# ZERO-DIVISOR AND IDEAL-DIVISOR GRAPHS OF COMMUTATIVE RINGS

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1 ZERO-DIVISOR AND IDEAL-DIVISOR GRAPHS OF COMMUTATIVE RINGS T. BRAND, M. JAMESON, M. MCGOWAN, AND J.D. MCKEEL Abstract. For a commutative ring R, we can form the zero-divisor graph Γ(R) or the ideal-divisor graph Γ I (R) with respect to an ideal I of R. We consider the diameters of direct products of zero-divisor and ideal-divisor graphs. 1. Introduction and Definitions The concept of a zero-divisor graph was first introduced in 1988 in [3] by Beck, who was interested in the coloring of the zero-divisor graph. However, we use the slightly altered definition of a zero-divisor graph offered by Anderson and Livingston in [1]. Anderson and Livingston also proved that the diameter of Γ(R) is less than or equal to three for all commutative rings R. This knowledge of small diameter of zero-divisor graphs led Axtell, Stickles, and Warfel to find necessary and sufficient conditions for the direct product of two commutative rings R 1 and R 2 to have various diameters in [2]. Given a commutative ring R, recall that the set of zero-divisors Z(R) is the set {x R there exists y R such that xy = 0}, where R = R {0}. Also, Z(R) = Z(R) {0}. Finally, we define regular elements to be reg(r) = R Z(R) and the annihilator of a zero-divisor x to be ann(x) = {y Z(R) xy = 0}. We can define the zero-divisor graph of R, Γ(R), as follows: x Γ(R) if and only if x Z(R), and x, y Γ(R) are adjacent if and only if xy = 0. Furthermore, we can define the diameter of any graph diam(γ) = max{d(x, y) x and y are distinct vertices of Γ(R)}. For the sake of generalization we can think of {0} as an ideal of the ring R. Given a commutative ring R and any ideal I of R, we define the ideal-divisors of R with respect to I as Z I (R) = {x R there exists y R I such that xy I}. Furthermore, we can define Z I (R) as Z I (R) I. Also, reg I (R) = R Z I (R) and ann I (x) = {y Z I (R) xy I}. We can define the ideal-divisor graph of R, Γ I (R), by letting x be an element of Γ I (R) if and only if x is an element of Z I (R) ; x and y in Γ I (R) are adjacent if and only if xy is an element of I. The ideal-divisor graph was first discussed by Redmond in [4]. He was able to generalize many of the concepts and theorems of the zero-divisor graph to the idealdivisor graph. In particular, Redmond showed that the diameter of Γ I (R) is less than or equal to three for all commutative rings R and I an arbitrary ideal of R. In this paper we complete the classification done by Axtell, Stickles, and Warfel in [2] to include direct products of commutative rings that have diameter zero and generalize the entire classification to ideal-divisor graphs. Date: February 27,

2 2 T. BRAND, M. JAMESON, M. MCGOWAN, AND J.D. MCKEEL 2. Ideal-Divisor Graphs of Direct Products We begin with some necessary lemmas generalized from similar lemmas used by Axtell, Stickles, and Warfel. Lemma 2.1. Let R 1 and R 2 be rings with ideals I 1 and I 2, respectively, with Z I1 (R 1 ) or Z I2 (R 2 ) (or both). If reg I1 (R 1 ) and reg I2 (R 2 ), then diam(γ I1 I 2 R 2 )) = 3. Proof. Without loss of generality, let a Z I1 (R 1 ). Then there exists b Z I1 (R 1 ) such that ab I 1. Let r 1 reg I1 (R 1 ), r 2 reg I2 (R 2 ). Then (r 1, 0) (0, r 2 ) (a, 0) (b, r 2 ) is a path of length 3. Assume there is a shorter path from (r 1, 0) to (b, r 2 ). Clearly, (r 1, 0)(b, r 2 ) / I 1 I 2, so there is no path of length 1. Assume there exists (x, y) Z I1 I 2 R 2 ) such that (x, y)(r 1, 0), (x, y)(b, r 2 ) I 1 I 2. Then xr 1 I 1 and yr 2 I 2, which gives x I 1 and y I 2, a contradiction. Hence, diam(γ I1 I 2 R 2 )) = 3. Lemma 2.2. If R 1 = I 1 and diam(γ I2 (R 2 )) > 0, then diam(γ I1 I 2 R 2 )) = diam(γ I2 (R 2 )). Proof. Let R 1 = I 1 and let diam(γ I2 (R 2 )) > 0. Suppose diam(γ I1 I 2 R 2 )) = n > diam(γ I2 (R 2 )) such that n = 2 or 3. Then (a 0, x 0 ), (a 1, x 1 ),..., (a n, x n ) Z I1 I 2 R 2 ) such that (a 0, x 0 ) (a 1, x 1 )... (a n, x n ) is a minimal path. Then x 0 x 1... x n, but since n > diam(γ I2 (R 2 )), x 0 x 1... x n must not be a minimal path. This can happen in two ways: If i, j such that 0 i < j n, j i + 1, and x i x j. Then (a i, x i ) (a j, x j ), a contradiction of (a 0, x 0 ) (a 1, x 1 )... (a n, x n )being a minimal path. If n = 3 and y Z I2 (R 2 ) such that y / {x i 0 i n} and x 0 y x n. Then (a 0, x 0 ) (0, y) (a n, x n ), a contradiction of (a 0, x 0 ) (a 1, x 1 )... (a n, x n )being a minimal path. So diam(γ I1 I 2 R 2 )) diam(γ I2 (R 2 )). Suppose diam(γ I1 I 2 R 2 )) < diam(γ I2 (R 2 )) = n such that 1 n 3. Then x 0, x 1,..., x n Z I2 (R 2 ) such that x 0 x 1... x n is a minimal path. Since R 1 = I 1, a 0, a 1,..., a n R 1, (a 0, x 0 ) (a 1, x 1 )... (a n, x n ) is a minimal path of length n, a contradiction. So diam(γ I1 I 2 R 2 )) diam(γ I2 (R 2 )). Thus diam(γ I1 I 2 R 2 )) = diam(γ I2 (R 2 )). Lemma 2.3. (Analogue to ASW Lemma 2.2) Let R be a commutative ring with ideal I. If R = Z I (R) and diam(γ I (R)) = 1, then R 2 I. Proof. Note that for all x, y R with x y, xy I because diam(γ I (R)) = 1. Assume that a 2 / I for some a R. Let b Z I (R) with b a. Observe that a + b a. Since R = Z I (R) and diam(γ I (R)) = 1, we have ab I. So, a(a + b) = a 2 + ab I, which implies a 2 I, a contradiction. Theorem 2.4. Let R 1, R 2 be commutative rings such that diam(γ(r 1 )) = diam(γ(r 2 )) = 0. Then

3 ZERO-DIVISOR AND IDEAL-DIVISOR GRAPHS OF COMMUTATIVE RINGS 3 (i) diam(γ R 2 )) = 0 if and only if (without loss of generality) R 1 = {0}. (ii) diam(γ R 2 )) = 1 if and only if R 1 = R 2 = 2 and R 1 = R2. (iii) diam(γr 2 )) = 3 if and only if R 1 Z(R 1 ), R 2 Z(R 2 ), and (without loss of generality) Z(R 1 ) = 1. (iv) diam(γ R 2 )) = 2 otherwise. Proof. (i) ( ) Assume for contradiction that diam(γ R 2 )) = 0 and R 1, R 2 {0}. Then x R1, y R2. Now (x, 0) and (0, y) are two adjacent vertices in Γ R 2 ), so diam(γ R 2 )) > 0. ( ) Assume without loss of generality that R 1 = {0}. Then R 1 R 2 = R2, so diam(γ R 2 )) = 0. (ii) ( ) Assume diam(γ R 2 )) = 1. Assume without loss of generality R 1 2. If R 1 = 1, then R 1 = {0}, so diam(γ R 2 )) = 0 by part (i). Thus assume R 1 3. Thus we have distinct x, y R1. Note that xy 0 because diam(γ(r 1 )) = 0. Also, note that R 2 > 1 by part (i) also. Let z R2. Now, we have path (x, 0) (0, z) (y, 0) which is minimal since xy 0. Thus diam(γ R 2 )) 2. Assume R 1 R 2. By argument above, R 1 = R 2 = 2. Recall that there are only two rings of order two: Z 2 and 2Z 4. Since diam(γ(z 2 2Z 4 )) = 2, we have a contradiction. Thus R 1 = R2. ( ) Assume R 1 = R 2 = 2 and R 1 = R2. Since there are only two rings of order two, and diam(γ(2z 4 2Z 4 )) = 1 = diam(γ(z 2 Z 2 )) = 1, then diam(γ R 2 )) = 1. (iii) ( ) Assume diam(γ R 2 )) = 3. Assume without loss of generality that Z(R 1 ) = R 1. Also R 1 > 1 by part (i). Thus, since diam(γ(r 1 )) = 0, R 1 = 2Z4. Then every vertex in Γ R 2 ) is adjacent to (x, 0), where x R1. Then diam(γ R 2 )) 2. This is a contradiction, of diam(γ R 2 )) = 3, so R 1 Z(R 1 ) and R 2 Z(R 2 ). Assume Γ(R 1 ) = Γ(R 2 ) = K 0, the graph with no vertices. Thus Z(R 1 ) = Z(R 2 ) =. Thus every vertex in Γ R 2 ) must be of the form (x, 0) or (0, y), where x reg(r 1 ), y reg(r 2 ). Thus Γ R 2 ) is complete bipartite, so diam(γ R 2 )) 2. ( ) By Lemma 2.1. (iv) Follows from (i), (ii), (iii). 3. Ideal-Divisor Graphs of Direct Products (Diameter Zero By N) Lemma 3.1. Let Γ I (R) be the ideal-divisor graph of R with respect to an ideal I such that Γ I (R) has exactly one vertex. Then I = {0}. Proof. Assume Γ I (R) as above. Then x Z I (R) such that x 2 I. Let i I. Now (x + i)x = x 2 + ix I. Thus, since there is only one vertex in Γ I (R), we know that x + i = x, so i = 0. Thus I = {0}. Lemma 3.2. (Analogue to ASW Lemma 2.3) Let R be a commutative ring with ideal I such that diam(γ I (R)) = 2. Suppose Z I (R) is a (not necessarily proper) subring of R. Then for all x, y Z I (R), there exists a z / I such that xz, yz I.

