The sum-annihilating essential ideal graph of a commutative ring 1. Introduction

19 Pages·2017·0.54 MB·English

CommunicationsinCombinatoricsandOptimization CCO
Vol. 1No. 2,2016pp.117-135
DOI:10.22049/CCO.2016.13555 Commun.Comb.Optim.
The sum-annihilating essential ideal graph of a
commutative ring
Abbas Alilou and Jafar Amjadi∗
DepartmentofMathematics,AzarbaijanShahidMadaniUniversity,Tabriz,I.R.Iran
[email protected]
Received: 10 March 2016; Accepted: 26 September 2016;
Available Online: 29 September 2016.
Communicated by Stephan Wagner
Abstract: Let R be a commutative ring with identity. An ideal I of a ring
R is called an annihilating ideal if there exists r∈R\{0} such that Ir=(0)
and an ideal I of R is called an essential ideal if I has non-zero intersection
with every other non-zero ideal of R. The sum-annihilating essential ideal
graph of R, denoted by AER, is a graph whose vertex set is the set of all
non-zero annihilating ideals and two vertices I and J are adjacent whenever
Ann(I)+Ann(J) is an essential ideal. In this paper we initiate the study of
thesum-annihilatingessentialidealgraph. Wefirstcharacterizeallringswhose
sum-annihilatingessentialidealgraphsarestarsorcompletegraphsandthenwe
establishsharpboundsonthedominationnumberofthisgraph. Furthermore,
wedetermineallisomorphismclassesofArtinianringswhosesum-annihilating
essentialidealgraphshavegenuszeroorone.
Keywords: Commutative rings, annihilating ideal, essential ideal, genus of a
graph
2010 Mathematics Subject Classification:13A15,16N40
1. Introduction
Thehistoryofstudyingagraphassociatedtoacommutativeringhasbeganby
thepaper[6,9,21],andthenitfollowedovercommutativeandnoncommutative
rings(seeforexample[4,5,18,22,23]). Sincethenahugenumberofworkshave
∗ Corresponding Author
(cid:13)c 2016 AzarbaijanShahidMadaniUniversity. Allrightsreserved.
118 The sum-annihilating essential ideal graph of a commutative ring
beenaddedtotheliteratureaboutgraphsassociatedtoalgebraicstructures. In
a recent study [11], the annihilating ideal graph, AG(R), is defined as follows:
The vertex set of this graph is the set of all non-zero annihilating ideals of
R, and two distinct vertices I and J are adjacent if and only if IJ = (0).
The interplay between the ring theoretic properties of a ring R and the graph
theoretic properties of its annihilating ideal graph has been investigated in [1–
3, 8, 10, 24]. In this paper, we continue the study of associating a graph to a
commutative ring.
Throughout this paper, all rings are assumed to be commutative rings with
identity that are not integral domain. If X is either an element or a subset of
R, then the annihilator of X is defined as Ann(X) = {r ∈ R | rX = 0}. An
ideal I of a ring R is called an annihilating ideal if Ann(I) (cid:54)= (0). We denote
the set of all maximal ideals, the set of all minimal prime ideals, and the set
of all associated prime ideals of a ring R by Max(R), Min(R) and Ass(R),
respectively. The ring R is said to be reduced if it has no non-zero nilpotent
element. AnidealI ofRiscalledanessentialidealifI hasnon-zerointersection
with every other non-zero ideal of R.
Let G be a simple graph with the vertex set V(G) and edge set E(G). The
degree of a vertex v ∈ V(G) is defined as d (v) = |{u ∈ V(G) | uv ∈ E(G)}|.
G
A graph G is regular or r-regular if d (v) = r for each vertex v of G. The
G
distance d (u,v) between two vertices u and v in a connected graph G is the
G
length of the shortest uv-path in G. The greatest distance between any pair
of vertices u and v in G is the diameter of G and is denoted by diam(G). A
universalvertexisavertexthatisadjacenttoallotherverticesofG. Ifagraph
Gcontainsauniversalvertexwithnoextraedge, thenGiscalledastar graph.
AcliqueofGisamaximalcompletesubgraphofGandthenumberofvertices
in the largest clique of G, denoted by ω(G), is called the clique number of G.
