THE CARDINALITY OF AN ANNIHILATOR CLASS IN A VON

20 Pages·2008·0.26 MB·English

InternationalElectronicJournalofAlgebra
Volume4(2008)63-82
THE CARDINALITY OF AN ANNIHILATOR CLASS
IN A VON NEUMANN REGULAR RING
John D. LaGrange
Received: 10October2007;Revised: 23January2008
CommunicatedbySarahGlaz
Abstract. One defines an equivalence relation on a commutative ring R by
declaring elements r1,r2 ∈ R to be equivalent if and only if annR(r1) =
annR(r2). If [r]R denotes the equivalence class of an element r ∈ R, then
it is known that |[r]R| = |[r/1]T(R)|, where T(R) denotes the total quotient
ringofR. Inthispaper,weinvestigatetheextenttowhichasimilarequality
will hold when T(R) is replaced by Q(R), the complete ring of quotients of
R. The results are applied to compare the zero-divisor graph of a reduced
commutativeringtothatofitscompleteringofquotients.
Mathematics Subject Classification (2000): 13A99
Keywords: von Neumann regular ring, zero-divisor graph, complete ring of
quotients
1. Introduction
Let R be a commutative ring. One easily checks that an equivalence relation
on R is given by declaring elements r ,r ∈ R to be equivalent if and only if
1 2
ann (r ) = ann (r ). The cardinalities of such equivalence (annihilator) classes
R 1 R 2
were considered in [13], where the authors were interested in ring-theoretic prop-
erties shared by von Neumann regular rings with identical zero-divisor structures.
In [3], the authors show that every ring has the same zero-divisor structure as its
total quotient ring. The proof of this result demonstrates that the cardinality of
the annihilator class of an element does not change when the element is regarded
as a member of its total quotient ring. We examine the degree to which this result
can be generalized to a particular extension of a reduced total quotient ring.
Throughout, R will always be a commutative ring with 1(cid:54)=0. Let Z(R) denote
the set of zero-divisors of R and T(R) = R its total quotient ring. A ring
R\Z(R)
R will be called reduced if nil(R) = (0). A commutative ring R with 1 (cid:54)= 0 is
von Neumann regular if for each x ∈ R, there is a y ∈ R such that x = x2y or,
equivalently, R is reduced with Krull dimension zero [9, Theorem 3.1].
64 JOHN D. LAGRANGE
A subset D ⊆ R is dense in R if ann (D) = (0). Let D and D be dense
R 1 2
idealsofR andletϕ ∈Hom (D ,R)(i=1,2). Notethatϕ +ϕ isan R-module
i R i 1 2
homomorphism on the dense ideal D ∩D , and ϕ ◦ϕ is an R-module homomor-
1 2 1 2
phism on the dense ideal ϕ−1(D ) = {r ∈ R | ϕ (r) ∈ D }. Then Q(R) = F/∼ is
2 1 2 1
a commutative ring, where F = {ϕ ∈ Hom (D,R) | D ⊆ R is a dense ideal} and
R
∼ is the equivalence relation defined by ϕ ∼ϕ if and only if there exists a dense
1 2
ideal D ⊆R such that ϕ (d)=ϕ (d) for all d∈D [12, Proposition 2.3.1]. In [12],
1 2
J. Lambek calls Q(R) the complete ring of quotients of R.
Letϕ∈Q(R)denotetheequivalenceclasscontainingϕ. Foralla/b∈T(R),the
ideal bR of R is dense and ϕ ∈Hom (bR,R), where ϕ (br)=ar. One checks
a/b R a/b
that the mapping a/b (cid:55)→ ϕ is a ring monomorphism, and that ϕ and ϕ are
a/b 0 1
the additive and multiplicative identities of Q(R), respectively. In particular, the
mapping R→Q(R) defined by r (cid:55)→ϕ is an embedding. However, these mappings
r
need not be onto (see [12]). If the mapping R → Q(R) is onto (i.e., r (cid:55)→ ϕ
r
is an isomorphism), then R is called rationally complete. Note that Q(R) is von
Neumann regular if and only if R is reduced [12, Proposition 2.4.1]. Thus every
reduced rationally complete ring is von Neumann regular.
