The k-Zero-Divisor Hypergraph of a Commutative Ring

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Volume 2007, Article ID 50875,15pages doi:10.1155/2007/50875

Research Article

The k-Zero-Divisor Hypergraph of a Commutative Ring

Ch. Eslahchi and A. M. Rahimi

Received 28 June 2006; Revised 15 October 2006; Accepted 27 February 2007 Recommended by Dalibor Froncek

The concept of the zero-divisor graph of a commutative ring has been studied by many authors, and thek-zero-divisor hypergraph of a commutative ring is a nice abstraction of this concept. Though some of the proofs in this paper are long and detailed, any reader familiar with zero-divisors will be able to read through the exposition and find many of the results quite interesting. LetRbe a commutative ring andkan integer strictly larger than 2. Ak-uniform hypergraphHk(R) with the vertex setZ(R,k), the set of allk-zero- divisors inR, is associated toR, where eachk-subset ofZ(R,k) that satisfies thek-zero- divisor condition is an edge inH_k(R). It is shown that ifRhas two prime idealsP1andP2

with zero their only common point, thenHk(R) is a bipartite (2-colorable) hypergraph with partition setsP1−ZandP2−Z, whereZis the set of all zero divisors ofRwhich are notk-zero-divisors inR. IfRhas a nonzero nilpotent element, then a lower bound for the clique number ofH3(R) is found. Also, we have shown thatH3(R) is connected with diameter at most 4 wheneverx²=/0 for all 3-zero-divisorsxofR. Finally, it is shown that for any finite nonlocal ringR, the hypergraphH3(R) is complete if and only ifRis isomorphic toZ2×Z2×Z2.

Copyright © 2007 Ch. Eslahchi and A. M. Rahimi. This is an open access article distrib- uted under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

1. Introduction

The notion of a zero-divisor graphΓ(R) of a commutative ringRwas first introduced by Beck in [1] and was further investigated in [2], where the authors were interested in color- ings ofΓ(R), though their vertex set included the zero element. In [3–9] the authors, using the set of nonzero zero divisors ofRas vertex set ofΓ(R), were interested in examining the interplay between the ring-theoretic properties ofRand the graph-theoretic properties

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ofΓ(R). In this paper, we extend the concept of a zero-divisor of a commutative ringRto that of ak-zero-divisor and investigate the interplay between the ring-theoretic properties ofRand the graph-theoretic properties of its associatedk-uniform hypergraphH_k(R). In this section, we define and study some examples ofk-zero-divisors and recall some definitions from graph theory. InSection 2, we define and study some basic properties of the k-uniform hypergraphH_k(R) andk-zero-divisors of a commutative ringR. Finally, in the last section, we merely concentrate on the properties of 3-zero-divisor hypergraphs.

Definition 1.1. LetRbe a commutative ring andk≥2 a fixed integer. A nonzero nonunit elementa1 in Ris said to be ak-zero-divisor inRif there existk−1 distinct nonunit elementsa2,a3,. . .,a_kinRdiﬀerent froma1such thata1a2a3···a_k=0 and the product of no elements of any proper subset ofA= {a1,a2,. . .,ak}is zero.

Clearly, a 2-zero-divisor inRis a zero divisor, but the converse is not true in general.

For example, 2 is a zero divisor inZ4, but it is not a 2-zero-divisor.

Remark 1.2. In the literature, on zero-divisor graphs, the edges are defined to be between the distinct nonzero zero-divisors in order to construct a graph with no loops. Here, we assume distinctness of the elements inDefinition 1.1fork-zero-divisors in order to have ak-uniform hypergraph, for any fixed integerk≥3. Note that the graph constructed by 2-zero-divisors is exactly the same as the zero-divisor graph of a ring.

Example 1.3. The element 2 inZ30is a 3-zero-divisor since 2·3·5=0, and the product of no elements of any proper subset of{2, 3, 5}is zero.

ByZ(R,k) we denote the set of all k-zero-divisors ofR. It is not diﬃcult to show that the statement “the product of no elements of any proper subset ofA is zero” or the statement “the product of no elements of any (k−1)-subset of Ais zero” can be used inDefinition 1.1equivalently. Clearly, fromDefinition 1.1, every element of the set {a2,a3,. . .,ak}is ak-zero-divisor inR. It is clear that everyk-zero-divisor inRis also a zero divisor inR, but, the converse is not true in general. For example, the element 2 is a zero divisor, but not a 3-zero-divisor inZ10.

