Table Of ContentCOMAXIMAL GRAPH OF COMMUTATIVE RINGS
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HAMIDREZAMAIMANI,MARYAMSALIMI,ASIYEHSATTARI,ANDSIAMAKYASSEMI
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Dedicated to Ju¨rgen Herzog on the occasion of his 65th birthday
n
a
J Abstract. LetRbeacommutative ringwithidentity. LetΓ(R)beagraph
withverticesaselementsofR,wheretwodistinctverticesaandbareadjacent
1 if and only if Ra+Rb = R. In this paper we consider a subgraph Γ2(R) of
3 Γ(R) whichconsists of non-unitelements. We lookat the connectedness and
the diameter of this graph. We completely characterize the diameter of the
C] graphΓ2(R)\J(R). Inaddition,itisshownthatfortwofinitesemi-localrings
RandS,ifRisreduced, thenΓ(R)∼=Γ(S)ifandonlyifR∼=S.
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.
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t
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m 1. Introduction
[ For the sake of completeness, first we state some definitions and notions used
throughoutto keepthis paper asself containedas possible. We define a coloring of
1
v a graph G to be an assignment of colors (elements of some set) to the vertices of
8 G, one color to each vertex, so that adjacent vertices are assigned distinct colors.
1 If n colors are used, then the coloring is referred to as an n-coloring. If there
9
exists an n-coloring of a graph G, then G is called n-colorable. The minimum n
1
for which a graph G is n-colorable is called the chromatic number of G, and is
0
7 denoted by χ(G). For a graph G, the degree of a vertex v in G is the number
0 of edges of G incident with v. Recall that a graph is said to be connected if for
/ each pair of distinct vertices v and w, there is a finite sequence of distinct vertices
h
t v =v1,··· ,vn =wsuchthateachpair{vi,vi+1}isanedge. Suchasequenceissaid
a
to be a path and the distance, d(v,w), between connected vertices v and w is the
m
length of the shortest path connecting them. The diameter of a connected graph
:
v is the supremum of the distances between vertices. The diameter is 0 if the graph
i consists of a single vertex and a connected graph with more than one vertex has
X
diameter 1 if and only if it is complete; i.e., each pair of distinct vertices forms an
r
a edge. An r-partite graph is one whose vertex set can be partitioned into r subsets
so that no edge has both ends in any one subset. A complete r-partite graph is
one in which each vertex is joined to every vertex that is not in the same subset.
The complete bipartite (i.e., 2-partite) graph with part sizes m and n is denoted
by K . A graph in which each pair of distinct vertices is joined by an edge is
m,n
called a complete graph. We use K for the complete graph with n vertices. A
n
clique of a graph is its maximal complete subgraph and the number of vertices in
2000 Mathematics Subject Classification. 05C75,13A15.
Key words and phrases. connected graph,diameter,completegraph,cleanrings.
*Correspondingauthor. DepartmentofMathematics,UniversityofTehran,P.O.Box13145–
448Tehran,Iran.
H.R.MaimaniwassupportedinpartbyagrantfromIPMNo. 85050117.
S.YassemiwassupportedwassupportedbyagrantfromIPMNo. 85130214.
1
2 HAMIDREZAMAIMANI,MARYAMSALIMI,ASIYEHSATTARI,ANDSIAMAKYASSEMI
the largest clique of graph G, denoted by clique(G), is called the clique number
of G. Obviously χ(G) ≥ clique(G) for general graph G (see [4, page 289]). Let
G = (V ,E ) and G = (V ,E ) be two graphs with disjoint vertices set V and
1 1 1 2 2 2 i
edges set E . The join of G and G is denoted by G = G ∨G with vertices set
i 1 2 1 2
V ∪V and the set of edges is E ∪E ∪{xy|x∈V and y ∈V }.
