Abstract
Let be a completely regular Hausdorff space and let be the ring of all continuous real valued functions defined on . In this paper, the line graph for the zero-divisor graph of is studied. It is shown that this graph is connected with diameter less than or equal to 3 and girth 3. It is shown that this graph is always triangulated and hypertriangulated. It is characterized when the graph is complemented. It is proved that the radius of this graph is 2 if and only if has isolated points; otherwise, the radius is 3. Bounds for the dominating number and clique number are also found in terms of the density number of .
1. Introduction
Let be a completely regular Hausdorff space and the ring of all continuous real valued functions defined on .
For each , let , , , and .
For all notations and undefined terms concerning the ring , the reader may consult [1].
If , then is a field isomorphic to . So we will assume that .
Let be a commutative ring. is the set of zero-divisors of , and . The zero-divisor graph of , , usually written as , is the graph in which each element of is a vertex, and two distinct vertices and are adjacent if and only if . For further details about this graph, see [2] and the survey [3] for a list of references.
The line graph of a graph , denoted by , is a graph whose vertices are the edges of and two vertices of are adjacent wherever the corresponding edges of are incident to a common vertex; see [4]. In this case, if are adjacent vertices in , then is a vertex in . For any undefined terms in graph theory, the reader may consult [5].
The zero-divisor graph for was introduced and studied in [6]. A more general study for reduced rings was done in [7].
In this paper we will study the line graph for the zero-divisor graph .
An element if and only if . Let . Then is a vertex in if . Since is an undirected graph, then . We will study when is connected and calculate its diameter and girth. We will show that is always triangulated and hypertriangulated and characterize when is complemented. We will find the radius and give bounds for the dominating and clique numbers.
2. Connectedness
Let be a graph and let and be two distinct vertices in . The distance between and is the length of the shortest path joining them in ; if no such path exists, we set . The associate number of a vertex of a graph is defined to be . A vertex is center in if for any vertex . The radius of is defined to be and the diameter of is . The graph is connected if any two of its vertices are linked by a path in ; otherwise is disconnected. In this section, we will show that is connected, and we will also calculate its diameter and radius.
It was shown in [6] that is connected with . We now show that a similar result also holds for .
Theorem 1. If holds, then is connected with
Proof. If , and are nonadjacent vertices in , then for . We may assume that and .
Thus we have the path . So is connected with .
Assume that holds, and and are nonadjacent vertices in . We have 3 cases:
Case I. for some . In this case, is a path in .
Case II. for all , but for some . In this case, there exists such that is a path in . So is a path in . Note that if there exist such that , then and for some and for some or for some and for some , which contradicts the fact that for all .
Case III. for all . This case will not occur since otherwise, if holds for and , then , and so . Thus for ; hence, , , which implies that for . Moreover, for , and so for . Hence for ; see Lemma 1.2 in [6], a contradiction.
Since , we conclude that is connected with .
Using Theorem 1 and Lemma 1.2 in [6], we can conclude the following.
Corollary 2. Assume that holds. Assume further that and are distinct elements in . Then(1) if and only if for some ;(2) if and only if for all , and for some ;(3) if and only if for all , and for all .
In fact, we can be more precise concerning the diameter of .
Theorem 3. Assume holds. Then is connected with if or ; otherwise .
Proof. Assume or . Let and be nonadjacent vertices in . Then for at least one . We may assume . But ; hence, for some ; that is, for some , which implies that by the above corollary. Since this is true for every two nonadjacent vertices in , then whenever or .
Assume that holds. Pick four distinct points , , , and in . Since is a completely regular Hausdorff space, there exist four mutually disjoint open sets , where and . Consider the functions such that , and for each . Clearly , whenever . Consider the functions , , and . Obviously for each and ; hence, , are two distinct vertices in with whenever and for every and every . By the above corollary, . Hence .
Now, we calculate the radius of . It was shown in [6] that if , then
A similar result holds for .
Theorem 4. If is a vertex in , then
Proof. It is clear that holds for any vertex in .
Assume . Let be a vertex in that is not adjacent to . Then for all . But and so for some . Thus and is a path in , and . Hence . Now assume that and . Let be mutually disjoint open sets in such that for each and let be an open set in such that for each . Define such that , , , and . Let . Then and for each . Similarly define such that , , , and . Let . Then and for each . Moreover, , so and it follows by Corollary 2 that , and since , we have .
Corollary 5. If holds, then if and only if has isolated points. Otherwise .
