The reduced ring order and lower semi-lattices
aa r X i v : . [ m a t h . R A ] F e b The reduced ring order and lower semi-lattices
W.D. Burgess and R. Raphael
Preprint Version
Abstract.
Every reduced ring R has a natural partial order defined by a ≤ b if a = ab ; it generalizes the natural order on a boolean ring. The articleexamines when R is a lower semi-lattice in this order with examples drawnfrom weakly Baer rings (pp-rings) and rings of continuous functions. Locallyconnected spaces and basically disconnected spaces give rings C( X ) which aresuch lower semi-lattices. Liftings of countable orthogonal (in this order) setsover a surjective ring homomorphisms are studied.
1. Introduction.
The subject of this article is an aspect of the reduced ringorder (here called the rr -order). This partial order is defined in all reduced rings(rings with no non-trivial nilpotent elements). The global assumption in the sequelwill be that the rings considered are unitary and reduced and, for some parts, alsocommutative. If R is a reduced ring the set of idempotents of R , which are here allcentral, is denoted B ( R ). A regular ring will always mean von Neumann regular . Definition . Let R be a reduced ring then for a, b ∈ R , a ≤ rr b if ab = a .This is readily seen to be a partial order which has been called variously the Abian order and the
Conrad order (see, for example, [ ] which dealt with orthogonalcompletions in this order). In a reduced ring R , if a = ab then, also, a = ba . Thisorder generalizes to all reduced rings the natural order on boolean rings.In a reduced ring R and r ∈ R the left and right annihilators of r coincide andare denoted ann r . The following well-known property of reduced rings, obvious inthe commutative case, will be used often. It was first shown in [ ] and [ ] and isfound in e.g. [ , Lemma 12.6]. Lemma . If R is a reduced ring and p is a minimal prime ideal of R then p is completely prime, i.e., R/ p is a domain. The rr -order can be a very coarse one but there are important classes of ringswhere it plays an interesting role. To illustrate one extreme, if R is a domain,then 0 is the unique minimal element in the order and no other pairs of elementsare related. In fact in any reduced ring 0 is the unique minimal element whilenon zero-divisors are maximal. At the other extreme, a reduced ring R is a lattice Mathematics Subject Classification. in the rr -order only when R is boolean (Proposition 2.8) but the main questionasked below is when a reduced ring is a lower semi-lattice in this order ( rr -lowersemi-lattice). Here there are many rings where the answer is positive. Moreover,some classes of rings are characterized by this property, for example weakly Baerrings within almost weakly Baer rings (Theorem 2.6). It is shown that if R is acommutative reduced ring it can be embedded, as an essential extension, in a weaklyBaer ring; this extension is proper if R is not already weakly Baer (Proposition 2.14and Corollary 2.20).If R → S is a surjective homomorphism of reduced rings when can a countableorthogonal family (in the rr -order) in S be lifted to an orthogonal family in R ?This is answered in two cases, Theorem 2.10 and Proposition 3.12.Among rings of continuous functions, two quite different classes of topologicalspaces yield rr -lower semi-lattices, locally connected spaces (Theorem 3.5) and ba-sically disconnected spaces ([ , Example 3.2]). A characterization of these spacesis available (Theorem 3.2) but requires looking at each pair of functions. However,even if X is a compact, metric and connected space it does not follow that C( X )is an rr -lower semi-lattice (Theorem 3.5(2)).Consider the case of R = Q α ∈ A D α where each D α is a domain. Here, any pairof elements has an infimum, noted ∧ rr . Let ( a α ) and ( b α ) be elements of R . Define( c α ) ∈ R by c α = a α when a α = b α and c α = 0 when a α = b α . It is readily seenthat ( c α ) = ( a α ) ∧ rr ( b α ).Such computations are not readily available in a general reduced ring R . Thequestion can be posed in the following way. Embed R in S = Q p ∈ Min R R/ p , whereMin R is the minimal spectrum. Given a, b ∈ R then c = a ∧ rr b can be foundin S but that element need not be in R . What is the “best approximation” to c which can be found in R ? This question will be considered in the sequel, sometimesdirectly, sometimes indirectly. See Corollary 2.2, below for an easy case. It is alsoshown that if | Min R | ≤ R is an rr -lower semi-lattice but not necessarily formore than three minimal prime ideals (Proposition 2.13).
2. The reduced ring order and lower semi-lattices.
Since the rr -order ona domain R is very easy to understand and makes R into a lower semi-lattice,the first place to look for examples is where rings are built out of domains in astraightforward way.Recall ([ , Chapter V2]) that a ring R can be represented as the ring of sectionsof a sheaf, called the Pierce sheaf over the boolean space Spec B ( R ), where B ( R )is the boolean algebra of central idempotents of R . (In a reduced ring, all theidempotents are central.) The stalks, for x ∈ Spec B ( R ), are the rings R x = R/Rx .For a ∈ R , put a x = a + Rx and define z( a ) = { x ∈ Spec B ( R ) | a x = 0 } , an openset in Spec B ( R ). Its complement is coz a .An interesting case in this study is where the stalks of this representation areall domains. For a reduced ring R , what is wanted is that the Rx , as above,are precisely the minimal prime ideals of R . In the commutative case these ringswere characterized in [ , Theorem 2.2] as the rings R in which the annihilatorof an element is generated by idempotents; these rings are called almost weaklyBaer rings . The name will here be abbreviated to awB rings . For the purposesof this article, the definition will be extended to all reduced rings . The awB ringsinclude commutative (Von Neumann) regular rings and, more generally, strongly EDUCED RING ORDER 3 regular rings, and Baer rings, among others. The characterization of rings whosePierce stalks are domains extends readily to general reduced rings as shown inProposition 2.5.The next proposition gives some information about the rr -order. If a reducedring R is viewed as a subring of Q p ∈ Min R R/ p , a product of domains, and r, s ∈ R ,to ask about r ∧ rr s is to ask to when there is an element t ∈ R which is zero when r + p and s + p differ, coincides with them when t + p = 0 and is the unique element rr -maximal among all such elements. Proposition . Let R and S be reduced rings. (i) Let a, b, c ∈ R where c commutes with a : if a ≤ rr b then ca ≤ rr cb and, in particular, if e = e then ae ≤ rr be .(ii) If a, b ∈ R , c = a ∧ rr b and u is a central unit in R , then uc = ua ∧ rr ub .(iii) If φ : R → S is a ring homomorphism then φ preserves the rr -order on R .(iv) If R and S are rr -lower semi-lattices and φ : R → S a surjective homo-morphism with kernel K , where K is generated by idempotents, then φ preservesthe lower semi-lattice structure. Proof.
