UFDC Home  Search all Groups  UF Institutional Repository  UF Institutional Repository  UF Theses & Dissertations  Vendor Digitized Files   Help 
Material Information
Subjects
Notes
Record Information

Full Text 
ON COMMUTATIVE fRINGS WHICH ARE RICH IN IDEMPOTENTS By SCOTT DAVID WOODWARD A DISSERTATION PRESENTED TO THE GRADUATE SCHOOL OF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY UNIVERSITY OF FLORIDA 1992 '.1 ^ ^ r I I ACKNOWLEDGEMENTS I would like to express my deep gratitude to my advisor, Dr. Jorge Martinez. His guidance, perseverance and patience have made this work possible. His concern for his students and his perspective of the human condition have been a personal inspiration. I would like to thank the members of my committee, Drs. Krishna Alladi, Roy Bolduc, Doug Cenzer and Chris Stark, for their time, interest and input. In addition I would like thank Dr. Neil White, for his consistent support, and Dr. Robert Long, for inspiring me to follow my bliss. My thanks go to the Department of Mathematics for its support during my years as a graduate student. In particular, I would like to thank the department secretaries for keeping me on top of things and making my time here pleasant, and Randy Fischer for helping me with my computer phobia. Special thanks go to my family; to my parents for encouraging me to be who I am; to my grandmother, the family's first mathematician; to my children for making what I do matter. Finally, a very special thank you to Rita, my wife and best friend. Without her belief in me, none of this would have happened. This work is lovingly dedicated to the memory of my father. TABLE OF CONTENTS ACKNOWLEDGEMENTS ............................ ABSTRACT .. .. ... .. .. .. .. ... .. .. .. .. .. .. . CHAPTERS 1 INTRODUCTION ............................... Lattice Ordered Groups and the Yosida Space ......... Commutative SemiPrime fRings and the Maximal Spectrum Tychonoff Spaces and SemiPrime fRings ........... Boolean Duality .......................... 2 LOCALGLOBAL fRINGS .......................... 2.1 Introduction . . . . 2.2 The Specker Subring of a SemiPrime fRing . . 2.3 A Characterization of LocalGlobal fRings . . 3 QUASISPECKER fRINGS AND fRINGS OF SPECKER TYPE . 3.1 Introduction . . . . 3.2 QuasiSpecker fRings and fRings of SpeckerType . . 3.3 QuasiSpecker Spaces and Specker Spaces . . 3.4 The Absolute of a Hausdorff Space. . ... 3.5 Specker Spaces and Absolutes ...................... 4 CONCLUSION ................................. REFERENCES ................................... BIOGRAPHICAL SKETCH................................. 1 8 11 14 19 19 23 28 44 Abstract of Dissertation Presented to the Graduate School of the University of Florida in Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy ON COMMUTATIVE fRINGS WHICH ARE RICH IN IDEMPOTENTS By SCOTT DAVID WOODWARD August 1992 Chairman: Dr. Jorge Martinez Major Department: Mathematics In this dissertation we compare algebraic properties of commutative semiprime f rings A having the bounded inversion property with topological properties of Max(A), the compact Hausdorff space of maximal ideals of A with the hullkernel topology. An fring A is localglobal if for every primitive polynomial f(t) E A[t], there is an a E A such that f(a) is a multiplicative unit. We prove that if A is a commutative semiprime fring with identity and bounded inversion, then each of the following implies the next. 1. A is localglobal. 2. For each primitive a bt2 E A[t] with 0 < a, b, there exists a c E A such that a bc2 is a unit. 3. Max(A) is zerodimensional. 4. Max(A) Max(S(A)), where S(A) is a subalgebra of A generated by the idempotents of A. Furthermore, the last three conditions are equivalent and if A has a strong unit, then all the conditions are equivalent. As a corollary, we show that for X a compact Hausdorff space that C(X) is localglobal if and only if X is zerodimensional. As a special case we show that if A is a Bezout ring or X is a quasiF space then the assumption of a strong unit or compactness can be dropped. We then take a closer look at the containment of S(A) in A. Speckertype and quasispecker rings are defined. It is shown that A is a speckertype ring if and only if A is an fsubring of Q(S(A)), the complete ring of quotients of S(A), if and only if Q(A) = S(A)L, the lateral completion of S(A). We then define specker spaces and quasispecker spaces. It is shown that a space X is a specker (quasispecker) space if and only if #X, the StoneCech compactification, is a specker (quasispecker) space. Finally we show that if X is a quasispecker space and EX, the absolute of X, is a specker space, then X is a specker space. The converse obtains when X is compact and has a countable 7rbase. CHAPTER 1 INTRODUCTION This dissertation is largely an exploration of relationships that exist between certain algebraic structures and topological spaces that can be associated with one another in a natural and sometimes functorial manner. The historical and methodological basis for this comparison goes back to the pioneering work of Marshall Stone, in particular his paper "Applications of the Theory of Boolean Rings to General Topology" [29] where he shows that the category of Boolean algebras with lattice homomorphisms is dual to the category of compact zerodimensional Hausdorff spaces with continuous maps. This idea of considering certain algebraic substructures to be the points of a topological space pervades much of what follows. 1.1 Lattice Ordered Groups and the Yosida Space A latticeordered group, denoted group is a group (G, +, 0) together with a par tial ordering < on G such that (G, <) is a lattice satisfying a compatibility condition between the ordering and the group operation: If a < b then a + c < b + c and c + a < c + b We will explicitly define what a lattice is in Section 1.4. Although in the general theory of latticeordered groups there is no assumption or necessity that the group operation be commutative, for the purposes of this dissertation we will assume that all groups are commutative. A latticeordered group which is a vector space over the reals such that scalar multiplication by positive real numbers preserves the order is called a vector lattice. For a, b E G we will denote the least upper bound (join) and greatest lower bound (meet) of a and b by a V b and a A b respectively. For infinite or arbitrary subsets of G we will denote respectively the least upper bound and greatest lower bound of these sets when they exist, by V g\ and A gx when they are indexed and V S and A S otherwise. G+ will denote the set of all g E G with 0 < g. Latticeordered groups have the following properties. A much more complete list of the properties of egroups as well as proofs of these results can be found in Chapter 1 of "LatticeOrdered Groups" [2]. Let G be an group, a, 6, c E G, then 1. a+ (bV c) = (a+b)V (a +c) and dually. 2. (a V b) = (a) A (b) and dually. 3. a+ b = (aVb)+(aAb). 4. (G, A, V) is a distributive lattice, that is aA (b6V c) = (a A b) V (a A c) and dually. 5. (G, +) is a torsion free group. Latticeordered groups also have the Riesz Interpolation Property; 6. If hi,..., h, E G+ and if 0 < g < h, +* + h, then there exists g,..., g E G+, with 0 < gi < hi for 1 < i For a, b E G we say that a is disjoint to b if a A b = 0. For g E G the positive part of g is g+ = g V 0 and the negative part of g is g = (g) V 0. Then g+ A g = 0 so that we can write any element uniquely as the difference of disjoint elements, namely g = g+ g. The absolute value of g is lgI = g+ + g. For (abelian) igroups we have the triangle inequality; 7. a + bl < al + Ibl. From (3) above we have that; 8. If aAb=Othen a+b=aVb. In the case that an infinite join or meet exists, we have the following infinite versions of (1), (2) and (4); 9. g + (V hA) = V(g + hx) and dually. 10. (V hA) = A(h>) and dually. 11. g A (V hA) = V(g A hA) and dually. We now turn our attention to the subgroups of an igroup. Definition 1.1.1 If G is an igroup, a subgroup H of G is called an isubgroup if H is a sublattice of G; that is, H is closed under finite meets and joins. An isubgroup H is said to be convex if whenever hi < g < h2 with hi, h2 E H, then g E H. A normal convex isubgroup is called an iideal. If G and H are igroups, a map q : G + H is called an ihomomorphism if it preserves both the group and lattice structure. Since we are dealing strictly with abelian igroups, normality of subgroups is not an issue. From now on we will not distinguish between convex isubgroups and iideals. We have the following theorem relating ihomomorphisms and iideals. Theorem 1.1.1 (1.2.1 [2]) Let G and H be igroups and let q : G > H be an & homomorphism from G onto H. 1. Ker4, the kernel of 0, is an fideal. 2. If N is an fideal, then there is an ordering on G/N such that G/N is an fgroup and the canonical group homomorphism 0 : G + G/N is an fhomomorphism. 3. G/Ker4 is fisomorphic to H. The ordering on GIN from (2) is induced by the ordering on G by defining g + N < h + N if there is a k E N such that g < h + k. That this is a lattice ordering with (g + N) V h + N = (g V h) + N and dually, is the content of Theorem 1.2.1 [2]. Let C(G) denote the set of eideals of G. For A, B E C(G), let A A B = A n B and A V B = the subgroup generated by A and B. By convexity and the Riesz Interpolation Property, these operations make C(G) a lattice. Definition 1.1.2 A lattice A is said to be complete if for every subset S of A, both V S and A S exist in A. A is called brouwerian if for a, b\ E A, aA (V b\) = V(a Ab\). We are now ready to state the following theorem which is a result from 1942 due to G. Birkhoff [6]. Theorem 1.1.2 C(G) is a complete brouwerian sublattice of the lattice of all subgroups of G. There are certain types of fideals which are distinguishable in C(G) that will be of particular interest. Definition .1.13 Let g E G. C E C(G) is said to be a value of g if C is maximal with respect to not containing g. C is said to be prime if G/C is totally ordered. We have the following theorems which allow us to distinguish values and prime fideals in C(G). See Chapter 2 [5]. Theorem 1.1.3 Let C E C(G). Then C c n{B E C(G) : C C B} if and only if there is a g E G such that C is a value of g. An application of Zorn's Lemma establishes the following theorem. Theorem 1.1.4 Let B E C(G) with g B. Then there is a value C of g with B C C. Definition 1.1.4 For a value C, C* = f{B E C(G) : B C} is called the cover ofC. The following provides several useful characterizations of prime fideals. Theorem 1.1.5 (Theorem 2.4.1 [5]) Let P E C(G). The following are equivalent. 1. P is prime. 2. If A, B E C(G) and An B = P then A = P or B = P. 3. {C E C(G) : P C C} is totally ordered. 