Uncharted Territory of Zero Divisor Graphs and Their Complements

Transcription

1 Uncharted Territory of Zero Divisor Graphs and Their Complements Amanda Phillips, Julie Rogers, Kevin Tolliver, and Frances Worek July, 004 Abstract Let Γ(Z n ) be a zero divisor graph whose vertices are nonzero zero divisors of Z n and whose edges connect two vertices whose product is zero modulo n. Then Γ(Z n ) represents the complement of Γ(Z n ). The authors explore the center of Γ(Z n ) and Γ(Z n ). Further study is done on planarity, independent sets and cliques, vertices of minimum degree, and connectivity of Γ(Z n ). Introduction The set of zero divisors in a ring are not easily studied using algebraic means because of their unique structure. Therefore, zero divisor graphs are studied to learn more about zero divisors. One relatively unexplored area in graph theory is that of zero divisor graphs and their complements. Our intention in this paper is to introduce and discuss the concept of zero divisor graphs with a focus on their complements by expanding upon the work done by Melody Brickel. An element a in a ring R is a zero divisor if there exists a nonzero element b in R such that ab = 0. We examine the ring Z n = {0,,..., n }, defined as the set of all residue classes modulo n. For ease of notation, we will omit the bars. For example, let us consider Z 8. The set of zero divisors of Z 8 is Z 8 = {0,,, 3, 4, 5, 6, 7}. Z(Z 8 ) = {0,, 4, 6} Since 4 and 6 4 are 0 modulo 8, the elements,4, and 6 along with 0 are zero divisors.

2 A simple graph G is a pair (V, E) where V is a set of vertices and E is a set of edges unordered pairs {v, w} of distinct elements of V. In order to construct the graph of zero divisors of Z n, denoted Γ(Z n ), we need to define the set of vertices and edges. The vertices of Γ(Z n ), denoted V (Γ(Z n )), are the nonzero zero divisors. The edges of Γ(Z n ) are pairs {x, y}, written xy for ease of notation, in V (Γ(Z n )) such that x y and xy = 0 under multiplication modulo n. The complement of a graph G, denoted G, is a simple graph in which all the vertices of G are the same as the vertices of G, but the edges xy E(G) if and only if xy E(G). Consider the zero divisor graph G = Γ(Z n ). The edges of the complement, Γ(Z n ), are the pairs xy in V (Γ(Z n )) such that x y and xy 0 under multiplication modulo n. We provide definitions and notation in a preliminary section, and a glossary of definitions is provided at the end of the paper. We will explore the centers of both zero divisor graphs and their complements. The remainder of our research concentrates on the complements of zero divisor graphs. We determine when these graphs are planar and when an associate class is an independent set. We find a vertex of minimum degree, and this result leads us to the connectivity of Γ(Z n ) for n of a specific type. Preliminaries Let R be a ring and a, b, and u be elements in R. An element a is called a zero divisor if there exists an element b 0 such that ab = 0. The set of zero divisors in R is denoted Z(R). Denote by Z(R) the set of nonzero zero divisors in R. An element a is called a unit if there exists an element b such that ab =. For our purposes in this paper we look at Z n which is a finite ring, so we can say that every element of Z n is either a zero divisor or a unit. Also, we say that two elements a and b are associates if there exists a unit u such that a = ub. We define a b to mean that a and b are associates. For every a, the annihilator of a, denoted ann(a), is the set of all elements x in R such that ax = 0. Note, ann(a) is an ideal of R. We also make use of the Chinese Remainder Theorem, which states that Z m m r = Z m Z m... Z mr when m,..., m r are pairwise relatively prime integers. Let S V (G). The subgraph induced by S is the graph H such that V (H) = S and E(H) = {v v E(G):v, v S}. A path is a sequence of vertices and edges in G such that no vertex is repeated. G is connected if for every u and v in V (G), with u v, there exists a u, v-path.

3 Throughout this paper, we write n = p e p er r, where p < p <... < p r are the prime factors of n. Let v be a vertex in the graph Γ(Z n ). Let us define v i = n p i = p e p e i i p er r, i r. Let N v represent the neighborhood set of v. In other words, N v consists of all vertices adjacent to v. Let A v represent the associate class of v; that is, A v consists of all the associates of v. Since we are considering Z n in this paper we state the following definition. Let n be an integer with x Z n. Pick x Z, 0 x n, such that the class of x (mod n) is x. We then say m x if m x as integers where m Z. 3 Centers In this section we will look at the centers of zero divisor graphs and their complements. To be able to understand what a center is we must first define distance and eccentricity. Let G be a graph and u, v V (G) where u v. Then the distance from u to v, denoted d(u, v), is the length of the shortest u, v path. If there is no u, v path then d(u, v) =. For example, in Γ(Z ), we can see from Figure?? that d(, 3) = 3 and d(4, 8) = Figure : Γ(Z ) The eccentricity of u V (G), denoted ε(u), is the maximum distance from u to any other vertex. Looking back at Γ(Z ), we see that ε(0) = 3 because d(0, v) is at most 3 for any vertex v. Finally we say that the center of G is the subgraph induced by the set of vertices of minimum eccentricity. Note that the center of Γ(Z ) consists of the vertices 4, 6, and 8 together with the edges from 4 to 6 and 6 to 8 because ε(4) = ε(6) = ε(8) =, which is the minimum eccentricity of Γ(Z ). Now let us explore the centers of zero divisor graphs in general. 3

