on the annihilator graph of a commutative ring

14 Pages·2013·0.15 MB·English

Communications in Algebra®, 42: 1–14, 2014
Copyright © Taylor & Francis Group, LLC
ISSN: 0092-7872 print/1532-4125 online
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DOI: 10.1080/00927872.2012.707262
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ON THE ANNIHILATOR GRAPH OF A COMMUTATIVE RING
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Ayman Badawi
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Department of Mathematics and Statistics, American University of Sharjah,
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Sharjah, UAE
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LetRbeacommutativeringwithnonzeroidentity,Z4R5beitssetofzero-divisors,and
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if a∈Z4R5, then let ann 4a5=8d∈R—da=09. The annihilator graph of R is the
14 R
(undirected) graph AG4R5 with vertices Z4R5∗=Z4R5\809, and two distinct vertices
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x and y are adjacent if and only if ann 4xy56=ann 4x5∪ann 4y5. It follows that
R R R
16 each edge (path) of the zero-divisor graph â4R5 is an edge (path) of AG4R5. In this
17 article, we study the graph AG4R5. For a commutative ring R, we show that AG4R5 is
18 connected with diameter at most two and with girth at most four provided that AG4R5
has a cycle. Among other things, for a reduced commutative ring R, we show that the
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annihilator graph AG4R5 is identical to the zero-divisor graph â4R5 if and only if R
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has exactly two minimal prime ideals.
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Key Words: Annihilator graph; Annihilator ideal; Zero-divisor graph.
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2000 Mathematics Subject Classification: Primary 13A15; Secondary 05C99. Q1
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1. INTRODUCTION
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29 Let R be a commutative ring with nonzero identity, and let Z4R5 be its set
30 of zero-divisors. Recently, there has been considerable attention in the literature
31 to associating graphs with algebraic structures (see [8, 11–14]). Probably the most
32 attention has been to the zero-divisor graph â4R5 for a commutative ring R. The
33 set of vertices of â4R5 is Z4R5∗, and two distinct vertices x and y are adjacent if
34 and only if xy = 0. The concept of a zero-divisor graph goes back to Beck [6],
35 who let all elements of R be vertices and was mainly interested in colorings. The
36 zero-divisor graph was introduced by David F. Anderson and Paul S. Livingston
37 in [3], where it was shown, among other things, that â4R5 is connected with
38 diam4â4R55 ∈ 801112139 and gr4â4R55 ∈ 83141ˆ9. For a recent survey article on
39 zero-divisor graphs, see [5]. In this article, we introduce the annihilator graph AG4R5
40 for a commutative ring R. Let a ∈ Z4R5 and let ann 4a5 = 8r ∈ R—ra = 09. The
R
41 annihilator graph of R is the (undirected) graph AG4R5 with vertices Z4R5∗ =
42 Z4R5\809, and two distinct vertices x and y are adjacent if and only if ann 4xy5 6=
R
43 ann 4x5∪ann 4y5. It follows that each edge (path) of â4R5 is an edge (path) of
R R
44 AG4R5.
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Received December 27, 2011; Revised June 14, 2012. Communicated by S. Bazzoni.
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Address correspondence to Ayman Badawi, Department of Mathematics and Statistics,
48 American University of Sharjah, P.O. Box 26666, Sharjah, UAE; E-mail: [e-mail]
1
2 BADAWI
49 In the second section, we show that AG4R5 is connected with diameter at most
50 two (Theorem 2.2). If AG4R5 is not identical to â4R5, then we show that gr4AG4R55
51 (i.e., the length of a smallest cycle) is at most four (Corollary 2.11). In the third
52 section, we determine when AG4R5 is identical to â4R5. For a reduced commutative
53 ring R, we show that AG4R5 is identical to â4R5 if and only if R has exactly two
54 distinct minimal prime ideals (Theorem 3.6). Among other things, we determine
55 when AG4R5 is a complete graph, a complete bipartite graph, or a star graph.
56 Let â be a (undirected) graph. We say that â is connected if there is a path
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between any two distinct vertices. For vertices x and y of â, we define d4x1y5 to be
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the length of a shortest path from x to y (d4x1x5 = 0 and d4x1y5 = ˆ if there is no
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path). Then the diameter of â is diam4â5 = sup8d4x1y5—x and y are vertices of â9.
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The girth of â, denoted by gr4â5, is the length of a shortest cycle in â (gr4â5 = ˆ if
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â contains no cycles).
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A graph â is complete if any two distinct vertices are adjacent. The complete
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graph with n vertices will be denoted by Kn (we allow n to be an infinite cardinal).
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A complete bipartite graph is a graph â which may be partitioned into two disjoint
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nonemptyvertexsetsAandB suchthattwodistinctverticesareadjacentifandonly
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if they are in distinct vertex sets. If one of the vertex sets is a singleton, then we call
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â a star graph. We denote the complete bipartite graph by Km1n, where —A— = m and
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—B— = n (again, we allow m and n to be infinite cardinals); so a star graph is a K11n
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and K11ˆ denotes a star graph with infinitely many vertices. Finally, let Km13 be the
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graph formed by joining â = Km13 (= A∪B with —A— = m and —B— = 3) to the star
71 1
graph â = K11m by identifying the center of â and a point of B.
