Electronic Journal of Graph Theory and Applications 12 .
, 169Ae180 On domination numbers of zero-divisor graphs of commutative rings Sarah E.
Andersona .
Michael C.
Axtella .
Brenda K.
Kroschela .
Joe A.
Stickles.
Jr.
Department of Mathematics.
University of St.
Thomas.
St.
Paul.
MN.
USA b School of Mathematics and Computational Sciences.
Millikin University.
Decatur.
IL.
USA ande1298@stthomas.
edu, mike.
axtell@stthomas.
edu, bkkroschel@stthomas.
edu, jstickles@millikin.
Abstract Zero-divisor graphs of a commutative ring R, denoted e(R), are well-represented in the literature.
In this paper, we consider domination numbers of zero-divisor graphs.
For reduced rings.
Vatandoost and Ramezani characterized the possible graphs for e(R) when the sum of the domination numbers of e(R) and the complement of e(R) is n Oe 1, n, and n 1, where n is the number of nonzero zero-divisors of R.
We extend their results to nonreduced rings, determine which graphs are realizable as zero-divisor graphs, and provide the rings that yield these graphs.
Keywords: zero divisor graph, commutative rings, domination, total domination Mathematics Subject Classification : 05C25, 13A70, 05C69 DOI: 10.
5614/ejgta.
Introduction The concept of the graph of the zero-divisors of a commutative ring was first introduced by Beck in .
when discussing the coloring of a commutative ring.
In his work all elements of the ring were considered vertices of the graph.
Since the seminal paper by D.
Anderson and Livingston .
, the standard is to regard only nonzero zero-divisors as vertices of the graph, and we adhere to this standard.
The zero-divisor graph of R, denoted e(R), is the graph with V .
(R)) = Z(R)O , and for distinct r, s OO Z(R)O , r Oe s OO E.
(R)) if and only if rs = 0.
Among other results.
Anderson and Livingston proved that e(R) is always connected and has diameter at most 3 (.
Received: 30 August 2023.
Revised: 12 April 2024.
Accepted: 15 May 2024.
On domination numbers of zero-divisor graphs of commutative rings Anderson et al.
Theorem 2.
This discovery of strong graphical structure in zero-divisor graphs has inspired researchers to continue exploring how the graphical structure of the zero-divisor graph might reveal information about the algebraic structure in the ring, a desire necessitated by the frequent lack of closure under addition in the set of zero-divisors of a ring.
For general surveys of e(R), see .
Domination has been extensively studied in the literature, though less frequently in regards to graphs constructed from rings.
(See .
, .
, and .
for some recent examples of papers on ) The main focus of this paper concerns the domination number of zero-divisor graphs.
In particular, we generalize theorems from Section 4 of .
The results presented in that paper focus on what graphs are possible, and we determine which of these graphs are actually realizable as zero-divisor graphs of commutative rings.
Further, whereas the results of .
were restricted to reduced commutative rings with identity, or rings that have no nontrivial nilpotent elements, we extend these results to non-reduced rings.
In some papers on zero-divisor graphs, e(R) is not a simple graph in the sense that a vertex v could have a loop if and only if v 2 = 0.
Since looped vertices do not impact the domination number of a graph, this paper will adopt the convention that all zero-divisor graphs are simple graphs.
For reference, we will make copious use of the results found in .
Below is a summary theorem outlining the main results of this paper.
Theorem 1.
Let R be a commutative ring with identity.
(R)) = n2 if and only if R O = Z3 y Z3 or R O = Z2 y Z2 y Z2 .
(R)) .
(R)) = n 1 if and only if R = Z2 yZ2 or Z(R) is an ideal with (Z(R))2 = 0 .
(R)) .
(R)) = n if and only if R O = Z6 .
Z8 .
Z2 .
3 ).
Z3 yZ3 , or Z4 .
x, x2 Oe.
(R)) .
(R)) = n Oe 1 if and only if R O = Z2 y F4 or R O = Z2 y Z2 y Z2 Definitions Throughout, by a ring we mean a commutative ring with identity, typically denoted by R.
use Z(R) to denote the set of zero-divisors of R and Z(R)O to denote the set of nonzero zerodivisors.