4 Proposition 3.. Let n = p e p er r, where r > or if r = then n p. Also let B = {x Z n : p i x for all i}. The center of Γ(Z n ) is the subgraph induced by ( r i= A v i ) B. Let v V (G) be any vertex. Then, by Lemma??, v = cp f... p fr r, where c is a unit, p v < n, and 0 f i e i for all i. Because v 0, there exists k such that f k e k. Fix j k. If v p j, then p j N v. If v = p j, then p j N v. Therefore d(p j, v) >, which implies that ε(v) >. Thus, no vertex has eccentricity. Let i r. Choose v A vi and t V (G), t v. If t is adjacent to v, then d(v, t) =. If not, then p i t and there exists j i such that t = lp j for some l Z. Note that t is adjacent to v j. We also know that all the vertices in A vi are adjacent to all the vertices in A vj. Therefore, if any vertex t is not adjacent to v, there still exists a v, t path through v j of length. Thus ε(v) =, which implies that ε(u) = for all u A vi and for all i. Similarly, for any vertex u = mp p r for some m Z, if t V (G) and t is adjacent to u, then the distance is. Otherwise, there is some j such that there exists a u, t path through v j. Therefore, ε(u) =. Now we will show that every other vertex in the graph is not in the center because they have eccentricity greater than. Choose u such that u is not associate to p i, for any i, and p p r u. Then there exists j such that p j u implies that u is not adjacent to any vertex in A vj. Note, p j is only adjacent to v j and its associates, but no other vertex. So p j is not adjacent to u. Thus, d(u, p j ) > because any u, p j path has to start at u and end at p j, and must contain an associate of v j as an internal vertex; since u is adjacent to neither p j nor A vj, this path must contain at least one more internal vertex. Hence d(u, p j ) 3 and from [?] we know that diam(γ(z n )) 3, so we can conclude that d(u, p j ) = 3. Therefore ε(u) = 3 implies that u is not in the center. Proposition 3.. If n = p, then the center of Γ(Z n ) is p. The vertex set of G is defined as V (G) = {, 4,..., (p ), p}. Since p is adjacent to all multiples of, ε(p) =. Also note that any multiple of is adjacent only to p, which means that ε = for all multiples of. Therefore, p is the only element in the center since it has the minimum eccentricity. Proposition 3.3. If n = p e, then the center of Γ(Z n ) is A p e. 4

5 Every vertex in A p e has ε = because V (G) = {p, p,..., p, p,..., (p )p,..., (p )p e } and all the vertices are multiples of p; therefore, every vertex in A p e is adjacent to all vertices in the graph. Every other vertex in the graph has some non-neighbor and therefore has ε and is not in the center. We now look at the center of Γ(Z n ). Lemma 3.4. Let n = p e p er r, r, and let v V (Γ(Z n )). For all i j, either p i N v or p j N v. Explore two cases, either v A vi or v A vi. First case v A vi. The vertex p i is not adjacent to v = up e p e i i p er r and v p j = up e p e i i p e j+ j p er r, which is not divisible by n. So v and p j are adjacent. Second case, v A vi. Then v = up f p fr r, where u is a unit and 0 f i e i for all i r. There exists f i < e i or there exists k such that f k e k. Therefore, v p i is not divisible by n, since f i < e i or f k e k. This implies that p j is adjacent to v. Proposition 3.5. The center of Γ(Z n ), denoted G, for n = p e p er r long as r 3. is Γ(Z n ) as Let x V (G). Consider three cases, either there exists an i such that x A vi, there exists an i such that x N vi, or x E = V (G) ( r i= A v i ) ( r i= N v i ). Suppose x A vi, x is adjacent to every vertex in N x. Note p j, p k N x, such that j, k are distinct and not equal to i. If w N x, either w is adjacent to p j or p k, by Lemma??. Then there is a x, w-path with length. Suppose x N vi. Note p j, p k N x, such that j, k are distinct and not equal to i. If w N x, either w is adjacent to p j or p k, by Lemma??. Then there is a x, w-path with length. Suppose x E. Note p j, p k N x, such that j, k are distinct and not equal to i. If w N x, either w is adjacent to p j or p k, by Lemma??. Then there is a x, w-path with length. Proposition 3.6. The center of Γ(Z n ), denoted G, for n = p e p e, where e > or e >, is the subgraph induced by V (G) (A v A v ). 5