AringextensionR⊆S iscalledaringofquotientsofRiff−1R={r ∈R | fr ∈
R}isdenseinS forallf ∈S. Inparticular,T(R)isaringofquotientsofR. IfS is
a ring of quotients of R, then there exists an extension of the mapping R→Q(R)
whichembedsSintoQ(R)[12,Proposition2.3.6]. Therefore,everyringofquotients
ofR canberegardedasasubringofQ(R). ItfollowsthatadensesetinR isdense
in every ring of quotients of R. Also, R has a unique maximal (with respect to
inclusion) ring of quotients, which is isomorphic to Q(R) [12, Proposition 2.3.6].
In recognition of this observation, we shall abuse notation and denote the maximal
ring of quotients of R by Q(R). It is not hard to check that Q(R) = Q(T(R)) for
anyringR. Infact,ifR⊆S ⊆Q,thenQisaringofquotientsofRifandonlyifQ
is a ring of quotients of S and S is a ring of quotients of R (e.g., see the comments
prior to Lemma 1.5 in [8]).
Let B(R) = {e ∈ R | e2 = e}, the set of idempotents of R. Then the relation
“≤” defined by a≤b if and only if ab=a partially orders B(R), and makes B(R)
a Boolean algebra with inf as multiplication in R, the largest element as 1, the
smallestelementas0,andcomplementationdefinedbya(cid:48) =1−a. Onechecksthat
a∨b = (a(cid:48) ∧b(cid:48))(cid:48) = a+b−ab, where “+” is addition in R. A set E ⊆ B(R) is
called a set of orthogonal idempotents if e e = 0 for all distinct e ,e ∈ E. For a
1 2 1 2
reference on the Boolean algebra of idempotents, see [12].
THE CARDINALITY OF AN ANNIHILATOR CLASS 65
ABooleanalgebraB iscomplete ifinfE existsforeverysubsetE ⊆B. IfB isa
complete Boolean algebra, then supE = inf{b | b ∈ B and b ≥ e for all e ∈ E}. It
is well known that every Boolean algebra B is a subalgebra of a complete Boolean
algebra D(B), where the infimum of a set in B (when it exists) is the same as its
infimum in D(B). Here, D(B) is the “so called” Dedekind-MacNeille completion
of B [12, c.f. Section 2.4]. Note that D(B(R))=B(Q(R)) for every von Neumann
regular ring R [8, Theorem 11.9]. In particular, B(Q(R)) is complete. Moreover,
B(R)=B(Q(R)) whenever B(R) is complete.
In this paper, we continue the investigations of [3] and [11]. We will denote the
annihilator class of an element r in R by [r] , i.e., [r] = {s ∈ R | ann (s) =
R R R
ann (r)}. As in [4], we define the zero-divisor graph of R, Γ(R), to be the (undi-
R
rected) graph with vertices V(Γ(R)) = Z(R) \ {0}, such that distinct v ,v ∈
1 2
V(Γ(R)) are adjacent if and only if v v =0. It is shown in [3, Theorem 2.2] that
1 2
Γ(R) ∼= Γ(T(R)) for any commutative ring R; the equality |[r] | = |[r] | for
R T(R)
all r ∈ R follows directly from the proof of this theorem (where we have identi-
fied R with its canonical image in T(R)). Both of these results fail when T(R) is
replaced by Q(R) (e.g., Examples 2.10 and 2.11). In Section 2, we give necessary
and sufficient conditions for the equality |[r] | = |[r] | to hold, where R is a
R Q(R)
vonNeumannregularringsuchthatB(R)iscompleteand2(cid:54)∈Z(R)(seeTheorem
2.15). If either B(R) is not complete or 2 ∈ Z(R), then the equality may or may
not hold (see Examples 2.11, 2.17, and Corollary 2.16). This result is applied in
Section 3 to give sufficient conditions for Γ(R) ∼= Γ(Q(R)) to hold when R is a
reduced ring. In particular, we provide a characterization of zero-divisor graphs
which satisfy Γ(R) ∼= Γ(Q(R)), where R is a reduced ring such that |Z(R)| < ℵ
ω
and 2(cid:54)∈Z(R) (see Theorem 3.3).
