring S simply a hyperideal when S is a multiplicative hyperring. The notion of a typical hyperideal in a multiplicative hyperring, called C-ideal, is introduced in [18] to study prime and primary hyperideals of multiplicative hyperrings. A hyperideal I of a multiplicative hyperring S is a C-ideal if for any A∈C,  A∩I≠ϕ⇒A⊆I, where C={s1∘s2∘⋯∘sn:si∈S, n∈N}⊆P*(S). Following is the definition of a C-ideal in an arbitrary multiplicative Hv-ring.Definition 14.
A left (resp., right)Hv-ideal I of a multiplicative Hv-ring (S,+,∘) is called a left (resp., right)  C-ideal if for any type l hyperproduct [s1∘s2∘⋯∘sn]l  (resp., type r hyperproduct [s1∘s2∘⋯∘sn]r) of elements si∈S, we have that [s1∘s2∘⋯∘sn]l∩I≠ϕ⇒[s1∘s2∘⋯∘sn]l⊆I  (resp., [s1∘s2∘⋯∘sn]r∩I≠ϕ⇒[s1∘s2∘⋯∘sn]r⊆I). An Hv-ideal I of a multiplicative Hv-ring S is called a C-ideal if it is a left as well as a right C-ideal in S.We writeC-ideal(S) (resp., C-ideal*(S)) to denote the set of all (resp., proper) C-ideals of a multiplicative Hv-ring S. In a commutative multiplicative Hv-ring S, every left C-ideal is a right C-ideal and vice versa, since commutativity implies the equality [s1∘s2∘⋯∘sn]r=[sn∘sn-1∘⋯∘s1]l, for any si∈R. Note that in a multiplicative hyperring (even if it is not commutative) a hyperideal is a left C-ideal (as an Hv-ideal) if and only if it is a right C-ideal.Proposition 15.
IfR is a strong hyperdomain, then the set C-Ideal*(R[x]) is nonempty.Proof.
For eachn∈N∪{0}, let Cn={f(x)∈R[x]:ordf(x)≥n}∪{0R[x]}. Then Cn is a subgroup of the group (R[x],+) (since for any f(x),g(x)∈R[x],  ord(-f(x))=ordf(x) and ord(f(x)+g(x))=min⁡{ordf(x),ordg(x)}, when f(x)+g(x)≠0R[x]). Let g(x)∈R[x] and f(x)∈Cn. If g(x)=0R[x] or f(x)=0R[x], then g(x)*f(x)={0R[x]}⊆Cn (since 0R[x] is absorbing in R[x]). So, let g(x)≠0R[x] and f(x)≠0R[x]. Then, by Proposition 11(ii), for all h(x)∈g(x)*f(x), ordh(x)=ord(g(x)*f(x))=ordf(x)+ordg(x)>n. Thus, g(x)*f(x)⊆Cn. Hence, Cn is an Hv-ideal of R[x] (since R[x] is a commutative multiplicative Hv-ring). Let [f1(x)*f2(x)*⋯*fn(x)]l be a type-l hyperproduct of elements of the Hv-ring R[x], such that [f1(x)*f2(x)*⋯*fn(x)]l⊈Cn. Then, fi(x)≠0R[x] for each i (since 0R[x]∈Cn is absorbing in R[x]). Thus, ord([f1(x)*f2(x)*⋯*fn(x)]l)=ordg(x), for all g(x)∈[f1(x)*f2(x)*⋯*fn(x)]l. Now, since [f1(x)*f2(x)*⋯*fn(x)]l⊈Cn, so there exists h(x)∈[f1(x)*f2(x)*⋯*fn(x)]l such that h(x)∉Cn. Then, for any g(x)∈[f1(x)*f2(x)*⋯*fn(x)]l, ordg(x)=ordh(x)<n. Hence, by definition of Cn, [f1(x)*f2(x)*⋯*fn(x)]l∩Cn=ϕ and so Cn is a left C-ideal and thus a C-ideal of R[x] (since R[x] is commutative). Hence, Cn∈C-Ideal*(R[x]), for all n>0.