6 Consider the five cases, either x is in A v, A v, N v, N v, or E = V (G) A v A v N v N v. Suppose x A v. The vertex x is adjacent to every vertex in N v. Say w V (G) A v. Note p N x, which is adjacent to every vertex w. Then there is a xw-path with length two. Say w V (G) N v E. Now N v and N v are disjoint sets because there is no vertex in G that is not divisible by p and not divisible by p. Also, p is adjacent to p. Note p N v. Therefore, there is a x, w-path with length 3, and every vertex in A v has eccentricity 3. Similarly, every vertex in A v has eccentricity 3. Next, let x N v. Note p N x. If w N x, w is adjacent to p. Then there is a x, w-path with length. So N v has eccentricity. Similarly, N v has eccentricity. Suppose x E. Note p N x. If w N x, w is adjacent to p. Then there is a x, w-path with length. So E has eccentricity. Let v V (G) be any vertex. Then v = cp f p f, where c is a unit and 0 f i e i for all i. There exists k such that f k e k. The vertex w = p e f p e f N v. Therefore, d(v, w) > implies ε(v) >. Thus, no vertex has eccentricity. It follows that every vertex in G A v A v has eccentricity of two. The subgraph induced by these vertices is the center. 4 Planarity In this section, we determine when Γ(Z n Z nr ) is planar where n,..., n r Z. A graph is called planar if it can be drawn in the plane with none of its edges overlapping. Using Kuratowski s Theorem, if there exists a subgraph of G which is isomorphic to K 5, the complete graph on 5 vertices, or K 3,3, the complete bipartite graph with 3 elements in each subset of the partition, then G is not planar. A complete graph, denoted K n, is the graph on n vertices with all possible edges. G is bipartite if its vertices can be partitioned into two nonempty independent subsets. Theorem 4.. Γ(Z n Z nr ) is planar if and only if Z n Z nr = Zp for some prime p or Z n Z nr is isomorphic to Z Z, Z Z 3, Z 8, Z Z 4, Z Z Z, Z 3 Z 3, Z Z 5, Z 4 Z 3, or Z 3 Z 5. Consider the four cases: r 4, r = 3, r =, and r =. 6

7 Figure : K 5 Figure 3: K 3,3 Case : Consider Z n Z nr with r 4. There is a subgraph isomorphic to K 3,3 induced by the bipartition {(,0, 0, 0,...), (0,, 0,,...), (,, 0, 0,...)} {(,,, 0,...), (,, 0,,...), (, 0, 0,,...)}. Thus Γ(Z n... Z nr ) is not planar. Case : Consider Z n Z n Z n3. If n i 3 for some i, then a K 3,3 subgraph may be induced. Without loss of generality, assume n 3 3. Then a K 3,3 subgraph may be induced by the bipartition {(, 0, ), (0, 0, ), (0,, )} {(0, 0, ), (0,, ), (, 0, )}. Therefore Γ(Z n... Z nr ) is not planar. If n = n = n 3 =, then Γ(Z Z Z ) is planar, as shown in Figure??. (0, 0, ) (0,, ) (, 0, ) (0,, 0) (, 0, 0) (,, 0) Figure 4: Γ(Z Z Z ) Case 3: Let r =. That is, consider Z n, where n = p e p es s. We shall examine three subcases, a) s 3, b) s =, c) s =. a) If s 3, then Z n = Zp e Z e p Z p es, and we may construct a subgraph s isomorphic to K 3,3 using the bipartition {(, 0,..., 0), (,,,..., 0), (,,,... )} {(,, 0,..., 0), (,,,..., 0), (,,,..., 0)}. Therefore Γ(Z n ) is not planar. 7

8 b) Consider n = p e p e. First let us look at p 7. In this case, there is a subgraph isomorphic to K 5 induced by the vertices {p, p,..., 5p}. Next, consider the case n = e 3 e. When e = and e =, Γ(Z Z 3 ) is planar as shown in Figure??. (0, ) (0, ) (, 0) Figure 5: Γ(Z Z 3 ) When e, the graph contains a K 5 subgraph induced by {, 4, 6, 8, 0}. Therefore Γ(Z n ) is not planar. When e = and e =, Γ(Z n ) is planar, as shown in Figure??. (0, ) (, ) (3, 0) (0, ) (, ) (, 0) (, 0) Figure 6: Γ(Z 4 Z 3 ) When e 3, a subgraph isomorphic to K 5 is induced by the vertices {, 4, 8, 0, 4}. Now consider the case where n = e 5 e. If e = and e =, the graph is planar, as shown in Figure??. When e, the vertices {5, 5, 5, 35, 45} induce a subgraph isomorphic to K 5. This may also be done when e, using the vertices {, 4, 6, 8, }. Consider the case n = 3 e 5 e. For the case e = and e =, the graph is planar, as shown in Figure??. 8

