How Many Ways Can You Divide Zero?

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1 How Many Ways Can You Divide Zero? Katherine Benson, Louis Cruz, Yesenia Cruz Rosado, Melissa Tolley, Bryant Watkins Abstract Let Σ(R) be the graph whose vertices are the nonzero zero divisors of a ring R and whose edges are pairs {u, v} where u v and u + v is a zero divisor. We explore basic properties of sum graphs including degree, planarity, coloring, cycles, and isomorphisms by studying examples and finding various patterns. 1 Introduction A graph is a mathematical structure consisting of points joined by lines. There are thousands of problems related to computer science that can be modeled by graphs, so the growth of computer use over the past 40 years has led to a corresponding development in graph theory. In this paper we focus on sum graphs. In essence, these are graphs whose vertices can be represented as a set S of positive integers, and have the property that vertex i and vertex j are adjacent if and only if i + j is in S. Apart from giving rise to combinatorial problems, sum graphs are fascinating in their own right. They have potential applications in data compression, specifically compressed storage of graphs in memory. Ring Theory Preliminaries In order to understand our work with sum graphs, some ring definitions are helpful. Definition.1. A ring is a nonempty set R with two binary operations, + and satisfying: (R, +) is an abelian group. (R, ) is associative, commutative, and has an identity of 1. For all a, b, c R, the distributive law a (b + c) = (a b) + (a c) holds. 1

2 We work with the set Z n = {0, 1,, n 1}, which is a ring with respect to addition and multiplication modulo n. For convenience of notation we often omit the overhead bars. Definition.. An integral domain is a ring R such that a, b R and ab = 0 implies that either a = 0 or b = 0. Equivalently, a 0 and b 0 implies that ab 0. Definition.3. In a ring R, a zero divisor is an element c R such that there exists x R, x 0 where cx = 0. Clearly 0 R is always a zero divisor. We denote the set of zero divisors by Z(R), and we let Z(R) = Z(R) {0} denote the set of nonzero zero divisors. Theorem.4. [4, Theorem 19.3] In the ring Z n, the nonzero zero divisors are precisely those nonzero elements that are not relatively prime to n. Other than zero divisors, another important class of elements are units, which are helpful and key elements in our work. Definition.5. A unit is an element u R that has a multiplicative inverse in R. That is, there exists v R such that uv = 1. Theorem.6. Let R be a finite ring. Then every element of R is either a zero divisor or a unit. In particular, every element of Z n is either a zero divisor or a unit. Theorem.7. (Chinese Remainder Theorem)[5, Corollary.6] If m = p k 1 1 p kt t (k i > 0; p i distinct primes), then Z m = Zp k 1 1 Z k p t. t 3 Graph Theory Preliminaries Our work is based on a graph-theoretic approach of studying zero divisors. Therefore, some basic definitions from graph theory are helpful. Definition 3.1. A graph G consists of a vertex set V (G), an edge set E(G), and an association to each edge, e E(G), of two vertices called the endpoints of e. Definition 3.. The complement of a graph G, denoted Ḡ, is the graph whose vertex set is V (G) but for which {x, y} E(Ḡ) if and only if {x, y} / E(G) An adjacent vertex of a vertex v in a graph is a vertex that is connected to v by an edge. The neighborhood of a vertex v, denoted N(v), is the set of all vertices adjacent to v. An edge is incident at a vertex if that vertex is one of its endpoints. The degree of a vertex v is the number of edges incident at it, denoted deg(v). A dominating vertex is a vertex in a graph that is adjacent to every other vertex of G.

3 Definition 3.3. An vertices. independent set in a graph is a set of pairwise nonadjacent Definition 3.4. A clique in a graph is a set of pairwise, adjacent vertices. Definition 3.5. The complete graph on n vertices, denoted K n, is a graph such that each pair of distinct vertices is adjacent. Definition 3.6. A graph G is bipartite if V (G) is the union of two disjoint independent sets called partite sets of G. The complete bipartite graph, denoted K n,m, is a bipartite graph whose independent sets have size n and m, respectively, and has the maximum number of edges possible. 3.1 Sum Graphs We focus our research on sum graphs. Definition 3.7. Let R be a ring. Define the sum graph Σ(R) as follows: V (Σ(R)) = Z(R) ; {x, y} E(Σ(R)) x + y Z(R). 4 Degrees Based on examples of sum graphs we can see patterns that lead to some interesting propositions. In this section we focus on propositions that deal with the degrees of vertices in sum graphs. Proposition 4.1. Let R = Z p e 1 1 Z p er r, where max {e 1,..., e r } and n = V (Σ(R)). Then (Σ(R)) = n 1. Proof. Let R and n be as above. Suppose that e 1 > 1. Consider x = (p 1, 0..., 0) V (Σ(R)). Now take another vertex y = (b 1, b,..., b r ). As y V (Σ(R)) at least one b j = cp j, c Z. If j = 1, then x + y = (p 1 (1 + c), b,..., b r ) This implies that x + y V (Σ(R)). Now if j > 1, then x + y = (p 1 + b 1, b,..., b j 1, cp j, b j+1,..., b r ). This implies that x + y V (Σ(R)). Therefore x is adjacent to all other vertices in Σ(R). This implies that deg(x) = n 1. Therefore (Σ(R)) = n 1 3