Notice that if a = ab then, also, a = ba since ( ab − ba ) = 0.(i) Clearly a = ab implies ( ca ) = cacb . Moreover, idempotents are central in R . (ii) By (i), uc ≤ rr ua and uc ≤ rr ub . Suppose d ≤ rr ua and d ≤ rr ub , then u − d ≤ rr a and u − d ≤ rr b . Hence, u − d ≤ rr c giving d ≤ rr uc and uc = ua ∧ rr ub .(iii) Obvious.(iv) If a, b ∈ R and c = a ∧ rr b it must be shown that φ ( c ) = φ ( a ) ∧ rr φ ( b ).Put s = φ ( a ) ∧ rr φ ( b ). Clearly φ ( c ) ≤ rr s , using (iii). Pick t ∈ φ − ( s ). There existidempotents e, f ∈ K and r, r ′ ∈ R so that t − ta = er and t − tb = f r ′ . Let g = (1 − e )(1 − f ). It follows that ( t − ta ) g = 0 = ( t − tb ) g which implies ( tg ) = tga and tg ≤ rr a and, similarly, tg ≤ rr b . Thus, tg ≤ rr c . Note that φ ( g ) = 1. Applying φ gives φ ( tg ) = φ ( t ) = s ≤ rr φ ( c ). This shows that φ ( c ) = s . (cid:3) Corollary . Let R be a reduced ring and P ( R ) = Q p ∈ Min R/ p . If for a, b ∈ R , a ∧ rr b = 0 in P ( R ) then a ∧ rr b = 0 in R. In particular, if ab = 0 then a ∧ rr b = 0 in both rings. Proof.
This follows immediately from Proposition 2.1 (iii) and how infs arecalculated in P ( R ). (cid:3) In Proposition 2.1 (iv) the hypotheses on φ are required. If, for example, φ is asurjective homomorphism from a domain R onto a domain S with non-zero kerneland if a = b are not in ker φ with φ ( a ) = φ ( b ), then a ∧ rr b = 0 but { a, b } is sent to { φ ( a ) = φ ( b ) } and φ ( a ) ∧ rr φ ( b ) = φ ( a ) = 0.In what follows the case of a not necessarily commutative reduced ring R whichis a Pierce sheaf of domains will be examined. To simplify notation, for a ∈ R ,ann B a = ann a ∩ B ( R ). The awB rings are those for which ann a = (ann B a ) R .A weakly Baer ring (wB ring) is a reduced ring such that for each r ∈ R there is e ∈ ann B r such that ann r = Re . This means that for x ∈ Spec B ( R ), r x = 0 x ifand only if e / ∈ x . (Such rings are also called pp-rings , for example in [ ], or Rickartrings in [ ].)To simplify notation the following definition is included. W.D. BURGESS AND R. RAPHAEL
Definition . A reduced ring R which is an rr -lower semi-lattice is called rr -good . If, for r, s ∈ R , r ∧ rr s = 0, the elements r and s are called rr -orthogonal.The property of being rr -orthogonal is what is called a Pierce property . Lemma . Let R be a reduced ring. Then, for r, s ∈ R , r ∧ rr s = 0 if andonly if for all x ∈ Spec B ( R ) , r x ∧ rr s x = 0 x . Proof.
Suppose first that r ∧ rr s = 0. If, for some x ∈ Spec B ( R ), there is a ∈ R with 0 x = a x ≤ rr r x and a x ≤ rr s x in R x . Then there is e ∈ Spec B ( R ) \ x such that a e = aer = aes and ae = 0, a contradiction. Suppose now that r x and s x are rr -orthogonal in each stalk. If there is 0 = a with a = ar = as then, forsome x ∈ Spec B ( R ), a x = 0 x leading to a contradiction. (cid:3) Before going on to characterize the rr -good rings among the awB rings it willbe shown that the Niefield-Rosenthal characterization of commutative rings whosePierce stalks are domains extends readily to all reduced rings. Proposition . Let R be a reduced ring. Then each of the stalks of thePierce sheaf of R is a domain if and only if R is an awB ring. Proof.
Recall the notation for the Pierce sheaf: the base space is Spec B ( R )and the stalks are the rings R/Rx , x ∈ Spec B ( R ). Suppose first that R is awB. Iffor r, s ∈ R and x ∈ Spec B ( R ), r x = 0 x and ( rs ) x = 0 x then there is e ∈ B ( R ) \ x such that rse = 0. Hence, by the awB property, there are idempotents e , . . . , e k ∈ ann r and t , . . . , t k ∈ R such that se = P ki =1 t i e i . However, the e i ∈ x showingthat se ∈ Rx and, thus, ( se ) x = s x = 0 x because e x = 1 x . Hence, R x is a domain.In the other direction, if each R x is a domain then, for r ∈ R , if s ∈ ann r then s x = 0 x if r x = 0 x . Hence, coz s ⊆ z( r ), where z( r ) is an open set. Since z( r ) isa union of clopen (closed and open) sets of the form coz e , e ∈ B ( R ) and er = 0.The compactness of the closed set coz s implies that coz s is covered by a finite setof the coz e . The union of this finite collection is also clopen, say coz f , f ∈ B ( R )and f ∈ ann r . From this, s = sf . Hence, ann r = (ann B r ) R . (cid:3) Theorem . Let R be an awB ring. Then R is rr -good if and only if R is awB ring. Proof.
Suppose first that R is a wB ring. Given non-zero a, b ∈ R with a = b ,let eR = ann( a − b ), e = e . The claim is that ae = be = a ∧ rr b . First, ae ≤ rr a and ae ≤ rr b since ae · a = a e = ( ae ) and ae · b = be · b = ( be ) = ( ae ) . If c ≤ rr a and c ≤ rr b then c ( a − b ) = 0 so that c = ce . Then, cae = c e = c showing that c ≤ rr ae and, similarly, c ≤ rr be .Now suppose that R is an awB ring which is rr -good. Consider r ∈ R . Itmust be shown that ann r = eR for some e ∈ B ( R ). Set a = r + 1 and b = 1. Byassumption there is c = a ∧ rr b . Since c = ca = cb , c ∈ ann( a − b ). Notice thatsince c = cb = c c , c ∈ ann B ( a − b ). Let f ∈ ann B ( a − b ). Set g = f − cf . Now ga = g ( r + 1) = g so that g ≤ rr a and g ≤ rr b , showing that g ≤ rr c . Hence, g = gc ,but gc = 0 giving g = 0 and thus f ≤ rr c . Hence, c is the unique maximal elementof ann B ( a − b ) showing that ann r = cR . (cid:3) The calculations above show the following in an awB ring R : r and s in R havea non-zero rr -lower bound if and only if there is 0 = e = e with e ( r − s ) = 0 and er = 0. EDUCED RING ORDER 5
Constructions of wB rings will be discussed below. Among rings of continuousfunctions, C( X ) is wB if and only if X is basically disconnected (see the remarkbefore Example 3.2 in [ ]). According to [ , Example 3.2], the ring of continuousreal valued functions C( β N \ N ) is awB but not wB. Rings of continuous functionsC( X, Z ), where X is a topological space and Z has the discrete topology, are easilyseen all to be wB. Notice that the property rr -good is not a Pierce property. If R is rr -good then each Pierce stalk is rr -good but the example C( β N \ N ) has stalksdomains and, hence, rr -good but the ring is not, by Theorem 2.6.Before pursuing consequences of Theorem 2.6 it is worthwhile to observe howrarely pairs of elements, even in a wB ring, can have an rr -supremum (denoted ∨ rr ). Moreover, it is shown below that R is an rr -upper semi-lattice if and only ifit is boolean.Assume for now that R is an awB ring and that r, s ∈ R . Consider the followingpairwise disjoint subsets of Spec B ( R ): A r,s = coz r ∩ coz s ∩ z( r − s ) , B r,s = coz r ∩ z( s ) C r,s = z( r ) ∩ coz s , D r,s = z( r ) ∩ z( s )If, in addition R is wB, each coz t is the cozero set of an idempotent. Forthe given elements r and s the following idempotents, e , e , e , e are defined by:coz e = A r,s , coz e = B r,s , coz e = C r,s and coz e = D r,s . Note that theseidempotents are orthogonal. This notation will be used in the following proposition. Proposition . Let R be an awB ring. (i) Given r, s ∈ R then r ∨ rr s existsonly if ( ∗ ) coz r ∪ coz s ⊆ A r,s ∪ B r,s ∪ C r,s . (ii) If, in addition, R is wB then ( ∗ ) implies that r ∨ rr s exists. Proof.