We review some basic graph-theoretic definitions, and for the necessary definitions and notations of hypergraphs, we refer the reader to standard texts of graph theory such as [10]. A hypergraph is a pair (V,E) of disjoint sets, where the elements ofEare nonempty subsets (of any cardinality) ofV. The elements ofV are the vertices, and the elements of Eare the edges of the hypergraph. The hypergraphH =(V,E) is calledk-uniform whenever every edgeeofHis of sizek. Ak-uniform hypergraphH is called complete if everyk-subset of the vertices is an edge ofH. The definition of a clique and the clique number of ak-uniform hypergraph are taken from [11,12] as follows.

LetH be ak-uniform hypergraph. A subsetAofV(H) is called a clique ofH if every k-subset ofAis an edge ofH. The clique number ofH, denoted byω(H), is defined to be

ω(H)₌max|A_{| |}Ais a clique

k−1 . (1.1)

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Anr-coloring of a hypergraphH=(V,E) is a mapc:V→ {1, 2,. . .,r}such that for every edgeeofH, there exist at least two verticesxand yinewithc(x)=/ c(y). The smallest integerrsuch thatH has anr-coloring is called the chromatic number ofH and is denoted byχ(H). In [11], it is shown that for anyk-uniform hypergraphH,χ(H)≥ ω(H). A path in a hypergraphH is an alternating sequence of distinct vertices and edges of the formv1,e1,v2,e2,. . .,v_ksuch thatv_i,v_i+1is ine_i for all 1≤i≤k−1. The number of edges of a path is its length. The distance between two verticesxandyofH, denoted by dH(x,y), is the length of the shortest path fromxtoy. If no such path betweenxand y exists, we setd_H(x,y)= ∞. The greatest distance between any two vertices inHis called the diameter ofHand is denoted by diam(H). The hypergraphHis said to be connected whenever diam(H)<∞. A cycle in a hypergraphHis an alternating sequence of distinct vertices and edges of the formv1,e1,v2,e2,. . .,v_k,e_k,v1 such thatv_i,v_i+1are ine_i for all 1≤i≤k−1 withvk,v1∈ek. The girth of a hypergraphHcontaining a cycle, denoted by gr(H), is the smallest size of the length of cycles ofH.

2.k-zero-divisor hypergraphs

In this section, we define and study some properties of thek-uniform hypergraphH_k(R), thek-zero-divisors of a commutative ringR, and provide some examples.

Definition 2.1. A ringRis said to be ak-integral domain wheneverZ(R,k), the set of all k-zero-divisors ofR, is the empty set.

Example 2.2. Let (R,M) be a local ring with maximal idealM /=0 such thatM²=0. Then Ris a 3-integral domain which is not an integral domain.

Example 2.3. For any integerk≥3, we have the following results.

(1) Letn=p₁^α¹p₂^α²···p^αr^rbe the prime decomposition ofn, wherepi=/ pjwhenever i /= jand 1≤αifor alli,j=1, 2,. . .,r. ThenZnis ak-integral domain whenever

i≤rα_i≤k−1.

(2) Let n_i=p^α₁_i¹ⁱp^α₂_i²ⁱ···p^αri^ri be the prime decomposition of n_i for distinct primes pji’s and 1≤αjifor all 1≤i≤tand j=1, 2,. . .,r. ThenZn1×Zn2× ··· ×Zntis a k-integral domain whenever

j≤r1

αj1+

j≤r2

αj2+···+

j≤rt

αjt≤k−1. (2.1)

(3) Let F be a field and let f(x) be a polynomial in F[x] such that f(x)= P1(x)^α¹P2(x)^α²_···P_r(x)^α^r, whereP_i(x)_∈F[x] are distinct irreducible polynomi- als and 1≤α_ifor all 1≤i≤r. ThenF[x]/(f(x)) is ak-integral domain whenever

i≤rαi≤k−1.

(4) LetRibe an integral domain for eachi=1, 2,. . .,n. ThenR=R1×R2× ··· ×Rn

is ak-integral domain whenevern≤k−1.