For the set of integers modulo n and the field with n elements, we use the notations Zn and Fn , respectively.
For a OO R, the annihilator of a is ann.
= .
OO R | ax = .
A ring is local if it has a unique maximal ideal, typically denoted by M .
For a general algebra reference, see .
For any graph G, we denote the set of vertices of G by V (G) and the set of edges by E(G).
We will write v Oe w when vertices v and w are adjacent, or are incident to the same edge edge.
a path between v and w, we mean a sequence of vertices and edges v Oe x1 Oe x2 Oe A A A Oe xn Oe w, and G is connected if there exists a path between any two distinct vertices.
The distance between v and w, denoted by d.
, .
, is the number of edges in a shortest path connecting v and w .
ote that d.
, .
= 0 and d.
, .
= O if no such path exist.
The diameter of G is diam(G) = sup.
, .
| v, w OO V (G)}.
For a general graph theory reference, see .
If every pair of distinct vertices are adjacent in a graph G, then G is said to be a complete graph, and a complete graph on n vertices is denoted as Kn .
A graph G is called complete bipartite if On domination numbers of zero-divisor graphs of commutative rings Anderson et al.
there exist sets A.
B OC V (G) such that A O B = V (G).
A O B = OI, for all vi , vj OO A and wi , wj OO B, we have vi Oe vj OO E(G), wi Oe wj OO E(G), and for all vi OO A and wj OO B, we have vi Oe wj OO E(G).
Finite complete bipartite graphs are denoted as Km,n , where |A| = m and |B| = n.
If |A| = 1, then the graph K1,n is called a star graph.
A graph in which at least one vertex is adjacent to every other vertex is called a star-shaped reducible.
The graph v1 Oev2 OeA A AOevn with no other edges or vertices is called the path graph on n vertices and is denoted Pn , while the graph v1 Oe v2 Oe A A A Oe vn Oe v1 with no other edges or vertices is called the cycle graph on n vertices and is denoted Cn .
To create the corona of graphs G and H, denoted G H, let V (G) = .
1 , v2 , .
, vn }.
Enumerate n copies of H as H1 .
H2 , .
Hn .
Then we create G H by joining vi to every vertex in Hi with an edge for i = 1, .
, n.
For a graph G, a set X OI V (G) is a dominating set of G if for every y OO V (G)\X there exists x OO X such that x Oe y OO E(G).
The domination number of G, denoted (G), is (G) = min{|X| | X is a dominating set of G}.
This paper will also focus on the complement of a zero-divisor graph of a commutative ring Given R, the complement of e(R) is denoted e(R) with V .
(R)) = V .
(R)), and a Oe b OO E.
(R)) if and only if a Oe b OO / E.
(R)).
, ab = 0 in R.
Throughout this paper, we only consider finite rings and will use n to denote |V .
(R))|.
equivalently, |Z(R)O |.
Domination numbers of zero-divisor graphs In .
Vatandoost and Ramezani investigated the domination number and signed domination number of reduced commutative rings.
A reduced commutative ring is a commutative ring in which x2 = 0 if and only if x = 0.
Given R, a reduced commutative ring with identity, the results in .
classified the realizable graphs for e(R) if .
(R)) .
(R)) OO .
Oe 1, n, n .
The results presented below make repeated use of the excellent paper by Redmond, .
this paper.
Redmond classifies all possible zero-divisor graphs with 14 or fewer vertices and their associated commutative rings.
Thus, given a graph with 14 or fewer vertices, it is possible to know whether it corresponds to a zero-divisor graph of a commutative ring and to which ring.
Our first theorem generalizes .
Theorem 4.
, which states that for R, a reduced commutative ring with identity, .
(R)) = n2 if and only if e(R) is C4 or K3 K1 .
The following result from .
is used in the proof.
Lemma 3.
, .
For a graph e with even order m and no isolated vertices, .
= n2 if and only if the components of e are the cycle C4 or the corona H K1 , where H is a connected graph.
Theorem 3.
Let R be a commutative ring with identity.
Then .