3. Cycles
A cycle in a graph is a closed walk such that no vertex, except the initial and the final vertex, appears more than once, while the girth of , which is denoted by , is the length of the shortest cycle in . If a graph has no cycles, then its girth . If and are two vertices in , by , we mean the length of the smallest cycle containing and , and we write if there is no cycle containing and . An edge which joins two vertices of a cycle but is not itself an edge of the cycle is a chord of that cycle. A graph is chordal if every cycle of length greater than three has a chord. A graph is called triangulated if each vertex in is a vertex of a triangle. A graph is called hypertriangulated if each edge in is an edge of a triangle.
In this section, we will calculate the girth of , find the shortest cycle containing two vertices, and show that is never chordal. We will also show that is always triangulated and hypertriangulated.
It was shown in [6] that if has at least points, then . A similar but yet stronger case holds for .
Theorem 6. If holds, then is triangulated and hypertriangulated.
Proof. Let be any vertex in .
Then is a triangle in . Thus is triangulated. Now let be any edge in . Then is a cycle in for some such that , and so is an edge in a triangle. Thus is hypertriangulated.
Corollary 7. If holds, then .
Example 8. Let with the discrete topology. Then is not triangulated nor hypertriangulated. Let
If is a triangle, then or ; hence, is not a vertex in any triangle in . For any such that , the edge is not an edge in any triangle in , but as a vertex in is a vertex in the triangle and the edge is an edge in the triangle for some .
We now find the length of the shortest cycle containing any two distinct vertices in .
Theorem 9. Let and be two distinct vertices in . Then(1) if and only if for some ;(2) if and only if for all and for some , for all or and where ;(3) if and only if for all and for only one , for only one ;(4) if and only if for all and for all .
Proof. (1) Assume holds. It is clear that if there is a triangle containing and , then for some .
For the converse assume that . Then there exists such that , and so
is a cycle of length in containing and . Thus .
(2) Assume that holds. Then it follows by (1) that holds for all . Hence we have the cycle
where and . Assume that , , , and , which implies that and hold. Now if , , and , then and . Finally, if , , then , a contradiction.
For the converse, assume for all . If holds for all , then
is a cycle of length in containing and , while if and hold, then
is a cycle of length in containing and . Hence in both cases we have .
(3) Assume that holds. Then for all . If for all , and we have the cycle
of length in , then and . But , contradicting the assumption. Similarly we will have a contradiction, if we have the cycle
Thus we must have for only one and only one .
For the converse, assume for all and for only one , for only one , say . By (1) and (2), there is no cycle of length or containing both and . There exists such that , and so the cycle
is of length in containing and . Hence we have .
(4) If holds, then by (1), (2), and (3) holds for all and for all .
For the converse, assume that for all and for all . By the previous steps . It follows by Corollary 2 that , and so there exists such that is a path in . There exists such that . The cycle
is of length in containing and . Hence we have .
Theorem 10. The graph is never chordal.
Proof. This is true since for there exists , with , and hence is a cycle of length in , where no chord can be added.
4. Dominating Sets
A proper prime ideal in a ring that contains no smaller prime ideal is called a minimal prime ideal, and the set of all minimal prime ideals in will be denoted by . For any subset of a ring , we define the hull of to be .
Let be a topological space. The density of , denoted by , is the smallest cardinal number of the form , where is a dense subset of . The weight of , denoted by , is the smallest cardinality of the form , where is a base for . The cellularity of the space is is a family of pairwise disjoint nonempty open subsets of . A zero-set in is said to be a middle zero-set if there exist two proper zero-sets and such that and . A space is middle P-space if every nonempty middle zero-set in has nonempty interior. Let denote the infinite countable cardinal number, and let denote the first infinite uncountable cardinal number. If is an infinite cardinal number, then let .
A simple graph in which all the vertices of are pairwise adjacent is called complete graph. A complete graph on n vertices is denoted by . A maximal complete subgraph of a graph is called a clique. The clique number of is is a complete subgraph of . In a graph , a dominating set is a set of vertices such that every vertex outside is adjacent to at least one vertex in . The dominating number of a graph , denoted by , is the smallest number of the form , where is a dominating set. For distinct vertices and in a graph , we say that and are orthogonal, written as if and are adjacent and there is no vertex of which is adjacent to both and ; see [8]. A graph is called complemented, if for each vertex of , there is a vertex of (called a complement of ) such that .
In this section we will estimate the dominating number and the clique number for and characterize when is complemented.
Note that if is a dominating set for , then for each vertex in , we have or or for some .
Lemma 11. Let be a dominating set in . Then .
Proof. Note first that , since if , then for each . Also note that if , then , since if , then for each .
Now, if for each , there exists such that , then , while if there exists such that for each , then since is a dominating set, we must have , for each and some . Thus .