4 4 T. BRAND, M. JAMESON, M. MCGOWAN, AND J.D. MCKEEL Proof. Let x, y Z I (R). If x I, then simply choose z such that yz I. We can similarly find z if y I or if x = y. Thus we assume x, y are distinct and not in I. Since diam(γ I (R)) = 2, if xy I, we know that there exists some z Z I (R) such that xz, yz I. So assume xy I. If x 2 I, then clearly we can choose z = x. A similar situation exists if y 2 I. Therefore we may also assume that x 2, y 2 I. Let X = {x Z I (R) xx I} and Y = {y Z I (R) yy I}. Note that since x Y and y X, these sets are not empty. Furthermore, if X Y, then choosing z X Y will suffice. Thus assume that X Y =. Consider the element x + y. Obviously x + y x and x + y y. If x + y I, then x(x + y) I which implies that x 2 I. Thus x + y / I. Thus x + y Z I (R) since Z I (R) is a subring. Furthermore, since x 2, y 2 I, we know that x + y X and x + y Y. But the diameter of Γ I (R) is 2, so we know that we can find some w X such that xw I, w(x + y) I. But then w(x+y) I which implies that wy I, so w X Y, which is a contradiction. Theorem 3.3. Let R 1, R 2 be commutative rings and I 1, I 2 be their respective ideals such that diam(γ I1 (R 1 )) = diam(γ I2 (R 2 )) = 0. Then R 2 )) = 0 if and only if (without loss of generality) I 1 = R 1 and either R 1 = {0} or I 2 is a prime ideal R 2 )) = 1 if and only if (without loss of generality) I 1 = {0} and either Z I1 (R 1 ) with I 2 = R 2 {0} or R 1 = R2 with R 1 = 2 R 2 )) = 3 if and only if R 1 Z I1 (R 1 ), R 2 Z I2 (R 2 ) and (without loss of generality) Z I1 (R 1 ) = 1. (iv) diam(γ I1 I 2 R 2 )) = 2 otherwise. Proof. (i) ( ) Assume diam(γ I1 I 2 R 2 )) = 0. Now, if reg I1 (R 1 ), reg I2 (R 2 ), then x reg I1 (R 1 ), y reg I2 (R 2 ). Then (x, 0)(0, y) = 0 I 1 I 2, which is a contradiction of diam(γ I1 I 2 R 2 )) = 0. Thus, without loss of generality, reg I1 (R 1 ) =, so R 1 = I 1. If Γ I1 I 2 R 2 ) = K 0, then I 1 I 2 is a prime ideal. Now, consider ab I 2. Then (0, a)(0, b) I 1 I 2. Since I 1 I 2 is prime, (0, a) I 1 I 2 or (0, b) I 1 I 2.Thus a I 2 or b I 2. Thus I 2 is a prime ideal. If Γ I1 I 2 R 2 ) has one vertex, then I 1 I 2 = {0}, by Lemma 3.1. Then I 1 = I 2 = {0}, and Γ I1 I 2 R 2 ) = Γ R 2 ). By Theorem 2.4, then without loss of generality, R 1 = {0}. ( ) Assume I 1 = R 1 and either R 1 = {0} or I 2 is a prime ideal. Let I 1 = R 1 = {0}. If diam(γ I1 I 2 R 2 )) > 0, (0, y 1 ), (0, y 2 ) / I 1 I 2 such that (0, y 1 )(0, y 2 ) I 1 I 2. Then y 1, y 2 / I 2 and y 1 y 2 I 2, so diam(γ I2 (R 2 )) > 0. This is a contradiction, so diam(γ I1 I 2 R 2 )) = 0. Let I 1 = R 1, and I 2 be a prime ideal of R 2. If diam(γ I1 I 2 R 2 )) > 0, then there exist distinct (x 1, y 1 ), (x 2, y 2 ) / I 1 I 2 such that (x 1, y 1 )(x 2, y 2 ) I 1 I 2. Since R 1 = I 1, we have y 1, y 2 / I 2 and y 1 y 2 I 2. Thus I 2 is not a prime ideal. This is a contradiction, so diam(γ I1 I 2 R 2 )) = 0. (ii) ( ) Assume diam(γ I1 I 2 R 2 )) = 1. Assume Γ I1 (R 1 ) = Γ I2 (R 2 ) = K 0. Since diam(γ I1 I 2 R 2 )) = 1, vertices (x, 0), (0, y) Z I1 I 2 R 2 ). Thus x reg I1 (R 1 ), y reg I2 (R 2 ). For all i I 1, (x + i, 0)(0, y) I 1 I 2, and (x + i, 0)(x, 0) /