(R)) = n2 if and only if RO = Z3 y Z3 or R O = Z2 y Z2 y Z2 .
Proof.
(N.
It is easy to check that |Z(Z3 y Z3 )O | = 4 and .
(Z3 y Z3 )) = 2 and that |Z(Z2 y Z2 y Z2 )O | = 6 and .
(Z2 y Z2 y Z2 )) = 3 .
ee Figure .
(N.
By Lemma 3.
1, e(R) is the cycle C4 or the corona H K1 , where H is a connected graph.
If e(R) is C4 , then R O = Z3 y Z3 by .
Now suppose e(R) is H K1 where H is a connected Let A = .
i OO Z(R)O | deg.
i ) > 1 in e(R)}.
A consists of the vertices from H.
On domination numbers of zero-divisor graphs of commutative rings .
, .
, .
Anderson et al.
, 0, .
, 1, .
, 1, .
, 0, .
, 0, .
, 1, .
, .
, .
Figure 1.
e(Z3 y Z3 ) and e(Z2 y Z2 y Z2 ) Since diam.
(R)) O 3, the induced subgraph on A is complete.
We consider two cases based on the size of A.
If |A| = 2, then e(R) is the path graph P4 with a Oe b Oe c Oe d.
However, by .
Example 1.
P4 is not the zero-divisor graph of any commutative ring with identity.
Suppose |A| > 3.
Let ai OO A and let ai OO Z(R)O with ann.
i ) O A = .
i }.
Consider a1 a2 .
Then a1 a2 = 0.
Otherwise, ann.
1 ) = ann.
2 ) = A O .
, a2 }, a contradiction.
Since a1 .
1 a2 ) = 0, we have a1 a2 OO Z(R)O .
Since a1 a2 OO ann.
1 )\.
1 }, we see that a1 a2 OO A.
Let b OO A\.
1 , a2 , a1 a2 } .
ince |A| > .
Since the subgraph induced by A is complete, b.
1 a2 ) = 0.
Thus, ba1 = 0 because ba2 = 0.
This is a contradiction.
Therefore, |A| = 3.
This implies e(R) O = K3 K1 .
By .
R O = Z2 y Z2 y Z2 .
The following theorem characterizes exactly when .
(R)) .
(R)) = n 1.
This theorem is a generalization of .
Theorem 4.
, which states for R, a reduced commutative ring with identity, .
(R)) .
(R)) = n 1 if and only if e(R) is the complete graph Kn .
Theorem 3.
Let R be a commutative ring with identity.
Then the following are equivalent.
(R)) .
(R)) = n 1.
e(R) is the complete graph Kn .
R O
= Z2 y Z2 or xy = 0 for all x, y OO Z(R).
R = Z2 y Z2 or Z(R) is an ideal with Z(R)2 = .
Proof.
The equivalence of 1 and 2 follows from the proof of .
Theorem 4.
, while the equivalences of 2, 3, and 4 follow from .
Corollary 2.
7 and Theorem 2.
Note that if R is Artinian, then statement .
of Theorem 3.
3 is equivalent to R O = Z2 y Z2 or R is local with maximal ideal M such that M 2 = .
In reference to .
, if R is reduced the following corollary holds.
Corollary 3.
Let R be a reduced commutative ring with identity.
Then .
(R)) .
(R)) = n 1 if and only if R O = Z2 y Z2 .
Proof.
If R O = Z2 y Z2 , then |Z(Z2 y Z2 )O | = 2.
By construction of e(Z2 y Z2 ) and e(Z2 y Z2 ), we see that .
(Z2 y Z2 )) .
(Z2 y Z2 )) = 1 2 = 2 1.
Conversely, by Theorem 3.
= Z2 y Z2 or xy = 0 for all x, y OO Z(R).
Thus, since R is reduced, we have R O = Z2 y Z2 .
On domination numbers of zero-divisor graphs of commutative rings Anderson et al.
We remind the reader of a useful graph theory result.
Lemma 3.
Theorem 13.
If a simple graph G has n vertices and no isolated vertices, then (G) O n2 .
Thus, since zero-divisor graphs of commutative rings are connected and in this paper simple graphs are considered, .