72 2 2
Throughout, R will be a commutative ring with nonzero identity, Z4R5 its set
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of zero-divisors, Nil4R5 its set of nilpotent elements, U4R5 its group of units, T4R5 its
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total quotient ring, and Min4R5 its set of minimal prime ideals. For any A ⊆ R, let
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A∗ = A\809. We say that R is reduced if Nil4R5 = 809 and that R is quasi-local if R
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has a unique maximal ideal. The distance between two distinct vertices a1b of â4R5
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is denoted by d 4a1b5. If AG4R5 is identical to â4R5, then we write AG4R5 = â4R5;
78 â4R5
79 otherwise, we write AG4R5 6= â4R5. As usual, ú and ún will denote the integers and
80 integers modulo n, respectively. Any undefined notation or terminology is standard,
81 as in [9] or [7].
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2. BASIC PROPERTIES OF AG4R5
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85 Inthissection,weshowthatAG4R5isconnectedwithdiameteratmosttwo.If
86 AG4R5 6= â4R5, we show that gr4AG4R55 ∈ 83149. If —Z4R5∗— = 1 for a commutative
87 ring R, then R is ring-isomorphic to either Z or Z 6X7/4X25 and hence AG4R5 =
4 2
88 â4R5. Since commutative rings with exactly one nonzero zero-divisor are studied in
89 [2, 3, 10], throughout this article we only consider commutative rings with at least
90 two nonzero zero-divisors.
91 We begin with a lemma containing several useful properties of AG4R5.
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Lemma 2.1. Let R be a commutative ring.
94
95 (1) Let x1y be distinct elements of Z4R5∗. Then x−y is not an edge of AG4R5 if and
96 only if ann 4xy5 = ann 4x5 or ann 4xy5 = ann 4y5.
R R R R
ONTHEANNIHILATORGRAPHOFACOMMUTATIVERING 3
97 (2) If x−y is an edge of â4R5 for some distinct x1y ∈ Z4R5∗, then x−y is an edge of
98 AG4R5. In particular, if P is a path in â4R5, then P is a path in AG4R5.
99 (3) If x−y is not an edge of AG4R5 for some distinct x1y ∈ Z4R5∗, then ann 4x5 ⊆
R
100 ann 4y5 or ann 4y5 ⊆ ann 4x5.
R R R
101 (4) If ann 4x5 * ann 4y5 and ann 4y5 * ann 4x5 for some distinct x1y ∈ Z4R5∗, then
R R R R
102 x−y is an edge of AG4R5.
103 (5) If d 4x1y5 = 3 for some distinct x1y ∈ Z4R5∗, then x−y is an edge of AG4R5.
â4R5
104 (6) If x−y is not an edge of AG4R5 for some distinct x1y ∈ Z4R5∗, then there is a
105 w ∈ Z4R5∗\8x1y9 such that x−w−y is a path in â4R5, and hence x−w−y is also
106 a path in AG4R5.
107
108 Proof. (1) Suppose that x−y is not an edge of AG4R5. Then ann 4xy5 =
R
109 ann 4x5∪ann 4y5 by definition. Since ann 4xy5 is a union of two ideals, we have
R R R
110 ann 4xy5 = ann 4x5 or ann 4xy5 = ann 4y5. Conversely, suppose that ann 4xy5 =
R R R R R
111 ann 4x5 or ann 4xy5 = ann 4y5. Then ann 4xy5 = ann 4x5∪ann 4y5, and thus x−
R R R R R R
112 y is not an edge of AG4R5.
113
114 (2) Supposethatx−y isanedgeofâ4R5forsomedistinctx1y ∈ Z4R5∗.Then
115 xy = 0andhenceann 4xy5 = R.Sincex 6= 0andy 6= 0,ann 4x5 6= Randann 4y5 6=
R R R
116 R. Thus x−y is an edge of AG4R5. The “in particular” statement is now clear.
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(3) Suppose that x−y is not an edge of AG4R5 for some distinct x1y ∈
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Z4R5∗.Thenann 4x5∪ann 4y5 = ann 4xy5.Sinceann 4xy5isaunionoftwoideals,
119 R R R R
we have ann 4x5 ⊆ ann 4y5 or ann 4y5 ⊆ ann 4x5.
120 R R R R
121 (4) This statement is now clear by (3).
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(5) Suppose that d 4x1y5 = 3 for some distinct x1y ∈ Z4R5∗. Then
123 â4R5
ann 4x5 * ann 4y5andann 4y5 * ann 4x5.Hencex−y isanedgeofAG4R5by(4).