A graph is planar if it has a drawing without crossings. The genus of a graph
G, λ(G), is the minimum integer k such that the graph can be drawn without
crossing itself on a sphere with k handles (i.e. an oriented surface of genus
k). Thus, a planar graph has genus 0, because it can be drawn on a sphere
without self-crossing. Let χ(G) denote the chromatic number of G, that is,
the minimal number of colors needed to color the vertices of G so that no two
adjacent vertices have the same color. Obviously, χ(G)≥ω(G). We write K
n
(resp. K ) for the complete graph of order n (resp. infinite complete graph),
∞
P for the path of length n−1, and K for the complete bipartite graph
n m,n
with partite sets of size m and n. For terminology and notation not defined
here, the reader is referred to [25].
A set D ⊆ V(G) is a dominating set of G if every vertex of V(G)−D has a
neighbor in D. The domination number γ(G) is the minimum cardinality of
a dominating set in G. The concept of domination in graphs, with its many
A. Alilou and J. Amjadi 119
variations,isnowwellstudiedingraphtheoryandtheliteratureonthissubject
has been surveyed and detailed in the two books by Haynes, Hedetniemi, and
Slater [12, 13].
Motivatedbytheworkdoneonannihilatingidealgraphofaring,inthispaper
we introduce the sum-annihilating essential ideal graph of a commutative ring
whichisnotanintegraldomainandisdefinedasfollows: Thevertexsetofthis
graph isthe set ofall non-zeroannihilating idealsand two vertices I and J are
adjacent whenever Ann(I)+Ann(J) is an essential ideal. For convenience we
denote this graph by AE .
R
TheaimofthisarticleistostudysomepropertiesofAE . Wefirstcharacterize
R
all rings whose sum-annihilating essential ideal graphs are stars or complete
graphs and then we determine all isomorphism classes of Artinian rings whose
sum-annihilating essential ideal graph has genus zero or one.
We make use of the following observations and results in this paper.
Observation 1. ([11]) Let R be ring. Then every descending chain (resp. as-
cending chain) of non-zero annihilating ideals terminates if and only if R is Artinian
(resp. Noetherian).
Observation 2. IfI,J areidealsofR suchthatI isessentialandI ⊆J, thenJ
is essential.
Observation 3. For any ideal I of R, I+Ann(I) is an essential ideal.
Proof. Let I be an ideal of R such that I +Ann(I) is not essential. Then
J ∩(I +Ann(I))=(0) for some non-zero proper ideal J of R. It follows that
IJ = (0) and so J ⊆ Ann(I) which is a contradiction. Thus I +Ann(I) is
essential.
Observation 4. Let R be a ring and let I,J be two arbitrary non-zero proper
ideals of R. Then
(1) If Ann(I) is essential, then I is a universal vertex in AE .
R
(2) If IJ =(0), then I and J are adjacent in AE ,
R
(3) If I +J = R, Ann(I) (cid:54)= (0) and Ann(J) (cid:54)= (0), then Ann(I) and Ann(J) are
adjacent in AE .
R
Proof. (1) LetJ beanarbitraryvertexofAE . ByObservation2,Ann(I)+
R
Ann(J) is an essential ideal of R and so I and J are adjacent in AE .
R
Hence, I is a universal vertex.
120 The sum-annihilating essential ideal graph of a commutative ring
(2) Let IJ = (0). Then I ⊆ Ann(J) and so I +Ann(I) ⊆ Ann(I)+Ann(J).
It follows from Observations 2 and 3 that I and J are adjacent in AE .
R
(3) Let I+J =R. Then Ann(I)∩Ann(J)=(0) and we deduce from (2) that
Ann(I) and Ann(J) are adjacent in AE .
R
By Observation 4 (part (2)), AG(R) is a spanning subgraph of AE . Hence,
R
we can use some of results obtained on the annihilating-ideal graph to obtain
the same results for the sum-annihilating essential ideal graph of a ring. The
following examples show that the graphs AG(R) and AE are not the same.
R
Example 1. If R=Z , then AE is K −e and AG(R) is P .
12 R 4 4
Example 2. If p is a prime number and R=Z , then AE is K and AG(R) is
p4 R 3
the path P .
3
Theorem A. ([11]) If R is an Artinian ring, then every non-zero proper ideal I
of R is a vertex of AG(R).
Corollary 1. If R is an Artinian ring, then every non-zero proper ideal I of R is
a vertex of AE .
R
Theorem B. ([11])LetR bearing. Thenthefollowingstatementsareequivalent.
(1) AG(R) is a finite graph.
(2) R has only finitely many ideals.
(3) Every vertex of AG(R) has finite degree.
Moreover, AG(R) has n≥1 vertices if and only if R has n non-zero proper ideals.
Theorem C. ([11]) For every ring R, the annihilating ideal graph AG(R) is con-
nected with diam(AG(R))≤3.