4. For a, b G, ifa A b P then aEP or b E P. 5. For a, bE G, if aAb = 0 then a E P or bE P. It is clear then that a value is prime, and from (4) above it appears that prime f ideals behave rather like prime ring ideals. We will eventually see in certain cases just how far this likeness goes. For now we have the following correspondence theorem. Theorem 1.1.6 (Proposition 2.4.7 [5]) Let G be an fgroup, H an fsubgroup of G. Then the map P i P n H is a onetoone correspondence between the prime fideals of G not containing H and the proper prime fideals of H. For a value V, V* is an egroup having V as an esubgroup. Since V is a prime eideal of V*, by the correspondence theorem we get that the quotient group V*/V is a totally ordered egroup with no proper tideals. The following will characterize a particular large class of totally ordered groups. Definition 1.1.5 An fgroup G is said to be archimedean if na < b for all n E N implies that a < 0. This is equivalent to, if 0 < a,b E G, then there is an n E N such that na : b. Therefore in the totally ordered case this becomes the familiar definition of archimedeaneity. The following theorem is due to O. H6lder [21]. It is from a paper published in 1901, and the proof is a generalization of the classical construction of the reals from the rationals using the cut completion. Theorem 1.1.7 Let G be a totally ordered group. The following are equivalent. 1. G is archimedean. 2. G is fisomorphic to an additive subgroup of R. 3. G has no proper eideals. We now turn our attention to one of the topological spaces which can be associated with an group, namely the Yosida space of an archimedean group. Let g E G and denote by G(g) the ideal generated in G by g. That is G(g) = {a e G : \al < nng\ for some n E N}. An element 0 < u E G is called a unit if for any a E G, if a A u = 0 then a = 0; u is called a strong unit if G(u) = G. Note that a strong unit is a unit. Suppose now that G is an archimedean fgroup with unit u. Let Yos(G, u) = {V E C(G) : V is a value of u} The sets of the form Ua = {V E Yos(G,u) : a V}, for all a E G, are a base for the open sets of a topology on Yos(G, u). We will call Yos(G, u), with this topology the Yosida space of G with respect to u. This is in fact the hullkernel topology on Yos(G, u). We have the following theorem. Theorem 1.1.8 (Corollary 10.2.5 [5]) Let G be an archimedean Igroup with unit u. Then Yos(G, u) is a compact Hausdorff space. For a topological space X, denote by D(X) the set of all continuous functions with values in the extended reals that are real valued on a dense subset of X. That is, D(X) = {f : X + RU {oo} : f is continuous and f1(R) is dense in X} It should be pointed out here that although D(X) inherits a lattice ordering from the ordering on R U {oo}, it is not generally true that D(X) is a group. The problem is that one can define f + g by (f + g)(x) = f(x) + g(x) on f'(R) n g'(R) and the domain of definition is a dense subset of X, but there is no guarantee that f + g so defined can then be extended to all of X. If such an extension does exist, by the density of f'(R) f g1(R) it is unique and in this case we define f + g to be this unique extension. By an egroup in D(X) we mean a sublattice that is a subgroup with the group operation as defined above. The following is the Yosida Embedding Theorem for archimedean fgroups. Theorem 1.1.9 Let G be an archimedean fgroup with unit u > 0. Then there exists i embedding q$: G  D(Yos(G, u)) such that G(G) is an fgroup and G is eisomorphic to f(G). Moreover, can be taken so that (u) = 1, the constant function 1, and if V, W E Yos(G, u) then there is a g E G such that (g)(V) # (g)(W). Theorem 1.1.9 originates in a 1942 paper by K. Yosida [34]. Hager and Robertson [19] show that the separation of points uniquely determines the Yosida space of an archimedean group. For later results we will need a precise description of the embedding 4 whose existence is guaranteed by Theorem 1.1.9. Let V E Yos(G, u). Since G is archimedean, by H6lder's theorem, V*/V is f isomorphic to an additive subgroup of R. Let Oy be such an isomorphism. Then 4: G + D(Yos(G, u)) is defined b, for g E G, Ov(g + V) if g EV* O(g)(V)= oo if g V* and g + V > V oo if g V* and g + V < V 1.2 Commutative SemiPrime fRings and the Maximal Spectrum For the purposes of this dissertation we will consider only commutative rings with identity. For more on frings, see chapter 9 in "Groupes et Anneaux R6ticul6s" [5]. Definition 1.2.1 A latticeordered ring (ring) is a ring (A, +, , 0, 1) together with an ordering < on the elements of A such that the following hold: 1. (A, <, +, 0) is an group. 2. For every a, b, c A, if a < b and 0 < c then ac < be and ca < cb. Definition 1.2.2 An fring is said to be an fring if A is order isomorphic to a sub direct product of totally ordered rings. This is equivalent to, 3. For every a, b, cA with 0 c, ifaAb=0 then acAb= 0 = ca A b. By an falgebra we mean an fring that is also a vector lattice, and by an fsubring we mean a subring that is also a sublattice. The following theorem lists some of the properties of frings. Theorem 1.2.1 (Theorem 9.1.10 [5]) Let A be an fring and let a, b, c A. Then, 1. If c 0, then c(a A b) = ca A cb and dually. 2. If a A b = 0, then ab = 0. 3. a2 > 0. 4. If a > 0 and ab > 0 then b > 0. 5. For a, b >0 ab = (a A b)(a V b). If we are not careful, confusion can result in the context of latticeordered rings, due to the overlapping of terminology from ring theory and the theory of lattice ordered groups. In an attempt to keep these ideas distinct and be consistent with acceptable usage, we repeat some old definitions and state some new ones. By an ideal of A we mean a convex tsubgroup of the group (A, +, <) and by an ideal of A we mean an ideal of the ring (A, +, ). An ideal P is a prime ideal if ab E P implies a E P or b E P. An ideal P is a prime 1ideal if a A b E P implies a E P or b E P. An element u E A will be called a multiplicative unit if it is an unit in the ring (A, +, ), an order unit or strong order unit if it is such in the Ggroup (A, +, <). The following example is meant to help make these distinctions clear. Example Let A be the set of all real sequences. If we define addition and multi plication pointwise, then A becomes a commutative ring. If we also order pointwise, that is {a,} 2 {bn} if a, > b, for all n E N, then A is an fring. In fact if we also define real multiplication pointwise, then A is an falgebra. Let B be the set of bounded rational sequences. With operations and ordering defined as for A, B is a commutative fring. B is an fsubring of A, and B is not an falgebra. Neither is B an ideal nor an eideal of A. Consider the sequences a = {1,1/2,1/3, } and b = {1,, 1,1, }. In B, a is a weak order unit, but not a strong order unit or a multiplicative unit. In A, a is a weak order unit and a multiplicative unit, but not a strong order unit. In fact, A has no strong order units. As an element of B, b is a strong (hence weak) order unit and a multipicative unit. In A, b is a weak order unit and a multiplicative unit. The following important concept is due to M. Henriksen et al.[20]. Definition 1.2.3 An fring A is said to have the bounded inversion property if every 1 < a E A is a multiplicative unit. Denote by Spec(A), Min(A), Max(A) the set of prime ideals, minimal prime ideals and maximal ideals of A respectively. These become topological spaces when endowed with their respective hullkernel topologies. In particular, the basic open sets of Spec(A) are of the form, for a E A, Ma = {P E Spec(A) : a < P}. A ring A is said to be semiprime if it has no nonzero nilpotent elements. This is equivalent to requiring the intersection of prime ideals to be zero. The following results are well known and will be used extensively. Lemma 1.2.1 (Theorem 9.3.1 [5]) Let A be a semiprime fring and let a, b A. Then ab = O if and only if a A b = 0. Lemma 1.2.2 (Theorem 9.3.1 [5]) Let A be a semiprime fring. Then P is a minimal prime ideal if and only if P is a minimal prime eideal. The following two results can be found in Henrikson et al.[20] where the notion of bounded inversion is introduced. Lemma 1.2.3 Let A be a semiprime fring. Then A has the bounded inversion prop erty if and only if every maximal ideal is an ideal. Lemma 1.2.4 Let A be a semiprime fring with identity and bounded inversion. Then Max(A) is a compact Hausdorff topological space. 1.3 Tychonoff Spaces and SemiPrime fRings We will now look at the ring of continuous real valued functions. Let X be a topological space. Denote by C(X) the set of all continuous real valued functions on X. For f,g E C(X), define 1. f + g by (f + g)(x) = f(x) + g(x) for all x E X. 2. f g by (f g)(x) = f(x)g(x) for all x E X. We will write fg for f g. These operations make (C(X), +, ) a commutative ring. If in addition we define scalar multiplication by, 3. For r E R, rf by (rf)(x) = r(f(x)) for all x E X. Then C(X) is a real vector space. We also define <, A and V by, 4. f < g if f(x) 5 g(x) for all x E X. 5. f A g by (f A g)(x) = min{f(x),g(x)}, for each x E X. 6. f V g by (f V g)(x) = max{f(x),g(x)}, for each x E X. With these, C(X) becomes a lattice ordered ring and a vector lattice. In addition, C(X) is an fring and C(X) has the bounded inversion property. To distinguish topological spaces by the algebraic properties of C(X), the next theorem will allow us to restrict our attention to the class of Tychonoff spaces. We first need the following definitions. Definition 1.3.1 Two subsets A and B of X are said to be completely separated if there is an f E C(X) such that f(A) = 0 and f(B) = 1. Definition 1.3.2 A Hausdorff space X is said to be a Tychonoff space if for any closed set K and any x K, {x} and K are completely separated. Theorem 1.3.1 (Theorem 3.9 [15]) For every topological space X, there exists a Ty chonoff space Y and a continuous mapping r from X onto Y such that the mapping g + g o r is an isomorphism from C(Y) onto C(X) This induced mapping is both a ring and a lattice isomorphism. For the remainder of this paper, unless explicitly stated otherwise, "space" will mean "Tychonoff space". For Y C X we will denote the closure of Y in X by clx(Y) or, when no ambiguity will result, by cl(Y). Similarly we will denote the interior of Y in X by intx(Y) or int(Y). Definition 1.3.3 For f E C(X) let Z(f) = x E X : f(x) = 0}. This is called the zero set of f. Let coz(f) = {x E X : f(x) $ 0}. This is called the cozero set of f. Let Z(X) = {Z(f) : f E C(X)}. Tychonoff spaces have the following useful characterization. Theorem 1.3.2 (Theorem 3.2 [15]) A space X is a Tychonoff space if and only if Z(X) is a base for the closed sets of X. It will occasionally be helpful to use this result as: X is Tychonoff if and only if the cozero sets are a base for the open sets of X. What follows is a rudimentary description of the construction and properties of the StoneCech compactification of a Tychonoff space. A more detailed account can be found in Chapter 6 [15], or Chapter 1 [33]. Let X be a Tychonoff space and let Z(X) denote the zerosets of X. F C Z(X) is called a zfilter if, 1. 