By [13], it is true that a nonintegral domain with a finite number of zero divisors is finite. Similarly, we pose the following question for the rings with a finite number of k-zero-divisors.

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Question 1. Does the finiteness ofk-zero-divisors in a non-k-integral domainRimply the finiteness of zero-divisors or, equivalently, finiteness ofR?

Definition 2.4. For any fixed integerk≥3, an idealP of a ringRis said to bek-prime whenever for any setA= {a1,a2,. . .,a_k}of nonzero, distinct, and nonunit elements ofR, a1a2···ak∈Pimplies that the product of the elements of a proper subset ofAis inP.

Note that by this definition, every prime ideal ofRis ak-prime ideal ofR.

Example 2.5. Let (R1,M1) and (R2,M2) be two local rings with nonzero maximal ideals M1andM2, respectively. We show thatM1×M2is a 3-prime ideal inR=R1×R2which is not a prime ideal inR. Let (a1,b1), (a2,b2), and (a3,b3) be arbitrary elements inR1×R2, where for each 1≤i≤3, (ai,bi) is a nonzero nonunit in R. Clearly, (a1,b1)·(a2,b2)· (a3,b3)=(a1a2a3,b1b2b3)∈M1×M2implies that at least one of the elementsai’s (bj’s) belongs toM1(M2) for somei(j) in{1, 2, 3}. In this case, there always exists a proper subset of{(a1,b1), (a2,b2), (a3,b3)}such that the product of its elements belongs toM1× M2. But since (1, 0)·(0, 1)∈M1×M2 and neither of the elements (1, 0) and (0, 1) is in M1×M2, thenM1×M2is a 3-prime ideal inR1×R2which is not a prime ideal inR.

The following theorem is similar to the well-known fact on the relationship between prime ideals and integral domains.

Theorem 2.6. LetP be an ideal in the ringR. ThenR/Pis ak-integral domain ifP is a k-prime ideal.

The proof follows directly from the definition, and we leave it to the reader.

The converse of the above theorem is not true in general. For example, the ideal8 generated by 8 inZ48is not a 3-prime ideal, butZ48/8is a 3-integral domain.

Next, we extend the concept of zero-divisor graph of a commutative ringRto that of ak-zero-divisor hypergraph.

Definition 2.7. LetRbe a commutative ring (with 1=/ 0) and letZ(R,k) be the set of all k-zero-divisors in R. Associate ak-uniform hypergraphH_k(R) to Rwith vertex set Z(R,k), and for distinct elementsx1,x2,. . .,xkinZ(R,k), the set{x1,x2,. . .,xk}is an edge ofHk(R) if and only ifx1x2···xk=0 and the product of elements of no (k−1)-subset of {x1,x2,. . .,x_k}is zero.

Clearly, from the above definition we can conclude that for anyk≥3,H_k(R) is the empty set if and only ifRis ak-integral domain.

Theorem 2.8. LetRbe a non-k-integral domain. If there exist prime idealsP1andP2inR such thatP1∩P2= {0}, thenχ(Hk(R))=2.

Proof. SinceP1∩P2= {0}, thenP1∪P2 is equal to the set of all zero divisors ofR. On the other hand, since eachk-zero-divisor is also a zero divisor, eachk-zero-divisor must belong to the prime idealsP1orP2. Consider the functionc:V(H_k(R))→ {1, 2}given by

c(x)=

⎧⎨

⎩1, x∈P1,

2, x∈P2. (2.2)

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In order to prove thatcis a 2-coloring ofHk(R), we need to show that there is no edge einHk(R) such that every vertex ofeobtains the same color. Without loss of generality, lete= {x1,x2,. . .,x_k}be an edge ofH_k(R) such thatc(x1)=c(x2)= ···=c(x_k)=1. Since x1x2···xk=0∈P2 andP2is a prime ideal ofR, thenxi∈P2for at least one 1≤i≤k, which is a contradiction. Therefore,χ(Hk(R))≤2. On the other hand, sinceRis not a k-integral domain, thenH_k(R) has at least one edge, which implies thatχ(H_k(R))≥2,

and the proof is complete.