1 2 1 2 1 2
From now on let R be a commutative ring with identity. In [3], Beck considered
Γ(R)asagraphwithverticesaselementsofR,wheretwodifferentverticesaandb
areadjacentifandonlyifab=0. Hestudiedfinitelycolorableringswiththisgraph
structure and showed that χ(Γ(R)) = clique(Γ(R)) for certain classes of rings. in
[2], Anderson and Naseer have made further study of finitely colorable rings and
have given an example of a finite local ring with 5=clique(Γ(R))<χ(Γ(R))=6.
In [7], Sharma andBhatwadekardefine another graphonR, Γ(R), with vertices
as elements of R, where two distinct vertices a and b are adjacent if and only if
Ra+Rb=R. They showed that χ(Γ(R))<∞ if and only if R is a finite ring. In
this case χ(Γ(R)) = clique(Γ(R)) = t+ℓ, where t and ℓ, respectively, denote the
number of maximal ideals of R and the number of units of R.
In this paper, we study further the graph structure defined by Sharma and
Bhatwadekar.
Let Γ (R) be the subgraph of Γ(R), generated by the units of R, and Γ (R) be
1 2
thesubgraphofΓ(R)generatedbynon-unitelements. Insection2,itisshownthat
the graph Γ (R)\J(R) is a complete bipartite if and only if the cardinal number
2
of the set Max(R) is equal 2 (see Theorem 2.2). Also we show that R is a finite
product of quasi-localrings if and only if R is clean and clique(Γ (R)\J(R))<∞
2
(see Theorem 2.5).
Insection3,themainresultsaysthatΓ (R)\J(R)isconnectedanddiam(Γ (R)\
2 2
J(R))≤3(see Theorem3.1). Inaddition, wecompletely characterizethe diameter
of the graph Γ (R)\J(R).
2
In the final section, it is shownthat for two finite semi-localrings R and S, if R
is reduced, then Γ(R)∼=Γ(S) if and only if R∼=S (see Corollary 4.6).
2. Bipartite graphs
Throughout this paper R will be a commutative ring with identity, U(R) its
groupofunits,J(R)itsJacobsonradical,andI(R)itssetofidempotents. AringR
issaidtobequasi-localifithasauniquemaximalideal;ifmistheuniquemaximal
ideal of R, we will often write (R,m).
LetΓ(R)bethegraphrepresentedbyRwithdefinitionofSharma-Behatwadekar.
LetΓ (R)=<U(R)>andΓ (R)=<R\U(R)>be the subgraphsofΓ(R). Then
1 2
it is easy to see that Γ(R)=Γ (R)∨Γ (R).
1 2
Lemma 2.1. The following hold:
(a) Γ (R) is a complete graph.
1
(b) a∈J(R) if and only if deg a=0.
Γ2(R)
Proof. Since (a) is clear we just prove (b). Suppose a ∈ J(R). Then for any
m ∈ Max(R), a ∈ m. If deg a 6= 0, then there exists b ∈ Γ (R) such that
Γ2(R) 2
Ra+Rb=R. On the other hand there exists n∈Max(R) with b∈n andso 1∈n
that is a contradiction.
Conversely,assume that deg a=0. Assume contrarya∈/ J(R). Then there
Γ2(R)
exists m ∈ Max(R) such that a ∈/ m. Thus Ra+m = R. Therefore there exists
COMAXIMAL GRAPH OF COMMUTATIVE RINGS 3
b ∈ m such that Ra+Rb = R. This contradicts our assumption. In the following
we study the cases where Γ (R)\J(R) is complete bipartite graph and where this
2
graph is n-partite.
(cid:3)
We knowthateachx∈U(R)is adjacenttoeveryvertexofΓ(R)anditisshown
thateachx∈J(R) isanisolatedvertexofΓ (R). Thus the mainpartofthe graph
2
Γ(R) is the subgraph Γ (R)\J(R). For this reason the main aim of this paper is
2
to study the structure of this subgraph.
Theorem 2.2. The following are equivalent:
(i) Γ (R)\J(R) is a complete bipartite graph.
2
(ii) The cardinal number of the set Max(R) is equal 2.
