ISSN 2148-838Xhttps://doi.org/10.13069/jacodesmath.938105 J. Algebra Comb. Discrete Appl. 8(2) • 119–138 Received: 12 October 2020 Accepted: 8 February 2021 Journal of Algebra Combinatorics Discrete Structures and Applications The exact annihilating-ideal graph of a commutative ring Research Article Subramanian Visweswaran, Premkumar T. Lalchandani Abstract: The rings considered in this article are commutative with identity. For an ideal I of a ring R, we denote the annihilator of I in R by Ann(I). An ideal I of a ring R is said to be an exact annihilating ideal if there exists a non-zero ideal J of R such that Ann(I) = J and Ann(J) = I. For a ring R, we denote the set of all exact annihilating ideals of R by EA(R) and EA(R)\{(0)} by EA(R)∗. Let R be a ring such that EA(R)∗ 6= ∅. With R, in [Exact Annihilating-ideal graph of commutative rings, J. Algebra and Related Topics 5(1) (2017) 27-33] P.T. Lalchandani introduced and investigated an undirected graph called the exact annihilating-ideal graph of R, denoted by EAG(R) whose vertex set is EA(R)∗ and distinct vertices I and J are adjacent if and only if Ann(I) = J and Ann(J) = I. In this article, we continue the study of the exact annihilating-ideal graph of a ring. In Section 2 , we prove some basic properties of exact annihilating ideals of a commutative ring and we provide several examples. In Section 3, we determine the structure of EAG(R), where either R is a special principal ideal ring or R is a reduced ring which admits only a finite number of minimal prime ideals. 2010 MSC: 13A15, 05C25 Keywords: Exact annihilating ideal, Exact annihilating-ideal graph, Connectedness, Reduced ring, Special principal ideal ring 1. Introduction The rings considered in this article are commutative with identity which are not integral domains. Let R be a ring. For an element a ∈ R, the annihilator of a in R, denoted by AnnR(a) or simply by Ann(a) is defined as Ann(a) = {r ∈ R | ra = 0}. Recall from [12] that an element x ∈ R is said to be an exact zero-divisor if there exists y ∈ R\{0} such that Ann(x) = Ry and Ann(y) = Rx. It is clear that any exact zero-divisor of R is a zero-divisor of R. We denote the set of all zero-divisors of a ring R by Subramanian Visweswaran (Corresponding Author); Department of Mathematics, Saurashtra University, Rajkot, India (Retired) (email: s_visweswaran2006@yahoo.co.in). Premkumar T. Lalchandani; Department of Mathematics, Dr. Subhash Science College, Junagadh, India (email: finiteuniverse@live.com). 119 https://orcid.org/0000-0002-4905-809X https://orcid.org/0000-0001-8938-7552 S. Visweswaran, P. T. Lalchandani / J. Algebra Comb. Discrete Appl. 8(2) (2021) 119–138 Z(R) and Z(R)\{0} by Z(R)∗. As in [15], we denote the set of all exact zero-divisors of R by EZ(R) and EZ(R)\{0} by EZ(R)∗. Let R be a ring such that EZ(R)∗ 6= ∅. Recall from [15] that the exact zero- divisor graph of R, denoted by EΓ(R) is an undirected graph whose vertex set is EZ(R)∗ and distinct vertices x and y are adjacent in EΓ(R) if and only if Ann(x) = Ry and Ann(y) = Rx. Several properties of the exact zero-divisor graph of a commutative ring were investigated in [15, 16]. Let R be a ring. Recall from [7] that an ideal I of R is said to be an annihilating ideal if there exists r ∈ R\{0} such that Ir = (0). As in [7], we denote the set of all annihilating ideals of R by A(R) and A(R)\{(0)} by A(R)∗. The concept of annihilating-ideal graph of a commutative ring was introduced and investigated by M. Behboodi and Z. Rakeei in [7]. Let R be a ring. Recall from [7] that the annihilating-ideal graph of R, denoted by AG(R) is an undirected graph whose vertex set is A(R)∗ and distinct vertices I and J are adjacent in this graph if and only if IJ = (0). Motivated by the interesting results proved on the annihilating-ideal graph of a ring in [7, 8], several researchers contributed to the study of annihilating-ideal graphs of commutative rings (for example, refer [1], [2], [11]). Inspired by the above mentioned work on annihilating-ideal graphs of rings and by the work on exact zero-divisor graphs of rings in [15, 16], in [17], P.T. Lalchandani introduced and studied the concept of the exact annihilating-ideal graph of a commutative ring. Let R be a ring. Recall from [17] that an ideal I of R is said to be an exact annihilating ideal if there exists a non-zero ideal J of R such that Ann(I) = J and Ann(J) = I, where for an ideal A of R, the annihilator of A in R, denoted by AnnR(A) or simply by Ann(A) is defined as Ann(A) = {r ∈ R | rA = (0)} [4, page 19]. As in [17], we denote the set of all exact annihilating ideals of a ring R by EA(R) and we denote EA(R)\{(0)} by EA(R)∗. It is clear that for any ring R, EA(R)∗ ⊆ A(R)∗. Let R be a ring such that EA(R)∗ 6= ∅. Recall from [17] that the exact annihilating-ideal graph of R, denoted by EAG(R) is an undirected graph whose vertex set is EA(R)∗ and distinct vertices I and J are adjacent in EAG(R) if and only if Ann(I) = J and Ann(J) = I. The graphs considered in this article are undirected and simple. For a graph G, we denote the vertex set of G by V (G) and the edge set of G by E(G). For a ring R with EA(R)∗ 6= ∅, it is clear that V (EAG(R)) = EA(R)∗ ⊆ A(R)∗ = V (AG(R)). Observe that if I,J ∈ EA(R)∗ are such that I and J are adjacent in EAG(R), then Ann(I) = J and Ann(J) = I. Hence, IJ = (0) and so, I and J are adjacent in AG(R). Therefore, EAG(R) is a subgraph of AG(R). The aim of this article is to continue the study of the exact annihilating-ideal graph of a commutative ring which was carried out in [17]. Throughout this article, we consider rings R such that EA(R)∗ 6= ∅ (it is noted in a remark which appears just preceding the statement of Corollary 2.2 that for a ring R, EA(R)∗ 6= ∅ if and only if R is not an integral domain) and study the interplay between the graph-theoretic properties of EAG(R) and the ring-theoretic properties of R. This article consists of three sections including the introduction. In Section 2 of this article, we discuss some results on the exact annihilating ideals of R, where R is a commutative ring which is not an integral domain. Let I ∈ A(R)∗. It is proved in Lemma 2.1 that the statements (1) I ∈ EA(R)∗; (2)I = Ann(J) for some non-zero ideal J of R; and (3) Ann(Ann(I)) = I are equivalent. For a ring R, we denote the set of all proper ideals of R by I(R) and I(R)\{(0)} by I(R)∗. Many examples of rings R are provided in Section 2 such that I(R)∗ = A(R)∗ = EA(R)∗ (see Examples 2.3, 2.8, Lemmas 2.4 and 2.6). We denote the cardinality of a set A by |A|. Whenever a set A is a subset of a set B and A 6= B, then we denote it by A ⊂ B. It is well-known that for a ring T , |A(T)∗| = 1 if and only if (T,Z(T))) is a special principal ideal ring (SPIR) with (Z(T))2 = (0)[7, Corollary 2.9(a)]. For a ring R, we denote the set of all prime ideals of R by Spec(R) and the set of all maximal ideals of R by Max(R). Let I be a non-zero proper ideal of a ring R. Motivated by [7, Corollary 2.9(a)], it is shown in Theorem 2.9 that the statements (1) EA(R)∗ = {I} and (2) I ∈ Spec(R),I2 = (0), and Z(R) = I are equivalent. In Example 2.14, a ring R is provided to illustrate that (2) ⇒ (1) of Theorem 2.9 can fail to hold if the assumption that I = Z(R) is omitted. It is verified that the ring R given in Example 2.14 is such that |EA(R)∗| = 2. Inspired by this example, it is natural to try to determine necessary and sufficient conditions on the ideals I,J of a ring R such that EA(R)∗ = {I,J}. It is well-known that the set of all nilpotent elements of a ring R is an ideal of R [4, Proposition 1.7] and is called the nilradical of R. We denote the nilradical of a ring R by nil(R). A ring R is said to be reduced if nil(R) = (0). We denote the set of all units of R by U(R). We denote the set of all minimal primes ideals of a ring R by Min(R). For non-zero proper ideals I,J of a reduced ring R which is not an integral domain, it is proved in Theorem 2.16 that the statements (1) EA(R)∗ = {I,J}; (2) J = Ann(I), I,J ∈ Spec(R); and (3) Min(R) = {I,J} are equivalent. Let R be a ring. Let p ∈ Spec(R) be such that p = Rp is principal, 120 S. Visweswaran, P. T. Lalchandani / J. Algebra Comb. Discrete Appl. 8(2) (2021) 119–138 n ≥ 2 is least with the property that pn = (0), and p = Z(R). Then it is shown in Proposition 2.10 that EA(R)∗ = {pi | i ∈{1, . . . ,n− 1}} and moreover, it is verified in Proposition 2.10 that A(R)∗ = EA(R)∗ if and only if p ∈ Max(R). Let R be a reduced ring. Let n ≥ 2 and let Min(R) = {pi | i ∈{1, 2, . . . ,n}}. Let C denote the collection of all non-empty proper subsets of {1, 2, . . . ,n}. It is proved in Proposition 2.15 that EA(R)∗ = { ∏ i∈A pi | A ∈ C}. Moreover, it is verified in Proposition 2.15 that A(R)∗ = EA(R)∗ if and only if pi ∈ Max(R) for each i ∈{1, 2, . . . ,n}. Let T be a unique factorization domain (UFD). It is shown in Theorem 2.17 that the statements (1) For each prime element p of T , A( T T p2 )∗ = EA( T T p2 )∗; (2) T is a principal ideal domain (PID); and (3) For each I ∈ I(T)∗ with I /∈ Max(T), A( T I )∗ = EA( T I )∗ are equivalent. Let T be a UFD with at least two non-associate prime elements. It is proved in Theorem 2.18 that the statements (1) For all non-associate prime elements p1,p2 of T , A( TT p1p2 ) ∗ = EA( T T p1p2 )∗; (2) T is a PID; and (3) For any I ∈ I(T)∗ with I /∈ Max(T), A( T I )∗ = EA( T I )∗ are equivalent. Let R be a von Neumann regular ring which is not a field. It is shown in Corollary 2.19 that |EA(R)∗| < ∞ if and