5 ZERO-DIVISOR AND IDEAL-DIVISOR GRAPHS OF COMMUTATIVE RINGS 5 I 1 I 2. This implies that (x+i, 0) = (x, 0), so i = 0 and thus I 1 = {0}. Similarly, I 2 = {0}. By Theorem 2.4, R 1 = R 2 = 2 and R 1 = R2. Assume Γ I1 (R 1 ) = K 1 and Γ I2 (R 2 ) = K 0. By Lemma 3.1, I 1 = {0}. Let x Z I1 (R 1 ) (x must exist since Γ I1 (R 1 ) = K 1 ). If R 2 I 2, y reg I2 (R 2 ). Then (x, 0)(x, y) I 1 I 2 and (x, 0)(0, y) I 1 I 2, but (x, y)(0, y) / I 1 I 2. Thus diam(γ I1 I 2 R 2 )) 2. So I 2 = R 2. Also note that I 2 {0} (since I 2 = {0} would imply diam(γ I1 I 2 R 2 )) = 0 by part (i) above). Assume Γ I1 (R 1 ) = Γ I2 (R 2 ) = K 1. By Lemma 3.1, I 1 = I 2 = {0}, so R 1 = R 2 = 2 and R 1 = R2 by Theorem 2.4. ( ) Assume that I 1 = {0} and R 1 = R2 with R 1 = 2. Thus diam(γ I1 I 2 R 2 )) = 1 by Theorem 2.4. Assume that I 1 = {0} and Z I1 (R 1 ) with I 2 = R 2 {0}. Let x Z I1 (R 1 ). Thus, for all vertices (x, y 1 ), (x, y 2 ) Z I2 I 2 R 2 ) (we know y 1, y 2 I 2, so the first entry must indeed be x), it s clear that (x, y 1 )(x, y 2 ) I 1 I 2. Also, there are at least two vertices since I 2 {0}, so diam(γ I1 I 2 R 2 )) = 1. (iii) ( )Assume diam(γ I1 I 2 R 2 )) = 3. Assume Γ I1 (R 1 ) = Γ I2 (R 2 ) = K 0. Then Z I1 (R 1 ) = Z I2 (R 2 ) =. Now, consider (x 1, y 1 ), (x 2, y 2 ) Z I1 I 2 R 2 ) such that d((x 1, y 1 ), (x 2, y 2 )) = 3. Since (x 1, y 1 ) / I 1 I 2, assume without loss of generality that x 1 reg I1 (R 1 ). Then y 1 I 2. Similarly, x 2 reg I1 (R 1 ) or y 2 reg I2 (R 2 ). - If x 2 reg I1 (R 1 ), then y 2 I 2 because (x 2, y 2 ) Z I1 I 2 R 2 ). Let (a, b) be a vertex adjacent to (x 1, y 1 ). We know that x 1 a I 1, so a I 1. Thus (a, b)(x 2, y 2 ) I 1 I 2 since a I 1 and y 2 I 2. Thus d((x 1, y 1 ), (x 2, y 2 )) = 2. - If y 2 reg I2 (R 2 ), then x 2 I 1. Thus (x 1, y 1 )(x 2, y 2 ) I 1 I 2, so d((x 1, y 1 ), (x 2, y 2 )) = 1. In either case, we have a contradiction. Assume Γ I1 (R 1 ) = K 1 and Γ I2 (R 2 ) = K 0. I 1 = {0} by Lemma 3.1. If R 2 = Z I2 (R 2 ), then R 2 = I 2. Let x Z I1 (R 1 ). Since R 2 = Z I2 (R 2 ), the set of vertices is Z I1 I 2 R 2 ) = {(x, i) i I 2 }, all of which are adjacent to each other by definition. So Γ I1 I 2 R 2 ) = K R2. This is a contradiction of diam(γ I1 I 2 R 2 )) = 3, so R 2 Z I2 (R 2 ). If R 1 = Z I1 (R 1 ), then (x, 0) is adjacent to all other vertices in Γ I1 I 2 R 2 ), where x Z I1 (R 1 ). Thus diam(γ I1 I 2 R 2 )) 2. Thus R 1 Z I1 (R 1 ). Assume Γ I1 (R 1 ) = Γ I2 (R 2 ) = K 1. Then I 1 = I 2 = {0} by Lemma 3.1, so R 1 Z I1 (R 1 ) and R 2 Z I2 (R 2 ) by Theorem 2.4. ( ) Assume R 1 Z I1 (R 1 ), R 2 Z I2 (R 2 ), Γ I1 (R 1 ) = K 1 and diam(γ I1 I 2 R 2 )) 3. Let x 1 reg I1 (R 1 ), x 2 Z I1 (R 1 ), and y reg I2 (R 2 ). Consider the distinct vertices (x 1, 0) and (x 2, y). Clearly, (x 1, 0)(x 2, y) / I 1 I 2 by construction. Thus (a, b) Z I1 I 2 R 2 ) such that (a, b)(x 1, 0) I 1 I 2 and (a, b)(x 2, y) I 1 I 2. Since x 1 a I 1 and yb I 2, then (a, b) I 1 I 2, which is a contradiction. Thus diam(γ I1 I 2 R 2 )) = 3. (iv) Follows from (i), (ii), (iii).

