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arXiv:1106.0350v1 [math.RA] 2 Jun 2011

On realizing zero-divisor graphs of po-semirings∗

Houyi Yu†and Tongsuo Wu‡

Department of Mathematics, Shanghai Jiaotong University, Shanghai, 200240, China

Abstract.

In this paper, we determine bipartite graphs and complete

graphs with horns, which are realizable as zero-divisor graphs of po-semirings.

As applications, we classify commutative rings R whose annihilating-ideal

graph AG(R) are either bipartite graphs or complete graphs with horns.

Key Words: Po-semirings; Graph properties; Zero-divisors; Annihilating-

ideals of a commutative ring

MSC(2000): 13A15; 05C75.

1Introduction

Throughout this paper, all rings are assumed to be commutative with identity. For a

ring R, let Z(R) be its set of zero-divisors. The zero-divisor graph of R, denoted by

Γ(R), is a simple graph (i.e., an undirected graph without loops and multiple edges) with

vertices Z(R)∗= Z(R)\{0}, such that distinct vertices x and y are adjacent if and only if

xy = 0. The concept for a ring was first introduced and studied by Beck in [8] and further

investigated by many authors, see [6, 5, 4]. Later, DeMeyer, McKenzie and Schneider [12]

extended the notion to commutative semigroups S with 0 in a similar manner. The idea

establishes a connection between graph theory and algebraic theory and will be beneficial

for those two branches of mathematics.

For a ring R, let I(R) be the set of ideals of R, A(R) the set of annihilating-ideals of R,

where a nonzero ideal I of R is called an annihilating-ideal if there exists a non-zero ideal

J of R such that IJ = 0. The annihilating-ideal graph AG(R) of R, first introduced and

∗The first author is partly supported by Shanghai Jiaotong University Innovation Fund for Postgrad-

uates (NO. AE071202).

†yhy178@163.com (H.Y. Yu)

‡Corresponding author, tswu@sjtu.edu.cn (T.S. Wu)

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studied in [9], is a graph with vertex set A(R)∗= A(R)\{0}, such that distinct vertices

I and J are adjacent if and only if IJ = 0. The graph provides an excellent setting for

studying some aspects of algebraic property of a commutative ring, especially, the ideal

structure of a ring. Clearly, the graph AG(R) is an empty graph if and only if R is an

integral domain.

In fact, I(R) admits a natural algebraic structure, called a po-semiring by Wu, Lu

and Li [18]. Recall that a commutative semiring is a set A which contains at least two

elements 0,1 and is equipped with two binary operations, + and ·, called addition and

multiplication respectively, such that the following conditions hold:

(1) (A,+,0) is a commutative monoid with zero element 0.

(2) (A,·,1) is a commutative monoid with identity element 1.

(3) Multiplication distributes over addition.

(4) 0 annihilates A with respect to multiplication, i.e., 0a = 0, ∀a ∈ A.

Recall the following definition from [18].

Definition 1.1. A partially-ordered semiring is a commutative semiring (A,+,·,0,1),

together with a compatible partial order ≤, i.e., a partial order ≤ on the underlying set A

that is compatible with the semiring operations in the sense that it satisfies the following

conditions:

(5) x ≤ y implies x + z ≤ y + z, and

(6) 0 ≤ x and y ≤ z imply that xy ≤ xz for all x,y,z in A.

If A satisfies the following additional condition, then A is called a po-semiring:

(7) The partially ordered set (A,≤,0,1) is bounded, i.e., 1 is the largest element and

0 is the least element of A.

Note that condition (7) is a very strong assumption. Under the assumption, a po-

semiring A is in fact a dioid, where a semiring is called a dioid if its addition is idempotent

(a+ a = a, ∀a ∈ A). Furthermore, the above defined partial order ≤ for a po-semiring A

is identical with the new partial order ≤1defined by the following

a ≤1b if and only if a + b = b.

In other words, (A,+,0,1) is a bounded join-semilattice. Clearly, any bounded, distribu-

tive lattice is a po-semiring under join and meet, where a ≤ b if and only if a ∧ b = a.

We remark that the class of po-semirings is much smaller than the class of semigroups.

For example, the five-element lattice M5depicted in Figure 1 is not a distributive lattice,

so it is not a po-semiring. But (M5,∧) is clearly a semigroup.

For a po-semiring A, denote by Z(A) the set of all multiplicative zero-divisors. The

zero-divisor graph of the multiplicative semigroup (A,·,1), denoted by Γ(A), is called the

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zero-divisor graph of the po-semiring A. Clearly, all known results on zero-divisor graphs

of semigroups hold for Γ(A). A nonzero element x ∈ A is called minimal, if 0 < y ≤ x

implies x = y for any y ∈ A. Note that each minimal element of A is a zero divisor, if

|Z(A)| ≥ 2. Refer to [18] for more details on po-semirings.