only if there exist n ≥ 2 and fields F1,F2, . . . ,Fn such that R ∼= F1 ×F2 ×···×Fn as rings. Let R be a ring such that EA(R)∗ 6= ∅. The aim of Section 3 of this article is to discuss some results regarding the properties of EAG(R). Let I,J ∈ EA(R)∗ be such that I 6= J. It is proved in Proposition 3.1 that there is a path in EAG(R) between I and J if and only if I and J are adjacent in EAG(R). If I−J is an edge of EAG(R), then for any A ∈ EA(R)∗\{I,J}, it is shown in Lemma 3.2 that I and A are not adjacent in EAG(R) and J and A are not adjacent in EAG(R). As a consequence of Lemma 3.2, it is deduced in Corollary 3.3 that if g is any component of EAG(R), then g is a complete graph with at most two vertices. Let R be a ring. Let p ∈ Spec(R)\{(0)} be such that p2 = (0), and Z(R) = p. It is noted in Proposition 3.4 that EA(R)∗ = {p} and moreover, it is verified in Proposition 3.4 that its conclusion holds for a SPIR (R,m) with m 6= (0) but m2 = (0). For a real number x, we denote the integer part of x by [x]. Let R be a ring. Let p ∈ Spec(R) be such that p = Rp is principal. Let n ≥ 3 be least with the property that pn = (0) and Z(R) = p. Then it is proved in Proposition 3.5 that the following statements hold: (1) If n is odd, then EAG(R) has exactly [ n 2 ] components and each component is a complete graph with two vertices. (2) If n ≥ 4 is even, then EAG(R) has exactly n 2 components g1, . . . ,gn 2 −1,gn 2 such that gj is a complete graph with two vertices for each j ∈ {1, . . . , n2 − 1} and gn2 is a complete graph on a single vertex. Moreover, it is noted in Proposition 3.5 that the statements (1) and (2) hold for a SPIR (R,m) with the property that mn = (0) but mn−1 6= (0). Let R,p = Rp = Z(R) be as in the statement of Proposition 3.5. Let n ≥ 2 be least with the property that pn = (0). Then it is shown in Theorem 3.7 that the statements (1) EAG(R) = AG(R) and (2) (R,p) is a SPIR and n ∈ {2, 3} are equivalent. It is verified in Example 3.8 that the ring R provided by D.D. Anderson and M. Naseer in [3, page 501] is such that EAG(R) 6= AG(R) which illustrates that (2) ⇒ (1) of Theorem 3.7 can fail to hold if the hypothesis that p is principal is omitted. Let R be a reduced ring which is not an integral domain. It is shown in Lemma 3.9 that each component of EAG(R) is a complete graph with two vertices. It is proved in Corollary 3.10 that EAG(R) is connected if and only if |Min(R)| = 2 and it is shown in Corollary 3.11 that EAG(R) = AG(R) if and only if R ∼= F1 × F2 as rings, where Fi is a field for each i ∈ {1, 2}. If |Min(R)| = n ≥ 2, then it is proved in Corollary 3.12 that EAG(R) has exactly 2n−1 − 1 components. Let R be a ring such that EA(R)∗ 6= ∅. It is shown in Theorem 3.14 that the statements (1) EAG(R) = AG(R) and (2) Either R ∼= F1 ×F2 as rings, where Fi is a field for each i ∈ {1, 2} or (R,m) is a SPIR and if n ≥ 2 is least with the property that mn = (0), then n ∈{2, 3} are equivalent. Let R be a ring. The Krull dimension of R is simply referred to as the dimension of R and is denoted by dimR. Let R be a ring such that dimR = 0. If EAG(R) is connected, then it is proved in Proposition 3.16 that |Max(R)| ≤ 2 and if |Max(R)| = 2 then it is shown in Corollary 3.17 that EAG(R) is connected if and only if R ∼= F1×F2 as rings, where Fi is a field for each i ∈{1, 2}. Let R be a ring such that EA(R)∗ 6= ∅. It is noted in Corollary 3.20 that girth(EAG(R)) = ∞ and EAG(R) is perfect. 2. Some basic properties of EA(R)∗ As mentioned in the introduction, the rings considered in this article are commutative with identity. Let R be a ring such that EA(R)∗ 6= ∅. The aim of this section is to discuss some basic properties of the 121 S. Visweswaran, P. T. Lalchandani / J. Algebra Comb. Discrete Appl. 8(2) (2021) 119–138 exact annihilating ideals of R. Let R be a ring which is not an integral domain. Let I ∈ A(R)∗. In Lemma 2.1, we provide a necessary and sufficient condition for I to be in EA(R)∗. Lemma 2.1. Let R be a ring and let I ∈ A(R)∗. The following statements are equivalent: (1) I ∈ EA(R)∗. (2) I = Ann(J) for some non-zero ideal J of R. (3) Ann(Ann(I)) = I. Proof. (1) ⇒ (2) We are assuming that I ∈ EA(R)∗. Hence, by definition, there exists a non-zero ideal J of R such that Ann(I) = J and Ann(J) = I. (2) ⇒ (3) We are assuming that I = Ann(J) for some non-zero ideal J of R. Note that Ann(Ann(I)) = Ann(Ann(Ann(J)) = Ann(J) = I. (3) ⇒ (1) We are assuming that Ann(Ann(I)) = I. Let us denote Ann(I) by J. Since I ∈ A(R)∗ by hypothesis, Ann(I) 6= (0). Thus J 6= (0) and is such that Ann(I) = J and Ann(J) = I. This proves that I ∈ EA(R)∗. Let R be a ring which is not an integral domain. Then A(R)∗ 6= ∅. Let A ∈ A(R)∗. Then Ann(A) 6= (0) and as A(Ann(A)) = (0), it follows that Ann(A) ∈ A(R)∗. It follows from (2) ⇒ (1) of Lemma 2.1 that Ann(A) ∈ EA(R)∗. The above arguments imply that for a ring R, EA(R)∗ 6= ∅ if and only if R is not an integral domain. Corollary 2.2. Let R be a ring such that A(R)∗ 6= ∅. The following statements are equivalent: (1) A(R)∗ = EA(R)∗. (2) If I ∈ A(R)∗, then I = Ann(J) for some non-zero ideal J of R. (3) For any I ∈ A(R)∗, Ann(Ann(I)) = I. Proof. The statements (1) ⇒ (2) and (2) ⇒ (3) follow respectively from (1) ⇒ (2) and (2) ⇒ (3) of Lemma 2.1. For any ring T , as EA(T)∗ ⊆ A(T)∗, the proof of (3) ⇒ (1) follows immediately from (3) ⇒ (1) of Lemma 2.1. We illustrate Corollary 2.2 with the help of the example provided by D.D. Anderson and M. Naseer in [3, page 501]. We verify that I(R)∗ = A(R)∗ = EA(R)∗ for the ring R provided in [3, page 501] in Example 2.3. For any n ≥ 2, we denote the ring of integers modulo n by Zn. Example 2.3. Let T = Z4[X,Y,Z] be the polynomial ring in three variables X,Y,Z over Z4. Let I be the ideal of T generated by {X2 − 2,Y 2 − 2,Z2,XY,Y Z − 2,XZ, 2X, 2Y, 2Z}. Let R = T I . Then I(R)∗ = A(R)∗ = EA(R)∗. Proof. It is convenient to denote X + I,Y + I,Z + I by x,y,z, respectively. It was already noted in [3, page 501] that R is local with m = Rx + Ry + Rz as its unique maximal ideal, m2 = {0 + I, 2 + I}, m3 = (0 + I), and |R| = 32. Observe that Z(R) = m and from m3 = (0), we get that each proper ideal of R is an annihilating ideal of R. Therefore, I(R)∗ = A(R)∗. The ring R was also considered in [8, Proposition 2.1] and it was noted there that I(R)∗ = {(2 +I), (x), (y), (z), (x+y), (y +z), (z +x), (x+y + z), (x,y), (y,z), (z,x), (x,y +z), (y,z +x), (z,x+y), (x+y,y +z), (x,y,z)}. From the multiplication table provided in [3, page 503], it follows that Ann(2+I) = m,Ann(x) = {0+I, 2+I,y,y+2,z,z+2,y+z,y+z+ 2},Ann(y) = {0+1, 2+I,x,x+2,y+z,y+z+2,x+y+z,x+y+z+2},Ann(z) = {0+I, 2+I,x,x+2,z,z+ 2,x+z,x+z+2},Ann(x+y) = {0+I, 2+I,y+z,y+z+2,z+x,z+x+2,x+y,x+y+2},Ann(y+z) = {0+ I, 2+I,x,x+2,y,y+2,x+y,x+y+2},Ann(z+x) = {0+I, 2+I,z,z+2,x+y,x+y+2,x+y+z,x+y+z+2}, and Ann(x + y + z) = {0 + I, 2 + I,x + z,x + z + 2,y,y + 2,x + y + z,x + y + z + 2}. Note that 122 S. Visweswaran, P. T. Lalchandani / J. Algebra Comb. Discrete Appl. 8(2) (2021) 119–138 x(x + z) = y(x + z) = z(y + z) = x(x + y) = x(x + y + z) = 2 + I. From the above given arguments, it is clear that Ann(m2) = m,Ann(m) = m2,Ann(Rx) = Ry + Rz,Ann(Ry + Rz) = Rx,Ann(Ry) = Rx + R(y + z),Ann(Rx + R(y + z)) = Ry,Ann(Rz) = Rx + Rz,Ann(Rx + Rz) = Rz,Ann(R(x + y)) = R(y + z) + R(z + x),Ann(R(y + z) + R(z + x)) = R(x + y),Ann(R(y + z)) = Rx + Ry,Ann(Rx + Ry) = R(y+z),Ann(R(z+x)) = Rz+R(x+y),Ann(Rz+R(x+y)) = R(z+x),Ann(R(x+y+z)) = Ry+R(x+z), and Ann(Ry + R(x + z)) = R(x + y + z). From the above discussion, we obtain that I(R)∗ = A(R)∗ and each proper A of R is such that Ann(Ann(A)) = A. Hence, we obtain from (3) ⇒ (1) of Corollary 2.2 that A(R)∗ = EA(R)∗. Therefore, I(R)∗ = A(R)∗ = EA(R)∗. Recall that a principal ideal ring R is said to be a special principal ideal ring (SPIR) if R has a unique prime ideal. If m is the unique prime ideal of R, then it follows from [4, Proposition 1.8] that nil(R) = m. Since m is principal, we get that m is nilpotent. Suppose that R is not a field. Then m 6= (0). Let n ≥ 2 be least with the property that mn = (0). Then it follows from the proof of (iii) ⇒ (i) of [4, Proposition 8.8] that {mi | i ∈{1, . . . ,n−1}} is the set of all non-zero proper ideals of R. If R is a SPIR with m as its only prime ideal, then we denote it by the notation (R,m) is a SPIR. Let (R,m) be a SPIR which is not a field. We verify in Lemma 2.4 that I(R)∗ = A(R)∗ = EA(R)∗. Lemma 2.4. Let (R,m) be a SPIR which is not a field. Then I(R)∗ = A(R)∗ = EA(R)∗. Proof. Let n ≥ 2 be least with the property that mn = (0). Note that I(R)∗ = {mi | i ∈{1, . . . ,n−1}}. From mn = (0), it follows that I(R)∗ = A(R)∗. Let i ∈ {1, . . . ,n − 1}. Observe that Ann(mi) = mn−i and so, Ann(Ann(mi)) = Ann(mn−i) = mi. Therefore, we obtain from (3) ⇒ (1) of Corollary 2.2 that A(R)∗ = EA(R)∗. This proves that I(R)∗ = A(R)∗ = EA(R)∗. Corollary 2.5. Let T be a PID which is not a field. Let m ∈ Max(T). Let n ≥ 2 and let R = T mn . Then I(R)∗ == A(R)∗ = EA(R)∗. Proof. Let m ∈ m be such that m = Tm. Observe that R is a principal ideal ring. Note that m mn ∈ Spec(R). Let P ∈ Spec(R). Then P = p mn for some p ∈ Spec(T) with p ⊇ mn. This implies that p ⊇ m and so, p = m. Therefore, P = m mn . Thus R is a principal ideal ring with Spec(R) = { m mn }. Hence, (R, m mn ) is a SPIR. Therefore, we obtain from Lemma 2.4 that I(R)∗ = A(R)∗ = EA(R)∗. We provide some more examples in Example 2.8 to illustrate Corollary 2.2. We use Lemmas 2.6 and 2.7 in the verification of Example 2.8. Lemma 2.6. Let n ≥ 2 and let Ri be a ring for each i ∈ {1, 2, . . . ,n}. Let R = R1 × R2 ×···× Rn. Suppose that for any i ∈ {1, 2, . . . ,n} and any ideal Ii of Ri, AnnRi (AnnRi (Ii)) = Ii. Then I(R)∗ = A(R)∗ = EA(R)∗. Proof. Let I ∈ I(R)∗. Then for each i ∈ {1, 2, . . . ,n}, there exists an ideal Ii of Ri such that I = I1 × I2 × ···× In. Since I 6= R, it follows that Ii 6= Ri for at least one i ∈ {1, 2, . . . ,n}. If Ii = (0), then I(Rei) = (0) × (0) × ···× (0), where ei is the element of R whose i-th coordinate equals 1 and whose j-th coordinate equals 0 for all j ∈ {1, 2, . . . ,n}\{i}. As Rei is a non-zero ideal of R and I(Rei) equals the zero ideal of R, it follows that I ∈ A(R)∗. Suppose that Ii 6= (0). Then from the hypothesis, AnnRi (AnnRi (Ii)) = Ii, we get that Ii ∈ A(Ri)∗ and so, I ∈ A(R)∗. This shows that I(R)∗ ⊆ A(R)∗ and so, I(R)∗ = A(R)∗. Let I = I1 × I2 × ···× In ∈ I(R)∗ = A(R)∗. Observe that Ann(Ann(I)) = AnnR1 (AnnR1 (I1))×AnnR2 (AnnR2 (I2))×···×AnnRn (AnnRn (In)) = I1×I2×···×In = I. Thus for each I ∈ A(R)∗, Ann(Ann(I)) = I. Hence, we obtain from (3) ⇒ (1) of Corollary 2.2 that A(R)∗ = EA(R)∗. Therefore, I(R)∗ = A(R)∗ = EA(R)∗. Lemma 2.7. Let R be a ring and let m ∈ Max(R). Let q be a m-primary ideal of R. Then R q ∼= Rmqm as rings. 