2. The Cardinality of [e]
Q(R)
The investigation in this section involves a set-theoretic treatment of elements
in a ring. The main theorems are numbered 2.4, 2.8, 2.15, and 2.16. The results
numbered 2.1 through 2.8 develop useful relations within Q(R), and ultimately
provide an interpretation of elements in Q(R) as subsets of a set. The results
numbered 2.9 through 2.17 provide answers regarding the cardinalities of [e] and
R
[e] .
Q(R)
Throughout this section, R will always be a von Neumann regular ring unless
statedotherwise. Ifr ∈R, sayr =r2s, thene =rsistheunique idempotentthat
r
66 JOHN D. LAGRANGE
satisfies [r] =[e ] (c.f. the discussion prior to Theorem 4.1 in [3], or Remark 2.4
R r R
of [11]). Moreover, r =ue for some unit u of R [9, Corollary 3.3].
r
The following proposition shows that a nonzero element of a ring of quotients of
R will map some idempotent of R into R nontrivially. Recall that f−1R is dense
in S whenever f is a nonzero element of a ring of quotients S of R. In particular,
there is an r ∈R such that fr ∈R\{0}.
Proposition 2.1. Let R be a von Neumann regular ring. If R ⊆ S is a ring of
quotients of R, then for all 0 (cid:54)= f ∈ S there exists an e ∈ B(R) such that e ≤ e
f
and 0(cid:54)=fe∈R.
Proof. Let 0 (cid:54)= f ∈ S. Choose r ∈ R such that 0 (cid:54)= fr ∈ R. There is a unit u of
R such that r =ue , and hence fe =u−1fr ∈R\{0}. Let e=e e (note that it
r r f r
makes sense to talk about e since S ⊆Q(R) and Q(R) is von Neumann regular).
f
Let s∈Q(R) and t∈R be elements such that f =f2s and r =r2t. Then
e=e e =(fs)(rt)=(fr)(st)=e ∈R.
f r fr
Moreover, e≤e and fe=fe ∈R\{0}. (cid:164)
f r
ForanysetA⊆R,letE ={e ∈B(R) | r ∈A}. Ife∈B(R),thenconsiderthe
A r
set R (R) = {∅ (cid:54)= A ⊆ R | e e = 0 for all distinct r ,r ∈ A, and supE = e}.
e r1 r2 1 2 A
Note that R (R) (cid:54)= ∅ since {e} ∈ R (R). Also, if supE = e and 0 (cid:54)= e(cid:48) ∈ B(R)
e e A
withe(cid:48) ≤e,thenthereexistsane(cid:48)(cid:48) ∈E suchthate(cid:48)e(cid:48)(cid:48) (cid:54)=0. Otherwise,e(cid:48)(cid:48) ≤1−e(cid:48)
A
for all e(cid:48)(cid:48) ∈ E , and thus e = supE ≤ 1−e(cid:48). But this implies that e(cid:48)e = 0, a
A A
contradiction. This fact is generalized in (1) of the following proposition.
Proposition 2.2. Suppose that E ⊆B(R) is a set of orthogonal idempotents in a
von Neumann regular ring R.
(1) Let e(cid:48) ∈ B(R). Then e(cid:48)supE = 0 if and only if E ∪ {e(cid:48)} is a set of
orthogonal idempotents. In particular, rsupE = 0 if and only if re(cid:48) = 0
for all e(cid:48) ∈E (r ∈R).
(cid:80)
(2) Suppose that E is finite; say E ={e ,...,e }. Then supE = n e .
1 n j=1 j
(3) Let e(cid:48) ∈B(R). If f ∈Q(R) such that e(cid:48) ≤e , then fe(cid:48) ∈[e(cid:48)] .
f Q(R)
(4) Let e(cid:48),e∈B(R) such that e(cid:48) ≤e and 2e(cid:48) ∈[e(cid:48)] . Then e(cid:48)+e∈[e] .
R R
Proof. Note that supE ∈B(Q(R)).