0 F. 2. IfZ Z, Z2 E F then Z1 n Z2 E F. 3. If Z F and Z C K C X then K E F. A maximal zfilter is called a zultrafilter. Let fX denote the set of all zultrafilters of X. For Z E Z(X) let Z = {a E fX : Z E a}. We topologize fX by taking {Z: z E Z(X)} as a base for the closed sets. For a E fX, if na # 0 then there is an x E X such that noa = {x}. The map x : {a E fX :x E a} is a dense embedding of X into 3X. It is a theorem due to Gelfand and Kolmogoroff [13] that the map p '* {f E C(X) : p E clpxZ(f)} is a one to one correspondence between the points of pX and the maximal ideals of C(X). In fact, this map is a homeomorphism from fX onto Max(C(X)). We will use this homeomorphism extensively. Denote by C*(X) the set of all bounded continuous real valued functions on X. Then C*(X) is an fsubring of C(X). We need to make the following definition. Definition 1.3.4 Let X C Y be topological spaces. X is said to be C*embedded in Y if every f E C*(X) can be extended to a function in C(Y). The following is most of Theorem 1.46 [33] and is a compilation of results char acterizing #X. Theorem 1.3.3 Every Tychonoff space X has a unique compactification O3X which has the following equivalent properties: 1. X is C*embedded in f3X. 2. Every continuous mapping of X into a compact space Y extends uniquely to a continuous map from fX into Y. 3. If Za and Z2 are zero sets in X, then clpxZ1 l clxZ2 = clx(Z1 n Z2). 4. Disjoint zero sets in X have disjoint closures in 3X. 5. Completely separated sets in X have disjoint closures in fiX. 6. 3X is the maximal compactification of X in the sense that if Y is any com pactification of X, then there is a continuous map from 3X onto Y that is the identity on X. Moreover, fX is unique in the sense that if T is a compactification of X .stisfying any of the above conditions, then there is a homeomorphism of 3X onto T that is the identity on X. For the purposes of this dissertation, the major significance of Theorem 1.3.3 is that in many cases, we can without loss of generality assume the space we are dealing with is compact. 1.4 Boolean Duality A lattice is a partially ordered set (B, <) such that for every pair {a, b} C B, the least upper bound of a and b exists in B as does the greatest lower bound of a and b. We will denote the least upper bound of a and b by a V b and call this the join of a and b, and we will denote the greatest lower bound of a and b by a A b and call this the meet of a and b. A lattice B is said to be bounded if it has a least and largest element. In this section we will denote these, when they exist, by 0 and 1 respectively. Definition 1.4.1 A bounded lattice B is said to be complemented if for every a E B there is a b E B such that a A b = 0 and a V b = 1. Such a b is called a complement of a. Definition 1.4.2 A lattice B is said to be distributive if for every a, b, c E B, a A (bV c) = (aAb) V (a A c). Definition 1.4.3 A bounded distributive complemented lattice B is said to be a boolean algebra. In a boolean algebra, the complement of any element is unique and for a E B we will denote the complement of a by a'. For a, b E B, we will denote the difference by a \ b = a A b'. Boolean algebras have the following properties. A more complete list can be found in Chapter 1 of "Boolean Algebras" [28] or Chapter 3 in [27]. Let B be a boolean algebra and let a, b, c E B. Then, 1. (a V b)' = a' A b' and dually. 2. 0' = 1, 1' = 0, (a')' = a. 3. a < b if and only if b' < a'. 4. a A b = 0 if and only if a b'; a V b if and only if a' < b. 5. For any finite {a : 1 < i < k} C B, a \ (V ai) = A(a \ a). Definition 1.4.4 Let A, B be boolean algebras. A map 0 : A  B is called a boolean homomorphism if 0 is a lattice homomorphism and for every a E A, O(a') = (0(a))'. We are now going to look at the construction of the Stone space of a boolean algebra. This construction was originally carried out in M. Stone's 1936 paper, "Ap plication of the theory of boolean rings to general topology" [29]. Let B be a boolean algebra and let 0 # F C B. F is called a filter on B or a Bfilter if the following hold; 1. 0 c F. 2. If a, b E then a A b E . 3. IfaE Fanda A Bfilter F is called a Bultrafilter if it is a maximal Bfilter; that is, if g is a Bfilter and F C G then F = G. Bultrafilters have the following characterization. See Section 2.3 [27]. Theorem 1.4.1 Let F be a Bfilter. The following are equivalent. 1. 7 is a Bultrafilter. 2. For every a E B, if a A b # 0 for every b F then a E F. 3. For every a E B, either a E T or a' E F. For a boolean algebra B, let St(B) denote the set of Bultrafilters. For b E B let 14 = {F E St(B) : b F}}. We topologize St(B) by taking {Ub : b E B} as a base for the open sets. Since we will see this kind of construction again, we will look closely here at the relationship between intersection, union and complementation of basic open sets in St(B), and meet, join and complementation of elements of B. Let a, b E B. The following can be easily verified. 1. UaUb = baVb. 2. U U Ub = UaAb. 3. St(B) \ U, = U,. 4. U1 = 0 and Uo = St(B). If F E St(B), since F is a Bultrafilter, for 0 = b E B either b E F or b' E F. Therefore either F Ub, or F E Ub. If F E Ua n b then a V b F so that F E Uavb* By (1) above, {Ub : b E B} is a base for the open sets of a topology on St(B). Let X be a Hausdorff space. A subset K of X is called clopen if K is both closed and open. Let B(X) denote the set of clopen sets of X. With the operations of set theoretic union (U), intersection (n) and complementation ('), (B(X), U, n,'), is a boolean algebra. Definition 1.4.5 A Hausdorff space X is said to be zerodimensional ifB(X) is a base for the open sets of X. Definition 1.4.6 A Hausdorff space X is said to be totally disconnected if for every x E X, the connected component of x is {x}. We will make use of the following result which is Theorem 16.17 in [15]. Theorem 1.4.2 Let X be a Tychonoff space. Each of the following conditions implies the next and if X is compact, all conditions are equivalent. 1. Disjoint zero sets are contained in disjoint clopen sets. 2. X is zerodimensional. 3. X is totally disconnected. We are now ready to state Stone's Representation Theorem. The following version is Theorem 3.2 (d) [27]. Theorem 1.4.3 Let B be a boolean algebra. Then, 1. St(B) is a compact zerodimensional Hausdorff space. 2. {Ub : b E B} = B(St(B)). 3. The map b F b Ub is a boolean isomorphism from B onto 3(St(B)). Categorically Let CZD denote the category having as objects compact zero dimensional Hausdorff spaces and continuous maps as morphisms. Let BA be the category of boolean algebras and boolean maps. Define F : CZD + BA by F(X) = B(X) for objects and if f : X + Y is a morphism, define F(f) : B(Y)  B(X) by: for K E B(Y), F(f)(K) = fl(K). It can then be shown that F is a contravariant functor from CZD to BA. Define G : BA + CZD by G(B) = St(B) for objects and if 0 : A  B is a morphism, define G(O) : St(B)  St(A) by: for F E St(B), G(O)(.) = 01'(). G is a contravariant functor from BA to CZD. Moreover, as (3) from Stone's Theorem may indicate, CZD and BA are dual categories. CHAPTER 2 LOCALGLOBAL fRINGS 2.1 Introduction Let A be a commutative semiprime fring with identity and bounded inversion. We first need to recall some ideas from commutative ring theory (see Chapters 2 and 4 in "Multiplicative Ideal Theory" [16]). A subset S of A is called a regular multiplicative closed subset if 0 0 S, S is closed under multiplication and S contains no divisors of zero. Denote by S1A the set of all fractions a/s where a E A and s E S. If we define equality, addition and multiplication as in the classical construction of the quotient field of an integral domain, then S1A is a commutative semiprime ring with identity and the map a F+ (as)/s is a monomorphism from A into SA. If 1 E S, we can take s = 1 and this map becomes a i a/I. S1A is called the ring of quotients of A with respect to S. If we take S to be the set of all regular elements of A, then S1A is called the classical ring of quotients of A and will be denoted by qA. We can define an ordering on qA by first observing that for a/s we can assume that s > 0, since a/s = as/s2. Then define (a/s) V 0 = (a V 0)/s. This extends the ordering on A in such a way that qA is an fring and the embedding of A into qA preserves order. If a/s > 1 in qA, then a > s and since s is regular so is a. Therefore, qA has bounded inversion. We now look at the localization of A at a prime ideal P. Let O(P) = {a E A : ab = 0 for some b 0 P}. Then it can be shown that O(P) is the intersection of the minimal prime Lideals contained in P and therefore that O(P) is an ideal and an ideal. We define Ap to be the ring of quotients of A/O(P) with respect 19 to S = {b + O(P) : b V P}. The point of factoring out O(P) is to insure that S contains no divisors of zero. In the case where M is a maximal ideal, since A is an fring with bounded inversion, A/O(M) is already a local ring having M/O(M) as its unique maximal ideal. Therefore the elements of S are already invertible and so AM A/O(M). In the context of commutative semiprime frings with bounded inversion, when we refer to the localization at a maximal ideal we will mean the quotient ring A/O(M). We need the following definitions. Definition 2.1.1 A polynomial f E A[zx, ..., xn] is said to represent a unit over A if there exists al,..., a, E A such that f(al,..., an) is a multiplicative unit in A. For f E A[x1, ..., Xn], let fM denote the polynomial in AM[Xl, ..., z,] whose coeffi cients are the respective images of the coefficients of f under the map a i a/1. Definition 2.1.2 A ring A is said to have the localglobal property if for each polyno mial f E A[xl,..., ,], whenever fM represents a unit over AM, for all M E Max(A), then f represents a unit over A. In their paper "Module Equivalences" [11], Estes and Guralnick look at many of the implications of the localglobal property. The authors first point out that historically, many of the results obtained for localglobal rings are extensions of results first obtained by R. S. Pierce for rings which are Von Neumann regular modulo their Jacobson radical [26]. In particular, for modules over localglobal rings, M1 N _ M2 E N implies M1 M2, and Mn N' implies M N. Recall that a polynomial f(x) E A[x] having coefficients ax,..., an is said to be primitive if there exists bl,..., bn E A such that albl + + anb = 1. Definition 2.1.3 A ring A is said to satisfy the primitive criterion if for any primitive polynomial f E A[x], f represents a unit over A. By a residue field we mean the field of quotients of the integral domain A/P for P a prime ideal. The following result, due to Estes and Guralnick [11], will allow us to determine when an fring is localglobal in terms of the primitive criterion. Lemma 2.1.1 A ring A satisfies the primitive criterion if and only if A is a localglobal ring with all residue fields infinite. Consequently, if A is a commutative fring then A is localglobal if and only if A satisfies the primitive criterion. We will use this characterization of localglobal frings to eventually characterize