123 S. Visweswaran, P. T. Lalchandani / J. Algebra Comb. Discrete Appl. 8(2) (2021) 119–138 Proof. This is well-known. We provide a proof of this lemma for the sake of completeness. Let f : R → Rm be the usual homomorphism of rings defined by f(r) = r1. Using the hypothesis that q is m-primary, it can be shown that f−1(qm) = q. Hence, f induces an injective ring homomorphism f : R q → Rm qm defined by f(r + q) = f(r) + qm. We verify that f is onto. Let Y be any element of Rm qm . Then there exist r ∈ R,s ∈ R\m such that Y = r s + qm. Since s ∈ R\m and m ∈ Max(R), we get that m + Rs = R. Hence, √ q + √ Rs = R and so, we obtain from [4, Proposition 1.16] that q + Rs = R. Therefore, there exist x ∈ R and q ∈ q such that q + xs = 1. Hence, r = rq + rxs and so, Y = r s + qm = rsx+rq s + qm = rx 1 + qm, since rq s ∈ qm. Thus Y = f(rx + q). This shows that f is onto. Hence, f : R q → Rm qm is an isomorphism of rings. Therefore, R q ∼= Rmqm as rings. Example 2.8. (1) Let n ≥ 2. Let R = F1 ×F2 ×···×Fn, where Fi is a field for each i ∈{1, 2, . . . ,n}. Then I(R)∗ = A(R)∗ = EA(R)∗. (2) Let T be a Dedekind domain and let I be a non-zero proper ideal of T such that I /∈ Max(T). Let R = T I . Then I(R)∗ = A(R)∗ = EA(R)∗. (3) Let T be a principal ideal domain. Let I be a non-zero proper ideal of T such that I /∈ Max(T). Let R = T I . Then I(R)∗ = A(R)∗ = EA(R)∗. Proof. (1) Let i ∈{1, 2, . . . ,n}. Observe that Fi and (0) are the only ideals of Fi and for each ideal Ii of Fi, AnnFi (AnnFi (Ii)) = Ii. Now, it follows from Lemma 2.6 that I(R) ∗ = A(R)∗ = EA(R)∗. (2) Since T is a Dedekind domain, T is Noetherian, dimT = 1, and T is integrally closed. Thus any non-zero prime ideal of T is maximal. It follows from [4, Corollary 9.4] that there exist distinct maximal ideals m1, . . . ,mn of T and positive integers k1, . . . ,kn such that I = n∏ i=1 mkii . Observe that for each i ∈{1, . . . ,n}, √ mkii = mi ∈ Max(T) and so, we obtain from [4, Proposition 4.2] that m ki i is a mi-primary ideal of T. We know from Lemma 2.7 that T m ki i ∼= Tmi (m ki i )mi as rings. We know from (i) ⇒ (iii) of [4, Theorem 9.3] that Tmi is a discrete valuation ring and so, it is a PID. Now, for all distinct i,j ∈{1, . . . ,n},√ mkii + √ m kj j = T and so by [4, Proposition 1.16] that m ki i + m kj j = T . Hence, we obtain from [4, Proposition 1.10] that T I ∼= T m k1 1 ×···× T m kn n . Note that for each i ∈{1, . . . ,n}, (mi)mi is the unique maximal ideal of Tmi and (m ki i )mi = ((mi)mi ) ki. Therefore, we obtain that T I ∼= Tm1 ((m1)m1 ) k1 ×···× Tmn ((mn)mn ) kn as rings. Let i ∈{1, . . . ,n} and let us denote the ring Tmi ((mi)mi ) ki by Ri and the unique maximal ideal (mi)mi of Tmi by ni. If ki = 1, then Ri is a field. If ki ≥ 2, then as Tmi is a PID, we obtain from the proof of Corollary 2.5 that (Ri, ni n ki i ) is a SPIR. From the proof of Lemma 2.4, we know that AnnRi (AnnRi (Ii)) = Ii for each ideal Ii of Ri. Now, R ∼= R1 ×···× Rn as rings. Suppose that n = 1. Since I /∈ Max(T) by hypothesis, k1 ≥ 2 and so, it follows from Lemma 2.4 that I(R)∗ = A(R)∗ = EA(R)∗. Suppose that n ≥ 2. As AnnRi (AnnRi (Ii)) = Ii for each ideal Ii of Ri and for each i ∈ {1, 2, . . . ,n}, we obtain from Lemma 2.6 that I(R)∗ = A(R)∗ = EA(R)∗. (3) If T is a PID which is not a field, then we know from [4, Example (1), page 96] that T is a Dedekind domain. Therefore, the conclusion of (3) follows immediately from (2). Let R be a ring such that A(R)∗ 6= ∅. It was shown in [7, Corollary 2.9(a)] that A(R)∗ = {I} if and only if (R,I) is a SPIR with I2 = (0) (see also, [19, Lemma 2.6]). Let R be a ring and let I ∈ A(R)∗. In Theorem 2.9, we determine necessary and suffcient conditions on I such that EA(R)∗ = {I}. Theorem 2.9. Let R be a ring and let I be a non-zero ideal of R. The following statements are equivalent: (1) EA(R)∗ = {I}. (2) I ∈ Spec(R),I2 = (0), and Z(R) = I. 124 S. Visweswaran, P. T. Lalchandani / J. Algebra Comb. Discrete Appl. 8(2) (2021) 119–138 Proof. (1) ⇒ (2) As I ∈ EA(R)∗, I ∈ A(R)∗ and Ann(I) ∈ A(R)∗. Hence, we obtain from (2) ⇒ (1) of Lemma 2.1 that Ann(I) ∈ EA(R)∗ = {I} and so, Ann(I) = I. This proves that I2 = (0). We next verify that I ∈ Spec(R). It is clear that I 6= R. If B is any non-zero ideal of R with Ann(B) 6= (0), then Ann(B) ∈ A(R)∗ and it follows from (2) ⇒ (1) of Lemma 2.1 that Ann(B) ∈ EA(R)∗ = {I}. Hence, Ann(B) = I. Let a,b ∈ R be such that ab ∈ I = Ann(I). This implies that Iab = (0). Suppose that a /∈ I = Ann(I). Then Ia 6= (0) and from I(Ia) = (0), it follows that Ann(Ia) 6= (0). Hence, Ann(Ia) = I. From b ∈ Ann(Ia), we get that b ∈ I. This proves that I ∈ Spec(R). As any member of A(R) is a subset of Z(R), it follows that I ⊆ Z(R). Let r ∈ Z(R)∗. Then there exists s ∈ R\{0} such that rs = 0. As Rs 6= (0) and Ann(Rs) 6= (0), it follows that Ann(Rs) = I. From rs = 0, we obtain that r ∈ I. This shows that Z(R) ⊆ I and so, Z(R) = I. (2) ⇒ (1) We claim that for any non-zero ideal J of R with J ⊆ I, Ann(J) = I. Let r ∈ Ann(J). Then rJ = (0). From J 6= (0), it follows that r ∈ Z(R) = I. This shows that Ann(J) ⊆ I. From I2 = (0) and J ⊆ I, we get that JI ⊆ I2 = (0). This shows that I ⊆ Ann(J). Therefore, Ann(J) = I and so, in particular, Ann(I) = I. Hence, I ∈ EA(R)∗. Let A ∈ EA(R)∗. Then there exists a non-zero ideal B of R such that Ann(A) = B and Ann(B) = A. This implies that AB = (0) and so, A∪B ⊆ Z(R) = I. Hence, Ann(A) = Ann(B) = I. From Ann(B) = A, we obtain that A = I. This proves that EA(R)∗ = {I}. Let R be a ring such that it admits p ∈ Spec(R) with p = Rp is principal, p 6= (0) but p is nilpotent. Let n ≥ 2 be least with the property that pn = (0). If Z(R) = p, then we prove in Proposition 2.10 that EA(R)∗ = {pi | i ∈{1, . . . ,n− 1}}. Proposition 2.10. Let R be a ring. Let p ∈ Spec(R) be such that p = Rp is principal, p 6= (0) but p is nilpotent, and Z(R) = p. Let n ≥ 2 be least with the property that pn = (0). Then EA(R)∗ = {pi | i ∈ {1, . . . ,n− 1}}. Moreover, A(R)∗ = EA(R)∗ if and only if p ∈ Max(R). Proof. Let i ∈ {1, . . . ,n − 1}. As p = Rp, we get that pi = Rpi. From pn = 0, it follows that pn−i ∈ Ann(Rpi). Hence, Rpn−i ⊆ Ann(Rpi). Let r ∈ Ann(Rpi). Then rpi = 0. As pi 6= 0, we obtain that r ∈ Z(R) = Rp. We claim that r ∈ Rpn−i. This is clear if r = 0. Suppose that r 6= 0. It is possible to find j ∈ {1, . . . ,n− 1} such that r ∈ Rpj\Rpj+1. Hence, there exists s ∈ R\Z(R) such that r = pjs. From rpi = 0, we get that spi+j = 0. As s ∈ R\Z(R), it follows that pi+j = 0. Since n is least with the property that pn = 0, we obtain that i + j ≥ n and so, j ≥ n − i. Therefore, r ∈ Rpj ⊆ Rpn−i. This proves that Ann(Rpi) ⊆ Rpn−i and so, we obtain that Ann(Rpi) = Rpn−i. As n− i ∈{1, . . . ,n− 1}, it follows that Ann(Rpn−i) = Rpi. Thus for any i ∈ {1, . . . ,n− 1}, Ann(pi) = pn−i and Ann(pn−i) = pi. This proves that {pi | i ∈{1, . . . ,n−1}}⊆ EA(R)∗. Let A ∈ EA(R)∗. Then there exists a non-zero ideal B of R such that Ann(A) = B and Ann(B) = A. From AB = (0), we get that A∪B ⊆ Z(R) = Rp. It is possible to find j ∈{1, . . . ,n− 1} such that B ⊆ Rpj but B 6⊆ Rpj+1. Note that Rpn−j ⊆ Ann(B) = A. Let b ∈ B\Rpj+1. As B ⊆ Rpj, it follows that b = spj for some s ∈ R\Z(R). From AB = (0), we obtain that for any a ∈ A, a(spj) = 0 and so, apj = 0. This implies that a ∈ Ann(Rpj) = Rpn−j. This shows that A ⊆ Rpn−j and so, A = Rpn−j. Hence, EA(R)∗ ⊆ {pi | i ∈ {1, . . . ,n− 1}}. Therefore, we obtain that EA(R)∗ = {pi | i ∈{1, . . . ,n− 1}}. We next verify the moreover part of this proposition. Assume that A(R)∗ = EA(R)∗. As EA(R)∗ = {pi | i ∈ {1, . . . ,n − 1}}, we obtain that A(R)∗ is finite. Hence, R satisfies descending chain condition (d.c.c.) on A(R)∗. Therefore, it follows from [7, Theorem 1.1] that R is Artinian and so, we obtain from [4, Proposition 8.1] that p ∈ Max(R). We also include a direct argument to show that p ∈ Max(R). Let m ∈ Max(R) be such that p ⊆ m. Let m ∈ m. If pm = 0, then m ∈ Z(R) = p. Suppose that pm 6= 0. Note that Rpm ∈ A(R)∗. Therefore, Rpm = pi = Rpi for some i ∈{1, . . . ,n−1}. If i = 1, then p = rpm for some r ∈ R. Hence, p(1 −rm) = 0. This implies that 1 −rm ∈ Z(R) = p ⊆ m and so, 1 ∈ m. This is impossible and therefore, i ≥ 2. From pn = 0 and Rpm = Rpi, it follows that pn−i+1m = 0. As i ≥ 2, it follows that pn−i+1 6= 0. Hence, m ∈ Z(R) = p. This proves that m ⊆ p and so, p = m ∈ Max(R). Conversely, assume that p ∈ Max(R). Let P ∈ Spec(R). Now, P ⊇ (0) = pn. This implies that P ⊇ p. Since p ∈ Max(R), it follows that P = p. Therefore, Spec(R) = Max(R) = {p}. Now, p = Rp is principal and n ≥ 2 is least with the property that pn = (0). Hence, we obtain from the proof of (iii) ⇒ (i) 125 S. Visweswaran, P. T. Lalchandani / J. Algebra Comb. Discrete Appl. 8(2) (2021) 119–138 of [4, Proposition 8.8] that {pi = Rpi | i ∈ {1, . . . ,n− 1}} is the set of all non-zero proper ideals of R. Therefore, it follows that (R,p) is a SPIR and so, I(R)∗ = A(R)∗ = EA(R)∗ = {pi | i ∈{1, . . . ,n−1}}. We provide Example 2.12 to illustrate Theorem 2.9 and Proposition 2.10. We use Lemma 2.11 in the verification of Example 2.12. Lemma 2.11. Let p be a prime element of an integral domain T. Let n ≥ 2. Let R = T T pn . Let p = T p T pn . Then EA(R)∗ = {pi | i ∈{1, . . . ,n− 1}}. Proof. By hypothesis, p is a prime element of T. Hence, Tp ∈ Spec(T) and so, p = T p T pn ∈ Spec(R = T T pn ). It is clear that p = R(p + Tpn) is principal. Observe that n ≥ 2 is least with the property that pn = (0 + Tpn). Note that Tpn is a Tp-primary ideal of T . Hence, the zero ideal (0 + Tpn) of R is a p-primary ideal of R. Therefore, we obtain from [4, Proposition 4.7] that Z(R) = p. Now, it follows from Proposition 2.10 that EA(R)∗ = {pi | i ∈{1, . . . ,n− 1}}. Example 2.12. Let T = Z[X] be the polynomial ring in one variable X over Z. Let n ≥ 2 and let R = T T Xn . Let p = T X T Xn . Then EA(R)∗ = {pi | i ∈ {1, . . . ,n− 1}}, A(R)∗ = {I ∈ I(R)∗ | I ⊆ p}, and A((R)∗ 6= EA(R)∗. Proof. Note that T is an integral domain. Indeed, T is a unique factorization domain and X is a prime element of T . Therefore, we obtain from Lemma 2.11 that EA(R)∗ = {pi | i ∈{1, . . . ,n−1}}. Let I ∈ I(R)∗ be such that I ⊆ p. Note that pn−1 6= (0 + TXn) and Ipn−1 = (0 + TXn). Hence, I ∈ A(R)∗. Let A ∈ A(R)∗. As any annihilating ideal of a ring is contained in its set of zero-divisors, we get that A ⊆ Z(R) = p. This proves that A(R)∗ = {I ∈ I(R)∗ | I ⊆ p}. Observe that I = R(2X + TXn) ⊆ p, I 6= (0 + TXn), and I /∈ {pi | i ∈ {1, . . . ,n − 1}}. Hence, I ∈ A(R)∗\EA(R)∗. Therefore, A(R)∗ 6= EA(R)∗. In Example 2.13, we illustrate that (2) ⇒ (1) of Theorem 2.9 can fail to hold if the assumption that I2 = (0) is omitted. Example 2.13. Let R be as in Example 2.3. In the notation of Example 2.3, R is a local Artinian ring with unique maximal ideal m = Rx + Ry + Rz, m3 = (0), and Z(R) = m. It is already verified in the verification of Example 2.3 that R has 16 non-zero proper ideals and I(R)∗ = A(R)∗ = EA(R)∗. In Example 2.14, we illustrate that (2) ⇒ (1) of Theorem 2.9 can fail to hold if the assumption that Z(R) = I is omitted. Example 2.14. Let T = K[X,Y ] be the polynomial ring in two variables X,Y over a field K. Let A = TX2 + TXY . Let R = T A . Let p = T X A . Then p = R(X + A) is principal, p2 = (0 + A), and |EA(R)∗| = 2. Proof. As X is a prime element of T , it follows that TX ∈ Spec(T) and so, p = T X A ∈ Spec( T A = R). It is clear that p = R(X + A) is principal and from X2 ∈ A, we obtain that p2 = (0 + A). Observe that T T X+T Y ∼= K as rings. From K is a field, it follows that TX + TY ∈ Max(T) and so, m = T X+T YA ∈ Max(R). It is convenient to denote X + A by x and Y + A by y. Note that m = Rx + Ry. Observe that A = TX ∩ (TX2 + TY ). As √ TX2 + TY = TX + TY ∈ Max(T), we obtain from [4, Proposition 4.2] that TX2 + TY is a TX + TY -primary ideal of T . Hence, A = TX ∩ (TX2 + TY ) is a minimal primary decomposition of A with TX is a TX-primary ideal of T and TX2 + TY is a TX + TY -primary ideal of T. Therefore, (0 + A) = T X A ∩ T X 2+T Y A is a minimal primary decomposition of the zero ideal of R with T X A is a p-primary ideal of R and T X 2+T Y A is a m-primary ideal of R. Hence, it follows from [4, Proposition 4.7] that Z(R) = p ∪ m = m, since p ⊂ m. Note that mp = (0 + A) and so, m ⊆ Ann(p). From m ∈ Max(R) and Ann(p) 6= R, it follows that Ann(p) = m. From pm = (0 + A), we get that 126 S. Visweswaran, P. T. Lalchandani / J. Algebra Comb. Discrete Appl. 8(2) (2021) 119–138 p ⊆ Ann(m). Let r ∈ Ann(m). Then rm = (0 + A) ⊂ p. As p ∈ Spec(R) and m 6⊆ p, we obtain that r ∈ p. This proves that Ann(m) ⊆ p and so, Ann(m) = p. Thus the non-zero ideals m,p of R are such that Ann(p) = m and Ann(m) = p. Hence, p,m ∈ EA(R)∗ and so, |EA(R)∗| ≥ 2. Let C ∈ EA(R)∗. We claim that C ∈ {p,m}. As C ∈ EA(R)∗, there exists a non-zero ideal D of R such that Ann(C) = D and Ann(D) = C. Observe that CD = (0 + A) ⊂ p. From p ∈ Spec(R), it follows that either C ⊆ p or D ⊆ p. Suppose that C ⊆ p. Then m = Ann(p) ⊆ Ann(C) = D. From D 6= R, we get that D = m and so, C = Ann(D) = p. If D ⊆ p, then it follows similarly that C = m. Therefore, C ∈{p,m}. This proves that EA(R)∗ = {p,m} and so, |EA(R)∗| = 2. Let R be a ring such that EA(R)∗ 6= ∅. Inspired by Theorem 2.9 and Example 2.14, we try to char- acterize ideals I,J of a ring R such that EA(R)∗ = {I,J}. In Theorem 2.16, we are able to characterize ideals I,J of a reduced ring R such that EA(R)∗ = {I,J}. We use Proposition 2.15 in the proof of (3) ⇒ (1) of Theorem 2.16. Let R be a reduced ring which is not an integral domain. Suppose that |EA(R)∗| < ∞. Let A ⊆ Z(R)∗ be such that xy = 0 for all distinct x,y ∈ A. Note that for each x ∈ A, x 6= 0, Ann(x) ∈ A(R)∗, and it follows from (2) ⇒ (1) of Lemma 2.1 that Ann(x) ∈ EA(R)∗. Let x,y ∈ A be such that x 6= y. Observe that y ∈ Ann(x). Since R is reduced and y 6= 0, it follows that y /∈ Ann(y). Hence, Ann(x) 6= Ann(y). From the assumption that |EA(R)∗| < ∞, we get that A is finite. Hence, we obtain from (4) ⇒ (3) of [6, Theorem 3.7] that Min(R) is finite. This shows that if |EA(R)∗| < ∞ for a reduced ring R, then |Min(R)| < ∞. Let R be a reduced ring with |Min(R)| = n ≥ 2. Then we prove in Proposition 2.15 that |EA(R)∗| = 2n − 2. Proposition 2.15. Let R be a reduced ring which is not an integral domain. Let |Min(R)| = n and let Min(R) = {p1, . . . ,pn}. Then |EA(R)∗| = 2n − 2. Moreover, A(R)∗ = EA(R)∗ if and only if pi ∈ Max(R) for each i ∈{1, 2, . . . ,n}. Proof. It is known that any prime ideal p of a ring T contains a minimal prime ideal of T [14, Theorem 10]. Since R is reduced, nil(R) = (0). We know from [4, Proposition 1.8] that (0) = nil(R) = ⋂ p∈Spec(R) p. Since any prime ideal of R contains at least one minimal prime ideal of R, we obtain that ⋂ p∈Min(R) p = (0). As Min(R) = {p1, . . . ,pn}, we obtain that n⋂ i=1 pi = (0). It is clear that n ≥ 2, since R is not an integral domain. Note that distinct minimal prime ideals of a ring R are not comparable under the inclusion relation and hence, it follows from [4, Proposition 1.11(ii)] that for any proper non-empty subset A of {1, 2, . . . ,n}, ⋂ i∈A pi 6= (0). Let A ⊂ {1, 2, . . . ,n} with A 6= ∅. Let us denote ⋂ i∈A pi by IA. Observe that for any A ⊂{1, 2, . . . ,n} with A 6= ∅, Ac ⊂{1, 2, . . . ,n} and Ac 6= ∅, where Ac = {1, 2, . . . ,n}\A and it is easy to verify that Ann(IA) = IAc. Hence, IA ∈ A(R)∗ and note that Ann(Ann(IA)) = IA. Therefore, we obtain from (3) ⇒ (1) of Lemma 2.1 that IA ∈ EA(R)∗. This proves that {IA |A ⊂{1, 2, . . . ,n},A 6= ∅}⊆ EA(R)∗. Let I ∈ EA(R)∗. As I ∈ A(R)∗, Ir = (0) for some r ∈ R\{0}. Since n⋂ i=1 pi = (0), r /∈ pi for at least one i ∈ {1, 2, . . . ,n}. From Ir = (0) ⊂ pi ∈ Spec(R), we get that I ⊆ pi. Since I 6= (0), there exists at least one j ∈{1, 2, . . . ,n} such that I 6⊆ pj. Thus there exists A ⊂{1, 2, . . . ,n},A 6= ∅ such that I ⊆ pi for each i ∈ A and I 6⊆ pj for any j ∈ {1, 2, . . . ,n}\A. From IAnn(I) = (0) ⊆ pj for any j ∈ Ac, we obtain that Ann(I) ⊆ pj for each j ∈ Ac. Thus I ⊆ IA and Ann(I) ⊆ IAc. From Ann(I) ⊆ IAc, it follows that IA = Ann(IAc ) ⊆ Ann(Ann(I)). Since I ∈ EA(R)∗, we obtain from (1) ⇒ (3) of Lemma 2.1 that Ann(Ann(I)) = I and so, IA ⊆ I. Hence, I = IA for some A ⊂ {1, 2, . . . ,n} with A 6= ∅. This proves that EA(R)∗ ⊆{IA | A ⊂{1, 2, . . . ,n},A 6= ∅} and so, EA(R)∗ = {IA | A ⊂{1, 2, . . . ,n},A 6= ∅}. If A1,A2 are distinct non-empty proper subsets of {1, 2, . . . ,n}, then it is clear that IA1 6= IA2. Since |{A ⊂{1, 2, . . . ,n},A 6= ∅}| = 2n − 2, it follows that |EA(R)∗| = 2n − 2. We next verify the moreover part of this proposition. Assume that A(R)∗ = EA(R)∗. Hence, |A(R)∗| = 2n − 2 < ∞. Therefore, R satisfies d.c.c. on A(R)∗ and so, we obtain from [7, Theorem 1.1] that R is Artinian. We know from [4, Proposition 8.1] that Spec(R) = Max(R). Therefore, pi ∈ Max(R) 127 S. Visweswaran, P. T. Lalchandani / J. Algebra Comb. Discrete Appl. 8(2) (2021) 119–138 for each i ∈ {1, 2, . . . ,n}. We also include a direct argument to show that pi ∈ Max(R) for each i ∈ {1, 2, . . . ,n}. First, we show that p1 ∈ Max(R). Let m ∈ Max(R) be such that p1 ⊆ m. Since distinct minimal prime ideals of a ring are not comparable under the inclusion relation, it follows from [4, Proposition 1.11(ii)] that there exists x ∈ ( n⋂ j=2 pj)\p1. Let m ∈ m. Suppose that xm 6= 0. As x ∈ Z(R)∗, it follows that xm ∈ Z(R)∗, and so, Rxm ∈ A(R)∗. It is shown in the previous paragraph that EA(R)∗ = {IA | A ⊂{1, 2, . . . ,n},A 6= ∅}, where for a non-empty proper subset A of {1, 2, . . . ,n}, IA = ⋂ i∈A pi. From the assumption A(R)∗ = EA(R)∗, it follows that Rxm = IA for some non-empty proper subset A of {1, 2, . . . ,n}. From xm ∈ ( n⋂ j=2 pj)\{0} and n⋂ i=1 pi = (0), we get that 1 /∈ A. Hence, A ⊆ {1, 2, . . . ,n}\{1}. It follows from the choice of x that x ∈ IA. Therefore, x ∈ Rxm. This implies that x(1 − rm) = 0 for some r ∈ R. As x /∈ p1, we obtain that 1 − rm ∈ p1 ⊆ m. This implies that 1 ∈ m and this is impossible. Therefore, xm = 0. Hence, m ∈ p1, since p1 ∈ Spec(R) and x /∈ p1. This proves that m ⊆ p1. Therefore, p1 = m ∈ Max(R). Similarly, it can be shown that pj ∈ Max(R) for each j ∈{2, . . . ,n}. Conversely, assume that pi ∈ Max(R) for each i ∈{1, 2, . . . ,n}. Note that pi +pj = R for all distinct i,j ∈ {1, 2, . . . ,n} and n⋂ i=1 pi = (0). Hence, we obtain from [4, Proposition 1.10(ii) and (iii)] that the mapping f : R → R p1 × R p2 ×···× R pn defined by f(r) = (r + p1,r + p2, . . . ,r + pn) is an isomorphism of rings. Let i ∈ {1, 2, . . . ,n}. Since pi ∈ Max(R), it follows that Rpi is a field. Let us denote the ring R p1 × R p2 ×···× R pn by T. It follows from Example 2.8(1) that I(T)∗ = A(T)∗ = EA(T)∗. Since R ∼= T as rings, we obtain that I(R)∗ = A(R)∗ = EA(R)∗. Theorem 2.16. Let R be a reduced ring which is not an integral domain. The following statements are equivalent: (1) EA(R)∗ = {I,J}. (2) J = Ann(I) and I,J ∈ Spec(R). (3) Min(R) = {I,J}. Proof. (1) ⇒ (2) As I ∈ EA(R)∗, it follows that I ∈ A(R)∗ and so, Ann(I) 6= (0). It is clear that Ann(I) ∈ A(R)∗. Observe that we obtain from (2) ⇒ (1) of Lemma 2.1 that Ann(I) ∈ EA(R)∗ = {I,J}. Since R is reduced, I2 6= (0) and so, Ann(I) 6= I and therefore, Ann(I) = J. Let B ∈ A(R)∗. Then Ann(B) ∈ A(R)∗. Therefore, we obtain from (2) ⇒ (1) of Lemma 2.1 that Ann(B) ∈ EA(R)∗. From the hypothesis EA(R)∗ = {I,Ann(I)}, it follows that if B ∈ A(R)∗, then either Ann(B) = I or Ann(B) = Ann(I). We next verify that I,J = Ann(I) ∈ Spec(R). Let a,b ∈ R be such that ab ∈ I. Then abAnn(I) = (0). We know from (1) ⇒ (3) of Lemma 2.1 that Ann(Ann(I)) = I. If aAnn(I) = (0), then a ∈ Ann(Ann(I)) = I. Similarly, if bAnn(I) = (0), then b ∈ I. Hence, we can assume that aAnn(I) 6= (0) and bAnn(I) 6= (0). Now, aAnn(I) 6= (0), Ann(aAnn(I)) 6= (0), bAnn(I) 6= (0), and Ann(bAnn(I)) 6= (0). Therefore, Ann(aAnn(I)),Ann(bAnn(I)) ∈ EA(R)∗ = {I,Ann(I)}. Observe that Ann(aAnn(I)) 6= Ann(bAnn(I)). For if Ann(aAnn(I)) = Ann(bAnn(I)), then from abAnn(I) = (0), it follows that b2Ann(I) = (0). Since R is reduced, we get that bAnn(I) = (0) and this contradicts our assumption. Hence, Ann(aAnn(I)) 6= Ann(bAnn(I)). Therefore, either Ann(aAnn(I)) = I or Ann(bAnn(I)) = I. If Ann(aAnn(I)) = I, then b ∈ I. If Ann(bAnn(I)) = I, then a ∈ I. This proves that I ∈ Spec(R). Similarly, it can be shown that Ann(I) ∈ Spec(R). (2) ⇒ (3) We are assuming that the ideals I,J of R are such that J = Ann(I), and I,J ∈ Spec(R). By hypothesis, R is not an integral domain. Hence, I 6= (0) and J 6= (0). As R is reduced, I2 6= (0), and so, it follows that I 6= Ann(I) = J. Note that IJ = (0). If r ∈ I ∩ J, then r2 ∈ IJ = (0) and since R is reduced, we obtain that r = 0 and so, I ∩J = (0). It is convenient to denote I by p1 and Ann(I) by p2. Note that p1 ∩ p2 = (0). We claim that Min(R) = {p1,p2}. If p ∈ Spec(R), then from p1 ∩ p2 = (0), it follows that p ⊇ pi for some i ∈{1, 2}. Since R is not an integral domain, we obtain that p1 and p2 are not comparable under the inclusion relation. The above arguments imply that Min(R) = {p1 = I,p2 = J}. 128 S. Visweswaran, P. T. Lalchandani / J. Algebra Comb. Discrete Appl. 8(2) (2021) 119–138 (3) ⇒ (1) We are assuming that Min(R) = {I,J}. It now follows from the proof of Proposition 2.15 that EA(R)∗ = {I,J}. Let T be a UFD. If A( T T p2 )∗ = EA( T T p2 )∗ for every prime element p of T , then we prove in Theorem 2.17 that T is a PID. Theorem 2.17. Let T be a UFD which is not a field. The following statements are equivalent: (1) For any prime element p of T, A( T T p2 )∗ = EA( T T p2 )∗. (2) T is a PID. (3) For any non-zero proper ideal I of T with I /∈ Max(T), I( T I )∗ = A( T I )∗ = EA( T I )∗. Proof. (1) ⇒ (2) Let p be a prime element of T . We claim that Tp ∈ Max(T). For the sake of convenience, let us denote T T p2 by R. Observe that Tp ∈ Spec(T) and let us denote T p T p2 by p. Note that p ∈ Spec(R), p = R(p + Tp2) is principal, p 6= (0 + Tp2) but p2 = (0 + Tp2) and we know from the proof of Lemma 2.11 that Z(R) = p. We are assuming that that A(R)∗ = EA(R)∗. Therefore, we obtain from the moreover part of Proposition 2.10 that p ∈ Max(R). As p = T p T p2 , we get that Tp ∈ Max(T). This is true for any prime element p of T. Let P ∈ Spec(T)\{(0)}. Since any non-zero non-unit of T can be expressed as the product of a finite number of prime elements of T, it follows that P ⊇ Tp for some prime element p of T . As Tp ∈ Max(T), we obtain that P = Tp ∈ Max(T). This shows that dimT = 1. Hence, any prime ideal of T is principal. Therefore, we obtain from [14, Exercise 10, page 8] that any ideal of T is principal. Therefore, T is a PID. (2) ⇒ (3) This follows from Example 2.8(3). (3) ⇒ (1) This is clear, since for any prime element p of T, Tp2 /∈ Max(T). Let T be a UFD which is not a field. If for every pair of non-associate prime elements p1,p2 of T , EA( T T p1p2 )∗ = A( T T p1p2 )∗, then we prove in Theorem 2.18 that T is a PID. Suppose that T has a prime element p such that any prime element of T is an associate of p in T. Let a be any non- zero non-unit of T . Then a = upn for some u ∈ U(T) and n ≥ 1. Hence, Ta ⊆ Tp. Therefore, Max(T) = Spec(T)\{(0)} = {Tp}. Let I be any non-zero proper ideal of T . Then I ⊆ Tp. From ∞⋂ n=1 Tpn = (0), we get that there exists n ∈ N such that I ⊆ Tpn but I 6⊆ Tpn+1. Let x ∈ I\Tpn+1. Then x = upn for some u ∈ U(T). This implies that pn = u−1x ∈ I. This proves that Tpn ⊆ I and so, I = Tpn. Thus any ideal of T is principal and so, T is a PID. Hence, in proving Theorem 2.18, we assume that T has at least two non-associate prime elements. Theorem 2.18. Let T be a UFD such that T has at least two non-associate prime elements. The following statements are equivalent: (1) For any non-associate prime elements p1,p2 of T, EA( TT p1p2 ) ∗ = A( T T p1p2 )∗. (2) T is a PID. (3) For any non-zero proper ideal I of T with I /∈ Max(T) , EA( T I )∗ = A( T I )∗. Proof. (1) ⇒ (2) We are assuming that for any two non-associate prime elements p1,p2 of T , EA(R)∗ = A(R)∗ with R = T T p1p2 . Let p be any prime element of T . By assumption, T has at least two non- associate prime elements. Let q be a prime element of T such p and q are non-associates in T. By (1), EA( T T pq )∗ = A( T T pq )∗. Observe that T T pq is a reduced ring with Min( T T pq ) = { T p T pq , T q T pq }. From EA( T T pq )∗ = A( T T pq )∗, we obtain from the moreover part of Proposition 2.15 that T p T pq , T q T pq ∈ Max(R) 129 S. Visweswaran, P. T. Lalchandani / J. Algebra Comb. Discrete Appl. 8(2) (2021) 119–138 and so, Tp,Tq ∈ Max(T). Thus for any prime element p of T , Tp ∈ Max(T). Now, it follows as in the proof of (1) ⇒ (2) of Theorem 2.17 that T is a PID. (2) ⇒ (3) This follows from Example 2.8(3). (3) ⇒ (1) This is clear, since for any non-associate prime elements p1,p2 of T , Tp1p2 /∈ Max(T). Recall from [10, Exercise 16, page 111] that a ring T is von Neumann regular if given a ∈ T , there exists b ∈ T such that a = a2b. If a is a non-zero non-unit of a von Neumann regular ring T , then from a = a2b, it follows that e = ab = a2b2 = e2. Hence, e is an idempotent element of T with e /∈ {0, 1}. It is known from (a) ⇔ (d) of [10, Exercise 16, page 111] that a ring T is von Neumann regular if and only if dimT = 0 and T is reduced. An idempotent element e of R with e /∈ {0, 1} is referred to as a non-trivial idempotent element. Let R be a von Neumann regular ring which is not a field. We verify in Corollary 2.19 that |EA(R)∗| < ∞ if and only if there exist n ≥ 2 and fields F1,F2, . . . ,Fn such that R ∼= F1 ×F2 ×···×Fn as rings. Corollary 2.19. Let R be a von Neumann regular ring which is not a field. The following statements are equivalent: (1) |EA(R)∗| < ∞. (2) There exist n ≥ 2 and fields F1,F2, . . . ,Fn such that R ∼= F1 ×F2 ×···×Fn as rings. Proof. (1) ⇒ (2) Since R is von Neumann regular, we obtain that Spec(R) = Max(R) = Min(R). Since R is reduced, we get that ⋂ m∈Max(R) m = (0). From R is not a field, it follows that |Max(R)| ≥ 2. We are assuming that |EA(R)∗| < ∞. Hence, we obtain from the remark which appears just preceding the statement of Proposition 2.15 that |Min(R) = Max(R)| < ∞. Let Max(R) = {mi|i ∈{1, 2, . . . ,n}}. Now, it follows as in the proof of the moreover part of Proposition 2.15 that R ∼= n∏ i=1 R mi as rings. Let i ∈ {1, 2, . . . ,n} and let us denote the field R mi by Fi. Thus there exist n ≥ 2 and fields F1,F2, . . . ,Fn such that R ∼= F1 ×F2 ×···×Fn as rings. (2) ⇒ (1) We are assuming that there exist n ≥ 2 and fields F1,F2, . . . ,Fn such that R ∼= F1×F2×···×Fn as rings. Let us denote the ring n∏ i=1 Fi by T . We know from Example 2.8(1) that I(T)∗ = A(T)∗ = EA(T)∗. Therefore, |EA(T)∗| = |I(T)∗| = 2n − 2. Hence, |EA(R)∗| = 2n − 2 < ∞. 3. Some results on EAG(R) Let G = (V,E) be a graph. G is said to be connected if for distinct vertices a,b ∈ V , there exists at least one path in G between a and b. Let G = (V,E) be a connected graph. Let a,b ∈ V with a 6= b. Recall from [5] that the distance between a and b, denoted by d(a,b) is defined as the length of a shortest path in G between a and b. We define d(a,a) = 0 and define the diameter of G, denoted by diam(G) as diam(G) = sup{d(a,b) | a,b ∈ V}. A simple graph G is said to be complete if every pair of distinct vertices of G are adjacent in G. Let n ≥ 1. A complete graph with n vertices is denoted by Kn [5, Definition 1.1.11]. Let R be a ring such that EA(R)∗ 6= ∅. The aim of this section is to discuss some results on EAG(R). First, we prove some results regarding the connectedness of EAG(R). Proposition 3.1. Let R be a ring such that EA(R)∗ 6= ∅. Let I,J ∈ EA(R)∗ be such that there is a path in EAG(R) between I and J. Then I and J are adjacent in EAG(R). In particular, if EAG(R) is connected and if |EA(R)∗| ≥ 2, then diam(EAG(R)) = 1. 