4 Proposition 4.. Let R = Z p1 Z pr, where p i p i+1, p i prime for 1 i r and n = V (Σ(R)). Then, (Σ(R)) = n ( r 1 i=1 (p i 1)) 1. Proof. Let p i and R be as above. Let w = (0,..., 0, 1) V (Σ(R)). Then the degree of w is n minus the number of vertices that are not adjacent to w, minus one. Now take x V (Σ(R)) a vertex that is not adjacent to w; this implies that x = (u 1,..., u r 1, 0), where u i is a unit in Z pi, 1 i r 1. Thus there are exactly r 1 i=1 (p i 1) vertices not adjacent to w. This implies that deg(w) = n ( r 1 i=1 (p 1)) 1. Now take y = (y 1,..., y r ) V (Σ(R)). Let J = {i y i 0}. Let z = (z 1,..., z r ) V (Σ(R)) not adjacent to y. This implies that z i 0 for i / J and z j y j for j J. As z V (Σ(R)), z j = 0 for some j J. Let T = {z such that z is not adjacent to y}. Take T j T, T j = {z T z j = 0}. Then, T T j = k j (p k 1). This implies that deg(y) n ( k j (p k 1)) 1 deg(w). Therefore (Σ(R)) = n ( r 1 s=1 (p s 1)) 1 Proposition 4.3. Σ(Z n ) is a complete graph n = p r where p is prime and r Proof. ( ) Suppose n = p r where p is prime. The vertices of Σ(Z p r) are multiples of p so any vertex in the sum graph can be written in the form v i = kp for some k N. Then v 1 + v = k 1 p + k p = (k 1 + k )p. Since any multiple of p is in V (Σ(Z p r)), then v i, v j V (Σ(Z p r)) i, j N. ( ) Let Σ(Z n ) be a complete graph and suppose n = p e 1 1 p e... p er r,p i distinct primes and r. Then by the Chinese Remainder Theorem, Z n = Zp e 1 1 Z p e... Z p er Suppose v 1, v V (Σ(Z n )), v 1 = (1, 0,..., 0) and v = (0, 1,..., 1). Then v 1 + v must have a zero divisor in one coordinate in order for there to be an edge between them, but v 1 + v = (1,..., 1) and 1 is a unit in Z pi 1 i r. This means that there is no edge between v 1 and v, which is a contradiction. r. Proposition 4.4. In Z n, let n = p 1 p where p i is prime and p 1 < p. The vertices with maximum degree will be of the form v = kp 1 k N. Proof. The number of vertices in the neighborhood of p 1 is equal to n p 1 = p and the size of the neighborhood of p is equal to p 1. Because p i is only adjacent to the vertices of the form v = kp i k N, N(p i ) = N(kp i ) [3]. Since p 1 < p, p > p 1 so N(p 1 ) > N(p ). This means that any vertex of the form v = kp 1 k N, will have maximum degree. 4