9 (0, ) (0, 3) (0, ) (0, 4) (, 0) Figure 7: Γ(Z Z 5 ) (0, ) (0, 3) (0, ) (0, 4) (, 0) (, 0) Figure 8: Γ(Z 3 Z 5 ) When e or e, Γ(Z n ) is not planar since a subgraph isomorphic to K 5 may be constructed from the vertices {5, 0, 0, 5, 30}. c) For n = p e when e =, Γ(Z p ) is the empty graph since Z(Z p ) =. For n = p e when p 7 and e 3 the graph is not planar since we can form a K 5 subgraph from the vertices p, p,..., 5p. The vertex set of Γ(Z p ) is {p, p,..., (p )p}. Thus the product of any two vertices is congruent to zero modulo p, resulting in Γ(Z p ) being completely disconnected. Thus our graph is planar. When p = and e 4, we can form a K 5 subgraph from the vertices {, 4, 6, 0, 4}. Therefore, Γ(Z n ) is not planar. When p = and e = 3, n = 8. Γ(Z 8 ) is shown to be planar in Figure??. When p = 3 and e 3, we can form a K 5 subgraph from the vertices {3, 6,, 5, }. Therefore Γ(Z n ) is not planar. When p = 5 and e 3, we can form a K 5 subgraph from the vertices {5, 0, 5, 0, 30}. Therefore the graph is not planar. Case 4: Consider r =. Z n Z n = Z p e...pr er Z q f...qs fs = Z e p... Z fr p r Z f... Z fs q s, with p,..., p r distinct primes and q,..., q s distinct primes. q 9

10 4 6 Figure 9: Γ(Z 8 ) When r or f, Z n is a direct product of three or more factors, thus reducing the problem to cases and. When r =, f =, and p q, Z e p Z f q = Zp e q f, reducing the problem to case 3. When r =, f =, p = q, and p e f, p are both greater than or equal to 4, a K 3,3 subgraph is induced by the bipartition {(0, ), (0, ), (0, 3)} {(, 0), (, 0), (3, 0)}. Thus Z e p Z f p is not planar. The previous case does not cover when p e < 4 or p f < 4. When p e < 4 and p f < 4, the only cases to consider are Γ(Z Z ) and Γ(Z 3 Z 3 ). Both graphs are planar as shown in Figures?? and??. (0, ) (, 0) Figure 0: Γ(Z Z ) (0, ) (0, ) (, 0) (, 0) Figure : Γ(Z 3 Z 3 ) When p f > 4 and p e < 4, a K 5 subgraph may be induced from the vertices {(, 0), (0, ), (0, 3), (0, 4), (0, 5)}. A similar argument holds for p e > 4 and p f < 4. If p e < 4 and p f = 4, it is the case Γ(Z Z 4 ). This graph is planar, as shown in Figure??. 0

11 (0, ) (, ) (0, ) (0, 3) (, 0) Figure : Γ(Z Z 4 ) 5 Independent Sets and Cliques A clique in a graph is a set of pairwise adjacent vertices. An independent set in a graph is a set of pairwise nonadjacent vertices. Our goal in this section is to show that the associate class of a zero divisor in Z n is either a clique or an independent set in the graph Γ(Z n ). For example let us examine n = p p for Γ(Z n ), where p, p are distinct primes. There are four associate classes: A p = {x Z n : p x}, A p = {x Z n : p x}, A p = {x Z n : p x}, A p p = {x Z n : p p x} A p A p A p A p p We can say that A p is a clique. For all x, x A p, x is adjacent to x since x x = up, where u is a unit, which is not divisible by n. Also, A p p is an independent set. For all x, x A p p, x is not adjacent to x since x x = up p, where u is a unit, which is divisible by n. Proposition 5.. A p a pa par r is an independent set if and only if a e, a e,..., a r er. ( ): Let A a p pa pa k be an associate class of the graph, where k r. k par r Choose x, x A a p pa pa k. Assume that the associate class is an independent k par r set and there exists some number a k < e k, where a a k a r. Now x x = up a p a p a k k p ar r, where u is a unit. Since a k < e k, then a k < e k, which means that x x it is not divisible by n. This means that x and x are adjacent which is a contradiction to the definition of an independent set.