130 S. Visweswaran, P. T. Lalchandani / J. Algebra Comb. Discrete Appl. 8(2) (2021) 119–138 Proof. Let I,J ∈ EA(R)∗ be such that there is a path in EAG(R) between I and J. We claim thatI and J are adjacent in EAG(R). Suppose that I and J are not adjacent in EAG(R). Let I0 = I − I1 − ···− In = J be a shortest path in EAG(R) between I and J. It is clear that n ≥ 2. Note that for all i ∈ {0, 1, . . . ,n − 1}, Ii and Ii+1 are adjacent in EAG(R). Hence, Ann(Ii) = Ii+1 and Ann(Ii+1) = Ii. If A ∈ EA(R)∗, then we know from (1) ⇒ (3) of Lemma 2.1 that A = Ann(Ann(A)). Therefore, I = I0 = Ann(Ann(I0)) = Ann(I1) = I2. This is a contradiction. Therefore, I and J are adjacent in EAG(R). We now verify the in particular statement of this proposition. Suppose that EAG(R) is connected and |EA(R)∗| ≥ 2. Let I,J ∈ EA(R)∗ be such that I 6= J. Since EAG(R) is connected, there exists a path in EAG(R) between I and J. Hence, we obtain from what is shown in the previous paragraph that I and J are adjacent in EAG(R). Therefore, it follows that diam(EAG(R)) = 1. Let G = (V,E) be a graph. Recall from [9, page 21] that a maximal connected subgraph of G is called a component of G. Let R be a ring such that EA(R)∗ 6= ∅. We prove in Corollary 3.3 that each component of EAG(R) is a complete graph with at most two vertices. We use Lemma 3.2 in the proof of Corollary 3.3. Lemma 3.2. Let R be a ring such that EA(R)∗ 6= ∅. Let I − J be an edge of EAG(R). Let A ∈ EA(R)∗\{I,J}. Then I and A are not adjacent in EAG(R) and J and A are not adjacent in EAG(R). Proof. Since I − J is an edge of EAG(R), we obtain that Ann(I) = J and Ann(J) = I. As A ∈ EA(R)∗, we know from (1) ⇒ (3) of Lemma 2.1 that Ann(Ann(A)) = A. As A /∈ {I,J}, it follows that Ann(A) /∈ {I,J}. Therefore, we obtain that I and A are not adjacent in EAG(R) and J and A are not adjacent in EAG(R). Corollary 3.3. Let R be a ring such that EA(R)∗ 6= ∅. If g is any component of EAG(R), then g is a complete graph with at most two vertices. In particular, if EAG(R) is connected, then EAG(R) is a complete graph with at most two vertices. Proof. Let g be any component of EAG(R). Suppose that |V (g)| ≥ 2. Let I,J ∈ V (g) with I 6= J. Then there exists a path in EAG(R) between I and J. Hence, we obtain from Proposition 3.1 that I and J are adjacent in EAG(R) and so, they are adjacent in g. Let A ∈ EA(R)∗\{I,J}. We know from Lemma 3.2 that I and A are not adjacent in EAG(R) and J and A are not adjacent in EAG(R). Therefore, A /∈ V (g) and so, V (g) = {I,J}. This proves that any component g of EAG(R) is a complete graph with at most two vertices. We next verify the in particular statement of this corollary. Suppose that EAG(R) is connected. Then EAG(R) is the only component of EAG(R) and so, EAG(R) is a complete graph with at most two vertices. Next, we assume that (R,m) is a SPIR and try to determine the structure of EAG(R). Proposition 3.4. Let R be a ring. Let p ∈ Spec(R) be such that p 6= (0) but p2 = (0). If p = Z(R), then EAG(R) is a graph with V (EAG(R)) = {p}. In particular, if (R,m) is a SPIR with m 6= (0) but m2 = (0), then EAG(R) is a graph with V (EAG(R)) = {m}. Proof. We know from (2) ⇒ (1) of Theorem 2.9 that EA(R)∗ = {p}. As V (EAG(R)) = EA(R)∗, we obtain that V (EAG(R)) = {p}. We next verify the in particular statement of this proposition. Let (R,m) be a SPIR with m 6= (0) but m2 = (0). As Z(R) = m, it follows that V (EAG(R)) = {m}. Proposition 3.5. Let R be a ring. Let p ∈ Spec(R) be such that p = Rp is principal, n ≥ 3 is least with the property that pn = (0), and Z(R) = p. Then the following statements hold: 131 S. Visweswaran, P. T. Lalchandani / J. Algebra Comb. Discrete Appl. 8(2) (2021) 119–138 (1) If n is odd, then EAG(R) has exactly [ n 2 ] components and each component is a complete graph with two vertices. (2) If n is even, then EAG(R) has exactly n 2 components g1,g2, . . . ,gn 2 such that gj is a complete graph with two vertices for each j ∈{1, . . . , n 2 − 1} and gn 2 is a complete graph on a single vertex. In particular, if (R,m) is a SPIR and n ≥ 3 is least with the property that mn = (0), then the statements (1) and (2) hold for EAG(R). Proof. Note that V (EAG(R)) = EA(R)∗ and we know from Proposition 2.10 that EA(R)∗ = {pi | i ∈ {1, 2, . . . ,n − 1}}. We know from the proof of Proposition 2.10 that for each i ∈ {1, 2, . . . ,n − 1}, Ann(pi) = pn−i and Ann(pn−i) = pi. Suppose that n ≥ 4. Let j ∈ {1, . . . , [ n 2 ] − 1}. As 2j < n and n is least with the property that pn = (0), it follows that pj 6= pn−j. Observe that pj and pn−j are adjacent in EAG(R). Let gj be the subgraph of EAG(R) induced by {pj,pn−j}. Then gj is a complete graph with two vertices and it follows from Corollary 3.3 that gj is necessarily a component of EAG(R). (1) Assume that n is odd. If n = 3, then V (EAG(R)) = {p,p2} and EAG(R) is a complete graph with two vertices. Let n ≥ 5. Note that p n−1 2 6= p n+1 2 . Observe that Ann(p n−1 2 ) = p n+1 2 and Ann(p n+1 2 ) = p n−1 2 . Hence, p n−1 2 and p n+1 2 are adjacent in EAG(R). Let g[ n 2 ] be the subgraph of EAG(R) induced by {p n−1 2 ,p n+1 2 }. Note that g[ n 2 ] is a complete graph with two vertices and it follows from Corollary 3.3 that g[ n 2 ] is necessarily a component of EAG(R). Observe that V (EAG(R)) = EA(R)∗ = [ n 2 ]⋃ j=1 {pj,pn−j} = [ n 2 ]⋃ j=1 V (gj). It is clear that for any distinct j1,j2 ∈ {1, 2, . . . , [ n2 ]}, V (gj1 ) ∩ V (gj2 ) = ∅. From the above arguments, it is clear that EAG(R) has exactly [ n 2 ] components and each component is a complete graph with two vertices. (2) Assume that n is even. It is clear that n ≥ 4. Observe that V (EAG(R)) = EA(R)∗ = ( n 2 −1⋃ j=1 {pj,pn−j}) ∪{p n 2 } = ( n 2 −1⋃ j=1 V (gj)) ∪{p n 2 }. Let gn 2 be the subgraph of EAG(R) induced by {p n 2 }. It is clear that for all distinct j1,j2 ∈{1, 2, . . . , n2}, V (gj1 )∩V (gj2 ) = ∅. From the above given arguments, it follows that EAG(R) has exactly n 2 components g1,g2, . . . ,gn 2 such that gj is a complete graph with two vertices for each j ∈{1, . . . , n 2 − 1} and gn 2 is a complete graph on a single vertex. We next verify the in particular statement of this proposition. Now, (R,m) is a SPIR and n ≥ 3 is least with the property that mn = (0). Let m ∈ m be such that m = Rm. Observe that Z(R) = m. Hence, the hypotheses of this proposition are satisfied and therefore, the statements (1) and (2) hold for EAG(R). Remark 3.6. Let R be a ring. Let p ∈ Spec(R) be such that p = Rp is principal, p 6= (0) but p is nilpotent. Let n ≥ 2 be least with the property that pn = (0). Then the following hold: (1) (R,p) is a SPIR if and only if p ∈ Max(R). (2) Suppose that Z(R) = p. Then EAG(R) is connected if and only if n ∈{2, 3}. Proof. (1) Assume that (R,p) is a SPIR. Then p ∈ Max(R) and indeed, it is the only prime ideal of R. Conversely, assume that p ∈ Max(R). Then it is shown in the proof of the moreover part of Proposition 2.10 that (R,p) is a SPIR. (2) If n ≥ 4, then [ n 2 ] ≥ 2. We know from Proposition 3.5 that EAG(R) has exactly [ n 2 ] components. Thus if EAG(R) is connected, then n ∈ {2, 3}. Assume that n ∈ {2, 3}. If n = 2, then we know from Proposition 3.4 that V (EAG(R)) = {p}. If n = 3, then we know from the proof of Proposition 3.5(1) that EAG(R) is a complete graph with two vertices. Therefore, if n ∈{2, 3}, then EAG(R) is connected. 132 S. Visweswaran, P. T. Lalchandani / J. Algebra Comb. Discrete Appl. 8(2) (2021) 119–138 Let R be a ring. Let p ∈ Spec(R) be such that p satisfies the hypotheses mentioned in the statement of Remark 3.6. If Z(R) = p, then in Theorem 3.7, we characterize R such that EAG(R) = AG(R). Theorem 3.7. Let R be a ring. Let p ∈ Spec(R) be such that p = Rp is principal. Let n ≥ 2 be least with the property that pn = (0). If Z(R) = p, then the following statements are equivalent: (1) EAG(R) = AG(R). (2) (R,p) is a SPIR and n ∈{2, 3}. Proof. (1) ⇒ (2) From the assumption EAG(R) = AG(R), we get that EA(R)∗ = V (EAG(R)) = V (AG(R)) = A(R)∗. We know from Proposition 2.10 that EA(R)∗ = {pi | i ∈ {1, . . . ,n − 1}}. Hence, A(R)∗ = {pi | i ∈ {1, . . . ,n− 1}}. We first verify that (R,p) is a SPIR. In view of the statement (1) of Remark 3.6, it is enough to prove that p ∈ Max(R). As A(R)∗ = EA(R)∗, we obtain from the moreover part of Proposition 2.10 that p ∈ Max(R). Therefore, (R,p) is a SPIR. It is known that AG(R) is connected and diam(AG(R)) ≤ 3 [7, Theorem 2.1]. Therefore, from EAG(R) = AG(R), we get that EAG(R) is connected. Hence, we obtain from Remark 3.6(2) that n ∈{2, 3}. (2) ⇒ (1) We are assuming that (R,p) is a SPIR and n ∈{2, 3}. If n = 2, then we know from the proof of Lemma 2.4 that EA(R)∗ = A(R)∗ = {p}. Hence, EAG(R) = AG(R) in this case. If n = 3, then we know from the proof of Lemma 2.4 that EA(R)∗ = A(R)∗ = {p,p2}. Thus V (EAG(R)) = V (AG(R)) = {p,p2}. We know from the proof of the statement (1) of Proposition 3.5 that EAG(R) is a complete graph with vertex set {p,p2}. For any ring T , EAG(T) is a subgraph of AG(T). Hence, AG(R) is a complete graph with vertex set {p,p2}. Therefore, EAG(R) = AG(R) in this case also. Therefore, if (R,p) is a SPIR and n ∈{2, 3}, then EAG(R) = AG(R). Let