(1)Ife(cid:48)e(cid:48)(cid:48) (cid:54)=0forsomee(cid:48)(cid:48) ∈E,thene(cid:48)e(cid:48)(cid:48) ≤e(cid:48)(cid:48) ≤supE impliesthate(cid:48)e(cid:48)(cid:48)supE =
e(cid:48)e(cid:48)(cid:48) (cid:54)= 0; in particular, e(cid:48)supE (cid:54)= 0. Conversely, suppose that e(cid:48)supE (cid:54)= 0. Since
e(cid:48)supE ≤ supE, the above comments show there exists an e(cid:48)(cid:48) ∈ E such that
THE CARDINALITY OF AN ANNIHILATOR CLASS 67
(e(cid:48)supE)e(cid:48)(cid:48) (cid:54)= 0; in particular, e(cid:48)e(cid:48)(cid:48) (cid:54)= 0. Thus E∪{e(cid:48)} is not a set of orthogonal
idempotents.
The “in particular” statement holds since [r] =[e ] for all r ∈R.
R r R
(cid:80)
(2) It is easy to check that e = n e ∈ B(R). Also, e e = e for all j ∈
j=1 j j j
{1,...,n}. Hence supE ≤ e. But e ≤ supE for all j ∈ {1,...,n}, and thus
j
esupE =e; that is, e≤supE. Therefore, e=supE.
(3) Clearly ann (e(cid:48)) ⊆ ann (fe(cid:48)). Let a ∈ ann (fe(cid:48)). Then ae(cid:48) ∈
Q(R) Q(R) Q(R)
ann (f)=ann (e ). Thus 0=ae(cid:48)e =ae(cid:48); that is, a∈ann (e(cid:48)). Hence
Q(R) Q(R) f f Q(R)
ann (e(cid:48))=ann (fe(cid:48)), i.e., fe(cid:48) ∈[e(cid:48)] .
Q(R) Q(R) Q(R)
(4) If r ∈ann (e), then re=0 and re(cid:48) =ree(cid:48) =0. Hence r ∈ann (e(cid:48)+e), and
R R
thereforeann (e)⊆ann (e(cid:48)+e). Toshowthereverseinclusion,letr ∈ann (e(cid:48)+e).
R R R
Note that 0 = re(cid:48)(e(cid:48) +e) = r(2e(cid:48)). Then 2e(cid:48) ∈ [e(cid:48)] implies that re(cid:48) = 0, and
R
therefore re=re(cid:48)+re=r(e(cid:48)+e)=0. Hence ann (e(cid:48)+e)⊆ann (e). (cid:164)
R R
In order to investigate cardinality, we shall translate the elements of an equiva-
lence class [e] into sets of elements of R (R). Such a correspondence is given
Q(R) e
in Theorem 2.4, and is motivated by the following example.
Example 2.3. LetF beaninfinitefieldandJ aninfiniteindexingset. LetF =F
j
(cid:81)
for all j ∈ J. Define R = {(r ) ∈ F | {r } ⊆ {s ,...,s } for some
j j∈J j j j∈J 1 n
{s ,...,s } ⊆ F, for some n ∈ N} (c.f. [11, Example 3.5]). Note that R is von
1 n
Neumann regular. Let D be the dense ideal of R generated by the minimal nonzero
idempotents of R (that is, the elements with a 1 in precisely one coordinate and 0
(cid:81) (cid:81)
elsewhere). Then D is contained in f−1R for all f ∈ F . Thus F is a
(cid:81) j∈J j j∈J j
ring of quotients of R. Moreover, F is rationally complete [12, Proposition
(cid:81) j∈J j
2.3.8]. Therefore, Q(R)= F .
j∈J j
Consider R from Example 2.3. Suppose that F = Q, J = N, and let e be the
multiplicative identity of R (the largest element of B(R)). Note that there is a
correspondence between R (R) and [e] , which is defined by taking the “sum”
e Q(R)
of the elements of a set in R (R). For example, the set
e
{(1,0,0,...),(0,2,0,...),(0,0,3,...),...}∈R (R)
e
corresponds to the element (1,2,3,...)∈Q(R). This correspondence is generalized
in the following theorem.
Theorem 2.4. Let R be a von Neumann regular ring and suppose that e∈B(R).
