Domination and Independence Numbers of Γ(Z n )

International Mathematical Forum, 3, 008, no. 11, 503-511 Domination and Independence Numbers of Γ(Z n ) Emad E. AbdAlJawad and Hasan Al-Ezeh 1 Department of Mathematics, Faculty of Science Jordan University, Amman 1194, Jordan e.abdaljawad@yahoo.com, alezehh@ju.edu.jo Abstract The zero divisor graph of a commutative ring with one ( say R) isa graph whose vertices are the nonzero zero-divisors of this ring, with two distinct vertices are adjacent in case their product is zero. This graph is denoted by Γ(R). We evaluate the domination number of Γ(Z n ). We find out that the domination number of Γ(Z p α 1 1 pα... pα ) is equal to. We compute the independence number of Γ(Z n ) for some cases of n. For instance we evaluate the the independence number of Γ(Z p ), Γ(Z p q ), and Γ(Z pq ). Mathematics Subject Classification: Primary: 13A99; Secondary: 05C69 Keywords: Zero-divisor graph, Domination number, Independence number 1 Introduction Zero-divisor graphs first introduced by Bec, see [4]. Bec were mainly interested in graph coloring. In his wor, he let the elements of a commutative ring R to be the vertices of the graph and two distinct vertices x and y are adjacent if xy = 0. In a subsequent wor, Anderson and Livingston introduce the zero-divisor graph of a commutative ring R, see []. In their definition, they let the nonzero zero-divisors of R to be the set of vertices for the graph and two distinct vertices x and y are adjacent if xy = 0. Usually the set of zero-divisors of R is denoted by Z(R) and the set of nonzero zero-divisors of R is denoted by Z (R) =Z(R) {0}. The zero-divisor graph of R, Γ(Z (R)), is usually written Γ(R). The definition of the zero-divisor graph that was given 1 This wor is a part of a doctoral dissertation done by the first author and supervised by the second author. Corresponding author: Hasan Al-Ezeh.

504 E. E. AbdAlJawad and H. Al-Ezeh by Anderson and Livingston is the one that have been used in the literature now. Also in this paper we will use their definition. Many articles have been done on zero-divisor graphs, the reader is advised to consult [1,, 3, 7] for more details. Some researchers generalized the idea to commutative semigroups, see [5, 6]. Others wored on the noncommutative case where they introduce the directed graph related to the zero divisors of noncommutative rings. For more information see [9, 10, 11]. A dominating set in a graph Γ is a subset of the vertex set of Γ with the property that every vertex in Γ is either in the dominating set or adjacent to a vertex that is in the dominating set. The domination number of Γ, denoted by Domn(Γ), is defined as the cardinality of a minimum dominating set of Γ. Next we compute the domination number of Γ(Z n ). The Domination Number of Γ(Z n ) First we give the Domn(Γ(Z n )) where n is a prime power or two multiplied by a prime. Lemma 1. For Γ(Z n ),ifn = p, where p is a prime number, then Domn(Γ(Z n )) = 1. Also if n =p then Domn(Γ(Z n ))=1. Proof. If n = p then x = p 1 is adjacent to every other element in Γ(Z n ) and hence Domn(Γ(Z n )) = 1. Now, if n =p then the vertex set of Γ(Z n )is {,,..., (p 1),p}. It is clear that the vertex p is adjacent to every other vertex and hence Domn(Γ(Z n )) is equal to one. Lemma. Let n = p α 1 1 p α...p α where p 1,p,...,p are distinct primes and the α i s are positive integers. Also suppose that n p and n is not a prime power. Let D be any dominating set in Γ(Z n ) with minimum cardinality among all dominating sets. Then, for every i {1...}, D must contain at least one vertex from G i = {sp α 1 1...p α i 1 i...p α : gcd(s, p i )=1} Proof. Tae F i = {rp i : gcd(r, p α 1 1...pα i 1 i...p α )=1} then F i is a set of null vertices for every i. We have every vertex in F i is only adjacent to every vertex in G i. Suppose that D doesn t contain any vertex from G i0 for some i 0. Since D is a dominating set then D will contain all the vertices of F i0. Since n p we get the size of F i0 is greater than or equal to two. Now tae K =(D F i0 ) {h} where h is an element of G i0. We have, K is a dominating set where the size of K is less than the size of D. This contradicts the fact that D is a dominating set with minimum cardinality. Hence D will contain at least one element from G i for every i. Next we use the previous lemmas to find Domn(Γ(Z n )).