(R)) O n2 .
This result will be utilized in the proof of Theorem 3.
Theorem 1.
In addition, we will use a result that relates the number of vertices, number of edges, and domination number of a graph.
The next lemma follows from a theorem in .
, which states for a simple graph G with n vertices and m edges, if (G) Ou 2, then .
Oe (G)).
Oe (G) .
mO Lemma 3.
Let G be a simple graph with n Ou 2 vertices.
Then (G) = n Oe 1 if and only if G has exactly one edge.
Proof.
(N.
Clear.
(N.
If G has exactly one edge, then precisely one vertex is dominated and, thus, (G) = n Oe 1.
We now consider when .
(R)) .
(R)) = n.
In the case when R is a reduced commutative ring with identity.
Vatandoost and Ramezani proved .
(R)) .
(R)) = n if and only if e(R) is C4 or P3 (.
Theorem 1.
Theorem 3.
Let R be a commutative ring with identity.
Then the following are equivalent.
(R)) .
(R)) = n.
e(R) is C4 or P3 .
R O
= Z6 .
Z8 .
Z2 .
3 ).
Z3 y Z3 , or Z4 .
x, x2 Oe .
Proof.
The equivalence of 2 and 3 follows from .
It is straightforward to verify .
Ne .
Ne .
If .
(R)) = n2 , then R O = Z3 y Z3 or R O = Z2 y Z2 y Z2 by Theorem 3.
Observe that .
(Z3 y Z3 )) = 2 and .
(Z2 y Z2 y Z2 )) = 2.
Thus, .
(R)) .
(R)) = n holds when RO = Z3 y Z3 but not for R O = Z2 y Z2 y Z2 .
Hence.
R O = Z3 y Z3 and e(Z3 y Z3 ) is C4 .
If .
(R)) < 2 , then .
(R)) > 2 .
Therefore, e(R) has an isolated vertex by Lemma 3.
Thus, .
(R)) = 1 and .
(R)) = n Oe 1.
By Lemma 3.
6, e(R) has exactly one edge.
We now consider possible values of n.
If n = 1, then .
(R)) = .
(R)) = 1.
If n = 2, then .
(R)) consists of two isolated vertices.
In both cases, .
(R)) = n Oe 1.
These contradictions imply that n Ou 3.
If n > 3, then e(R) consists of n Oe 2 isolated vertices and two vertices that are incident to a single edge.
Without loss of generality, say a1 Oea2 OO E.
(R)).
Let Z(R)O = .
1 , a2 , .
, an } with ai aj = 0 whenever i = j and .
, .
= .
, .
For any ai , aj OO Z(R)O , there exists ak OO Z(R)O such that ak ai = 0 and ak aj = 0.
Thus, ak .
i aj ) = 0.
Hence.
Z(R) is closed under addition.
Since |Z(R)O | > 3, there exists ai OO Z(R)O \.
1 , a2 } such that a1 ai OO / .
1 , a2 }.
We see On domination numbers of zero-divisor graphs of commutative rings Anderson et al.
0 = a1 .
1 ai ) = a21 a1 ai = a21 , which yields 0 = a1 .
1 a2 ) = a21 a1 a2 = a1 a2 , a Thus, it must be that n = 3.
In this case the graph on the left in Figure 2 shows the only possiblity for e(R).
This implies that e(R) is as shown on the right in Figure 2.
Hence, e(R) is P3 .
Figure 2.
e(R) for n = 3, and its associated e(R).
We now discuss necessary and sufficient conditions for .
(R)) .
(R)) = n Oe 1.
Note that when R is a reduced commutative ring with identity, .
Theorem 1.
(R)) .
(R)) = nOe1 if and only if e(R) is isomorphic to K1,3 or K3 K1 .
First, we provide two observations that will be helpful when classifying these rings.
Observation 3.
If m, n > 1, then (Km,n ) (Km,n ) = 4.
If m = 1 or n = 1, then (Km,n ) (Km,n ) = 3.
In addition, (Kn ) (Kn ) = 1 n.
Observation 3.
For a commutative ring R with identity, the equation .
(R)) .