The prototype of a po-semiring is the po-semiring I(R) of a commutative ring R. The

multiplication is the ideal multiplication, the addition is the addition of ideals, the partial

order is the usual inclusion. Therefore, the annihilating-ideal graph of R is the zero-divisor

graph of the po-semiring I(R), i.e., AG(R) = Γ(I(R)).

All throughout, let G be a finite or an infinite simple graph. The vertex set of G is

denoted by V (G). The core of G, denoted by C(G), is the largest induced subgraph of G

in which every edge is an edge of a cycle in G. For a vertex x of G, let N(x) be the set of

vertices adjacent to x, and call it the neighborhood of x. A vertex is called an end vertex

if its degree is 1. All end vertices which are adjacent to a same vertex of G together with

the edges is called a horn. We adopt more graph theoretic notations from [11].

We recall some notation used in [14]. Let X and Y be disjoint nonempty subsets of

the vertex set of a graph. We use the notation X −Y to represent the complete bipartite

graph with parts X and Y . In particular, if u ?∈ X, then u − X represents a star graph,

that is, u is adjacent to every vertex in X and no two distinct vertices in X are adjacent.

The graph depicted in Figure 2 is called a complete bipartite graph with a horn, where

the induced subgraphs on nonempty sets X,Y,U are discrete. We denote the graph by

X − U − v − Y . In particular, G is called a two-star graph if |U| = 1.

0

1

a

b

c

Figure 1

XU

v

Y

Figure 2

a1

X

a2

a3

Y

Figure 3

Let Λ be an index set, and let m,n be two finite or infinite cardinal numbers such that

n = |Λ| ≥ 1, 0 ≤ m ≤ n, and let Knbe a complete graph with V (Kn) = {ai| i ∈ Λ}.

We denote the complete graph Kn together with m horns X1,X2,··· ,Xm by Kn(m),

where a1−X1,a2−X2,··· ,am−Xm. For example, the graph in Figure 3 above is K3(2).

Clearly, K1(0) is an isolated vertex, K1(1), K2(0) (i.e., K2) and K2(1) are star graphs,

K2(2) is a two-star graph, K3(0) is a triangle, while K3(m) is a triangle with m horns

(1 ≤ m ≤ 3). In this paper, we mainly study the case of m ≤ 3. So we always assume

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a1− X,a2− Y,a3− Z for brevity.

As in [13], a graph will be called realizable (for po-semirings) if it is isomorphic to

Γ(A) for some po-semiring A. In this paper, we investigate the realization problem of

graphs as zero-divisor graphs of po-semirings. In Section 2, it is shown that a bipartite

graph G is realizable if and only if G is either a complete bipartite graph or a complete

bipartite graph with a horn. Realizable complete graphs with horns are then completely

determined in Section 3. In particular, it is shown that Kn(m) is the zero-divisor graph

of some po-semiring if and only if either 0 ≤ m ≤ min{2,n} or m = n = 3. The final

section is devoted to the study of annihilating-ideal graph AG(R) of a commutative ring

R. It is proved that AG(R) is a complete bipartite graph with a horn if and only if

R∼= D×S, where D is an integral domain and S is a ring with a unique non-trivial ideal.

AG(R)∼= K3(3) if and only if R∼= F1×F2×F3, where F1,F2,F3are fields. We also show

that there exists no ring R such that AG(R)∼= Kn(2) for any n ≥ 3.

Throughout the paper, set

X = {xi| i ∈ Γ}, Y = {yk| k ∈ Θ}, Z = {zp| p ∈ Ω}, U = {us| s ∈ Φ}.

2Bipartite graphs which are realizable for

po-semirings

In this section, we give a complete classification of all bipartite graphs which can be

realized as po-semiring graphs.

Lemma 2.1. Any complete bipartite graph G is realizable for po-semirings.

Proof. Assume that G has X and Y as its two parts. Set A = {0,1,w} ∪ X ∪ Y and

define a partial order ≤ by

0 < x1< x2< ··· < w < 1, 0 < y1< y2< ··· < w < 1.

Define a commutative addition by

b + c =

?

max{b, c}

w

if b and c are comparable in A,

otherwise.

(1)

Define a commutative multiplication on A by

0a = 0, 1a = a (∀a ∈ A), w2= w,

xixj= x1, xiyk= 0, xiw = x1, ykyl= y1, ykw = y1(i,j ∈ Γ,k,l ∈ Θ).

Clearly, A is a po-semiring such that Γ(A)∼= G.

?

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Lemma 2.2. Let G be a complete bipartite graph with a horn and set G : X −U −v−Y .