Since the intersection of leftC-ideals of a multiplicative Hv-ring S is also a left C-ideal of S and S is itself a left C-ideal, so the smallest left C-ideal containing a subset A of S, being naturally called the left C-ideal generated by A, exists and is, in fact, the intersection of all left C-ideals containing A. The left C-ideal generated by a left Hv-ideal I of S is called the left C-closure of I and is denoted by Cl(I). Clearly, Cl(Cl(I))=Cl(I) for a left Hv-ideal I. Cr(I) and C(I), respectively, denote the right C-closure of a right Hv-ideal I and the C-closure of an Hv-ideal I of S. For an Hv-ideal I of a commutative multiplicative Hv-ring S, Cl(I)=Cr(I)=C(I). The following lemma presents a description of the set Cl(I), for a left Hv-ideal I of a multiplicative Hv-ring S. The set Cr(I) for a right Hv-ideal I can be described dually.Lemma 16.
LetS be a multiplicative Hv-ring with an i-set E and let L denote the set of all left Hv-ideals of S. F:P(S)→P(S) is a mapping defined by
(7)F(A)=⋃{∑i=1n[ri1∘ri2∘⋯∘rimi]l:n,mi∈N,ri1∈Ssatisfying[ri1∘ri2∘⋯∘rimi]l∩A≠ϕ∑i=1n[ri1∘ri2∘⋯∘rimi]l},
for any A∈P(S). Then for any I,I1,I2∈L, we have the following: (i)  F(I)∈L, (ii)  I⊆F(I), (iii)  I1⊆I2⇒F(I1)⊆F(I2),(iv)  F(I)=I if and only if I is a left C-ideal of S, and(v)  left C-closure of I is Cl(I)=⋃k=1∞‍Fk(I), where for any k∈N, Fk denotes the (k-1)-times mapping composition of F.Proof.
All the assertions made in this lemma can be established by adopting the arguments that are applied in proving well-known analogous results on “complete closure of a set” in semihypergroup theory (see [3, 8]).Proposition 17.
LetR be a strong hyperdomain with an i-set. Then, for any Hv-ideal I of the multiplicative Hv-ring R[x], ord(C(I)*)≥ordI*, where for any Hv-ideal J of R[x], J*=J∖{0R[x]}.Proof.
SinceR is a (strong) hyperdomain, the multiplicative Hv-ring is commutative. So, for any Hv-ideal I of R[x], C(I)=Cl(I)=⋃k=1∞‍Fk(I) (by Lemma 16(v)). For any n∈N, f(x)∈Fn(I)*⇒f(x)∈∑i=1m‍Ai, for some type-l hyperproducts Ai=[fi1(x)*fi2(x)*⋯*fimi(x)]l of elements of R[x], satisfying Ai∩Fn-1(I)≠ϕ. Since here R[x] is a strong Hv-domain (by Proposition 13) with absorbing zero, we may assume that 0R[x]∉Ai for each i (since {0R[x]}+A=A for any A∈P*(R[x])). Then, for each i(=1,2,…,m) and ji(=1,2,…,mi), fiji(x)≠0R[x]. Thus, by Corollary 12, ordAi=ordgi(x) for any gi(x)∈Ai. Now since, for each i, Ai∩Fn-1(I)≠ϕ, so there exists gi(x)∈Ai such that gi(x)∈Fn-1(I)*. Then, for any f(x)∈Fn(I)*, ordf(x)≥∑i=1m‍ordAi=min.{ordgi(x):i}≥ord(Fn-1(I)*). Consequently, ord(Fn(I)*)≥ord(Fn-1(I)*), for any n∈N. Now, let f(x)∈C(I)* be arbitrary. Then f(x)∈Fn(I)* for some n∈N. So, we have that ordf(x)≥ord(Fn(I)*)≥ord(Fn-1(I)*)≥⋯≥ord(F(I)*)≥ordI*. Thus, ord(C(I)*)≥ordI*.