Next result is an immediate consequence of Theorems B and C and the fact
that AG(R) is a spanning subgraph of AE .
R
Corollary 2. Let R be a ring. Then
(1) AE is a connected graph and diamAE ≤3.
R R
(2) The degree of each vertex in AE is finite if and only if the number of non-zero
R
proper ideals of R is finite.
A. Alilou and J. Amjadi 121
If R=F ×F ×F where F is a field for i=1,2,3, then AE is a connected
1 2 3 i R
graph with diamAE = 3 (see Figure 1), and hence the bound of Corollary 2
R
is sharp.
2. Properties of sum-annihilating essential ideal graphs
In this section, we investigate the basic properties of the sum-annihilating es-
sential ideal graphs. We recall that a ring R is decomposable if there exist
non-zero rings R and R such that R=R ×R , otherwise it is indecompos-
1 2 1 2
able.
FirstweclassifyallcommutativeringsRwhosesum-annihilatingessentialideal
graphs AE are stars. We start with the following lemma.
R
Lemma 1. Let R be a decomposable ring. Then AE is a star if and only if
R
R=F ×D where F is a field and D is an integral domain.
Proof. LetR=R ×R andletAE beastar. IfR isnotafieldfori=1,2
1 2 R i
andI isanon-zeroproperidealofR ,then(R ×(0),(0)×R ,I ×(0),(0)×I )
i i 1 2 1 2
is C which is a contradiction. Henceforth, we may assume that R is a field.
4 1
If R is an integral domain, then we are done. Suppose R is not an integral
2 2
domainandJ isanon-zeroannihilatingidealofR . Thenclearly(0)×R and
2 2
(0)×J are adjacent to R ×(0). Since AE is a star, R ×(0) is the center
1 R 1
of AE and so R ×(0) is adjacent to R ×J. It follows that Ann (R ×
R 1 1 R 1
(0))+Ann (R ×J) = (0)×R +(0)×Ann (J) = (0)×R is essential, a
R 1 2 R2 2
contradictionwith((0)×R )∩(R ×(0))=(0). ThusR isanintegraldomain.
2 1 2
Conversely, let R = F ×D where F is a field and D is an integral domain.
Then V(AE )={F ×(0),(0)×I |(0)(cid:54)=I(cid:2)D}. It is easy to see that AE is
R R
a star with center F ×(0).
Corollary 3. Let R be an Artinian ring with at least two non-zero annihilating
ideals. Then AE is a star if and only if AE (cid:39)K .
R R 2
Proof. Since K is a star graph, if part of the Corollary is clear.
2
Conversely, assumethatAE isastar. EitherR islocalorR isnotlocal. IfR
R
is local, then it follows from Lemma 12 that AE is isomorphic to K . If R is
R 2
not local, then R is decomposable. We deduce from Lemma 1 that R=F ×D
whereF isafieldandDisanintegraldomain. SinceRisArtinian,weconclude
that D is Artinian and so D is a field. Then clearly AE (cid:39)K .
R 2
Theorem 1. LetRbearingwithatleasttwonon-zeroproperideals. ThenAE
R
is a star if and only if one of the following holds:
122 The sum-annihilating essential ideal graph of a commutative ring
(a) R has exactly two non-zero proper ideals.
(b) R=F ×D where F is a field and D is an integral domain which is not a field.
(c) R has a minimal ideal I such that I2 = (0), I is not essential and I is the
annihilator of every non-zero proper ideal different from I.
Proof. If R has exactly two non-zero proper ideals, then R is Artinian and
so |V(AE )|=2, by Corollary 1. Since AE is connected, we have AE (cid:39)K
R R R 2
as desired. If R =F ×D where F is a field and D is an integral domain that
is not a field, then it follows from Lemma 1 that AE is a star. Now, let R
R
have a minimal ideal I such that I2 = (0), I is not essential and Ann(J) = I
for each non-zero proper ideal different from I. Since I ⊆ Ann (I), Ann (I)
R R
is essential by Observation 3 and so I is a universal vertex. Let J,K be two
arbitrary distinct vertices of AE different from I. If J is adjacent to K then
R
Ann (J)+Ann (K)=I is essential, a contradiction. Hence J and K are not
R R
adjacent and this implies that AE is a star.
R
Conversely, let AE be a star. If |V(AE )| < ∞, then by Observation 1, R
R R
is Artinian and it follows from Corollary 3 that R is isomorphic to F ×F .
1 2
Hence, R satisfies (a).