The mapping σ :R (R)→[e] defined by
e e Q(R)
σ (A)=f if and only if f ∈[e] with fe =r for all r ∈A
e Q(R) r
68 JOHN D. LAGRANGE
is a well-defined function. Moreover, σ (A) ∈ R if and only if σ (A) = σ (A(cid:48)) for
e e e
some A(cid:48) ∈R (R) with |A(cid:48)|<∞.
e
Proof. Fix e ∈ B(R). To show that σ is well-defined, we first show that every
e
element of R (R) corresponds to some element in [e] . Let A ∈ R (R). Note
e Q(R) e
thatD =(1−e,E )isadenseidealofR: Anyelementr ∈R\{0}thatannihilates
A
1−esatisfiesre=r (cid:54)=0,andthereforedoesnotannihilateallofE byProposition
A
2.2 (1). Define ϕ∈Hom (D,R) by
R
(cid:161) (cid:88) (cid:162) (cid:88)
ϕ t(1−e)+ t e = t r.
r r r
er∈EA er∈EA
(Indeed, ϕ is well-defined since multiplication by the appropriate idempotent will
show that equal elements of D have equal “like terms,” and clearly t e = t(cid:48)e
r r r r
implies that t r =t(cid:48)r.) Then ϕ(1−e)=0 and ϕ(e )=r for all r ∈A. Therefore,
r r r
there exists an element f ∈Q(R) such that f(1−e)=0 and fe =r for all r ∈A.
r
It follows that e ≤ e (in B(Q(R))). To prove the reverse inequality, let r ∈ A.
f
Then
r =fe =e fe =e r.
r f r f
Thus r(1−e ) = 0, which implies that e ≤ e . Hence e = supE ≤ e , and
f r f A f
therefore e=e . This shows that f ∈[e] , and therefore σ (A)=f. It remains
f Q(R) e
to show that σ is single-valued. Suppose that A maps to both f and g. Then
e
(f −g) annihilates D. But D is dense in Q(R), and thus f −g = 0, i.e., f = g.
Therefore, σ is well-defined.
e
To see that the “moreover” statement is true, suppose that σ (A) ∈ R. Then
e
σ (A) = σ (A(cid:48)), where A(cid:48) = {σ (A)}. Conversely, suppose that A(cid:48) ∈ R (R) with
e e e e
|A(cid:48)|<∞; say A(cid:48) ={r ,...,r }. Then
1 n
(cid:88)n (cid:88)n (cid:88)n
σ (A(cid:48))=σ (A(cid:48))e=σ (A(cid:48))( e )= (σ (A(cid:48))e )= r ∈R,
e e e rj e rj j
j=1 j=1 j=1
where the second equality follows from Proposition 2.2 (2) (c.f. the last paragraph
prior to the statement of this theorem). (cid:164)
By the last part of the previous proof, we have
Corollary 2.5. Let R be a von Neumann regular ring. If A ∈ R (R) is a finite
e
(cid:80)
set, then σ (A)= r ∈[e] .
e r∈A R
Let e∈B(R) and define the set
E (R)={E | A∈R (R)}.
e A e
THE CARDINALITY OF AN ANNIHILATOR CLASS 69
We shall write f ≺ E whenever E ∈ E (R) and σ (A) = f for some A ∈ R (R)
e e e
with E =E. By Proposition 2.2 (3), this is equivalent to declaring f ≺E if and
A
only if E is a set of orthogonal idempotents such that supE =e =e and fe(cid:48) ∈R
f
for all e(cid:48) ∈ E (indeed, let A = {fe(cid:48)} , c.f. the second paragraph in the proof
e(cid:48)∈E
of Theorem 2.8). In particular, if r ∈ [e] , then r ≺ E for all E ∈ E (R), i.e.,
R e
{r ∈[e] | r ≺E}=[e] for all E ∈E (R).
R R e
Corollary 2.6. Let R be a von Neumann regular ring and suppose that e∈B(R).
If E ∈E (R), then
e
|{A∈R (R) | E =E}|=|{f ∈[e] | f ≺E}|.
e A Q(R)
Proof. The mapping {A∈R (R) | E =E}→{f ∈[e] | f ≺E} defined by
e A Q(R)
A(cid:55)→σ (A) is a well-defined surjection by Theorem 2.4 and the definition of ≺. It
e
is injective since if A ,A ∈{A∈R (R) | E =E} with σ (A )=σ (A ), then
1 2 e A e 1 e 2
A ={σ (A )e(cid:48)} ={σ (A )e(cid:48)} =A .