Notice that condition ( ∗ ) implies that A r,s ∪ B r,s ∪ C r,s ∪ D r,s =Spec B ( R ).(i) Suppose ( ∗ ) fails but that r and s have an rr -upper bound c . If, for example, x ∈ coz r and x / ∈ A r,s ∪ B r,s ∪ C r,s then r x = 0, s x = 0 and r x = s x . But then r x = r x c x shows r x = c x but s x = s x c x gives s x = c x , a contradiction. Hence, r and s do not have an upper bound. Similarly if x ∈ coz s and x / ∈ A r,s ∪ B r,s ∪ C r,s .(ii) Suppose that coz r ∪ coz s ⊆ coz( e + e + e ). Define c = ( e + e ) r + e s .Then, r − rc = r − r ( e + e ) r − re s = 0 since re = 0 and r ( e + e ) = r , by ( ∗ ).Similarly sc = s since c is also ( e + e ) s + e r . Thus, c is an upper bound of r and s . For any t ∈ R , c + te is also an upper bound but t = 0 gives a smallest oneof the form c + te . It will be shown that any upper bound d has this form. If d isan upper bound and x ∈ coz e , r x = r x d x giving d x = r x and, similarly d x = s x .Thus c and d are equal over coz e . Similarly for coz e and coz e showing that c and d can only differ over D r,s . Hence, c = r ∨ rr s . (cid:3) Proposition . A ring R is an rr -upper semi-lattice if and only if it is aboolean ring. Proof.
Each boolean ring is an upper semi-lattice in its natural order, whichcoincides with the rr -order. On the other hand, suppose R is any reduced ringwhich is an rr -upper semi-lattice. If a ∈ R , a = 1, put c = a ∨ rr
1. Then, 1 = 1 c so that c = 1. Then a = a . (cid:3) The next result looks at surjective homomorphisms from a wB ring. The fol-lowing lemma will be useful.
W.D. BURGESS AND R. RAPHAEL
Lemma . Let R be a wB ring and a, b ∈ R . There is a largest e ∈ B ( R ) such that a ∧ rr eb = 0 . Proof.
By Theorem 2.6, R is rr -good. Let c = a ∧ rr b and e ∈ B ( R ) be suchthat ann c = eR . Then for x ∈ Spec B ( R ), c x = 0 x if and only if e x = 1 x . Inother words, c x = a x = b x = 0 x if and only if e x = 0 x . Then there is no x so that a x = ( eb ) x = 0 x . This shows that a ∧ rr eb = 0. Now suppose that f ∈ B ( R ) issuch that e < f . Thus, there is x ∈ Spec B ( R ) such that f x = 1 x while e x = 0 x .Then, for each such x it follows that c x = a x = b x = ( f b ) x = 0 x which shows that a ∧ rr f b = 0 by Lemma 2.4. (cid:3) The following result about lifting rr -orthogonal sets also gives another familyof examples in the study of lifting orthogonal sets in [ ]. See also Proposition 3.12,below. Theorem . Let R and S be reduced rings where R is wB and φ : R → S is a surjective ring homomorphism such that φ ( B ( R )) = B ( S ) . If { s n } n ∈ N is an rr -orthogonal set in S then it lifts, via φ , to an rr -orthogonal set in R . Proof.
Let r ∈ R be any lifting of s . Suppose for n ≥ s , . . . , s n havebeen lifted to r , . . . , r n , respectively, where { r , . . . , r n } is rr -orthogonal in R . Let u be any lifting of s n +1 . Notice that for any pair 1 ≤ i < j ≤ n , r i and r j nevercoincide and are non-zero in a Pierce stalk. Assume { x ∈ Spec B ( R ) | for some 1 ≤ i ≤ n, ( r i ) x = u x = 0 x } 6 = ∅ for otherwise r n +1 = u can be used. As in Lemma 2.9there are maximal idempotents e , . . . , e n such that r i ∧ rr e i u = 0, and, by whathas been observed, { − e , . . . , − e n } are orthogonal, in the sense of idempotents.This is because it is not possible to have, for i = j , ( r i ) x = u x = ( r j ) x = 0 x ,which would happen if ( e i ) x = ( e j ) x = 0 x , i.e., if ((1 − e i )(1 − e j )) x = 0 x . Put ε = 1 − ((1 − e ) + · · · + (1 − e n )). Then, for x ∈ Spec B ( R ), ε x = 0 x exactlywhere one of the ( e i ) x = 0 x . Now put r n +1 = εu . It first needs to be shownthat r i ∧ rr εu = 0 for i = 1 , . . . , n or, by Lemma 2.4, that ( r i ∧ rr εu ) x = 0 x forall x ∈ Spec B ( R ). If ( εu ) x = 0 x then ε x = 1 x and then ( e i ) x = 1 x for all i .This means that ( r i ∧ rr εu ) x = ( r i ∧ rr e i u ) x = 0 x . If ( εu ) x = 0 x then, of course,( r i ∧ rr εu ) x = 0 x . By Lemma 2.4, each r i ∧ rr εu = 0.It must now be shown that φ ( r n +1 ) = s n +1 . The restriction of φ to B ( R ) is,by hypothesis, onto B ( S ); call the restriction β . The maximal ideals of B ( S ) are inone-to-one correspondence with those of B ( R ) containing ker β . It is only necessaryto look at those x ∈ Spec B ( R ) where ker β ⊆ x . Consider now x ∈ Spec B ( R ) withker β ⊆ x and β ( x ) = y ∈ Spec B ( S ). (i) If ε x = 0 x then, for some i , 1 ≤ i ≤ n ,( e i ) x = 0 x . This means that ( r i ∧ rr u ) x = 0 and then ( r i ) x = u x = 0 x . Then, φ ( r i ) y = ( s i ) y = φ ( u ) y = ( s n +1 ) y = 0 y , since s i and s n +1 are orthogonal. However, φ ( εu ) y = 0 y as well. (ii) If ε x = 1 x then for all i , 1 ≤ i ≤ n , ( r i ) x = u x or ( r i ) x = u x = 0 x . In any case φ ( r i ) y = ( s i ) y = φ ( u ) i = ( s n +1 ) y or φ ( r i ) y = φ ( u ) y = 0 y .Here, φ ( u ) y = φ ( εu ) y . Hence, in all cases, φ ( u ) y = φ ( εu ) y = ( s n +1 ) y . In otherwords, φ ( r n +1 ) = s n +1 . (cid:3) Notice that in Theorem 2.10 that the hypotheses do not guarantee that φ preserves the rr -order infima but that, in any case, countable rr -orthogonal setsin S lift to rr -orthogonal sets in R . No example is known to date of a surjectivehomomorphism of reduced rings where the lifting of countable rr -orthogonal setsfails. EDUCED RING ORDER 7
A strongly regular ring (see e.g., [ , page 199]) R is a wB ring where the stalksare division rings. In this case Proposition 2.1 (iii) can be strengthened. If r ∈ R then there is r ′ ∈ R with r r ′ = r and ( r ′ ) r = r ′ . The idempotent rr ′ will bedenoted e ( r ). If φ : R → S is a surjective ring homomorphism with kernel K then S is also strongly regular, φ ( e ( r )) = e ( φ ( r )) and K is generated by idempotents.Moreover, if r ∈ K then e ( r ) ∈ K and vice versa . Corollary . Let R and S be strongly regular rings equipped with the rr -order, and φ : R → S a surjective ring homomorphism. Then, φ is an rr -lowersemi-lattice homomorphism. Moreover, a countable rr -order orthogonal set in S can be lifted to an rr -order orthogonal set in R . Proof.
