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Sheaf methods applied to coherent rings Carson, Andrew Bruce 1971

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SHEAF METHODS APPLIED TO COHERENT RINGS by ANDREW BRUCE CARSON B . S c , C a r l e t o n U n i v e r s i t y . O ttawa, O n t a r i o 1966 A THESIS SUBMITTED IN PARTIAL FULFILMENT OF THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY i n t h e Department o f MATHEMATICS We a c c e p t t h i s t h e s i s as c o n f i r m i n g t o t h e r e q u i r e d s t a n d a r d The U n i v e r s i t y o f B r i t i s h C o l u m b i a F e b r u a r y 1971 In present ing th i s thes is in pa r t i a l f u l f i lmen t o f the requirements fo r an advanced degree at the Un ivers i ty of B r i t i s h Columbia, I agree that the L ibrary sha l l make it f r ee l y ava i l ab le for reference and study. I fu r ther agree tha permission for extensive copying of th i s thes is fo r scho la r l y purposes may be granted by the Head of my Department or by his representat ives . It is understood that copying or pub l i ca t ion of th i s thes is fo r f inanc ia l gain sha l l not be allowed without my wr i t ten permiss ion. Department The Univers i ty of B r i t i s h Columbia Vancouver 8, Canada Supervisor: Dr. S.S. Page ABSTRACT A commutative r i n g i s c a l l e d coherent i f the i n t e r s e c t i o n of any two f i n i t e l y generated i d e a l s i s f i n i t e l y generated and the a n n i h i l a t o r i d e a l of an a r b i t r a r y element of the r i n g i s f i n i t e l y generated. Pierce's representation of a r i n g R as the rin g of a l l g l o b a l sections of an appropriate sheaf of rin g s , k , i s described. Some theorems are deduced r e l a t i n g the coherence of the r i n g R to c e r t a i n p roperties of the sheaf k . The sheaves from the above representa-t i o n f o r RfX~I and R \TG + ~n > where R i s a commutative von Neumann regular r i n g and G i s a l i n e a r l y ordered abelian group, are cal c u l a t e d . A p p l i c a t i o n s of the above theorems now show that RfX~l i s coherent and y i e l d necessary and s u f f i c i e n t conditions f o r RTlG +~n to be coherent. i i i . TABLE OF CONTENTS Page I n t r o d u c t i o n 1 0 R e p r e s e n t a t i o n o f r i n g s by s e c t i o n s o f sh e a v e s . 7 1 Cohe r e n t r i n g s . 17 2 Coherence o f R\X~] . 34 3 R e p r e s e n t a t i o n o f RfTG T I by s e c t i o n s o f s h e a v e s . 36 4 The s t r u c t u r e o f R when R r r G + _ l l i s n o t c o h e r e n t and G j- I , t h e i n t e g e r s . 60 5 A c o n d i t i o n f o r RrTXTl t o be c o h e r e n t . 64 6 A n e c e s s a r y c o n d i t i o n f o r RlTX~ n t o be c o h e r e n t . 68 7 Example of a B o o l e a n r i n g R t h a t i s A - s e l f - i n j e c t i v e b u t i s n o t - c o m p l e t e . 76 i v . ACKNOWLEDGEMENTS The author wishes to express h i s thanks to his research supervisor, Dr. S.S. Page, for encouragement and advice given during the preparation of t h i s t h e s i s . The f i n a n c i a l support of the H.R. MacMillan family and the Univ e r s i t y of B r i t i s h Columbia i s g r a t e f u l l y acknowledged. Introduction In t h i s thesis a l l r i n g have unity, a l l modules are u n i t a l , and a l l non-zero ri n g homomorphisms preserve the i d e n t i t y . Except i n the i n t r o d u c t i o n a l l rings are commutative. Some terminology i s now introduced. ( D e f i n i t i o n ) Let R be a r i n g . A l e f t R-module M i s f i n i t e l y  presented i f f there e x i s t s an exact sequence of l e f t R-modules 0-»-K-»-F-»-M-»-0 where F and K are f i n i t e l y generated and F i s free. ( D e f i n i t i o n ) A r i n g R i s l e f t coherent i f f every f i n i t e l y generated l e f t i d e a l i s f i n i t e l y presented. Corresponding d e f i n i t i o n s may be made with respect to r i g h t modules and r i g h t i d e a l s . A l e f t coherent commutative r i n g i s said to be coherent. The following d e f i n i t i o n allows an i n t e r n a l descrip-t i o n of coherent rings to be given. ( D e f i n i t i o n ) Let R be a r i n g . a) R has property a) i f f the i n t e r s e c t i o n of any two f i n i t e l y generated l e f t i d e a l s i n R i s f i n i t e l y generated. b) R has property b) i f f the i n t e r s e c t i o n of any two p r i n c i p a l i d e a l s i s p r i n c i p a l . c) R has p r o p e r t y c) i f f f o r any r E R the l e f t - a n n i h i l a t o r i d e a l of r ( l . a n n ( r ) = {s e R : s r = 0} ) i s f i n i t e l y g e n e r a t e d . d) R has p r o p e r t y d) i f f f o r any r e R l . a n n (r ) i s generated by an idempotent . e) R has p r o p e r t y e) i f f any f i n i t e l y generated l e f t i d e a l i n R i s p r i n c i p a l . The concept of a coherent r i n g was i n t r o d u c e d by Chase i n [4]. He showed t h a t f o r any r i n g R d i r e c t products of f a m i l i e s of f l a t r i g h t R-modules are f l a t i f f R i s l e f t coherent . The f o l l o w i n g i s p a r t of theorem 2.1 from tha t paper : Theorem: A r i n g R i s l e f t coherent i f f i t has p r o p e r t i e s a) and c ) . I n v iew of t h i s v a r i o u s combinat ions of p r o p e r t i e s a) - e) can be used e i t h e r to g e n e r a l i z e , d e f i n e , or s p e c i a l i z e the concept of coherence. Some elementary p r o p e r t i e s of f i n i t e l y presented modules and ( l e f t ) coherent r i n g s appear as e x e r c i s e s i n B o u r b a k i [1]. The f o l l o w i n g are examples of l e f t coherent r i n g s : i ) Any l e f t N o e t h e r i a n r i n g . i i ) Any l e f t s e m i - h e r e d i t a r y r i n g . i i i ) As a p a r t i c u l a r case of i i ) , any von Neumann-regular r i n g . (A r i n g R i s von Neumann r e g u l a r i f f f o r each r e R there e x i s t s r ' e R such tha t r r ' r = r .) i v ) L e t {R } be a d i r e c t e d system of l e f t coherent r i n g s such a t h a t i f a < a 1 then R , i s a r i g h t f l a t R -module . Then — a — — l i m (R ) = R i s a l e f t coherent r i n g , a -y v) Let R be a l e f t N o e t h e r i a n r i n g and {X } be a f i n i t e or i n f i n i t e se t of i n d e t e r m i n a t e s commuting w i t h themselves and elements 4 3. of R . Then FpfX F\ , the r i n g of polynomials i n the indeterminates X with c o e f f i c i e n t s from the r i n g R , i s l e f t coherent, a v i ) Let the notation be as i n v ) . Then RtTCX VYl . the r i n g of formal power serie s i n the indeterminates X with c o e f f i c i e n t s from the r i n g R , i s l e f t coherent. Examples i ) , i i ) , and i i i ) are easily, v e r i f i e d . In addition, any semi-hereditary r i n g has property d). Example iv) i s from Bourbaki (ex 11, p. 63 of [1]). Example v) follows from iv) and the H i l b e r t basis theorem since RRXV1 = Um ({RrX,,...Xl : n i s a n a t u r a l number and'(X n,...X } Q {X }}) a I n 1 n a -»-Examples v) and v i ) are s i m i l a r . There i s some evidence to i n d i c a t e that, from a homological point of view, l e f t coherent rings are a reasonable g e n e r a l i z a t i o n of l e f t Noetherian rings i n the sense that they, rather than l e f t Noetherian rings, are the appropriate concept. For example Chase's c h a r a c t e r i z a t i o n of l e f t coherent rings i n terms of r i g h t f l a t modules may be viewed as a g e n e r a l i z a t i o n of exercise 4 p. 122 of [3] which states that the d i r e c t product of a family of r i g h t f l a t modules over a l e f t Noetherian r i n g i s r i g h t f l a t . The f a c t that l.gl.dim (R) = r.gl.dim (R) where R i s l e f t and r i g h t Noetherian i s a s p e c i a l case of the following r e s u l t about coherent r i n g s : i f R i s a l e f t (right) coherent r i n g and M i s a f i n i t e l y presented l e f t ( r i g h t ) R-module then Pd(M) = w.dim (M) . (l.gl.dim (R) denotes the l e f t g l o b a l dimension of R , r.gl.dim (R) denotes the r i g h t g l o b a l dimension of R , Pd(M) denotes the p r o j e c t i v e dimension of M and w.dim (M) denotes the weak dimension of M .) In p a r t i c u l a r c y l i c l e f t (right) modules of a l e f t (right) Noetherian r i n g are f i n i t e l y presented. To specia-l i z e t h i s to the Noetherian case the following r e s u l t i s used: for any r i n g R l.gl.dim (R) = sup (Pd(M) : M i s a c y c l i c l e f t R-module} A s i m i l a r r e s u l t holds on the r i g h t . Thus i f R i s l e f t and r i g h t Noetherian we have l.gl.dim (R) = sup {Pd(M) : M i s a c y c l i c l e f t R-module} = sup {w.dim (M) : M i s a c y c l i c l e f t R-module} <^w.dim (R) <_ r.gl.dim (R) By symmetry we also have r.gl.dim (R) <_ l.gl.dim (R) In view of the proceeding i t i s reasonable to enquire as to whether or not coherent rings are a g e n e r a l i z a t i o n of Noetherian rings with respect to non-homological p r o p e r t i e s . I t i s known that i f R i s l e f t Noetherian so are RrXl and RrrXll . This suggests the following two questions. F i r s t , f o r what coherent rings R i s RFX1 coherent? Second, f o r what coherent rings R i s RrTGn~l coherent where RrTG +~n denotes the r i n g of formal power serie s with c o e f f i c i e n t s from R and p o s i t i v e indices from the l i n e a r l y ordered abelian group G ? These questions are answered i n t h i s thesis for R i n the category of commutative von Neumann regular r i n g s . For the r e s t of the i n t r o d u c t i o n R denotes a commutative von Neumann regular r i n g . Let S be a commutative r i n g . The following f a c t i s c r u c i a l i n proving a l l theorems i n t h i s t h e s i s . There e x i s t s a Boolean space X and a sheaf K over X of indecomposable rings such that S i s isomorphic to the r i n g of a l l g l o b a l sections of K over X . This representation of S i s described i n more d e t a i l i n §0. In §1 (1.7 and 1.13 i n p a r t i c u l a r ) properties a), b), c ) , d), and e) f o r S and the weak global dimension are r e l a t e d to c e r t a i n a l g e b r a i c properties of the s t a l k s of the sheaf K and to c e r t a i n properties of the g l o b a l sections of K over X . In §2 the sheaf associated with RrX~l i s ca l c u l a t e d . I t then follows from §1 that R r X l has proper-t i e s a), b), c ) , d), and e) (and thus i s coherent) and that w.gl.dim (RfXl) = 1 . Now l e t S = Rff G+_in - In §3 a preliminary study of the structure of K i s made and used to show that S has property c) (or d)) i f f the Boolean r i n g of idempotents of R i s X -complete (see 3.3 of t h i s thesis) where X - i s an o r d i n a l depending on G . In sections 4,5, and 6 the s t a l k s of K are ca l c u l a t e d f o r various R and G i n s u f f i c i e n t d e t a i l to e s t a b l i s h ( v i a 1.7 and 1.13) that the following conditions are equivalent: i ) R|TG +~n has property a) . i i ) RfTG + rTl has property b). i i i ) R (TG + " n has property e) . iv) w.gl.dim (RlTG +71) <_ 1 . If G i s not isomorphic to the integers the above conditions are s a t i s f i e d i f f R i s a f i n i t e d i r e c t sum of f i e l d s , (See §4). I f G i s isomorphic to the integers then the above conditions are s a t i s f i e d i f f R has a p a r t i a l form of s e l f - i n j e c t i v i t y called. s e l f - i n j e c t i v i t y . (See sections 5 and 6.) In sections 6 and 7 i t i s shown by examples that both of the following implications are f a l s e : f i r s t , Rrrx"]"] has property a) (or an equivalent) -> RCTX~ll has property c ) . Second, RffXl'] has property c) -> RrfX~n has property a). I f R i s a Boolean r i n g the second i m p l i c a t i o n does hold while the f i r s t does not. I t should be noted that these r e s u l t s are not i n t o t a l agree-ment with the following theorem of Jensen (pp. 238 and 239 of [6]): For a Boolean r i n g R the following are equivalent: 1) R i s s e l f - i n j e c t i v e . 2) w.gl.dim ( R r r G + l - j ) = 1 and R f T G + T l i s coherent, f o r any l i n e a r l y ordered group G 3) RffG+~"]~] i s coherent f o r any l i n e a r l y ordered group G I f R i s atomic (as a Boolean algebra) then 4) i s equivalen to the preceeding conditions. 4) w.gl.dim (RrrG+~n) = 1 f o r any l i n e a r l y ordered group G Soublin ( i n [10] ) gives an example of a commutative coherent r i n g T such TTX~1 f a i l s to have property a) and thus f a i l s to be coherent. Soublin's example and the r e s u l t s of this thesis on R)TX17 suggest that i n non-homo l o g i c a l s e t t i n g s i t i s misleading to think of coherent rings as being a g e n e r a l i z a t i o n of Noetherian r i n g s . §0 Representation of rings by sections of sheaves A l l r e s u l t s i n this thesis (except f o r 0.11, 0.17, and 0.18) are paraphrased from part one of [8]. In t h i s section R denotes a commutative r i n g with unity. It should be noted that i n t h i s thesis a l l rings are commutative with unity. Let X be an index set and {M : x e X} be a family of • x i d e a l s i n R such that f\ (M ) = 0 . Then R may be represented xeX as a subdirect product of the k where k = R/M . However such x x x a representation gives l i t t l e information about R unless there i s a reasonable way of determining which subring of (© IT) (k ) i s xeX X isomorphic to R . Pierce i n [8] shows that i f X and {M : x e X} are appropriately chosen t h i s can be accomplished by topologizing X and k = (J (k ) i n such a way that {a c ( $ i r ) ( k ) : X ° k i s xeX xeX continuous} i s a subring of ( ©TT) (k ) isomorphic to R . In this xeX X s i t u a t i o n the k^ are a l l indecomposable rings. This can also be expressed by saying that there e x i s t s a sheaf k(R) of indecomposable rings over a t o p o l o g i c a l space X(R) such that R i s isomorphic to the r i n g of a l l global sections of k(R) over X(R) . The construc-t i o n of the t o p o l o g i c a l space X(R) and the sheaf k(R) , along with some b a s i c d e f i n i t i o n s and r e s u l t s , i s o u t l i n e d below. 8. 