Remark 2.9. From the above theorem, it is clear thatHk(R) is a bipartite hypergraph with partition setsV(Hk(R))∩P1andV(Hk(R))∩P2. Note that in [4], it is shown that for any reduced ringR, the zero-divisor graphΓ(R) is bipartite if and only if there exist two distinct prime idealsP1 andP2ofRsuch thatP1∩P2= {0}. In addition, ifΓ(R) is bipartite, then it is a complete bipartite graph.

Remark 2.10. By considering the ringR=Z2×Z2×Z2, we see thatχ(H3(R))=2. But there are no prime idealsP1andP2inRsatisfying the condition ofTheorem 2.8. There- fore, the converse ofTheorem 2.8is not true in general.

Theorem 2.11. LetR=R1×R2× ··· ×R_n, whereR_iis an integral domain for eachi= 1, 2,. . .,n.

(1) Ifn=k, thenχ(Hk(R))=2.

(2) Ifn=k+t, thenχ(H_k(R))≤2 +tfor allt≥0.

Proof. Letk=n. We claim that Z(R,k)= a1,a2,. . .,ak

|exactly one of theai’s is zero for 1≤i≤k. (2.3) It is obvious that anyk-zero-divisor must have at least one zero component. Letx1= (a11,a12,. . .,a1k) be ak-zero-divisor with at least two zero components. Without loss of generality, assume thata11=a12=0. Consequently, there existx2,x3,. . .,x_k_∈V(H_k(R)) such that{x1,x2,. . .,x_k} ∈E(H_k(R)), wherex_i=(a_i1,a_i2,. . .,a_ik) for all 1≤i≤k. Thus,

i≥1ai j=0 for eachj≥3. Now sinceRjis an integral domain, then for each fixed j≥3, there exists at least oneij with 1≤i≤ksuch thataijj=0. LetIbe the set of allij’s such thata_i_j_j=0 for the smallestiin the set{1, 2,. . .,k}. Thus, we havex1

i∈Ix_i=0 and since

|I| ≤k−2, we have a contradiction. Now letx1=(a1,a2,. . .,ak)∈Rsuch that exactly one and only one of the components is zero. Without loss of generality, assume thata1=0.

Letx_i=(1, 1,. . ., 1, 0, 1, 1,. . ., 1), where theith component is the only zero component ofx_i for 2≤i≤k. It is obvious that{x1,x2,. . .,xk} ∈E(Hk(R)) and the claim is true. Consider the functionc:V(Hk(R))→ {1, 2}given by

c(x)=

⎧⎨

⎩

1 the first component ofxis zero,

2 otherwise. (2.4)

It is easy to see thatcis a 2-coloring ofHk(R), and sinceHk(R) has at least one edge, χ(H_k(R))=2.

For the proof of part 2, assumen=k+twitht≥0 a fixed integer. The proof is by induction ont. From part 1, the first step of induction fort=0 is true. Now, assume that

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t≥1 and the result is true fork+t. Letc:V(Hk(R1×R2× ··· ×Rk+t))→ {1, 2,. . .,t+ 2} be a t+ 2-coloring ofHk(R1×R2× ··· ×Rk+t). Consider the functionc:V(Hk(R1× R2× ··· ×R_k+t+1))→ {1, 2,. . .,t+ 3}given by

c(x)=

⎧⎨

⎩

c(x) the last component ofxis zero,

t+ 3 otherwise. (2.5)

From this, it is not diﬃcult to show thatc is a (t+ 3)-coloring ofHk(R1×R2× ··· ×

Rk+t+1), and the proof is complete.

As a very special case of the above theorem, it is easy to show that the chromatic number ofH3(Z₂⁴) andH3(Z₂⁵) is 3. Note that the chromatic number ofH3(Z₂⁵) is strictly less than 2 + (5−3), and the chromatic number ofH3(Z⁴₂) equal to 3 shows that the bound is sharp.

3. 3-zero-divisor hypergraphs

In this section, we only focus on some graph-theoretic properties ofH3(R). We show that H3(R) is connected with diameter at most 4 provided thatx²=/ 0 for all 3-zero-divisorsx inR. We find a necessary and suﬃcient condition for its completeness, and we also find a lower bound for its clique number.