the localglobal property in terms of Max(A). Recall that for an fring A, B C A is said to be an fsubring if B is a subring and a sublattice of A. We have the following theorem relating an fsubring of an fring and their respective maximal spectra. Theorem 2.1.1 Let A be a commutative semiprime fring with identity and bounded inversion. If B is an fsubring of A with bounded inversion and 1 E B then the map 0: Max(A)  Max(B) defined by O(M) = the unique maximal ideal containing M n B, is a continuous sur jection. PROOF We first note that for M E Max(A), M n B is a prime ideal and a prime tideal of B. Therefore M n B C M' E Max(B) for some M' and as this is also a prime eideal and the prime ideals form a root system, M' is unique. Therefore, 0 is well defined. To show that 0 is onto, let N E Max(B). Since 1 N, there is a value V of 1 in B with N C V. Let W be the ideal generated by V in A; that is W = {a E A : lal < v for some v E V}. Then 1 W so that W C W', a value of 1 in A. Let P be a minimal prime ideal contained in W'. Then P is a minimal prime ideal, so that P C M for some M E Max(A). Since M is a prime fideal, P C M C W'. Now, W' B is a prime eideal of B containing V with 1 W' n B. As V is a value of 1, V = W' nB and V D M B. Since MnfB is a prime ideal and a prime #ideal of B it is contained in a unique maximal ideal O(M) of B with O(M) C V. If N 0 O(M), then B = N + )(M) C N V O(M) C V. This is a contradiction. Therefore, O(M) = N, so that 4 is onto. To show that ) is continuous, we will consider 4 as the composite po o p where L : Spec(B)  Max(B) is defined by p(P) = the unique maximal ideal containing P and p : Max(A) + Spec(B) is defined by p(M) = Mn B. To see that p is continuous, let x E B and let A/a = {P E Spec(B) : x P}. This is a basic open set for the hull kernel topology on Spec(B). Since x E B C A, M = {M E Max(A) : x A} is a basic open set for the hullkernel topology on Max(A). It is easy to see that M E Mx if and only if M n B E Af, so that p1(A',) = M, and p is continuous. To show that p is continuous, we will show continuity at an arbitrary point. This part of the proof is modeled on Lemma 10.2.3 [5]. Let P E Spec(B) and suppose that U is an open neighborhood of p(P) in Max(B). Without loss of generality, we may assume that U is a basic open set in Max(B). Say U = M,= {M E Max(B) : x M}. Then U = Max(B)nAN, where NA = {P E Spec(B) : x ( P} is a basic open set in Spec(B). Let Q E Max(B) \A' Then Q and p(P) are distinct maximal ideals. Since Q and j(P) are also tideals, we can find 0 < x E Q \ L(P) and 0 < y E p(P) \ Q with x A y = 0, so that Q and t(P) are contained in disjoint open sets in Spec(B). Suppose Q E UQ and p(P) E VQ, both open with Ug n VQ = 0. Consider {UQ : Q E Max(B) \ NAf}, where the UQ are as above. Since any neighborhood of a maximal ideal M in Spec(B) is a neighborhood of any prime P C M, {UQ : Q E Max(B) \ AJ'} is an open cover of Spec(B) \ Now, Spec(B) \ is a closed subspace of of Spec(B) and so is compact. Therefore there exists {Qi, 1 < i < n}, such that Spec(B) \ANx C U='i UQi. Let V = nf=1 VQ,, where the VQ, are as above. Then p(P) E V so that P E V and V C .A. Then p(V) C (.NA) = .MA and therefore, p is continuous. We have shown that = p o p and that this is a continuous surjection from Max(A) to Max(B). QED Of particular interest in what follows will be the occasions when the map 0 defined in the proceeding theorem is onetoone. To describe these occurences we make the following definition. Definition 2.1.4 Let A and B be as in the statement of Theorem 2.1.1, and let q be the continuous surjection guaranteed by the theorem. If 0 is a homeomorphism, we say that B separates the points of A. 2.2 The Specker Subring of a SemiPrime fRing For A an fring, let A(1) denote the set of bounded elements of A. That is A(1) = {a E A : Ja < n 1 for some n E N}. Then A(1) is an fsubring and an ideal of A. If A = A(1) we say that A is bounded. We have the following corollary to Theorem 2.1.1. Corollary 2.2.1 Let A be a commutative semiprime fring with identity and bounded inversion. Then Max(A) 2 Max(A(1)). PROOF We first note that if A has bounded inversion then so does A(1), so that both Max(A) and Max(A(1)) are compact Hausdorff. Since A(1) is an fsubring of A with 1 E A(1), by Theorem 2.1.1, it suffices to show that the map 0 : Max(A) + Max(A(1)) is onetoone. Recall that for M E Max(A), O(M) is the unique element of Max(A(1)) containing M n A(1). Suppose that N, M E Max(A) with N i M. If O(N) = O(M), since A(1) E C(A) and C(A) is distributive, we have that O(M) A()) (NA()) A()) = (NVM)nA(1) = (N+M)nA(1) = AnA(1) = A(1) This is a contradiction. Therefore, O(N) j O(M) so that is onetoone. QED For A a semiprime fring with bounded inversion, if n E N, then n 1 > 1 in A so that (n 1)1 E A. In particular, A is divisible, so that we may consider A as an algebra over Q. Let S(A) denote the Q subalgebra generated by the idempotents of A. We will call this the speaker subring of A. If A is a vector lattice, the specker subring of A is S3(A) the real subalgebra generated by the idempotents of A. Definition 2.2.1 An group G is called hyperarchimedean if every ihomomorphic image is archimedean. We will call an fring A hyperarchimedean if its group structure is hyper archimedean. Most of following theorem was obtained originally for vector lattices, and is due in this more general form to P. Conrad [8]. Theorem 2.2.1 Let G be an (group. The following are equivalent. 1. G is hyperarchimedean. 2. Each proper prime iideal is maximal and hence minimal. 3. G = G(g) E g9 for each g E G. 4. G is fisomorphic to an subgroup G* of nJ Ri and for each 0 < x, y E G* there exists an n > 0 such that nxi > yi for all xi = 0. 5. IfO < a, bEG then aA nb = aA (n + 1)b for some n > 0. We also have the following characterization which will be useful in the present context. Theorem 2.2.2 (Theorem 14.1.7 [5]) An group G is hyperarchimedean if and only if G is isomorphic to an subgroup of C(X) such that G separates the points of X and the support of each g E G is compact and open. Since S(A) is archimedean and has a strong order unit, by Yosida embedding, S(A) embeds as an subgroup of C(Yos(S(A), 1)) and S(A) separates the points of Yos(S(A)). Since each a E S(A) is a finite sum of idempotents and idempotents must map to the characteristic functions of clopen sets, the support of a is clopen hence compact and open. We then have most of the following lemma. Lemma 2.2.1 For A a commutative semiprime fring with identity and bounded in version, S(A) is hyperarchimedean. If A is also a vector lattice then SR(A) is hyper archimedean. Theorem 2.2.3 Let A be a commutative semiprime fring with identity and bounded inversion. If S(A) is convex, then A is hyperarchimedean and A = S(A). PROOF Suppose that S(A) is convex in A. By Theorem 1.1.6, the map P * P f S(A) is a one to one order preserving correspondence between the prime ideals of A not containing S(A) and the proper prime ideals of S(A). Since S(A) is hyper archimedean, the prime ideals of S(A) are trivially ordered and consequently, so are the prime eideals of A which do not contain S(A). We will show that no prime ideal of A contains S(A). Let P be a prime eideal of A. Then P C V a value of 1. V D Q a minimal prime fideal and since A is semiprime, Q is also a prime ideal and so is contained in a maximal ideal M. Since A has bounded inversion, M is an prime tideal and so, Q c M C V and Q c P C V. We will show that P C M. Now consider the quotient ring A/M. Since M is both a maximal ideal a prime eideal, A/M is an ordered field. Since S(A) and M are convex, (S(A)+ M)/M is a convex fsubring of A/M. Also, as 1 ~ M, M h S(A) so that M n S(A) is a proper prime fideal of S(A). Since S(A) is hyperarchimedean, S(A)/(S(A)nM) is archimedean. By Theorem 2.3.9 [5], we have that S(A)/(S(A) n M) (S(A) + M)/M as igroups and hence as frings. Suppose now that x > 1 in A/M. Then x1 < 1, but 1 c (S(A)+M)/M and this is convex and archimedean; a contradiction. Therefore, A/M is a real field, in particular VIM = 0 so that V = M. Then P C M and since M 6 S(A), we have that P ; S(A). We have shown that any prime iideal does not contain S(A) and therefore the prime t ideals of A are trivially ordered. By Theorem 2.2.1, A is hyperarchimedean. Clearly, S(A) C A(1). Since 1 E S(A) and S(A) is convex, A(1) C S(A). Since A is hyper archimedean with identity, by Lemma A [8], A = A(1). Therefore A = S(A). QED We should note that if A is, in addition, a vector lattice, by the above and Proposition 1.2 [8], the following are equivalent. 1. A is hyperarchimedean. 2. A = S(A). 3. Sa(A) is convex in A. We will also need the following lemma. Lemma 2.2.2 If A is a commutative semiprime fring with identity and bounded inversion, then so is S(A). If A is a vector lattice this holds for S,(A). PROOF That S(A) is a commutative semiprime fring with identity follows from S(A) being an fsubring of A and 12 = 1. We need only show that S(A) has bounded inversion. Let a E S(A). Then a = E?=1 qjei where for 1 < i < n, qi E Q and e, E A is idempotent. We first note that a can be written as a = 71i rifi where the ri E Q are distinct and more importantly, the fi E A are idempotents such that fifj = 0 for i 0 j. Suppose now that 1 < a E S(A). Then there is a b E A such that ba = 1. That is b(ET=I rifi) = ETi brifi = 1. For each 1 < i < m we get that br;f, = fi, so that Ei1 fi = 1. Let c = 1(())f/. Then c E S(A) and ac = E 1 fi = 1. Therefore, S(A) has bounded inversion. The proof is similar for Si(A), taking real coefficients instead of rational. QED 2.3 A Characterization of LocalGlobal fRings The main result of this section is the following theorem. Theorem 2.3.1 Let A be a commutative semiprime fring with identity and bounded inversion. Each of the following conditions implies the next one. 1. A is localglobal. 2. Ifa bt2 E A[t] is primitive with positive coefficients, then a bt2 represents a multiplicative unit in A. 3. Max(A) is zerodimensional. 