0.1) (D e f i n i t i o n ) Sheaves of Rings: Let X be a t o p o l o g i c a l space. Suppose that f o r each x e X a r i n g k with zero 0 and X x i d e n t i t y 1^ i s given. Assume k^ (~\ k^ = (j> f o r x =j= y Let k = U (k ) . Let IT: k -> X denote the map such that xeX X i f r e k then i r(r) = x . Assume that k i s topologized i n X such a way that the following three axioms are s a t i s f i e d . i ) I f r e k there e x i s t open sets (J i n k and N i n X such that r e U and ir maps (J homeomorphically onto N i i ) Let k + k , denoting {(r,s) e k x k : iT(r) = -rr(s)} , have the topology induced by the product topology i n k x k Then the mapping r ->• - r on k to k and the mappings (r,s) ->- r-s and (r,s) r + s on k + k to k are continuous. i i i ) The mapping x 1 on X to k i s continuous. X When these axioms are s a t i s f i e d k i s c a l l e d a sheaf of rings  over X . The rings k^ are c a l l e d the st a l k s of the sheaf k The p a i r (X,k) i s c a l l e d a ringed space. 0.2) (D e f i n i t i o n ) Isomorphisms of Ringed Spaces: Let (X,k) and (Y,S) be ringed spaces. An isomorphism of (X,k) onto (Y,S) i s a p a i r (A,p) where X i s a homeomorphism of X onto Y and y i s a homeomorphism of S onto k such that u maps S, . . isomorphically onto k for each x e X . (X,k) and A (x; x (Y,S) are isomorphic i f f there e x i s t s an isomorphism of (X,k) onto (Y,S) . 0.3) ( D e f i n i t i o n ) Subsheaves: Let k be a sheaf of rings over the to p o l o g i c a l space X . A subset k' of k i s c a l l e d a sub- sheaf of k i f k' i s open i n X and f o r each x e X 9. k ' n k i s a subring of k . I t i s e a s i l y v e r i f i e d that x x k' i s also a sheaf of rings over X when given the topology induced by k The concepts required to determine a subring of ir) (k ) xeX using the topologies on X and k are now introduced. I t should be noted that ( © T r ) ( k ) = {a : X 5- k and f o r each xeX X x e X Tr(a(x)) = x} 0 . 4 ) ( D e f i n i t i o n ) Sections: Let (X,k) be a ringed space and l e t Y be a subspace of X i ) A section of k over Y i s a continuous map a : Y ->- k such that TT(C(X)) = x f o r a l l x e Y . The set of a l l sections of k over Y i s denoted r(.Y,k) . The elements of T(X,k) are c a l l e d the glo b a l sections of k over X . To say that an element a e r(Y,k) can be extended to a global s e c t i o n means that there e x i s t s a' e r(X,k) such that a' = a i i ) Define pointwise addition and m u l t i p l i c a t i o n on r(Y,k) using the addi t i o n and m u l t i p l i c a t i o n i n the s t a l k s . Then r(Y,k) i s a r i n g . i i i ) For any a e r(X,k) l e t S(a) = {x e X : a(x) =f 0^} and l e t Z(CT) = {x e X : a(x) = 0 } . iv) Let U be a subset of X that i s both open and closed i n X . Define ii> : X ->• k by ^ T,(x) = 1 when x e U and U U x ty (x) = 0 when x e X - TJ . Then ik T e F(X,k) . This TU x U notation w i l l be used frequently i n this t h e s i s . The topologies on X and k allow the following r e l a t i o n -ship to be established amongst elements of T(X,k) 1 0 . 0.5) (lemma) Let (X,k) be a ringed space. Suppose that x e X and a, x e T(X,k) are such that a(x) = T (X ) . Then there e x i s t s N , a neighborhood of x , such that x N This r e s u l t i s used to show that c e r t a i n properties of a s t a l k k hold " l o c a l l y " as w e l l , x 0.6) ( C o r o l l a r y ) : Let (X,k) be a ringed space and l e t a e r(X,k) . Then Z(a) i s open and S(a) i s closed i n X 0.7) ( D e f i n i t i o n ) Boolean Spaces: Let X be a t o p o l o g i c a l space. i ) A subset U of X i s clopen i f f i t i s both open and closed i n X . i i ) X i s t o t a l l y disconnected i f f i t has a basis c o n s i s t i n g of clopen sets. i i i ) X i s a Boolean space i f f i t i s compact, Hausdorff, and t o t a l l y disconnected. Sheaves used i n t h i s thesis w i l l be over Boolean spaces. The next p r o p o s i t i o n asserts that Boolean spaces have a very s p e c i a l form of compactness. 0.8) (Proposition) Let X be a Boolean space. Then X has the p a r t i t i o n property. That i s to say i f * s a covering of X by open sets there e x i s t s {P^,...Pn} , a f i n i t e c o l l e c t i o n of clopen subsets of X , such that: 11. i ) For 1 < i < n there e x i s t s an a. such that P. cr N — — i x — a. i i ) P ± n ?. = * for i =)= j i i i ) U (P.) = X . 1=1 x The c o l l e c t i o n {P^,...P^ i s c a l l e d a p a r t i t i o n of X r e f i n i n g the cover {N } a 0.9) ( D e f i n i t i o n ) Reduced Ringed Spaces: A ringed space (X,k) i s a reduced ringed space i f f i ) X i s a Boolean space i i ) For each x e X i s an indecomposable r i n g . It should be noted that i f (X,k) i s a reduced ringed space and {P ,...,P } i s a p a r t i t i o n of X then the map n r ( X , k ) -> ® I r ( P . , k ) such that 1=1 1 n 1=1 i s an isomorphism. P. x This, along with 0.8, allows one to show that c e r t a i n " l o c a l " properties are a c t u a l l y " g l o b a l " i n the sense that they are also properties of r(X ,k) . We w i l l frequently derive properties of r(X ,k) from those of the k using 0.5 and 0. x The following two r e s u l t s w i l l also be us e f u l . 0.10) (lemma) Let (X,k) be a reduced ringed space and l e t Y be a closed subset of X . Then each element of T(Y,k) can be extended to a global s e c t i o n . 12. In §3 we s h a l l see that c e r t a i n p a r t i a l forms of s e l f - i n j e c t i -v i t y f o r the ri n g T(X,k) may be defined by asserting that an analogue of 0 .10 holds f o r c e r t a i n open subsets Y of X 0.11) (lemma) Let (X,k) be a reduced ringed space and x E X Then k =/lim (T ( U,k)) VXEU and U i s clopen i n X, Proof: I t i s a standard f a c t about sheaves that k - /lim ( r ( U , k ) ) \. The lemma follows from this \XEU and U i s open i n X since X i s a Boolean space and thus the set of clopen neigh-borhoods of x i s c o f i n a l i n the set of a l l neighborhoods of 0.12) ( D e f i n i t i o n ) Regular Ringed Spaces: A reduced ringed space (X,k) i s regular i f f f o r each x e X k i s a f i e l d . — — — — j£ 0.13) (lemma) Let (X,k) be a regular ringed space and l e t a e r(X,k) . Then. S(o~) and Z(a) are both clopen i n X . Pierce's construction of the reduced ringed space (X(R),k(R)) such that R i s isomorphic to T(X(R),k(R)) i s now outlined. 9 0.14) ( D e f i n i t i o n ) i ) Define- B(R) = {e e R : e = e} . Note that <B(R), +',•> i s a Boolean r i n g where e +' f = e + f - 2 e f i i ) Let X(R) be the set of maximal id e a l s i n the Boolean r i n g B(R). For any e E B(R) l e t X(e) =' {M E B(R) : e { M) . 13. Let X(R) have the topology induced by {X(e) : e e B(R)} I t i s w e l l known that X(R) i s a Boolean space. In f a c t (X(e) : e e B(R)} i s a clopen bas i s f o r the open sets i n X(R) i i i ) I f M e X(R) l e t M = {re : r e R and e e M} . I t may be v e r i f i e d that M i s an i d e a l i n R . Thus R/M i s a r i n g . Further, (R/M) Pi (R/N) = ty i f M =j= N . iv) For any M e X(R) l e t k^(R) = (R/M) . Let k(R) = U(k M(R)) and l e t TT : k(R) X(R) be defined such that MeX(R) ir _ 1(M) = k^(R) . v) For any r e R l e t cr^ : X(R) ->- k(R) be defined by c (M) = r + M e ^ ( R ) f o r any M e X(R) . Topologize k(R) by l e t t i n g {o"r(U) : r e R and U i s open i n X} be a basis f o r the open sets. I t i s r e a d i l y seen that f o r each r e R 0 r e T(X(R),k(R)) . v i ) Let R° denote (X(R),k(R)) . v i i ) For any reduced ringed space, (X,k) , l e t (X,k) denote r(x,k) . v i i i ) Define "\ R : R -> r(X(R),k(R)) by * \ R ( r ) = . 0.15) (Proposition) i ) ^ R : R + r(X(R),k(R)) = (R°)* i s a r i n g isomorphism. i i ) The correspondances ° and * are inverse, one-one, and onto between the family of isomorphism classes of rings and the family of isomorphism classes of reduced ringed spaces. i i i ) The r i n g R i s von Neumann regular i f f (X(R),k(R)) i s a regular ringed space. S i m i l a r l y a reduced ringed space (X,k) 14. i s regular i f f (X,k) i s a von Neumann regular r i n g , iv) Let (X,k) be a reduced ringed space. For each x e X l e t = {a E B((X,k)*) : a(x) = 0} . The map X + X((X,k)*) given by x -> i s a homeomorphism. It i s po s s i b l e to define morphisms of ringed spaces. The correspondances ° and * may then be extended to c a t e g o r i c a l equivalences. However these morphisms are i n general somewhat complicated. They may be avoided i n t h i s thesis through use of the following p r o p o s i t i o n . 0.16) (Proposition) Let R be a subring of the r i n g S such that B(R) = B(S) . Then c l e a r l y X(R) = X(S) . Let k = (r) : r e R} Then k i s a subsheaf of k(S) such that : R (X,k) R i s an isomorphism. Thus R° and (X,k) may be i d e n t i f i e d . 0.17) Let (X,k) = R° . Let x E X . i ) The map R -> k^ given by r -> o ^ x ) i s an epimorphism. i i ) As an Pv-module under t h i s epimorphism k i s f l a t . x Proof: i ) This i s e a s i l y checked. i i ) I t follows from the remark immediately a f t e r 0.9 that r(U,k) i s p r o j e c t i v e i f U i s clopen. The r e s u l t now follows from 0.11. Q.E.D. 0.18) (Theorem) Let (X,k) = R° . Then w.gl.dim (R) = sup {w.gl.dim (k ) : x e X} 1 5 . Proof: To show that sup {w.gl.dim (k ) : x e X } <_w.gl.dim ( R ) the following fact (paraphrased from ex. 1 0 p. 1 2 3 of [ 3 ] ) will be used: Let S -> T be a ring homomorphism such that T is flat as an S-module and let A be an S-module. Then w.dim T(A® s T) <_ w.dinig(A) where any T-module is given the S-module structure induced by the homomorphism S -> T . When this is applied in this proof A will be a T-module and S -> T will be onto so that we will have A® T = A as T-modules. Thus w. dim (A) = w.dim (A® T) < w.dim (A) . It follows from this that w.gl.dim(T) <_w.gl.dim(S) . By 0 . 1 7 and the preceeding comment i t thus follows that w.gl.dim(k ) < w.gl.dim(R) for x — each x E X . To complete the proof suppose that n is an integer such that w.gl.dim(k^) <_ n for each x e X . Let A and B be arbitrary R-modules and m > n an integer. Two facts will be used. First, for any x £ X (Tor (A,B))® k k = Tor X (A® k , B ® k ) = 0 . Second i f C is an R-module m ^ x ' x such that C ® k = 0 for a l l x e X then C = 0 . The lemma x is completed by using these two facts with C = TorM(A,B) . The fir s t fact follows from the hypothesis on m and a paraphrase of exercise 1.1 p. 1 2 3 of [ 3 ] . The paraphrase is true essentially because each k is flat as an R-module. The second fact is x actually just a restatement of the proposition ( 1 . 7 in [ 8 ] ) that A ( M C ) = 0 . This implies that the natural map ME X ( R ) C (<©JT) (C/MC) is a monomorphism. ME X ( R ) 16. L e t M E X(R) . Then . C ® 1^ = C® ( l i m ( r ( U , k ) ) ) = U m (C ® r ( U , k ) ) = (C/MC) . (xeU and U i s c l o p e n ) ( x e U and U i s c lopen) Thus i f C $ k = 0 f o r each x e X we must have C = 0 . x Q . E . D . The p r e c e e d i n g theorem i s t r u e e s s e n t i a l l y because each s t a l k k i s a d i r e c t l i m i t of d i r e c t summands of R and the x R f u n c t o r s T o r M ( > ) commute w i t h d i r e c t l i m i t s . S ince the M f u n c t o r s Ext ( , ) do not n e c e s s a r i l y commute w i t h d i r e c t R l i m i t s i t i s not s u p r i s i n g that there i s no analogue of theorem 0.18 f o r the o r d i n a r y g l o b a l dimension of R Theorem 0.18 p r o v i d e s a means of c a l c u l a t i n g the weak g l o b a l d imension of R from those of the s t a l k s . No analogue of t h i s h o l d s f o r a r b i t r a r y s u b d i r e c t p r o d u c t s . For i n s t a n c e the r i n g I of i n t e g e r s i s a s u b d i r e c t product of the f i e l d s I / M such tha t M i s a maximal i d e a l i n I . However the I / M a l l have weak g l o b a l dimension zero y e t w . g l . d i m ( I ) = 1 An elementary method f o r c o n s t r u c t i n g examples of r i n g e d spaces i s now g i v e n . 0.19) ( D e f i n i t i o n ) S imple Sheaves: L e t X be a t o p o l o g i c a l space and l e t S be a r i n g . L e t S have the d i s c r e t e t o p o l o g y . L e t k = X x S have the p r o d u c t t o p o l o g y . For each x e X l e t k^ = {x} x s . Then k i s a sheaf of r i n g s over X c a l l e d the s imple S sheaf over X 17. §1 Coherent rings In t h i s section R w i l l denote a commutative Ring with unity and (X,k) w i l l denote R° . Some r e s u l t s w i l l be obtained r e l a t i n g properties a), b), c), d), and e) f o r the r i n g R with cer-t a i n properties of the reduced ringed space (X,k) 1.1) (Proposition): Consider the following conditions: i ) R has property c ) . i i ) R has property d). i i i ) For any T e r(X,k) , S ( T ) i s clopen. i v ) k i s a Hausdorff space. Then i i ) i ) and i i i ) < — > i v ) . I f each k i s an i n t e g r a l X domain then each of i ) , i i ) , i i i ) , and i v ) are equivalent. Proof: i i ) -*• i ) : This i s obvious. i i i ) ->iv): By 0.14.v) f o r any a e T(X,k) and x e X , sets of the form a(N ) where i s a clopen neighborhood of x form a basis f o r the neighborhoods of a(x) i n k . Suppose that k i s not Hausdorff. Thus, since X i s Hausdorff, there e x i s t two points on the same s t a l k , say k , that can not be separated by open sets. Thus there e x i s t a and x (see 0.17) i n r(X,k) and x £ X such that a(x) f x(x) yet f o r any clopen neighborhood of x,N , there e x i s t s y £ N such that X X o(y) = x(y) . Thus S(a - x) i s not open. iv) -> i i i ) : Suppose there e x i s t s T Z T(X,k) such that S(x) i s not clopen. Then S(x) i s not open (see 0.6) so there 18. e x i s t s x e S(x) such that f o r each clopen neighborhood of x, N , N O (X - S(x)) {= <j> • Then 0 (x) = 0 f- T(X) yet f o r each X X OT-V X K N there e x i s t s y e (an (N ) f\ x(N )) . Thus k i s not Hausdorff. x 0^ x x Now suppose that each k^ i s an i n t e g r a l domain. Since R - r(X,k) R can be replaced with T(X,k) f o r the r e s t of t h i s proof. n i ) -y i i i ) : Let x e r(X,k) be such that l.ann(x) = I r(X,k)x. n i = l S(x)£ x - U (S(x.)) since the k are domains. Let i = l 1 X n x e (X - U (S(x.))) and suppose x(x) = 0 . .Find N , a , _ 1 X x 1=1 n clopen neighborhood of x , such that N Q (X - U S(x.)) X i = l 1 (see 0.6) and y e N implies x(y) = 0 . Then \> e l.ann(x) x y IN x n yet \. E T(X,k)x^ . This contradicts the choice of the x^ x i = l n n and y i e l d s X - U ( S ( x . ) ) C S ( x ) . Thus S ( x ) = X - U ( S ( T . ) ) . i = l 1 i = l 1 Since the support of a s e c t i o n i s always closed this establishes that S(x) i s clopen. i i i ) -y i i ) : Now l e t x e r(X,k) be such that S(x) i s clopen. Since the k^ are i n t e g r a l domains, a e l.ann(x) <—> S(a) C l X - S(x) . Thus, i f a e l.ann(x) , a = ^ ( X _ S ( ^ ) ) £ r ( x > k ) * ^ ( X - s ( T ) ) ^/v qr ^ e r( x>^) since (X - S(x)) i s clopen. Hence l.ann(x) vX-a {.x)) i s generated by an idempotent. Q.E.D. In order to r e l a t e properties a), b), and e) f o r the r i n g R with c e r t a i n properties of the sheaf (X,k) i t i s necessary to have some lemmas allowing i d e a l s i n R to be studies i n terms of the k 19. 