6 6 T. BRAND, M. JAMESON, M. MCGOWAN, AND J.D. MCKEEL Theorem 3.4. Let diam(γ I1 (R 1 )) = 0 and diam(γ I2 (R 2 )) = 1. (i) diam(γ R 2 )) > 0. (ii) diam(γ R 2 )) = 1 if and only if both R 2 1 I 1 and R 2 2 I 2 or R 1 = I 1. (iii) diam(γr 2 )) = 2 if and only if R 1 I 1 and (without loss of generality) R 2 1 I 1 and R 2 2 I 2. (iv) diam(γ R 2 )) = 3 if and only if R 2 1 I 1 and R 2 2 I 2. Proof. Let diam(γ I1 (R 1 )) = 0 and diam(γ I2 (R 2 )) = 1. (i) Let x, y Z I 2 (R 2 ) such that x y and xy I 2. Then (0, x)(0, y) = (0, xy) I 1 I 2, so diam(γ I1 I 2 R 2 )) > 0. (ii) ( ) Let diam(γ I1 I 2 R 2 )) = 1. Suppose R1 2 I 1, then a, b R 1 such that ab / I 1. Clearly a, b / I 1. Let c ZI 2 (R 2 ). Then (a, c), (b, 0) ZI 1 I 2 R 2 ), but (a, c)(b, 0) = (ab, 0) / I 1 I 2, so d((a, c), (b, 0)) > 1, a contradiction. Suppose R2 2 I 2, then a, b R 2 such that ab / I 2. Clearly a, b / I 2. If ZI 1 (R 1 ), let c ZI 1 (R 1 ). Then (c, a), (0, b) ZI 1 I 2 R 2 ), but (c, a)(0, b) = (0, ab) / I 1 I 2, so d((c, a), (0, b)) > 1, a contradiction. Thus Z I1 (R 1 ) =, so consider R 1 I 1. If R 1 I 1, then R1 2 I 1, a contradiction from earlier. Thus R 1 I 1 =, so R 1 = I 1. ( ) Suppose R 1 = I 1. By Lemma 2.2, diam(γ I1 I 2 R 2 )) = diam(γ I2 (R 2 )) = 1. Suppose R1 2 I 1 and R2 2 I 2. Then for all (a, b), (c, d) Z I1 I 2 R 2 ), since ac I 1 and bd I 2, (ac, bd) I 1 I 2, so diam(γ I1 I 2 R 2 )) = 1. (iii) ( ) Let diam(γ I1 I 2 R 2 )) = 2. By contrapositive of Lemma 2.2, R 1 I 1. Suppose Z I1 (R 1 ). Assume reg I1 (R 1 ), reg I2 (R 2 ). Let a reg I1 (R 1 ), b reg I2 (R 2 ), c Z I1 (R 1 ), d Z I2 (R 2 ). Since a is regular, we need an element from I 1 in the first component to kill a, and since b is regular, we need an element from I 2 in the second component to kill b. Then ann I1 I 2 ((a, d)) = {(i, l) dl I 2 and i I 1 } and ann I1 I 2 ((c, b)) = {m, i) cm I 1 and i I 2 }. Since ann I1 I 2 ((a, d)) ann I1 I 2 (c, b) = I 1 I 2, only trivial elements (elements in I 1 I 2 ) can kill both (a, d) and (c, b), so d((a, d), (c, b)) > 2, a contradiction. Thus either reg I1 (R 1 ) = or reg I2 (R 2 ) =. Suppose reg I2 (R 2 ) =. Then R 2 = Z I2 (R 2 ) and by Lemma 2.3, R2 2 I 2. By ii), R1 2 I 1. Now suppose reg I1 (R 1 ) =. Since diam(γ I1 (R 1 )) = 0 and Z I1 (R 1 ), Γ I1 (R 1 ) = K 1. Then Z I1 (R 1 ) = {a}, a 2 I 1, and R 1 = {0, a}. Then R1 2 I 1, and by ii), R Suppose Z I1 (R 1 ) =. Then since R 1 I 1, reg I1 (R 1 ), and R1 2 I 1. Assume reg I2 (R 2 ). Let a reg I1 (R 1 ), x Z I2 (R 2 ), r reg I2 (R 2 ). Then (a, x), (0, r) Z I1 I 2 R 2 ). Since a is regular, we need an element from I 1 in the first component to kill it a, and since r is regular, we need an element from I 2 in the second component to