Then there exists a po-semiring A such that Γ(A)∼= G.

Proof. Let A = {0,1,v,w} ∪ X ∪ Y ∪ U. We define a partial order ≤ on A by

0 < v < x1< x2< ··· < w < 1,

0 < v < y1< y2< ··· < w < 1,

0 < u1< u2< ··· < y1< y2< ··· < w < 1,

Define a commutative addition by ui+ v = y1for any ui∈ U, and other additions are

defined by (1). Clearly, (A,+) is a commutative semigroup.

Now we define a commutative multiplication on A such that (A,+,·,0,1) is a po-

semiring. For any a ∈ A, let 0a = 0, 1a = a. Furthermore, for any i,j ∈ Γ, k,l ∈ Θ and

s,t ∈ Φ, let

xixj= xiw = x1, xiyk= xiv = v, xius= 0,

ykyl= ykus= u1, ykv = 0, ykw = y1,

usut= usw = u1, usv = v2= 0, vw = v, w2= w.

A direct checking shows that the associativity holds for (A,·). Since the multiplication

is commutative, we only need to verify that the left distributivity holds for (A,+,·,0,1).

In fact, take any a,b,c ∈ A and it is obvious, by the addition and multiplication defined

above, that b < c implies b + c = c and ab ≤ ac, so a(b + c) = ac = ab + ac. Thus, we

only need to assume that b and c are incomparable and show that a(b + c) = ab + ac.

If a = us, then

us(ut+ v) = usy1= u1= usut+ usv,

us(ut+ xi) = usw = u1= usut+ usxi,

us(xi+ yk) = usw = u1= usxi+ usyk.

Similarly, we can show that a(b + c) = ab + ac always holds when a ∈ {v,xi,yk,w} and

b,c ∈ A.

Hence (A,+,·,0,1) is a posemiring. Clearly, Γ(A)∼= G and the result follows.

?

In order to check that (A,+,·,0,1) defined above is a po-semiring, what is the most

difficult is to check the two associative laws and one distributive law hold.

each examination concerns three elements at most. By the definition of addition and

multiplication in Lemma 2.2, it is enough to take |X| = |Y | = |U| = 3 so that |A| = 13,

and hence we can do the examination by taking the advantage of a computer.

In fact,

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In general, a po-semiring corresponding to a given graph need not be unique up to

isomorphism. For example, for the given complete bipartite graph with a horn X − U −

v − Y , we can define another po-semiring A′such that A′is not isomorphic to the one

defined in Lemma 2.2.

Example 2.3. Set X +Y = {x+y | x ∈ X, y ∈ Y }, and set A′= {0,1,v}∪X ∪Y ∪U ∪

(X + Y ). Let (A′,+) be an upper semilattice with the Hasse diagram in Figure 4.In

v

0

x1

1

y1

us

y2

u1

yk

x2

xi

xi+ yk

Figure 4

particular, x1, y1and u1are the unique minimal elements in X, Y , and U, respectively.

Now we define a commutative multiplication on A such that (A′,+,·,0,1) is a po-semiring.

For any i,j ∈ Γ, k,l ∈ Θ, s,t ∈ Φ, we define a multiplication on A′by

0a = 0,1a = a (∀a ∈ A′),

xixj= min{xi,xj}, xiyk= xiv = v, xius= 0, xi(xj+ yk) = xixj,

ykyl= ykus= u1, ykv = 0, yk(xi+ yl) = y1, usut= us(xi+ yk) = u1, usv = 0,

v2= 0, v(xi+ yk) = v, (xi+ yk)(xj+ yl) = xixj+ y1.

Then (A′,+,·,0,1) is a po-semiring with Γ(A′) is isomorphic to the given graph. Clearly,

A′is not isomorphic to the po-semiring A constructed in Lemma 2.2, if |X| ≥ 2.

The following is the main result of this section.

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Theorem 2.4. Let G be the zero-divisor graph of a po-semiring with |V (G)| ≥ 2. Then

the following conditions are equivalent.

(i) G is triangle-free.

(ii) G is a bipartite graph.

(iii) G is either a complete bipartite graph or a complete bipartite graph with a horn.

Proof. Note that any star graph is a complete bipartite graph and any two-star graph is

a complete bipartite graph with a horn, so if G = Γ(S) for some zero-divisor semigroup S,

then the equivalence of (i) and (ii) follows from [14, Theorem 2.1] while the equivalence of

(ii) and (iii) follows from [14, Theorem 2.10]. Therefore, we only need to show that both

complete bipartite graphs and complete bipartite graphs with a horn can be realized for

po-semiring, which follows from Lemmas 2.1 and 2.2.

?