1.2) ( D e f i n i t i o n ) : For I an i d e a l i n R and x e X l e t I = {o (x) : a e 1} . X d I t i s clear that each I i s an i d e a l i n k and that i f x x I i s f i n i t e l y generated as an R-ideal then each I i s f i n i t e l y X generated as a k ^ - i d e a l . (1.3) (lemma): Let I and J be two id e a l s i n R and x e X Then (I f] J) = I f l J x x x Proof: C l e a r l y (I P\ J) C I r\ J . Now l e t a (x) e I O J X X X C X X Thus there e x i s t a e I and b e J such that a (x) = a(x) = c a a, (x) . Find N , a clopen neighborhood of x such that D X = a = cr. a b N N X X Find e e R such a = iKT ™ e N N x x Then a = \p a = ^„ a. = a , so ea = eb E I C\ J ea N a N b eb x x Thus a (x) = i/> (x)a (x) = o (x)a (x) = a (x) e (I O J ) c N a e a ea x x Thus I H J — (I n J ) . Hence I f\ J = (I H J) x x x x x x Q.E.D. 1.4) (lemma): Let I and J be two i d e a l s i n R i ) Let c , f o r some r e R , be such that a (x) e I f o r r r x a l l x £ X . Then r e I . i i ) I f I = J f o r a l l x e X then I = J . x x 20. Proof: i) For each x £ X find a e I such that x oa (x) = a (x) and find N , a clopen x r x neighborhood of x such that aa x = a N x N x {N : x e X} x is an open cover of X so that by the partition property there exists {P^ ,...P^ } , a partition of X into clopen sets, such that each P. C some N from the cover. For each i 3 - x. ( 1 < i < n) let e. e R be such that a = ty^ . For any — — l e. P. x 1 x e X x E some P. so J n n (a Z e -a ) (x) = Z ty M'a (x) = a (x) = a (x) . . . i x . . . . J r . a a r 1=1 l 1=1 l x. x. 1 3 n Thus r = Z e.'a e l . . , i x . 1=1 l ii ) Suppose that I = J for a l l x z X . Let a E I x x Then for a l l x e X a (x) e l = J so by i) a e I . Thus a x x J I C J . Similarly J £ I . Q.E.D. It is clear that if I is a finitely generated R-ideal then each I is a finitely generated k -ideal. The converse x x is not in general true. For example suppose that X contains a non-isolated point, y , and that each k^ is a field. Let I = {r E R : a (y) = 0 } . Then each I is a principal ideal r y x in k^ yet I is not finitely generated as an R-ideal. The next two lemmas are partial converses. 21. 1.5) (lemma) Let n be a positive integer and I an ideal in R such that for each x e X there exists T , . . . T e T(X,k) 1, x n, x and N a neighborhood of x such that for any y E N X X {T, (y),...T (y)} generates I . Then there exists a sub-x y x n j x y set of R containing n or fewer elements that generates I Proof: {N : x e X} where the N are as described in the x x hypothesis is an open cover of X so that by the partition property there exists {P,,...P } , a partition of X into clopen sets 1 m • such that each P.C some N from the cover. Find a. e R i — x. 1 l m n (1 <_ i <_ n) such that a = E T . ty . Let J = E R-a. i j=i J . J 1=1 for any x e X , x e some P.' so that J m J x = * ( k x ( . E n T i , x . ' V ) 0 0 ) 1=1 J=l J J = E (k * T. (x)) = I ... X 1,X . , X 1=1 -]' n Thus by 1.14 I = J = E R-a. . i - l 1 Q.E.D. 1.6) (lemma) Let I be an ideal in R such that for each x e X there exists a positive integer n(x) . T, ,...T , n e T(X,k) l,x n(x),x and N , a neighborhood of x , such that for any y e N X X {"rn (y)»-..T , N (y)} generates I . Then I is finitely 1, x n(x),x x generated as an ideal in R 22. Proof: {N : x E X} where the N are as described in the x x hypothesis forms an open cover of X so that by the partition property there exists {P^ ,...P^ } , a partition of X into clopen sets, such that P. C some N from the cover. Let J m n = sup (n(x.) : 1 < i < m} . Define T. = Ex.. i — — i . , i,x. P. J=l J J (1 < i < n) where i t is understood that x. =0 for - - x,x. n(x.) < i < n . Then for each x E X {x,(x),...x (x)} generates J JL I n I so by 1.5 I is generated by some subset of R containing X n or fewer elements. Q.E.D. The next proposition applies the foregoing to obtain various algebraic conditions on the that are sufficient to imply that R has some of properties a), b) and e). 1.7) (Proposition) Suppose that for each x E X k has property X e . Then: i) R has property e). ii ) If each k^ is an integral domain then R also has proper-ties a) and b) and w.gl.dim(R) 1 i i i ) If R has property d) then R is semi-hereditary. Proof: i) It suffices to show that the ideal I = Ra + Rb is principal where a and b E R are arbitrary. Fix arbitrary x E X . Then I = k a (x) + k o, (x) is a finitely generated x x a x b ideal in k so there exists c e l (for simplicity c will x x denote c ) such that k cr (x) + k a, (x) = k a (x) . Thus x x a x b x c 23. there exists r, s, t^, and t^ e R such that I a (x)a (x) + a (x)a, (x) = a (x) ; and ~C 3. S D C II c (x) = a (x)a (x) and a, (x) = a (x)a (x) a t^ c b t£ c Since there are only finitely many equations above there exists N , a neighborhood of x such that for any y e N X X I' a (y)o (y) + o (y)a (y) = a (y) ; and i ' cL o D C II' a (y) = a_ (y)cr (y) and a, (y) = a. (y)a (y) a t^ c b t2 c Conditions I' and II' combine to show that whenever y e N , k a (y) + k a, (y) = k a (y) . Thus, by lemma 1.5, J x y a J y b  J y c J J I is principal. i i ) The following fact will be used in this proof: If D is an integral domain with property e) and x,y,w e D are such that 0 = D + D = D then - , ^  , and — e D x y w w w w and D O D = D . (— etcetera denotes the element x y (|Z) w x«(w L) in the classical ring of quotients for D .) This is analogous to the situation in Euclidean domains. Let R have property e). Then properties a) and b) are equivalent for R . Therefore i t suffices to show that for any a,b e R , i n J is principal where I = Ra and J = Rb . Let C = S(cr ) C\ S(a, ) . C is closed in X . a b By i) find c e R such that Ra + Rb = Rc . Since k a (x) + k a. (x) = k a (x) for a l l x e X i t follows from x a x b x c the opening remark that x e C implies that 24. a (x) a (x) a (x)a, (x) a b , a b , , 7—r- , 7—r- , and 7—r e k and cr (x) a (x) a (x) x a (x)o, (x) (k x a a (x))n ( V b ( x ) ) = k x ( a o ) c Now I s h a l l show that the maps a , a : C •+ k g i v e n by ct p a a ( x ) a (x) = —7—T-a a (x) a b ( x ) and a (x) = 7 — r are c o n t i n u o u s . For B a (x) a a ( x ) a r b i t r a r y x e C f i n d r e R such t h a t a (x) = —7-^- = a (x) a a (x) r Thus cr (x) = a (x) *a (x) . P i c k a neighborhood of x , N , a r c x such t h a t y e N i m p l i e s a (y) = a (y)a ( y ) . Hence y e C O N i m p l i e s x a r c x ° (y) = a (y) . Thus a r  J a N A C x S ince a N A C x i s cont inuous at x so i s a a S ince C i s c l o s e d i n X we may assume w i t h o u t l o s s of g e n e r a l i t y t h a t e r ( X , k ) (See 0 . 1 0 ) . S i m i l a r l y cr^ i s cont inuous and may be viewed as a member of r ( X . k ) . Observe tha t a o\ a b B a S ince C and S(a a. - ana ) are d i s j o i n t c l o s e d s e t s i n the a b g a J Boolean space X t h e r e e x i s t s a c l o p e n se t P such that C £ P and S(a a, - ana ) £ X - P . a b B a Thus a a, a b a.o 3 a D e f i n e a e r ( X , k ) by a (x) = a (x)a, (x) = a D (x )a (x) m m a b B a f o r x e P , and a (x) = 0 f o r x \. P T h i s i s cont inuous m s i n c e P i s c l o p e n . Observe t h a t by i t s c o n s t r u c t i o n a a ( x ) a b ( x ) o" (x) = 7—r f o r x e C and a (x) = 0 f o r x | C m a (x) ™ T m Thus, u s i n g the opening d i s c u s s i o n and 1 .3 , f o r x e C 25. (R ) = k a (x) = k (— ) v m x x m x a (x) k a (x) A k o, (x) = 1 A J = ( I r\ J) x a x b x x x For x i C (R ) = 0 = 1 A J = ( I A J) . Thus by 1.4 T m x x x x R = i n J = R P * R 1 _ . m a b To see t h a t w . g l . d i m ( R ) <_ 1 the f o l l o w i n g f a c t (from thm. 4 .1 of [4] w i l l be used : I f S i s a s e m i - h e r e d i t a r y r i n g then w . g l . d i m ( S ) <_ 1 . S ince f o r each x e X k i s an i n t e g r a l domain i t f o l l o w s that p r i n c i p a l i d e a l s i n a k are p r o j e c t i v e so each X k i s s e m i - h e r e d i t a r y . Hence w . g l . d i m ( R ) = sup { w . g l . d i m ( k ) : X X x e X} <_ 1 . i i i ) Suppose that R has p r o p e r t y d ) . S ince by i ) i t has p r o p e r t y e) i t s u f f i c e s to show t h a t the p r i n c i p a l i d e a l s of R are p r o j e c t i v e . Le t a e R be a r b i t r a r y and f i n d e e B(R) such tha t l . a n n ( a ) = Re. Then 0 -> Re Re © R / n N = R Ra + 0 (1-e) (where r ->- r . a ) i s a s p l i t exact sequence so t h a t Ra i s p r o j e c t i v e . Q . E . D . 1.8) ( D e f i n i t i o n ) A r i n g S i s l o c a l i f f i t has a unique maximal i d e a l . In p r o p o s i t i o n 1.13 we use 1.7 to f i n d a h o m o l o g i c a l c o n d i t i o n on R t h a t , i f each k^ i s a l o c a l r i n g , i s s u f f i c i e n t f o r R to have c o n d i t i o n s a ) , b) and e ) . 26. First three lemmas about local rings are needed. 1.9) (lemma) Let S be a commutative ring. The following condi-tions are equivalent: i) S is a local ring, i i ) All nonunits of S are contained in a proper ideal M i i i ) The nonunits of S form an ideal, iv) For r and s e S r + s is a unit implies that either r is a unit or s is a unit. Proof: i) <—> i i ) <—> i i i ) is proposition 5 from §2.2 of [7]. i)'<—> iv) is essentially the same as exercise 7 from §2.2 of [7]. Q.E.D. 1.10) (Corollary) Let S be a commutative local ring. Let 0 = a e S and x c S be such that xa = a . Then x is a unit. Proof: Rewrite xa = a as (1-x)*a = 0 . Since a =j= 0 this shows that 1-x is not a unit. Then by 1.9 iv) x is a unit since x + (1-x) = 1 Q.E.D. 1.11) (lemma) Let S be a commutative local ring with zero divisors. Then w.gl.dim(S) > 1 . 27. Proof: Find a, b non-zero in S such, that a.b = 0 I shall use exercise 5 from chapter 6 of [3] to show that the canonical map S (7) S, — ! g ) S - S is not a monomor-3. o D 3. o 3. phism so w.dinu (S ) > 0 . Since 0 -> S -> S ->(S/S ) ->• 0 is S a a a an exact sequence of modules this would show w.dim (S/S ) > 1 o Si and hence w.gl.dim(S) > 1 Suppose that S ® S - > S ® S is a monomorphism. a b b a b Thus, since a.b = 0 , a e S and b e S, , there exists a b (by the exercise) r^a e S^  and s j e ^ (1 <_ j < some integer n ) n such that: i) a = £ (r.a s\ ; and j = i J y i i ) s.b = 0 for 1 < i < n . 3 n n from i) get a = ( Z r.s.) a so that by 1.10 ( E r.s.) is a unit. Thus by 1.9 iv) some r.,s., is a unit and hence J 3 s., is a unit. This is a contradiction since s.,b = 0 yet 3 3 b =f 0 . This contradiction establishes that S ( $ Q S->-S $ S is not a monomorphism so w.gl.dim(S) > 1 Q.E.D. 1.12) (lemma) Let S be a commutative local ring. Then: i) w.gl.dim(S) <_ 1 implies that S is an integral domain. i i ) w.gl.dim(S) <_ 1 implies that S has property e). Now suppose that S is also semi-prime. That is to say s n = 0 -> s = 0 for each positive integer n and s e S Then: i i i ) S has property e) implies that w.gl.dim(S) <_ 1 Proof: i) This is just a restatement of 1.11. ii ) Let w.gl.dim(S) _< 1 In theorem 4.2 of [4] Chase has 28. shown that i f D i s an i n t e g r a l domain then D i s semi-heredi-tary i f f w.gl.dim(S) <_ 1 . It thus follows from i ) that S i s semi-hereditary. It i s w e l l known that i f a r i n g i s l o c a l a l l i t s f i n i t e l y generated p r o j e c t i v e modules are f r e e . Let 0 =|= I be a f i n i t e l y generated i d e a l i n S . Then I i s pro-j e c t i v e and hence fre e . Thus there e x i s t s {a.,...a } , a free - I n basis f o r I . I f n = 1 we are done. Suppose n > 1 Then a^ • a^ - a^ •,a^ = 0 . This contradicts our choice of {a.,...a } . This c o n t r a d i c t i o n establishes n = 1 . Thus 1 n 1 i s p r i n c i p a l so that S has property e). i i i ) Let S be semi-prime with property e). I t s u f f i c e s to show that S i s an i n t e g r a l domain since then every p r i n c i p a l i d e a l would be p r o j e c t i v e . Since S has property e) t h i s would e s t a b l i s h that S i s semi-hereditary and thus w.gl.dim(S) <_ 1 . Let a,b e S be such that a =j= 0 , b ={= 0 , yet ab = 0 Find c e S such that S + S. = S . Hence f i n d x,y,r,s e S a b c , J such that ra + sb = c , a = xc, and b = yc. Hence (rx + sy)c = ra + sb = c so by 1.10 rx + sy i s a unit. Thus by 1.9 e i t h e r rx or sy i s a un i t . Suppose without loss of g e n e r a l i t y that rx i s a u n i t . Then so i s x 2 xb = xycb = yxcb = yab = 0 . Thus, since x i s a un i t , 2 b = 0 so b = 0 . This c o n t r a d i c t i o n establishes that S i s an i n t e g r a l domain. Q,E.D. 29 . 1 .13) (Proposition) Let each k be a local ring. Then: X i) w.gl.dim(R) <_ 1 implies that R has properties a), b) and e). ii ) If R has property d) then w.gl.dim(R) <_ 1 implies that R is semi-hereditary. Now suppose that R is semi-prime. Then: i i i ) If R has property e) then w.gl.dim(R) <_ 1 iv) If R has property e) then R also has properties a) and b). Proof: i) Suppose that w.gl.dim(R) <_ 1 . By 0 . 1 8 w.gl.dim(R) = sup .{w. gl.dim(k ) : x e X} . Thus for each x E X w.gl.dim(k ) < 1 so that by 1 .12 k is an integral domain X X with property e). By 1 . 7 this establishes that R has properties a) , b) and e) . ii ) This follows from i) and 1 . 7 . i i i ) . i i i ) Fix an arbitrary x e X . First note that k^ is semi-prime. To see this suppose there exists 0 =j= a (x) e k and a Q. X positive integer n such that (o (x)) n = 0 . Choose N , a 3. X clopen neighborhood of x , such that (a (y)) n = 0 for any 3. y e N . Choose e e R such that a - 4 , . Then x e N x (a „(y))n = (<kT (y)°0(y))n = 0 f o r a n y y e x a n d a Q , ( x ) = X (x)a (x) = a (x) so that 0 =f ea e R yet (ea) n = 0 x This contradicts the hypothesis that R is semi-prime and thus establishes that k^ is semi-prime. Now let R have property e). Then for each x e X k^ also has property e) and hence, by 1 .12 i i i ) , w.gl.dim(k ) <_ 1 . Thus w.gl.dim(R) = sup {w.gl.dimk : X X x e X} < 1 . 