## 5. Polynomials over Multiplicative Hyperfield

A nonzero elementa of a multiplicative Hv-ring (S,+,∘) with an i-set E={e1,e2,…,en} is referred to be an E-invertible element (or an E-unit) of S if, for each i=1,2,…,n, (n∈N), there exist Aij∈Hl(a)={[sp∘sp-1∘⋯∘s1∘a]l:si∈S, p∈N} and Bik∈Hr(a)={[a∘s1∘s2∘⋯∘sq]r:si∈S, q∈N} such that ei∈(∑j=1m‍Aij)∩(∑k=1l‍Bik). An element a(≠0S) of the multiplicative Hv-ring S with a hyperidentity e is said to be e-hyperinvertible (or an e-hyperunit) in S if there exist A∈Hl(a) and B∈Hr(a) such that e∈A∩B.Ifa∈S is an E-unit (resp., e-hyperunit) in a multiplicative Hv-ring (S,+,∘) with two i-sets E and E′ (resp., with two hyperidentities e and e′), then one can easily verify that a is also an E′-unit (resp., an e′-hyperunit) in S. We thus call an E-unit (resp., an e-hyperunit) of a multiplicative Hv-ring S simply a unit (resp., a hyperunit). Denote by U(S) and Uh(S), respectively, the sets of units and hyperunits of a multiplicative Hv-ring S.AnHv-ideal I(≠S) of a multiplicative Hv-ring S is maximal in S if, for any Hv-ideal J of S, I⊊J⊆S⇒J=S. For a commutative multiplicative Hv-ring S with an i-set, this is immediate to observe that a∈U(S) if and only if a∉M, for any maximal Hv-ideal M of S.Proposition 18.
Let the multiplicative hyperringR be commutative and contain a hyperidentity e. Then, for a polynomial f(x)=∑‍akxk∈R[x], f(x)∈Uh(R[x]) if and only if a0∈Uh(R).Proof.
Sincee is a hyperidentity in the multiplicative hyperring R, ex0∈R[x] is a hyperidentity in the multiplicative Hv-ring R[x]. Now let f(x)=∑‍akxk∈Uh(R[x]). Then, there exist gi(x)=∑‍bi,kxk∈R[x] (i=1,2,…,n; n∈N) such that ex0∈[f(x)*g1(x)*⋯*gn(x)]r. So, e∈a0(b1,0b2,0…bn,0) and thus there exists b∈b1,0b2,0…bn,0⊆R such that e∈a0b whence a0∈Uh(R) (since R is commutative).
Conversely, leta0 be hyperinvertible in R with respect to the hyperidentity e. Then there exists b0∈R such that e∈a0b0. We assert that there is a sequence {bk}k∈N in R whose nth term bn (n∈N) is inductively defined so as to satisfy the relation that
(8)0R∈a0bn+a1bn-1+a2bn-2+⋯+anb0⋯(#).
In fact, we see that0R∈0Rb0=(-a1+a1)b0⊆(e(-a1)+a1)b0⊆((a0b0)(-a1)+a1)b0⊆(a0b0)(-a1)b0+a1b0=a0((-b0)a1b0)+a1b0⇒ there exist b1∈(-b0)a1b0 such that 0R∈a0b1+a1b0 (the relation # for n=1). Suppose, for some m≥1, the terms bk(1≤k≤m-1) are defined in such a way that each bk(1≤k≤m-1) satisfies the relation (#) for k. Then (a1bm-1+a2bm-2+⋯+am-1b1+amb0) is defined to be a nonempty subset of R. Let t∈a1bm-1+a2bm-2+⋯+am-1b1+amb0. Then 0R=-t+t∈e(-t)+t⊆(a0b0)(-t)+t⊆a0(-b0)(a1bm-1+a2bm-2+⋯+am-1b1+amb0)+(a1bm-1+a2bm-2+⋯+am-1b1+amb0)⇒ there exists bm∈(-b0)(a1bm-1+a2bm-2+⋯+am-1b1+amb0) such that 0R∈a0bm+a1bm-1+a2bm-2+⋯+amb0 (the relation # for k=m). Hence the assertion is true for all n∈N. Thus consider the polynomial g(x)=∑‍bkxk∈R[x]. Then from the definition of *, f(x)*g(x)={∑‍ckxk:c0∈a0b0 and ck∈a0bk+a1bk-1+a2bk-2+⋯+akb0 for k∈N}. Thus, ex0∈f(x)*g(x) (due to relation # and since e∈a0b0). So, f(x) is hyperinvertible in R[x]; that is, f(x)∈Uh(R[x]).
The (left, right)Hv-ideal of a multiplicative Hv-ring S generated by A∈P(S) is the smallest (resp., left, right) Hv-ideal of S containing A which is denoted by (resp., 〈A〉l,〈A〉r) 〈A〉. The principal (left, right) Hv-ideal of the multiplicative Hv-ring S generated by an element a of S, denoted by (resp., 〈a〉l, 〈a〉r) 〈a〉, is the (resp., left, right) Hv-ideal (resp., 〈{a}〉l, 〈{a}〉r) 〈{a}〉 of the multiplicative Hv-ring S.