Since the Pierce stalks of strongly regular rings are division rings, theserings are awB but, in addition, each principal ideal is generated by an idempotentso that they are, in fact, wB rings. Thus, R and S are rr -good (Theorem 2.6)and φ preserves the lower semi-lattice structure since all ideals are generated byidempotents (Proposition 2.1 (iv)). The lifting follows from Theorem 2.10. (cid:3) Notice that when φ : R → S is a ring surjection of wB rings, φ need not bea rr -lower semi-lattice morphism. Consider φ : Z → Z / Z ∼ = Z / Z × Z / Z . The rr -orthogonal set { , } goes to the pair ((1 , , (1 , rr -orthogonal.Theorem 2.6 gives a class of reduced rings which are rr -good and shows thatthere are reduced rings which are not. The following example is a reduced ringwhich is not rr -good but which has a wB subring S with B ( R ) = B ( S ). Section 3presents examples of rings which are rr -good but which are not awB, for exampleC( R ), which has no non-trivial idempotents. Example 2.12 will show up again inExample 2.16 Example . Fix a prime p and let R be the ring of sequences of integerswhich are eventually constant modulo p . Then, R lies between the rr -good rings S of eventually constant integer sequences and P = Q N Z with B ( R ) = B ( S ) .However, R is not rr -good. Proof.
The rings S and P are even wB. Notice that an idempotent in R hasfinite or cofinite support. Consider r which is constantly 1 and s = (1 , p + 1 , , p +1 , , . . . ). Thsn, r and s have non-zero rr -lower bounds, for example (1 , , , , . . . )or (1 , , , , , . . . ) but no largest one. (cid:3) In [ , Corollary 1.6] it is pointed out that in a commutative noetherian ring,the conditions awB and wB coincide. A commutative noetherian reduced ring R is an order in a finite product of fields and it is natural to ask if such a ring is rr -good. When R has three or fewer minimal prime ideals the answer is “yes”(neither commutativity nor the noetherian property is needed) but not in general.The proof for three minimal primes will show what can go wrong with four or more. Proposition . Let R be a reduced ring.(i) Suppose R has three or fewer minimal primes. Then R is rr -good.(ii) If R has four minimal prime ideals it need not be rr -good. Proof. (i) The case of three minimal primes, p , p , p , will be considered.Embed R in R/ p × R/ p × R/ p = S via φ : R → S . The ring R will be viewedas a subring of S . Consider a, b ∈ R and their images in S , say ( a , a , a ) and( b , b , b ). If a i = b i for i = 1 , , a and b have no non-zero lower bounds in W.D. BURGESS AND R. RAPHAEL S and so, a ∧ rr b = 0 in R . If a and b differ in precisely two places, say, a = b and a = b then the inf in S is ( a , , a , , , ) ∈ R then that is the inf in R otherwise a ∧ rr b = 0. Finally suppose a and b coincide in the first two placesbut not in the third. The inf in S is ( a , a , R then it is a ∧ rr b . Ifnot, there are three possibilities: ( a , , ∈ R , (0 , a , ∈ R and neither is in R .In each case a ∧ rr b exists in R .(ii) Let K be a field. Let Q be a product of four copies of the field K ( x, y, z ).The ring R is generated as a K [ x, y, z ]-subalgebra of Q by the following elements: r = ( x + x, y, y, y ) b = (0 , z, , s = ( x, y, y, y ) b = (0 , , z, a = (0 , y, y, b = (0 , , , z ) b = (0 , , y, y ) = (1 , , , R is noetherian with four minimal prime ideals and will be seen not to be rr -good.Note first that Q cl ( R ) = Q since r − s, b , b , b give the four idempotents of Q in Q cl ( R ). The ring R is noetherian with 4 minimal primes by Goldie’s Theorem.Note that, in R , a and b are common lower bounds of r and s , but a and b arenot rr -related. The only candidate for a lower bound of r and s which is largerthan both a and b is (0 , y, y, y ). The aim is to show that (0 , y, y, y ) is not in R .Suppose on the contrary that it is. Then, there is a polynomial expression F , withcoefficients in K [ x, y, z ], in 7 variables with zero constant term which is evaluatedat the vector v = ( r, s, a, b, b , b , b ) to yield F ( v ) + C = (0 , y, y, y ), where C is aconstant from K [ x, y, z ].Since z does not appear in an essential way in the main equation and is onlyused to obtain idempotents in Q cl ( R ) it is possible to reduce the problem by setting z = 0. Once this has been done there is a polynomial f in four variables u , u , u , u with coefficients in K [ x, y ] and zero constant term. Set w = ( r, s, a, b ) and then f ( w ) + c = (0 , y, y, y ) for some constant c ∈ K [ x, y ]. The aim is to show that suchan equation leads to a contradiction. Subscripts will indicate the four components.The polynomial f can be written f = h + f a + f b + f ab where h has monomialsonly in u and/or u , f a has monomials with factor u but not u , f b has monomialswith factor u but not u , and f ab has monomials with factors both u and u .Recall that all these polynomials have 0 constant term.Note first that only h influences the first component. Set h ( w ) = l . Then, c =( − l, − l, − l, − l ). Moreover, l is divisible by x . Now consider the other components.Notice the following: f a ( w ) = f a ( w ) = 0, f b ( w ) = f b ( w ) = 0, f ab ( w ) = f ab ( w ) = f ab ( w ) = 0, and( i ) h ( w ) = h ( w ) = h ( w ) ( ii ) f a ( w ) = f a ( w ) ( iii ) f b ( w ) = f b ( w ) The elements h ( w ) , f a ( w ) , f b ( w ) are divisible by y and f ab ( w ) is divisible by y . The components can now be computed.( f ( w ) + c ) = h ( w ) + f a ( w ) − l = y ( f ( w ) + c ) = h ( w ) + f b ( w ) − l = y ( f ( w ) + c ) = h ( w ) + f a ( w ) + f b ( w ) + f ab ( w ) − l = y EDUCED RING ORDER 9
It follows that h ( w ) + f a ( w ) = l + y and h ( w ) + f b ( w ) = l + y and, using (i), f b ( w ) = f a ( w ) . In the expression for ( f ( w ) + c ) the sum of the first two termsequals l + y showing that f b ( w ) + f ab ( w ) = 0. Since f ab ( w ) is divisible by y ,so is f b ( w ) and, from that, so is f a ( w ) . This last is because f ( w ) = f ( w ) andthe various equalities with (ii) and (iii) yield f a ( w ) = f b ( w ) .In ( ∗ ) h ( w ) + f a ( w ) + f b ( w ) + f ab ( w ) = l + y all the terms on the leftare divisible by y . Hence, l is divisible by y . To proceed, a more detailed look at h ( u , u ) is needed. Let A be a finite set of pairs of non-negative integers i, j with i + j ≥ i, j ∈ A , a ij ∈ K [ x, y ]. Each a ij can be written a ′ ij + a ′′ ij , where a ′ ij contains all the terms of a ij with factor y and a ′′ ij all the remaining terms. Write( ∗∗ ) l = h ( x + x, x ) = X A a ij ( x + x ) i x j = X A a ′ ij ( x + x ) i x j + X A a ′′ ij ( x + x ) i x j . Since y | l , the last term in ( ∗∗ ) is zero.From the expression for f ( w ) , it follows that f a ( w ) + f b ( w ) + f ab ( w ) = l + y − h ( w ) is divisible by y . Since x | l and y | l , one can write l = l ′ xy . Now, h ( w ) = X A a ′ ij y i + j + X i + j ≥ a ′′ ij y i + j + ( a ′′ + a ′′ ) y . The first two terms on the right of this expression are divisible by y and, hence, y divides l ′ xy + y − ( a ′′ + a ′′ ) y and y divides l ′ x + 1 − ( a ′′ + a ′′ ). It followsthat a ′′ + a ′′ = px + 1 where p ∈ K [ x ]. Now return to ( ∗∗ ) to look at the lastterms which are a ′′ ( x + x ) + a ′′ x = a ′′ x + ( px + 1) x . There are no other degree1 terms in x . This contradicts the fact that P A a ′′ ij ( x + x ) i x j = 0 and concludesthe proof. (cid:3) The remainder of this section is devoted to showing how to embed each com-mutative reduced ring into a wB ring in various ways. Every such ring will be seento have an essential wB extension by using the complete ring of quotients in thenext proposition. It does not require commutativity.