Lemma 2.5. Any isolated vertex G is realizable for po-semirings.

Proof. Let G = {a}. Set A = {0,1,a}, and define a partial order < by 0 < a < 1. Define

a commutative addition by x + y = max{x,y} for any x,y ∈ A. Define a commutative

multiplication by

0x = 0, 1x = x(∀ x ∈ A), a2= 0.

Then it is routine to check that A is a po-semiring such that Γ(A)∼= G.

?

Recall that a tree is a simple connected graph G without a cycle, i.e., the core of G is

empty. By Theorem 2.4 and Lemma 2.5 , we have the following corollary.

Corollary 2.6. Let G be a tree. Then G is the graph of a po-semiring if and only if G is

one of the following graphs: an isolated vertex, a star graph, a two-star graph.

3Complete graphs with horns which are realizable

for po-semirings

In this section we classify all po-semiring graphs which are complete graphs or complete

graphs with horns. Note that the following facts were already obtained in section 2:

Kn(m) is realizable for po-semirings, for any m,n with 1 ≤ n ≤ 2, 0 ≤ m ≤ n. Thus we

only need to consider the case of n ≥ 3 in this section.

Lemma 3.1. For any n ≥ 3 and m ∈ {0,1,2}, there exists a po-semiring A such that

Kn(m)∼= Γ(A).

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Proof. It suffices to show that each graph of the given type has a corresponding po-

semiring.

Case 1. Suppose that m = 0. Then G∼= Knis a complete graph where V (Kn) =

{ai | i ∈ Λ}. Set A = {0,1} ∪ {ai | i ∈ Λ}, and define a partial order ≤ on A by

0 < a1< a2< ··· < 1. Define an addition on A by x + y = max{x,y} for any x,y ∈ A.

Define a commutative multiplication on A by

0x = 0, 1x = x (∀x ∈ A) and aiaj= 0 (∀i,j ∈ Λ).

Then it is routine to check that A is a po-semiring with Γ(A)∼= Kn.

Case 2. Suppose that m = 1. Set A = {0,1} ∪ {ai | i ∈ Λ} ∪ X and define a

partial order ≤ on A by 0 < a1< a2< ··· < x1< x2< ··· < 1. Define an addition by

x + y = max{x,y} for any x,y ∈ A. Define a commutative multiplication by

0a = 0, 1a = a (∀a ∈ A),

aiaj= a1xk= 0, arxk= ar, xkxl= x1(∀i,j,r ∈ Λ,r ?= 1,∀k,l ∈ Γ).

Then (A,+,·,0,1) is a po-semiring such that Γ(A)∼= Kn(1), where a finite or an infinite

horn X is adjacent to the vertex a1.

Case 3. Suppose that m = 2. Denote X + Y = {x + y | x ∈ X, y ∈ Y }, and set

A = {0,1} ∪ {ai| i ∈ Λ} ∪ X ∪ Y ∪ (X + Y ). Let (A,+) be a upper semilattice with the

Hasse diagram in Figure 5.

In particular, an is the unique maximal element of {ai | i ∈ Λ}. Now we define

a commutative multiplication on A such that (A,+,·,0,1) is a po-semiring. For any

i,j ∈ Λ, k,l ∈ Γ, p,q ∈ Θ, we define a multiplication on A by

0a = 0,1a = a (∀a ∈ A),

aiaj= 0 (j ?= n), a2

n= an,

a1xk= 0, a1yp= a1(xk+ yp) = a1,

a2xk= a2, a2yp= 0, a2(xk+ yp) = a2,

anxk= anyp= an(xk+ yp) = an,

aixk= a2, aiyp= a1, ai(xk+ yp) = a3, (i ?= 1,2,n),

xkxl= x1, xkyp= an, xk(xl+ yp) = x1,

ypyq= yp(xk+ yq) = y1, (xk+ yp)(xl+ yq) = x1+ y1.

Then (A,+,·,0,1) is a po-semiring with Γ(A)∼= Kn(2).

?

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a1

1

0

a2

a3

ai

x1

x2

y1

y2

an

xk

yp

xk+ yp

Figure 5

Lemma 3.2. There exists a posemiring A such that Γ(A)∼= K3(3).

Proof. Set A = {0,1,a1,a2,a3,w} ∪ X ∪ Y ∪ Z. Define a partial order on A by

0 < a1,a2< z1< z2< ··· < w < 1,

0 < a2,a3< x1< x2< ··· < w < 1,

0 < a1,a3< y1< y2< ··· < w < 1.