(R)) = nOe1 where |Z(R)O | = n can only hold if n > 3.
Note that Observation 3.
9 follows from the fact that if n = 1 or n = 2, then the equation fails to hold as .
(R)) Ou 1 and .
(R)) Ou 1.
If n = 3, then we see that e(R) is either K1,2 or K3 since e(R) is connected.
In both cases, .
(R)) .
(R)) = n Oe 1.
In the following proposition, two more possibilities for n are eliminated when .
(R)) .
(R)) = n Oe 1.
Proposition 3.
Let R be a commutative ring with identity.
If .
(R)) .
(R)) = n Oe 1 and e(R) is star-shaped reducible, then n OO / .
, .
Proof.
If n = 5, then by .
we have R O = Z2 y Z5 .
Since e(Z2 y Z5 ) is K1,4 , .
(R)) .
(R)) = 1 2 = 5 Oe 1.
If n = 6, then by again by .
, e(R) is K6 .
Hence, .
(R)) .
(R)) = 1 6 = 6 Oe 1.
We now build upon the above results.
Theorem 3.
Let R be a commutative ring with identity.
Then the following are equivalent.
(R)) .
(R)) = n Oe 1.
e(R) is K1,3 or K3 K1 .
On domination numbers of zero-divisor graphs of commutative rings Anderson et al.
R O
= Z2 y F4 , or R O = Z2 y Z2 y Z2 .
Proof.
By .
, we have .
Ni .
Ne .
By Observation 3.
8, if e(R) is K1,3 , then .
(R)) .
(R)) = 3 = 4 Oe 1 = |Z(R)O | Oe 1 since |Z(R)O | = 4.
Similarly, if e(R) is K3 K1 , then, as shown in Figure 3, .
(R)) = 3 while .
(R)) = 2.
Hence, .
(R)) .
(R)) = 3 2 = 6 Oe 1 = |Z(R)O | Oe 1 since |Z(R)O | = 6.
Figure 3.
K3 K1 and K3 K1 .
Ne .
If .
(R)) .
(R)) = n Oe 1, then n Ou 4 by Observation 3.
As before, since e(R) is connected, we have .
(R)) O n2 by Lemma 3.
Three cases are considered: .
(R)) = n2 , .
(R)) = n2 Oe 1, and .
(R)) < n2 Oe 1.
Case 1.
Suppose that .
(R)) = n2 .
Then R O = Z3 y Z3 or R O = Z2 y Z2 y Z2 by Theorem From Theorem 3.
7, if R O = Z3 y Z3 , then .
(R)) .
(R)) = n.
Thus.
R O = Z2 y Z2 y Z2 and e(R) is K3 K1 .
Case 2.
Suppose that .
(R)) = n2 Oe 1.
Thus .
(R)) = n2 .
By .
, we have .
(R)).
(R)) O n.
So, ( n2 Oe .
( n2 ) O n, which implies that n2 Oe 1 O 2.
Therefore, n O 6.
Observation 3.
9, n = 4, 5, or 6.
However, n2 Oe 1 OO Z, so n = 4 or n = 6.
If n = 4, then e(R) is one of K2,2 .
K4 , or K1,3 by .
Since (K2,2 ) = 2 = n2 Oe1, e(R) is not K2,2 .
Since (K4 ) = 4, we see that e(R) is not K4 .
Since .
(K1,3 )) .
(K1,3 )) = 1 2 = 4Oe1, we see that e(R) is K1,3 .
By .
R O = Z2 y F4 .
If n = 6, then e(R) is one of K6 .
K3 K1 .
K2,4 , or K3,3 by .
Since (K6 ) = 1 = n2 Oe 1 and (K3 K1 ) = 3 = n2 Oe 1, e(R) is not K6 or K3 K1 .
By Observation 3.
K2,4 and K3,3 do not satisfy .
(R)) .
(R)) = n Oe 1.
Case 3.
Suppose that .
(R)) < n2 Oe 1.
Since .
(R)) > n2 , by Lemma 3.
5 e(R) has an isolated vertex, say w.
Thus, in e(R) the vertex w is adjacent to all other vertices.
Hence, e(R) is star-shaped reducible.