Domination and independence numbers 505 Theorem 1. (a) If n =p, where p is any prime number, then Domn(Γ(Z n )) = 1. (b) If n = p α 1, where p is any prime number and α 1 is a positive integer, then Domn(Γ(Z n )) = 1. (c) Let n = p α 1 1 pα...pα where p 1,p,...,p are distinct primes and the α i s are positive integers. Also suppose that n p and n is not a prime power. In this case Domn(Γ(Z n )) = Proof. The proof of (a) and (b) is in Lemma 1. For (c), let D be a set that contain exactly one element of G i = {sp α 1 1...p α i 1 i...p α : gcd(s, p i )=1} for every i {1,,...,}, say this element is s i p α 1 1...pα i 1 i...p α. We want to show that D is a dominating set with minimum cardinality. If x is any zero divisor of Z n then x is divisible by p i0 for some i 0 {1,,...,} and so x is adjacent to s i0 p α 1 1...pα i 0 1 i 0...p α. Hence D is a dominating set. Using Lemma, any dominating set of Γ(Z n ) has at least one element of G i for every i {1,,...,n} and hence D has a minimum cardinality. 3 The Independence Number of Γ(Z n ) An independent set in a graph Γ is a subset I of the vertex set of Γ such that no two vertices of I are adjacent i.e. I is a subset of null vertices. The independence number of Γ, denoted by Indep(Γ), is defined as the cardinality of a maximum independent set of Γ. In this section we will compute the independence number of Γ(Z n ) for some cases of n. We present some nown algorithms to compute the zero-divisor graph of Z n for some cases of n. Some of these algorithms were presented by Joan Krone. For more details see [8]. Case 1: If n = p for some prime p and positive integer. One can find the zero-divisors of Z p by taing the numbers 1,,p 1 1 then multiply those numbers by p. One can divide the zero-divisors into 1 sets according to how many factors of p each divisor has. These sets are S p i = {sp i : gcd(s, p i )=1} where i {1,,..., 1}. Connect the vertices of the set S p i to the vertices of the set S p j when i + j. This way one can build the zero-divisor graph of Z p. The use of the Euler s phi-function gives the sizes of the S p i s and one will get S p i = p i p (i+1). Case : If n = p q r where p and q are distinct primes and and r are positive integers. One can divide the zero-divisors into three families. The first family is S p i = {sp i : gcd(s, p i q r )=1} where i {1,,...,}. The second family is S q i = {sq i : gcd(s, p q r i )=1} where i {1,,...,r}. The third family is S p i q j = {spi q j : gcd(s, p i q r j )=1} where i {1,,...,}, j {1,,...,r} and we can not have both i = and j = r. Connect elements

506 E. E. AbdAlJawad and H. Al-Ezeh of S p i to elements of S p i q j if i + i and j = r. Similarly, connect elements of S q j to elements of S p i q j if j + j r and i =. Connect elements of S p i q j to elements of S p i q j if i + i and j + j r. The next lemma gives the sizes of these sets. Lemma 3. 1) S p i = q r 1 p i 1 (p 1)(q 1) for i {1,..., 1} and S p = q r 1 (q 1). ) S q i = p 1 q r i 1 (q 1)(p 1) for i {1,...,r 1} and S q r = p 1 (p 1). 3) S p i q j = p i 1 q r j 1 (p 1)(q 1) for i {1,..., 1} and j {1,...,r 1}. S p q j = qr j 1 (q 1) and S p i q r = p i 1 (p 1). Proof. We will show that S p i = q r 1 p i 1 (p 1)(q 1). For the other sets the proof is similar. Tae T i = {p i, p i, 3p i,...,(p i q r )p i } {q, q, 3q,..., (p i q r 1 )q}. Note that T i is the set of elements that are divisible by p i and not divisible by any power of q. Tae T i+1 = { p i+1, p i+1, 3p i+1,..., (p i 1 q r )p i+1 } {q,q, 3q,..., (p i 1 q r 1 )q}. Also T i+1 is the set of elements that are divisible by p i+1 and not divisible by any power of q. We have S p i = T i T i+1 and hence S p i = T i T i+1 = p i q r p i q r 1 (p i 1 q r p i 1 q r 1 ) = q r 1 p i (q 1) q r 1 p i 1 ( 1+q) = q r 1 p (q 1)(p i p i 1 ) = q r 1 p i 1 (q 1)(p 1) The next theorem gives the independence number for Γ(Z n ) when n is a prime power. Theorem. Let p be any prime number and let be any positive integer that is greater than or equal to two then 1) If =then Indep(Γ(Z p )) = 1. ) If is an odd integer that is greater than two then we have Indep(Γ(Z p )) = 1 (p 1)p (i+1) = p 1 (p 1 1). 3) If is an even integer that is greater than two then we have 1 Indep(Γ(Z p )) = 1 + (p 1)p (i+1) =1+p 1 (p 1).