124 R R R R
125 (6) Suppose that x−y is not an edge of AG4R5 for some distinct x1y ∈
126 Z4R5∗. Then there is a w ∈ ann 4x5∩ann 4y5 such that w 6= 0 by (3). Since xy 6= 0,
R R
127 we have w ∈ Z4R5∗\8x1y9. Hence x−w−y is a path in â4R5, and thus x−w−y is
128 a path in AG4R5 by (2). ƒ
129
130 In view of (6) in the preceding lemma, we have the following result.
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132 Theorem 2.2. Let R be a commutative ring with —Z4R5∗— ≥ 2. Then AG4R5 is
133 connected and diam4AG4R55 ≤ 2.
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135 Lemma 2.3. Let R be a commutative ring, and let x1y be distinct nonzero elements.
136 Suppose that x−y is an edge of AG4R5 that is not an edge of â4R5 for some distinct
137 x1y ∈ Z4R5∗. If there is a w ∈ ann 4xy5\8x1y9 such that wx 6= 0 and wy 6= 0, then x−
R
138 w−y is a path in AG4R5 that is not a path in â4R5, and hence C 2 x−w−y−x is a
139 cycle in AG4R5 of length three and each edge of C is not an edge of â4R5.
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141 Proof. Suppose that x−y is an edge of AG4R5 that is not an edge of â4R5. Then
142 xy 6= 0. Assume there is a w ∈ ann 4xy5\8x1y9 such that wx 6= 0 and wy 6= 0. Since
R
143 y ∈ ann 4xw5\4ann 4x5∪ann 4w55, we conclude that x−w is an edge of AG4R5.
R R R
144 Since x ∈ ann 4yw5\4ann 4y5∪ann 4w55, we conclude that y−w is an edge of
R R R
4 BADAWI
145 AG4R5. Hence x−w−y is a path in AG4R5. Since xw 6= 0 and yw 6= 0, we have
146 x−w−y is not a path in â4R5. It is clear that x−w−y−x is a cycle in AG4R5 of
147 length three and each edge of C is not an edge of â4R5. ƒ
148
149 Theorem 2.4. Let R be a commutative ring. Suppose that x−y is an edge of AG4R5
150 that is not an edge of â4R5 for some distinct x1y ∈ Z4R5∗. If xy2 6= 0 and x2y 6= 0, then
151 there is a w ∈ Z4R5∗ such that x−w−y is a path in AG4R5 that is not a path in â4R5,
152 and hence C 2 x−w−y−x is a cycle in AG4R5 of length three and each edge of C is
153 not an edge of â4R5.
154
155 Proof. Suppose that x−y is an edge of AG4R5 that is not an edge of â4R5.
156 Then xy 6= 0 and there is a w ∈ ann 4xy5\4ann 4x5∪ann 4y55. We show w y 8x1y9.
R R R
157 Assume w ∈ 8x1y9. Then either x2y = 0 or y2x = 0, which is a contradiction. Thus
158 w y 8x1y9. Hence x−w−y is the desired path in AG4R5 by Lemma 2.3. ƒ
159
160 Corollary 2.5. Let R be a reduced commutative ring. Suppose that x−y is an edge
161 of AG4R5 that is not an edge of â4R5 for some distinct x1y ∈ Z4R5∗. Then there is a
162 w ∈ ann 4xy5\8x1y9 such that x−w−y is a path in AG4R5 that is not a path in â4R5,
R
163 and hence C 2 x−w−y−x is a cycle in AG4R5 of length three and each edge of C is
164 not an edge of â4R5.
165
166 Proof. Suppose that x−y is an edge of AG4R5 that is not an edge of â4R5 for
167 some distinct x1y ∈ Z4R5∗. Since R is reduced, we have 4xy52 6= 0. Hence xy2 6= 0
168 and x2y 6= 0, and thus the claim is now clear by Theorem 2.4. ƒ
169
In light of Corollary 2.5, we have the following result.
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Theorem 2.6. Let R be a reduced commutative ring, and suppose that AG4R5 6=
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â4R5. Then gr4AG4R55 = 3. Furthermore, there is a cycle C of length three in AG4R5
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such that each edge of C is not an edge of â4R5.
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In view of Theorem 2.4, the following is an example of a nonreduced
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commutative ring R where x−y is an edge of AG4R5 that is not an edge of â4R5
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for some distinct x1y ∈ Z4R5∗, but every path in AG4R5 of length two from x to y
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is also a path in â4R5.
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180
Example 2.7. Let R = ú . Then 2−6 is an edge of AG4R5 that is not an edge
181 8
of â4R5. Now 2−4−6 is the only path in AG4R5 of length two from 2 to 6
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and it is also a path in â4R5. Note that AG4R5 = K3, â4R5 = K112, gr4â4R55 = ˆ,
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gr4AG4R55 = 3, diam4â4R55 = 2, and diam4AG4R55 = 1.
184
185
The following is an example of a nonreduced commutative ring R such that
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AG4R5 6= â4R5 and if x−y is an edge of AG4R5 that is not an edge of â4R5 for some
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distinct x1y ∈ Z4R5∗, then there is no path in AG4R5 of length two from x to y.