Theorem 3.1. LetH3(R) be the 3-zero-divisor hypergraph of a ringRsuch thatx²=/ 0 for every 3-zero-divisorx∈R. ThenH3(R) is connected and

diam H3(R)≤4. (3.1)

Proof. For the proof of the theorem, it is enough to show that for each two edgese1= {a1,a2,a3}ande2= {b1,b2,b3}ofH3(R), there exist edgese3ande4which satisfy one of the following conditions:

e3∩e1= ∅/ , e3∩e2= ∅/ , (∗1) or

e3∩e1= ∅/ , e4∩e2= ∅/ , e4∩e3= ∅/ . (∗2) Consequently, for the rest of the proof, we can always assume thata_i=/b_janda_i= −/ b_jfor alli,j∈ {1, 2, 3}. LetGbe the bipartite graph constructed as follows:V(G)=e1∪e2and aibj∈E(G) if and only ifaibj=0 in the ringR.

Suppose G has two isolated vertices, one in e1 and the other in e2. For example, deg_G(a3)=deg_G(b3)=0. If there exists an elementc∈ {a1,a2,b1,b2}such thata3b3c=0, thene3= {a3,b3,c}satisfies (∗1). Suppose that this is not the case. Ifa3b3∈ {/ a1,a2,b1, b2}, then e3= {a1,a2,a3b3}ande4= {b1,b2,a3b3}satisfy (∗2). Otherwise without loss of generality, assume thata3b3=a1. Thene3= {a1,b1,b2}satisfies (∗1). The rest of our proof depends on the number of edges ofG.

Case 1. Suppose|E(G)| ≤2. ThenGhas two isolated vertices, one ine1and the other in e2.

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Case 2. Suppose|E(G)| =3. We study this case for four diﬀerent subcases as follows.

Case 2.1. Assume the degree of each vertex ofGis one and E(G)=

a1b1,a2b2,a3b3

. (3.2)

Consider the set{a1,a2b3,b1+b2}. Ifa1=a2b3, thena1b2=0 is a contradiction. Ifa1= b1+b2, then b1a2a3=0, and e3= {b1,a2,a3} satisfies (∗1). If b1+b2=a2b3, then a1b2a3=0, ande3= {a1,b2,a3} satisfies (∗1). Otherwise,e3= {a1,a2b3,b1+b2} is an edge. Similarly if we consider the set{b1,a2b3,a1+a3}, then we find an edgee3 which satisfies (∗1) ore4= {b1,a2b3,a1+a3}is an edge withe3ande4satisfying (∗2).

Case 2.2. Assume that the degree of exactly one of the vertices ofGis one. Without loss of generality, suppose that

E(G)=

a1b1,a1b2,a2b3

. (3.3)

Consider the set{a2,a3b1,a1+b3}. Ifa2=a3b1, thena1a2=0 implies a contradiction. If a2=a1+b3, thena2b2b1=0, ande3= {a2,b2,b1}satisfies (∗1). Ifa1+b3=a3b1, then a3b1b2b1=0. In this case ifa3=b1b2, thena1a3=0, also,b1=b1b2implies thatb1b3= 0, which in both cases we have a contradiction. Therefore,e3= {a3,b1b2,b1}is an edge which satisfies (∗1). If none of the above conditions holds, then the sete3= {a2,a3b1,a1+ b3}is an edge. Now consider the set{b2,a3b1,b3}. Similarly, we find an edgee3 which satisfies (∗1), ore4= {a2,a3b1,a1+b3}is an edge wheree3ande4satisfy (∗2).

Case 2.3. Let the degree of two vertices ofGbe two. Without loss of generality, suppose that

E(G)=

a1b1,a1b2,a2b2

. (3.4)

In this case, deg_G(a3)=deg_G(b3)=0, and the proof is complete.

Case 2.4. Assume that the degree of one vertex ofGis three. Without loss of generality, suppose

E(G)=

a1b1,a1b2,a1b3

. (3.5)

Suppose thata²₁a2=/ 0. Consider the set{a1a2−b1,a1,a3}. Ifa1a2−b1=a1, thenb2b1=0 is a contradiction. If a1a2−b1=a3, then a3b3b2=0, and therefore e3= {a3,b2,b3} is an edge satisfying (∗1). In the other case,e3= {a1a2−b1,a1,a3}is an edge. Similarly, if we consider the set{a1a2−b1,b2,b3}, we will find an edgee3that satisfies (∗1), ore4= {a1a2−b1,b2,b3}is an edge withe3ande4that satisfy (∗2). Now leta²1a2=0. Consider the set{a1−b1,a1,a2}. Ifa1−b1=a2, thena2b3b2=0, and thereforee3= {a2,b2,b3}is an edge satisfying (∗1). In the other case,e3= {a1−b1,a1,a2}is an edge. Similarly, if we consider the set{a1−b1,b2,b3}, we will find a contradiction, ore4= {a1a2−b1,b2,b3}is an edge withe3ande4that satisfy (∗2).