7 ZERO-DIVISOR AND IDEAL-DIVISOR GRAPHS OF COMMUTATIVE RINGS 7 kill r. Then ann I1 I 2 ((a, x)) = {(i, y) xy I 2 and i I 1 } and ann I1 I 2 ((0, r)) = {(b, i) b R 1 and i I 2 }. Since ann I1 I 2 ((a, x)) ann I1 I 2 ((0, r)) = I 1 I 2, only trivial elements (elements in I 1 I 2 ) can kill both (a, x) and (0, r), so d((a, x), (0, r)) > 2, a contradiction. Thus reg I2 (R 2 ) =, so R 2 = Z I2 (R 2 ) and by Lemma 2.3, R2 2 I 2. ( ) Assume R1 2 I 1, R2 2 I 2, and R 1 I 1. Let c R 1 I 1. Then (c, 0) is adjacent to every vertex in Γ I1 I 2 R 2 ), so diam(γ I1 I 2 R 2 )) 2. Then from i), diam(γ I1 I 2 R 2 )) 1, so diam(γ I1 I 2 R 2 )) = 2. Assume R1 2 I 1 and R2 2 I 2. Let c R 2 I 2. Then (0, c) is adjacent to every vertex in Γ I1 I 2 R 2 ), so diam(γ I1 I 2 R 2 )) 2. Then from i), diam(γ I1 I 2 R 2 )) 1, so diam(γ I1 I 2 R 2 )) = 2. (iv) Follows from (i), (ii), and (iii). Corollary 3.5. Let diam(γ(r 1 )) = 0 and diam(γ(r 2 )) = 1. (i) diam(γ R 2 )) > 0. (ii) diam(γ R 2 )) = 1 if and only if R 2 1 = 0 = R 2 2 or R 1 = {0}. (iii) diam(γ R 2 )) = 2 if and only if R 1 {0} and (without loss of generality) R 2 1 = 0 and R (iv) diam(γ R 2 )) = 3 if and only if R 2 1, R Theorem 3.6. Let R 1 and R 2 be rings with ideals I 1 and I 2, respectively, such that diam(γ I1 (R 1 )) = 0 and diam(γ I2 (R 2 )) = 2. Then: R 2 )) > 1. R 2 )) = 2 if and only if R 1 = Z I1 (R 1 ) or R 2 = Z I2 (R 2 ). R 2 )) = 3 if and only if R 1 Z I1 (R 1 ) and R 2 Z I2 (R 2 ). Proof. (i) Since diam(γ I2 (R 2 )) 2, there exist distinct y 1, y 2 Z I2 (R 2 ) with y 1 y 2 / I 2. Then (0, y 1 )(0, y 2 ) = (0, y 1 y 2 ) / I 1 I 2. Therefore diam(γ I1 I 2 R 2 )) > 1. (ii) ( ) Let R 1 = Z I1 (R 1 ) (the proof of Theorem 3.13 satisfies the case where R 2 = Z I2 (R 2 )). Since diam(γ I1 (R 1 )) = 0, either R 1 = I 1 or R 1 = I 1 {a}, where a 2 I 1. Thus, R 2 1 I 1. Therefore, (r, 0)(x, y) I 1 I 2 for all r R 1, (x, y) Z I1 I 2 R 2 ). Hence, diam(γ I1 I 2 R 2 )) 2. By (i), diam(γ I1 I 2 R 2 )) = 2. ( ) By the contrapositive of Lemma 2.1. (iii) By (i) and (ii). Corollary 3.7. Let R 1 and R 2 be rings with diam(γ(r 1 )) = 0 and diam(γ(r 2 )) = 2. Then: (i) diam(γ R 2 )) > 1. (ii) diam(γ R 2 )) = 2 if and only if R 1 = Z(R 1 ) or R 2 = Z(R 2 ). (iii) diam(γ R 2 )) = 3 if and only if R 1 Z(R 1 ) and R 2 Z(R 2 ). Theorem 3.8. Let R 1 and R 2 be commutative rings. If diam(γ I1 (R 1 )) = 0 and diam(γ I2 (R 2 )) = 3, then: R 2 )) > 1. R 2 )) = 2 if and only if R 1 = Z I (R 1 ) I 1. R 2 )) = 3 if and only if R 1 = I 1 or R 1 Z I (R 1 ).