188
189
Example 2.8.
190
191 (1) Let R = ú ×ú and let a = 401151b = 41125, and c = 40135. Then a−b and
2 4
192 c−b are the only two edges of AG4R5 that are not edges of â4R5, but there is
ONTHEANNIHILATORGRAPHOFACOMMUTATIVERING 5
193 no path in AG4R5 of length two from a to b and there is no path in AG4R5 of
194 length two from c to b. Note that AG4R5 = K213, â4R5 = K113, gr4AG4R55 = 4,
195 gr4â4R55 = ˆ, diam4AG4R5 = 2, and diam4â4R55 = 3.
196 (2) Let R = ú ×ú 6X7/4X25. Let x = X+4X25 ∈ ú 6X7/4X25, a = 401151b =
2 2 2
197 411x5, and c = 4011+x5. Then a−b and c−b are the only two edges of
198 AG4R5 that are not edges of â4R5, but there is no path in AG4R5 of length
199 two from a to b and there is no path in AG4R5 of length two from c to b.
200 Again, note that AG4R5 = K213, â4R5 = K113, gr4AG4R55 = 4, gr4â4R55 = ˆ,
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diam4AG4R5 = 2, and diam4â4R55 = 3.
202
203
Theorem 2.9. Let R be a commutative ring and suppose that AG4R5 6= â4R5. Then
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the following statements are equivalent:
205
206 (1) gr4AG4R55 = 4;
207 (2) gr4AG4R55 6= 3;
208 (3) If x−y is an edge of AG4R5 that is not an edge of â4R5 for some distinct x1y ∈
209 Z4R5∗, then there is no path in AG4R5 of length two from x to y;
210 (4) There are some distinct x1y ∈ Z4R5∗ such that x−y is an edge of AG4R5 that is
211 not an edge of â4R5 and there is no path in AG4R5 of length two from x to y;
212 (5) R is ring-isomorphic to either ú ×ú or ú ×ú 6X7/4X25.
2 4 2 2
213
214 Proof. 415 ⇒ 425. No comments.
215
216 425 ⇒ 435. Suppose that x−y is an edge of AG4R5 that is not an edge of
217 â4R5 for some distinct x1y ∈ Z4R5∗. Since gr4AG4R55 6= 3, there is no path in AG4R5
218 of length two from x to y.
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435 ⇒ 445. Since AG4R55 6= â4R5 by hypothesis, there are some distinct x1y ∈
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Z4R5∗ such that x−y is an edge of AG4R5 that is not an edge of â4R5, and hence
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there is no path in AG4R5 of length two from x to y by (3).
222
223 445 ⇒ 455. Suppose there are some distinct x1y ∈ Z4R5∗ such that x−y is
224 an edge of AG4R5 that is not an edge of â4R5 and there is no path in AG4R5 of
225 length two from x to y. Then ann 4x5∩ann 4y5 = 809. Since xy 6= 0 and ann 4x5∩
R R R
226 ann 4y5 = 809, by Lemma 2.3 we conclude that ann 4xy5 = ann 4x5∪ann 4y5∪
R R R R
227 8y9 such that y2 6= 0 or ann 4xy5 = ann 4x5∪ann 4y5∪8x9 such that x2 6= 0 (note
R R R
228 that if 8x1y9 ⊆ ann 4xy5, then x−xy−y is a path in AG4R5 of length two).
R
229 Without lost of generality, we may assume that ann 4xy5 = ann 4x5∪ann 4y5∪8y9
R R R
230 and y2 6= 0. Let a be a nonzero element of ann 4x5 and b be a nonzero element
R
231 of ann 4y5. Since ann 4x5∩ann 4y5 = 809, we have a+b ∈ ann 4xy5\4ann 4x5∪
R R R R R
232 ann 4y55, and hence a+b = y. Thus —ann 4x5— = —ann 4y5— = 2. Since xy2 = 0, we
R R R
233 have ann 4x5 = 801y29 and ann 4y5 = 801xy9. Since y2 +xy = y, we have 4y2 +
R R
234 xy52 = y2. Since xy3 = 0 and xy2 = x2y2 = 0, we have 4y2 +xy52 = y2 implies that
235 y4 = y2. Since y2 6= 0 and y4 = y2, we have y2 is a nonzero idempotent of R. Hence
236 ann 4xy5 = ann 4x5∪ann 4y5∪8y9 = 801y21xy1y9. Thus ann 4xy5 ⊆ yR and since
R R R R
237 yR ⊆ ann 4xy5, we conclude ann 4xy5 = yR = 801y21xy1y9. Since y2 +xy = y and
R R
238 y4 = y2, we have 4y2 +xy53 = y3 and hence y3 = y2. Thus y2R = y4yR5 = 801y29.