Case 3. Suppose|E(G)| =4. We study this case using four diﬀerent subcases as follows.

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Case 3.1. Assume the degree of one vertex ofGis three. Without loss of generality, suppose that

E(G)=

a1b1,a1b2,a1b3,a2b3

. (3.6)

Consider the set {a3b1,a2,a1+b3}. If a3b1=a2, then a3b3b1=0, and therefore e3= {a3,b1,b3}is an edge satisfying (∗1). Ifa3b1=a1+b3, thena²₁=0 is a contradiction.

Ifa2=a1+b3, thenb²3=0 is a contradiction. In the other case,e3= {a3b1,a2,a1+b3}is an edge. Similarly, if we consider the set{a3b1,b2,b3}, we will find an edgee3that satisfies (∗1), ore4= {a3b1,b2,b3}is an edge withe3ande4that satisfy (∗2).

Case 3.2. Assume that the degree of four vertices ofGis two. Without loss of generality, suppose that

E(G)=

a1b1,a1b2,a2b1,a2b2

. (3.7)

In this case, deg_G(a3)=deg_G(b3)=0, and the proof is complete.

Case 3.3. Let the degree of three vertices ofGbe two. Suppose without loss of generality that

E(G)=

a1b1,a1b2,a2b2,a2b3

. (3.8)

Consider the set{a3b3,a1,a2}. Ifa3b3=a1 ora2, thena3b3b2=0, and therefore e3= {a3,b2,b3}is an edge that satisfies (∗1). In the other case,e3= {a3b3,a1,a2}is an edge.

Similarly, if we consider the set{a3b3,b1,b2}, we will find an edgee3that satisfies (∗1), or e4= {a3b3,b1,b2}is an edge withe3ande4that satisfy (∗2).

Case 3.4. Assume that the degree of two vertices ofGis two. In this case, there might be two diﬀerent nonisomorphic cases. Without loss of generality, for one case we can assume that

E(G)₌a1b1,a1b2,a2b2,a3b3

, (3.9)

and in the other case

E(G)=

a1b1,a1b2,a2b3,a3b3

. (3.10)

In the first case, consider the set {a3b1,a2,a1+b2}. Ifa3b1=a2, thena3b1b2=0, and thereforee3= {a3,b1,b2}is an edge that satisfies (∗1). Ifa3b1=a1+b2, thena²1=0 is a contradiction. Also,a2=a1+b2implies thatb²2=0, which is a contradiction. In the other case,e3= {a1+b2,a2,b1a3}is an edge. Similarly, if we consider the set{a3b3,b1,a1+b2}, we will find an edgee3that satisfies (∗1), ore4= {a3b3,b1,a1+b2}is an edge withe3and e4that satisfy (∗2).

Similarly, for the second case, by considering the sets {a1 +b1,a2,a3} and {a1+b1,b2,b3}, we find an edgee3that satisfies (∗1), or two edgese3 ande4 that satisfy (∗2).

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Case 4. Suppose|E(G)| =5. We continue our investigation for five diﬀerent nonisomorphic subcases as follows.

Case 4.1. Without loss of generality, we can assume that

E(G)=

a1b1,a1b2,a1b3,a2b1,a2b2

. (3.11)

Consider the set {a3b3,a2,a1+b2}. If a3b3=a2, then a3b3b2=0, and therefore e3= {a3,b2,b3}is an edge that satisfies (∗1). Ifa3b3=a1+b2, then a²₁=0, which is a contradiction. Ifa1+b2=a2, thenb1b2=0 is a contradiction. In the other case,e3= {a1+ b2,a2,a3b3}is an edge. Similarly, if we consider the set{a3b3,b1,b2}, we will find an edge e3which satisfies (∗1), ore4= {a3b3,b1,b2}is an edge withe3ande4that satisfy (∗2).