If the multiplicativeHv-ring S has an i-set, then for any a∈S, 〈a〉l=Hl(a)-=⋃{∑i=1nAi:Ai∈Hl(a),n∈N} and 〈a〉r=Hr(a)-=⋃{∑i=1nAi:Ai∈Hr(a),n∈N}.Definition 19.
A commutative multiplicativeHv-ring S with an i-set E is called a multiplicative Hv-field (resp., an inversive multiplicative Hv-field) if S∖〈0S〉⊆U(S) (resp., S∖〈0S〉⊆Uh(S)). If a multiplicative Hv-field (resp., an inversive multiplicative Hv-field) is a multiplicative hyperring, then we call it a multiplicative hyperfield (resp., an inversive multiplicative hyperfield).Proposition 20.
LetF be a polynomially structured inversive multiplicative hyperfield. Then a polynomial f(x)=∑‍akxk∈F[x]∖{0F[x]} is hyperinvertible in F[x] if and only if  ordf=0.Proof.
IfF is an inversive multiplicative hyperfield, then by Proposition 18, any polynomial f(x)=∑‍akxk∈F[x]∖{0F[x]} is hyperinvertible in F[x] if and only if a0≠0F. Hence, the result follows.Definition 21.
A commutative multiplicativeHv-ring S is called a principal C-ideal Hv-ring if every C-ideal of S is a principal Hv-ideal. A principal C-ideal Hv-ring which is a (strong)  Hv-domain is called a principal C-ideal (strong)  Hv-domain.Proposition 22.
LetF be a polynomially structured inversive multiplicative hyperfield. Then, the multiplicative Hv-ring F[x] of polynomials over F is a principal C-ideal Hv-ring.Proof.
Lete be a hyperidentity of the inversive multiplicative hyperfield F. Then, the polynomial ex0 is a hyperidentity in F[x] and thus F[x]=〈ex0〉 (since F is commutative). So, let I be any proper C-ideal of F[x]. If I={0F[x]}, then I is the principal hyperideal 〈0F[x]〉 (since, for F having absorbing zero, 0F[x] is absorbing in F[x]). Suppose that I≠{0F[x]}. Then, take a nonzero polynomial f(x)=∑‍akxk∈I such that ordf(x)≤ordt(x) for any t(x)∈I*. Let us write ordf(x)=m. Then, am≠0F and ak=0F for any 0≤k<m. Consider then the polynomial g(x)=∑k∈N0*‍bkxk∈F[x], where bk=am+k. Then, clearly f(x)∈(exm)*g(x). Also, by Proposition 18, g(x) is hyperinvertible in F[x]. Thus there exists h(x)∈F[x] such that ex0∈g(x)*h(x). Now, f(x)∈(exm)*g(x)⇒f(x)*h(x)⊆[exm*g(x)*h(x)]r. Again, f(x)*h(x)⊆I (since f(x)∈I). So, [exm*g(x)*h(x)]r∩I≠ϕ. Hence, [exm*g(x)*h(x)]r⊆I (since I is a C-ideal and every C-ideal is a right C-ideal). Now, [exm*g(x)*h(x)]l∩[exm*g(x)*h(x)]r≠ϕ, whence [exm*g(x)*h(x)]l∩I≠ϕ (since [exm*g(x)*h(x)]r⊆I). Consequently, [exm*g(x)*h(x)]l⊆I (since I is a left C-ideal). Then, exm∈(exm)*(ex0)⊆exm*(g(x)*h(x))=[exm*g(x)*h(x)]l⊆I. Thus, 〈exm〉⊆I.
Now, leth(x)=∑‍bkxk∈I* be arbitrary. Suppose that ordh=n. Then, bn≠0F and bk=0F for any 0≤k<n. By choice of f(x) from I, here m≤n. So, one can define a polynomial g(x)=∑‍ckxk∈F[x], where ck=0F for all 0≤k≤n-m-1 and ck=bm+k for all k≥n-m. Clearly then h(x)∈exm*g(x)⊆〈exm〉, whence I⊆〈exm〉. Thus, I=〈exm〉.Remark 23.