Proof. (ii)⇒(i). Let Max(R)={m ,m }. Thus the vertices set of Γ (R)\J(R) is
1 2 2
equal to the set (m \m )∪(m \m ). Let a ∈ m \m and b ∈ m \m . Thus
1 2 2 1 1 2 2 1
Ra+Rb*m ∪m and so Ra+Rb=R.
1 2
(i)⇒(ii). Suppose Γ (R)\J(R) is a complete bipartite graph with two part V
2 1
and V . Set M = V ∪J(R) and M = V ∪J(R). We show that M and M are
2 1 1 2 2 1 2
two maximal ideals of R and Max(R) = {M ,M }. Let x,y ∈ M = V ∪J(R).
1 2 1 1
Consider the following three cases:
Case 1. Assume that x,y ∈J(R). Then x−y ∈J(R) and so x−y ∈M .
1
Case 2. Assume thatx∈J(R)andy ∈V . Then x−y∈/ J(R). If x−y∈U(R),
1
thenRx+Ry=Randsoweobtainacontradiction. Ifx−y ∈M ,thenx−y ∈V
2 2
andsoR(x−y)+Ry=R. Thus Rx+Ry=R whichis a contradiction. Therefore
x−y ∈V ⊆M .
1 1
Case 3. Assume that x,y ∈V . If x−y ∈ J(R) then there is nothing to prove.
1
Therefore we assume x−y ∈/ J(R). With the same proof as case 2, the assertion
holds.
Now suppose that r ∈ R and x ∈ M . If x ∈ J(R), then clearly rx ∈ M .
1 1
Therefore suppose that x ∈/ J(R). Also rx is not unit. Suppose that rx ∈ M .
2
Then rx ∈ V and so R(rx)+Rx = R. Thus x is a unit element of R which is a
2
contradiction. So rx∈M .
1
To now we showed that M is an ideal of R. By the structure of Γ(R), for any
1
x∈R\M , we have M +Rx=R. This implies that M is a maximal ideal.
1 1 1
With the sameargumentM is amaximalidealofR. NowifN ∈Max(R)then
2
N ⊆M ∪M and so N =M or N =M .
1 2 1 2
This finishes the proof.
(cid:3)
Proposition 2.3. Let n>1. Then the following hold:
(a) If |Max(R)|=n<∞, then the graph Γ (R)\J(R) is n-partite.
2
(b) If the graph Γ (R)\J(R) is n-partite, then |Max(R)| ≤ n. In this case if
2
the graph Γ (R)\J(R) is not (n−1)-partite, then |Max(R)|=n.
2
Proof. (a). LetMax(R)={m ,··· ,m }andsetV =m \J(R)andforeachi≥2,
1 n 1 1
Vi =mi\∪tt==i1−1mt. Using Prime Avoidence Theorem,Vi 6=∅ for eachi. It is easy
to see that any two vertices belong to V are not adjacent.
i
(b). Let V ,··· ,V be the n parts of vertices of Γ (R)\J(R). Assume contrary
1 n 2
|Max(R)|>nandletm ,··· ,m ∈Max(R). Foranyi,choosex ∈m \∪ m .
1 n+1 i i j6=i j
Thenitiseasytoseethat{x ,··· ,x }isacliqueinΓ (R)\J(R). BythePigeon
1 n+1 2
4 HAMIDREZAMAIMANI,MARYAMSALIMI,ASIYEHSATTARI,ANDSIAMAKYASSEMI
Hole Principal, two of x ’s should belong to one of V ’s, that is a contradiction.
i i
Therefore|Max(R)|≤n. NowsupposethatΓ (R)\J(R)isnot(n−1)-partiteand
2
|Max(R)|=m<n. By(a) the graphwillbe m-partiteandthis is a contradiction.
(cid:3)
Proposition 2.4. Let R be a ring with |Max(R)|≥2. Then the following hold:
(a) If Γ (R)\J(R) is a complete n-partite graph, then n=2.
2
(b) If there exists a vertex of Γ (R)\J(R) which is adjacent to every other vertex
2
then R∼=Z ×F, where F is a field.