4. Max(A) ^ Max(S(A)). (If A is a vector lattice, Max(A) Max(Sa(A)).) Furthermore, (2), (3) and (4) are always equivalent and if A is bounded, then all four conditions are equivalent. We proceed with a series of lemmas which will be used in the proof of this theorem. Lemma 2.3.1 (Proposition 2.1 [9]) Min(A) is a zero dimensional Hausdorff space. For a ring A, let J(A) denote the Jacobson radical of A. Then J(A) = ({M: M E Max(A)}. We have the following lemma. Lemma 2.3.2 Let A be a commutative semiprime fring with identity and bounded inversion. Then Max(A) Max(A/J(A)). PROOF We first observe that since Max(A) C C(A), J(A) is an Gideal of A so that the canonical map A + A/J(A) is an homomorphism as well as a ring homo morphism. Consequently, A/J(A) is a commutative semiprime fring with iden tity and bounded inversion. The canonical map A i+ A/J(A) induces a oneto one correspondence between the maximal ideals of A and the maximal ideals of A/J(A). Let 0 : Max(A) Max(A/J(A)) be this induced map. Since Maz(A) and Max(A/J(A)) are both compact Hausdorff spaces, we need only show that 0 is continuous. Let M = {M E Max(A) : x M} be a basic open set of Max(A). Since M D J(A) for every M E Max(A), we have that M E M, x M = 7 9 0(M) O(M) E M, where x E A and 7 E Max(A/J) are such that x i 7 under the canonical map, and My = {M E Max(A/J) : T M M}. Thus 01(MA) = M.. Therefore, 0 is continuous. QED We have seen that the maximal spectra of A and A/J(A) are homeomorphic. We now consider some algebraic properties of A and A/J(A). Lemma 2.3.3 Let A be a commutative semiprime fring with identity and bounded inversion. Then A is localglobal if and only if A/J(A) is localglobal. PROOF Let p(t) = ao + alt + + ant" E A[t] and let p(t) = 0 + at + ** + Ent" E (A/J(A))[t] where ai and ai are such that ai H+ di under the canonical map A  A/J. By Lemma 2.1.1, we need only show that p(t) is primitive if and only if p(t) is. It is clear that if p(t) is primitive, then so is p(t). Suppose now that p(t) is primitive. Then there exists bo,1, , b A/J(A) such that aobo + albl + + ab = 1 + j for some j E J(A). If aobo + alb + + a,b, = 1 + j E M for some M E Max(A), then 1 = aobo + albl + + a,b, j E M as j E J(A) C M. This is a contradiction. Therefore, aoboa + + +a,bn is a unit, so that p(t) is primitive and the result follows. QED Lemma 2.3. Let A be a commutative semiprime fring with identity and bounded inversion. Then Yos(A, 1) 2 Max(A). If in addition A is bounded then Yos(A, 1) = Max(A). PROOF For each M E Max(A), M is a prime eideal with 1 ( M and so it is contained in a unique value V E Yos(A, 1). For V E Yos(A, 1), V D P a minimal prime fideal and a minimal prime ideal. Then there is an M E Max(A) with P C M C V. If Mi = M2 Max(A) with M, M2 C V then A = Mi + M2 C M1 V M2 C V, so that each value of 1 in A contains exactly one maximal ideal. As in the proof of Theorem 2.1.1, this correspondence induces the required homeomorphism. Suppose that A is bounded. Since the topologies on both spaces are their respec tive hullkernel topologies, it suffices to show that Yos(A, 1) = Max(A) as sets. Let Y e Yos(A, 1). Y D P a minimal prime ideal, hence a minimal prime ideal. Then P C M E Max(A) with P C M C Y. Now consider the quotient ring AIM. Since M is a maximal ideal, AIM is a field. Since M is also a prime ideal, AIM is an or dered field. Since A is bounded, AIM is as well. Therefore, AIM is orderisomorphic to a subfield of R. Now, Y/M is an eideal of A/M and so by Holder's Theorem, Y/M = 0. Therefore, Y = M, so that Y is a maximal ideal of A. We then have that Yos(A, 1) C Max(A). By the above homeomorphism, Yos(A, 1) = Max(A). QED For the next result we will need the following definition. Definition 2.3.1 Let G be an group. For 0 < a, b E G we say that a is infinitesimal to b, denoted a < b, if na < b for all n E N. Lemma 2.3.5 Let A be a bounded commutative semiprime fring with identity and bounded inversion. Then J(A) = {x E A : lz < 1} and A is archimedean if and only if J(A) = 0. PROOF Let x E J(A). Since J(A) is convex, we may assume that x > 0. Suppose now that for some n E N, nx 1. Then (nx 1)+ > 0, so there is a minimal prime ideal P with (nx 1)+ > 0 mod P. Since P is prime, (nx 1) = 0 mod P, so that (nx 1) > 0 mod P and nx > 1 mod P. Now, P is a minimal prime ideal so there is an M E Max(A) with P C M. The canonical map a + P a + M is a lattice homomorphism from A/P to A/M so that nx > 1 mod M. But, x E J(A) C M and M is convex so that 1 E M; a contradiction. Therefore, x < 1. Suppose now that lzx < 1. Since each M E Max(A) is a prime iideal, nixI < 1 mod M for all M E Max(A), n E N. By Lemma 2.3.4, each M E Max(A) is a value of 1 in A. Since A is bounded, A/M = M*/M is archimedean so that lx = 0 mod M, whence x E M. Since this holds for each M E Max(A), we have that x E J(A). Clearly, if A is archimedean, then J(A) = 0. Suppose now that J(A) = 0. Let 0 < a, b E A and suppose that na < b for all n E N. Then na < b < bV1 = c for all n E N. Since c > 1 and A has bounded inversion, c1 E A. Then (na)c1 = n(ac') < 1 for all n E N. Therefore, ac1 E J(A) = 0, so that a = 0, and A is archimedean. QED It should be pointed out for later use that the containment J(A) C {x E A : lxi << 1} obtains without the assumption that A is bounded. We will need the following lemma which characterizes the clopen subsets of Max(A). Lemma 2.3.6 Let A be a commutative semiprime fring with identity and bounded inversion. Then KC C Max(A) is clopen if and only if KC = MA where x E A is idempotent. PROOF Suppose first that KS = M. where x E A is idempotent. Then Mx1 is a basic open set disjoint to M. with Mx1 U M. = Max(A). Therefore Mx is clopen. Suppose now that KS C Max(A) is clopen. Since KS is open, K U = UJxB M, for some B C A. Since K) is a closed subset of Max(A), it is compact and therefore there exists {x( : 1 < i < n} such that K = UF= M,,. Since each M E Max(A) is convex, we may assume that each xi > 0. Then KS = U=l M,= = M, where x = V=i xi. By similar argument, since KS is clopen, Max(A) \ K = My for some 0 < y E A. Now M, n My = MA = 0 so that xA y E M for every M E Max(A). Let ' = x (xA y) and y'= y (xA y). Then M = MT,, My = My and A y'= 0. Since Mx, U My, = Mxvy, = Max(A), x' V y' is a multiplicative unit. Now let e = x'(x' V y')' and f = y'(x' V y')1. Then e V f = 1, e A f = 0 and Me = Mx. Since eAf = 0, ef = 0 so that e2 = e2V = e2Vef = e(e V f) = e() = e. Therefore, K = Me where e is idempotent. QED We now proceed to the proof of the main theorem. PROOF The proof will consist of two parts. We first will show that (1) implies (2); that (2) implies (3); that (3) is equivalent to (4), and that (3) implies (2). We will then show that if A is bounded that (3) implies (1). Clearly, (1) implies (2). Suppose now that (2) holds. Since Max(A) is compact, by Theorem 1.4.2, it suffices to show that Max(A) is totally disconnected. Let N, M E Max(A) with N 5 M. Since A has bounded inversion, both M and N are ideals. Pick 0 < x E M \ N. By the maximality of N, the ideal generated by x and N, (N, x) = A. In particular, there exists 0 5 y E N, a E A such that y + ax = 1. Then y M, and (y V 0) + ax = (y + ax) V (0 + ax) 2 y + ax = 1 so that by bounded inversion, (y V 0) + ax = u is a multiplicative unit in A. By the convexity of N and M, y V 0 E N \ M. By a suitable relabelling, we have produced 0 < x E M \ N, 0 < y E N \ M with (x,y) = 1. Now consider x yt2 E A[t]. This satisfies the hypothesis of (2) and so there is an a E A so that xya2 = u is a multiplicative unit in A. Let u+ = uV0 and u = (u)V0. Then u = u+ u and u+ Au = 0. Recall that the basic open sets of Max(A) are of the form M.x = {M E Max(A) : x M} for x E A. We now note that since the maximal ideals are tideals, if 0 < y E A, then Mx n My = MaAy and M, UMy = MV2y. Now, u+ Vu > u and as A has bounded inversion u+ V u is a multiplicative unit. Then .M+ U Mu = Mu+v = Max(A) and M/+ n M. = M,+Au = 0. Therefore {M,+,.Mu} is a clopen partition of Max(A). Now, ya2 < x ya2 < x so that 0 < u+ < x and 0 < u < ya2. Then as x E M and M is convex, u+ E M so that u 4 M and therefore M E MA. Similarly, since y E N, it follows that N E M,+. Thus, for any two distinct points in Max(A) there is a partition of Max(A) into two clopen subsets each containing exactly one of the points. Therefore, Max(A) is totally disconnected. Suppose now that (3) holds. That is, Max(A) is zerodimensional. Let : Max(A)  Max(S(A)) be the map defined in Theorem 2.1.1. Then is continuous and onto, and since S(A) is hyperarchimedean, O(M) = M M S(A) for all M E Max(A). Since both of Max(A) and Max(S(A)) are compact Hausdorff, it suffices to show that 4 is onetoone. Let M $ N E Max(A). Since Max(A) is zero dimensional, hence totally disconnected, there is a clopen set /C C Max(A) with M E KC, N )C. Since CK is clopen, by Lemma 2.3.6, there is an e E A, idempotent such that CK = Me. Since e is idempotent, e E S(A) so that e n M n S(A) and e E N n S(A). Therefore O(M) 7 O(N), and therefore is onetoone. If Max(A) Max(S(A)) then, since S(A) is hyperarchimedean, Max(S(A)) = Min(S(A)), and Min(S(A)) is zerodimensional, so that Max(A) is as well. That (3) implies (2) is a special case of the second part of the proof. Suppose it has already been shown that if A = A(1) then all four conditions are equivalent. Suppose now that (3) holds, that is Max(A) is zerodimensional and A is not necessarily bounded. Let a bt2 E A[t] be primitive with positive coefficients. Then there are c, d E A such that ac + bd = 1. Then I1 = Jac + bdl < ac + IbdI = aic + bldl < (a+b)lcl+(a+b)ldl = (a+b)(lcl +ld). Since A has bounded inversion, (a+b)(Icl +dl) is a unit, so that (a + b) is as well. Let a' = a(a + b)1 and b' = b(a + b)1. Then a' b't2 e A(1)[t] and a'+ b' = 1 so that a'b't2 is primitive with positive coefficients in A(1). If Max(A) is zerodimensional, by Corollary 2.2.1, Max(A(1)) is also. Therefore there is an x e A(1) such that a' b'x = u is a multiplicative unit in A(1). Since A(1) C A, x E A and u is a multiplicative unit in A. Therefore abx2 = u(a+b) is a multiplicative unit in A. Now let us prove that (4) implies (1) for bounded rings. By applications of Lemmas 2.3.1, 2.3.2, 2.3.4, we may without loss of generality assume that A is archimedean. By Lemma 2.3.4, Yos(A, 1) = Max(A) and by the preliminaries on the Yosida embedding of an archimedean fring, A embedds as an fsubring A C D(Max(A)), where A is an fring such that 1 I 1. Since A is bounded, the Yosida embedding is actually into C(Max(A)). Let Max(A) = X. Identifying A with its image in C(X), we have that A is an fsubring of C(X) with 1 E A. Let ao + alt + + at" E A[t] be primitive. If we consider the ai as elements of C(X), we have that nfl=L Z(ai) = 0. Then {coz(ai) : 1 < i < n} is a basic cover of X (in the sense of Chapter 16 [15]). Since X is compact and zerodimensional, there exists a refinement of this basic cover by a partition of X. That is, there ex ists {KIi : 1 < i < n}, a pairwise disjoint collection of clopen subsets of X, with Ki C coz(ai) and such that U!I, K; = X. Define f : Ki x R + R by f(x, r) = ao(x) + ai(x)r + + an(x)r'. Fix xo E KIi. Then as Ki C coz(ai), ai(x) 0, so that x E ai1(R \ {0}). Let fro : R + R be defined by fxo(r) = f(xo,r). Then fo is continuous, so that f,0j(R \ {0}) is a nonempty open subset of R. Therefore there is an 0 / ro E f, 1(R \ {0}) n Q. Now define fo : KCi R by fro() = f(x, ro). Then fro is continuous, and by construction, fro(xo) 0 O. Therefore, coz(fr) # 0. Recall that ro is determined by xo, so for ease of notation, let coz(fro) = N~o. Then {N x : x E K:i} is an open cover of Ki. Since Ki is compact, there exists {N,. : 1 < j < m} such that Ki = UTj NZ,. Since KC; is a clopen subset of a compact zerodimensional space X, Ki is a compact zerodimensional space. Therefore there is a refinement of the basic cover {N., : 1 < j < m} by a partition of Ki. Let {j : 1 < j < m} be this partition. Recall that X = Max(A). Since each j is clopen in X by Lemma 2.3.6, there is in ej E A idempotent such that Ej = Me,. In the identification of A with A, ej = XMe Define si : X + R by si(x) = rlei(x) +. +rmem(x). Then si is continuous. Repeat this construction for each Ki, and put s = sl + + s,. Then s E C(X) and s is a linear combination of rational multiples of idempotents of A. Since A has bounded inversion, s E A. Now consider ao + als + + ans" as an element of C(X). By the construction of s, Z(ao + als + .