5 e Proposition 4.5. In Z n, let n = p 1 e 1 p e... p r r where p i is prime, e i > 1 for some i, r and p 1 < p <... < p r. The vertices with maximum degree will be of the form v = kp 1 p... p r k N. Proof. If we let v = kp 1 p r, then v will be adjacent to every vertex in the sum graph because it is a multiple of all the p i and v n. If w is not of the form kp 1 p r, then there exists p i such that p i w. Then it is not adjacent to vertices of the form w = tp i t N. So v is a dominating vertex if and only if v is a multiple of all p i. 5 Connectedness Lemma 5.1. Let R 1,..., R r be rings and R = R 1 R r. Then for any i, 1 i r the set of vertices of Σ(R) with zero in the i th coordinate forms a clique. Proof. Let R = R 1 R r be a ring with S the set of all vertices with 0 in the i th coordinate. Let a, b S. Then, a + b will also have a zero in the i th coordinate and thus a + b S. Therefore, S is a clique. Theorem 5.. Let R = R 1 R where each R i is an integral domain. Then Σ(R) is disconnected and is the union of two cliques with respective size R 1 1 and R 1. Proof. Since R 1 and R are integral domains, R 1 and R have no nonzero zero divisors. Thus Z(R) = S 1 S where S 1 = {(a, 0) : a 0} and S = {(0, b) : b 0}. By Lemma 5.1 S 1 and S are cliques. To show that Σ(R) is disconnected, suppose there exist adjacent vertices (a, 0) and (0, b). It follows that (a, 0) + (0, b) = (a, b) Z(R). Since R 1 and R are integral domains, nonzero zero divisors take the form (a, 0) or (0, b). Therefore (a, b) / Z(R), and thus S 1 and S are connected components of Σ(R). Now we need to show S 1 = R 1 1 and S = R 1. Every vertex in S 1 is of the form (a, 0). There are R 1 choices for a, but 0 R 1, and (0, 0) / S 1. So there are R 1 1 vertices in S 1. Similarly, there are R 1 vertices in S. Proposition 5.3. Let R = R 1 R R r where each R i is a ring and R 1 = n 1,..., R r = n r. If r 3, then diam(σ(r)). In particular, Σ(R) is connected. Proof. We give a proof in the case r = 3; the general case follows similarly. Let r = 3, so R = R 1 R R 3 and let x, y Z(R) be distinct vertices where x = (a 1, a, a 3 ) and y = (b 1, b, b 3 ). If x y then d(x, y) = 1. Suppose x y. In order for d(x, y), there must be a z V (Z(R)) such that x z y. 5

6 Choose z = ( a 1, 0, b 3 ). If x = z or y = z, then d(x, y) = 1. Now assume z x and z y. If a 1 0 or b 3 0, then x z y. If a 1 = 0 and b 3 = 0, x = (0, a, a 3 ) and y = (b 1, b, 0). Now we can choose z = (0, 1, 0) such that x z y and d(x, y) =. Therefore, when r = 3, diam(r). 6 Isolated and Pendant Vertices Proposition 6.1. Let R = Z Z p, p prime, then Σ(R) is the disjoint union of a clique and an isolated vertex. Proof. Since and p are primes we know that V (R) = {(1, 0)} {(0, b) : b 0 Z p }. Also all the vertices of the form (0, b) such that b Z p are pairwise adjacent; therefore, they form a clique. The sum (1, 0) + (0, b) gives us a unit (1, b), therefore (1, b) / V (Z(R)). Hence Σ(R) is the disjoint union of a clique and the isolated vertex, (1, 0). Proposition 6.. Let R = Z 3 Z p, p 3 prime, then Σ(R) contains pendant vertices, namely (1, 0) and (, 0). Proof. Since 3 and p are primes we know that V (R) = {(1, 0), (, 0)} {(0, b) : b 0 Z p }. Also all the vertices of the form (0, b) are pairwise adjacent, therefore they form a clique. The sum (a, 0) + (0, b) where b Z p and a Z 3 gives us units (, b) or (1, b), b Z p, therefore (, b) and (1, b) / V (Z(R)). But (, 0) + (1, 0) V (Z(R)), so (1, 0) and (, 0) are adjacent pendant vertices and are not adjacent to any of the vertices of the clique. Proposition 6.3. Let R = Z Z p, where p is prime. Then Σ(R) is a star with V (R) 1 pendant vertices. Proof. By Proposition 6.1 we know that the vertex (1, 0) is an isolated vertex in the sum graph Σ(Z Z p ) where p is a prime, and that the rest of the vertices are pairwise adjacent. Therefore, in the complement of the sum graph we have that the vertex (1, 0) is adjacent to the rest of the vertices. Hence there are V (R) 1 pendant vertices. Proposition 6.4. Let R = Z 3 Z p, p 3 prime, then Σ(R) contains vertices of degree V (R) and the rest of the vertices are of degree. 6