2
Proof. Let m ,m be two maximal ideals of R. Since the elements of m \J(R) are
1 2 i
not adjacent, and at least one element of m \J(R) is adjacent to one element of
1
m \J(R), so m \J(R) and m \J(R) are subsets of two distinct parts of Γ (R).
2 1 2 2
Thatmeans(m \J(R))∩(m \J(R))=∅. WeclaimthatJ(R)=m ∩m . Inother
1 2 1 2
case, J(R) ( (m ∩m ) and so there exists x ∈ (m ∩m )\J(R). This elements
1 2 1 2
belongs to m \J(R) and m \J(R), that is a contradiction. Therefore we obtain
1 2
J(R)=m ∩m and so |Max(R)|=2. Now by theorem 2.2 we have n=2.
1 2
(b). Let x be a non-unit element of R which is adjacent to every other vertex
of Γ (R)\J(R). Since x is comaximal with each nonunit outside the Jacobson
2
radical, x is idempotent, J(R) = (0) and m = {0,x} is a maximal ideal. Thus for
each nonunit s ∈ R\m, having xR+sR = R implies (1−x)sR = (1−x)R and
this implies (1−x)R=F is a field. Hence R∼=Z ×F.
2
(cid:3)
A ring is said to be clean if each of its elements can be written as the sum of
a unit and an idempotent cf. [6] (see also [1]). For example, a quasi-local ring is
clean. The following result gives an application of Sharma-Bhatwadegar graph to
characterize clean rings.
Theorem 2.5. For the ring R, the following are equivalent:
(a) R is a finite product of quasi-local rings.
(b) R is clean and clique(Γ (R)\J(R)) is finite.
2
Proof. (a)⇒(b). Let R = R ×···×R where each R is quasi-local with unique
1 n i
maximal ideal m . Set N = R × ··· × R × m × R × ··· × R for any
i i 1 i−1 i i+1 n
i=1,··· ,n. TheneachN belongstoMax(R). Foranyichoosex ∈N \∪ N .
i i i ℓ6=i ℓ
Then it is easy to see that Rx +Rx = R for all i 6= j. In addition by using
i j
the Pigeon Hole Principal, there is no any n+1 family elements of ∪n N which
i=1 i
pairwise adjacent. Thus clique(Γ (R)\J(R))=n<∞.
2
On the other hand, each R is clean and so by [1, Proposition 2(3)], R is clean.
i
(b)⇒(a). Supposethatclique(Γ (R)\J(R))isfinite. AssumecontrarythatI(R)
2
has infinitely many idempotent elements then by [7, Lemma 2.1] there exists an
infinite sequencee ,e ,··· ofnon-trivialidempotentsinΓ (R)\J(R)suchthatthe
1 2 2
set S consisting of elements e (i≥1) is an infinite clique. This is a contradiction.
i
(cid:3)
3. Diameter of the graph
In this section we completely characterize the diameter of Γ (R)\J(R). The
2
following result shows that Γ (R)\J(R) is a connected graph and its diameter is
2
not greater than 3.
COMAXIMAL GRAPH OF COMMUTATIVE RINGS 5
Theorem 3.1. The graph Γ (R)\J(R) is connected, and diam(Γ (R)\J(R))≤3.
2 2
Proof. Let a,b∈(R\U(R))\J(R). We consider two cases:
Case 1 Assume that ab ∈/ J(R). There exists x ∈ (R\U(R))\J(R) such that
Rab+Rx=R. Thus Ra+Rx=Rb+Rx=R. So we have the path a—x—b, and
so d(a,b)≤2.
Case 2 Assume that ab ∈ J(R). Set S = {m|m ∈ Max(R),a ∈ m} and S =
a b
{m|m ∈ Max(R),b ∈ m}. Clearly, Max(R) = S ∪S . Now suppose that x is
a b
adjacent to a in Γ (R). Then x ∈/ J(R). If a ∈ m, then x ∈/ m and so x ∈ n ∈
2
Max(R), where n∈S \S . Thus bx∈/ J(R). Therefore by Case 1, d(b,x)≤2 and
b a
so d(a,b)≤3.