R be a ring. Let p ∈ Spec(R) be such that p = Rp is principal, p2 6= (0) but p3 = (0), and Z(R) = p. Then we know from the proof of Proposition 3.5(1) that EAG(R) is a complete graph with V (EAG(R)) = {p,p2}. We provide Example 3.8 to illustrate that in the above result, if the hypothesis that p is principal is omitted, then the conclusion can fail to hold. Example 3.8. Let R be the ring considered by D.D. Anderson and M. Naseer in [3, page 501]. Then I(R)∗ = A(R)∗ = EA(R)∗, AG(R) is connected with diam(AG(R)) = 2, and EAG(R) has exactly eight components and each component is a complete graph with two vertices. Proof. The ring R is also considered in Example 2.3 of this article. In the notation of Example 2.3, R = T I , where T = Z4[X,Y,Z], the polynomial ring in three variables X,Y,Z over Z4, and I is the ideal of T generated by {X2 − 2,Y 2 − 2,Z2,XY,XZ,Y Z − 2, 2X, 2Y, 2Z}. Let us denote X + I by x, Y + I by y, and Z + I by z. Observe that R is a local Artinian ring with m = Rx + Ry + Rz as its unique maximal ideal, m2 = {0 + I, 2 + I}, m3 = (0 + I), and |R| = 32. It is already noted in Example 2.3 that I(R)∗ = {m,m2,Rx,Ry,Rz,R(x + y),R(y + z),R(x + z),R(x + y + z),Rx + Ry,Ry + Rz,Rx + Rz,Rx + R(y + z),Ry + R(z + x),Rz + R(x + y),R(x + y) + R(y + z)}. It is verified in Example 2.3 that I(R)∗ = A(R)∗ = EA(R)∗. Let A,B ∈ A(R)∗ with A 6= B. Suppose that A and B are not adjacent in AG(R). From m3 = (0), we obtain that A−m2 −B is a path of length two between A and B in AG(R). This proves that diam(AG(R)) ≤ 2. Observe that (Ry)(Rz) 6= (0) and so, Ry and Rz are not adjacent in AG(R). This shows that diam(AG(R)) ≥ 2 and so, diam(AG(R)) = 2. We next verify that EAG(R) has exactly eight components and each component is a complete graph with two vertices. Let A1 = {A11 = m,A12 = m2},A2 = {A21 = Rx,A22 = Ry + Rz},A3 = {A31 = Ry,A32 = Rx + R(y + z)},A4 = {A41 = Rz,A42 = Rx + Rz},A5 = {A51 = R(x + y),A52 = R(y + z) + R(z + x)},A6 = {A61 = R(y + z),A62 = Rx + Ry},A7 = {A71 = R(z + x),A72 = Rz + R(x + y)}, and A8 = {A81 = R(x + y + z),A82 = Ry + R(x + z)}. Let gi be the subgraph of EAG(R) induced by Ai for each i ∈ {1, 2, . . . , 8}. We know from the proof of Example 2.3, that Ann(Ai1) = Ai2 and Ann(Ai2) = Ai1 for each i ∈ {1, 2, 3, . . . , 8}. It is clear that gi is a complete graph with two vertices for each i ∈ {1, 2, 3, . . . , 8} and it follows from Corollary 3.3 that each gi is a component of EAG(R). As 133 S. Visweswaran, P. T. Lalchandani / J. Algebra Comb. Discrete Appl. 8(2) (2021) 119–138 Ai = V (gi) for each i ∈{1, 2, 3, . . . , 8}, EA(R)∗ = 8⋃ i=1 Ai, Ai∩Aj = ∅ for all distinct i,j ∈{1, 2, 3, . . . , 8}, it follows that {gi | i ∈{1, 2, 3, . . . , 8}} is the set of all components of EAG(R). Let R be a reduced ring which is not an integral domain. If T is a ring which is not an integral domain, then it is already noted in the paragraph which appears just preceding the statement of Corollary 2.2 that EA(T)∗ 6= ∅. Hence, EA(R)∗ 6= ∅. In Corollary 3.10, we answer when EAG(R) is connected. In Corollary 3.11, we prove that EAG(R) = AG(R) if and only if R ∼= F1 ×F2 as rings, where Fi is a field for each i ∈{1, 2}. We use Lemma 3.9 in the proof of Corollary 3.10. Lemma 3.9. Let R be a reduced ring which is not an integral domain. Then EA(R)∗ 6= ∅ and any component g of EAG(R) is a K2. Proof. If R is not an integral domain (whether it is reduced or not), then it is already noted that EA(R)∗ 6= ∅. Let g be a component of EAG(R). Let I ∈ V (g). It follows from (1) ⇒ (3) of Lemma 2.1 that Ann(Ann(I)) = I. Since R is reduced, I2 6= (0) and so, I 6= Ann(I). With J = Ann(I), it follows that Ann(J) = I. Hence, I and J are adjacent in EAG(R). Therefore, J ∈ V (g). Also, I and J are adjacent in g. It follows from Corollary 3.3 that g is a complete graph with two vertices. Corollary 3.10. Let R be a reduced ring which is not an integral domain. The following statements are equivalent: (1) EAG(R) is connected. (2) |Min(R)| = 2. Proof. (1) ⇒ (2) Assume that EAG(R) is connected. We know from Lemma 3.9 that EAG(R) is a complete graph with two vertices. Hence, |EAG(R)| = 2. As V (EAG(R)) = EA(R)∗, we get that |EA(R)∗| = 2. Hence, it follows from (1) ⇒ (3) of Theorem 2.16 that |Min(R)| = 2. (2) ⇒ (1) Let Min(R) = {pi | i ∈ {1, 2}}. We know from the proof of Proposition 2.15 that EA(R)∗ = {pi | i ∈ {1, 2}}, Ann(p1) = p2, and Ann(p2) = p1. Therefore, p1 and p2 are adjacent in EAG(R). This shows that EAG(R) is a complete graph with two vertices and so, we obtain that EAG(R) is connected. Corollary 3.11. Let R be a reduced ring which is not an integral domain. The following statements are equivalent: (1) EAG(R) = AG(R). (2) R ∼= F1 ×F2 as rings, where Fi is a field for each i ∈{1, 2}. Proof. (1) ⇒ (2) We are assuming that EAG(R) = AG(R). We know from [7, Theorem 2.1] that AG(R) is connected. Therefore, EAG(R) is connected. Hence, we obtain from the proof of (1) ⇒ (2) of Corollary 3.10 that |Min(R)| = 2 and EA(R)∗ = Min(R). Let Min(R) = {p1,p2}. Now, A(R)∗ = V (AG(R)) = V (EAG(R)) = EA(R)∗. In such a case, we obtain from the proof of moreover part of Proposition 2.15 that R ∼= F1 ×F2 as rings, where Fi is a field for each i ∈{1, 2}. (2) ⇒ (1) Assume that R ∼= F1 × F2 as rings, where Fi is a field for each i ∈ {1, 2}. Let us denote the ring F1 × F2 by T. We know from Example 2.8(1) that I(T)∗ = A(T)∗ = EA(T)∗. Note that I(T)∗ = {m1 = (0) × F2,m2 = F1 × (0)}. Observe that Min(T) = {mi | i ∈ {1, 2}}. Now, it follows from the proof of (2) ⇒ (1) of Corollary 3.10 that EAG(T) is a complete graph with two vertices. Since EAG(T) is a subgraph of AG(T), we get that EAG(T) = AG(T) is a complete graph with two vertices. Since R ∼= T as rings, we obtain that EAG(R) = AG(R). Corollary 3.12. Let R be a reduced ring with |Min(R)| = n ≥ 2. Then EAG(R) has exactly 2n−1 − 1 components and each component is a K2. 134 S. Visweswaran, P. T. Lalchandani / J. Algebra Comb. Discrete Appl. 8(2) (2021) 119–138 Proof. We know from Proposition 2.15 that |V (EAG(R)) = EA(R)∗| = 2n − 2. Let t be the number of components of EAG(R). Let {gi | i ∈ {1, . . . , t}} be the set of all components of EAG(R). We know from Lemma 3.9 that gi is a K2 for each i ∈ {1, . . . , t}. Now, EA(R)∗ = t⋃ i=1 V (gi), |V (gi)| = 2 for each i ∈ {1, . . . , t}, V (gi) ∩V (gj) = ∅ for all distinct i,j ∈ {1, . . . , t}. Therefore, 2n − 2 = |EA(R)∗| = 2t and so, t = 2n−1 − 1. This proves that EAG(R) has exactly 2n−1 − 1 components and each component is a K2. Corollary 3.13. Let n ≥ 2 and let Ri be an integral domain for each i ∈ {1, 2, . . . ,n}. Let R = R1 ×R2 ×···×Rn. Then EAG(R) has exactly 2n−1 − 1 components and each component is a K2. Proof. Note that R is a reduced ring and {p1,p2, . . . ,pn} is the set of all minimal prime ideals of R, where for each i ∈ {1, 2, . . . ,n}, pi = I1 × ··· × Ii × ··· × In with Ii = (0) and Ij = Rj for all j ∈ {1, 2, . . . ,n}\{i}. Thus |Min(R)| = n and so, we obtain from Corollary 3.12 that EAG(R) has exactly 2n−1 − 1 components and each component is a complete graph with two vertices. Let R be a ring which is not reduced. We are not able to determine I,J ∈ EA(R)∗ such that EA(R)∗ = {I,J}. However, as a consequence of Corollary 3.3 and [7, Theorem 2.7], we characterize in Theorem 3.14 rings R with EA(R)∗ 6= ∅ such that EAG(R) = AG(R). Theorem 3.14. Let R be a ring such that EA(R)∗ 6= ∅. The following statements are equivalent: (1) EAG(R) = AG(R). (2) Either R ∼= F1 ×F2 as rings, where Fi is a field for each i ∈{1, 2} or (R,m) is a SPIR satisfying the property that if n ∈ N is least such that mn = (0), then n ∈{2, 3}. Proof. (1) ⇒ (2) We are assuming that EAG(R) = AG(R). We know from [7, Theorem 2.1] that AG(R) is connected. Therefore, EAG(R) is connected. Hence, we obtain from Corollary 3.3 that EAG(R) is complete. Therefore, AG(R) is complete and so, it follows from [7, Theorem 2.7] that one of the following holds: (a) R ∼= F1 × F2 as rings, where Fi is a field for each i ∈ {1, 2}; (b) Z(R) is an ideal of R with (Z(R))2 = (0); (c) (R,m) is a SPIR with m3 = (0) but m2 6= (0). Assume that (b) holds. As Z(R) is an ideal of R, Z(R) is necessarily a prime ideal of R. Let us denote Z(R) by p. Now, p 6= (0), p2 = (0), and p = Z(R). Therefore, we obtain from (2) ⇒ (1) of Theorem 2.9 that EA(R)∗ = {p}. From EAG(R) = AG(R), it follows that A(R)∗ = EA(R)∗ = {p}. Hence, we obtain from [7, Corollary 2.9(a)] that (R,m) (with m = p) is a SPIR with m2 = (0). Therefore, we obtain that either R ∼= F1 × F2 as rings, where Fi is a field for each i ∈ {1, 2} or (R,m) is a SPIR satisfying the property that if n ∈ N is least such that mn = (0), then n ∈{2, 3}. (2) ⇒ (1) Suppose that R ∼= F1 × F2 as rings. Then we obtain from (2) ⇒ (1) of Corollary 3.11 that EAG(R) = AG(R). Suppose that (R,m) is a SPIR satisfying the property that if n ∈ N is least such that mn = (0), then n ∈{2, 3}. Then we know from (2) ⇒ (1) of Theorem 3.7 that EAG(R) = AG(R). Let n ≥ 2 and let T = K[X1,X2, . . . ,Xn] be the polynomial ring in n variables X1,X2, . . . ,Xn over a field K. Let R = T T X1X2 . Observe that R is a reduced ring with Min(R) = { T X1 T X1X2 , T X2 T X1X2 } and thus |Min(R)| = 2. Hence, we obtain from (2) ⇒ (1) of Corollary 3.10 that EAG(R) is connected. We know from [18, Theorem 3, page 281] that each maximal ideal of T is of height n and hence, it follows that dimR = n − 1. It follows from [18, Corollary 1, page 279] that TX1 is the intersection of all maximal ideals of T that contain TX1. Therefore, it follows that Max(R) is infinite. If a ring is zero-dimensional which admits at least one non-zero exact annihilating ideal and if its exact annihilating-ideal graph is connected, then we prove in Proposition 3.16 that there is a bound on the number of its maximal ideals. We use Lemma 3.15 in the proof of Proposition 3.16. Lemma 3.15. Let R be a ring such that dimR = 0. Let n ≥ 2. If |Max(R)| ≥ n, then there exist zero-dimensional rings R1,R2, . . . ,Rn such that R ∼= R1 ×R2 ×···×Rn as rings. 