Theorem 12. If holds, then .
Proof. Let be any dominating set for .
For every vertex , choose , , and . Let .
Claim. is dense in .
It is sufficient to prove that for any vertex in , . Let be a vertex in . Then we have 3 cases.
Case I. Consider . Then clearly, .
Case II. Consider , but for some . Then again .
Case III. Consider , but for some . Then , since . But ; hence, .
Thus is dense in , and so , because . Since this is true for any dominating set, we have .
Theorem 13. If holds, then .
Proof. Let be any dominating set for .
Let , , , . Let be any non-empty proper open set in , and let , . There exists an open sets and in such that and . Define such that and . This implies that , and so and . Let . Then . Define such that and . Then clearly and .
We have 3 cases.
Case I. Consider . Then clearly, .
Case II. Consider , but for some . Then again .
Case III. Consider , but for some . Then .
Thus is a base for , and so . Since this is true for any dominating set, we have .
Using the argument in the last proof, one can deduce easily the following corollary.
Corollary 14. If holds, and is a dominating set in , then the set , , , is a base for open sets in .
We now relate dominating sets in with dominating sets in .
Theorem 15. If holds, and is a dominating set in , then the set is a dominating set for .
Proof. Let . Then for all . Since dominates , then for each , there exists such that . Thus and is adjacent to . Hence dominates .
Theorem 16. Let be a dominating set in and let and . Then dominates .
Proof. Let be a vertex in that is not in . Then . Since dominates , there exists such that . Thus and is adjacent to . Hence dominates .
Let be a dominating set in such that , and let and . It was shown in [6] that , and so and , since . Thus .
Theorem 17. Let be an infinite space with .
Then .
Proof. We have shown in Theorem 12 that , and by the remark after Theorem 16, . But it is known that , see [9], and , see [10], and so .
Corollary 18. Let be an infinite space with . If , then .
Proof. Since , then using Theorem 13, we have .
Corollary 19. If is an infinite space, then .
Example 20. Using Lemma 11 and Theorem 16, we get that , since . Similarly .
Example 21. is dense in and so . But note that . So .
We now investigate cliques and the clique number in .
Theorem 22. If is a finite clique in , then .
Proof. If is a clique, then there exists such that . Thus is adjacent to both and , a contradiction. So . If , then clearly is a complete subgraph of . If is a vertex in that is adjacent to all vertices in , then implies that or . If , then or , since is adjacent to . So or and in both cases we have . A similar result will also be obtained if . Thus is a clique with . Now assume that is a clique in with and let and with and . So we may assume that since is adjacent to . Let with and . Then we may assume that and , since is adjacent to both and . Thus . But as shown earlier any vertex adjacent to all vertices in must be one of them, contradicting the fact that . Thus or . Hence , and therefore is an infinite set.
The following proposition was proved in [6].
Proposition 23. The graph is triangulated if and only if has no isolated points, while is hypertriangulated if and only if is a connected middle P-space.
Corollary 24. (1) Every vertex in belongs to a finite clique if and only if is a connected middle P-space.
(2) For every there exists such that is a vertex in a finite clique if and only if has no isolated points.
Proof. The result follows by the above theorem together with Proposition 23.
Using Corollary 24 together with Proposition 2.2 and Corollary 2.5 in [6], we get the following.
Theorem 25. The following are equivalent.(1) is complemented.(2)For all , there exists such that and .(3)For all , there exists such that .(4)For all , there exists such that is not a vertex in any finite clique in .(5) is compact.
Corollary 26. If is an infinite clique in , then for some .
Thus if such that is maximal, then .
Let be a family of pairwise disjoint non-empty open subsets of with maximum cardinality. For every there exists such that .
Let be the power set of . For each , define as follows:
Lemma 27. The functions defined above are well-defined continuous functions.
Proof. The function is well defined, since for each , on the boundary of . Since the functions and are continuous on the closed sets and , then it follows that the function is continuous.
Theorem 28. If holds, then .
Proof. Let be a family of pairwise disjoint non-empty open subsets of with , and for all let be a nonzero continuous function with . Let be the power set of . Then for every the function defined by is a well-defined continuous function; see Lemma 27. Clearly and for every ; hence, the induced subgraph of in is a complete subgraph of . Since this is true for every family of pairwise disjoint non-empty open subsets of , then . The second inequality comes from the fact and the previous corollary; that is, .
Example 29. If is a discrete space with , then . Also , since .
Acknowledgment
This paper is a part of the doctoral dissertation of the first author under supervision of the second author at the mathematics department at The University of Jordan.