12 ( ): Assume that a i e i for all i. Choose x, x A a p pa up a p a p a k k p ar r, where u is a unit. Since a i e i. Now x pa k k par r x = for all i, then a i e i, which means the product is divisible by n. This means that x and x are not adjacent and the associate class is an independent set. Proposition 5.. Every non-independent associate class is a clique. Assume A p f A f p p f k k p f k k p fr r is not an independent set. This implies f k < e k. Let x, y. We know that x y = up f p fr p f k r k p fr r, where u is a unit. Since < e k, the product will not be divisible by n and all vertices will be adjacent. f k Therefore, the associate class will be a clique. 6 Minimum Degree We now look at vertices of minimum degree in Γ(Z n ). First we will discuss some definitions and give some examples referring back to Figure??. Let G be a finite graph. The degree of a vertex v in V (G), denoted deg v, is the number of edges containing v as an endpoint. For example in Γ(Z ), deg (4) = 3 and deg (9) =. The minimum degree of G is defined as δ(g) = min{deg(v) : v V (G)}, and the vertices which have the minimum degree are called vertices of minimum degree. In Γ(Z ), and 0 are both vertices of minimum degree because deg () = deg (0) =. We will prove in Theorem?? that a vertex of minimum degree in Γ(Z n ) is p e p e p er r, but first we must establish the following lemmas. Lemma 6.. Every nonzero zero divisor in Z n is associate to a zero divisor of the form p f p fr r where 0 f i e i. Let x be a nonzero zero divisor. Then there exists some p i such that x = c p i. If c is a unit in Z n, then x and p i are associates. If not, then there exists some p i such that x = c p i p. If c is a unit in Z n, then x and p i p i are associates. If not, we continue this process inductively until c k is a unit in Z n. Then we can say that x is associate to a zero divisor of the form p f p fr r where 0 f i e i. Lemma 6.. If u and v are associates, then N u {u, v} = N v {u, v}.

13 ( ): Let x N u {u, v}. We can say that n xu. By the previous lemma and because u and v are associates, this implies that n xv which implies x N v {u, v}. Also, we must exclude {u, v} because u could be in N v and vice versa but neither u nor v can be in their own neighborhood sets. ( ): A similar argument can be made. Lemma 6.3. Let u = p n p n p r n r and x i = p n p i n i + p r n r be vertices in G. Then N xi N u. Suppose n,..., n r are integers such that for all k we have n k e k, and such that n n i < e i and n j < e j for some i < j r. Let u = p n n p p r r and x i = p n p i n i + p r n r. Since x i p e n p e n p e i n i i p er nr r = 0, Similarly, N xi = {w V (G) : p i r j= p e j n j j w}. If y N xi then p r i j= pe j n j j N u = {w V (G) : r j= p e j n j j w}. y which implies that r j= pe j n j j y meaning y N u. However, z = p e n p i e i n i p r e r n r N u and z N xi. Therefore, N xi N u. Lemma 6.4. For all u V (G) where G = Γ(Z n ), there exists some k such that N vk {v k, u} N u {v k, u}. Choose an arbitrary vertex u. By Lemma?? we know that u is associate to a vertex of the form w = p f p fr r. By Lemma?? we can say that N u {u, w} = N w {u, w}. If there exists a k such that w = v k then N w = N vk which implies N vk {v k, u} = N u {v k, u}. If no such k exists, then there exists an i {,..., r} such that x = p i w 0. Then N x {x, w} N w {x, w} N w {w} by Lemma??. If there exists a k such that x = v k then stop. If no such k exists, then there exists an i {,..., r} such that x = p i x. Then N x {x, x, w} N x {x, x, w} N w {x, w} N w {w} by Lemma??. If there exists a k such that x = v k then stop. If no such k exists, then continue this process inductively until x m = v k for some k. Let 3