135 S. Visweswaran, P. T. Lalchandani / J. Algebra Comb. Discrete Appl. 8(2) (2021) 119–138 Proof. We prove this lemma using induction on n. Suppose that |Max(R)| ≥ 2. Let us denote the ring R nil(R) by T . From |Max(R)| ≥ 2, it follows that |Max(T)| ≥ 2. Note that dimT = 0 and we know from [4, Proposition 1.7] that T is reduced. Therefore, T is a von Neumann regular ring which is not a field. Hence, T admits a non-trivial idempotent element a + nil(R). Since nil(R) is a nil ideal of R, we obtain from [13, Proposition 7.14, page 405] that there exists an idempotent element e of R such that a + nil(R) = e + nil(R). It is clear that e /∈{0, 1}. Note that the mapping f : R → Re×R(1−e) defined by f(r) = (re,r(1−e)) is an isomorphism of rings. Let us denote the ring Re by R1 and R(1−e) by R2. Let i ∈{1, 2}. Since Ri is a homomorphic image of R, it follows that dimRi = 0. Thus there exist zero- dimensional rings R1,R2 such that R ∼= R1 ×R2 as rings. Let n ≥ 3 and assume by induction that the lemma is true for n−1. Now, |Max(R)| ≥ n > n−1. By induction hypothesis, there exist zero-dimensional rings R′1, . . . ,R ′ n−1 such that R ∼= R′1 ×···×R′n−1 as rings. Since |Max(R)| ≥ n, |Max(R′i)| > 1 for at least one i ∈ {1, . . . ,n− 1}. Without loss of generality, we can assume that |Max(R′1)| > 1. Hence, by the case n = 2, there exist zero-dimensional rings R′11,R ′ 12 such that R ′ 1 ∼= R′11×R′12 as rings. Therefore, R ∼= R′11 × R′12 × R′2 ×···× R′n−1 as rings. Let R1 = R′11,R2 = R′12,R3 = R′2, . . . ,Rn = R′n−1. Then dimRi = 0 for each i ∈{1, 2, . . . ,n} and R ∼= R1 ×R2 ×R3 ×···×Rn as rings. Proposition 3.16. Let R be a zero-dimensional ring such that EA(R)∗ 6= ∅. If EAG(R) is connected, then |Max(R)| ≤ 2. Proof. We are assuming that dimR = 0, EA(R)∗ 6= ∅, and EAG(R) is connected. Suppose that |Max(R)| ≥ 3. Then it follows from Lemma 3.15 that there exist zero-dimensional rings R1,R2, and R3 such that R ∼= R1 ×R2 ×R3 as rings. Let us denote the ring R1 ×R2 ×R3 by T . Since R ∼= T as rings, we obtain that EAG(T) is connected. Hence, we obtain from Proposition 3.1 that for any I.J ∈ EA(T)∗ with I 6= J, I and J are adjacent in EAG(T). Let e1 = (1, 0, 0),e2 = (0, 1, 0), and e3 = (0, 0, 1). Observe that for all distinct i,j ∈ {1, 2, 3}, eiej = (0, 0, 0) and so, Tei ∈ A(T)∗. Moreover, for all i ∈ {1, 2, 3}, Ann(Ann(Tei)) = Tei and so, we obtain from (3) ⇒ (1) of Lemma 2.1 that Tei ∈ EA(T)∗. Observe that Ann(Te1) = Te2 + Te3 6= Te2 and so, Te1 and Te2 are not adjacent in EAG(T). This is a contradiction. Therefore, |Max(R)| ≤ 2. Let R be a ring such that dimR = 0 and |Max(R)| = 2. In Corollary 3.17, we characterize R such that EAG(R) is connected. Corollary 3.17. Let R be a ring such that dimR = 0 and |Max(R)| = 2. The following statements are equivalent: (1) EAG(R) is connected. (2) R ∼= F1 ×F2 as rings, where Fi is a field for each i ∈{1, 2}. Proof. (1) ⇒ (2) By hypothesis, dimR = 0 and |Max(R)| = 2. We know from Lemma 3.15 that there exist rings R1,R2 such that dimRi = 0 for each i ∈ {1, 2} and R ∼= R1 × R2 as rings. Since |Max(R)| = 2, it follows that |Max(Ri)| = 1 for each i ∈ {1, 2}. Let mi denote the unique maximal ideal of Ri for each i ∈ {1, 2}. Let us denote the ring R1 × R2 by T. As EAG(R) is connected, it follows that EAG(T) is connected. Note that {(0) ×R2,R1 × (0)} ⊆ EA(T)∗. We know from Corollary 3.3 that EAG(T) is a complete graph with two vertices. Therefore, EA(T)∗ = {(0) × R2,R1 × (0)}. We next verify that Ri is a field for each i ∈ {1, 2}. Suppose that R1 is not a field. Then m1 6= (0). Since Spec(R1) = {m1}, it follows from [4, Proposition 1.8] that nil(R1) = m1. Let x ∈ m1,x 6= 0. Let n ≥ 2 be least with the property that xn = 0. Then xn−1 ∈ Ann(x) and xn−1 6= 0. Observe that Ann(x) × (0) ∈ EA(T)∗ = {(0) ×R2,R1 × (0)}. This is impossible. Therefore, R1 is a field. Similarly, it can be shown that R2 is a field. Hence, R ∼= F1 ×F2 as rings, where Fi = Ri is a field for each i ∈{1, 2}. (2) ⇒ (1) We are assuming that R ∼= F1 ×F2 as rings, where Fi is a field for each i ∈ {1, 2}. Note that R is reduced and |Min(R)| = 2. Hence, we obtain from (2) ⇒ (1) of Corollary 3.10 that EAG(R) is connected. 136 S. Visweswaran, P. T. Lalchandani / J. Algebra Comb. Discrete Appl. 8(2) (2021) 119–138 Let T be a Dedekind domain. Let I be a non-zero proper ideal of T I /∈ Max(T) and let R = T I . It is shown in Example 2.8(2) that I(R)∗ = A(R)∗ = EA(R)∗. In Corollary 3.18, we characterize R such that EAG(R) is connected. Corollary 3.18. Let I be a non-zero proper ideal of a Dedekind domain T such that I /∈ Max(T). Let R = T I . The following statements are equivalent: (1) EAG(R) is connected. (2) |Max(R)| ≤ 2 and EAG(R) is complete. (3) Either R ∼= F1 ×F2 as rings, where Fi is a field for each i ∈{1, 2} or (R,M) is a SPIR and if k ≥ 2 is least with the property that Mk = (0 + I), then k ∈{2, 3}. Proof. (1) ⇒ (2) Since dimT = 1, it follows that dimR = 0. Hence, we obtain from Proposition 3.16 that |Max(R)| ≤ 2 and we obtain from Corollary 3.3 that EAG(R) is a complete graph with at most two vertices. (2) ⇒ (3) Suppose that |Max(R)| = 2. Since dimR = 0 and EAG(R) is connected, we obtain from (1) ⇒ (2) of Corollary 3.17 that R ∼= F1 ×F2 as rings, where Fi is a field for each i ∈ {1, 2} and in this case, I = m1m2 for some distinct m1,m2 ∈ Max(T). Suppose that |Max(R)| = 1. We know from the proof of Example 2.8(2) that I = mk for some k ≥ 2 and (R,M = m mk ) is a SPIR. It follows from [4, Corollary 9.4] that mi 6= mj for all distinct i,j ∈ N. Hence, k is least with the property that Mk = (0 +I). From the assumption that EAG(R) is connected, we obtain from Remark 3.6(2) that k ∈{2, 3}. (3) ⇒ (1) If R ∼= F1 ×F2 as rings, where Fi is a field for each i ∈ {1, 2}, then it follows from the proof of (2) ⇒ (1) of Corollary 3.11 that EAG(R) is a complete graph with two vertices. Suppose that (R,M) is a SPIR and if k ≥ 2 is least with the property that Mk = (0 + I), then k ∈ {2, 3}. Then it follows from the proof of Remark 3.6(2) that EAG(R) is a complete graph with at most two vertices. Therefore, EAG(R) is connected. Let n ≥ 2 be such that n is not a prime number. Since Z is a PID and hence, a Dedekind domain, and Zn ∼= ZnZ as rings, the following corollary is an immediate consequence of Corollary 3.18. Corollary 3.19. Let n ≥ 2 be not a prime number. Let R = Zn. Then EAG(R) is connected if and only if either n = p1p2 for some distinct prime numbers p1,p2 or n ∈{p2,p3} for some prime number p. Let G = (V,E) be a graph. Suppose that G admits a cycle. Recall from [5, page 159] that the girth of G, denoted by girth(G) is defined as the length of a shortest cycle in G. If G does not contain any cycle, then we define girth(G) = ∞. Recall from [5, Definition 1.2.2] that a clique of G is a complete subgraph of G. Suppose that there exists k ∈ N such that any clique of G is a clique on at most k vertices. Then the clique number of G, denoted by ω(G) is defined as the largest integer n ≥ 1 such that G contains a clique on n vertices [5, Definition, page 185]. We set ω(G) = ∞ if G contains a clique on n vertices for all n ≥ 1. Let G = (V,E) be a graph. Recall from [5, page 129] that a vertex coloring of G is a map f : V → S, where S is a set of distinct colors. A vertex coloring f : V → S is said to be proper if adjacent vertices of G receive distinct colors of S; that is, if a and b are adjacent in G, then f(a) 6= f(b). The chromatic number of G, denoted by χ(G) is the minimum number of colors needed for a proper vertex coloring of G [5, Definition 7.1.2]. It is well-known that for any graph G, ω(G) ≤ χ(G). Recall from [5] that a graph G is said to be weakly perfect if χ(G) = ω(G). A graph G is said to be perfect if any induced subgraph H of G is weakly perfect; that is, for any induced subgraph H of G, χ(H) = ω(H). Corollary 3.20. Let R be a ring such that EA(R)∗ 6= ∅. Then the following hold: (1) girth(EAG(R)) = ∞. (2) EAG(R) is perfect. 137 S. Visweswaran, P. T. Lalchandani / J. Algebra Comb. Discrete Appl. 8(2) (2021) 119–138 Proof. 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Algorithms Appl. 8(3) (2016) Article ID: 1650043 22 pages. 138 https://doi.org/10.1142/S1005386714000200 https://doi.org/10.1142/S1005386714000200 https://doi.org/10.1016/j.disc.2011.10.020 https://doi.org/10.1016/j.disc.2011.10.020 https://doi.org/10.1006/jabr.1993.1171 https://doi.org/10.1016/0021-8693(88)90202-5 https://doi.org/10.1142/S0219498811004896 https://doi.org/10.1142/S0219498811004896 https://doi.org/10.1142/S0219498811004902 https://doi.org/10.1142/S0219498811004902 https://doi.org/10.1080/00927872.2011.587489 https://doi.org/10.1080/00927872.2011.587489 https://doi.org/10.1007/s00209-009-0639-z https://doi.org/10.1007/s00209-009-0639-z https://dx.doi.org/10.22124/jart.2017.2400 https://dx.doi.org/10.22124/jart.2017.2400 https://doi.org/10.1142/S1793830916500439 https://doi.org/10.1142/S1793830916500439 https://doi.org/10.1142/S1793830916500439 Introduction Some basic properties of EA(R)* Some results on EAG(R) References