ZERODIVISOR AND IDEALDIVISOR GRAPHS OF COMMUTATIVE RINGS


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1 ZERODIVISOR AND IDEALDIVISOR GRAPHS OF COMMUTATIVE RINGS T. BRAND, M. JAMESON, M. MCGOWAN, AND J.D. MCKEEL Abstract. For a commutative ring R, we can form the zerodivisor graph Γ(R) or the idealdivisor graph Γ I (R) with respect to an ideal I of R. We consider the diameters of direct products of zerodivisor and idealdivisor graphs. 1. Introduction and Definitions The concept of a zerodivisor graph was first introduced in 1988 in [3] by Beck, who was interested in the coloring of the zerodivisor graph. However, we use the slightly altered definition of a zerodivisor 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 zerodivisor 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 zerodivisors 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 zerodivisor x to be ann(x) = {y Z(R) xy = 0}. We can define the zerodivisor 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 idealdivisors 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 idealdivisor 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 idealdivisor graph was first discussed by Redmond in [4]. He was able to generalize many of the concepts and theorems of the zerodivisor 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 idealdivisor graphs. Date: February 27,
2 2 T. BRAND, M. JAMESON, M. MCGOWAN, AND J.D. MCKEEL 2. IdealDivisor 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 ZERODIVISOR AND IDEALDIVISOR 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. IdealDivisor Graphs of Direct Products (Diameter Zero By N) Lemma 3.1. Let Γ I (R) be the idealdivisor 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 ZERODIVISOR AND IDEALDIVISOR 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 ZERODIVISOR AND IDEALDIVISOR 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 ZERODIVISOR AND IDEALDIVISOR 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).
11 ZERODIVISOR AND IDEALDIVISOR GRAPHS OF COMMUTATIVE RINGS 11 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 )) = 3. Then diam(γ I1 I 2 R 2 )) = 3. Proof. Since diam(γ I1 (R 1 )) = 3, there exist vertices x 1, x 2, x 3, x 4 Γ I1 (R 1 ) such that there is a minimal path x 1 x 2 x 3 x 4 and d(x 1, x 4 ) = 3. Similarly, we can find vertices y 1, y 2, y 3, y 4 Γ I2 (R 2 ) such that there is a minimal path y 1 y 2 y 3 y 4 and d(y 1, y 4 ) = 3. Now consider the following path in Γ I1 I 2 R 2 ); (x 1, y 1 ) (x 2, y 2 ) (x 3, y 3 ) (x 4, y 4 ) Assume that d((x 1, y 1 ), (x 4, y 4 )) < 3. Then there are two cases: (x 1, y 1 ) is adjacent to (x 4, y 4 ). But this would imply that x 1 was adjacent to x 4 in Γ I1 (R 1 ), which would contradict the fact that d(x 1, x 4 ) = 3. (x 1, y 1 ) is adjacent to some (a, b) which is adjacent to (x 4, y 4 ). Assume a I 1. But then y 1 b I 2 and by 4 I 2 which would imply that in Γ I2 (R 2 ), we would have the path y 1 b y 4 which contradicts d(y 1, y 4 ) = 3. We get a similar contradiction if b I 2. Thus d((x 1, y 1 ), (x 4, y 4 )) 3. Since the diameter of a idealdivisor graph is always bounded by 3, we get that diam(γ I1 I 2 R 2 )) = 3. References [1] Anderson, D. and Livingston, P., The ZeroDivisor Graph of a Commutative Ring, J. Algebra 217 (1999), [2] Axtell, M., Stickles, J., and Warfel, J., ZeroDivisor Graphs of Direct Products of Commutative Rings, Houston Journal of Mathematics(to appear). [3] Beck, I., Coloring of Commutative Rings, J. Algebra 116 (1988), [4] Redmond, S., An IdealBased ZeroDivisor Graph of a Commutative Ring, Comm. Alg. Vol. 31, No. 9 (2003), Harvey Mudd College address: Harvey Mudd College address: John Carroll University address: University of Evansville address:
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