30. i v ) T h i s f o l l o w s from i i i ) and i ) . Q.E.D. In o r d e r to apply 1.7 and 1.13 to a p a r t i c u l a r R i t w i l l be necessary to compute the sheaf ( X , k ) . The f o l l o w i n g lemmas w i l l be used f o r such a computat ion i n §5 . 1.14) ( D e f i n i t i o n : ) L e t A be a c a r d i n a l , i ) A s e t D Q X i s a A - s e t i f f there e x i s t s a f a m i l y of c l o p e n subsets of X {U } such that U = U(U ) and I{U }I < A . a a a ' a 1 i i ) (X,k) has the A - e x t e n s i o n p r o p e r t y i f f f o r any A - s e t U and a e r(U,k) there e x i s t s a ' e r (X ,k) such tha t o' U " ° i i i ) (X ,k ) has the unique A - e x t e n s i o n p r o p e r t y i f f i t has the A - e x t e n s i o n p r o p e r t y and f o r any A - s e t U and a, a ' e r ( X , k ) , a U U U U 1.15) (lemma) Let (X ,k) have the unique A - e x t e n s i o n p r o p e r t y and U = U(U ) where {U } i s a d i s j o i n t f a m i l y of c l o p e n a a a se t s such t h a t l^ a^ l < A • ^ e t ^ be a f i n i t e l y generated i d e a l i n R and r e R be such that a (U) e I f o r a l l u e U r u Then a (x) e I f o r a l l x e U r x P r o o f : L e t a , , . . . a generate I . Then f o r each x e X 1 n ° a ( x ) , . . . o (x) generates I . For each a l e t e e R a , a ° x a 1 n be such tha t a = i i i T T . For each a I i s generated e U e a a a by a . e ' , . . . a e . A l s o a (x) = a (x)a (x) = a ( x H T , (x) l a n a re r e r r U a a a e ( I )a (x) = ( I )x f o r any x e X so by 1.4 re e l x e e J 3 a e a a a 31. Thus there e x i s t s s . , . . . s E R such that l , a n , a n re = E s . a .e a 1 > a 1 a For each i (1 < i < n) d e f i n e x . : U -> k — — l by x . ( x ) = a (x) where a i s chosen such that x e U i s . a OL i» S i n c e the U are d i s j o i n t the x . are w e l l - d e f i n e d . S i n c e the U are open, f o r a r b i t r a r y x E U , x e some U and a a U i s a neighborhood of x such that x . a  b x U x , a so x i s cont inuous a t x . Thus x.^  e r ( U , k ) For a r b i t r a r y x e U f i n d U such tha t x e U . Then a a n a (x) = a (x)t|;7T (x) = a (x )a (x) = a (x) = E a (x) -a (x) r r ^ U r e re . , s . a .e a a a 1=1 i , a x a n a n = E a (x)a ( x H T T (x) = E x . ( x ) a (x) . Thus . ., s . a . U . ., x a . x=l x , a x a x=l x n = ( E x . a ) TT - i x a.' U x=l x U S i n c e (X ,k ) has the A - e x t e n s i o n p r o p e r t y we may assume that a l l the s e c t i o n s i n the above e q u a t i o n are g l o b a l , S ince ( X , k ) has the unique A - e x t e n s i o n p r o p e r t y n n = ( E T.O ) . , x a. x=l x x , so f o r x e U a (x) = ( E x . ( x ) o (x) e I r . . x a . x=l x Q.E.D. 1.16) (lemma) L e t (X ,k ) , I , U , and { i O be as i n the h y p o t h e s i s of 1 .15 . Let b . , . . .b e R be such tha t a, (u) , . . . a , (u) 1 m b - D 1 m generates 1^ f o r a l l u e .U . Then a ( x ) , . . . ( x ) generates U _ 1 m I f o r a l l x e U x m P r o o f : L e t J = E R, 1=1 \ Then by h y p o t h e s i s = I f o r a l l u e U . L e t a e J be a r b i t r a r y . Then f o r a l l u e U 32. a (U) E I . S ince I i s f i n i t e l y generated we have, by 1 .15 , a u a (x) e I f o r a l l x e "u . Thus J p i I f o r a l l x e U . a x x x S ince J i s f i n i t e l y generated we a l s o have I r\ J f o r a l l X X x e U Q . E . D . In o r d e r to a p p l y 1.16 i t i s necessary to o b t a i n some i n f o r m a -t i o n about the e x i s t e n c e of d i s j o i n t f a m i l i e s of c l o p e n s e t s . 1.17) (lemma) i ) L e t U be an JX?, - s e t i n X . Then U can be expressed as the u n i o n of a c o u t a b l e d i s j o i n t f a m i l y of c l o p e n s e t s , i i ) L e t U be open i n X . Then there e x i s t s a d i s j o i n t f a m i l y of c l o p e n subsets of X •> {V^} such tha t V = U where V = U (V ) . a a P r o o f : i ) T h i s i s a s tandard c o n s t r u c t i o n . S ince U i s an _7sv1 - s e t we have U = U(U. ) where {U.} i s an a p p r o p r i a t e i countab le f a m i l y o f c l o p e n s e t s . D e f i n e U ' = U and f o r o o 0 < n E I l e t U ' = U - U(U! ) . Then U = U(U!) and {U!} n n _ . x x x 0<_x<n x i s a d i s j o i n t countab le f a m i l y of c lopen s e t s . i i ) L e t S = { F : F i s a d i s j o i n t f a m i l y of c l o p e n se ts such tha t W e F -> W U} . 5 i s p a r t i a l l y ordered by i n c l u s i o n , C . I f C i s a c h a i n i n 5 i t i s e a s i l y checked that U ( F ) E 5 F E C so that by Z o r n ' s lemma S has a maximal e lement , say M = {V } • a Then V r> U f o r i f there e x i s t e d x e U - V there would e x i s t , s i n c e U i s open, ^ C U - V C U - V a neighborhood of x Thus M U {N .^} would b e l o n g to S , c o n t r a d i c t i n g the m a x i -m a l i t y of M . Thus we have V C U <^  V . Hence V = U . Q . E . D . 33. I t should be noted that i f U i s a X-set the set V obtained i n 1.17 i i ) need not be a X-set. Since X i s clopen i t i s a 2-set. I s h a l l show that i f X i s i n f i n i t e there e x i s t s an open set V of the form V = U(V ) (where a _ _ {V^} i s a d i s j o i n t family of clopen sets) such that V = X = X yet V i s not a 2-set. Suppose X i s i n f i n i t e . Since X i s compact there e x i s t s x e X such that {x} i s not open. Let U = X - {x} . Then U = X . Obtain {V } a d i s j o i n t a family of clopen sets and V = U(V') U such that V = U = X = X a a V i s not a 2-set f o r i f i t were i t would be closed and we would have V = V = U = X , a c o n t r a d i c t i o n . §2 Coherence of Rf X~l Throughout this section R will denote a commutative von Neumann regular ring and S will denote the polynomial ring RfX"! (X,K) will denote S° . Clearly B(R) = B(S) . Thus by 0.16) R° may be identified with (X,k) where k is a sub-sheaf of K such that ~\ : R T(X,k) is an isomorphism. Since R is von Neumann S R regular (X,k) is a regular ringed space. (see 0.15). 2.1) (lemma) i) For each x e X , K = k fxl . X X i i ) For each x z X K is an integral domain with property e) X n i i i ) For ( E a.X1) z S i=l 1 n . n (a £ a.-X1)(x) = E a (x)-X1 , where 1=1 1=1 l K and kJTXl are identified, x X Proof: i) Refer to the basic definitions given in 0.14. An arbitrary point in X is actually a maximal proper ideal in B(R) . Then = Rrxl/Rrxl'M = (R/R-M)fX> k ^ R ) ^ = ^ f x l . i i ) Since each k is a field (R is von-Neumann regular) x this result is immediate from i ) . i i i ) It is clear that under the identification = kJ"X~] a .(x) = X 1 is identified with 1 • X 1 = 1 • X1 . The result X 1 now follows from the fact that a,.N(x) : S -*• K = k rx~\ is a (*) x x ' homomorphism. Q . E . D . 35. 2.2) (Proposition) w.gl.dim(S) = 1 . Proof: It is well known that i f T is a Noetherian ring then w.gl.dim(T) = gl.dim(T) . (see ex. 3 p. 122 of [3].) It is also well known (Chapter IX Theorem 7.11 of [3]) that i f F is a field then gl. dim(FfX"~j) = 1 , and FfXl is Noetherian. Thus, by 2.1. i) and 0.18 w.gl.dim(S) = sup {w.gl.dim(K^) : x e X} = 1 2.3) (Theorem) R-TX1 = S is a semi-hereditary ring with properties a), b), c), d), and e). Proof: It follows from 2.1 i i ) and 1.7 that S has properties n a), b), and e). Let ( E a.-X ) = s e S be arbitrary. Since i=l 1 R is von Neumann regular and the a. e R , therefore each S(a ) l a, n . n i i s clopen. Thus S (a ) = S ( E a .X1) = (~\ (S(a )) is clopen s . , a. . .. a. i=l l i=l l so by 1.1 S has property d) (and therefore c)). The result now follows from 1.7 i i i ) . 36. §3 Representation of RrrG~*"~]~] by sections of sheaves In this section R will denote a commutative von Neumann regular ring and G an abelian linearly ordered group. The next lemma will be needed to define and work with RTrG + ~ n , the ring of formal power series with coefficients from R and indices from G+ = .{g e G : g >_ 0G} . (lemma) Let U and V be well-ordered as subsets of G+ Then: i) U + V is well-ordered as a subset of G+ i i ) U u V is well-ordered as a subset of G+ i i i ) For any g e G there are only finitely many u E U and v e V such that u + v = g Proof: i) Let <J> =j= S C TJ + V . Let u^ = the least element of { u e U ; u + v e S for some v E V} . Let v^ = the least element of {v E V : u^ + v E S} . If u^ + v^ is the least element of S we are done. Otherwise let u^ = the least element of { U E U : U + V E S and u + v u^ + v^ } and let v^ = the least element of {v e V : u^ + v E S} . By the construction of u l * U 2 ' V l ' a n C^ V 2 W e n a v e u i < u 2 a n c ^ v 2 < V l ' ^ u^ + is not the least element of S we continue as above and either obtain u + v as the least element of S for some n n natural number n or v ^ + - j _ < v i ^ o r a ^ n a t u r a ± numbers i The latter would yield a subset °f v with no least element contradicting the fact that V is well-ordered. Thus S has a least element so U + V is well-ordered. 37. i i ) Let U' = U X {0} and V = V Y {0} . Then U u V c u ' + V' Since U' and V are obviously well-ordered the result follows from i ) . A direct proof without reference to G+ is also easy. i i i ) Suppose that there are infinitely many u e U such that u + v = g for some v e V . Let u^ = the least element of { u e U : u + v = g for some v e V} . Let v^ be the element of V such that u^ + v^ = g . By supposition {u e U : u > u^ and u + v = g for some v E V} ^  <J> so we may choose its least element and v^ e V such that + v^ = g . By construc-tion U£ > u^ so v2 < v i * Continuing this way u^ and v^ are defined for each positive integer i such that u.,., > u. r & l+l 1 v.,, < v. . Thus {v.} V contains no least element. This l+l 1 1 contradicts the fact that V is well-ordered. Q.E.D. (Definition): The ring Rl"rG+~p of formal power series with coefficients from R and indices from G+ + a) RfrG " 1 1 consists of a l l formal sums of the form E r^ • X where the r e R , the g E G+ , no g appears twice, and a a a {g } is a well-ordered subset of G*" a , ga ha + b) Two elements, E r -X and Z s • X e RlTC ~ H , are a a equal i f f i) r 4= 0 -> there exists h. such that h. = g and a 1 3 3 a r = s„ . a 3 i i ) s„ I 0 •> there exists g such that g = h. and 3 a a 3 38. c) (Remark) In view of 3.1 i i ) any two elements of RrTG +1"l , Sct h8 g y Z r X and Z s„ X , can be rewritten as Z r' X and a , 8 y u + Z s' -X respectively, where ^ is a common set of indices that is well-ordered. Thus, given any finite subset F of RTrG +Tl we may choose {g } , a well-ordered subset of G+ such that a 8a each element of F has the form Z r "X for appropriate a r e R This will be done without comment when required. Since a a union of infinitely many well-ordered subsets of G+ need not be well-ordered no similar simplification is possible when F is infinite. g g In d), e), and f) let Z r ' X a and Z s X a z RTTG TI . a a g g g d) Addition: Let (Z r X a) + (Z s X a) = Z(r + s )X a . a a a a e) Multiplication: Let {h } = {g } + {g } . Then let S a g a hB (Z r X )(Z s X ) = Z t • X p where a a 8 each t„ = /z r * s a . y g a + g y = h 8 By 3.1 i i i ) the definition of any t D involves only a finite p sum. By 3.1 i) {h } is well-ordered as a subset of G + so u P H R + £ t 0 X P e RVrG TI • p f) Notation: i) Since ^Sa^ l s well-ordered we may assume without loss of generality that i t is indexed by an ordinal in an order preserving way. That is to say there exists an ordinal A such that g : A -»- G+ given by a -> g^ for a < A is a strictly monotonically increasing map. It will often be assumed without comment that {g } is indexed this way. a 39, ii ) If G is the group of integers then G+ = 7^ is already well-ordered. Thus any element of RrrG +1~\ can be written in the form E r. • X 1 1 o<i< js? i i i ) To conform with tradition, i f G is the group of integers then R r r G + - n will be denoted R(TXT1 Throughout the section S will denote RffG +~n and (X,K) will denote S° . Clearly B(R) = B(S) . Thus by 0.16 R may be identified with (X,k) where k is the subsheaf of K such that ~* : R ->• r(X,k) is an isomorphism. Since R is von Neu-R mann regular (X,k) is a regular ringed space. The analogue of 2.1 i) for (X,K) does not hold. In fact i t will be shown that there is a canonical epimorphism K ->- k \"rG+~n and that this is an isomorphism i f x is an x x isolated point in X . Lemmas 3.1 and 3.2 are working lemmas for studying this and other structural facts about the K x 8ou „ „,cr „ g a x 3.1) (Lemma) For (E r X ) e S , S( E r X ) = U(S(o )) a a o< r a Proof: I shall prove the equivalent fact: Z( UE r X % = (n(Z(a )))" r a From the definitions (see 0.14) a point in X is actually a maximal proper ideal in B(R) = B(S) and for M e X g = RTrG +TI./RrrG +Tl .M and = R/R.M . Let M e Z(°E r a X ") Then E r X § a e RlTG+~T7.M so E r X § a = (E r X S a)-e = .a a a S a E(r e)X for some e e M . Thus each r (= r e)e RM a a • a 40. X ( l - e ) (see 0.14) i s a neighborhood of M such tha t M ' E X ( l - e ) <-> e e M * . Thus M ' E X ( l - e ) each r = r e e R - M ' a a + each a ( M ' ) = 0 -> M ' e f l ( Z ( o )) . Thus M E (p i (Z(a r ) ) ) l n t ' a a a i n t a 0 1 Converse ly l e t M E (0(2(a ) ) ) ' . Then there e x i s t s a a a neighborhood X ( l - e ) of M ( f o r a p p r o p r i a t e e e B ( R ) ) such tha t M ' e X ( l - e ) -> M ' £ f\(Z(a )) . In terms of s e c t i o n s t h i s a a says tha t a ( M ' ) = 0 -> a ( M ' ) = 0 f o r each r . Thus f o r J e r a a any M ' e X , a ( M * ) = o ( M ' ) a ( M ' ) = a ( M ' ) f o r each a a g r . Thus r = r e f o r each r . Hence £ r • X a = a a . a a a g g E ( r e)X a = (Z r X a ) .e £ R I T G + T ! - M . Thus M E Z(cr v 8 a ) . a a L r -X a Q.E.D. 8 a 8 a 3.2) (Lemma) L e t £ r X and Z s X £ S and l e t x e X be a a a r b i t r a r y but f i x e d . Then (cr v 8 a ) (x) = (a_ v S a ) (x) i f f Er • A Es X a a there e x i s t s N , a neighborhood of x , such t h a t cr x r a as a f o r each a N x N x P r o o f ; S i n c e : S -> r(X,K) i s an isomorphism i t s u f f i c e s to show tha t (cr „ a) (x) = 0 i f f there e x i s t s N , a neighborhood Er X x a. of x , such tha t a r a = 0 f o r each r . The lemma now a N x f o l l o w s from 3.1. Q.E.D. The a l g e b r a i c s t r u c t u r e of the K . i s r e l a t e d to the x s t o p o l o g i c a l s t r u c t u r e s of X and K and these are r e l a t e d to the a l g e b r a i c s t r u c t u r e of R To e s t a b l i s h these r e l a t i o n s h i p s more p r e c i s e l y some concepts , r e l a t e d to 1.14, must be i n t r o d u c e d . From 3.3 to 3.8 T w i l l denote a commutative r i n g , (Z,n) w i l l 41. denote T°, Z' an a r b i t r a r y B o o l e a n s p a c e , and A an a r b i t r a r y c a r d i n a l . 