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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Sheaf methods applied to coherent rings Carson, Andrew Bruce 1971
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Title | Sheaf methods applied to coherent rings |
Creator |
Carson, Andrew Bruce |
Publisher | University of British Columbia |
Date Issued | 1971 |
Description | A commutative ring is called coherent if the intersection of any two finitely generated ideals is finitely generated and the annihilator ideal of an arbitrary element of the ring is finitely generated. Pierce's representation of a ring R as the ring of all global sections of an appropriate sheaf of rings, k , is described. Some theorems are deduced relating the coherence of the ring R to certain properties of the sheaf k . The sheaves from the above representation for R⌈X⌉ and R⌈⌈G⁺⌉⌉ , where R is a commutative von Neumann regular ring and G is a linearly ordered abelian group, are calculated. Applications of the above theorems now show that R⌈X⌉ is coherent and yield necessary and sufficient conditions for R⌈⌈G⁺⌉⌉ to be coherent. |
Genre |
Thesis/Dissertation |
Type |
Text |
Language | eng |
Date Available | 2011-03-30 |
Provider | Vancouver : University of British Columbia Library |
Rights | For non-commercial purposes only, such as research, private study and education. Additional conditions apply, see Terms of Use https://open.library.ubc.ca/terms_of_use. |
DOI | 10.14288/1.0080486 |
URI | http://hdl.handle.net/2429/33106 |
Degree |
Doctor of Philosophy - PhD |
Program |
Mathematics |
Affiliation |
Science, Faculty of Mathematics, Department of |
Degree Grantor | University of British Columbia |
Campus |
UBCV |
Scholarly Level | Graduate |
AggregatedSourceRepository | DSpace |
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