Case 4.2. Assume without loss of generality that

E(G)=

. (3.12)

Consider the set{a1+b1,a2,b2}. Ifa1+b1=a2, thenb₁²=0 is a contradiction. Ifa1+b1= b2, thena²1=0 implies a contradiction. In the other case,e3= {a1+b2,a2,a3b3}is an edge that satisfies (∗1).

E(G)=

. (3.13)

Consider the set{a1+b1,a2,b2}. Ifa1+b1=a2, thenb₁²=0 is a contradiction. Ifa1+b1= b2, thena2a3b2=0, ande3= {a2,b2,a3}is an edge which satisfies (∗1). In the other case, e3= {a1+b1,a2,b2}is an edge that satisfies (∗1).

E(G)=

. (3.14)

Consider the set{a3+b1,a1,b3}. Ifa3+b1=a1 ora3+b1=b3, then a1a2b3=0, and e3= {a1,a2,b3}is an edge that satisfies (∗1). In the other case,e3= {a3+b1,a1,b3}is an edge which satisfies (∗1).

E(G)=

. (3.15)

Consider the set{a1+b2,a2,b1}. Ifa1+b2=a2, thenb2²=0. Ifa1+b2=b1, thena²1=0, which is a contradiction. Thereforee3= {a1+b2,a2,b1}is an edge that satisfies (∗1).

Case 5. Suppose|E(G)| =6. We study three diﬀerent nonisomorphic subcases as follows.

E(G)=

a1b1,a1b2,a1b3,a2b1,a2b2,a3b1

. (3.16)

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Consider the sets{a1+b1,a2,a3}and{a1+b1,b2,b3}. Ifa1+b1=a2, thenb1b2=0. Ifa1+ b1=a3, thenb²1=0. Also,a1+b1=b2ora1+b1=b3 implies thata²1=0, and in either case, we have a contradiction. Therefore,e3= {a1+b1,a2,a3}ande4= {a1+b1,b2,b3} are two edges that satisfy (∗2).

E(G)=

. (3.17)

Consider the set{a1+b3,a3,b1}. Ifa1+b3=a3, thenb²3=0. Alsoa²1=0 whenevera1+ b3=b1, which is a contradiction. Therefore,e3= {a1+b3,a3,b1}is an edge that satisfies (∗1).

E(G)=

. (3.18)

In this case, similar to the above subcase,e3= {a1+b3,a3,b1}is an edge which satisfies (∗1).

Case 6. Suppose that 7≤ |E(G)| ≤9. In this case, there always exist two vertices with degree three, one frome1and the other frome2. Letd_G(a1)=d_G(b1)=3. Consider the sets{a1+b1,a2,a3}and{a1+b1,b2,b3}. Ifa1+b1=a2ora3, thenb₁²=0; and ifa1+b1= b2orb3, thena²1=0, which is a contradiction in all cases. Therefore,e3= {a1+b1,a2,a3} ande4= {a1+b1,b2,b3}are two edges that satisfy (∗2).

Remark 3.2. From the above theorem and the fact that

gr H3(R)≤2 diam H3(R)+ 1, (3.19) we can conclude that the diameter and girth of any hypergraphH3(R) containing a cycle and satisfying the conditions in the above theorem are bounded by 4 and 9, respectively.

Note that a similar result for a zero-divisor graphΓ(R) is studied in [5,8,9,14] as follows.

(1)Γ(R) is connected and diam(Γ(R))≤3.

(2) IfΓ(R) contains a cycle, then gr(Γ(R))≤4.

Lemma 3.3. LetRbe a finite ring with|R| ≥4. ThenR^∼₌Z2×Z2, or there exist two distinct elementsxandyinR− {0, 1}such thatxy /=0.

Proof. For the case|R| =4, it is clear thatRis isomorphic to eitherZ2×Z2,Z2[x]/x² orZ4, which implies the desired result. Next, we study the case for|R_{| ≥}5 by a contrary method. SupposeR− {0, 1} = {a1,a2,. . .,a_m},m≥3, anda_ia_j=0 for all 1≤i /= j≤m.

It is clear thata2+ 1 is diﬀerent from 0 and 1. Otherwise,a1=0 ora2=0, which is a contradiction to the choice ofa1anda2. Ifa2+ 1=/ a1, thena1(a2+ 1)=0, and we have a1=0, which is a contradiction. Thus,a2+ 1=a1. Similarly,a1a3=0, anda3+ 1=a1

implies thata3=a2, which is a contradiction.