In a ring, an invertible element can never be a divisor of zero. This not true in general for a multiplicative hyperring. In fact, on the commutative group of integers(Z,+), if we define a hyperoperation ∘ by stating that x∘y={0,xy}, for all x,y∈Z, then (Z,+,∘) is a commutative multiplicative hyperring with a hyperidentity 1. Every nonzero element of (Z,+,∘) is a zero divisor and 1,-1∈Z are, in particular, hyperunits of (Z,+,∘). To get a parity with the ring theory in this regard, we perceive the notion of strong hyperinvertibility of an element of a multiplicative hyperring.Definition 24.
A hyperinvertible element of a multiplicative hyperring with a hyperidentity is said to be strongly hyperinvertible (or a strong hyperunit), if it is not a zero divisor in that multiplicative hyperring. A commutative multiplicative hyperring with absorbing zero and a hyperidentitye is said to be a strongly inversive multiplicative hyperfield if each of its nonzero elements is a strong hyperunit.Example 25.
LetR={a2+b3:a, b∈Q} and A={2,3}. Then, with respect to usual addition + of reals, (R,+) is a commutative group, with identity 0. On R, ∘ is a hyperoperation defined by
(9)(a2+b3)∘(c2+d3)={(a2+b3)·t·(c2+d3):t∈A}={p2+2q3,3q2+p3},
where p=2ac+3bd and q=bc+ad.
Then,(R,+,∘) is a strongly inversive multiplicative hyperfield which is polynomially structured.Definition 26.
A local (C-local) multiplicative Hv-ring is a commutative multiplicative Hv-ring with an i-set, which has a unique maximal Hv-ideal (resp., C-ideal).Proposition 27.
LetF be a polynomially structured strongly inversive multiplicative hyperfield. Then, (i)  F[x] is a principal C-ideal strong Hv-domain, (ii)  for any f∈F[x]∖{0F[x]}, there exist a hyperinvertible element g∈F[x] and m∈N∪{0} such that f∈(exm)*g, where e is a hyperidentity in F, and (iii)  F[x] is a local as well as a C-local multiplicative Hv-ring.Proof.
(i) HereF is a strongly inversive multiplicative hyperfield. So, F is a strong hyperdomain. Thus, by Proposition 13, the multiplicative Hv-ring F[x] is a strong Hv-domain. Again, by Proposition 22, F[x] is a principal C-ideal Hv-ring (since F is an inversive multiplicative hyperfield). So, by Definition 21, F[x] is a principal C -ideal strong Hv-domain.
(ii) Letf∈F[x]∖{0F[x]} and J=〈f〉. Then, J=Hr(f)¯. So, for any h∈J*, there exist Ai=[f*fi1*fi2*⋯*fimi]r∈Hr(f)(i=1,2,…,k;  k,mi∈N) such that h∈∑i=1k‍Ai. Thus, h=∑i=1k‍ti for some ti∈Ai. Since F[x] is a strong Hv-domain (by Proposition 13) with absorbing zero, so 0F[x]∈Ai=[f*fi1*fi2*⋯*fimi]r⇒[f*fi1*fi2*⋯*fimi]r={0F[x]}. Hence, without any loss of generality (since {0F[x]}+A=A for any A∈P*(F[x])), we may assume that 0F[x]∉Ai for any i (since h≠0F[x]). Then, for any i=1,2,…,k and j=1,2,…,mi, fij≠0F[x]. So, by Proposition 11(ii), ordti=ord(Ai)≥ordf and so ordh=min.{ordti:i}≥ordf. Thus ordJ*≥ordf. So, by Proposition 17, f is a nonzero polynomial in the C-ideal C(J), such that ordf≤ordh for all h∈C(J)*. So, there exists an invertible element g∈F[x] and an integer m≥0 such that f∈(exm)*g, where e is a hyperidentity in F (see proof of Proposition 22).
(iii) For any integerm≥1, since exm∈(ex)*(exm-1), so 〈exm〉⊆〈ex〉. Again, for any Hv-ideal I of the multiplicative Hv-ring F[x], if f∈I* is such that ordf≤ordg for all g∈I*, then I⊆〈exm〉, where m=ordf. So, 〈ex〉 is the unique maximal Hv-ideal in F[x]. Thus, F[x] is a local multiplicative Hv-ring. Now, ex∈C1 (as defined in Proposition 15). So, ord(ex)≤ordf for any f∈C1*. Thus, C1 being a C-ideal of F[x], we have C1=〈ex〉 (see proof of Proposition 22). Hence, F[x] is a C-local multiplicative Hv-ring.



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