1 e 1 e(cid:48)∈E e 2 e(cid:48)∈E 2
Therefore,
|{A∈R (R) | E =E}|=|{f ∈[e] | f ≺E}|.
e A Q(R)
(cid:164)
SupposethatRisareducedring. Thenthemappingann (J)(cid:55)→ann (J∩R)
Q(R) R
(J ⊆Q(R))isawell-definedbijectionofAnn(Q(R))ontoAnn(R),whereAnn(R)=
{ann (J) | J ⊆ R} [12, Proposition 2.4.3]; in particular, [r] ⊆ [r] for all
R R Q(R)
r ∈R. Alternatively, suppose that R is a von Neumann regular ring. Then [e] =
R
{r ∈[e] | r ≺E} for all E ∈E (R). Since σ (R (R))⊆[e] , we have
R e e e Q(R)
Proposition2.7. LetRbeavonNeumannregularringandsupposethate∈B(R).
Then [e] ⊆{f ∈[e] | f ≺E} for all E ∈E (R). In particular, [e] ⊆[e] .
R Q(R) e R Q(R)
Ofcourse,the“inparticular”statementoftheabovepropositioncanbejustified
bythesimplerargumentthatr =ueforsomeunituofR (andhenceofQ(R)), for
all r ∈[e] . However, we will apply the first part of the proposition in the proof of
R
Lemma 2.12.
Note that Theorem 2.4 implies that some of the elements of [e] correspond
Q(R)
to elements of R (R). The next theorem shows that every element in [e] is of
e Q(R)
this type.
Theorem 2.8. Let R be a von Neumann regular ring. Suppose that e ∈ B(R).
Then σ is surjective. In particular, |[e] |≤|R (R)|.
e Q(R) e
70 JOHN D. LAGRANGE
Proof. Fix e∈B(R). The result is trivial for the case e=0. Suppose that e(cid:54)=0.
Toshowthatσ isonto,chooseanyf ∈[e] . LetC ={∅(cid:54)=E ⊆B(R) | e(cid:48)e(cid:48)(cid:48) =0
e Q(R)
foralldistincte(cid:48),e(cid:48)(cid:48) ∈E,e(cid:48) ≤eforalle(cid:48) ∈E,andfe(cid:48) ∈Rforalle(cid:48) ∈E}. Notethat
C (cid:54)= ∅ since {0} ∈ C. Let C be partially ordered by inclusion; then an application
of Zorn’s lemma shows that C has a maximal element, call it E. We will show
that supE = e. If not, then consider 0 (cid:54)= e(cid:48) = e−supE ∈ B(Q(R)). Note that
fe(cid:48) ∈ [e(cid:48)] by Proposition 2.2 (3). Hence Proposition 2.1 implies that there
Q(R)
exists an e(cid:48)(cid:48) ∈ B(R) such that e(cid:48)(cid:48) ≤ e(cid:48) and fe(cid:48)(cid:48) = fe(cid:48)e(cid:48)(cid:48) ∈ R\{0}. Also, e(cid:48) ≤ e
implies e(cid:48)(cid:48) ≤e, and thus
e(cid:48)(cid:48)supE =e(cid:48)(cid:48)(e−e(cid:48))=e(cid:48)(cid:48)−e(cid:48)(cid:48) =0.
But then E∪{e(cid:48)(cid:48)}∈C by Proposition 2.2 (1), contradicting the maximality of E.
Therefore, supE =e.
Let A = {fe(cid:48) | e(cid:48) ∈ E}. Then Proposition 2.2 (3) implies E = E, and thus
A
A ∈ R (R). Also, e = e(cid:48) implies that fe = fe(cid:48) for all fe(cid:48) ∈ A. Hence
e fe(cid:48) fe(cid:48)
σ (A)=f.
e
The “in particular” statement is clear. (cid:164)
We now turn our attention to the cardinality of [e] . The previous theorem
R
allowsonetoderiveinformationaboutthecardinalityof[e] fromthesetR (R).