14 B = {v k, w, x, x,..., x m }. By construction, N vk B = N xm B N w {v k, w}. It is easy to see that if x k w = 0 then x k v k = 0. It follows that if x k N vk i.e. x k v k 0 then x k N w i.e. x k w 0. Therefore, N vk {v k, w} N w {v k, w} which implies N vk {v k, u, w} N w {v k, u, w} = N u {v k, u, w} by Lemma??. Using a similar argument if w N vk then w N u. Therefore, N vk {v k, u} N u {v k, u}. Lemma 6.5. Let n be an integer. Then Z(Z n ) = n φ(n) is the cardinality of the set of nonzero zero divisors of Z n, i.e the number of vertices in Γ(Z n ) and Γ(Z n ). There are n elements in Z n, and the number of units of Z n is given by φ(n). Since every element of a finite ring is either a zero divisor or a unit, n φ(n) will tell us the number of zero divisors in Z n. Since Z(Z n ) denotes nonzero zero divisors we have the formula Z(Z n ) = n φ(n). e e Lemma 6.6. Let v i = p p i e i p r r be an element of V (Γ(Z n )). Then { n φ(n) ann(vi ) if vi N vi = 0 n φ(n) ann(v i ) + if vi = 0. The annihilator of an element a in Z n is denoted ann(a) and consists of all elements r in Z n such that ra = 0. Therefore r = 0 is contained in every ann(a). Also, a is contained in ann(a) if and only if a = 0. In Γ(Z n ) since v i is adjacent to all other vertices w such that aw = 0, and no vertex can be adjacent to itself or 0, { ann(vi ) if vi deg (v i ) = 0 ann(v i ) if vi = 0. Now looking at Γ(Z n ), we know that edges are replaced with non-edges and vice versa. Therefore, N vi will be equal to the total number of vertices in the graph minus the degree of v i in the original graph. Using the previous lemma, { n φ(n) N vi = Z(Z n ) ann(vi ) if vi deg (v i ) = 0 n φ(n) ann(v i ) + if vi = 0. Lemma 6.7. Let n be an integer then Z n = ann(a) (a). Fix a in a ring R and a mapping m a : R R where m a (x) = ax. Then m a is a group homomorphism and Ker (m a ) = {r R : m a (r) = 0}. Then we can say R = ker(m a ) im(m a ) = ann(a) (a). Lemma 6.8. Let n be an integer then (a) = 4 n. gcd(a,n)

15 We can say that (a) is equal to the order of a in Z n. Let j = a and let d = gcd(a, n). We can say. ja = rn for some r Z and n. a = n a d d 0(mod n). We also know that j n which implies that for some k Z, then jk = n. This implies d d that jkd = n, which gives us kd n. We can also say that jakd = na which leads to rnkd = na. By cancellation, rkd = a which implies that kd a. Now we have kd a and kd n and since gcd(a, n) = d this implies that k =. Since jk = n, this implies d that j = n. This leads us to n = a where d was defined as gcd(a, n). Therefore, d d a = n. gcd(a,n) e e Theorem 6.9. Let G = Γ(Z n ) where n = p p r r e minimum degree is v = p e e p p r r. and e i. Then a vertex of For each i, i r, and each integer f i, 0 f i e i, define A f p = {x Z p fr n : r f f p p r r x}. Then the classes A f p partition V (G) into equivalence classes as p fr r a consequence of Lemma??. From Lemma?? we can say that for all u in V (G), N vi {v i, u} N u {v i, u} so there exists some i such that N vi has the smallest cardinality of all the vertices in G. { n φ(n) ann(vi ) if vi The cardinality of N vi = 0 n φ(n) ann(v i ) + if vi = 0 as shown in Lemma?? and Lemma??. Therefore, since we are looking at the vertices of the form v i and if we choose i such that ann(v i ) is maximized, then N vi will be minimized. We know from Lemma n?? that Z n = ann(a) (a) for all a V (G) and by Lemma?? (a) =. gcd(a,n) n Therefore, n = ann(v i ) ann(v gcd(v i,n) i) = gcd(v i, n). Since n = p i v i, this implies that ann(v i ) = v i. e e We claim that v i v i+ + for all i r. Since v i = p p i e i p r r and e e v i+ = p p i e i p i+ e i+ p r r, we can write v i+ = p i p i+ v i. Because p i < p i+ for all i, we can say v i+ < v i which leads to v i+ + v i. Now, N vi n φ(n) v i + = n φ(n) (v i ) 5

16 n φ(n) v i+ N vi+ and therefore N vi N vi+. Now we can say N v has the minimum cardinality of all the vertices in G and e v = p e e p p r r is a vertex of minimum degree. 7 Connectivity Using the vertex of minimum degree that we found in the previous section, we will now look at the connectivity of Γ(Z n ) when n = p e p e. A vertex cut of a graph G is a subset S V (G) such that G S has more than one component or is a single vertex. The connectivity of G, denoted κ(g), is the cardinality of the smallest set S such that S is a vertex cut. Let us establish the following lemma to use in our proofs. e Lemma 7.. Let G = Γ(Z n ) where n = p p er r and e >. Then N v = δ(g) = p e (p )p e p er r p e (p )p e (p ) p er r (p r ). Recall v = p e p e p er r. Identifying Z n with Z e p Z p er using the Chinese Remainder Theorem, we can write v = (p e r, 0,..., 0). Then N v contains all vertices w such that the first component contains a unit in Z e p and the other components can be anything as long as one of the components is a zero divisor. The number of choices for the first component can be found using the Euler phi function to give us the number of units in Z e p. This will give us p e (p ) choices for the first component. Since the other components can be anything, there are p e i i choices for each of the other components i r. We must also subtract off the cases where all the components are units which will be p e (p ) pr er (p r ). Therefore, there are p e (p )p e p er r p e (p )p e (p ) p er r (p r ) neighbors of v, and since v is a vertex of minimum degree it is also the minimum degree of G, denoted δ(g). e Theorem 7.. Let G = Γ(Z n ), where n = p e p κ(g) is equal to the minimum degree δ(g). and e >. Then the connectivity 6