* + ans") = 0. If as an element of A, ao + als + + ans" E M for any M E Max(A), then the Yosida embedding of A in C(X) would give us an x E X, namely M, with ao(x) + al(x)s(x) + ..* + an(x)s"(x) = 0. Therefore ao + als + + ans" M for every M E Max(A), so that ao + als + + a,sn is a unit in A. Therefore A is localglobal. In the case that A is a vector lattice, nothing is changed by taking Sa(A) in the place of S(A). QED We should note here that one interesting aspect of Theorem 2.3.1 is that condition (2) is a first order condition in the language of commutative semiprime rings with identity; that is, condition (2) can be written in terms of universal and existential quantifiers and a finite number of elements, operations and relations. We should also note that in the above proof for A archimedean, bounded, the substitution s E A such that ao + als + .* + ans" is a unit was an element of S(A). This still holds if we drop the assumption of archimedeaneity. Corollary 2.3.1 If A is a bounded commutative semiprime fring with identity and bounded inversion which is localglobal and ao + alt + + ant' E A[t] is primitive, then there is an s E S(A) such that ao + als + + ans" is a unit in A. PROOF It suffices to show that an idempotent in A/J(A) lifts to an idempotent in A. That is, if x2 + J(A) = x + J(A), then there is an e E A idempotent such that x + J(A) = e + J(A). So, suppose that x E A is such that x2 + J(A) = x + J(A). Then x2 x = x(x 1) E J(A), so that x(x 1) E M for every M E Max(A). Since M is a proper prime ideal this gives us that x E M or x 1 E M, but not both. Therefore, Mx and Mx1 partition Max(A), and M, is clopen. By Lemma 2.3.6, M. = Me where e E A is idempotent. Suppose that x + J(A) = e + J(A). Then there exists an M E A such that x e 0 M. Since M., = Me we have that x 0 M and e ( M. But x(x 1),e(e 1) E M, so x 1,e 1 E M. A contradiction as x e = (x 1) (e 1) M. Therefore x + J(A) = e + J(A). QED Corollary 2.3.2 Let A be a bounded commutative semiprime fring with identity and bounded inversion and let B be an fsubring of A with bounded inversion and 1 E B. If A is local global and S(A) C B, then B is localglobal. PROOF Since A is bounded and B is an fsubring of A, B is bounded. By Theorem 2.3.1, Max(A) Max(S(A)). Let 0 be this homeomorphism. By Theorem 2.1.1, and Lemma 2.2.1, the maps 0 : Max(A)  Max(B) and 0 : Max(B) + Max(S(A)) are continuous surjections with 0 o 0 = 0. Therefore i is onetoone and hence a homeomorphism. Then Max(B) is zerodimensional and so, by Theorem 2.3.1, B is localglobal. QED Corollary 2.3.3 If A is a commutative semiprime fring with identity and bounded in version which is in addition archimedean, bounded and localglobal, then C(Max(A)) is the largest bounded archimedean localglobal extension B of A such that A separates the points of B. PROOF We first note that for A as above, C(Max(A)) is a commutative semiprime fring with identity and bounded inversion which is in addition archimedean, bounded and localglobal. Since Max(A) is compact Hausdorff, Max(C(Max(A))) 2 /(Max(A))  Max(A), so that A separates the points of C(Max(A)). Suppose now that B is a bounded, archimedean, localglobal extension of A with Max(A) Max(B). Then as in the proof of Theorem 2.3.1, B is an fsubring of C(Max(B)). Since Max(A) 2 Max(B), the result follows. QED Now let A = C(X) be the ring of continuous real valued functions on X. We will denote S~(C(X)) by S(X). We make the following observations. 1. C(X) is a commutative semiprime fring with identity and bounded inversion. 2. If X is compact, then C(X) is bounded. 3. If X is compact Hausdorff, then X O 3X 4 Max(C(X)). These, together with Theorem 2.3.1, give us the following corollary. Corollary 2.3.4 If X is a compact Hausdorff space, the following are equivalent: 1. C(X) is localglobal. 2. If fgt2 E C(X)[t] is primitive with positive coefficients, then fgt2 represents a unit in C(X). 3. X is zerodimensional. 4. X Max(S(X)). One of the troublesome aspects of the above results is that we get equivalence of the stated conditions for bounded rings or in the case of C(X), for X compact. We now consider a class of rings for which the assumption of boundedness can be dropped. We need the following definitions. Definition 2.3.2 A ring A with identity is said to be a Bezout ring if every finitely generated ideal is principal. Definition 2.3.3 A Tychonoff space X is said to be an Fspace if every dense cozero is C* embedded. Bezout rings and Bezout domains are discussed extensively throughout R. Gilmer's "Multiplicative Ideal Theory" [16]. The following is due to Gillman and Henriksen [14] and in the present form is Theorem 14.25 [15]. Theorem 2.3.2 Let X be a Tychonoff space. Then the following are equivalent. 1. X is an Fspace. 2. pX is an Fspace. 3. C(X) is a Bezout ring. 4. 0 < a < b E C(X) implies that a = bf for some f E C(X). 5. Every ideal in C(X) is convex. 6. Every ideal in C(X) is an tideal. 7. For < a, b C(X), the ideal (a,b) = (a + b). 8. For f E C(X), there exists a k E C(X) such that f = kIf . 9. The localization of C(X) at any maximal ideal is a valuation ring. We recall here that an integral domain D is called a valuation ring if the ideals of D form a chain. Recall also that for A a commutative semiprime fring with identity and bounded inversion we have identified the localization of A at a maximal ideal M with the quotient ring A/O(M). It is in this sense that (9) above should be interpreted. The following is a partial generalization of the above to semiprime frings with bounded inversion [25]. Theorem 2.3.3 Let A be a semiprime fring with bounded inversion. Then the fol lowing are equivalent. 1. A is a Bezout ring. 2. 0 < a < b E A implies that a = bf for some f E A. 3. Every ideal of A is convex. 4. Every ideal of A is an eideal. 5. For 0 < a, b A, the ideal (a,b) = (a + b). 6. The localization of A at any maximal ideal is a valuation ring. We should note that if A is Bezout, then by (5) above, (f+, f) = (f+ + f) = (If ). So that in particular, f = kifI for some f E A. For Bezout rings, Theorem 2.3.1 improves to the following. Theorem 2.3.4 Let A be a commutative semiprime fring with identity and bounded inversion. If in addition, A is a Bezout ring then the following are equivalent. 1. A is a localglobal ring. 2. If a bt E A[t] is primitive, then a bt represents a unit. 3. Max(A) is zerodimensional. PROOF Clearly (1) implies (2). Suppose now that (2) holds. Since Max(A) is compact, it suffices to show that Max(A) is totally disconnected. Suppose that M f N E Max(A). As in the proof of Theorem 2.3.1, there exists 0 < a E M \ N and 0 < b N \ M with a A b = 0. Let f = a b. Then f+ = a and f = b. By Theorem 2.3.3, since A is Bezout, there is a k E A such that f = kifl. Then a b = k(a + b) = ka + kb, so that (1 k)a = (1 + k)b. Since a A b = 0, we have that (1 k)a = (1 + k)b = 0, and so a = ka and b = kb. Let g = 1 k2. Then ga = gb = 0, so that g E N and g E M. Also, g + k = 1, so by the hypothesis there is a v E A such that k gv = u is a multiplicative unit. Consider now the basic open sets Mu+ and Ms. Since u is a multiplicative unit, these sets partition Max(A). Now, u+a= (u V )a= ua V = (k gv)a VO= kaVO= aVO=a,so that u+ N lest a E N. Similarly, ub = b, so that u V M. Therefore N E M,+ and M E Ms, and as these are disjoint, Max(A) is totally disconnected. Suppose now that (3) holds. Let ao + a+t + + a4t" E A[t] be primitive. Then there exists bo,bl,...,bn E A such that aobo + albl + gg + a,b, = 1. Then 1 = laobo + alb + + anbl (laol + la + + lan)(Ibo + bi + + Ibnl). Since A has bounded inversion, aol + lal + + lan = a is a multiplicative unit. Then, aoa1 + ala't + + analt' E A(1)[t]. Since A is Bezout, for each ai, 0 < i < n, there is a ki such that ai = klai. But then kiai = lail and we may assume without loss of generality that Ikl < 1 (take k' = (kAl)V1). Then koaoal+klala+...* knaa1 = 1 and therefore aoa1 a+aat +.. + analt" E A(1)[t] is primitive. Since Max(A) is zerodimensional, by Corollary 2.2.1, Max(A(1)) is as well. Applying Theorem 2.3.1 to Max(A(1)) we have that A(1) is localglobal. Therefore, there exists a v E A(1) such that aoa1 + ala1v + + analv' = u is a multiplicative unit in A(1) C A. Therefore ao + alv + .. + a,v" = au is a multiplicative unit in A. Since v E A(1) C A, A is localglobal. QED For X a Tychonoff space we have the following corollary. Theorem 2.3.5 If X is an Fspace, then the following are equivalent 1. C(X) is localglobal. 2. If f gt E C(X)[t] is primitive, then it represents a unit. 3. X is strongly zerodimensional. PROOF By Theorem 2.3.2, X is an Fspace if and only if C(X) is a Bezout ring. That (1) implies (2) and that (2) implies (3) then follows directly from Theorem 2.3.4 and the observation that Max(C(X)) OX. To prove that (3) implies (1), suppose that X is strongly zerodimensional; that is OX is zerodimensional. Then C(PX) is local global. But, C(f3X) C*(X) the ring of bounded continuous real valued functions on X. The proof now follows as in the proof of Theorem 2.3.4 taking C(X) = A and C*(X) = A(1). QED Recall in the statement of Theorem 2.3.1, that conditions (2), (3) and (4) are always equivalent and that (1) implies (2). The assumption of boundedness was needed only in the proof that (3) implies (1). The problem that prevented this implication in the general case is one of "cutting down" a primitive polynomial in f(t) E A[t] to a polynomial f(t) E A(1)[t] and maintaining primitivity. The most general result in this context is the following theorem. Theorem 2.3.6 If A is a commutatuve semiprime fring with identity and bounded inversion, then the following are equivalent. 