Then A is a partially-ordered set with a unique maximal element w ?= 1, where a1, a2and

a3are the only nonzero minimal elements. Define a commutative addition by

0 + a = a + a = a, 1 + a = 1, (∀a ∈ A),

a1+ a2= z1, a1+ a3= y1, a2+ a3= x1,

and for any b,c ∈ A such that {b,c} ? {a1,a2,a3}, we define

b + c =

?

max{b, c}

w

if b and c are comparable in A,

otherwise.

(2)

For any k ∈ Γ, p ∈ Θ and s ∈ Ω, define a commutative multiplication by

0a = 0, 1a = a (∀a ∈ A),

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a2

1= a1,a1a2= a1a3= a1xk= 0,a1yp= a1zs= a1w = a1,

a2

2= a2,a2a3= a2yp= 0,a2xk= a2zs= a2w = a2,

a2

3= a3,a3zs= 0,a3xk= a3yp= a3w = a3,

X2= {x1},XY = {a3},XZ = {a2},wX = {x1},

Y2= {y1},Y Z = {a1},wY = {y1},Z2= {z1},wZ = {z1},w2= w.

It is easy to check that (A,+,·,0,1) is a po-semiring such that Γ(A)∼= K3(3).

?

The following is an improvement of [16, Theorem 2.2].

Lemma 3.3. For any n ≥ 4 and any m with n ≥ m ≥ 3, Kn(m) has no corresponding

semigroups.

Proof. Assume to the contrary that S = {0} ∪ {ai| i ∈ Λ} ∪ X1∪ X2∪ ··· ∪ Xmis a

zero-divisor semigroup such that Γ(S)∼= Kn(m), where n ≥ 4 and n ≥ m ≥ 3.

First we claim that for any xj∈ Xj, we must have aixj= ai, where 1 ≤ i ?= j ≤ m.

In fact, if aixj= xkfor some xk∈ Xk, then ajxk= ai(ajxj) = 0 so that j = k whence

xixj= xi(aixj) = 0, a contradiction. Thus, aixj∈ {ai| i ∈ Λ}. Let aixj= ar, then for

any x′

Take x2 ∈ X2, x3 ∈ X3. Then a1(x2x3) = (a1x2)x3 = a1x3 = a1 so that x2x3 ?∈

{ai | i ∈ Λ,i ?= 1}. On the other hand, a2(x2x3) = 0 = a3(x2x3) which means that

x2x3∈ {ai| i ∈ Λ} and hence x2x3= a1. So (a4x2)x3= a4a1= 0, which implies that

a4x2 = a3. Substituting a3,x3,a1,x1 by a1,x1,a3,x3 respectively, we can obtain that

a4x2= a1, a contradiction.

i∈ Xi, arx′

i= (aix′

i)xj= 0 and hence r = i, that is, aixj= ai.

?

Note that a po-semiring is certainly a multiplicative commutative semigroup. So we

have

Corollary 3.4. For any n ≥ 4 and any m with n ≥ m ≥ 3, Kn(m) has no corresponding

po-semirings.

In view of Lemmas 2.5, 3.1, 3.2 and 3.3, we obtain the main result of this section.

Theorem 3.5. Let G = Kn(m) be a complete graph Kntogether with m horns, where

n and m could be any cardinal numbers such that 0 ≤ m ≤ n and n ≥ 1. Then G is a

po-semiring graph if and only if either 0 ≤ m ≤ min{2,n} or n = m = 3.

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4Some results on annihilating-ideal graphs of com-

mutative rings

The annihilating-ideal graph AG(R) of a commutative ring R was introduced in [9] and

further studied in [10, 1, 2, 3]. In this section, we add more results on the graph.

Lemma 4.1. Let A be a po-semiring and let x−u−y be a path in Γ(A). If it is contained

in neither triangle nor quadrilateral, then u is a minimal element of A.

Proof. Assume to the contrary that u is not a minimal element of A. Assume further

0 < v < u for some v ∈ A. If v = x, then xy = vy ≤ uy = 0 whence there is a triangle

x − u − y − x, a contradiction. Similarly, v ?= y. If v ?= x,y, then there is a square

x − u − y − v − x, another contradiction.

?

Lemma 4.2. Let R be a ring, I a minimal ideal of R. Then ann(I) is a maximal ideal

of R.

Proof. Since I is minimal, I is a principal ideal. Suppose that I = Rx for some x ∈ I.

Then ann(x) = ann(I) whence Rx∼= R/ann(I) is a simple R-module, so ann(I) is a

maximal ideal.

?

In view of [10, Theorem 2.3] and [3, Corollary 23], we known that AG(R) is a complete

bipartite graph if and only if either AG(R) is a star graph or R is a reduced ring with

|Min(R)| = 2. For a complete bipartite graph with a horn, we have

Theorem 4.3. The graph AG(R) is a complete bipartite graph with a horn if and only

if R∼= D × S, where D is an integral domain and S is a ring with a unique non-trivial

ideal. In this case, AG(R)∼= K1+Dr+K1+Drwhere r is either 1 or an infinite cardinal

number.