Now, let |V(AE )|=∞. Let I be the universal vertex of AE . We claim that
R R
I is a minimal ideal of R. Assume, to the contrary, that J is a non-zero ideal
of R such that J (cid:36) I. Then Ann (I)+Ann (J) = Ann (J). Since I and J
R R R
areadjacentinAE , Ann (J)isanessentialidealofR andsoJ isauniversal
R R
vertex of AE by Observation 4 and this leads to a contradiction. Hence, I is
R
a minimal ideal of R. Consider two cases.
Case 1. I2 (cid:54)=(0).
Then I2 = I. By Brauer’s Lemma ([15], p. 172, Lemma 10.22), R is decom-
posable. Since |V(AE )|≥3, we have R=F ×D, where F is a field and D is
R
an integral domain that is not a field. Hence R satisfies (b).
Case 2. I2 =(0).
Let J (cid:54)= I be an arbitrary vertex of AE . Then Ann(J) (cid:54)= J, for otherwise ,
R
Ann(J) is essential by Observation 3 and this implies by Observation 4 (part
(1)) that J = I, a contradiction. Since J.Ann(J) = (0), we deduce from
Observation4(part(2))thatJ andAnn(J)areadjacentinAE . SinceJ (cid:54)=I,
R
we have Ann (J)=I and I is not essential. Thus, R satisfies (c).
R
Next, we characterize all Artinian rings R whose sum-annihilating essential
ideal graphs are complete.
Lemma 2. If (R,m) is an Artinian local ring, then AE is a complete graph.
R
A. Alilou and J. Amjadi 123
Proof. Let I be an arbitrary vertex of AE . Then Ann (m) ⊆ Ann (I).
R R R
Since R is an Artinian local ring, m is nilpotent ([7], p.89, Proposition 8.4).
Hence, mn = (0) and mn−1 (cid:54)= (0) for some positive integer n and this implies
thatmn−1 ⊆Ann(I). LettbethesmallestpositiveintegersuchthatmtI =(0)
and mt−1I (cid:54)= (0). It follows that mt−1I ⊆ Ann(m) ∩ I and so Ann(m) is
essential. Since Ann(m) ⊆ Ann (I), we have that Ann (I) is essential. It
R R
follows from Observation 4 that I is a universal vertex. Since I is an arbitrary
vertex, we conclude that AE is a complete graph and the proof is complete.
R
Theorem 2. Let R be an Artinian ring. Then AE is a complete graph if and
R
only if one of the following holds:
1. R=F ×F where F and F are fields.
1 2 1 2
2. R is a local ring.
Proof. One side is clear. Let AE be a complete graph. Since R is Artinian,
R
R=R ×R ×···×R where R is an Artinian local ring for each 1≤i≤n.
1 2 n i
If n ≥ 3, then the vertices R ×(0)×···×(0) and R ×R ×(0)×···×(0)
1 1 2
are not adjacent in AE which is a contradiction. If n = 1, then R is an
R
Artinian local ring and we are done. Henceforth, we have n = 2. We claim
that R and R are fields. Assume, to the contrary, that R is not a field. Let
1 2 1
I be a non-zero proper ideal of R . By Theorem A, Ann (I ) (cid:54)= (0) and so
1 1 R1 1
Ann ((0)×R )+Ann (I ×R )=R ×(0). Since R ×(0) is not essential,
R 2 R 1 2 1 1
the vertices (0)×R and I ×R are not adjacent in AE , a contradiction.
2 1 2 R
Hence R and R are fields. This completes the proof.
1 2
Next result classifies all rings with at least one minimal ideal whose sum-
annihilating essential ideal graphs are complete bipartite graph.
Theorem 3. Let R be a ring with at least one minimal ideal. Then AE is a
R
complete bipartite graph with at least two vertices in each partition if and only if
R=D ×D where D and D are integral domains which are not fields.
1 2 1 2
Proof. LetR=D ×D whereD andD areintegraldomainswhicharenot
1 2 1 2
fields. Let X ={I×(0)|(0)(cid:54)=I(cid:2)D } and Y ={(0)×J|(0)(cid:54)=J(cid:2)D }. Since
1 2
D and D are integral domains, V(AE ) = X ∪Y. Clearly, |X|,|Y| ≥ 2, X
1 2 R
and Y are independent sets and every vertex of X is adjacent to every vertex
of Y. Thus, AE is a complete bipartite graph with desired property.