(cid:3)
Lemma 3.2. diam(Γ (R)\J(R))=1 if and only if R∼=Z ×Z .
2 2 2
Proof. Ifdiam(Γ (R)\J(R))=1,thenΓ (R)\J(R)iscompletegraph. Thusthere
2 2
exists a vertex of Γ (R)\J(R) which is adjacent to every other vertex. Therefore
2
R∼=Z ×F,whereF isafieldbyProposition2.4(b). SinceΓ (R)\J(R)iscomplete,
2 2
we have that F ∼=Z . Thus R∼=Z ×Z .
2 2 2
It is easy to see that for R∼=Z ×Z , diam(Γ (R)\J(R))=1.
2 2 2
(cid:3)
Our next result characterizes the graphs where diam(Γ (R)\J(R))=2.
2
Proposition 3.3. Assume that R is not local. The diameter of the graph Γ (R)\
2
J(R) is equal 2 if and only if one of the following holds:
(a) J(R) is a prime ideal.
(b) |Max(R)|=2 and R≇Z ×Z .
2 2
Proof. Note that if J(R) is prime and R is semi-local (i.e. has finite number of
maximal ideals), then R will be local. Let J(R) be a prime ideal and a,b ∈/ J(R).
Then ab ∈/ J(R), and so by the same argument as Theorem 3.1, there exists x ∈
(R\U(R))\J(R), such that a—x—b is a path. Thus diam(Γ (R)\J(R))≤2. If
2
diam(Γ (R)\J(R)) = 1, then by previous result R ∼= Z ×Z . But J(Z ×Z ) is
2 2 2 2 2
not a prime ideal. That is a contradiction.
Nowlet|Max(R)|=2andR≇Z ×Z ,then byTheorem2.2,Γ (R)\J(R)is a
2 2 2
complete bipartite graphwhere at least one of the parts has at least two elements.
Therefore diam(Γ (R)\J(R))=2.
2
Conversely, suppose that diam(Γ (R)\J(R)) = 2 and J(R) is not prime. let
2
a,b ∈/ J(R) but ab ∈ J(R). We claim that a and b are adjacent. Otherwise, there
exists t in Γ (R) such that Ra+Rt = Rb+Rt = R. Thus Rab+Rt = R and
2
so ab ∈/ J(R) which is a contradiction. Therefore Ra+Rb = R and so for some
p,q∈R,pa+qb=1. SetS =R/J(R)anda =pa+J(R)andb =qb+J(R). Then
1 1
a b =0anda +b =1 . Thereforea andb areidempotentelementsinS,andso
1 1 1 1 S 1 1
S =Sa ⊕Sb . We willshowthatSa isafield. Let06=x∈Sa and06=y ∈Sb .
1 1 1 1 1
Thenthereexistsα,β suchthatαx+βy =1 andsoα(a +b )x+β(a +b )y =1 .
S 1 1 1 1 S
Thus (αa )x+(βb )y = 1 . On the other hand a +b = 1 and so (αa )x = a
1 1 S 1 1 S 1 1
and (βb )y = b . Therefore x is a unit in Sa . Therefore Sa and Sb are fields
1 1 1 1 1
and so |Max(S)|=2. therefore |Max(R)|=2.
(cid:3)
Example 3.4. Let R=Zn where n=pℓ11···pℓrr.
6 HAMIDREZAMAIMANI,MARYAMSALIMI,ASIYEHSATTARI,ANDSIAMAKYASSEMI
Assume r ≥ 3. Let x = pℓ1···pℓr−1 and y = pℓ2···pℓr. Then x and y are not
1 r−1 2 r
adjacent. Also if x,y are adjacent z, then (z,x) = (z,y) = 1, which is impossible.
We have Rx+Rpℓr =R=Rpℓr+Rpℓ1 =Rpℓ1+Ry. Hence there is path x—pℓr—
r r 1 1 r
pℓ11—y. So diam(Γ2(Zn)\J(Zn))=3.