Domination and independence numbers 507 Proof. 1) It is clear that Indep(Γ(Z p )) = 1. This is because Γ(Z p )isa complete graph. ) According to the algorithm for calculating Γ(Z p ), we have 1 sets. These sets are S p,s p,...,s p 1, where S p i = {sp i : gcd(s, p i )=1}, and as we explain before S p i = p i p (i+1). As we explain in the beginning of this section, elements in S p i are adjacent to elements of S p i if i + i. Tae I = S p i. Observe that I is a set of null vertices and this is because 1 i {1,,..., 1 }. Hence I is an independent set. Since the set i= +1 S p i forms a complete subgraph of Γ(Z p ) we get I is an independent set with maximum cardinality. The size of I, which is the independence number, is equal to 1 1 1 I = S p i = p i p (i+1) = (p 1)p (i+1) = p 1 (p 1 1). 3) The proof is similar to that one of ). Theorem 3. If p and q are two distinct primes and is a positive integer with p <qthen Indep(Γ(Z p q)) = p 1 (q 1). Proof. According to the algorithm for calculating Γ(Z p q r) we get Γ(Z p q)= 1 S p i S q S p i q. We will show that I = S p i is an independent set with maximum cardinality among all independent sets in Γ(Z p q). We will show 1 that any independent set that intersects S q or S p i q will have a size that is 1 less than the size of I. Assume M is an independent set with M S p i q and M S q. Since the elements of S q are only connected to the elements of S p and M contains elements of S q then M S p =. Let T = {i : 1 M S p i q } and observe that T is nonempty because M S p i q. Since S p i q is connected to S p i we get M S p i = for all i T. Hence M S q S p i q S p i where T c = {1,,...,} T. We will estimate the size of i T i T c M. We have M S q + S p i q + S p i < S p + S p i + S p i. i T i T c i T i T c This is because S p i q = p i p i 1 < (q 1)(p i p i 1 ) = S p i and S q = p p 1 <q 1=S p. These less than because p <q. Hence we have

508 E. E. AbdAlJawad and H. Al-Ezeh M < S p + S p i + S p i = S p i = I. If M is a independent i T i T c 1 set with M S p i q and M S q = or M is an independent set with 1 M S p i q = and M S q then in the same way we can show that M < S p i. Thus S p i is an independent set with maximum cardinality among all independent sets in Γ(Z p q). The independence number in this case 1 is S p i = S p i = (q 1)(p 1)p i 1 + q 1=p 1 (q 1). Theorem 4. If p and q are two distinct primes and is a positive integer with p<qthen Indep(Γ(Z pq )) = { q 1 (q 1) + p(q 1 q 1 ) if is an odd integer, q 1 (q 1) + p(q 1 q )+1 if is an even integer. Proof. We show the result when is an odd integer. When is even the proof is similar. According to the algorithm for calculating Γ(Z p qr) we get 1 1 1 Γ(Z pq )=S p S q i S pq i. We will show that I = S p S q i S pq i is a an independent set with maximum cardinality among all independent sets in Γ(Z pq ). Let M be any independent set in Γ(Z pq ). We will show that if 1 M intersects S pq i or intersects S q i then the cardinality of M is less i= +1 i= +1 than the cardinality of I. Suppose that M 1 i= +1 S pq i. Since 1 i= +1 complete subgraph and M is an independent set then M will contain only one 1 element from S pq i. This element, say w, is adjacent to all the elements in 1 i= 1 S pq i i= +1 i= 1 S q i. Hence M is a subset of the size of M is less that the size of I. S pq i 3 3 S pq i is a S q i S p {w}. Hence