Proposition . Let R ⊆ S be reduced rings where S is wB. Let T ⊆ S be the ring generated by R and those idempotents from B ( S ) which generate theannihilators, in S , of the elements of R . Then, T is wB. Proof.
The following notation will be used. For r ∈ R , let ann S r = e ( r ) S ,where e ( r ) ∈ B ( S ). Set S = { e ( r ) | r ∈ R } and T = R [ S ]. It needs to be shown thatthe annihilator in T of an element of T is generated by an idempotent. A typicalnon-zero element t ∈ T can be written t = P ki =1 r i e i where the 0 = r i ∈ R andthe e i are idempotents which, after the standard manipulations on idempotents,can be assumed to be orthogonal and derived from elements of S (and are thusin T ). Put ε = 1 − ( e + · · · + e k ). The claim is that ann T t is generated by e = P ki =1 e ( r i ) e i + ε . It is clear that te = 0. Let a ∈ ann T t . Then a can bewritten in the form a = P mj =1 s j f j , with each s j ∈ R and the f j a set of orthogonalidempotents derived from those in S . For each i = 1 , . . . , k , the orthogonality of the e i shows that 0 = r i e i P mj =1 s j f j so that for each j = 1 , . . . , m , s j f j e i ∈ e ( r i ) T and s j f j e i ∈ e i e ( r i ) T . From this, 0 = ( P ki =1 r i e i )( P mj =1 s j f j ) = P mj =1 P ki =1 r i s j e i f j = P mi =1 r i e i a . By the orthogonality of the e i , each r i e i a = 0, which shows that e i a = e i ae ( r i ). From this, a = a ( P ki =1 e i e ( r i )) + aε ∈ ( P ki =1 e i e ( r i ) + ε ) T , asrequired. (cid:3) Before applying this proposition it should be noted that neither the category ofcommutative wB rings, WB , nor the category of commutative rr -good rings, RG ,is a reflective subcategory of the category of commutative reduced rings CR . (See[ , Chapter V].) Both WB and RG are closed under products but neither is closedunder equalizers as would be required for a reflective subcategory. Hence, in bothcases, the inclusion of it into CR cannot have a left adjoint. This is seen usingthe following examples. The first simple example, pointed out by J. Kennison,illustrates this for WB . Example . Let R = Z × Z and S = Z with φ : R → S defined by φ ( a, b ) =¯ a and ψ : R → S by ψ ( a, b ) = ¯ b . Then while R and S are wB, the equalizer E of φ and ψ is not wB. Proof.
Note that (0 , ∈ E but its annihilator is not generated by an idem-potent in E , which has no non-trivial idempotents. (cid:3) The same remark applies to the category RG in CR . Here the example is a bitmore complex. Example . Let R = Q N Z and S = Z . Let U be the set of all freeultrafilters on N . For each U ∈ U , let the ideal I U of Q N Z be the set of thoseelements 0 over a set in U . Define φ U : R → S by ( a n ) ( a n ) ( a n ) + I U . Thenthe equalizer of all the φ U is not rr -good. Proof.
The ring in Example 2.12 with p = 2, which is not rr -good, will used.(Any prime p could be used.) Suppose r = ( a n ) ∈ R is eventually constant modulo2. Then, since the ultrafilters are free, r will be in the equalizer. On the other hand,if r = ( a n ) ∈ R is not eventually constant modulo 2 the set, X , of components where r is even and the set, Y , where they are odd are disjoint infinite sets. The sets X and Y are not both in all U ∈ U . Thus, r will not have the same image under allof the φ U . (cid:3) In the case of a commutative reduced ring R (i.e., commutative semiprime rings)there are various standard ways of embedding R into a regular ring. Each will yielda wB ring according to Proposition 2.14. The first, which is mentioned in passing,is the universal regular ring studied in [ ] and [ ]. The functor T : CR → VN R ,where
VN R is the category of commutative regular rings, embeds each R ∈ CR in T ( R ) ∈ VN R so that the T is the left adjoint of the inclusion functor VN R → CR .If Proposition 2.14 is used, it can be shown that the construction is functorial, say U : CR → WB . However, U does not preserve rings which are already wB and,hence, details will be omitted.Another choice of regular ring would be the complete ring of quotients, Q ( R )of R ∈ CR , always regular in this case. This works well but is not functorial. Thereis another important regular ring between R and Q ( R ), namely the epimorphichull , the intersection of all regular rings between R and Q ( R ) (see [ , Satz 11.3]).However, the next lemma will show that the construction of Proposition 2.14 givesthe same result for any regular ring S , R ⊆ S ⊆ Q ( R ). EDUCED RING ORDER 11
Definition . Let R ∈ CR and Q ( R ) its complete ring of quotients. ThewB ring constructed from R and Q ( R ) using Proposition 2.14 is denoted W ( R ).It will be seen that an annihilator which is generated by an idempotent behaveswell when passing from a ring to an essential extension ring. If R ⊆ S are reducedrings then the extension is called (right) essential if every non-zero right ideal of S has non-zero intersection with R . A right essential extension is also called a rightintrinsic extension in [ ]. (Note that this is weaker than saying that R R is essentialin S R .) Lemma . Let S ⊆ T be reduced rings where T is a right essential extensionof S . Suppose that, for s ∈ S , there are idempotents e ∈ S and f ∈ T such that ann S s = eS and ann T s = f T . Then, e = f . Proof.