3.3) ( D e f i n i t i o n ) i ) Z' i s A - e x t r e m a l l y d i s c o n n e c t e d i f f U i s c l o p e n f o r each A - s e t U i i ) Z' i s e x t r e m a l l y d i s c o n n e c t e d i f f i t i s u - e x t r e m a l l y d i s c o n n e c t e d f o r each c a r d i n a l p . That i s to say Z' i s e x t r e m a l l y d i s c o n n e c t e d i f f U i s c l o p e n f o r each open UcZ. Z 1 i i i ) Z' has t h e A - d i s j o i n t n e s s p r o p e r t y i f f f o r U and V A - s e t s i n Z' , U n V = < j > - > - U r \ V = < j > . i v ) Z' has t h e d i s j o i n t n e s s p r o p e r t y i f f i t has t h e y - d i s j o i n t -n e s s p r o p e r t y f o r each c a r d i n a l u . That i s t o s a y , Z' has the d i s j o i n t n e s s p r o p e r t y i f f f o r any open s u b s e t s o f Z' , U and V , unv=<j) + uov=<j> .. v) An i d e a l I i n T i s a A - i d e a l i f f t h e r e e x i s t s { t } £ T g e n e r a t i n g I s u c h t h a t U^-H < ^ v i ) T i s A - s e l f - i n j e c t i v e i f f f o r each A - i d e a l I and f e Hom T(I,T) t h e r e e x i s t s f e Hom^(T,T) such t h a t f I = f v i i ) (Remark) T i s s e l f - i n j e c t i v e i f f i t i s u - s e l f - i n j e c t i v e f o r e a c h c a r d i n a l p v i i i ) Suppose t h a t T i s a B o o l e a n r i n g and t h a t T has t h e p a r t i a l o r d e r i n g i n d u c e d by i t s B o o l e a n r i n g s t r u c t u r e . T h at i s to s a y f o r s , t e T s<t <—> s . t = s . Then T i s A-complete i f f sup { t } e T ( i . e . e x i s t s ) f o r any { t } C- T s u c h t h a t a a | { t }| < A 42. The elementary r e l a t i o n s h i p s amongst the above concepts w i l l be found i n lemmas 3.4 - 3 . 9 . 3.4) (Lemma) L e t Z ' be A - e x t r e m a l l y d i s c o n n e c t e d . Then Z ' has the A - d i s j o i n t n e s s p r o p e r t y . P r o o f : Suppose that U and V are open subsets of Z ' such that U i s a A - s e t , U A V = ij> , ye t there e x i s t s x e U A V . S i n c e U i s open f i n d N , a neighborhood of x , such that N c. U . x x S i n c e x e V there e x i s t s v e N r v V C l . U P v V . S ince v e V and x ~ N and V are open there e x i s t s N <^  N f\ V , an open n e i g h b o r -x r v — x hood of v . S ince v e U t h e r e e x i s t s u e U A N . T h i s v c o n t r a d i c t s U n V = <j> . Thus U n V = <j> . In p a r t i c u l a r t h i s w i l l be t r u e i f V i s a A - s e t . Q . E . D . 3.5) (Lemma) Let Z ' have the d i s j o i n t n e s s p r o p e r t y . Then Z ' i s e x t r e m a l l y d i s c o n n e c t e d . P r o o f : Le t U be an open set i n Z ' . Then U and ( Z ' - U ) are open s e t s i n Z ' such tha t U Pi (Z* - U) = <j> . Hence U (~\ (X - U) = <f> . Thus, g i v e n x e U there e x i s t s a n e i g h b o r -hood N of x such tha t N f\ (X - U) = 6 . Thus N C U x x x — so U i s open and hence c l o p e n . Q . E . D . The above proof can not be g e n e r a l i z e d to show that i f Z ' has the A - d i s j o i n t n e s s p r o p e r t y then Z ' i s A - e x t r e m a l l y d i s c o n n e c t e d s i n c e a se t D C Z ' may be a A - s e t w h i l e (Z ' - U) 43. i s n o t . In §7 i t w i l l be shown t h a t f o r each c a r d i n a l u there e x i s t s a Boolean space w i t h the u - d i s j o i n t n e s s p r o p e r t y that i s not j-V - e x t r e m a l l y d i s c o n n e c t e d . 3.6) (Lemma) i ) L e t (Z , r i ) have the A - e x t e n s i o n p r o p e r t y . Then Z has the A - d i s j o i n t n e s s p r o p e r t y . i i ) L e t ( Z , n ) have the e x t e n s i o n p r o p e r t y . Then Z i s e x t r e m a l l y d i s c o n n e c t e d . P r o o f : i ) Suppose tha t U and V are A - s e t s i n Z such that TJ n V = <J> y e t there e x i s t s x e U r\ V . D e f i n e x e r ( U ^ V ,n) by x ( x ' ) = 0 f o r x* e U and x ( x ' ) = 1 f o r x ' e V . S ince ( Z , n ) has the A - e x t e n s i o n p r o p e r t y f i n d x ' e r ( Z , n ) such tha t T ' t t t = x . x(x) = 0 s i n c e x e U . A l s o x(x) = 1 s i n c e U o V x e V . T h i s c o n t r a d i c t i o n e s t a b l i s h e s tha t U O V = ty i i ) This f o l l o w s from i ) and 3 . 5 . Q . E . D . 3.7) (Lemma) i ) T i s A - s e l f - i n j e c t i v e ->• (Z,TI) has the A - e x t e n s i o n p r o p e r t y . i i ) L e t T be von Neumann r e g u l a r . Then T i s A - s e l f - i n j e c -t i v e <—> ( Z , n ) has the A - e x t e n s i o n p r o p e r t y . i i i ) Le t T be a Boolean r i n g . Then Z has the A - d i s j o i n t n e s s p r o p e r t y <—> ( Z , n ) has the A - e x t e n s i o n p r o p e r t y <—> T i s A - s e l f - i n j e c t i v e . 44. Proof: i) Suppose that T is A-self-injective. Let U = U(U ) where the U are clopen in Z' and I{U }I < A . Let a a 1 a 1 a T E r(U,n) be given. Let I = {a e r(Z,n) : S(a) c. some union of finitely many U } . Then I is a A-ideal in r(Z,n) generated by {ty } . Define f £ Horn (I, r ( Z , n ) ) as follows: a p ( Z , n ) for any a e I let f(cr)(x) = cr(x)x(x) when x £ U and f(a)(x) = 0 when x £ U . Since S(a) is a closed subset of the open set U there exists a clopen set V such that S(a)C V ^  U f(cO y is continuous because a y and x are. f(a) (Z-V) 0 so that is also continuous. Since V is clopen this establishes f(o~) e r(Z,n) . It is easily checked that f is a homomorphism. Since T = r(Z,n) is A-self-injective find f £ Horn (r(Z,n), r(Z,n)) r ( z , n ) such that f Then f'(l) e r(Z,n) . For any x E U find a such that x E . Then ty e I and f'(1)(x) f ' C D ( x ) • ty^ (X) = ( f ' C l ) • *v-)(X) = ( f ' ( l • tyy ))(X) = a a a tyv (X)T(X) = T(X) . Thus f'(l) a U = T Let (Z,n) have the. A-extension property. Let I be an ideal in T generated by {e } c T where |{e }| < A . Since R is von Neumann regular we may assume without loss of generality that the e^ are idempotents. Let f e Hom^(I,T) be given. Let U = U(S(ae )) . u is a A-set. Define T : U ->- n by a T(X) = a_, s(x) where x e S(ae ) . To see that T is well-defined suppose that x e S(ae ) A S(aeJ . Then a,, N (x) = r a 8 f (e ) a,, N (x) • a (x) = o,, N (x) = o r, N (x) = a £ / N (x) f C e a } e 3 f ( 6 a ) ' V f ( V e g } f ^ ' e a = a., . (x) .a (x) = a,, x . To see that x is continuous on f ( e ) e a ±(.e ) 45. U n o t e t h a t i f x e U t h e n x e some S(ae ) and S ( a e ) °f(e ) a a T e r ( u , n ) . „, . , so T i s c o n t i n u o u s a t x and S(ae ) a Choose x' e r ( Z , n ) such t h a t x' = x . Choose t e T su c h t h a t a = x . D e f i n e f e Hom^,(T,T) by f ' ( s ) = s . t f o r any s e T . Then f o r any e , a r , , N = a _ = c x J a f ( e ) e t e . a a a Thus i f x e S ( a e ) , ° f t ( e ) ( x) = ° e ( x ) x ( x ) = a& ( x ) a f ^ g ^ ( x ) a a a a a = a f , v (x) and i f x j: s ( ° e ) > a f ( e ) ^  = °e ^ X ^ T ^ X ^ = 0 a a a a = a e ( x ) ' ° f ( e ) ( x ) = CTf(e ) ( X ) ' T h e r e f o r e a f ' ( e ) = a f ( e ) a a a a a so f ' ( e ) = f ( e ) . S i n c e {e } g e n e r a t e s I t h i s e s t a b l i s h e s a a a j = f . Thus T i s A - s e l f - i n j e c t i v e . f 1 The c o n v e r s e was e s t a b l i s h e d i n i ) . i i i ) S i n c e T i s a B o o l e a n r i n g each f o r x e Z i s a 2 f i e l d s a t i s f y i n g t h e p o l y n o m i a l i d e n t i t y X - X = 0 . Thus each n i s the two element f i e l d { 0 , 1 } . I n v i e w o f i i ) x x x and 3 . 6 . i ) i t s u f f i c e s t o show t h a t i f Z has t h e A - d i s j o i n t n e s s p r o p e r t y t h e n (Z,n) has the A - e x t e n s i o n p r o p e r t y . L e t Z have t h e A - d i s j o i n t n e s s p r o p e r t y . L e t U = U(U ) where t h e U a r e c l o p e n i n Z and a a a |{u" a}| < A . L e t x e r ( U , n ) be g i v e n . By t h e i n t r o d u c t o r y remark, x e U i m p l i e s t h a t x ( x ) = 0 o r 1 F o r each a l e t V = {x e U : x ( x ) = 0} and W = {x e U : x ( x ) = 1} . a a a a Then V and W a r e open complements i n U . Thus t h e y a a a a r e c l o p e n s u b s e t s o f the space U where U has the r e l a t i v e a a Topology i n d u c e d by Z . Thus, s i n c e the a r e c l o p e n i n Z so a r e t h e V and W . L e t V = U(V ) and W = U(W ) . a a a a a a Then V and W a r e A - s e t s i n Z such t h a t V ^ W = <j> . 4 6 . Thus, by h y p o t h e s i s , V r> W = ty . Hence there e x i s t s a c l o p e n se t C such that V £ C and W n C = ty . L e t x ' = ty . Then x ' e r(Z,n) i s such that x ' U = T Q.E.D. The f o l l o w i n g lemma i s c r u c i a l i n a p p l y i n g the concept of the A - d i s j o i n t n e s s p r o p e r t y to i n v e s t i g a t e the s t r u c t u r e of (X,K) 3.8) (Lemma) Let^ Z ' have the A - d i s j o i n t n e s s p r o p e r t y . Then f o r any A - s e t s U and V , U r > V = U n V P r o o f : L e t U = U(U ) and V = U(V ) where the U and a 3 V are c l o p e n i n Z ' and I{U }I < A and |{VD}| < A . 3 ' a 1 1 3 ' D ft V c u ft V i s t r u e i n g e n e r a l . I f U A V = ty there i s n o t h i n g to p r o v e . Suppose x e U r> V . L e t F denote the X s e t of a l l c lopen neighborhoods of x . F i s d i r e c t e d under x O and i s a b a s i s f o r the neighborhoods of x . L e t d = N e F . Observe that U ^ N = U(U n N ) and x x x a x a V A N = U(V„ O N ) so t h a t U A N and V A N are A - s e t s . x „ 3 x x x C l e a r l y x e ( U n N ) n ( V A N ) so that by the h y p o t h e s i s there e x i s t s an element, say x , , i n ( U / ^ N ) A ( V A N ) d x x {x : d e F } i s a ne t on U n V converg ing to x . Thus Q. X x e U A V . Hence U O V 2 U A V so U r\ V = U A V Q.E.D. The f o r e g o i n g lemmas are summed up i n the f o l l o w i n g p r o p o -s i t i o n about R and (X ,k ) 47. 9_) (Proposition) i) R is A-self-inj ective <—> (X,k) has the A-extension property. i i ) R is A-self-injective -> B(R) is A-self-injective. i i i ) B(R) is A-self-injective <—> X has the A-disjointness property. iv) B(R) is A-self-injective ->-UAV = UAV whenever U and V are A-sets in X v) If R is self-injective so is B(R) vi) B(R) (as a Boolean ring) is A-complete <—> X is A-extremally disconnected. vii) B(R) is A-complete -> B(R) is A-self-injective. Proof: i) This is just a restatement of 3.7.ii). i i ) Let R A-self-injective. Then by 3.7.i) (X,k) has the A-extension property so that by 3.6.i) X has the A-disjointness property. Since B(B(R)) = B(R) i t follows from the basic de-finitions (see 0.14) that X = X(R) = X(B(R)). Thus by 3.7.iii) B(R) is A-self-injective. i i i ) As in i i ) , X(B(R)) = X . Thus this follows from 3.7.iii). iv) As in i i ) , X(B(R)) = X . Let B(R) be A-self-injective. Then by i i i ) X has the A-disjointness property. The result now follows from 3.8. v) This follows from i i ) since a ring is self-injective i f f i t is u-self-injective for each cardinal u vi) This result is standard. It follows via a trivial altera-tion of 22.4 in [9]. This alteration is required because the definition of A-completeness is slightly different in this thesis. 48. v i i ) Th is f o l l o w s from v i ) , 3 .4 , and 3 . 9 . i i i ) . Q.E.D. The next lemma g i v e s some i n s i g h t i n t o the d i f f i c u l t i e s i n v o l v e d i n s t u d y i n g the a l g e b r a i c s t r u c t u r e of the K , X s 3.10) (Lemma) i ) There i s a ( c a n o n i c a l ) epimorphism f x : k_jY"G +~l~l f ° r e a c h x e X . • i i ) For x e X the epimorphism f i s an isomorphism i f x i s an i s o l a t e d p o i n t i n X i i i ) I f X i s i n f i n i t e there e x i s t s x e X such t h a t f x i s not an i somorphism. P r o o f : i ) For each x e X d e f i n e f by f (a g ) = x x „ v a g a Ea -X Ea (x ) "X . I t f o l l o w s from 3.2 t h a t each f i s w e l l -a " x a d e f i n e d . i i ) L e t x be an i s o l a t e d p o i n t i n X . Then {x} i s a neighborhood of x . Suppose f (o g ) = 0 . Then X Ea -X a a ^ each a (x) = 0 and, s i n c e {x} i s open, x e (f\(Z(a ) ) ) a a a a a Thus by 3 .1 (a g ) (x) = 0 . Ea -X a a i i i ) L e t X be i n f i n i t e . F i r s t I must show tha t s i n c e X i s i n f i n i t e there e x i s t s an jxf - s e t that i s not c l o s e d . S ince X i s compact and i n f i n i t e there e x i s t s y e X such t h a t {y} i s not open. Thus X - {y} i s not c l o s e d . By 1 . 