Proof. (⇒) Let AG(R) be a complete bipartite graph with a horn X − U − c − Y .

Set p =

?

u ∈ U.

If |U| = 1, then AG(R) is a two-star graph. By [2, Theorem 2], AG(R)∼= P4and

R∼= F × S where F is a field and S is a ring with a unique non-trivial ideal. So we

assume that |U| ≥ 2 in next discussion.

Now, we show the following claims:

Claim 1. q is a maximal ideal. By Lemma 4.1, c is a minimal ideal. Note that

cq = 0, so we only need to show that ann(c) = q by Lemma 4.2. In fact, first we have

pi∈Xpi, q =

?

qj∈Yqj and u =

?

uk∈Uuk. Clearly, cq = 0. But for any

pi∈ X,uk∈ U, qpi?= 0,quk?= 0, which means that q ∈ Y . Similarly, we have p ∈ X and

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pq ?= 0. On the other hand, for any uj ∈ U, uj(pq) = 0 = c(pq) which together with

|U| ≥ 2 yield that pq = c. Thus, c2= (cq)p = 0 whence c(c + q) = 0, which together

with u(c + q) = uq ?= 0 and p(c + q) ?= 0 imply that c + q ∈ Y and hence c ⊆ q. Since

c(u + q) = 0, but p(u + q) = pq = c ?= 0 so that u + q ∈ Y , which means that u ⊆ q.

Therefore, we obtain that ann(c) = c ∪ u ∪ q = q.

Claim 2. p is a prime ideal and p + q2= R. Take any x ∈ u. Then p(Rx) ⊆ pu =

0 = c(Rx), so Rx ∈ U. Clearly, x2?= 0. Otherwise, (Rx)(Rx + c) = 0 = c(Rx + c). Since

c is minimal, it follows that Rx + c ?= c, so Rx + c ∈ U and hence Rx + c = Rx. Thus

c ⊆ Rx and hence cp ⊆ p(Rx) = 0, a contradiction. So x2?= 0 whence ann(x) = c ∪ p.

Since c is minimal, it follows that cp = c whence c ⊆ p and hence ann(x) = p. If

p is not a maximal ideal contained in Z(R), then there exists y ∈ Z(R)\p such that

p + Ry ⊆ Z(R). By the proof of Claim 1, we obtain that u ⊆ q and c ⊆ q, so we have

Z(R) = p ∪ u ∪ c ∪ q = p ∪ q whence y ∈ q\p. Take any z ∈ p such that cz ?= 0. Then

Rz ∈ X and hence c(y+z) = cz ?= 0 so that y+z ?∈ q, at the same time, u(y+z) = uy ?= 0

which implies that y + z ?∈ p, that is, y + z ?∈ Z(R), a contradiction. This proves that p

is maximal among all ideals of R that are annihilators of elements, so p is a prime ideal.

Clearly, p and q are incomparable. Otherwise, we have p ⊆ q, so c = pc ⊆ qc = 0, a

contradiction. Therefore, p + q2= R, as required.

Claim 3. p ∩ q2= 0. Note that for any uk∈ U, we have uk(p ∩ q2) ⊆ ukp = 0 and

c(p ∩ q2) ⊆ cq = 0 which together with |U| ≥ 2 yield that p ∩ q2= c or p ∩ q2= 0. If

p ∩ q2= c, then c ⊆ q2, hence c = pc ⊆ pq2= cq = 0, a contradiction. So, p ∩ q2= 0.

By Chinese Remainder Theorem, we have R∼= R/p × R/q2.

integral domain, R/q2is an Artinian local ring. Thus all non-trivial ideals of R/q2are

annihilating-ideals. If there exist two non-trivial ideals, say m,n, in R/q2, then m,n ⊆ q/q2

and hence mn = 0. So we have a triangle (R/p,0) − (0,m) − (0,n) − (R/p,0) in AG(R),

which contradicts the fact that AG(R) is a bipartite graph. This complete the proof of

necessity.

(⇐) Let R = D × S where D is an integral domain, S has a unique non-trivial ideal

m. If D is not a field, then D has infinitely many ideals, and AG(R) is the complete

bipartite graph with a horn X − U − c − Y , where

Clearly, R/p is an

c = (0,m), X = {(0,S)},

Y = {(b,m) | b ∈ I(D)∗}, U = {(b,0) | b ∈ I(D)∗}.

So AG(R)∼= K1+Dr+K1+Dr, where r is the cardinality of I(D)∗. In this case, r = ∞.