17 It is well-known that κ(g) δ(g). We will show that κ(g) δ(g), leading to κ(g) = δ(g). Let n = p e p e. By the Chinese Remainder Theorem, Z n = Zp e Z e p. By Theorem e?? a vertex of minimum degree in Z n is v = p e p. Since we are using the Chinese Remainder Theorem, this vertex is identified with (p e, 0). The neighborhood set of v contains all vertices w such that w = (u, a) where u is a unit in Z e p and a is not a unit in Z e p. Therefore, we have two cases. Case : w = (u, 0) Case : w = (u, k p n ) where k Z, k 0, and n < e In Case, w is not adjacent to z = (0, b) where b 0. In Case, w is not adjacent to z = (0, k p e n ) where k Z, k 0. We would like to find a vertex y such that y is adjacent to w and z in both of the cases above. We choose y = (x, u ) where x is a nonzero zero divisor in Z e p and u is a unit in Z e p. We now look at G, a subgraph of G where V (G ) = V (G) N v +. By construction, we know that at least one vertex of the form w will remain in G. We want to prove that for every vertex u in G there exists a u, w-path. If u is adjacent to w, then we are done. If u is not adjacent to w, then it is of the form z described above. If one vertex of the form y exists in G, then a u, w-path exists through y. We know N v = p e (p )p e p e (p )p e (p ) by Lemma??. If we have more than N v vertices like y, then we will have κ(g) δ(g). To find the number of vertices of the form y, we multiply the number of nonzero zero divisors in Z e p by the number of units in Z e p. The number of nonzero zero divisors in Z e p is p e p e (p ) by Lemma??, and the number of units in Z e p is p e (p ). Therefore, the total number of possibilities for y is (p e p e (p ) )(p e (p )). We now want to prove that the following inequality is true (p e p e (p ) )(p e (p )) > p e (p )p e p e (p )p e (p ) which can be reduced to p e (p e p p p e + ) >. We know that p e will always be positive, so if we prove p e p p p e + 0 the inequality will hold. Now p (p e ) p e p pe p e 7

18 p pe p + p p e p p + p p e p Since when e p e >, the inequality holds when p p +. Since p < p by construction, p p + will always hold. This proves that κ(g) δ(g), and we can conclude that κ(g) = δ(g). e Theorem 7.3. Let G = Γ(Z n ), where n = p p and e >. Then κ(g) = δ(g) A v +. We will once again use the Chinese Remainder Theorem to say Z n = Zp Z e p and a minimum degree vertex is v = (, 0). Knowing that Z p is a field, we can partition the vertices of G into four sets as follows. X = {(a, 0) : a 0} X = {(a, x) : a 0, x Z(Z e p ) } X 3 = {(0, x) : x Z(Z e p ) } X 4 = {(0, u) : u is a unit in Z e p } Note that the sets above will be nonempty as long as e >. We know that all vertices in X are adjacent to all vertices in X. Also, all vertices in X and X 3 are adjacent to all vertices in X 4. The vertices in X 3 and X 4 will not be adjacent to the vertices in X. Note that v X and there are p vertices in X. Also, there are (p )(p e p e (p ) ) vertices in X, derived similarly to the number of vertices of the form y in Theorem??. Since X = A v is a clique and v is not adjacent to itself this allows us to conclude that deg v = X + X = p + (p )(p e p e (p ) ). In addition, there are p e (p ) vertices in X 4. Now we will show that κ(g) δ(g) A v +. Examine G X. If we remove X from the graph, then X will be isolated from the rest of the graph, i.e. the graph will be disconnected. The number of vertices in X is (p )(p e p e (p ) ) which is equivalent to δ(g) A v +. Therefore, we can conclude that κ(g) δ(g) A v +. Since all vertices in X and X 3 are adjacent to all vertices in X 4 if we prove X 4 deg v A v + we can conclude that κ(g) δ(g) A v +. Therefore, we want to prove the following inequality p e (p ) (p )(p e p e (p ) ) 8