1 7 . i i ) f x n d {V } , a d i s j o i n t f a m i l y of c l o p e n se t s such that U(V ) a a a X - {y} = X . S ince X - {y} i s not c l o s e d there are i n f i n i t e l y many {V } . P i c k {U\} , a countab ly i n f i n i t e s u b - f a m i l y of {V } . L e t U = U(U. ) . Then U i s an - s e t . S i i a . i i >ince 49. the V are open U is not compact and thus is not closed in X . Thus there exists x e X such that x e U yet x £ any U\ . Let ^ : j \ ( -> G be a strictly monotoni-cally increasing function and for each i let e. e R be such g i that a = il)T, . Then let s = (Ze.'X ) e S . Hence e. U. 1 i l 8 i f x(x) = (x)-X = 0 yet ag(x) =f 0 (by 3.1) since i x e U(U.) = U(S(ae.)) . Thus f is not an isomorphism, . l . i x r l l Q.E.D. If each f in 3.10 had been an isomorphism then each x kxfrG+~\1 = Kx would immediately be an integral domain for which the lattice of.ideals was linearly ordered under inclusion. (Each k is a field.) Thus each K would have been an integral x x domain with property e) so that S , by 1.17.ii), would have been a ring with properties a), b), and e). However, i t will be possible to show that under certain circumstances each K x is an integral domain (even though some f are not isomorphisms) for which the lattice of ideals is linearly ordered, and that under other circumstances S has none of properties a), b) and e). The following concept will be used to investigate the algebraic structure of the K , 6 x s 3.11) (Definition) Let s = Z a X £ S . For any x £ U(S(oa )) a a a let val (a ) = the least element in {g : oa (x) + 0} x s a a 1 50. g g 3.12) (Lemma) Let s = (Ea X a) e S , t = (Eb X a) e S , a a U = U(S(a )) , V = U(S(a, )) , a = a , T = a . and x e U . a b s t a a a a i) val (a) = g implies that there exists N , an open x 3 x neighborhood of x , such that y e N ->- val (a) < g x y 3 i i ) If G is the group of integers then val (a) = g implies x p that there exists N , an open neighborhood of x , such that X y e N ->• val (a) = g Q . x y a i i i ) Let z e X and g E G + be such that for each open neigh-borhood of z, N , there exists u e N n U such that z z val^(a) = g . Then z e U and val^o) = g iv) Suppose that x e U r\ V . Then val (ox) = val (a) + val(x) X X X v) Let z e S(x) . Then for any N^  , an open neighborhood of z , there exists v e N ^ V z vi) Let z e U . Then a(z) is not a zero-divisor in K z vii) Let z e S(a) . Then for each N , an open neighborhood z of z , there exists u e N such that a(u) is not a zero z divisor in K u Proof: i) Let val (a) = g . Then oa (x) =j= 0 . Since x p 3 R is von Neumann regular find N^  , an open neighborhood of x such that y £ N •+ oa (x) 4= 0 . By the definition of val (a) x 3 1 y this shows that y e N ->• val (a) < g„ x y — 3 i i ) Let G be the group of integers. Then s may be written in the form s = Ea^-X1 where the i E G+ . Let val (a) = n . Let N = S(aa ) - ( U (S(aa.))) . Clearly N x x n . i x i<n is a neighborhood of x such that y e N •+ val (a) = n x y i i i ) By hypothesis z e S(aa ) . Since S(cra ) is closed p p this yields aa (z) 4= 0 so z e U and val (a) < g„ 8 1 z — g Suppose val (a) < g • Then by i) obtain , an open neighborhood of z , such that y e N -> val (a) < val (a) < g z y — z g This contradicts the hypothesis that there exists y' e such that val (a) = g y 3 h iv) Let s*t = Z c "X ^  . Let val (a) = g , and val (x) = y x a x Then g A < g at -> oax (x) = 0 and g, < g0 t °b, (x) = 0 . Also, A p A oa , (x) =1= 0 and ab . (x) =h 0 . Note that ot 1 g 1 ac (x) = (a u 0 ) ( x ) = ^ ° a (x)-ob (x) . y Z aa'bg a g g a + g g=h M g a + g g = \ Thus ac (x) = 0 for h < g . + g„ , and y y a g' ac (x) 4= 0 for h = g , + g„ , • Consequently y ' y a g val (ax) = g , + g o t = val (a) + val (x) x a g x x v) This is trivial since by 3.1) S(T) = V vi) Since x is aribtrary i t suffices to show that i f x (z) =f then a (z) x( z) =j= 0 . Let z e S(x) . Since z e U find g e G + such that val^Ca) = g . By i) find N^  , a neighborhood of z , such that y e N z val (a) <_ g . Let N^  be an arbitrary neigh-borhood of z . Then by v) there exists y e N^A N^  and g' e G+ such that val (x) = g' . Since y e N find ° y ° J z g" e G+ such that val^(a) = g" . Hence (by iv) v a l <0t) g* 'v g " ;>r t h a t (cr*T>(y} I1 0 . • TfewS r e 3 ( 0 ' r ) 30- 0 4= (<?«t) U'} ^ c ( z ) T U . ) , v i i ) S i n c e -S(:;) =- l i t h i s f o l i o s i ^ v M i a t a i y ft-aw v i } . . ( D e f i n i t i o n ) L-vt "' be< a cstfd&nal.. 2 i s a X-group i f f t h e r e does r ? t a s t r i c t Ay non.-M-onicaily i n c r e a s i n g f a n c c x o s Seraark: • The -^diniv*? group of ±n'..e:se~s i s rNC^-gwup bot r;.ot' an J ' \ • - g r u - . n . " The f o r r H - ^ i - i g conc^'p^s ••v-i'O. nov be. x-s&d t o - t l ^ t e ^ i i n ^ e K ^ ' - I ? "»h«».rt- -the-' K . . s.f.: j nt - j t f t H l ef-^K;i a?:, i- v i . : 1 -c»ed jiiovu;!-v i t g l y • :.o*..d-£-:p.>-iii:t;:'-;: s;-«ac£Z:r:&foa?; 5-' ha? p r e U-er*t.ias <Q . . i s . a • A - ^ r " •••r-. "'h'^c ; i;;t£s,r-'-*- . i f f • . ::*: P r o o f : I n v i e w .vf 3 - 9 . i : i > i s st 'ffiV.fts ta . - ' shcv t h a t e«;cv-. «f ; ^ i n t e g r a l -v-:',^.:•. : : . f£ X '• h.;ia ::..IJ A - - d L G J o i p - n e s F ^ / ' c p e r t y %, s * {Z& X ; a To shau- t h a t 3:H •a a a n d . t h e hvpo-clx;--. h a * t h e .:±iii:iX--2S p r o p e t t y . Lei . c* tea (x L e t 3 J t c - X ' i \ si . / . r i - : : i;-.v t o ehax^ th.-jt • -a' 53. so that, by 3.8, U(S(aa )) AU(S(ob )) = U(S(aa )) HU(S(ab )) . a a a a a a a a Thus i t suffices to show that U(S(aa )) AU(S(ab ))C,U(S(oc )) . a a u But this follows immediately from 3.12.iv). Conversely, suppose that each is an integral domain yet X does not have the A-disjointness property. That is to say there exists a cardinal A' < A and families of clopen sets, {U : a < A' } and {V : a < A'} such U A V = <j> yet a a U A V 4= <j) where U = U (U ) and V = U (V ) . I n view a<A a<A of the hypothesis on G there exists a strictly monotonically increasing function g^ ^ : A' -> G+ . For each a < A' let e and f e R be such that \b„ = a and ib„ = a^ a a U e V f a a a a g g Let s = £ e X 01 and t = £ f X a . Let x e U C\ V . Since a a U o V = (j) i t is easily seen that a s ( x ) 0 t ( x ) = 0 . Since x e U and x e V i t follows from 3.2 that a (x) + 0 and s 1 a (x) =1= 0 . Hence K is not an integral domain. This contra-diets the assumption that K is an integral domain. Thus X x has the A-disjointness property. Q.E . D . 3.15) (Proposition) Let A be the smallest cardinal such that G is a A-group. Then the following are equivalent: i) S has property c). i i ) S has property d). i i i ) B(R) is A-complete. 54. Proof: i i ) ->- i ) : This is obvious. i i i ) i i ) : Suppose B(R) is A-complete. Then by 3.9.vi) X is A-extremally disconnected. Hence by 3.4 X has the A-disjointness property so that by 3.14 each K is an integral domain. Let s = ( E a^ -X ) e S . Then by the hypothesis on A ]{g }| < A so that U = U(S(cr )) is a A-set. Hence S(cr ) = U ex a s. a a is clopen in X . Thus, by 1.1, S has property d). i) -»- i i i ) : Suppose that X is not A-extremally disconnected (i.e. B(R) is not A-complete) yet S has property c). Since X is not extremally disconnected find {U : a < A ' } , a a family of clopen sets indexed by some cardinal A ' < A , such that there exists x e U r\ (X - U) where U = U (U ) . For a < A ' each a < A ' let e e S be such that d> = a . By the a rU e J a a hypothesis on A there exists a strictly monotonically increasing + 8 a function g, s : A ' G . Let s = E e X e S . I shall ( > a < A ' a show that l.ann (s) is not finitely generated as an ideal in S . Suppose that there exists t , . . . t n e S such that n l.ann (s) = E S't. . Let N be a neighborhood of x i=l 1 X Since x e (X - U) there exists y e such that y e (X - U) . That is to say there exists y e N such that a (y) = 0 x s Thus w e l.ann (s) where W is a clopen set chosen such that y e W and o W 0 and w is chosen such that ii r 7 = a Hence, there exists i'(l<i'<n) such that val (a ) = 0 ^ i ' Since there are only finitely many t^ i s there exists j (l_<j<n) such that for each neighborhood, , there exists 55. y' e N' such that val .(a ) = 0 . Thus by 3.12.iii) x y t. J val (a ) = 0 . Hence (by 3.12.vi) a (x) is not a zero x v t . ^ t. 3 J divisor in K . Since x e U , a (x) ¥ 0 . This is a x s ' contradiction since a (x) = a (x) = a (x)-o (x) o t. s t. s s 1- J Therefore l.ann (s) is not finitely generated. Q.E.D. The following lemma will be used to investigate the ideal structure of the K . x s 3.16) (Lemma) Let A be an ordinal and let s = ( E a X ) e S a and x E X be such that val (a ) = 0_ . That is to say x s G , h g = 0^, and a (x) + 0 . Then there exists t = (£ b X y) e S. o G a 1 x u o such that a (y)-a (y) = ly for a l l y e S(o ) S L S o Proof: Let N = S(o ) . Since R is von Neumann regular this ao h is clopen in X . Define t = (E b^ X y) e S as follows: Let (y) = (a g (y)) 1 for y e N and (y) = 0 for o ao o y y i N . Let h = 0^  . For some ordinal <5 assume that J r o G a, and h are defined for a l l ordinals y < 6 in such a \ way that: h i) ( E b X y) E S . u<6 y i i ) The smallest g e G+ - {0} such that E cr^ (y')aa (y1) ={= 0 for some y' E N is an upper bound to {h^ : p < 6} 56, If no such g exists, since {h : y < 6} + {g : a < A} y a is a well-ordered subset of G , i t follows that: i i i ) 0 (y)o (y) = 1 for a l l y e N ; and D a o o iv) £ o, (y)*a (y) = 0 for a l l y e N and any g > 0 b a V a h In this case let t = E b «X V . It follows from y<6 P i i i ) , iv), 3.2 and the fact that N is open that a (y)'a (y) = ly for a l l y E N t s If such a g does exist let a (y) = ( - E a. (y)a <y)).(a (y)) 1 •u r D a a bo y a o h +g =g y a 43 and y<6 h for y e N and 0, (y) = 0 for y i N . Then E b X y  J b y r y y<6+l satisfies i) and i i ) above with 6 + 1 in place of 6 Continue by transfinite induction until the g of i i ) does not exist. g g 3.17) (Corollary) Let s = (E a X a ) , t = (E b X °) e S, g , g e G, d -I C£ r\ a a 1 I x e X , and the clopen neighborhood of x, N , be such that y E N -> val (o ) = g • •% = val (a ) . Then there exists J x y s t s' E S such that 0 s,(x)a s(x) = o^Cx) 57. Proof: Since a < a ->- a (y) = 0 f o r y e N we may (by 3.1) JL a x a assume without los s of ge n e r a l i t y that a = 0 f o r a < a, a 1 S i m i l a r l y we may assume that b = 0 whenever ot < a, . Let a 1 s" = E a -X . Then v a l (a „) = 0 . Thus, by 3.16, a x s f i n d t' e S such that a , (x)a ,, (x) = 1 . Let t s x s' = (£ b^'X ) * t ' . I t may then be v e r i f i e d that o g t ( x ) c g ( x ) = a t ( x ) . Q .E.D. The following r e s u l t i s standard. 3.18) (Corollary) Let R be a f i e l d . Then S i s an i n t e g r a l domain and the l a t t i c e of ideals i n S i s l i n e a r l y ordered. That i s to say, i f I and J are id e a l s i n S then e i t h e r I C J or J Cl I Proof: Since R i s a f i e l d B(R) = {0,1} so X(R) = {x} where x = {0} . Then k = (R/R-0) = R and K=(S/S-0) = S . We x x s h a l l show that i s an i n t e g r a l domain with a l i n e a r l y ordered l a t t i c e of i d e a l s . Since X contains only one point i t c l e a r l y has the di s j o i n t n e s s property so that by 3.14 and 3.9 i t i s an i n t e g r a l domain. To show that the l a t t i c e of i d e a l s i n K i s x l i n e a r l y ordered i t s u f f i c e s to show that for any s,t e S such that a (x) and a (x) 4= -0 , e i t h e r : i ) K a (x) C. K o (x) or s v t 1 x s — x t i i ) K a (x) C K a (x) . x t — x s 58. Since X is discrete there exists g and g„ e G such that a 8 val (a ) = g and val (a ) = gn . Suppose g < g„ . Then x s &a x t toB a — &8 by 3.17, since {x} is open, there exists s 1 e S such that a ,(x)*o (x) = a (x) . Thus i) holds. If g n < g then simi-s' s t to8 — a larly i i ) would hold. Q.E.D. 3.19) (Theorem) Let R be a finite direct sum of fields. Then S has properties a), b), c ) , d), and e), S is semi-hereditary and w.gl.dim(S) <_ 1 Proof: Since R is a finite direct sum of fields B(R) and hence X = X(R) are finite. Since X is Hausdorff this implies that each x e X is isolated. Thus by 3.10 each K = k C r G + - f l x x 1 Since R is von Neumann regular each k is a field so by 3.18 each K is an integral domain for which the lattice of ideals x is linearly ordered. Such domains have property e). Since X is finite i t is extremally disconnected. The theorem now follows from 3.9, 3.15, and 1.7. Q.E.D. The next lemma will allow 1.13 to be used to investigate the algebraic structure of S 3.20) (Lemma) i) S is semi-prime. ii ) For each x e X K is a local ring, 59. Proof: i ) This i s obvious since R i s semi-prime. g g i i ) Let x e X and s = (£ a X a ) and t = (E b X ") e S be a a such that a (x) + a (x) = 1 . Then a (x) + a, (x) = 1 where s t x r a b x o o g =0^ so e i t h e r a (x) 4= 0 or a, (x) 4= 0 . Suppose with-o G a 1 x b ' x o o out l o s s of g e n e r a l i t y that a (x) 4= 0 . Then v a l (a ) = Q „ so a ' x a G o that by 3.16 a (x) i s a unit i n K . Thus K i s a l o c a l r i n g , a x x Q.E.D. 60. §4 The s t r u c t u r e of R when R\TG ~)~\ i s not coherent and G f I , the i n t e g e r s . In t h i s s e c t i o n R w i l l denote a commutative von Neumann r e g u l a r r i n g , G a l i n e a r l y ordered a b e l i a n group tha t i s not i s o m o r p h i c to the a d d i t i v e group of i n t e g e r s , and S the r i n g RrrG + - \ 1 . (X,K) w i l l denote S° and as i n §3 k , a subsheaf of K , i s chosen such : R -> T ( X , k ) i s an i somorphism. that \ s R 4.1) (Lemma) R i s a f i n i t e d i r e c t sum of f i e l d s i f f X i s f i n i t e . P r o o f : L e t R be a f i n i t e d i r e c t sum of f i e l d s . Then c l e a r l y B(R) i s f i n i t e ' so tha t X = X(R) = the set of maximal proper i d e a l s of B(R) i s a l s o f i n i t e . Now l e t X = { x ^ , . . - x } be f i n i t e . Thus each {x^} i s c l o p e n i n X so t h a t i t i s e a s i l y checked that the map n n R -> E k g i v e n by r -> E a ( x . ) i s a r i n g i somorphism. . , x . . T r l i = l I i = l Q . E . D . The s t r u c t u r e of S when R i s a f i n i t e d i r e c t sum of f i e l d s was d i s c u s s e d i n 3.19. There fore i t w i l l be assumed f o r the r e s t of t h i s s e c t i o n t h a t X i s i n f i n i t e . I t w i l l be shown that S i s not s e m i - h e r e d i t a r y , w . g l . d i m ( S ) > 1 , and S has none of p r o p e r t i e s a ) , b ) , and e ) . 4.2) (Lemma) There e x i s t s {U : i e j\?o } , a d i s j o i n t f a m i l y of c l o p e n se ts i n X , such that there e x i s t s x e U - U where U = U (U.) . i 61. Proof: This was established in the proof of 3.10.iii) Q.E.D. 4.3) (Lemma) There exists g^ . ^  : j \ 0 •+ G + , a strictly monotonically increasing function, and g e G such that each g^ < g Proof: Either G+ - {0} contains a least element or i t does not. First suppose that i t does contain a least element, say g Q Then, by the group structure of G , T^ = {g e G : g > h} contains a least element for any h e G . For any i e j \ ^ 0 let §-^ +^  be the least element in T . The map g. . : jX 5^ G+ is 8 i ^ ' not cofinal for i f i t were then + i -»- + g. would be an isomor-— — &x phism between the additive group of integers and G . By construc-tion the map i -»- g is strictly monotonically increasing. Now suppose that G+ - {0} does not contain a least element. Pick g and g^ e G such that 0 < g^ < g . For any i e j Y ^ assume that g! is defined and pick g' such that 0 < g' < g'. . x l+l x+1 x Let g^ = g _ g^ • Then each g^ < g and the map i -»• g^ is strictly monotonically increasing. Q.E.D. For the rest of this section let g , the g^, U , the U\ , and x be as described in 4.2 and 4.3. For each i e j \ ^ 0 8 i let e. E R be such that a = $>„ . Let s. = Z e.*X and x e. rU. 1 x x x s 2 = X 8 . Let I = S.s^  , and J = S.s^ . Then I and J are principal ideals in S . We shall show that I ^  J is not finitely generated. Lemma 4.2 asserts that since X is infinite i t contains a point x that, in a fairly particular way, fails to be isolated. 1 shall use this fact to show that i f I A j is finitely generated then there exists an element t e S such that 62. some i n f i n i t e subset of {g - g^} i s a subset of i t s set of in d i c e s . Since such a subset contains no l e a s t element t h i s would contradict t e S . 