R
Conversely, let AE (cid:39) K where m,n ≥ 2. Suppose that I is a minimal
R m,n
ideal of R. If I2 =(0), then Ann (I) is essential by Observation 3 and so I is
R
a universal vertex, a contradiction. Thus I2 (cid:54)=(0). Since I is minimal, I2 =I.
124 The sum-annihilating essential ideal graph of a commutative ring
By Brauer’s Lemma, R = R ×R . We claim that R and R are integral
1 2 1 2
domains. AssumethatR isnotanintegraldomainandI isanon-zeroproper
2 2
ideal of R such that Ann (I ) (cid:54)= (0). As above, we have ((0)×I )2 (cid:54)= (0)
2 R2 2 2
and this implies that I (cid:54)= Ann (I ). By Observation 3, (R ,0), (0,I ) and
2 R2 2 1 2
(0,Ann (I )) induced a triangle in AE which is a contradiction. Hence, R
R2 2 R 2
is an integral domain. Similarly, R is an integral domain. Since m,n ≥ 2,
1
it follows from Theorem 1 that R and R are not fields. This completes the
1 2
proof.
Theorem 4. If R is an Artinian ring, then AE (cid:39) H ∨K where H is a
R m
multipartite graph and m∈N∪{∞}.
Proof. SinceR isArtinian, thenR=R ×···×R wheren=|Max(R)|. Let
1 n
X = {I ×···×I |I (cid:1)R for1≤i≤n}−{(0)×···×(0)}
1 n i i
X = {(R ×I ×···×I )|I (cid:2)R , 2≤i≤n}−{R ×···×R } and
1 1 2 n i i 1 n
X = {I ×···×I ×R ×I ×···×I |I (cid:1)R for1≤j ≤i−1
i 1 i−1 i i+1 n j j
andI (cid:2)R fori+1≤j ≤n}
j j
for each 2≤i≤n.
ItiseasytoverifythatV(AE )=X∪X ∪...∪X andthatthesetsX ,...,X
R 1 n 1 n
areindependentsets. LetH denotetheinducedsubgraphAE [X ∪...∪X ].
R 1 n
Then H is a n-partite graph. Assume now that I ×···×I ∈ X. Since R
1 n i
is an Artinian local ring, Ann (I ) is an essential ideal of R for each i. It
Ri i i
follows that Ann (I ×···×I ) is an essential ideal of R which implies that
R 1 n
I ×···×I is adjacent to all vertices of AE . In particular, the subgraph
1 n R
inducedbyX isaclique. ThusAE (cid:39)H∨K andtheproofiscomplete.
R |X|
Corollary 4. LetR=R ×···×R (n≥2)betheproductofArtinianlocalrings
1 n
R ,...,R . IfR hasonlyfinitelymanyidealsforeachi,thenχ(AE )=ω(AE )=
1 n i R R
n−1+(cid:81)n n , where n is the number of proper ideals of R .
i=1 i i i
Proof. Using notation of Theorem 4, we have |X|=(cid:81)n n −1 and AE =
i=1 i R
H∨K . SinceX isanindependentsetforeachi,weconcludethatω(AE )≤
|X| i R
χ(AE )≤n−1+(cid:81)n n .
R i=1 i
To prove the inverse inequality, we observe that the subgraph induced by the
set
X∪{R ×(0)×···×(0),(0)×R ×···×(0),...,(0)×···×(0)×R }
1 2 n
A. Alilou and J. Amjadi 125
is a clique which implies that χ(AE )≥ω(AE )≥n+|X|=n−1+(cid:81)n n ,
R R i=1 i
as desired. This completes the proof.
Corollary 5. If n=pα1pα2...pαm (m≥2) is the decomposition of the positive
1 2 m
integer n into primes, then χ(AEZn)=ω(AEZn)=m+α1α2...αm−1.
Corollary 6. If R = F ×···×F is the product of fields F ,...,F , then
1 n 1 n
χ(AE )=ω(AE )=n.
R R
Theorem 5. Let R be a ring. Then AE is a k- regular graph if and only if
R
R=F ×F whereF andF arefieldsandk=1orR isanArtinianlocalringwith
1 2 1 2
exactly k+1 non-zero proper ideals.
Proof. Onesideisclear. LetAE beak-regulargraph. ByCorollary2(part
R
2), R is Artinian and so R=R ×···×R where R is an Artinian local ring
1 n i
for each 1 ≤ i ≤ n . If R is a field for each i, then n = 2. For otherwise,
i
deg((0)×R ×···×R ) (cid:54)= deg(R ×(0)×···×(0)) contradicting regularity
2 n 1
of AE .