239 Since y2 is a nonzero idempotent of R and y2R is a ring with two elements, we
240 conclude that y2R is ring-isomorphic to ú . Let f ∈ ann 4y25. Then y2f = y4yf5 =
2 R
6 BADAWI
241 0, and thus yf ∈ ann 4y5. Hence either yf = 0 or yf = yx. Suppose yf = 0. Since
R
242 ann 4y5 = 801xy9, either f = 0 or f = xy. Suppose yf = yx. Then y4f −x5 = 0,
R
243 and thus f −x = 0 or f −x = xy. Hence f = x or f = x+xy. It is clear that
244 01x1xy1x+xy are distinct elements of R and thus ann 4y25 = 801x1xy1x+xy9.
R
245 Since ann 4y25 = 41−y25R, we have 41−y25R = 801x1xy1x+xy9. Since 41−y25R
R
246 is a ring with four elements, we conclude that 41−y25R is ring-isomorphic to either
247 ú or ú ×ú or F or ú 6X7/4X25. Since x−y is an edge of AG4R5 that is not an
4 2 2 4 2
248 edge of â4R5 and there is no path in AG4R5 of length two from x to y by hypothesis,
249 we conclude that R is non-reduced by Corollary 2.5. Since R is ring-isomorphic to
250 y2R×41−y25R and non-reduced, we conclude that R is ring-isomorphic to either
251 ú ×ú or ú ×ú 6X7/4X25.
2 4 2 2
252
455 ⇒ 415. See Example 2.8. ƒ
253
254
Corollary 2.10. Let R be a commutative ring such that AG4R5 6= â4R5, and assume
255
that R is not ring-isomorphic to ú ×B, where B = ú or B = ú 6X7/4X25. If E is an
256 2 4 2
edge of AG4R5 that is not an edge of â4R5, then E is an edge of a cycle of length three
257
in AG4R5.
258
259
Corollary 2.11. Let R be a commutative ring such that AG4R5 6= â4R5. Then
260
gr4AG4R55 ∈ 83149.
261
262
Proof. This result is a direct implication of Theorem 2.9. ƒ
263
264
265 3. WHEN IS AG4R5 IDENTICAL TO â4R5?
266
Let R be a commutative ring such that —Z4R5∗— ≥ 2. Then diam4â4R55 ∈
267
8112139 by [3, Theorem 2.3]. Hence, if â4R5 = AG4R5, then diam4â4R55 ∈ 81129 by
268
Theorem 2.2. We recall the following results.
269
270
Lemma 3.1.
271
272
(1) [3, the proof of Theorem 2.8] Let R be a reduced commutative ring that is not an
273 integral domain. Then â4R5 is complete if and only if R is ring-isomorphic to ú ×
2
274 ú .
2
275
(2) [10, Theorem 2.6(3)] Let R be a commutative ring. Then diam4â4R55 = 2 if and
276
only if either (i) R is reduced with exactly two minimal primes and at least three
277
nonzero zero divisors, or (ii) Z4R5 is an ideal whose square is not {0} and each
278
pair of distinct zero divisors has a nonzero annihilator.
279
280
We first study the case when R is reduced.
281
282
Lemma 3.2. Let R be a reduced commutative ring that is not an integral domain, and
283
let z ∈ Z4R5∗. Then:
284
285 (1) ann 4z5 = ann 4zn5 for each positive integer n ≥ 2;
R R
286 (2) If c+z ∈ Z4R5 for some c ∈ ann 4z5\809, then ann 4z+c5 is properly contained
R R
287 in ann 4z5 (i.e., ann 4c+z5 ⊂ ann 4z5). In particular, if Z4R5 is an ideal of R
R R R
288 and c ∈ ann 4z5\809, then ann 4z+c5 is properly contained in ann 4z5.
R R R
ONTHEANNIHILATORGRAPHOFACOMMUTATIVERING 7
289 Proof. (1) Let n ≥ 2. It is clear that ann 4z5 ⊆ ann 4zn5. Let f ∈ ann 4zn5. Since
R R R
290 fzn = 0 and R is reduced, we have fz = 0. Thus ann 4zn5 = ann 4z5.
R R
291
292 (2) Let c ∈ ann 4z5\809, and suppose hat c+z ∈ Z4R5. Since z2 6= 0, we have
R
293 c+z 6= 0,andhencec+z ∈ Z4R5∗.Sincec ∈ ann 4z5andRisreduced,wehavec y
R
294 ann 4c+z5. Hence ann 4c+z5 6= ann 4z5. Since ann 4c+z5 ⊂ ann 4z4c+z55 =
R R R R R
295 annR4z25 and annR4z25 = annR4z5 by (1), it follows that annR4c+z5 ⊂ annR4z5. ƒ
296
297 Theorem 3.3. Let R be a reduced commutative ring that is not an integral domain.
298 Then the following statements are equivalent:
299
300 (1) AG4R5 is complete;
301 (2) â4R5 is complete (and hence AG4R5 = â4R5);
302 (3) R is ring-isomorphic to ú ×ú .