4.4) (Lemma) I A J i s not f i n i t e l y generated. Proof: Suppose that I r\ J i s generated by { t ^ , . . . t n ) C S By 1.3 f o r each x e X (I A J) = I n J so that x x x {a ( x ) , . . . a (x)} generates I A J . Suppose that y e some 11 t X X 1 n U. . Then v a l (as.) = g. g = v a l (os.) and IL- i s a neigh-l y 1 I y 2 1 borhood of y so that by 3.1 and 3.17 I f\ J = (K a (y)) Pi (K a (y)) = K a (y) . y y y s w " y s 2 K J / J y s ^ ' Hence there e x i s t s an i 1 (1 < i 1 < n) such that v a l (a„ ) = g - — y t ± I & Since x e U and there are only f i n i t e l y many t ^ , g there e x i s t s j (1 <_ j <_ n) such that f o r each neighborhood of x there e x i s t s y e N such that v a l (a ) = g • For each p o s i t i v e m integer m C\ (X-U.) i s a neighborhood of x so there e x i s t s 1=1 1 i > m and y e U. such that v a l (a ) = g . But t. = m— ' m i y t . J m m j t.s„ f o r some t e S since t. e I = S.s, . Then 1 J 1 v a l (a ) = v a l (a ) - v a l (a ) = g - g. f o r each m y t y t. y s ' & & i m m j m l m Thus {g - g. : m e i s a subset of the set of indices f o r m t . Since i t has the inverse order of an i n f i n i t e w e l l ordered-set i t i s not well-ordered. This contradicts t e S and thus establishes that I A J i s not f i n i t e l y generated. Q.E.D. (Theorem) S is not semi-hereditary, w.gl.dim(S) > 1 , and S has none of properties a), b) and e). Proof: It follows from 4.4 that S has neither property a) nor property b). The rest follows from 1.13. Q,E.D. §5 A condition for RffX"P to be coherent In this section R will denote a commutative von Neumann i regular ring that is _/\^  -self-injective and S will denote RiTXll Let (X,K) denote S° and choose k , a subsheaf of K , such that : R -* F(X,k) is a ring isomorphism. R Since R is -self-injective i t follows from 3.9 that (X,k) has the jN^ -extension property, X has the jV", -disjointness property and B(R) is jK^ -self-injective. Since the additive group of integers is an jN^ -group but not an jV^ -group i t follows from 3.14 that each K is an integral domain. These facts and the results x of §1 will be used to show that for each x e X the lattice of ideals of K is linearly ordered so that has property e). It will then follow from 1.7 that w.gl.dim(S) <_ 1 and S has properties a), b), and e). 5.1) (Lemma) i) (X,K) has the jN^, -extension property. i i ) (X,K) has the unique jN^ -extension.property. f: i ) Let U be an -T^, -set and l e t T e T(U,K) be Proo given. Let i >_ 0 be a fixed but arbitrary integer. Define x . : U -> k as follows: For x e X find s = (I a. - X 3 ) e S 1 x j ,x such that T(x) = a (x) and let T.(x) = (a )(x) . It s x a. x 1, x follows from 3,1 that T^ (X) is independant of the particular s e S such that x(x) = a (x) x s x Temporarily fix arbitrary x E U . To see that each is continuous at x note that since a is continuous at x s x 65, there exists N C. U , a neighborhood of x , such that y E N ->- x(y) = a (y) . Hence y e N ->- T . (y) = (a ) (y) X S X 1 3 . X 1,X Since a is continuous at x so is x. . Since (X,k) a. l i,x has the J\\ -extension property find a_^  E R such that = x. . Perform the above construction for each integer a. 1 U i > 0 Let s = £ a.'X1 l Now I shall show that a Fix arbitrary x e X and choose N Cl TJ a neighborhood of x such that x — y E N -> x(y) = o o (y) . Thus y e N ->• (a ) (y) = (a ) (y) X S X 3 . a- . x x,x i,y for a l l i . Thus a N i,x x for a l l i N Hence by 3.2 x(x) = O g( x) • Since x e U was arbitrary this establishes a = x . Thus (X,K) has the -extension property. i i ) Note that for any a e T(X,K) S(o) = V for some set V . Let U be an -set in X and let a , o e T(X,K) be s' t such that a = o. U X, -Find an J~\i set V such that U S(o - a ) = V . U n V s t since a = a U Therefore U U C\ V = tj) since X has the JSC*J -disjointness property. Thus In view of i) this shows that (X,K) has the U U unique j \ * -extension property Q.E.D. 66. (Lemma) i) Each K is an integral domain. i i ) For each x e X the lattice of ideals in K is linearly x ordered. Proof: i) This was established in the opening remarks of this section. i i ) It suffices to show that for any s,t e S and x e X either K a (x) c. K a (x) or K a (x) Cl K a (x) . Fix arbitrary x s x t x t — x s s,t e S and let I = S.s + S.t . It clearly suffices to show that for any x e (S (a ) P> S ( a )) I is either generated by S t a (x) or by a (x) . s u Suppose s = E a.-X 1 , t = E b.- X 1 , V = U(S (a b )) and i i U = U(S(o )) . U and V are -sets. For any integers i i m > 0 and n > 0 let C = (S(o ) - U (S ( a ) ) ) O (S (a, ) — — m,n a „ . a. b m 0<_x<m I n U (S(o, ))) . Since R is von Neumann Regular each. C _ . b. m, n 0<j<n j is clopen in X . Let W = {x e U n V : val ( a ) < val ( a ) } s x s x t and W = {x e U n V : val ( a ) < val ( a )} . Then t x t — X s W = U(C ) and W = U(C ) so that W and W_ are s m,n t m,n s t 0<m<n<°° Oj^n^nK00 -sets. Clearly Wg P i Wfc = $ and Wg u W = U r\ V . Thus Wg O W = <j) ( X has the jV", disjointness property) and by 3 . 8 w u w = w u W = u o v = u n v = S(o ) n S ( a _ ) . Let S t S t S t x e S ( a ) P i S ( a ) . Then x belongs to exactly one of W or S t s W . First suppose x e W . For any y e W y e C t s s m,n for some integers m and n such that 0 < m < n . C is — m,n 67. a neighborhood of y such that y' e C implies that b J m,n val .(a ) = m < n = val ,(a ) . Thus by 3.17 there exists y s y t s' e S such that a ,(y)o (y) = a (y) . Hence a (y) generates s s t s I = K a (y) + K a (y) for any y c W . Since W is aii y y s J y t s s 1 set i t can, by 1.17,be expressed as union of a countable disjoint family of clopen sets. Since x e W i t now follows from 1.16 s that a (x) generates I . Now suppose x e W" . It similarly s x t follows that a (x) generates I t X Q.E.D. 5.3) (Theorem) S has properties a), b), and e), and w.gl.dim(S) <_ 1 Also, the following properties are equivalent: i) B(R) is j N ^ ( -complete. i i ) S has property c) i i i ) S has property d) iv) S is semi-hereditary v) S is coherent. Proof: By 5.2 each K is an integral domain whose lattice of x ideals is linearly ordered. Such rings have property e). Thus by 1.7 S has properties a), b), and e) and w.gl.dim(S) <_ 1 i) <—> i i ) <—> i i i ) : Since the additive group of integers is an. JN^  -group but not an /X^ -group this follows from 3.15. iv) —> v): This is trivial since every semi-hereditary ring is coherent. i i i ) —> iv): This follows from 1.7.iii). v) —> i i ) : This is trivial since every coherent ring has property c). Q.E.D. 68. §6 A necessary condition for Rr\"X~n to be coherent In this section R will denote a commutative von Neumann regular ring that is not self-inj ective and S will denote R f f X l l . (X,K) will denote S° and k the subsheaf of K such that ^ g R r(X,k) is a ring isomorphism. Since R is not R f-injective i t follows from 3.9 that there exists U , an jX*! -set, and T e r(U,k) such that x can not be extended to a global section. U and x will retain this meaning throughout §6. 6.1) (Lemma) i) There does not exist x' E r(U,k) such that U = x i i ) Suppose that Y c; U is a set of points such that for any x e Y there exists x e T(U u {x},k) such that x Then there exists x' e T(U u Y,k) such that x' U U = x i i i ) There exists x E U - U such that there does not exist x e r(U u {x},k) such that x_ = x x U Proof: i) Suppose that such a x' exists. Then since U is closed in X there exists x" £ T(X,k) such that x" Hence x" This contradicts our choice of x ii ) Note that i f x e U and x , x' £ T(U u {x},k) are such x x that x U U then x = x' since x £ U and k is x x Hausdorff. (It follows from 0.15 and 1.1) that k is Hausdorff) Define x' : U u Y -> k by x'(u) = x(u) for u E U and 69. x'(x) = T x ( x ) f° r x e Y - U . Clearly x' U x . It must now be shown that x' is continuous. If x z U then x' is con-tinuous at x because x is and U is open in X . Let x e Y - U . Find a e R such that x'(x) = a (x) . Thus x a x x (x) = x'(x) = a (x) has a neighborhood basis consisting of sets of the form a (U ) where U is a clopen neighborhood a x x x of x . Thus, since x^  is continuous, there is a clopen neighborhood of x, N , such that x (N f t ( U u {x})) C o (N ) x x x — a x x Hence y e N A u implies that x ( y ) = x ( y ) = a ( y ) . Then x x a x for any y e Y A N o X 3. is a continuous map ((U u {y}) r\ N ) X extending x Since y e U o N and k is Hausdorff U n N J x x i t follows that x (y) = x (y) = a (y) . Since a is continuous J y a J a x x this establishes that x' is continuous at x x' e T(U u Y,k) . Thus i i i ) Suppose that for each x e U there exists x e T(U u {x},k) X such that x x = x . Then apply i i ) with Y = U and obtain r . This contradicts i ) . Thus U x' e r(U,k) such that x' the lemma is established. Q.E.D. For the rest of this section let x E U be such that x can not be extended to an element in r (U <J {x},k) . Since U is an -set find (by 1.17) {U : i >_ 0 is an integer} , a dis-joint family of clopen sets, such that U = U(IL) . For each i 70. find e. e R such that a = iK* and r. e R such that 1 e. 1U. I l l a (x) = T(x) when x e U. and o (x) = 0 when x i U. r. I r. T i l l The e. and r. exist since the U. are clopen. Let s, = i i l r 1 I r.X1 and s = Z e.X1 . Let I = S.s, and J = S.s„ . -l 2 l 1 2 I and J are principal ideals. It will be shown that I A J is not finitely generated by showing that I n J , = (I ^  J), is not. 6.2) (Lemma) Let c-]_'c2 e ^ ' t^ i e ^ n t e § e r n > 0 , and y e U be such that a (y)a (y) = a (y)a (y) and val (a ) = n c l S l c2 S2 Y °1 Then there exists T e V (U \j {y},k) such that x y y u Proof: Find N , a neighborhood of y , such that y a (y')cr' (y1) = a (y ' ) a (y') and val , (a ) = n for any c l S l c2 S2 y c l i ± y' e N . Let c = Z a.X and c = Z b.X . Let y' e U n N y l i 2 i x be arbitrary but fixed. Find m such that y' e U . Since m U is a neighborhood of y' i t follows from 3.1 and the defini-m b tions of the r. and e. that a (y') = (a )(y') and l i s. J m J 1 r X m 0^ Cy ' ) = (a )(y') = avm(y') . Hence 2 vm e X m (a )(y') = (a s )(y*) = (a s )(y') = (a )(y') . Za.rX 1 4™ 1 1 2 2 Zb.-X ± + m l m . 1 l l Thus (a o r )(y') = o. (y1) . Since y' e S(o ) a r D a n m n n (val , ( a ) = n) and x(y') = a (y1) this establishes 1 m 71. \ ( y , ) x(y') = . Since y' e U A N is arbitrary and c a (y«) y n y e U r\ N this implies that T can be extended to a member y of r(U u {y},k) by setting Ty( u) = T(U) for u e U and T (y) = — — y aa (y) n Q.E.D. 6.3) (Lemma) I r\ J =)= 0 x x Proof: Since s *s„ E I A J therefore a (x)a (x) = 1 2 s i s2 a (x) e (I o J) =1 O J . Thus i t suffices to show that s^s^ x x x a (x)a (x) =j= 0 . For any y e U , y e some U. so S l S2 1 val (a ) = i = val (a ) and by 3.12.vi) a (y)a (y) 4= 0 y S l . y S2 S l S2 Since x e U and any section has closed support this establishes o o (x)a (x) f 0 . S l S2 Q.E.D. Let t.s_, ...t s. e I n J be such that each of 11 n 1 a (x)a (x), ...a (x)a (x) is none zero. It will be shown t, s_ v . t ' s, 1 1 n 1 in lemmas 6.4 - 6.6 that the elements a (x)o (x), ...a (x)a (x) t- s. t s. 1 1 n 1 do not generate I A J . x x For each integer m > 0 let V = {y e X : for some j(1 < j < n) val (a ) = m + 1 and m J J — J — y t val (o ) <_ m holds for no i (1 <_ i < n)} y i *~ 72. Since there are only finitely many t^, g i t follows from 0.15 that each V is clopen. m Let Y = U u (U r\ (U(V )) . It follows from 6.2 and 6.1.ii) m m that x may be viewed as a member of T(Y,k) . Let m >_ 0 be an integer. Since U AV is closed T ° m — TT may be viewed U A V m as a global section. Pick f^ e R such that (a )(x) = m (T - v )(x) for x e Vm and (a f )(x) =0 for x { V m m m Pick v e R such that a = \b„ . It may be verified that m v V m m for any integer i > 0 and m >_ 0 °f °e. = ( T , J V > V = ( T V } = % ' a r . m i m i m i m i so that f me. = v r. . This may be used to show that m l m l J (Z f Xm)(Z e.X1) = (Z v Xm)(Z r.X1) . Let s = (Z f Xm) (Z e.X1) m i m i j m i . Then s. e I n J . Thus a (x) e I A J 3 s^ x x It will be shown in lemma 6.5 that a (x) is not in the ideal S3 generated by the o (x)o (x), ... a (x)a (x) t 1 s± t]L s± 6.4) (Lemma) Let N be a neighborhood of x . Then there exists an integer m > 0 such that N A U H V 4= <j> ° — x m 1 Proof: Suppose that for each m N A, U A V = d> . Let i f x m Y M = {y e X : val (a ) = 0 for some i (1 <_ i <_ n) } . Since y i 73. there are only finitely many s > ^ l s clopen. It follows from 6.2 and the choice of x that x i M . Let N' = (X-M) C\ N . r x x Then N' is a neighborhood of x such that val (a ) = m fails, x y t. J 1 for each integer m >_ 0 and i ( l <_ i <_ n) , to hold. Since N' A U is open i t thus follows from 3.12.v) that N' O U £ Z(a ) X X t . 1 for each i (1 < i < n) . Clearly (X - U) A N C. Z(a ) . X S l Thus U A (X - U) A N C Z(a a ) so that by 3.1 a (x)a (x) = 0 X t . S - t . S -1 1 1 1 for each i (1 <_ i <_ n) . This is contrary to the hypothesis so the lemma-is established. Q.E.D. n 6.5) (Lemma) a (x) i £ K a (x)a (x) s„ T . , x t. s 3 i=l l 1 Proof: Suppose that the lemma is false. Then find X,, ... A e S I n and N , a neighborhood of x , such that y e N implies that X X n o (y) = E (a (y)a (y)a (y)) . By 6.4 find an integer m 3 i=l l i 1 and a y e N A U A V . Then y e U. for some j so by 3.12.iv) x m J and the definitions of s^ and s^ and the above equations i t follows that m + j = val (a ) > mf{val (a a ): l<i<n} = m + 1 + i J y s„ — y t. s/ J 17 3 l 1 This contradiction establishes the lemma. Q.E.D. 74. 6.6) (Corollary) i) I n. J is not finitely generated X X i i ) K is not coherent. x i i i ) If B(R) is -self-injective then is an integral domain that is not coherent. iv) I r\ J is not finitely generated. Proof: i) This follows immediately from 6.5. i i ) This follows immediately from i ) . i i i ) This follows from i i ) and 3.15. iv) There is an epimorphism S ->- K given by s -»- o (x) . The X s image of I n J under this map is (I A J) = I f\ J • The result X X X now follows from i ) . 