Since es = 0, e = ef . If f = ef then there is t ∈ T such that0 = ( f − ef ) t ∈ S . However, s ( f − ef ) t = sf (1 − e ) t = 0 so that f (1 − e ) t = f (1 − e ) te = 0, a contradiction. Hence, e = f . (cid:3) Corollary . Let R ∈ CR and S ∈ CR an essential extension of R whichis wB. Let V be the wB ring generated in S as in Proposition 2.14. If r ∈ R issuch that ann R r = eR for some idempotent e ∈ R then eS = ann S r . Thus, in theconstruction of V , the idempotent for r ∈ R is already in R . Proof.
This is a direct application of Lemma 2.18. (cid:3)
Everything is now in place to show that the construction of W ( R ) not onlygives an efficient way of constructing a wB ring from R but also that if R is alreadywB then W ( R ) = R . Corollary . If R ∈ CR is a wB ring then W ( R ) = R . Proposition . Let R , S and V be as in Corollary 2.19. Then, V ∼ = W ( R ) . Proof.
There is a copy of Q ( R ) in Q ( S ) ([ , Satz 10.1]). The wB ring W ( R )can be calculated inside this copy of Q ( R ). Now consider r ∈ R and the idempotents e and f where eV = ann V r and f Q ( R ) = ann Q ( R ) r . Then, ann Q ( S ) r = eQ ( S )and Lemma 2.18 shows that e = f . Hence, V and W ( R ) are generated over R bythe same set of idempotents. (cid:3) The construction of W ( R ) can be seen, using Lemma 2.18, to be an exampleof a wB absolute to Q ( R ) (right) ring hull of R for any commutative reduced ring,in the language of [ , Definition 8.2.1(i)]. In the case of a general reduced ring R ,the construction of Proposition 2.14 along with Lemma 2.18 will yield a wB rightring hull of R ([ , Definition 8.2.1(iii)]) if there is any essential right wB extension T of R .The process of constructing W ( R ) is illustrated by the following. Proposition . Suppose R ∈ CR is such that its classical ring of quotients Q = Q cl ( R ) is regular. Then W ( R ) is the ring generated over R by B ( Q ) . Proof.
In this case for r ∈ R , ann Q r = (ann R r ) Q and for each a/b ∈ Q ,ann Q a/b = (ann R a ) Q . Hence, in Proposition 2.14, all the idempotents of Q areadjoined. (cid:3)
3. Rings of continuous functions.
In this section examples of rr -good ringswill be found which are quite unlike wB rings. The basic reference is [ ].Suppose X is a completely regular space and C( X ) is considered with the rr -order. What does it mean for C( X ) to be rr -good? Let f, g ∈ C( X ) and suppose h = f ∧ rr g exists. Then where h is non-zero it must coincide with f and g . Thismeans that coz h ⊆ z( f − g ) ∩ coz f . (Here coz f and z( f ) have their usual meaningsin rings of functions.) Moreover, on coz h , h coincides with f and with g . Theseconditions only make h ≤ rr f and h ≤ rr g . To have the infimum there must be a unique largest cozero set U ⊆ z( f − g ) ∩ coz f and h with (i) coz h = U , (ii) h and f (and g ) coincide on U , and (iii) for every net Λ = { λ α } in U converging to a point y on the boundary of U , { f ( λ α ) } → f ( y ) = 0 and { g ( λ α ) } → g ( y ) = 0. Notation:For Y ⊆ X the boundary of Y is denoted bd Y = Y \ Y . Definition . A completely regular topological space X is called rr - good ifthe ring C( X ) is rr -good.As an example consider a pair of functions from C( R ). (It turns out that R isan rr -good space.) Let f ( x ) = sin x and g ( x ) = | sin x | . The function h ( x ) whichis f ( x ) where the two functions coincide and zero where they differ is continuous.Hence, h = f ∧ rr g .The purpose here is to find examples of rr -good spaces. It will be seen laterthat it is easy to construct spaces which are not rr -good. Before looking at detailsit is worthwhile to note a class of rr -good spaces; more will follow. Recall that aspace X is basically disconnected if the closure of a cozero set is open. As alreadymentioned these are precisely the spaces where C( X ) is wB and hence rr -goodby Theorem 2.6. These spaces include the P-spaces ([ , 4J]) where C( X ) is vonNeumann regular. Note that a space can be totally disconnected without being rr -good: the space X = β N \ N is an example where C( X ) is awB but not wB([ , Example 3.2]) and, by Theorem 2.6 again, X is not rr -good.A complete characterization of all rr -good spaces as a class of well-known spacesis elusive since examples include basically disconnected spaces and, as will be seen,locally connected spaces. However, Theorem 3.2 does characterize rr -good spacesalthough by properties not easily translated into global language. It makes moreexplicit what it means for f and g in C( X ) to have a non-zero infimum. Thedescription will later be used to get sufficient conditions for X to be rr -good. Theorem . Let X be a completely regular topological space. Fix f, g ∈ C ( X ) , f = g , and suppose that f and g have a non-zero rr -lower bound. Let { h α } α ∈ A be the set of non-zero rr -lower bounds of f and g , U α = coz h α , α ∈ A and U = S α ∈ A U α (i) Then, f ∧ rr g exists if and only if U has the property: ( ∗ ) if x ∈ bd U and N is any neighbourhood of x then, for some β ∈ A , N ∩ bd U β = ∅ .(ii) Moreover, f ∧ rr g exists if and only if U has the property: ( ∗∗ ) if x ∈ bd U then for each ε > there is a neighbourhood N ε of x such that for all α ∈ A , N ε ∩ U α ⊆ { y ∈ X | | h α ( y ) | < ε } . Proof. (i) Note, with f and g as above, that if bd U = ∅ then U is clopenand there is e = e ∈ C( X ) such that coz e = U . In this case f ∧ rr g = ef = eg and( ∗ ) is vacuously satisfied.Notice that, as in the statement, h α coincides with f and with g on coz h α ;hence, coz h α ⊆ z( f − g ) ∩ coz f . Moreover, if α = β in A , then ( bd U α ) ∩ U β = ∅ . EDUCED RING ORDER 13
This is because for y ∈ bd U α , h α ( y ) = f ( y ) = 0 while for x ∈ U β , f ( x ) = h β ( x ) =0. First assume property ( ∗ ) for f and g which have a non-zero lower bound. Theproof will show that, among the lower bounds, there is a maximal one. With the h α as in the statement, define h by h ( x ) = f ( x ) for x ∈ U and h ( x ) = 0 for x ∈ X \ U .The continuity of h on bd U must be shown. The condition ( ∗ ) says that for everyneighbourhood N of x ∈ bd U there is some β ∈ A with N ∩ bd U β = ∅ . Thusthere is y ∈ N with f ( y ) = 0. By continuity of f it follows that f ( x ) = 0; the factthat h is zero on X \ U shows that h is continuous at x and that h coincides with f (and with g ) on U . By construction h ≤ rr f and h ≤ rr g and, also, for each α ∈ A , h α ≤ rr h .In the other direction, if h = f ∧ rr g exists it is one of the h α showing thatcoz h = U . By its continuity, the property ( ∗ ) follows automatically.The proof of part (ii) is along the same lines as that of (i). (cid:3) The next result gives a sufficient condition on a space X to be an rr -good space.For convenience, the condition will be given a name. Definition . Let X be a completely regular space. Let { U α } α ∈ A be anyfamily of non-empty cozero sets in X with the following property: for α = β in A ,( bd U α ) ∩ U β = ∅ . The space X is said to satisfy the B-property (for boundary prop-erty ) if the following holds for each such family of cozero sets. Let z ∈ bd ( S α ∈ A U α )then for every neighbourhood N of z there is β ∈ A such that N ∩ bd U β = ∅ . Corollary . Let X be a completely regular topological space which satisfiesthe B-property. Then X is an rr -good space. Proof.