7 Proof. By Proposition 6. we know that the vertices (1, 0) and (, 0) are pendant vertices in the sum graph Σ(Z Z p ) where p 3 is a prime, and that the rest of the vertices are pairwise adjacent. Therefore, in the complement of the sum graph we have that the vertex (1, 0) is adjacent to the rest of the vertices except (, 0). The same situation holds for the vertex (, 0). Hence there are V (R) vertices of degree and vertices of degree V (R). Proposition 6.5. Let R = Z p1 Z pr, where r 3 and p i is prime, 1 i r. Then Σ(R) and Σ(R) do not contain isolated vertices. Proof. Let n = V (Σ(R)). We will first examine Σ(R). By Proposition 5.3 we know that Σ(R) is connected therefore, it does not have an isolated vertex. Now, by Proposition 4., we know that the max degree is given by n ( r 1 s=1 (p s 1)) 1 but since p i for all i, we know that the maximum degree of Σ(R) is at most n therefore none of the vertices are isolated in Σ(R). Proposition 6.6. Let R = Z m1 Z mr Z p n, where m i, n, r N and p is a prime. Then Σ(R) has at least one isolated vertex given by v 0 = (0 1, 0,, 0 r, p n 1 ). Proof. We prove this by contradiction. Let R and v 0 be as above. Now suppose that v 0 is not a dominating vertex in Σ(R), which implies there exists v V (Σ(R)) such that v 0 + v = v 1 where v 1 = (u 1, u,, u r, u r+1 ) with u j is a unit for 1 j r + 1. Hence v = (u 1, u,, u r, u r+1 p n 1 ). Since u r+1 is a unit in Z p n, p u r+1 and p n 1 p n, therefore we have that p u r+1 p n 1. This implies that u r+1 p n 1 is a unit in Z p n and therefore v / V (Σ(R)) which is a contradiction. Hence v 0 is a dominating vertex of Σ(R) and an isolated vertex in the complement Σ(R). 7 Coloring Definition 7.1. Let G be a graph. A k coloring of G is a map f : V (G) {1,,, k} = [k]. A k-coloring is called proper if adjacent vertices are assigned different colors. G is called k colorable if there exists a proper k-coloring. The chromatic number of G is denoted by χ(g) and is given by χ(g) = min{k : G is k-colorable}. Lemma 7.. For any graph G, χ(g) ω(g). 7

8 Proposition 7.3. Let R = Z 3 Z p, p prime, then χ(σ(r)) = p 1. Proof. By Proposition 6. we know that the Σ(R) is a disjoint union of a clique with p 1 vertices and a single edge. Therefore χ(σ(r)) p 1. In a proper coloring of Σ(R), p 1 distinct colors must be used on the clique, but any two distinct colors used there can be used to color the remaining vertices. Hence χ(σ(r)) = p 1. Proposition 7.4. Let R = Z Z p, p prime, then χ(σ(r)) = p 1. Proof. By Proposition 6.1 we know that the Σ(R) is a disjoint union of a clique with p 1 vertices and a vertex. Therefore χ(σ(r)) p 1. In a proper coloring of Σ(R), p 1 distinct colors must be used on the clique, but any color can be used to color the remaining vertex. Hence χ(σ(r)) = p 1. Proposition 7.5. Let R = Z p Z p, p prime, then χ(σ(r)) = p 1. Proof. We know Σ(R) is a graph composed of two components that are cliques, since p is prime each clique contains p 1 vertices and therefore we need at least p 1 colors for a proper coloring. Since the two cliques are disjoint we have that χ(σ(r)) = p 1. By studying the complement graphs of the results just mentioned, we are able to see some patterns that led us to the following general proposition. Proposition 7.6. Let G be a graph which is the disjoint union of two cliques. Then χ(g) =. Proof. Let G be a graph which is the disjoint union of two cliques. Then G is complete bipartite since nonedges become edges and vice versa. Now since V (G) consists of two independent sets, S 1 and S, we can assign each vertex in S i the same color. Now since all the vertices in S are adjacent to those in S 1, in order to obtain a proper coloring we need at least two colors. Hence, χ(g) =. 8 Planarity Definition 8.1. A graph G is planar if G can be drawn in the plane with no edge crossings. Definition 8.. An elementary subdivision of a graph G is a graph obtained by splitting an edge of G. Two graphs are homeomorphic if one can be obtained from the other by elementary subdivisions. 8

9 Figure 1: K 5 and K 3,3 Theorem 8.3. (Kuratowski s Theorem) A graph G is planar if and only if G contains no subgraph homeomorphic to K 5 or K 3,3 See Figure 1. [] Proposition 8.4. Let R = Z n Z m. If max{n, m} 6, Σ(R) is nonplanar. Proof. The set of nonzero zero divisors includes S 1 = {(a, 0) : a 0} and S = {(0, b) : b 0}. The order of S 1 is n 1. Similarly, there are m 1 vertices in S. Without loss of generality, assume n m. Therefore, S 1 has more vertices than S. By Lemma 5.1, S 1 is a clique. Since n 6, V (S 1 ) 6 1 = 5. Therefore, S 1 contains a 5-clique and thus S 1 contains a subgraph isomorphic to K 5. By Kuratowski s Theorem, Σ(R) is nonplanar. Proposition 8.5. Let R = Z n1 Z n Z n3. If max{n 1, n, n 3 } 3, then Σ(R) is nonplanar. Proof. Let R be a ring such that R = Z n1 Z n Z n3 and assume without loss of generality n 1 n n 3. The set of nonzero zero divisors will include S = {(0, a, a 3 ) a Z n, a 3 Z n3 } {(0, 0, 0)}. By Lemma 5.1, S is a clique. Since n 3 3, there are at least three options for a 3 and two options for a. Thus, any vertex in S is adjacent to at least 3 1 = 5 other vertices in S. Therefore S contains a 5- clique and thus S contains a subgraph isomorphic to K 5. By Kuratowski s Theorem, since S is nonplanar, then Σ(R) is nonplanar. Proposition 8.6. Let R = Z n1 Z nr be a ring. If r 4, then Σ(R) is nonplanar. 9