6.7) (Theorem) i) S has neither properties a) nor b). i i ) S does not have property e). i i i ) S is not coherent. iv) S is not semi-hereditary. v) w.gl.dim(S) > 1 Proof: i) This was established in 6.6iv). i i ) , i i i ) , iv), v): These now follow from i ) , either directly or via 1.13. Q.E.D. We now construct a von Neumann regular ring R" such that R"\7\~X~]~1 has properties c) and d) but not property a). Let X' be an infinite Boolean space. Find x E X' and an j\^-set U in X' such that x E U - U . Let S' be the 75. s i m p l e GF(4) sheaf over X ' where GF(2 ) i s the f i e l d w i t h 2 e lements . Let S" be the subsheaf of S ' such t h a t S" = GF(2) x and S" = S ' f o r x + y . L e t R" = r ( X \ S " ) . I t i s e a s i l y y y 1 seen t h a t S" does not have the J N ^ - e x t e n s i o n p r o p e r t y so that by 3.9 and 6.7 R'TVx~0 does not have p r o p e r t y a ) . I f X ' i s chosen to be jX^ - e x t r e m a l l y d i s c o n n e c t e d i t f o l l o w s from 3.15 tha t R"t\XT\ does have p r o p e r t i e s c) and d) . 76. § 7 Example of a Boolean ring R that is X -self-injective but is not -complete, In this section let j be a fixed but arbitrary Cardinal such that A >_ jN* . Identify \ with the least ordinal of cardinality A In this section we construct a Boolean space X with the A-disjointness property that is not JN^1 -extremally disconnected. R is constructed by letting R = r ( X , k ) where GF(2) is the two element field and k is the simple GF(2) sheaf over X . It follows from 3 . 9 , and 5 .3 that R is the required Boolean ring, w. gl. dim(R\TX"\l) < 1 . Rr\"Xll has properties a), b), and e), yet R f T X l l has neither property c) nor d). X is constructed to be a one point union of the form X = (Y v W/ ) where Y and W are Boolean spaces and the fixed p~q points p e Y and q e W are identified. Y and W are constructed such that they have the A-disjointness property and q k (dTT(V) - V) w for any A-set V in W This results in X having the A-disjoint-ness property. Y and W are also constructed such that q is not isolated in W and there exists an jN^ -set N in Y such that p e (c!y(N) - N) . Thus X can not be -extremally disconnected for N is an ^ -set in X such that q e (X-c^(N)) p> cl X(N) so that cl„(N) is not open in X X First the space W is constructed. If W = (A + 1) then q = A would be as described above. However (X + 1) does not even have the X-disjointness property. Let S = r((X + 1)>S) where 5 is 77. the simple GF(2) sheaf over (X + 1) W is actually constructed to be X(T) where T turns out to be the X-completion of the Boolean ring S . Let Q be the complete ring of quotients of S . Q is a self-injective Boolean ring so that by 2.4 of [6] i t is complete. For any subset F of Q let S/F denote the least upper bound of F in . Q and let f\F denote the greatest lower bound of F in Q For any, q,r e S let q v r denote y{q,r} and q A r denote A{q,r} . The following facts will be used. i) For T I , T 2 e S , T ± 1 T 2 < — > T± V T 2 = T 2 <—> T± A ^ = T± <—> T 1'T 2 = T ; L <—> S ( T L ) £ S ( T 2 ) <—> Z ( T 2 ) C Z ( T L ) . i i ) Let I be an index set, {a : a e 1} be a subset of Q and b e Q a Then \/{a v b : a e 1} = (V{a : a e I) V b . Also, 1 - V{a : a e 1} a a a = AU " a : a e 1} . a i i i ) Let I and J be index sets and {a „ : a e I and 8 e J} be a a, 8 subset of Q . Then V{a : a e I and 8 e J> = VMa : a e 1} : 3 e J} . o t , p a, p iv) Let r e Q . Then there exists subsets of S,F and F' , such that r = VF = A F ' • The first of the above facts is obvious and the second and third are from [8]. The fourth is from 2.4 of [6]. Let T = {t e Q : there exists {T : a < X} C , s such that a t = V d : a < X}} . 78. 7.1) (Lemma) Let {t : 6 < X} be a subset of T . Then y{t : g < X} e T . Proof: By the definition of T there exists {T a < X and J a,3 8 < X} , a subset of S , such that t = \/{T a < X} for 8 ex, p each 8 < A . Thus \/{t„ : 8 < X} = \/{V{T „: a < A} : g < A}-v 3 u a, g = o- A < A and g < X} e T . v a, 8 Q.E.D. In particular i f s,t e T then s y t e T 7.2) (Lemma) Let {C : a < X} be a family of closed subsets of a (X + 1) such that, for each a < X, X i C . Then there exists 1 a C , a clopen subset of (X + 1) , such that C C C for each a — a < X yet X \ C . Proof: For each a < X let d be the least upper bound for a C in the ordinal X + 1 . Since each C is closed we have a a each d < X . Let d be the least upper bound for {d : a < X} It follows from the choice of the ordinal X that d < X . Let C = {g e (X + 1) : 0 <_ g <_ d} . Then C is the required clopen subset of (X + 1) Q.E.D. 7.3) (Lemma) Let {T : a < X} be a subset of S such that a T , ( X ) = 0 for some a' < A . Then (A(x : a < A}) e T . Ct GC 79. Proof: Let L = { p e S : p < each T } — a It follows from fact iv) that y/L = /\(T : a < A} . It remains to show that \/L e T . Let U = U (S(p)). It follows from fact peL i) in the opening remarks to this section that U c. S(T ,) . Since S(T ,) is closed and does not contain A i t follows that U < A . a Thus there exists {N : a < A} , a family of clopen subsets a of (A + 1) , such that U = U (N ) . It thus follows from \^ or a<A fact i) that {a e S : a >_ each \\i } ={o e S : a >_ each p e L} a This establishes via fact iv) and the definition of T that A {x : a < A } = V L = V^A, : a < A} e T . a Q.E.D. 7.4 (Lemma) Let {T : a < A} be a subset of S such that T a(^) = 1 for each a < A . Then A { x : a < A } e T a Proof: By hypothesis ^ ^ T a ^ : o: < A} is a family of closed subsets of (A + 1) such that for each a < A A | Z(T ) By 7.1 find C , a clopen subset of (A + 1) , such that each Z(x ) C C yet A J: C . Let D = (A + 1) - C . Then for each T , T = (T v tyn • Each (T ' \\i ) (A) = 0 . Thus by e t a a C D a C 7.3 and the comment after 7.1 /\{T : a < A} = AUT -ib ) y <IV a < A} = -ty: a < \}) y ib e T . 80. 7.5) (Lemma) Let {x^  : a < A} be a subset of S . Then /\{x : a < A} e T . a Proof: This follows immediately from 7.3 and 7.4. Q.E.D. 7.6) (Lemma) If t e T then 1 - t e T . Proof: Suppose that {x : a < A} is a subset of S such a that \/{x : cx < A} = t . Then v a 1 - t = 1 - (V(T : a < A}) = A(l - x : a < A} e T . a a Q.E.D. 7.7) T is a A-complete Boolean ring. Clearly S is a subring of T Proof: It is well known that this follows from 7.1, the comment immediately after 7.1, and 7.6. Q.E.D. Let W = X(T) . Let f : W -> X(S) be defined by f (M) = M r\ S for each maximal proper ideal M in T Then f is a continuous onto map. (This is a well known fact about Boolean algebras. The map f is actually the function X(T) -> X(S) induced by the inclusion map S -> T . See §11 of [9].) Since T is A-complete, W is A-extremally disconnected, and thus W has the A-disjointness property. 8 1 . 7.8) (Lemma) There exists exactly one q e W such that f(q) = A Proof: Since f is onto i t suffices to show that at most one such q exists. Under the identification (A + 1) = X(S) the point A corresponds to the ideal K = {a z S : a(A) = 0} A (See 0.15.iii.) Thus i t suffices to show that the elements of M are uniquely determined where M is an ideal in T such that M r\ S = M, . Let t = V{T : a < A} where {x : a < A} is a A a a subset of S First suppose that for each a < A A \ s ( T a ) • Then by 7.2 find C , a clopen subset of (A + 1) , such that each S(x ) 9 C yet A \ C . Then f e M c M . Since t = V{T : a < ^ } _ S * l V w e have t = t* i j j e M Now suppose that there exists a' < A such that A e S(x ) Then t i M for i f t e M we would have x , = (x ,*t) z M n S = M . This would contradict x , (A) 4= 0 A • Q.E.D. For the rest of this section let q denote the unique point in W such that f(q) = A 7.9) (Lemma) i) q is not isolated in W ii ) There does not exist a A-set V in W such that q e (cl w(V) - V) . Proof: i) Suppose that q is isolated in W . Then W - {q} is compact. Since f : W -> (A + 1) is onto and f ^(A) = {q} it follows that f(W - {q}) = A (A + 1) . Thus A is a 82. compact subset of (A + 1) . This is a contradiction since "X is a limit ordinal. i i ) Suppose that the lemma is false. Thus there exists {Ca : a < A} , a family of clopen subsets of W , such that for each a < A q i C yet a e c 1 ( U (C )) . Then each a W , a a<A f(C^) is compact in (A + 1) and thus is closed. It follows from 7.8 that for each a < A, A (j; f C C ^ ) . Then A = f(q) e f ( c l w ( U (C ))) a<A C cl (f( U (C )) ( X + 1 ) a<A a = V ) ( a Vf ( c « ) ) } ' Since for each a < A f(C ) is closed and A <t f(C ) this a T a contradicts 7.2. Q.E . D . Let Y be the Stone-Czech compactification of the set N of natural numbers where N has the discrete topology. For the rest of this section let p denote a fixed but arbitrary element in (cl^CN) - N) . Let X = (Y \J W/ ) be the one point union of Y and W in which p and q are identified. Y and W may each be topologically identified with a subspace of X in such a way that Y U W = X and Y ° W =' {p} =' {q} . 7.10) (Lemma) i) i f x e (Y - {p}) then {N : N is a neighborhood X X of x in Y} is a neighborhood basis for x in X If x e (W - {q}) then {N : N is a neighborhood of x in W} X X is a neighborhood basis for x in X . If x = p = q then { N U N : N is a neighborhood of p in Y and N is a p q p q neighborhood of q in W} is a neighborhood basis for x in X i i ) If U C X then cl_(U) = cl v(U A Y) \J cl T 7(U A W) . — x y w i i i ) X is a Boolean space. iv) The set of points isolated in X is dense in X Proof: i) This is straightforward to check. i i ) This can readily be checked using i ) . i i i ) That X is Hausdorff and totally disconnected follows from i ) . X is compact because i t is a continuous image of the disjoint union of Y and W . The disjoint union of two compact spaces is clearly compact. iv) The following fact (paraphrased from p. 28 example A of [9]) will be used: a Boolean ring A is atomic i f and only i f X(A) has a dense subset consisting only of isolated points. It follows from this that S is atomic for {a + 1 : a < A} consists only of isolated points and is dense in (A + 1) It follows readily from this and the definition of T that T also is atomic. Thus cl^(N') = W where N' denotes the set of points that are isolated in W . Since N consists only of points that are isolated in Y i t follows from i) that N u N' consists only of points that are isolated in X . Note that c.1 (N U N ' ) = cl (N) u cl (N') = Y U W = X . X Y W Q.E.D. 84. 7.11) (Lemma) X is not -extremally disconnected. Proof: It follows from 7.10i) that N is an j \ * -set in X It follows from 7.10ii) that c l (N) = Y . Since p = q is X not isolated in W it follows from 7.10i) that Y is not open in X . Q.E.D. 7.12) (Lemma) X has the A-disjointness property. Proof: W has the A-disjointness property since T is A-complete and W = X(T) . It is established in [5] that the Stone-Czech compactification of an extremally disconnected space is extremally disconnected. Thus Y is extremally disconnected so that by 3.4 i t has the A-disjointness property. Let {U : a < A} and {V : a < A} be families of clopen a a subsets of X such that U A V = ty where U = U (U ) and a<A V = U (V ) . For each a < A a<A a let A = U A Y , B = V A Y , a a a a C = U P i W , and D = V O W a a a a Let A = U(A ) , B = U(B ) , C = U(C ) , and D = U (D ) a<A a a<A a . a<A a a<A a Then A and B are A-sets in the space Y such that A H B = ty and C and D are A-sets in the space W such that C 0 D = ty . Thus cl y(A) f\ c l (B) = ty and c y c ) 0 cl w(D) = ty . Note that by 7.10 i i ) cl (U) = cl x(A U C) = cl y(A) U cl w(C) and cl x(V) = cl x(B \J D) = 85. cl y(B) U cl w(D) . Thus c l X ( U ) C\ cl x(V) = (cl Y ( A ) n c l Y(B ) ) U(cl Y ( A ) n c l w(D ) ) U(cl w(C ) r v:l Y(B))t;(cl w(C)rNcl w(D)) Hence c l X ( U ) H cl x(V) = (cl y ( A ) C\ c l w ( D ) ) u <C1W(C) A cl y(B)) . It now suffices to show that cl (A) A cl (D) = <J> and JL W cl (C) C\ cl (B) = (J) . There are three cases to be considered. For the first case suppose that p £ U and p £ V . Then p \. C and p j; D so by 7.9 i i ) p \. clTT(C) and "p A cl (D) . w w (Recall that Y and W are identified with subspaces of X and in X p = q .) Hence cl (C) C\ Y = clTT(C) p\ Y = <j> and X w cl v(D) n Y = cltl(D) r\ Y = (j) . Thus X w cl (A) n clTT(D) £ Y n clT7(D) = <j) and Y W W cl w(C) H cl y(B) c cl w(C) r\ Y = <fc . For the second case suppose that p e U . Then p £ V since D r\V = | . Suppose without loss of generality that p e U Q . Then B C Y - U Q and Y - UQ is clopen in Y so that cl^B) C y - U Q and hence cl (B) A w = $ . Similarly clT1(D) O Y = 4, . Thus cL(A) r\ clrT(D) 9 Y A clT7(D) W . Y W w = (fi and cl w(C) Ocl Y(B) c W n cl y(B) = <(. For the third case suppose that p e V . This is similar to the first case. Q.E.D. 86. .13) (Theorem) i) There exists a Boolean space X that has the A-disjointness property but is not jX^ -complete. i i ) There exists a Boolean ring R such that R is atomic, >-self-injective, but not j\ -complete. i i i ) Let R be as in i i ) . Then w.gl. dim(RrrXYi) <_ 1 , RfTXTl has properties a), b), and e) , yet RITXT* has neither, property c) nor d). Proof: i) X has already been constructed. i i ) The construction of R given X was described in the opening remarks to this section. That R is atomic follows from 7.10 iv) and the fact that X(R) = X . The rest of i i ) follows from i) and 3.9. i i i ) This follows from i i ) and 5.3. Q.E.D. T. Crammer has observed in private communication that (Y - N) in the relative topology as a subspace of Y is a Boolean space with the -disjointness property that is not j V j -extremally disconnected. Since (Y - N) has no isolated points i t follows, similar to 7.13, that there exists an atomless Boolean ring R' such that R' is Xc^ -self-injective but is not JN?, -complete. Again, w.gl. dim(R' f rX"H ) <_ 1 , RTptTI has properties a), b), and e), yet R'fTx"]-) has neither property c) nor d). 87. Bibliography 1. Bourbaki, N. "Algebre Commutative," Herman, Paris, 1961. 2. Bredon, G.E. "Sheaf Theory," McGraw-Hill, (1967). 3. Cartan, H. and Eilenberg, S. "Homological Algebra," Princeton University Press (1956). 4. Chase, S.U. "Direct Product of Modules," Trans. Amer. Math. Soc. vol. 97 (1960) pp. 457-473. 5. Jensen, CU. "Some' Cardinality Questions for Flat Modules and Coherence," J. Algebra vol. 12 pp. 231-241. 6. Lambek, J. "Lectures on Rings and Modules," Blaisdell (1966). 7. Pierce, R.S. "Modules over Commutative Regular Rings," Memoirs of the Amer. Math. Soc. Number 70. 8. Sikorski, "Boolean Algebras," (second edition) Springer-Verlag, Berlin (1964). 9. Soublin, J.P. "Un Anneau Coherent dont l'Anneau des Polynomes n'est pas Coherent," C.R., Acad. Sci. Paris ser A 267 (1968), pp. 241-243, Gilman, L. and J e r i s o n , M. "Rings of Continuous Functions, Van Nostrand, (1960). 

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