Q(R) e
We will be able to relate the cardinalities of [e] and [e] if we can find a way
Q(R) R
to use the set R (R) to reveal information about |[e] |. The next three lemmas
e R
accomplish this by considering elements of the subset E (R) of R (R).
e e
Lemma 2.9. Let R be a von Neumann regular ring. Suppose that E ⊆B(R)\{0}
is a set of orthogonal idempotents with supE =e. Moreover, assume that B(R) is
complete and 2e(cid:48) ∈[e(cid:48)] for all e(cid:48) ∈E. Then |[e] |≥2|E|.
R R
Proof. Define the mapping ρ:P(E)→[e] by
R
ρ(E(cid:48))=supE(cid:48)+e,
where P(E) is the “power set” of E. Let E(cid:48) ⊆ E. It is clear that ann (supE(cid:48)) ⊆
R
ann (2supE(cid:48)). Conversely, let r ∈ ann (2supE(cid:48)). Then 2r ∈ ann (supE(cid:48)), and
R R R
hence2re(cid:48) =0foralle(cid:48) ∈E(cid:48)byProposition2.2(1). Thusr ∈ann (2e(cid:48))=ann (e(cid:48))
R R
for all e(cid:48) ∈ E(cid:48), and therefore Proposition 2.2 (1) implies that r ∈ ann (supE(cid:48)).
R
This shows that ann (supE(cid:48)) = ann (2supE(cid:48)), i.e., 2supE(cid:48) ∈ [supE(cid:48)] . Hence
R R R
ρ is well-defined by Proposition 2.2 (4). To show that ρ is injective, suppose that
E ,E ⊆E with E (cid:54)=E ; say 0(cid:54)=e(cid:48) ∈E \E . Then
1 2 1 2 1 2
e(cid:48)supE =e(cid:48) (cid:54)=0=e(cid:48)supE ,
1 2
THE CARDINALITY OF AN ANNIHILATOR CLASS 71
wherethelastequalityholdsbyProposition2.2(1). ItfollowsthatsupE (cid:54)=supE .
1 2
Thus E (cid:54)=E implies that ρ(E )(cid:54)=ρ(E ). Therefore, ρ is injective, and hence
1 2 1 2
|[e] |≥|P(E)|=2|E|.
R
(cid:164)
For the remainder of this section, it will be necessary to recall some facts from
set theory. In what follows, we will assume the generalized continuum hypothesis.
Given any cardinal m, let cf(m) denote the cofinality of m. Note that cf(m) ≤ m,
and cf(m) is infinite whenever m is infinite (e.g., see [15, Theorem 21.10]). An
infinite cardinal m is called regular if m = cf(m). If m is not regular, then it is
called singular. Note that every successor cardinal is regular. Recall that mm(cid:48) is
defined to be the cardinal number |AB|, where A and B are sets of cardinality m
and m(cid:48), respectively, and AB is the set of all functions from B into A. If ℵ and
α
ℵ are infinite cardinals, then
β
ℵ , ℵ <cf(ℵ )
(cid:169) α β α
ℵℵαβ = ℵα+1, cf(ℵα)≤ℵβ ≤ℵα
ℵ , ℵ <ℵ
β+1 α β
[15, Theorem 23.9]. Also, mℵβ = ℵβ+1 for every 2 ≤ m < ∞ [15, Theorem
(cid:80)
22.13]. The notation m is used to express the cardinality of the disjoint
(cid:96) i∈I i
union A , where |A |=m for each i∈I. If I is an infinite indexing set with
i∈I i i i (cid:80)
m infinite for some i ∈ I, then m = |I|sup m . A detailed exposition of
i i∈I i i∈I i
cardinal numbers can be found in chapter four of [15].
It is our goal to find conditions that ensure the equality |[e] | = |[e] |. We
Q(R) R
will see that it suffices to impose restrictions on the elements of the set E (R). The
e
next two examples motivate such restrictions.