1. Every primitive polynomial f(t) E A[t] having coefficients that are comparable to zero represents a multiplicative unit. 2. Max(A) is zerodimensionl. PROOF It suffices to show that a primitive polynomial having coefficients that are compa rable to zero in A[t] differs from a primitive polynomial in A(1)[t] by a multiplicative unit of A. We can then apply Theorem 2.3.1. The proof now proceeds exactly as in the proof of Theorem 2.3.4 taking the ki to be 1 if a, > 0 or 1 if a, < 0. QED CHAPTER 3 QUASISPECKER fRINGS AND fRINGS OF SPECKER TYPE 3.1 Introduction In the previous chapter, we have seen that if S(A) (respectively Si(A)) is a convex fsubring of the commutative semiprime fring A, then (i) S(A) = A (respectively SS(A) = A), and (ii) A is localglobal. In this chapter we will explore two other distinct types of containments of S(A) in A and the effects of these containments on the structure of A. One of these containments is based on the ring structure and the other on the order structure of A. We need the following definitions. Definition 3.1.1 Let R be a commutative ring. A subring S of R is said to be large if for every nonzero Ssubmodule I of R, I n S 5 0. R is also called an Sessential extension of S. Definition 3.1.2 Let H be an tgroup, G an esubgroup of H. G is said to be large in H if for every nonzero convex tsubgroup C of H, C n G $ 0. H is also called an essential extension of G. Since the above definitions become ambiguous in the context of frings, we will adopt the convention that whenever using definitions dependent on the lattice struc ture we will precede them by an "o". Thus if A is an fring and B is an fsubring such that B is large in A as an tsubgroup, we will say that B is olarge in A, or that A is an oessential extension of B. We will make extensive use of the following well known characterization of essen tial extensions. Theorem 3.1.1 Let R be a commutative ring, S a subring of R. Then, S is large in R if and only if for every 0 $ a E R, there is a b E S such that 0 5 ab E S. PROOF Suppose that S is large in R. Let 0 # a E R and let < a >= {sa : s E S} be the Ssubmodule of R generated by a. Then < a > nS 0 0, so there is a b E S such that 0 : ab E S. Conversly, if 0 : I is an Ssubmodule of R, then there is an 0 $ a E I C R. By the hypothesis, there is a b E S such that 0 : ab E S. Since I is an Ssubmodule, ab E I. QED Recall now from Chapter One that for an igroup G, C(G) is the collection of iideals of G, and that C(G) is a complete, distributive, brouwerian sublattice of the lattice of all subgroups of G. The next characterization of the olarge fsubrings of an fring will depend on a certain meet subsemilattice of C(G). We first need to set up some machinery. Definition 3.1.3 Let G be an igroup. For any subset T C G, let T' = {a G : Ial A It = 0 Vt E T}. T' is called the polar of T in G. We will use the following notational conventions. For g E G, g' = {g}1 and for any subset T C G, T = {TI}. g" is called the principal polar of g. Definition 3.1.4 Let G be an igroup. P E C(G) is said to be a polar of G if P = T' for some subset T C G. If we consider I as a unary operation on C(G) then the following results are well known and can be found in particular in Chapter 1 of "LatticeOrdered Groups" [2]. 1. BC B . 2. If B cC, then C CB . 3. B' = BI. 4. B' n C = (B V C)' where V is the supremum in the lattice C(G). 5. P E C(G) is a polar if and only if P = P". 6. For any subset T C G, T' E C(G). Let P(G) = {P E C(G) : P" = P}. The following is Theorem 1.2.5 [2]. That P(G) is a complete boolean algebra is originally due to F. Sik [31] (1960). The mapping context is part of a more general lattice theoretic result due to V. Glivenko [18] (1929). Theorem 3.1.2 P(G) is a complete boolean algebra when equipped with the meet n, the join U defined by B U C = (B V C)" and complementation I. Furthermore the map B B B" is a lattice epimorphism from C(G) to P(G). With this, the characterization we need is the following generalization of Theorem 11.1.15 [5]. These results are originally due to P. Conrad [7]. Theorem 3.1.3 Let G be an esubgroup of the group H. Then each of the following implies the next. 1. G is olarge in H. 2. For every 0 < h H, there is a 0 < g G and an n N such that g < nh. 3. For each nonzero P E P(H), P n G 54 0. 4. For each 0 5 h E H there exists a 0 : g E h" f G. 5. The map P + P n G is a boolean isomorphism from P(H) to P(G). Furthermore, 1 and 2 are always equivalent; 3,4 and 5 are always equivalent, and if H is archimedean then 3 implies 1. Now that we have some useful characterizations of essential and oessential ex tensions we need to look at some particular extensions in both contexts. We will first deal with the oessential extensions. The following depends only on the egroup structure. For an group G we will be interested in several oextensions of G; the lat eral completion, the orthocompletion and, in the case where G is archimedean, the Dedekind completion and the oessential closure. We need the following definitions and notation. Definition 3.1.5 Let G be an esubgroup of the Igroup H. G is dense in H if for every 0 < h E H, there is a 0 Clearly G is dense in H implies that G is olarge in H. Definition 3.1.6 An tgroup is Dedekind complete if every collection of elements which is bounded above has a supremum. For G archimedean, an Igroup G^ is a Dedekind completion of G if G is an Isubgroup of G^, G^ is Dedekind complete and each element of G" is the supremum of elements of G. Definition 3.1.7 An Igroup is laterally complete if every collection of pairwise dis joint elements has a supremum. An Igroup GL is the lateral completion of G if G is a dense isubgroup of GL, GL is laterally complete and no proper isubgroup of GL contains G and is laterally complete. Definition 3.1.8 Let G be an igroup. H E C(G) is said to be a cardinal summand of G if there is a K E C(G) with H A K = 0 and H V K = G. We denote this by G=H K. Cardinal summands, in addition to being direct summands are unique. Also, if G = H 1 K then both H and K are ggroups, and if g = h + k with h E H, k E K, then 0 < g if and only if 0 < h and 0 < k. Definition 3.1.9 An lgroup G is called strongly projectable if for each P E P(G), G = P P'. G is called projectable if for each g E G, G = g' l g". Definition 3.1.10 An lgroup is orthocomplete if it is laterally complete and pro jectable. An lgroup G is an orthocompletion of G if G is dense in Go, Go is orthocomplete and no proper lsubgroup of Go contains G and is orthocomplete. Definition 3.1.11 An lgroup is oessentially closed if it admits no proper oessential extensions. An oessential closure of an lgroup is an oessentially closed oessential extension. It should be pointed out that unless we restrict our attention to archimedean e groups an lgroup always admits a proper oessential extension because any ggroup can be lexicographically extended and will be large in this extension. For this reason, the idea of an oessential closure only makes sense in the category of archimedean fgroups. The lateral completion of an lgroup is always unique, if in addition G is an archimedean lgroup then G^ and G are unique as well. Recall, that for an lgroup G, P(G) is a complete boolean algebra. Let X = St(P(G)) be the Stone dual of P(G). Since P(G) is complete, X is compact, Haus dorff and extremally disconnected. Recall that D(X) = {f : X  RU {oo} : f is continuous and f(IR) is dense open} For f,g E D(X), Y = f'(R) fng'(R) is a dense subset of X. Since X is extremally disconnected, Y is C* embedded in X. Therefore fY = fX = X. We have f + g : Y  IR U {oo} with R U {do} compact. Therefore there exists a unique extension of f +g to /Y = X. We then define the group element f +g to be this extension. We then have that, for X extremally disconnected, D(X) with the pointwise ordering is an igroup. We can also define pointwise multiplication as above making D(X) a ring. In fact, D(X) is a complete vector lattice. The following theorem is due to S. Bernau [4]. The earlier work of Pinsker and Vulich [32] obtains a similar result for complete vector lattices. Theorem 3.1.4 If G is an archimedean igroup then there is an tisomorphism 7r of G onto a large isubgroup of D(X) that preserves all joins and meets. If G is a vector lattice, 7r also preserves scalar multiplication. Furthermore, if {ex : E A} is a maximal disjoint subset of G, r can be chosen so that {7r(ex) : A A} is a set of characteristic functions of a family of pairwise disjoint clopen subsets of X whose union is dense. In particular, if e is a weak order unit, then 7r can be chosen so that 7r(e) is the characteristic function of X and if e is a strong order unit, then r(G) C C(X). Finally if rli and r2 are any two such embeddings, then there is a d E D(X) such that drl(g) = 72(g) for all g E G. The following is Theorem 3.4 [7] and it provides us with an "umbrella" extension. Theorem 3.1.5 Each archimedean igroup G admits a unique oessential closure G". G is the Dedekind complete, laterally complete, divisible igroup in which G is olarge and Ge is iisomorphic to D(X) where X is the Stone dual of the boolean algebra of polars of G. We now turn our attention to the ring theoretic aspects of essential extensions. Recall that the classical ring of quotients of A is qA = S1A, where S is the set of regular elements of A. We need the following concept which is due to Y. Utumi [30]. Definition 3.1.12 Suppose that A is a subring of the ring B. B is called a ring of quotients of A if for every bl, b2 E B, with b2 : 0, there is an a E A such that abl E A and ab2 5 0. With this definition, it is clear that qA is a ring of quotients of A, and that any ring of quotients of A is an essential extension of A. It is not necessarily the case that qA is a maximal essential extension of A; the maximal object is the socalled complete ring of quotients, Q(A). What follows are two quite different constructions of Q(A). The first is due to J. Lambek and can be found in Chapter 2 [22]. Definition 3.1.13 An ideal I in a commutative ring A is said to be dense if for every a E A, al = 0 implies a = 0. Let A be a commutative ring and let Hom(D, A) be the set of A homomorphisms of D into A. Let F(A) = {f E Hom(D,A) : D is a dense ideal of A}. We call f E F(A) a fraction. We can define addition and multiplication on F(A) as follows. For fi e Hom(Di, A), f2 e Hom(D2, A), 1. fi + f2 E Hom(D n D2, A) is given by (fi + f2)(d) = f,(d) + f2(d). 2. fif2 E Hom((f21(D1)) n D1, A) is given by f1f2(d) = fi(f2(d)). F(A) is closed under the above operations since for D1, D2 dense, both D1 n D2 and f21(Di) n D are dense. Now define an equivalence relation on F(A) by f, = f2 if fi = f2 on some dense ideal D. For f e F(A) let [f] denote the equivalence class of f. Let Q(A) = {[f] : f E F(A)}. For a E A, let fa be multiplication by a. If [fa] = 0 then aD = 0 for some dense ideal D, so that a = 0. Therefore a F+ fa is an monomorphism of A in Q(A). This mapping is called the canonical monomorphism. With the above operations suitably altered for equivalence classes we have the following theorem which is Proposition 1, 2.3 [22]. Theorem 3.1.6 If A is a commutative ring, then Q(A) is also a commutative ring. It extends A and will be called its complete ring of quotients. The following useful results can also be found in Lambek [22]. Theorem 3.1.7 [Prop.6 2.3[22]] Let A be a subring of the commutative ring B. Then the following are equivalent. 1. B is a ring of quotients of A (in the sense of Utumi). 2. For all 0 f b E B, blA is a dense ideal of A and b(b'A) 0. 