10 Proof. Let R = Z n1 Z nr be a ring with r 4. When r = 4, the set of nonzero zero divisors will include the vertices (0, 0, 0, 1), (0, 0, 1, 0), (0, 1, 0, 0), (1, 0, 0, 0), (1, 1, 0, 0). These vertices form a 5-clique. Thus Σ(R) contains a subgraph isomorphic to K 5 and therefore, by Kuratowski s Theorem, Σ(R) is nonplanar. A similar argument holds for r 4 by simply adding zeros to the additional coordinates. This leaves only a finite number of rings of the form Z n1... Z nr that could give rise to planar sum graphs. Let R 1 = Z 3 Z 4, R = Z 4 Z 4, and R 3 = Z 5 Z 5 be rings. The sum graphs of these rings contain subgraphs isomorphic to K 5 and therefore Σ(R 1 ), Σ(R ), and Σ(R 3 ) are nonplanar. Proposition 8.7. The only sum graphs of rings of the form Z n1... Z nr where r that are planar are the following: Z Z, Z 3 Z, Z 3 Z 3, Z 4 Z, Z 5 Z, Z 5 Z 3, Z 5 Z 5, and Z Z Z. Proof. See Figures - 5. To check for the planarity of Σ(Z n ) where n = p e 1 1 p er r and p i are distinct primes, by the Chinese Remainder Theorem and our work above, we can reduce to the case where n = p e for some prime p. By the definition for sum graphs given in [3], any nonplanar graph will still be nonplanar by our definition because the graph will only have more edges. Hence by [3, Corollary 5.16], we only have to check when n = 4, 8, 9, 5, 49. For n = p e, Σ(Z n ) is a complete graph by Proposition 4.3. When n = 49, Σ(Z n ) contains a subgraph isomorphic to K 5 and therefore is nonplanar. When n = 4, 8, 9, 5, Σ(Z n ) has 1, 3,, 4 vertices, respectively, and therefore does not contain a subgraph isomorphic to K 5. Therefore, when n = 4, 8, 9, 5, Σ(Z n ) is planar. The same arguments show that for a finite direct product of any finite rings, if the number of factors is greater than or equal to 4, if there are three factors, R 1, R, R 3 where max{ R 1, R, R 3 } 3, and if there are two factors with one having size of at least 6, then the sum graphs of these rings are nonplanar. Proposition 8.8. Let R = R 1 R r where R i is a finite local ring. Let m i be the unique maximal ideal of R i. If some m i has six elements or more, Σ(R) is nonplanar. Proof. Let R = R 1 R r where R i is a finite local ring. Let m i be the unique maximal ideal of R i such that m i 6; m i consists of all the zero divisors of R i. 10

11 Figure : (a) Σ(Z Z ) and (b) Σ(Z 3 Z ) Figure 3: (a) Σ(Z 3 Z 3 ) and (b) Σ(Z 4 Z ) 11

12 Figure 4: (a) Σ(Z 5 Z ) and (b) Σ(Z 5 Z 3 ) Figure 5: (a) Σ(Z 5 Z 5 ) and (b) Σ(Z Z Z ) 1

13 Therefore, V (R i ) = m i {0}. Since every ideal is an Abelian group under addition, m i is closed under addition. Therefore, for every a, b V (R i ), a + b V (R i ). Since there are at least six vertices in m i, five of them are nonzero. Therefore Σ(R i ) contains a 5-clique and thus contains a subgraph isomorphic to K 5 which implies that Σ(R) is nonplanar. The finite local rings whose maximal ideal has size less than six as follows: Rings with m = : Z 4 and Z [x] (x ). Rings with m = 3: Z 9 and Z 3[x] (x ). Rings with m = 4: Z 8, Rings with m = 5: Z 5 and Z 5[x] (x ). Z 4 [x] (x, x ), Z [x, y] (x, y, xy), Z [x] (x 3 ), Z 4[x] (x ), Z [x, y] (x, y + y + 1). Definition 8.9. A graph G is outerplanar when G is planar and in some planar drawing every vertex of G is on the boundary of the unbounded face. Corollary (To Kuratowski s Theorem) A graph G is outerplanar if and only if G contains no subgraph homeomorphic to K 4 or K,3 See Figure 6. Figure 6: K 4 and K,3 13