Example 2.10. Let F be a field such that |F|=ℵ and set J =N. Suppose that
ω
R is the ring in Example 2.3. Choose an infinite subset I of N, and let e be the
idempotent with 1 in all coordinates i∈I and 0 elsewhere. Then
|[e] |=ℵ <ℵ =ℵℵ0 =|[e] |,
R ω ω+1 ω Q(R)
where the second equality holds since cf(ℵ )=ℵ [15, Theorem 22.11].
ω 0
(cid:81) (cid:76)
Example 2.11. Let K = Z (X), and define the ring R = Z + K. As
2 (cid:81) N 2 N
in the Example 2.3, we have Q(R) = K. Choose an infinite subset I of N,
N
and let e be the idempotent with 1 in all coordinates i ∈ I and 0 elsewhere. Then
|[e] |=ℵ <ℵ =|[e] |.
R 0 1 Q(R)
72 JOHN D. LAGRANGE
In Example 2.10, we found an element e∈B(R) with an infinite set E ∈E (R)
e
such that cf(|[e(cid:48)] |) ≤ |E| < |[e(cid:48)] | for some e(cid:48) ∈ E (namely, E was the set of
R R
minimal nonzero idempotents less than e, and e(cid:48) could have been any element of
E). In Example 2.11, we found an element e ∈ B(R) with a set E ∈ E (R) such
e
that 2e(cid:48) (cid:54)∈ [e(cid:48)] for some e(cid:48) ∈ E (as before, E was the set of minimal nonzero
R
idempotents less than e, and e(cid:48) could have been any element of E). As a result,
Lemma2.9failsfortheelemente. WhenRisavonNeumannregularringsuchthat
B(R) is complete, the desired equality will necessarily be obtained in the absence
of such scenarios.
WeshallsaythatanelementE ∈E (R)isregular iftherelation|E|<sup{|[e(cid:48)] | |
e R
e(cid:48) ∈E}impliesthateithersup{|[e(cid:48)] | | e(cid:48) ∈E}isfiniteor|E|<cf(sup{|[e(cid:48)] | | e(cid:48) ∈
R R
E}). As a special case, E ∈ E (R) is regular if |E| < sup{|[e(cid:48)] | | e(cid:48) ∈ E} implies
e R
that sup{|[e(cid:48)] | | e(cid:48) ∈E} is either finite or a regular cardinal. Clearly E is regular
R
if it is finite.
Lemma 2.12. Let R be a von Neumann regular ring, e ∈ B(R), and E ∈ E (R).
e
Assume that B(R) is complete and 2e(cid:48) ∈ [e(cid:48)] for all e(cid:48) ∈ E. If E ∈ E (R) is
R e
regular, then |[e] |=|{f ∈[e] | f ≺E}|.
R Q(R)
Proof. If E is finite, then {f ∈ [e] | f ≺ E} ⊆ [e] by Theorem 2.4. The
Q(R) R
reverse inclusion holds by Proposition 2.7, and hence the result follows. Suppose
that E is infinite; say |E|=ℵ for some ordinal α. Let sup{|[e(cid:48)] | | e(cid:48) ∈E}|=m.
α R
Define
(cid:161) (cid:162)
F :{A∈R (R) | E =E}→ ∪{[e(cid:48)] | e(cid:48) ∈E} E
e A R
by the rule
F(A)(e)=r if and only if e∈E with e=e for some r ∈A.
r
Note that if r ,r ∈A with r (cid:54)=r , then e e =0. In particular, r (cid:54)=r implies
1 2 1 2 r1 r2 1 2
that e (cid:54)= e , and therefore F is well-defined by definition. Also, F is injective
r1 r2
since if F(A )=F(A ), then r =F(A )(e )∈A for all r ∈A , and similarly we
1 2 1 r 1 2
have A ⊆A so that A =A . Hence
1 2 1 2
(cid:161) (cid:162)
|{A∈R (R) | E =E}|≤| ∪{[e(cid:48)] | e(cid:48) ∈E} E|=(ℵ m)ℵα,
e A R α
where the equality holds since the union is disjoint. Therefore,
(cid:169) m, m>ℵ
|{f ∈[e] | f ≺E}|≤(ℵ m)ℵα = α ,
Q(R) α
ℵ , m≤ℵ
α+1 α
where the inequality follows by Corollary 2.6, and the equality follows since E is
regular.