8 8 T. BRAND, M. JAMESON, M. MCGOWAN, AND J.D. MCKEEL Proof. (i) Same as Theorem 3.6 (i). (ii) ( ) Let diam(γ I1 I 2 R 2 )) = 2. Assume that R 1 Z I (R 1 ). Note that since diam(γ I1 I 2 (R 2 )(R 2 ) = 3), we can find a minimal path of length 3, y 1 y 2 y 3 y 4 We also know there exists r 1 reg I (R 1 ). But then consider the path (0, y 1 ) (r 1, y 2 ) (0, y 3 ) (r 1, y 4 ) Clearly none of the elements in the path are adjacent. If there exists some (i, b) where i I 1 such that (0, y 1 ) (i, b) (r 1, y 4 ) is a path, this contradicts the fact that d(y 1, y 4 ) = 3. Thus that path must be minimal as well. But this implies d((0, y 1 ), (r 1, y 4 )) = 3, a contradiction. If Z I (R 1 ) = I 1, then if j I 1, the path (j, y 1 ) (j, y 2 ) (j, y 3 ) (j, y 4 ) is clearly minimal, so we get a contradiction here as well. ( ) Let R 1 = Z I (R 1 ) I 1. Then there exist a, b R 1 (not necessarily distinct) where ab I 1. We know that we have the minimal path in Γ I2 (R 2 ) y 1 y 2 y 3 y 4 therefore the path (a, y 1 ) (b, y 2 ) (0, y 3 ) is minimal, so diamγ I1 I 2 R 2 )) 2. Note that since diam(γ I1 (R 1 )) = 0, if pq I 1, then p = q. Since R 1 I 1, we know that p exists. Thus all elements in Γ I1 I 2 R 2 ) are of the form (j, r 2 ) or (p, r 2 ) for some r 2 R 2. Thus the element (p, 0) is a bridge between any two other elements, so we know the diameter of Γ R 2 ) is exactly 2. (iii) Follows from (i) and (ii). Corollary 3.9. Let R 1 and R 2 be commutative rings. Then if diam(γ(r 1 )) = 0 and diam(γ(r 2 )) = 3 then: (i) diam(γ R 2 )) > 1. (ii) diam(γ R 2 )) = 2 if and only if R 1 = Z(R 1 ) {0}. (iii) diam(γ R 2 )) = 3 if and only if R 1 = {0} or R 1 Z(R 1 ). Theorem Let R 1 and R 2 be commutative rings with ideals I 1 and I 2, respectively, such that diam(γ I1 (R 1 )) = diam(γ I2 (R 2 )) = 1. Then R 2 )) = 1 if and only if R 2 1 I 1 and R 2 2 I 2. R 2 )) = 2 if and only if without loss of generality R 2 1 I 1 and R 2 2 I 2. R 2 )) = 3 if and only if R 2 1 I 1 and R 2 2 I 2. Proof. (i) ( ) Without loss of generality, let diam(γ I1 I 2 R 2 )) = 1 and R1 2 I 1. Then there exist x 1, x 2 R 1 such that x 1 x 2 / I 1. Therefore (x 1, 0)(x 2, 0) / I 1 I 2, so diam(γ I1 I 2 R 2 )) > 1. ( ) Let R1 2 I 1 and R2 2 I 2. Then for all (x 1, y 1 ), (x 2, y 2 ) R 1 R 2, we know that x 1 x 2 I 1 and y 1 y 2 I 2, so (x 1, y 1 )(x 2, y 2 ) = (x 1 x 2, y 1 y 2 ) I 1 I 2.

9 ZERO-DIVISOR AND IDEAL-DIVISOR GRAPHS OF COMMUTATIVE RINGS 9 (ii) ( ) Assume diam(γ I1 I 2 R 2 )) = 2. If R 2 1 I 1 and R 2 2 I 2, then diam(γ I1 I 2 R 2 )) = 1 by part (i), which forms a contradiction. Thus assume R 2 1 I 1 and R 2 2 I 2. By Lemma 2.3, there must exist x 1 reg I1 (R 1 ), and y 1 reg I2 (R 2 ). Let x 2 Z I1 (R 1 ), y 2 Z I2 (R 2 ) and consider the two elements (x 1, y 2 ), (x 2, y 1 ) Z I1 I 2 R 2 ). Clearly (x 1, y 2 )(x 2, y 1 ) = (x 1 x 2, y 1 y 2 ) / I 1 I 2 by choice of x 1 and y 1. Since diam(γ I1 I 2 R 2 )) = 2, there must exist an element (a, b) Z I1 I 2 R 2 ) such that (x 1, y 2 )(a, b) I 1 I 2 and (x 2, y 1 )(a, b) I 1 I 2. Thus x 1 a I 1, so a I 1. Similarly, y 1 b I 2, so b I 2. Thus (a, b) I 1 I 2, so we have a contradiction. ( ) Assume without loss of generality that R 2 1 I 1, R 2 2 I 2. Then diam(γ I1 I 2 R 2 )) > 1 by part (i). Also, since R 2 1 I 1, for all x 1, x 2 R 1, x 1 x 2 I 1. Thus (x 1, 0) is adjacent to every vertex in Γ I1 I 2 R 2 ), so a path of length 2 can be found between any two vertices in Γ I1 I 2 R 2 ) by way of the vertex (x 1, 0). Thus diam(γ I1 I 2 R 2 )) = 2. (iii) Follows directly from parts (i) and (ii). Theorem Let R 1 and R 2 be commutative rings with ideals I 1 and I 2, respectively, such that diam(γ I1 (R 1 )) = 1 and diam(γ I2 (R 2 )) = 2. Then: R 2 )) > 1. R 2 )) = 2 