If D is a field, then r = 1 and AG(R)∼= P4. This complete the proof of the theorem. ?

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Remark 4.4. Let R be a ring with only one non-trivial ideal. Then, by [15], either

R∼= K[x]/(x2) where K is a field or R∼= V/p2V , where V is a discrete valuation ring of

characteristic zero and residue field of characteristic p, for some prime number p. In the

latter case, the characteristic of R is p2.

By [1, Theorem 3], the graph AG(R) is a complete graph if and only if either AG(R)∼=

K2or Z(R) is an ideal of R with Z(R)2= 0. Moreover, in the first case, either R∼= F1×F2,

where F1,F2are fields, or (R,m) is a local ring with exactly two non-trivial ideals m and

m2. For complete graphs with horns, K1(0) is an isolated vertex, K1(1),K2(0) and K2(1)

are star graphs, while K2(2) is a two-star graph. By [1, Theorem 2] or Theorem 4.3 above,

AG(R) is a two-star graph if and only if AG(R)∼= P4, if and only if R∼= F × S, where

F is a field and S is a ring with a unique non-trivial ideal. For Kn(m) where n ≥ 3 and

1 ≤ m ≤ 3, we have the following results.

Lemma 4.5. Let AG(R)∼= Kn(1), where n ≥ 3. Then Z(R) is a maximal ideal. If

further R is Artinian, then Z(R)5= 0.

Proof. Let c be the only center of AG(R), that is, the only horn is adjacent to c. Then,

by Lemma 4.1, c is a minimal ideal. Take distinct a,b ∈ V (Kn)\{c}. Then a(c + b) = 0

so that c + b ∈ V (Kn). Obviously, c + b ?= c and hence c(c + b) = 0, that is, c2= 0.

Put m = Z(R). Clearly, m = ann(c), which means that m is maximal by Lemma 4.2.

If R is an Artinian ring, then m is a nilpotent ideal. If the nilpotency index is n ≥ 6,

then {mn−2,mn−1} ⊆ N(m2) ∩ N(m3), so m2and m3can not be end vertices, Hence

m2,m3∈ V (Kn) so that m5= 0, a contradiction. Therefore, n ≤ 5, as required.

?

The structure of R with AG(R)∼= Kn(1) seems to be rather complicated and hard to

determine completely. For n = 1 or n = 2 where the unique horn contains exactly one

vertex, see [17] for the detailed structure theorems on commutative rings with at most

three nontrivial ideals. We will discuss the problem for n ≥ 3 in a separate paper.

Theorem 4.6. There exists no ring R such that AG(R)∼= Kn(2) for any n ≥ 3.

Proof. Let R be a ring such that AG(R)∼= Kn(2) for some n ≥ 3. Suppose that Knis

the complete graph with V (Kn) = {ai| i ∈ Λ} where n = |Λ| ≥ 3, and the two horns are

a1− X, a2− Y . Put a =?

Claim 1. p and q are maximal ideals. In view of Lemma 4.1, we get that a1,a2are

minimal, so a1+ a2?∈ {a1,a2}. Since a3(a1+ a2) = 0, it follows that a1+ a2∈ {ai| i ∈

Λ,i ?= 1,2} and hence a1(a1+ a2) = 0, so a2

a2

2= 0,a2a = 0, which means that a ∈ {ai| i ∈ Λ}. Note that a1(a + p) = 0, it follows

i∈Λai, p =?

pj∈Xpjand q =?

qk∈Yqk. Then p ∈ X, q ∈ Y .

Now, we have the following claims:

1= 0, which yields that a1a = 0. Similarly,

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that a + p ∈ A(R). On the other hand, a2(a + p) = a2p ?= 0, so a + p ∈ X and hence

a + p ⊆ p whence a ⊆ p. Thus ann(a1) = a ∪ p = p. Consequently, p is a maximal ideal

of R by Lemma 4.2. Similarly, q is also maximal.

Claim 2. a3= 0. By the proof of Claim 1, a ⊆ p∩q. On the other hand, a1(p∩q) =

a2(p∩q) = 0 implies that p∩q ∈ V (Kn) and hence p∩q ⊆ a so that p∩q = a. Now, we

show that a2?= a. Assume to the contrary that a2= a. Since a1+a2⊆ a, it follows that a

is not minimal. So we can take 0 ?= x ∈ a such that a ?= Rx. Then there exist u,v ∈ a such

that x = uv. This implies that (Ru)(Rv) = Rx ?= 0. If a = Ru = Rv, then Rx = a2= a,

a contradiction. If at least one of Ru,Rv is not equal to a, without loss of generality,

suppose that Ru ?= a, then Ru ∈ V (Kn) whence (Ru)(Rv) ⊆ (Ru)a = 0, that is, Rx = 0,

another contradiction. Thus, a2?= a. Clearly, a2∈ V (Kn), hence a3= aa2= 0.