2 2
303
304 Proof. 415 ⇒ 425. Let a ∈ Z4R5∗. Suppose that a2 6= a. Since ann 4a35 = ann 4a5
R R
305 by Lemma 3.2(1) and a3 6= 0, we have a−a2 is not an edge of AG4R5, a
306 contradiction. Thus a2 = a for each a ∈ Z4R5. Let x1y be distinct elements in Z4R5∗.
307 We show that x−y is an edge of â4R5. Suppose that xy 6= 0. Since x−y is an edge
308 of AG4R5, we have ann 4xy5 6= ann 4x5, and thus xy 6= x. Since x2 = x, we have
R R
309 ann 4x4xy55 = ann 4x2y5 = ann 4xy5, and thus x−xy is not an edge of AG4R5, a
R R R
310 contradiction. Hence xy = 0 and x−y is an edge of â4R5.
311
312 425 ⇒ 435. It follows from Lemma 3.1(1).
313
435 ⇒ 415. It is easily verified.
ƒ
314
315
Let R be a reduced commutative ring with —Min4R5— ≥ 2. If Z4R5 is an
316
ideal of R, then Min4R5 must be infinite, since Z4R5 = ∪8Q—Q ∈ Min4R59. For the
317
construction of a reduced commutative ring R with infinitely many minimal prime
318
ideals such that Z4R5 is an ideal of R, see [10, Section 5 (Examples)] and [1,
319
Example 3.13].
320
321
Theorem 3.4. Let R be a reduced commutative ring that is not an integral domain,
322
and assume that Z4R5 is an ideal of R. Then AG4R5 6= â4R5 and gr4AG4R55 = 3.
323
324
Proof. Let z ∈ Z4R5∗, c ∈ ann 4z5\809, and m ∈ ann 4c+z5\809. Then m ∈
325 R R
ann 4c+z5 ⊂ ann 4z5 by Lemma 3.2(2), and thus mc = 0. Since c2 6= 0, we have
326 R R
m 6= c, and hence c+z 6= m+z. Since 8c1m9 ⊆ ann 4z5 and z2 6= 0, we have c+z
327 R
and m+z are nonzero distinct elements of Z4R5. Since 4m+z54c+z5 = z2 6= 0, we
328
have 4c+z5−4m+z5 is not an edge of â4R5. Since c2 6= 0 and m2 6= 0, it follows
329
that 4c+m5 ∈ ann 4z25\4ann 4c+z5∪ann 4m+z55, and thus 4c+z5−4m+z5 is
330 R R R
an edge of AG4R5. Since 4c+z5−4m+z5 is an edge of AG4R5 that is not an edge
331
of â4R5, we have AG4R5 6= â4R5. Since R is reduced and AG4R5 6= â4R5, we have
332
333 gr4AG4R55 = 3 by Theorem 2.6. ƒ
334
335 Theorem 3.5. Let R be a reduced commutative ring with —Min4R5— ≥ 3 (possibly
336 Min4R5 is infinite). Then AG4R5 6= â4R5 and gr4AG4R55 = 3.
8 BADAWI
337 Proof. If Z4R5 is an ideal of R, then AG4R5 6= â4R5 by Theorem 3.4. Hence assume
338 that Z4R5 is not an ideal of R. Since —Min4R5— ≥ 3, we have diam4â4R55 = 3 by
339 Lemma 3.1(2), and thus AG4R5 6= â4R5 by Theorem 2.2. Since R is reduced and
340 AG4R5 6= â4R5, we have gr4AG4R55 = 3 by Theorem 2.6. ƒ
341
342 Theorem 3.6. Let R be a reduced commutative ring that is not an integral domain.
343 Then AG4R5 = â4R5 if and only if —Min4R5— = 2.
344
345 Proof. Suppose that AG4R5 = â4R5. Since R is a reduced commutative ring that
346 is not an integral domain, —Min4R5— = 2 by Theorem 3.5. Conversely, suppose that
347 —Min4R5— = 2. Let P , P be the minimal prime ideals of R. Since R is reduced, we
1 2
348 have Z4R5 = P ∪P and P ∩P = 809. Let a1b ∈ Z4R5∗. Assume that a1b ∈ P .
1 2 1 2 1
349 Since P ∩P = 809, neither a ∈ P nor b ∈ P , and thus ab 6= 0. Since P P ⊆ P ∩
1 2 2 2 1 2 1
350 P = 809, it follows that ann 4ab5 = ann 4a5 = ann 4b5 = P . Thus a−b is not an
2 R R R 2
351 edge of AG4R5. Similarly, if a1b ∈ P , then a−b is not an edge of AG4R5. If a ∈ P
2 1
352 and b ∈ P , then ab = 0, and thus a−b is an edge of AG4R5. Hence each edge of
2
353 AG4R5 is an edge of â4R5, and therefore, AG4R5 = â4R5. ƒ
354
355 Theorem 3.7. Let R be a reduced commutative ring. Then the following statements
356 are equivalent:
357
(1) gr4AG4R55 = 4;
358
(2) AG4R5 = â4R5 and gr4â4R55 = 4;
359
(3) gr4â4R55 = 4;
360
(4) T4R5 is ring-isomorphic to K ×K , where each K is a field with —K— ≥ 3;
361 1 2 i i
(5) —Min4R5— = 2andeachminimalprimeidealofRhasatleastthreedistinctelements;
362
(6) â4R5 = Km1n with m1n ≥ 2;
363
(7) AG4R5 = Km1n with m1n ≥ 2.