14 Since a graph must be planar to be outerplanar, there are only a finite number of rings of the form Z n1... Z nr where r that could have outerplanar sum graphs. As seen in Figures 4 and 5, of the eight planar sum graphs we have found, Σ(Z 5 Z ), Σ(Z 5 Z 3 ), Σ(Z 5 Z 5 ), and Σ(Z Z Z ) all have subgraphs isomorphic to K 4. Therefore, these graphs are not outerplanar. This leaves Σ(Z Z ), Σ(Z 3 Z ), Σ(Z 3 Z 3 ), and Σ(Z 4 Z ) as the only sum graphs of this form that are outerplanar, which can be seen in Figures and 3. For the case where r = 1, Σ(Z 4 ), Σ(Z 8 ), and Σ(Z 9 ) are outerplanar. 9 Cycles and Girth Definition 9.1. A path is a sequence of vertices and edges with no repeated internal vertices. A path is called closed if the initial and final vertices conicide. A cycle is a closed path. The length of a cycle is the number of edges in the cycle. The smallest possible length of a cycle is 3. Lemma 9.. Let R = Z n Z m. Then Σ(R) has a cycle if and only if max{n, m} 4. Proof. ( ) Let R = Z n Z m be a ring with max{n, m} 4. The nonzero zero divisors of Σ(R) will include S 1 = {(a, 0) : a 0} and S = {(0, b) : b 0}. The order of S 1 is n 1 and the order of S is m 1. Without loss of generality, assume n m. By Lemma 5.1, S 1 is a clique and has 3 or more vertices. Thus, there are 3 pairwise adjacent vertices in Σ(R) which shows that Σ(R) contains a 3-cycle. ( ) As can be seen by Figures and 3 (a), Σ(Z Z ), Σ(Z 3 Z ), and Σ(Z 3 Z 3 ) do not have cycles. Definition 9.3. The girth of a graph G is the length of the shortest cycle in G or if there are no cycles in G. The girth of G is denoted gr(g). Proposition 9.4. Let R = Z n1 Z nr. If Σ(R) is nonplanar, then gr(σ(r)) = 3. Proof. Let R = Z n1 Z nr with Σ(R) a nonplanar graph. The results from Section 8 show if Σ(R) is nonplanar, then Σ(R) contains a subgraph isomorphic to K 5. Since K 5 is a 5-clique, it contains a 3-cycle. Therefore, gr(σ(r)) = 3. Proposition 9.5. Let R = Z n1 Z nr. Then gr(σ(r)) is either 3 or. Proof. By Proposition 9.4, if Σ(R) is nonplanar, then gr(σ(r)) = 3. This leaves finitely many planar sum graphs to check. By Lemma 9., Σ(Z Z ), Σ(Z 3 Z ), and 14

15 Σ(Z 3 Z 3 ) do not have any cycles and therefore gr(σ(r)) =. Also, gr(σ(z 4 )) = and gr(σ(z 9 )) = because they are isomorphic to K 1 and K, respectively. Since Σ(Z 8 ) is isomorphic to K 3 and Σ(Z 5 ) is isomorphic to K 4, these graphs have girth of 3. The rest of the planar graphs, namely Σ(Z 4 Z ), Σ(Z 5 ΣZ ), Σ(Z 5 Z 3 ), Σ(Z 5 Z 5 ), and ΣZ Z Z ), have girth of 3 as can be seen by Figures 3 (b), 4, and 5. Lemma 9.6. [7] Let G be a graph. Then gr(g) diam(r) + 1 Proposition 9.7. Let R = R 1 R R r where each R i is a ring and r 3. Then gr(σ(r)) 5. Proof. By Proposition 5.3, diam(σ(r)). Now by Lemma 9.6, gr(σ(r)) + 1 = 5 By considering different examples, we find gr(σ(r)) is not as easily determined as gr(σ(r)). Proposition 9.8. Let R = Z Z p. Then gr(σ(r)) =. Proof. By Proposition 6.3 we know that Σ(R) is a star, therefore it does not have cycles. Hence gr(σ(r)) =. Proposition 9.9. Let R = Z 3 Z p, with p 3 is prime. Then gr(σ(r)) = 4. Proof. By Proposition 6.4 we know that Σ(R) is bipartite. From this, it follows that gr(σ(r)) 3. There is a 4-cycle (1, 0), (0, 1), (, 0), (0, ), (1, 0) in Σ(R)). This shows that gr(σ(r)) = 4. We find Σ(Z Z 6 ) is an interesting case because gr(σ(z Z 6 ) = 6 as can be seen by Figure Isomorphisms Definition Let G and H be graphs. An isomorphism is a bijective function f : V (G) V (H) such that xy E(G) f(x)f(y) E(H). Definition 10.. The clique number, denoted ω(g), is the max{ S : S V (G), where S is a clique}. 15