This implies that .
(R)) = 1 and .
(R)) = n Oe 2.
By Observation 3.
9, n Ou 4, and by Proposition 3.
10, we have n = 5, 6.
Therefore, we have either n = 4 or n Ou 7.
Since .
(R)) Ou 1 and .
(R)) < n2 Oe 1, n = 4 is not possible.
Thus, n Ou 7.
The remainder of the proof will show that n Ou 7 is not possible.
Pick a minimum dominating set D of e(R).
Since .
(R)) = nOe2, there are vertices a, b OO and Z(R) = D O .
, .
Also, there exists di , dj OO D such that a Oe di , b Oe dj OO E.
(R)).
Let Ca and Cb be connected components of e(R) containing a and b, respectively.
We show two things:
On domination numbers of zero-divisor graphs of commutative rings Anderson et al.
D\(Ca O Cb ) consists solely of isolated vertices in e(R).
There are five possible graph configurations for Ca and Cb , and hence for e(R).
First, we show D\(Ca OCb ) consists solely of isolated vertices in e(R).
Pick d OO D\(Ca OCb ), and suppose d Oe x OO E.
(R)) for some x OO Z(R)O .
Clearly x OO / .
, .
Since d OO / Ca O Cb we have d OO / .
i , dj }.
Since d Oe x, a Oe di , b Oe dj OO E.
(R)) and Z(R)O = D O .
, .
DA = D\.
is a dominating set of e(R).
This is a contradiction since |DA | < |D|.
Hence, each vertex of D\(Ca O Cb ) is an isolated vertex of e(R).
The above shows that if Ca and Cb are the same component, then .
(R)) = n Oe 2 = (Ca ) n Oe |V (Ca )|.
In addition, if Ca and Cb are disjoint components, then .
(R)) = n Oe 2 = (Ca ) (Cb ) n Oe |V (Ca )| Oe |V (Cb )|.
The possible graph configurations for Ca and Cb are now investigated based on whether or not Ca and Cb are the same component or disjoint components.
Subcase 1.
Assume that Ca and Cb consist of the same connected component in e(R).
If |V (Ca )| = m Ou 5, then (Ca ) O m2 by Lemma 3.
Thus, .
(R)) O m2 n Oe m = n Oe m2 by Equation 1.
However, n Oe m2 > n Oe 2 since m Ou 5, a contradiction.
Thus, |V (Ca )| O 4.
Note that |V (Ca )| > 2 since a, b OO / D.
If |V (Ca )| = 4, then (Ca ) = 2 since, by Equation 1, n Oe 2 = (Ca ) n Oe 4.
Hence.
Ca is either C4 or P4 .
If |V (Ca )| = 3, then (Ca ) = 1 since, by Equation 1, n Oe 2 = (Ca ) n Oe 3.
Hence.
Ca is either C3 or P3 .
Subcase 2.
Assume Ca and Cb are disjoint components of e(R).
Clearly, |V (Ca )|, |V (Cb )| Ou 2.
Assume, without loss of generality, that |V (Ca )| = m Ou 3.
Then (Ca ) O m2 by Lemma 3.
Hence, by Equation 2, n Oe 2 = (Ca ) (Cb ) n Oe |V (Ca )| Oe |V (Cb )| O (Cb ) n Oe m Oe |V (Cb )| which simplifies to |V (Cb )| Oe (Cb ) O2 This inequality is impossible since |V (Cb )| Oe (Cb ) Ou 1 and m2 Ou 32 .
Similarly, |V (Cb )| cannot be greater than or equal to 3.
Thus, |V (Ca )| = |V (Cb )| = 2, and hence Ca and Cb are P2 .
The above work shows there are 5 possible configurations for e(R), as shown in Figure 4.
We show that none of these configurations for e(R) are possible.
Recall that n Ou 7 as stated at the beginning of Case 3.
Configuration 1.
The graphs of e(R) and e(R) for Configuration 1 are shown in Figure 5.
Let a, b, c, and d be as shown in Figure 5.
On domination numbers of zero-divisor graphs of commutative rings Anderson et al.