Assume that r = 2. In this case we have two maximal ideals M =< p >
1 1
and M2 =<p2 >. Then Γ(R) is a complete bipartite graph and so diam(Γ2(Zn)\
J(Zn))=2.
Assume that r =1. Then R is local and so Γ2(Zn)\J(Zn) is empty graph.
Example 3.5. Let R be an infinite PID. Then for any two non-unit elements a,b,
there exists a prime element p such that p does not divide a and b. Therefore
Ra+Rp=Rb+Rp=1. So d(a,b)≤2 and hence diam(Γ (R)\J(R))=2.
2
4. isomorphisms
Recall that two graphs G and H are isomorphic, denoted by G∼=H, if there is
a bijection ϕ:G→H of vertices such that the vertices x and y are adjacent in G
if and only if ϕ(x) and ϕ(y) are adjacent in H.
In this section, we consider the following question:
If R and S are two rings with Γ(R)∼=Γ(S), then do we have R∼=S?
The following examples show that the above question is not valid in general.
Example 4.1. Let R = Z and S = Z [x]/(x2). Then by simple computation we
4 2
can see that Γ(R)∼=Γ(S)(∼=K ∨K¯ ). But Z and Z [x]/(x2) are not isomorphic.
2 2 4 2
Example 4.2. Let R = Z and S = Z [x]/(x3). Then Γ(R)∼= Γ(S)(∼= K ∨K¯ ).
8 2 4 4
But R and S are not isomorphic.
Example 4.3. Let R = Z [x]/(x3) and S = Z [x,y]/(x2,y2,xy). Then Γ(R) ∼=
2 2
Γ(S)(∼=K ∨K¯ )). But R and S are not isomorphic.
4 4
In the following theorem we give a partial answer to the above question.
Theorem 4.4. Let {(R ,m )}m and {(S ,n )}n be two finite families of finite
i i i=1 j j j=1
quasi-local rings, and let R = R1 ×···×Rm and S = S1 ×···×Sn. If Γ(R) ∼=
Γ(S) then m = n and there is a permutation σ on the set {1,2,...,m} such that
|Ri/mi| = |Sσ(i)/nσ(i)| for each i = 1,··· ,m, and hence Ri/mi ∼= Sσ(i)/nσ(i). In
particular, if Γ(R)∼=Γ(S) and each Ri is a finite field, then each Sj is also a finite
field and Ri ∼=Sσ(i) for each i=1,··· ,m, and thus R∼=S.
Proof. First note that since |Max(R)| = n and |Max(S)| = m, and Γ(R) ∼= Γ(S),
we have that m = n. Set M = R ×···×R ×m ×R ×···×R and
i 1 i−1 i i+1 m
N =S ×···×S ×n ×S ×···×S foreachi=1,··· ,m. Foranyi=1,··· ,m
i 1 i−1 i i+1 m
letx ∈M \∪ M . Clearly{x ,··· ,x }isacliqueinΓ (R). Supposethatν(x )
i i j6=i j 1 m 2 i
is equal to the number of vertices of Γ(R) which are not adjacent to x . Then
i
ν(x )=|R−{y|y =x or y adjacent to x }|=|M |.
i i i i
For each i = 1,··· ,m, let y ∈ N be the image of x under the graph iso-
σ(i) σ(i) i
morphism. Then {y ,··· ,y } is a clique in Γ (S) and y ∈ N \∪ N .
1 m 2 σ(i) σ(i) j6=σ(i) j
It is easy to see that ν(y ) = |N |. Thus |M | = |N | and so |R /m | =
σ(i) σ(i) i σ(i) i i
|Sσ(i)/nσ(i)|. Therefore R/mi ∼=Sσ(i)/nσ(i).
In particular, if Γ(R)∼=Γ(S) and each Ri is a finite field. Thus J(R)=(0) and
so J(S)=(0) and hence n =(0) for each i. Therefore each S is also a finite field
i j
and Ri ∼=Sσ(i) for each i∈I, and thus R∼=S.