In the next theorem, we give a necessary and suﬃcient condition for a hypergraph H3(R) to be complete. In the process of the following proof, we consider the obvious fact

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that H3(Z2×Z2×Z2) has only one edge, and necessarily it is a complete hypergraph.

Note that for a detailed study of the completeness of a zero-divisor graphΓ(R), the reader is referred to [5].

Theorem 3.4. LetRbe a finite nonlocal ring. ThenH3(R) is complete if and only ifR= Z2×Z2×Z2.

Proof. The suﬃcient part of the theorem is trivial, becauseH3(R) has only one edge, and therefore is complete wheneverR=Z2×Z2×Z2. Suppose thatH3(R) is complete. It is a well-known fact that any finite ringRis isomorphic to the product of local rings. Thus, assume that R=R1×R2× ··· ×Rn, where eachRi is a local ring for alli=1, 2,. . .,n.

Now, we study the following cases for diﬀerent values ofn.

Case 1. Supposen≥4. It is clear thate1= {x1,x2,x3}ande2= {y1,y2,y3}with x1=(1, 1, 0, 0,. . ., 0), x2=(1, 0, 1, 0,. . ., 0), x3=(0, 1, 1, 0,. . ., 0),

y1=(1, 0, 0, 1,. . ., 0), y2=(1, 1, 0, 0,. . ., 0), y3=(0, 1, 0, 1,. . ., 0) (3.20) are two edges ofH3(R). Clearly,H3(R) is not complete since{x1,x2,y1}is not an edge of H3(R).

Case 2. LetR=R1×R2×R3. Without loss of generality, suppose that|R1| ≥3. Letx∈ R1 − {0, 1}. Obviously, {(1, 1, 0), (1, 0, 1), (0, 1, 1)} ∈ E(H3(R)) and {(x, 1, 0), (1, 0, 1), (0, 1, 1)}∈E(H3(R)). But{(x, 1, 0), (1, 0, 1), (1, 1, 0)}∈/E(H3(R)), which implies thatH3(R) is not complete. Hence, we can conclude that|Ri| ≤2 andR=Z2×Z2×Z2.

Case 3. LetR=R1×R2. IfH3(R) does not have any vertices, we do not have anything to prove. Therefore, first we assume that|Ri| ≥4 for each 1≤i≤2 and investigate the following subcases.

Case 3.1. The square of one of the components of some 3-zero-divisor ofRis zero. Let (a,b) be a 3-zero-divisor inRwitha²=0 and lete= {(a,b), (c,d), (f,g)}be an edge of H3(R). SinceZ2×Z2 is not a local ring, by Lemma 3.3 there exist distinct elementsx and yin R2− {0, 1}such that xy /=0. Now, from the fact that{(a, 1), (a,x), (1, 0)}and {(a, 1), (a,y), (1, 0)}are in E(H3(R)) and{(a,x), (a,y), (a, 1)}∈/ E(H3(R)), we can conclude thatH3(R) is not complete.

Case 3.2. The square of none of the components of any 3-zero-divisor ofRis zero. Sup- pose thate= {(a1,b1), (a2,b2), (a3,b3)}is an edge ofH3(R). In this case, there always exists i∈ {1, 2, 3}, sayi=1, such thata1a2=/0 anda1a3=/ 0, or similarly,b1b2=/0 andb1b3=/ 0.

Otherwise, the product of two elements ofewill be zero, which contradicts the definition foreto be an edge inH3(R). Without loss of generality, we assume thata1a2=/0 and a1a3=/ 0. By usingLemma 3.3, similar toCase 3.1, there exist distinct elementsxandyin R2− {0, 1}such thatxy /=0. Since{(a1, 0), (a2,x), (a3, 1)}and{(a1, 0), (a2,y), (a3, 1)}are the edges ofH3(R), and{(a2,x), (a2,y), (a3, 1)}is not an edge ofH3(R), thenH3(R) is not complete.

Next, we assume that the size of one of the ringsRi’s is 2, wherei=1, 2. Without loss of generality, assume thatR2=Z2. It is clear thatRdoes not have any 3-zero-divisors