Domination and independence numbers 509 Now, suppose that M i= +1 {i : M S q i }. Since M S q i. Let t be the maximum of the set i= +1 S q i we get t> +1. We will assume that t. Indeed if t = then the proof is similar. Since M S q t and M is an t t 1 independent set we get M S pq i =. Hence M S p S q i S pq i. i= t For any i { +1,..., 1} we have S q i =(p 1)q i 1 (q 1) <q i 1 (q 1) = q ( i) 1 (q 1) = S pq i and the less than because p<qand i> +1. Using this we get M S p + = S p + < S p + = S p + = I. t 1 1 1 t 1 t i= +1 1 i= t 1 + t 1 t 1 Hence the size of M is less than the size of I. Thus the independence number of Γ(Z pq )is Indep(Γ(Z pq )) = S p + 1 = q 1 (q 1) + 1 1 q i 1 (q 1)(p 1) + = q 1 (q 1) + p(q 1 q 1 ) q i 1 (q 1) 1 Theorem 5. If p and q are two distinct primes with p<qthen Indep(Γ(Z p q )) = (q 1)(p + pq p)+1

510 E. E. AbdAlJawad and H. Al-Ezeh Proof. According to the algorithm for calculating Γ(Z p q r) we get Γ(Z p q )= S p S q S p S q S pq S p q S pq. The set I = S p S q S p {a}, where a S pq, is an independent set in Γ(Z p q). We will show that I has a maximum cardinality among all independent sets in Γ(Z p q ). Assume that M is any independent set in Γ(Z p q ). Observe that T = S pq S p q S pq is a complete subgraph of Γ(Z p q). Since M is an independent set then M intersects T in at most one element. If M contains no elements from T then M is a subset of either S p S q S p or S p S q S q. Observe that S p S q S q < S p S q S p < I and this is because p<q. Hence the size of M is less than the size of I. IfM contain an element of T,sayw, and w S p q or w S pq then M is a subset of either S p S p {w} or S q S q {w} and this is because M is an independent set. In both cases we get M < I. If w S pq then M is a subset of either S p S q S p {w} or S p S q S q {w} and this is because M is an independent set. We have S p S q S q {w} < S p S q S p {w} and this is because p<q. Hence an independent set with maximum cardinality is S p S q S p {w}. Thus I is an independent set with maximum cardinality among all independent sets in Γ(Z p q ). The size of I, which is the independence number of Γ(Z p q), is equal to (p 1)(q 1)q+(q 1)p(p 1)+q(q 1)+1 = (q 1)(p +pq p)+1. Acnowledgements. The authors wishes to express deep gratitude to Omar A. AbuGhneim for his useful ideas and helpful suggestion that improved this wor considerably. References [1] E. E. AbdAlJawad, Some Properties of the Zero-Divisor Graph of the Ring of Dual Numbers of a Commuatitive Ring, Doctoral Dissertation, Jordan University, 007. [] D. F. Anderson and P. S. Livingston, The zero-divisor graph of a commutative ring, J. Algebra, 17 no. (1999), 434-447. [3] D. F. Anderson, A. Frazier, A. Lauve, and P. S. Livingston, The Zero- Divisor Graph of a Commutative Ring II, 61-7, Ideal theoretic methods in commutative algebra (Columbia, MO, 1999), 61-7, Lecture Notes in Pure and Appl. Math., 0, Deer, New Yor, 001. [4] I. Bec, Coloring of commutative rings, J. Algebra, 116 no. 1 (1988), 08-6. [5] F. DeMeyer, L. DeMeyer, Zero divisor graphs of semigroups, J. Algebra, 83 no. 1 (005), 190-198.

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