R
Thus, we may assume that R is not a field. Suppose that m is the maximal
1 1
ideal of R . Then m (cid:54)= (0) and so m ×(0)×···×(0) ∈ V(AE ). Since
1 1 1 R
Ann(m )isanessentialidealinR (seetheproofofLemma2),wededucethat
1 1
m ×(0)×···×(0)isadjacenttoeveryvertexofAE . SinceAE isk-regular,
1 R R
AE is a complete graph of order k+1. It follows from Theorem 2 that R is
R
an Artinian local and the proof is complete.
3. Domination number of AE
R
In this section, we investigate the domination number of the sum-annihilating
essential ideal graph of a ring. The first observation shows that γ(AE ) can
R
be arbitrary large.
Observation 5. If R=F ×F ×···×F (n≥3), where F is a field for each
1 2 n i
i∈{1,2,...,n}, then γ(AE )=n.
R
Proof. Let D = {F ,F ,...,F }. We show that D is a dominating set of
1 2 n
AE . Assume I ×···×I is a vertex of AE . Since I ×···×I (cid:54)= R, we
R 1 n R 1 n
have I = (0) for some i. Then Ann(I ×···×I )+Ann(F ) = R and so
i 1 n i
I ×···×I and F are adjacent in AE . Therefore D is a dominating set and
1 n i R
hence γ(AE )≤n.
R
Toshowthatγ(AE )≥n,letD beanyγ(AE )-setandFˆ =F ×···×F ×
R R i 1 i−1
(0)×F ×···×F for each i. Obviously deg(Fˆ)=1 and Fˆ is only adjacent
i+1 n i i
126 The sum-annihilating essential ideal graph of a commutative ring
to F for each i. Hence |D∩{F ,Fˆ}|≥1 foreach i which impliesthat |D|≥n
i i i
and so γ(AE )=n.
R
Next we provide some sufficient conditions for a ring R to have γ(AE )=1.
R
Proposition 1. LetRbearing. Thenγ(AE )=1ifandonlyifRhasanon-zero
R
proper ideal I such that Ann(I) is essential.
Proof. If R has a non-zero proper ideal I such that Ann(I) is essential, then
it follows from Observation 4 (1) that γ(AE )=1.
R
Conversely, let γ(AE ) = 1 and let {I} be a dominating set of AE . If I is
R R
not maximal, then there is a maximal ideal m such that I (cid:36)m. Since I and m
are adjacent, we deduce that Ann(I)+Ann(m) = Ann(I) is essential and we
aredone. Henceforth,weassumethatI isamaximalidealofR. FirstletRbe
local. If R has exactly one non-zero proper ideal, then clearly I = Ann(I) is
essential. Let J be a non-zero proper ideal of R different from I. Since J and
I are adjacent and since J (cid:36)I, we conclude that Ann(J)+Ann(I)=Ann(J)
is essential and we are done again.
Now R is not local. We consider two cases.
Case 1. R has exactly two maximal ideals.
LetI andmbetwodistinctmaximalidealsofRandletJ(R)=I∩m. Clearly,
J(R)(cid:36)I. If J(R)=(0), then by Chinese remainder theorem ([7], Proposition
1.10,pp7)wehaveR(cid:39) R×R whichcontradictsObservation5. ThusJ(R)(cid:54)=
I m
(0). Since I and J(R) are adjacent, we deduce that Ann(I)+Ann(J(R)) =
Ann(J(R)) is essential and we are done.
Case 2. R has at least three maximal ideals.
Let I,m and m be three distinct maximal ideals of R. Then we have Im (cid:54)=
1 2 1
(0), for otherwise we have Im ⊆ m and this implies that either I ⊆ m or
1 2 2
m ⊆ m which is a contradiction. Since Im (cid:36) I and since I and Im are
1 2 1 1
adjacent, we conclude that Ann(I)+Ann(Im ) = Ann(Im ) is essential and
1 1
the proof is complete.
Corollary 7. Let R=R ×···×R (n≥2) where R is not an integral domain
1 n i
for some i. Then γ(AE )=1 if and only if γ(AE )=1 for some i.
R Ri
Proof. Let γ(AE ) = 1 for some i, say i = 1. By Proposition 1, R has a
Ri 1
non-zero proper ideal I such that Ann(I ) is essential. Then clearly
1 1
Ann (I ×(0)×···×(0))=Ann (I )×R ×···×R
R 1 R1 1 2 n
is an essential ideal of R and hence γ(AE )=1 by Proposition 1.
R