16 Figure 7: Σ(Z Z 6 ) Proposition Let p 1, p, p 3, and p 4 be distinct primes. Then Σ(Z p1 p ) is not isomorphic to Σ(Z p3 p 4 ). Proof. Assume Σ(Z p1 p ) is isomorphic to Σ(Z p3 p 4 ) and each p i is distinct. This means that each sum graph has the same number of vertices and edges. Using the Euler Phi Function, the number of vertices of Σ(Z p1 p ) is p 1 + p. Using the same method to find the number of vertices of Σ(Z p3 p 4 ) and the fact that the number of vertices in each sum graph is equal, p 1 + p = p 3 + p 4. (1) Also, each sum graph has the same number of edges. Since Z p1 p = Zp1 Z p, the sum graph will have two disjoint cliques (5.) one with p 1 1 vertices and the other with p 1 vertices. Since each is complete, the first component has (p 1 1)(p 1 ) edges and the second has (p 1)(p ) for a total of (p 1 1)(p 1 ) + (p 1)(p ) edges in the graph. Since the graphs are isomorphic, = p 1(p 1 3) + p (p 3) + 4 p 1 (p 1 3) + p (p 3) + 4 = p 3 (p 3 3) + p 4 (p 4 3) + 4.[] Using (1), p 1 = p 3 + p 4 p. Substituting this into the second equation: (p 3 + p 4 p )(p 3 + p 4 p 3) + p (p 3) + 4 = p 3 (p 3 3) + p 4 (p 4 3) + 4 () p 3 + p 3 p 4 p p 3 3p 3 + p 4 p p 4 3p 4 + p = p 3 3p 3 + p 4 3p 4 16

17 p 3 p 4 p p 3 p p 4 + p = 0 p 3 p 4 + p = p (p 3 + p 4 ) Therefore, p divides p 3 p 4 + p. Since p divides p, p must also divide p 3 p 4, but since p, p 3, p 4 are primes, either p = p 3 or p = p 4, so the primes are not distinct and this is a contradiction. Therefore the sum graphs of Z p1 p and Z p3 p 4 cannot be isomorphic. Proposition Let p 1, p be distinct primes and r, s integers. Then Σ(Z p r 1 ) is not isomorphic to Σ(Z p s ). Proof. Assume that Σ(Z p r 1 ) = Σ(Z p s ) and p 1 and p are distinct. Applying the Euler Phi Function, Z(Z p r 1 ) = p r 1 p r 1 1 (p 1 1) 1 and Z(Z p s ) = p s p s 1 (p 1) 1. Since the sum graphs are isomorphic, they have the same number of vertices, so p r 1 p r 1 1 (p 1 1) 1 = p s p s 1 (p 1) 1 p r 1 1 = p s 1 p 1 p r 1 = p s 1 This implies that p 1 divides p s 1. Since p 1, p are primes, p 1 must divide p, so p 1 = p. This is a contradiction since we assumed the two primes were distinct. Therefore Σ(Z p r 1 ) Σ(Z p s ). Theorem Let p < q < r be primes and R = Z p Z q Z r. Then ω(σ(r)) = qr 1. Proof. First, we note that V (Σ(R)) is partitioned into three subsets as follows: S 1 = {(0, y, z) : y, z not both zero } S = {(x, 0, z) : x 0} S 3 = {(x, y, 0) : x 0, y 0} By Lemma 5.1, each S i is a clique. By an elementary counting argument, S 1 = qr 1, S = (p 1)r, S 3 = (p 1)(q 1). Clearly S 1 > S > S 3. 17