By the proof of Claim 2, p∩q = a, so pq ⊆ a and hence p3q3= 0. By Claim 1, p,q are

maximal, hence R is an Artinian ring with exactly two maximal ideals. By [7, Theorem

8.7], there exist two Artinian local rings, say (R1,m) and (R2,n), such that R = R1×R2.

Hence we can assume that p = (m,R2), q = (R1,n) and hence a = (m,n). Note that

(R1,0)−(0,R2)−(m,0) is a path in AG(R), it follows that (0,R2) is not an end vertices,

that is, (0,R2) ∈ V (Kn) and hence a(0,R2) = (0,n) = 0 so that n = 0. Similarly,

m = 0. This implies that R1 and R2 are fields, and AG(R)∼= K2, a contradiction.

The contradiction followed from the assumption that there exists a ring R such that

AG(R)∼= Kn(2) for some n ≥ 3.

?

Theorem 4.7. AG(R) is K3(3) if and only if R∼= F1× F2× F3, where F1,F2,F3are

fields.

Proof. Let AG(R)∼= K3(3). Suppose that V (K3) = {a,b,c}, the three horns are a−X,

b − Y and c − Z. By Lemma 4.1, a,b,c are minimal ideals. Denote m =?

n ∈ Y , p ∈ Z. Now, we show the following claims:

Claim 1. a2= a,b2= b,c2= c. Since a is minimal, it follows that a2∈ {0,a}. If

a2= 0, then a(a + b) = 0, which together with c(a + b) = 0 yields that a + b ∈ {a,b,c},

which contradicts the fact a,b,c are minimal ideals. Similarly, b2= b,c2= c.

Claim 2. m,n,p are maximal ideal of R. Since am = 0, so, by Lemma 4.2, we only

need to show that m = ann(a). Note that a(m + b) = 0, b(m + b) ?= 0, c(m + b) ?= 0,

hence m + b ∈ X whence m + b ⊆ m, that is, b ⊆ m. Similarly, c ⊆ m. Therefore,

ann(a) = b ∪ c ∪ m = m so that m is a maximal ideal. The other two results can be

obtained similarly.

Claim 3. m ∩n∩p = 0. Take any x ∈ m∩n∩p. If x ?= 0, then Rx ∈ N(a) ∩N(b) ∩

N(c) = ∅, a contradiction.

mi∈Xmi,

n =?

nj∈Ynj, p =?

pk∈Zpk. Clearly, am = 0, but bm ?= 0, cm ?= 0, so m ∈ X. Similarly,

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By Chinese Remainder Theorem, we have R∼= R/m × R/n × R/p, that is, R∼=

F1× F2× F3, where F1,F2,F3are fields. The converse is trivial.

?

Corollary 4.8. If AG(R) is K3(3), then |AG(R)| = 6.

It is interesting to make a comparison between the results of sections 2, 3 and those

in section 4. Lemma 2.2 shows that for any complete bipartite graph with a horn, there

exists a corresponding po-semiring. However, in view of Theorem 4.3, only a few of

complete bipartite graphs with a horn can ba realized as annihilating-ideal graph. We

can obtain a similar result for complete graphs with horns by Theorems 3.5, 4.6 and 4.7.

This observation indicates that the class of po-semirings I(R) of rings R is a very small

subclass of po-semirings. So, we have the following question.

Question 4.9. Classify the po-semirings A such that A∼= I(R) for some commutative

ring R.

References

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The classification of the annihilating-ideal graph of a commutative ring, Algebra

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[2] G. Aalipour, S. Akbari, R. Nikandish, M.J. Nikmehr and F. Shaveisi. Minimal prime

ideals and cycles in annihilating-ideal graphs, Rocky Mountain J. Math. (ac-

cepted).

[3] G. Aalipour, S. Akbari, R. Nikandish, M.J. Nikmehr and F. Shaveisi. On the coloring

of the annihilating-ideal graph of a commutative ring, Preprint.

[4] S. Akbari, H.R. Maimani, S. Yassemi, When a zero-divisor graph is plannar or a

complete r-partite graph, J. Algebra, 270(2003) 169 − 180.

[5] D.F. Anderson and P.S. Livingston. The zero-divisor graph of a commutative ring,

J. Algebra 217(1999) 434 − 447.

[6] D.D. Anderson and M. Naseer. Beck’s coloring of a commutative ring, J. Algebra

159(1993) 500 − 514.

[7] M.F. Atiyah and I.G. MacDonald. Introduction to Commutative Algebra, Addison-

Wesley, Reading, MA, 1969.

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