if and only if R 1 = Z I1 (R 1 ) or R 2 = Z I2 (R 2 ). R 2 )) = 3 if and only if R 1 Z I1 (R 1 ) and R 2 Z I2 (R 2 ). Proof. (i) Same as Theorem 3.6 (i). (ii) ( ) Let R 1 = Z I1 (R 1 ) (the case where R 2 = Z I2 (R 2 ) is addressed by the proof of Theorem 3.13). Thus, we have R1 2 I 1 by Lemma 2.3. Let a R1. Since (a, 0)(x, y) I 1 I 2 for all (x, y) Z ( I 1 I 2 R 1 R 2 ), we have diam(γ I1 I 2 R 2 )) 2. It follows from (i) that diam(γ I1 I 2 R 2 )) = 2. ( ) Assume that diam(γ I1 I 2 R 2 )) = 2, R 1 Z I1 (R 1 ), and R 2 Z I2 (R 2 ). Let x Z I1 (R 1 ), y Z I2 (R 2 ), m reg I1 (R 1 ), n reg I2 (R 2 ). Then (x, n)(m, y) / I 1 I 2. Since diam(γ I1 I 2 R 2 )) = 2, there exists (a, b) Z I1 I 2 R 2 ) such that (x, n)(a, b), (m, y)(a, b) I 1 I 2. Then ma I 1 and nb I 2, so (a, b) I 1 I 2, a contradiction. (iii) Follows from (i) and (ii). Theorem Let R 1 and R 2 be commutative rings with ideals I 1 and I 2, respectively, such that diam(γ I1 (R 1 )) = 1 and diam(γ I2 (R 2 )) = 3. Then: R 2 )) > 1. R 2 )) = 2 if and only if R 1 = Z I1 (R 1 ). R 2 )) = 3 if and only if R 1 Z I1 (R 1 ). Proof. (i) Same as Theorem 3.6 (i). (ii) ( ) Same as Theorem 3.11 (ii). ( ) Assume that diam(γ I1 I 2 R 2 )) = 2 and R 1 Z I1 (R 1 ). Let m reg I1 (R 1 ). Since diam(γ I2 (R 2 )) = 3, there exist distinct y 1, y 2 Z I2 (R 2 ) with y 1 y 2 / I 2, and there is no y 3 Z I2 (R 2 ) such that y 1 y 3, y 2 y 3 I 2. Now (m, y 1 ), (m, y 2 ) Z I1 I 2 R 2 ), and (m, y 1 )(m, y 2 ) / I 1 I 2.

10 10 T. BRAND, M. JAMESON, M. MCGOWAN, AND J.D. MCKEEL Since diam(γ I1 I 2 R 2 )) = 2, there exists (a, b) Z I1 I 2 R 2 ) such that (m, y 1 )(a, b), (m, y 2 )(a, b) I 1 I 2. Then ma I 1, so we have a I 1. Also, y 1 b, y 2 b I 2. Hence, b I 2. Thus, (a, b) I 1 I 2, a contradiction. (iii) By (i) and (ii). Theorem Let R 1 and R 2 be commutative rings with ideals I 1 and I 2, respectively, such that diam(γ I (R 1 )) = diam(γ I (R 2 )) = 2. Then: R 2 )) > 1 R 2 )) = 2 if and only if R 1 = Z I1 (R 1 ) or R 2 = Z I2 (R 2 ) R 2 )) = 3 if and only if R 1 Z I1 (R 1 ) and R 2 Z I2 (R 2 ) Proof. (i) Same as Theorem 3.6 (i). (ii) ( ) Without loss of generality, let R 1 = Z I1 (R 1 ). Since R 1 = Z I1 (R 1 ), by Lemma 3.2, for all x 1, x 2 Z I1 (R 1 ), there exists x 3 R 1 I 1 such that x 3 x 1, x 3 x 2 I 1. So for all (x 1, y 1 ), (x 2, y 2 ) Z I1 I 2 R 2 ), there exists (x 3, 0) Z I1 I 2 R 2 ) such that (x 1, y 1 )(x 3, 0) I 1 I 2. If without loss of generality (x 2, y 2 ) = (x 3, 0) then (x 1, y 1 )(x 2, y 2 ) I 1 I 2. Thus diam(γ I1 I 2 R 2 )) 2 and by (i), diam(γ I1 I 2 R 2 )) = 2. ( ) Same as Theorem 3.11 (ii). (iii) By (i) and (ii). Theorem Let R 1 and R 2 be commutative rings with ideals I 1 and I 2, respectively, such that diam(γ I1 (R 1 )) = 2 and diam(γ I2 (R 2 )) = 3. Then: R 2 )) > 1. R 2 )) = 2 if and only if R 1 = Z I1 (R 1 ). R 2 )) = 3 if and only if R 1 Z I1 (R 1 ). Proof. (i) Same as Theorem 3.6 (i). (ii) ( ) Same as Theorem 3.13 (ii). ( ) Let diam(γ I1 I 2 R 2 )) = 2. Assume that R 1 Z I1 (R 1 ). Let k reg I1 (R 1 ). Since diam(γ I2 (R 2 )) = 3, we know that we can find a minimal path y 1 y 2 y 3 y 4 where d(y 1, y 4 ) = 3. Similarly, we know that that since Γ I1 (R 1 ) has diameter 2, we can find x 1 and x 2 such that x 1 is adjacent to x 2. Then consider the following path in Γ I1 I 2 R 2 ): (k, y 1 ) (0, y 2 ) (x 1, y 3 ) (x 2, y 4 ) Note that (k, y 1 ) can t be adjacent to (x 2, y 4 ) since k reg I1 (R 1 ). Then there must exist some (a, b) such that (k, y 1 )(a, b) I 1 I 2 and (a, b)(x 2, y 4 ) I 1 I 2. Note that this forces a I 1. Then b can t be in I 2. But then d(y 1, y 4 ) = 2, which is a contradiction. Thus R 1 = Z I1 (R 1 ). (iii) By (i) and (ii).

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