19 which can be reduced to p e (p p ) p. We can say that p and p e (p p ). Therefore, the inequality is always true and we can conclude that X 4 deg v A v +. The vertices in X 4 correspond to our vertices of the form y used in the proof of Theorem?? and the vertices in X correspond to the vertices of the form w. By a similar argument, we can conclude κ(g) δ(g) A v + leading us to κ(g) = δ(g) A v +. 8 Conclusion We conclude this paper with a suggestion for further research. We have looked at connectivity for Γ(Z n ) when n = p e p e, but these findings could be generalized to the case where n is the product of three or more powers of primes. Our conjecture for this case is presented below. Conjecture 8.. Let G = Γ(Z n ) where n = p e p e p er r for primes p <... < p r. Then κ(g) = δ(g) A v + when e = and κ(g) = δ(g) when e >. 9 Acknowledgements A special thanks to our advisor, Dr. Reza Akhtar of Miami University, for his instruction, guidance, and sense of humor throughout our research experience. Our gratitude goes to Leigh Cobbs, our graduate assistant from Rutgers University, for her dedication and the encouragement she has given us to succeed. We thank Dr. Thomas Farmer of Miami University for assistance with the writing of our paper. In addition, we would like to acknowledge Dr. Vasant Waikar, Dr. Dennis Davenport, and all others at Miami University who have helped make the Summer Undergraduate Mathematical Science Research Institute possible. Our appreciation also goes to the National Security Agency and the National Science Foundation for their support. 9

20 0 Glossary adjacent Two vertices u and v are adjacent if there is an edge connecting u and v in the graph. annihilator Let R be a ring and a R. Define the annihilator of a to be the set ann(a) = {x R : ax = 0}. associates Let R be a ring. Two elements a, b R are called associates if there exists a unit u R such that a = ub. bipartite A graph is bipartite if its vertex set can be partitioned into two non-empty independent subsets. center Let G be a graph and u, v V (G), u v. The center of G is the subgraph induced by the set of vertices of minimum eccentricity. clique A clique is a subset C V (G) such that the subgraph induced by C contains all the possible edges. complement The complement of G, denoted Ḡ is the graph such that V (Ḡ) = V (G) and xy E(Ḡ) xy / E(G). complete graph The complete graph on n-vertices, denoted K n, is the graph with n vertices and all possible edges. connected Let G be a graph. G is connected if for every u, v V (G), u v, there exists a u, v path. connectivity The connectivity of G is κ(g) = min{k : there exists a vertex cut S, S = k}. degree The degree of a vertex v V (G) is the number of edges containing v as an endpoint. distance Let G be a graph and u, v V (G), u v. The distance from u to v is d(u, v) = min{length(p ) : P is a u, v path}. If there is no u, v path then d(u, v) =. eccentricity Let G be a graph. The eccentricity of u V (G) is ε G (u) = ε(u) = max d(u, v) where v V (G) and v u. independent set An independent set is a subset S V (G) such that the subgraph induced by S has no edges. path A path is a sequence of vertices and edges in G such that no vertex is repeated, unless this path is a cycle where the first and last vertices are the same. 0

21 planar A graph is planar if it can be drawn in the plane without any two edges crossing. simple graph A simple graph is a pair (V, E) where V = V (G), a vertex set, and E = E(G), an edge set, and each e E is an unordered pair {u, v} of distinct vertices {u, v} V. subgraph Let G be a graph. A subgraph of G is a graph H such that V (H) V (G) and E(H) E(G) and for every e E(H), the endpoints of e are in V (H). subgraph induced by S Let G be a graph, S V (G). The subgraph induced by S is the graph H such that V (H) = S and E(H) = {v v E(G) : v, v S}. unit Let R be a ring. An element a R is called a unit if there exists a b R such that ab =. vertex cut A vertex cut of graph G is a subset S V (G) such that G S has more than one component or is a single vertex. vertices of minimum degree Let G be a finite graph. The vertices of minimum degree are defined as δ(g) = min{deg(v) : v V (G)}. zero divisor Let R be a ring. An element z R is called a zero divisor if there exists a b R, b 0 such that zb = 0.

22 References [] R. Akhtar, L. Lee. Zero-divisor graphs via ideals, preprint. [] D. F. Anderson, A. Frazier, A. Lauve, P. Livingston. The zero-divisor graph of a commutative ring. II Lect. Notes Pure and Appl. Math. 0 (00), 6-7. [3] D. F. Anderson and P. Livingston. The zero-divisor graph of a commutative ring. J. Algebra. 7 (999), no., [4] M. Brickel. Complements of Zero Divisor Graphs. Plan B Project, Miami University, 004. [5] A. Lauve. Zero-divisor graphs of finite commutative rings. Senior Thesis, University of Oklahoma, 999. Amanda Phillips Franklin College Franklin, IN aphillips@franklincollege.edu Julie Rogers Marymount University Arlington, VA jfr30@marymount.edu Kevin Tolliver Morehouse College Atlanta, GA kptco06@hotmail.com Frances Worek The Pennsylvania State University University Park, PA few@psu.edu