364
365
Proof. 415 ⇒ 425. Since gr4AG4R55 = 4, AG4R5 = â4R5 by Theorem 2.6, and thus
366
gr4â4R55 = 4. 425 ⇒ 435. No comments. 435 ⇔ 445 ⇔ 455 ⇔ 465 are clear by [2,
367
Theorem 2.2]. 465 ⇒ 475. Since (6) implies —Min4R5— = 2 by [2, Theorem 2.2], we
368
conclude that AG4R5 = â4R5 by Theorem 3.6, and thus gr4AG4R55 = gr4â4R55 = 4.
369
475 ⇒ 415. This is clear since AG4R5 is a complete bipartite graph and n1m ≥ 2. ƒ
370
371
Theorem 3.8. Let R be a reduced commutative ring that is not an integral domain.
372
Then the following statements are equivalent:
373
374 (1) gr4AG4R55 = ˆ;
375 (2) AG4R5 = â4R5 and gr4AG4R55 = ˆ;
376 (3) gr4â4R55 = ˆ;
377 (4) T4R5 is ring-isomorphic to Z ×K, where K is a field;
2
378 (5) —Min4R5— = 2 and at least one minimal prime ideal ideal of R has exactly two
379 distinct elements;
380 (6) â4R5 = K11n for some n ≥ 1;
381 (7) AG4R5 = K11n for some n ≥ 1.
382
383 Proof. 415 ⇒ 425. Since gr4AG4R55 = ˆ, AG4R5 = â4R5 by Theorem 2.6, and
384 thus gr4â4R55 = ˆ. 425 ⇒ 435. No comments. 435 ⇔ 445 ⇔ 455 ⇔ 465 are clear by
ONTHEANNIHILATORGRAPHOFACOMMUTATIVERING 9
385 [2, Theorem 2.4]. 465 ⇒ 475. Since (6) implies —Min4R5— = 2 by [2, Theorem 2.4], we
386 conclude that AG4R5 = â4R5 by Theorem 3.6, and thus gr4AG4R55 = gr4â4R55 = ˆ.
387 475 ⇒ 415. It is clear. ƒ
388
389 In view of Theorem 3.7 and Theorem 3.8, we have the following result.
390
391 Corollary 3.9. Let R be a reduced commutative ring. Then AG4R5 = â4R5 if and only
392 if gr4AG4R55 = gr4â4R55 ∈ 841ˆ9.
393
394 For the remainder of this section, we study the case when R is nonreduced.
395
396 Theorem 3.10. Let R be a nonreduced commutative ring with —Nil4R5∗— ≥ 2 and let
397 AG 4R5 be the (induced) subgraph of AG4R5 with vertices Nil4R5∗. Then AG 4R5 is
N N
398 complete.
399
400 Proof. Suppose there are nonzero distinct elements a1b∈Nil4R5 such that
401 ab6=0. Assume that ann 4ab5=ann 4a5∪ann 4b5. Hence ann 4ab5 = ann 4a5 or
R R R R R
402 ann 4ab5 = ann 4b5. Without lost of generality, we may assume that ann 4ab5 =
R R R
403 ann 4a5. Let n be the least positive integer such that bn = 0. Suppose that abk6=0
R
404 foreachk,1≤k<n.Thenbn−1 ∈ ann 4ab5\ann 4a5,acontradiction.Henceassume
R R
405 that k, 1 ≤ k < n is the least positive integer such that abk = 0. Since ab 6= 0, 1 <
406 k < n. Hence bk−1 ∈ ann 4ab5\ann 4a5, a contradiction. Thus a−b is an edge of
R R
407 AG 4R5. ƒ
N
408
409 In view of Theorem 3.10, we have the following result.
410
411 Corollary 3.11. Let R be a nonreduced quasi-local commutative ring with maximal
412 ideal Nil4R5 such that —Nil4R5∗— ≥ 2. Then AG4R5 is complete. In particular, AG4ú 5
2n
413 is complete for each n ≥ 3 and if q > 2 is a positive prime number of ú, then AG4ú 5
qn
414 is complete for each n ≥ 2.
415
416 The following is an example of a quasi-local commutative ring R with
417 maximal ideal Nil4R5 such that w2 = 0 for each w ∈ Nil4R5, diam4â4R55=2,
418 diam4AG4R55=1, and gr4AG4R55 = gr4â4R55 = 3.