The B-property is a stronger form (strictly stronger by Example 3.8,below) of the property ( ∗ ) in Theorem 3.2. This guarantees the existance of f ∧ rr g whenever f and g have a non-zero lower bound. Otherwise, f ∧ rr g = 0. (cid:3) The next task is to find examples of spaces satisfying the conditions of Corol-lary 3.4. Recall the definition of a locally connected space (e.g., [ , page 374]): X is locally connected if for each x ∈ X and a neighbourhood N of x there is aconnected subset C ⊆ N such that x ∈ int ( C ). Theorem . Let X be a completely regular space. (1) If X is locally connectedthen X is an rr -good space. (2) A connected space X need not be rr -good even when X is a compact metric space. Proof. (1) It will be shown that X satisfies the B-property. Suppose { U α } α ∈ A is a family of non-empty open sets in X as in the first part of Definition 3.3. Let z ∈ bd ( S α ∈ A U α ). It may be supposed that z / ∈ bd U β for any β ∈ A . Choosea neighbourhood N of z . By [ , 6.3.3], there is a connected subset C of N with z ∈ int C . Then for some β ∈ A , int C ∩ U β = ∅ . If C ∩ bd U β = ∅ then there isnothing to prove. Otherwise, C ∩ U β = C ∩ U β . This means that C ∩ U β is clopenin C . Moreover, C ∩ U β = ∅ and C ∩ U β = C since z / ∈ U β . This contradicts thefact that C is connected. Hence, N ∩ bd U β = ∅ .(2) For the construction, [ , Example 27.8(b)] is used but truncated to makeit compact. Define X ⊆ R as follows. It consists of the vertical line segments { (0 , y ) | − ≤ y ≤ } and { (1 , y ) | − ≤ y ≤ } along with horizontal line segments { ( x, /n ) | ≤ x ≤ } , for n ∈ Z \ { } , and { ( x, | ≤ x ≤ } . Elements f, g ∈ C( X ) will be defined so that they do not have an rr -inf. f ( x, y ) = x if 0 ≤ x ≤ / , y = 0 or y = 1 /n, n ∈ Z \ { }− x + 2 if 1 / ≤ x ≤ , y = 0 or y = 1 /n, n ∈ Z \ { } g ( x, y ) = f ( x, y ) if x = 0 , x = 1 , or y = 0 or y = 1 /n, n ∈ Z \ { } , n even | n || n | +1 x if 0 ≤ x ≤ / , y = 1 /n, n ∈ Z odd − | n || n | +1 x + | n || n | +1 if 1 / ≤ x ≤ , y = 1 /n, n ∈ Z oddA typical rr -lower bound of f and g coincides with f and g on finitely many ofthe horizontal lines, say with y ∈ { /n , . . . , /n k } , the n i = 0 and even, andzero elsewhere. However, continuity on the line where y = 0 precludes a greatest rr -lower bound. (cid:3) The next result shows how to construct examples of spaces which are not rr -good. Proposition . Let X be a completely regular topological space with thefollowing property: there is a sequence of disjoint clopen subsets { C n } n ∈ N suchthat U = S n ∈ N C n is not closed in X and, for some x ∈ bd U , every neighbourhoodof x meets all but finitely many of the C n . Then, X is not an rr -good space. Proof.
The proof of Theorem 3.5 is a model. Define two elements of C( X ) asfollows: f is constantly 1 on X , and g is constantly 1 except on C n , n odd, whereit is constantly 1 + 1 /n . It will be shown that g is continuous. Put U = S n ∈ N C n and V = int ( X \ U ). Clearly g is continuous on U ∪ V . Let z ∈ bd U . Given ε > m ∈ N such that for all n > m , 1 /n < ε . Then for x ∈ X \ S ≤ p ≤ m C p , | g ( x ) − g ( z ) | < ε . This shows continuity at all points of X .It will be shown that f ∧ rr g does not exist. There are lower bounds. In fact,any function h which is 0 everywhere except on finitely many C n , n even, where itis 1, is continuous and a lower bound. However, since, for each n ∈ N , bd C n = ∅ ,condition ( ∗ ) of Theorem 3.2 fails at x (as does condition ( ∗∗ )). (cid:3) Proposition 3.6 has many corollaries. Some of them are now listed.
Examples . The following are examples of spaces X which are not rr -good.(1) Let X be the one-point compactification of a disjoint union of an infinite collec-tion of compact spaces; e.g., the one-point compactification of an infinite discretespace. (2) Let X be any dense subset of R such that R \ X has an accumulationpoint in X . If X is as stated then X n is also not rr -good. (3) Let { F n } n ∈ N be a setof closed subsets of R n which are pairwise separated by open sets, U = S n ∈ N F n is not closed, and for x ∈ bd U there is a sequence { x n } n ∈ N converging to x with x n ∈ F n ; then let X = S n ∈ N F n as a subspace of R n . Proof.
These can be seen to be special cases of Proposition 3.6. (cid:3)
It is next seen that the B-condition is not necessary for a space to be rr -good. EDUCED RING ORDER 15
Example . Let Y be an uncountable discrete space and X = Y ∪ { s } wherethe neighbourhoods of s are co-countable in X . Then, X is an rr -good space whichdoes not satisfy the B-condition. Proof.
According to [ , 4N], X is a P-space (C( X ) is regular) and, hence, rr -good. Now partition Y into countable subsets { Y α } α ∈ A and, for each α ∈ A ,let e α be the idempotent whose support is Y α . Each Y α is thus a cozero set in X with empty boundary. However, S A Y α = Y whose boundary is { s } . Hence, theB-condition fails. (cid:3) Recall that for a completely regular space X , C ∗ ( X ) is the ring of boundedreal-valued continuous functions on X . Proposition . Let X be a completely regular topological space. Then X is rr -good if and only if C ∗ ( X ) is an rr -good ring. Proof. If X is rr -good and f, g ∈ C ∗ ( X ) and h = f ∧ rr g , calculated in C( X ),then h ∈ C ∗ ( X ).In the other direction, let f, g ∈ C( X ). Put f b = f / (1+ f ) and g b = g/ (1+ g ).Then f b , g b ∈ C ∗ ( X ), as are f bb = f b / (1 + g ) and g bb = g b / (1 + f ). Let h = f bb ∧ rr g bb . Then h = hf bb = hg bb . Hence, h (1 + f ) (1 + g ) = h (1 + f )(1 + g ) f = h (1 + f )(1 + g ) g showing that h (1 + f )(1 + g ) is a lower bound for f and g . Let k ∈ C( X ) be a lower bound for f and g . Then, k = kf = kg and, hence, k/ (1 + f )(1 + g ) is a lower bound for f bb and g bb . It follows that k/ (1+ f )(1+ g ) ∈ C ∗ ( X ) and k/ (1+ f )(1+ g ) ≤ rr h . Then, by Proposition 2.1 (i), k ≤ rr h (1 + f )(1 + g ) showing that h (1 + f )(1 + g ) = f ∧ rr g in C( X ). (cid:3) Since for any space X , C ∗ ( X ) ∼ = C ( βX ), where βX is the ˇCech-Stone com-pactification of X , Proposition 3.9 shows that if X is rr -good so is βX . It is nothard to show that if X is rr -good so is any space V , X ⊆ V ⊆ βX . Moreover,since for any space Y between X and its realcompactification υX , C ( Y ) = C( X ),if X is rr -good so is V . (See [ , Chapter 8].)It is next shown that cozero sets in rr -good spaces are also rr -good, for example,any open set in an rr -good metric space. Proposition . Let X be a completely regular topological space which is rr -good. Let Y = coz k where k ∈ C ( X ) . The space Y is rr -good. Proof.