18 It remains to show that every clique has size at most S 1 = qr 1. Let C be an arbitrary clique in Σ(R). Suppose C contains v = (x, 0, z) S with z 0. Any of the other (p 1)r 1 elements of S is adjacent to v; moreover, if w = (0, y, z ) S 1 is adjacent to v, then either y = 0 and z 0 or z = z and y is arbitrary. This gives at most r + q possibilities for w. Finally, if u = (x, y, 0) S 3 is adjacent to v, then x = x. This gives at most q 1 possibilities for u. Thus, C (p 1)r + r + q 1 + q 1 = pr + q. We claim that pr + q qr 1. If not, then pr + q > qr 1; that is q < pr 1 r = pr p + p 1 r = p(r ) + p 1 r = p+ p 1 r p+p 1 q < p+. Thus, q < p +, so q p + 1, which is a contradiction unless p =, q = 3. In the case, the given upper bound for C becomes pr + q = r + 4 qr 1 = 3r 1. Now suppose C contains a vertex v = (x, 0, 0) S but no vertices of type (x, 0, z) with z 0. Thus, C contains at most p other vertices in S. If w = (0, y, z ) S 1 is adjacent to v, then either y = 0 or z = 0, which accounts for q +r 1 possibilities. Now any vertex of S 3 is adjacent to v; thus in this case C p 1 + q + r 1 + (p 1)(q 1) = pq + r 1 (q 1)r + r 1 = qr 1. Finally, suppose C contains no vertex of S but contains some vertex v = (x, y, 0) S 3. Evidently any of the other (p 1)(q 1) 1 vertices of S 3 is adjacent to v. In order for w = (0, y, z ) S 1 to be adjacent to v, either y = y or z = 0, which gives r + q 1 possibilities for w. Thus, C (p 1)(q 1) + r + q 1 = pq + r p 1 pq + r 1 qr 1 as in the previous paragraph. Therefore, any clique in Σ(R) containing a vertex of S or a vertex of S 3 has size at most S 1 = qr 1, and hence S 1 is a clique of maximum size. Theorem Let R 1 = Z p1 Z p Z p3 and R = Z q1 Z q Z q3 where R 1 and R are rings and p i, q j primes where p i q j for some i and j. Then Σ(R 1 ) Σ(R ). Proof. Assume Σ(R 1 ) = Σ(R ), so the clique number must be equal for each graph. By Theorem 10.5, ω(σ(r 1 )) = p p 3 1 and ω(σ(r )) = q q

19 . p p 3 = q q 3 For any q i, q i divides p p 3, so q i = p j for some i and j. This hold for each q and p. Therefore the sum graphs cannot be isomorphic unless the rings are equal. 11 Open Problems and Conjectures Conjecture In Z n, let n = p 1 p... p r where p is a distinct prime p 1 < p <... < p r, then N(p 1 ) > N(p ) N(p 3 )... N(p r ). The motivation for our conjecture is the following: the neighborhood of any vertex has to have size at least n/p i since it has to be adjacent to all the vertices of the form v = kp i where k N that do not include zero and itself (hence the subtraction of two). This is only a lower bound because it is possible that the vertex could be adjacent to a vertex of the form w = kp j where k N and i j. Conjecture 11.. Let R = Z p1 Z pr where p i is prime and p 1 < p <... < p r. The clique number is p p 3 p n 1. Proof. This proof should be similar to Lemma The problem arises when showing that the only maximal cliques are the S i. Corollary Let R 1 = Z p1 Z pr and R = Z q1 Z qk where R 1 and R are rings and p i, q j primes, where p i q j for some i and j. Then Σ(R 1 ) Σ(R ). Proof. This proof follows from the previous lemma and uses the same technique as the proof of Theorem Conclusions We have explored various aspects of sum graphs, but there is still room for further research. We suggest looking into the edge coloring, more investigation into the complements, and continued work on isomorphisms of sum graphs. 19

20 13 Acknowledgments We thank Dr. Reza Akhtar for his teaching and guidance throughout our research. We also appreciate all the help, encouragement and suggestions given to us by Amanda Phillips, our graduate assistant from Purdue University. Our thanks goes to Dr. Farmer who instructed us and gave us advice on writing this paper. We would also like to thank the other participants of SUMSRI, especially Team Sweet Action Score, and graduate assistants for their support and encouragement throughout this research process. Our gratitude goes to Dr. Davenport and Dr. Waikar, directors of SUMSRI, and all others who made this experience in SUMSRI possible for us. We would also like to thank the National Security Agency, National Science Foundation, and Miami University for their support of this program. References [1] D. Anderson, P. Livingston. The Zero-divisor Graph of a Commutative Ring, J. Algebra 17 (1999), no., [] R. Balakrishnan., K. Ranganathan. A Textbook of Graph Theory. Springer, 000. [3] C. Bicket, S. Graffeo, W. Ross, E. Washington. The Structure of Zero Divisor Sum Graphs. SUMSRI, Miami University, 006. [4] Fraleigh, John B. A First Course in Abstract Algebra. 7th ed. New York: Addison Wesley, 003. [5] Hungerford, Thomas W. Algebra. New York: Springer-Verlag, [6] A. Phillips, J. Rogers, K. Tolliver and F. Worek. Uncharted Territory of Zero Divisor Graphs and Their Complements. SUMSRI, Miami University, 004. [7] West, Douglas, Introduction to Graph Theory, Second Edition. Prentice Hall, 001. Katherine Benson Luther College Decorah, IA benska01@luther.edu 0

21 Louis Cruz University of Puerto Rico at Humacao Humacao, Puerto Rico Yesenia Cruz Rosado University of Puerto Rico at Humacao Humacao, Puerto Rico Melissa Tolley Agnes Scott College Decatur, Georgia Bryant Watkins Howard University Washington, D.C. 1