419
420 Example 3.12. Let R = ú 6X1Y7/4X21Y25, x = X+4X21Y25 ∈ R, and y = Y +
2
421 4X21Y25 ∈ R. Then R is a quasi-local commutative ring with maximal ideal Nil4R5 =
422 4x1y5R. It is clear that w2 = 0 for each w ∈ Nil4R5 and diam4AG4R55 = 1 by
423 Corollary 3.11. Since Nil4R52 6= 809 and xyNil4R5 = 809, we have diam4â4R55 = 2 by
424 Lemma3.1(2).Sincex−xy−4xy+x5−x isacycleoflengththreeinâ4R5,wehave
425 gr4AG4R55 = gr4â4R55 = 3.
426
427 Theorem 3.13. Let R be a nonreduced commutative ring with —Nil4R5∗— ≥ 2, and let
428 â 4R5 be the induced subgraph of â4R5 with vertices Nil4R5∗. Then â 4R5 is complete
N N
429 if and only if Nil4R52 = 809.
430
431 Proof. If Nil4R52 = 809, then it is clear that â 4R5 is complete. Hence assume
N
432 that â 4R5 is complete. We need only show that w2 = 0 for each w∈Nil4R5∗.
N
10 BADAWI
433 Let w∈Nil4R5∗ and assume that w2 6= 0. Let n be the least positive integer such
434 that wn = 0. Then n ≥ 3. Thus w1wn−1+w are distinct elements of Nil4R5∗. Since
435 w4wn−1+w5 = 0andwn = 0,wehavew2 = 0,acontradiction.Thusw2 = 0foreach
436 w ∈ Nil4R5. ƒ
437
438 Theorem 3.14. LetRbeanonreducedcommutativering,andsupposethatNil4R52 6=
439 809. Then AG4R5 6= â4R5 and gr4AG4R55 = 3.
440
441 Proof. Since Nil4R52 6= 809, AG4R5 6= â4R5 by Theorem 3.10 and Theorem 3.13.
442 Thus gr4AG4R55 ∈ 83149 by Corollary 2.11. Let F = ú ×B, where B is ú or
2 4
443 ú 6X7/4X25. Since Nil4F52 = 809 and Nil4F5 6= 809, we have gr4AG4R55 6= 4 by
2
444 Theorem 2.9. Thus gr4AG4R55 = 3. ƒ
445
446 Theorem 3.15. Let R be a nonreduced commutative ring such that Z4R5 is not an
447 ideal of R. Then AG4R5 6= â4R5.
448
449 Proof. Since R is nonreduced and Z4R5 is not an ideal of R, diam4â4R55 = 3 by
450 [10, Corollary 2.5]. Hence AG4R5 6= â4R5 by Theorem 2.2. ƒ
451
452 Theorem 3.16. Let R be a nonreduced commutative ring. Then the following
453 statements are equivalent:
454
(1) gr4AG4R55 = 4;
455
(2) AG4R5 6= â4R5 and gr4AG4R55 = 4;
456
(3) R is ring-isomorphic to either ú ×ú or ú ×ú 6X7/4X25;
457 2 4 2 2
113
(4) â4R5 = K ;
458
(5) AG4R5 = K213.
459
460
Proof. 415 ⇒ 425. Suppose AG4R5 = â4R5. Then gr4â4R55 = 4, and R is ring-
461
isomorphic to D×B, where D is an integral domain with —D— ≥ 3 and B = ú
462 4
or ú 6X7/4X25 by [2, Theorem 2.3]. Assume that R is ring-isomorphic to D×ú .
463 2 4
Since —D— ≥ 3, there is an a ∈ D\80119. Let x = 401151y = 411251w = 4a125 ∈ R.
464
Then x1y1w are distinct elements in Z4R5∗, w4xy5 = 40105, wx 6= 40105, and wy 6=
465
40105. Thus x−w−y−x is a cycle of length three in AG4R5 by Lemma 2.3,
466
a contradiction. Similarly, assume that R is ring-isomorphic to D×ú 6X7/4X25.
467 2
Again, since —D— ≥ 3, there is an a ∈ D\80119. Let x = X+4X25 ∈ ú 6X7/4X25. Then
468 2
it is easily verified that 40115−4a1x5−411x5−40115 is a cycle of length three in
469
AG4R5, a contradiction. Thus AG4R5 6= â4R5. 425 ⇒ 435. It is clear by Theorem 2.9.
470
435 ⇔ 445. It is clear by [2, Theorem 2.5]. 445 ⇒ 455. Since (4) implies (3) by [2,
471
Theorem 2.5], it is easily verified that the annihilator graph of the two rings in (3)
472
is K213. 445 ⇒ 455. Since AG4R5 is a K213, it is clear that gr4AG4R55 = 4. ƒ
473
474 We observe that gr4â4ú 55 = ˆ, but gr4AG4ú 55 = 3. We have the following
475 8 8
result.
476
477
Theorem 3.17. Let R be a commutative ring such that AG4R5 6= â4R5. Then the
478
following statements are equivalent:
479
480 (1) â4R5 is a star graph;