First note that k may be assumed to be bounded. If f ∈ C ∗ ( V ) then f k can be extended to ˜ f ∈ C( X ) by defining ˜ f ( x ) = 0 if x ∈ X \ Y (see [ ,Proposition 1.1]). Notice also that an rr -lower bound of two bounded functions isnecessarily bounded.It will first be shown that C ∗ ( Y ) is an rr -lower semi-lattice; the result for C( Y )follows from Proposition 3.9. Consider f, g ∈ C ∗ ( V ). Extend f k and gk as aboveto get ˜ f and ˜ g in C( X ). Let H = ˜ f ∧ rr ˜ g . Then, in particular, H = ˜ f H = ˜ gH . Set h = H | V showing that h = f kh = gkh . Consider 1 /k ∈ C( Y ). Then, ( h/k ) = f ( h/k ) = g ( h/k ). Thus h/k is a lower bound for f and g . Since f, g ∈ C ∗ ( V ), itfollows that h/k ∈ C ∗ ( V ).Suppose next that a is also a lower bound for f and g . Then, ak ≤ rr f k and ak ≤ rr gk . Consider ˜ a . It follows that (˜ a ) = ˜ a ˜ f = ˜ a ˜ g since the equality holds on Y and all the functions are 0 on X \ Y . Therefore, ˜ a ≤ H and ˜ a | Y is a lower bound of f k and gk , showing that ak ≤ rr h . Then, upon multiplication by 1 /k , a ≤ h/k .Thus, h/k = f ∧ rr g . (cid:3) It is not true that the restriction homomorphism C( X ) → C( Y ) in Proposi-tion 3.10 preserves ∧ rr . Example . Let X = R and Y = (0 , , a cozero set, and ρ : C ( X ) → C ( Y ) .Then, ρ does not preserve rr -infima. Proof.
Consider two function f and g in C( X ) which are never 0 except at x = 1 / , / f ∧ rr g = 0 in C( X ) but ρ ( f ) ∧ rr ρ ( g ) = h = 0 where h coincides with f and g on (0 , /
2) and is 0 on [1 / , (cid:3) In the context of C( X ) it is possible to construct other examples of ring surjec-tions where countable rr -orthogonal sets lift to rr -orthogonal sets. (See also Theo-rem 2.10.) A subspace V of X is C-embedded if every element of C( V ) extends toan element of C( X ), i.e., the restriction map ρ : C( X ) → C( V ) is surjective. Proposition . Let X be an rr -good space, V a C-embedded subspace and ρ : C ( X ) → C ( V ) the restriction homomorphism. If { f n } n ∈ N is an rr -orthogonalset in C ( V ) then there is an rr -orthogonal set { F n } n ∈ N in C ( X ) such that for all n ∈ N , ρ ( F n ) = f n . Proof.
The proof will be by induction. For F , pick any pre-image of f . Let n ≥ rr -orthogonal set F , . . . , F n has been found with, for1 ≤ i ≤ n , ρ ( F i ) = f i . Choose G ∈ C( X ) to be any pre-image of f n +1 .For 1 ≤ i ≤ n , let U i = coz( G ∧ rr F i ). If all the U i are empty then put G = F n +1 . In any case, notice that U i ∩ V = ∅ since G | V ∧ rr F i | V = 0 for 1 ≤ i ≤ n .It will be shown that, in any case, the U i are pairwise disjoint. Now supposethat for some 1 ≤ i < j ≤ n that U i ∩ U j = ∅ . On U i ∩ U j , all of G, F i and F j coincide. Suppose that bd ( U i ∩ U j ) = ∅ and that all three functions are 0 onit. Then let H be such that H coincides with G, F i and F j on U i ∩ U j but is 0elsewhere. This function is continuous. Moreover, H ≤ rr F i and H ≤ rr F j . This isimpossible since H = 0 and F i ∧ rr F j = 0. If follows that for some x ∈ bd ( U i ∩ U j )that G ( x ) = F i ( x ) = F j ( x ) = 0 (they coincide by continuity). In this case x ∈ U i ∩ U j . To see this consider a net { λ α } α ∈ A in U i ∩ U j which converges to x .Think of this as a net in U i ; since F i ( x ) = 0, the limit of the net, x , is in U i andnot in bd U i . Similarly, x is in U j . Thus, x ∈ U i ∩ U j . This is impossible by thedefinition of the boundary. The only remaining possibility is that bd ( U i ∩ U j ) = ∅ .This means that U i ∩ U j is clopen and thus the cozero set of an idempotent, say e ∈ C( X ). But then eG will be a lower bound for F i and for F j . This cannothappen unless e = 0 and then U i ∩ U j = ∅ .Hence, for each 1 ≤ i < j ≤ n , U i ∩ U j = ∅ . Notice also that each U i ⊆ X \ V .Now G can be modified as follows: Define G ′ by G ′ ( x ) = ( , if x ∈ S ni =1 U i G ( x ) , otherwiseThe function G is zero on each bd U i which shows that G ′ ∈ C( X ). Since G and G ′ coincide on V , they have the same image in C( V ). It remains to be shown that F i and G ′ are rr -orthogonal for i = 1 , . . . n . Suppose, for some i , that H is an EDUCED RING ORDER 17 rr -lower bound of F i and of G ′ . Put W = X \ S nj =1 U j . Since G ′ is 0 on S nj =1 U i ,coz H ⊆ W . Recall that coz( F i ∧ rr G ) = U i . Define a new function K as follows: K ( x ) = ( F i ∧ rr G )( x ) , x ∈ U i H ( x ) , x ∈ coz H , otherwiseThe function K is continuous and is an rr -lower bound of F i and of G . However, F i ∧ rr G ≤ rr K showing that K = F i ∧ rr G . But then, coz H = ∅ and H = 0. (cid:3) As examples where Proposition 3.12 can be used, recall that any compactsubspace V of X is C-embedded ([ , 6J]) and, hence, if X is rr -good the propositionapplies. It is also easy to see that if, in the above, the subspace V is only C ∗ -embedded then countable rr -orthogonal sets of bounded elements of C( V ) lift to rr -orthogonal sets in C( X ). References
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