COMAXIMAL GRAPH OF COMMUTATIVE RINGS 7
(cid:3)
The following example shows that the condition “R is a field” is necessary in
i
Theorem 4.4.
Example 4.5. Let R = Z ×Z , and S = Z ×Z . Then J(R) ∼= {0}×Z and
2 8 4 4 4
J(S) ∼= Z ×Z . Also U(R) = {1}×{1,3,5,7} and U(S) = {1,3}×{1,3}. Since
2 2
|Max(R)| = |Max(S)| = 2, then Γ2(R)\J(R) ∼= Γ2(S)\J(S) ∼= K4,4. Therefore
Γ(R)∼=Γ(S)∼=(K4,4∪K¯4)∨K4. But it is clear that R≇S.
Corollary 4.6. Let R and S be two finite semi-local rings and let R be reduced.
Then Γ(R)∼=Γ(S) if and only if R∼=S.
Proof. It is clear that R = F ×···×F , where F is a field for any i = 1,··· ,n.
1 n i
Now the assertion holds from Theorem 4.4.
(cid:3)
ThefollowingresultshowsthatthereexistsacopyofΓ(R/J(R))inthestructure
of Γ(R). This result obtains that for two rings R and S if Γ(R) ∼= Γ(S), then
R/J(R)∼=S/J(S).
Proposition 4.7. The following hold:
(a) If a is adjacent to b in Γ(R), then every element of a+J(R) is adjacent to
b+J(R).
(b) The elements of a+J(R) are adjacent if and only if a is an unit. In this
case, each element of a+J(R) is unit too.
(c) There exists a copy of Γ(R/J(R)) in the structure of Γ(R). In particular,
if Γ(R)∼=Γ(S), then R/J(R)∼=S/J(S).
Proof. (a). Suppose that Ra + Rb = R. Let x = a + r and y = b + r for
1 2
r ,r ∈J(R). Then there exists elements s,t∈R such that sa+tb=1. So
1 2
sx+ty =sa+tb+sr +tr
1 2
=1−(−sr −tr ).
1 2
Since −sr −tr ∈J(R), so sx+ty is a unit and hence Rx+Ry =R.
1 2
(b). Let (a+r ) be adjacent to a+r . Then R(a+r )+R(a+r ) = R and
1 2 1 2
so there exist s,t ∈ R such that s(a + r ) + t(a + r ) = 1. This implies that
1 2
(s+t)a=1−(r s+r t) and so (s+t)a is invertible. Therefore a is invertible.
1 2
(c). Choose a distinct representation {a } from the cosets of R/J(R). By parts
i
(a) and (b), we have < {ai} >∼= Γ(R/J(R)). Let ϕ : Γ(R) → Γ(S) be an isomor-
phism. Then ϕ(< {a } >) = {b } and < {b } >= S/J(S) and hence the assertion
i i i
is easily obtained. (cid:3)
Acknowledgments
This paper was finalized when H.R. Maimani and S. Yassemi were visiting the
Tata Institute of Fundamental Research (TIFR) under TWAS-UNESCO Associ-
ateship Scheme. It is a pleasure to thank both TWAS and TIFR for financial
support and hospitality. The authors would like to thank S.M. Bhatwadekar for
the stimulating discussions. The authors wish to thank an anonymous referee,
whose comments have improved this paper.
8 HAMIDREZAMAIMANI,MARYAMSALIMI,ASIYEHSATTARI,ANDSIAMAKYASSEMI
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Hamid Reza Maimani, Department of Mathematics, University of Tehran, Tehran,
Iran,andInstitute forTheoreticalPhysics andMathematics(IPM).
E-mail address: [email protected]
MaryamSalimi,Departmentof Mathematics,University of Tehran,Tehran,Iran.
Asiyeh Sattari,DepartmentofMathematics,University of Tehran,Tehran,Iran.
Siamak Yassemi, Department of Mathematics, University of Tehran, Tehran, Iran,
andInstituteforTheoretical Physics andMathematics(IPM).
E-mail address: [email protected]