Configuration 1 Configuration 2 Configuration 3 Configuration 4 Configuration 5 Figure 4.
The five configurations for e(R) Consider a b.
Since ann.
= ann.
, a = Oeb.
Thus, a b = 0.
Since there exists l OO Z(R)O with la = lb = 0, l.
= 0.
This implies a b OO Z(R)O .
Clearly, a b = a and a b = b.
Observe from e(R) that c.
= ca cb = ca = 0.
Thus, the element a b is not in the complete subgraph portion of e(R) which means a b OO .
, b, c, .
We see that a b = d since cd = 0.
Thus, a b = c.
However, 0 = dc = d.
= da db = db = 0, a contradiction.
So, e(R) cannot take this configuration.
Figure 5.
Configuration 1, e(R) on left and e(R) on right.
Configuration 2.
The graphs of e(R) and e(R) for Configuration 2 are shown in Figure 6.
Let a, b, and c be as shown in Figure 6.
Let l1 , l2 be distinct vertices in the complete subgraph portion of e(R) as in Figure 6.
Consider the elements l1 b and l2 b.
Neither element is 0 since ann.
1 ) = ann.
2 ) = ann.
implies li = Oeb.
It can then be seen from e(R) that .
1 b, l2 .
OI ann.
\ ann.
OI .
, .
Clearly.
On domination numbers of zero-divisor graphs of commutative rings Anderson et al.
Figure 6.
Configuration 2, e(R) on left and e(R) on right.
l1 b and l2 b are not equal to b.
Thus, l1 b = a = l2 b, which implies l1 = l2 , a contradiction.
So, e(R) cannot take this configuration.
Configuration 3.
The graphs of e(R) and e(R) for Configuration 3 are shown in Figure 7.
Let a, b, and c be as shown in Figure 7.
Figure 7.
Configuration 3, e(R) on left and e(R) on right.
Let l1 , l2 , l3 , l4 be distinct vertices in the complete subgraph portion of e(R) as in Figure 7.
Then a.
= ab = 0.
Thus, li b is not in the complete subgraph portion of e(R).
Also, since li OO ann.
for 1 O i O 4 but b.
Oeb OO / ann.
, we have li b = 0 for 1 O i O 4.
When i = j, we have lj .
= 0, so li b OO Z(R)O for 1 O i O 4.
Clearly, li b = b.
This implies that for 1 O i O 4 we have .
1 b, l2 b, l3 b, l4 .
OI .
, .
Without loss of generality, l1 b = l2 b, implying that l1 = l2 , a contradiction.
So, e(R) cannot take this configuration.
Configuration 4.
The graphs of e(R) and e(R) for Configuration 4 are shown in Figure 8.
Let a, b, c, and d be as shown in Figure 8.
Figure 8.
Configuration 4, e(R) on left and e(R) on right.
None of the zero-divisor graphs with 7 vertices are isomorphic to e(R) in Figure 8 since none of the realizable graphs in .
have exactly 4 vertices of degree 4.
Hence, n Ou 8.
Let l1 , l2 , l3 , l4 be distinct vertices in the complete subgraph portion of e(R).
As in the argument above for Configuration 3, .
1 b, l2 b, l3 b, l4 .
OI .
, c, .
This yields the same contradiction as in Configuration 3.
So, e(R) cannot take this configuration.
Configuration 5.
The graphs of e(R) and e(R) for Configuration 5 are shown in Figure 9.
Let a, b, c, and d be as shown in Figure 9.
On domination numbers of zero-divisor graphs of commutative rings Anderson et al.
Figure 9.
Configuration 5, e(R) on left and e(R) on right.
Again, none of the zero-divisor graphs with 7 vertices are isomorphic to e(R) in Figure 9 since none of the realizable graphs in .
have exactly 2 vertices of degree 4 as exhibited by a and b.
Hence n Ou 8.
Let l1 , l2 , l3 , l4 be distinct vertices in the complete subgraph portion of e(R).
in the argument above, we obtain .
1 b, l2 b, l3 b, l4 .
OI .
, c, .
This yields the same So, e(R) cannot take the configuration in Configuration 5.
We have now shown the four results given in Theorem 1.
References