3. There exists a monomorphism of B into Q(A) that extends the canonical monomor phism a F fa of A into Q(A). Recall that an Rmodule M is said to be injective if A and B are Rmodules and S: A 4 B is a monomorphism, then for any homomorphism 0 : A + M there is a t : B  M such that c o a = 4. R is called selfinjective if it is injective as an Rmodule. The following are all found in Chapter 4 [22]. Lemma 3.1.1 Q(Q(A)) = Q(A). If we restrict ourselves to the case where A is semiprime, we have the following theorems. Lemma 3.1.2 If A is semiprime, then Q(A) is selfinjective. Lemma 3.1.3 If A is semiprime, then Q(A) is the injective hull of A. Equivalently, Q(A) is the maximal essential extension of A. This gives us the following lemma. Lemma 3.1.4 If A is an essential extension of B, then Q(A) is an essential extension of Q(B). We now look at Banaschewski's construction of Q(A) [3]. This is a generalization of the construction carried out in "Rings of Quotients of Rings of Functions" [12] where it is shown that if X is a Hausdorff space, then Q(X)= Q(C(X))= lim{C(U) : U C X is dense, open} Let A be a commutative semiprime ring with identity. Let f C Spec(A) be such that n Q = 0. Call such a family of prime ideals a separating family. Topologize Q with the hullkernel topology. Let D = {U C Q : U is dense, open}. For U E D, let Au = {f E IpEu q(A/P) : for every Q E U there is a neighborhood V with Q E V C U and there are a,b E A with f(P') = (a + P')/(b+ P') for every P' E V}. For each U E Q, dense, open, Au is a subring of I1Pe q(A/P) containing A. If U C V are dense, open in Q, then the map rvu : Av  Au given by irv,u(f) = flu is a ring homomorphism. It can then be shown that the rings Au together with the maps 7v,u form a directed system indexed by I= {U E D} where V < U if U C V. Banaschewski's result is the following Theorem 3.1.8 Let A be a commutative semiprime ring, and Q a separating family of primes. Then, Q(A) = lim(Au, rv,u) In particular, since A is semiprime, if we take Q = Min(A), then the canonical map A F A/P + q(A/P) induces an imbedding A + A/P i q(A/P) PEMin(A) PEMin(A) Moreover, in the case that A is an fring, since each minimal prime ideal is an i ideal, each A/P is an ordered ring and hence each q(A/P) is an ordered field and the canonical maps are all fhomomorphisms. If we take the pointwise ordering on HPEMin(A) A/P and HPEMin(A) q(A/P), then these too are frings and the induced maps are fhomomorphisms. That a ring A has a unique maximal essential exten sion was first shown by Y. Utumi [30]. That there is a canonical ordering on Q(A), extending the ordering on A, for which Q(A) is an fring was shown by F. Anderson [1]. The following theorem bringing together these results with Banaschewski's con struction of Q(A) is due to J. Martinez [24]. Recall that a ring A is a Von Neumann regular ring if for every a E A there is an x E A such that axa = a. Theorem 3.1.9 For each semiprime fring A, the maximal ring of quotients Q(A) admits a latticeordering making it a semiprime fring and containing A as an f subring. f E Q(A) is positive if for each dense open subset W of Min(A), and each P E W there exists an open set U and positive elements a and b in A, so that P E U C W and for each Q E U, f(Q) = (a + Q)/(b + Q). This is the unique latticeordering on Q(A) making it an fring containing A as an fsubring. Finally, Q(A) is a Von Neumann regular ring. 3.2 QuasiSpecker fRings and fRings of SpeckerType Let A be a commutative semiprime fring with identy and bounded inversion. Recall that S(A) is the Q subalgebra of A generated by the idempotents of A, and that if A is an falgebra, Si(A) is the real subalgebra generated by the idempotents of A. Definition 3.2.1 We say that A is a speckertype ring if S(A) is a large subring of A. If A is an falgebra such that SR(A) is a large subring, then A is a speckertype algebra. We say that A is a quasispecker ring if S(A) is olarge in A. When A is an falgebra, the distinction between being a speckertype ring and a speckertype algebra is unimportant in terms of the method and content of the proofs that follow. The significant difference is that the objects which characterize the two specker properties are possibly different. We don't need to make this distinction for the definition of quasispecker, since in the case that A is an falgebra, S(A) is order large in Sn(A). Since many of the results are derived for the special case A = C(X) we will call a Tychonoff space X a specker space or a quasispecker space if C(X) is a speckertype algebra or a quasispecker ring respectively. We will need the following definition. Definition 3.2.2 A commutative semiprime ring A is locally inversion closed if for every a E A, and every P E Min(A) with a ( P, there is an open neighborhood U, with P E U C fa, and there exists b E A such that ab = 1 mod Q for every Q E U. The following characterization is due to J. Martinez [24]. Theorem 3.2.1 A semiprime fring A is locally inversion closed if and only if for each 0 < a E A, there exists {c : i E I} such that 0 < ci < a and axi = ci for some xi E Ac, and a1 = nci'. Theorem 3.2.2 Let A be a commutative semiprime fring with 1 and bounded inver sion. If A is localglobal and J(A) = 0 then A is locally inversion closed. PROOF Let 0 < a E A. Then Ma is a nonempty open subset of Max(A). Since A is localglobal, by Theorem 2.3.1, Max(A) is zerodimensional and so has a base of clopen sets. Let M E Ma. Then there is a clopen KM C Max(A) with M E KM C Ma. Since KM is clopen, by Lemma 2.3.6, there is an eM E A idempotent with KM = AeM. We claim that eM E a"1; otherwise, there is 0 < x E A with x A a = 0 and x A eM 7 0. Since J(A) = 0, there is an N E Max(A) with x A eM N. Since N is an 1ideal, we have that x 0 N and eM N. Then N E Mem C Ma, so that a 0 N. Since x A a = 0 and N is a prime 1ideal, x E N; a contradiction. Therefore eM E a". In particular, eM A a 7 0. Since eM is idempotent we can write A as A = AeM ED A(1 eM). Considering AeM we have that this is both an ideal and an eideal of A. Moreover, any ideal of AeM is an ideal of A. Now let P E Max(AeM). Since eM V P, the ideal generated in A by P and 1 eM is proper and so is contained in some N E Max(A) with eM V N. Then N C MeM C Ma so that a V N as well. Therefore aeM 0 N as N is a prime ideal, and P C N so that aeM V P. We have shown that aeM misses every maximal ideal in Max(AeM) so that aeM is a multiplicative unit in AeM. Therefore there is an XM E AeM such that aOfc IA = eM, the multiplicative identity in AeM. Now for each M E M,, let eM and XM be as above and let CM = aeM We have 1. 0 < CM < a. This holds since 0 < eM 1 and eM A a : 0. 2. a(aeMXM) = aeM = cM and aeMXM = CMXM E ACM. It remains to show that a1 = cM,. Clearly ac C fcM' as CM is a multiple of a. Suppose 0 < b a a'. Then b A a 0. Since J(A) = 0, there is a maximal ideal M with b A a 4 M. By convexity, b A M and a V M. Therefore M E Ma. As shown above, there is an idempotent eM with M E MeM C M,. Let cM = aeM. If bA cM = 0, since b M and M is prime, CM E M. Then CMeM = a(eM2) = aeM c M so that a M or eM e M. This is a contradiction since M E MeM C MA. Therefore bA CM A 0 and so b f cM QED Lemma 3.2.1 Let A be a commutative semiprime fring with 1 and bounded inver sion. Then S(A)L = Q(S(A)). If A is an falgebra, then S(A)L = Q(Sg(A)) PROOF Since S(A) is an archimedean fring with bounded inversion, by Theorem 1.8 [24], Q(S(A)) = S(A)o the orthocompletion of S(A). Since S(A) is archimedean, S(A)o = S(A)L (Theorem 8.2.5 [2]). The proof is identical if A is a vector lattice. QED It is in this context of lateral completions that the differences between S(A) and SR(A) first becomes apparent as the following example indicates. Example Let A be the ring of eventually constant real valued sequences. Then A is an falgebra. The idempotents of A consist of {0, 1} sequences that are eventually constant. Then S(A) consists of the eventually constant rational sequences and S(A)L = I Q, whereas, SR(A) consists of the eventually constant real sequences and Sa(A)L = FR. The first main result is the following theorem which characterizes speckertype frings in terms of the complete ring of quotients. An analogous result holds for falgebras. Theorem 3.2.3 Let A be a commutative semiprime fring with 1 and bounded inver sion. Then the following are equivalent. 1. A is an speckertype ring. 2. A is a subring of Q(S(A)). 3. S(A)L = Q(A). PROOF That (1) implies (2) follows directly from the definition since Q(S(A)) is the maximal S(A)essential extension of S(A). If S(A) C A C Q(S(A)) then A is an S(A)essential extension of S(A) so that (2) implies (1). We will now show that (2) implies (3). Suppose that A C Q(S(A)). For 0 a E Q(S(A)), by Theorem 3.1.1 there is a 0 6 b E S(A) such that 0 5 ab E S(A). Since S(A) C A, Q(S(A)) is an Aessential extension of A. By Lemma 3.1.1 and Theorem 3.1.8, Q(A) C Q(Q(S(A))) = Q(S(A)). Since A is a subring of Q(S(A)), A is an S(A)essential extension of S(A) so that Q(S(A)) C Q(A). Therefore Q(A) = Q(S(A)). By Lemma 3.2.1, S(A)L = Q(A). Since A is a subring of Q(A), that (3) implies (2) also follows directly from Lemma 3.2.1. QED We will now consider how speckertype rings, quasispecker rings and localglobal rings relate to one another. Recall that for any commutative semiprime fring with identity and bounded inversion, J(A) C {x E A : Ix << 1}, and therefore if A is archimedean then J(A) = 0. Corollary 3.2.1 Let A be a commutative semiprime fring with identity and bounded inversion. If A is an speckertype ring, then A is archimedean. The same conclusion obtains for falgebras. PROOF Since A is a speckertype ring, A is an fsubring of S(A)L. Since S(A) is archimedean, by Theorem 8.2.5 [2] its lateral completion S(A)L is as well. An fsubring of an archimedean fring is archimedean, so that A is archimedean. QED Theorem 3.2.4 Let A be a commutative semiprime fring with identity and bounded inversion. If A is a speckertype ring then A is a quasispecker ring. The same conclusion obtains for falgebras. PROOF Let P E 'P(A). Since A is an fring, P is a ring ideal and therefore an S(A) submodule of A. Since A is a speckertype ring, P n S(A) 4 0. Therefore, by Theorem 3.1.3, A is a quasispecker ring. QED Theorem 3.2.5 Let A be an archimedean commutative semiprime fring with identity and bounded inversion. If A is a localglobal ring, then A is a quasispecker ring. PROOF Let P e P(A) and let 0 7 a E P. As in the proof of Theorem 3.2.2, there is an e E A idempotent with e E a". Since a E P, a" C P, so that e E P n S(A) and A is a quasispecker ring. QED We will eventually see examples in the case A = C(X) to show that none of the two previous implications reverse. 3.3 QuasiSpecker Spaces and Specker Spaces In this section we will apply the preceding results for the case A = C(X). Follow ing the exposition in Rings of Continuous Functions by Gillman and Jerison [15], we consider for a space X, the ring of continuous real valued functions C(X). All spaces will be assumed to be Tychonoff. We recall the following notation. For f E C(X), coz(f) = {x E X : f(x) # 0} and Z(f) = { E X : f(x) = 0}. Some observations and recollections are in order. 1. For X Tychonoff, {coz(f) : f E C(X)} is a base for the open sets of X. 2. If we define a partial ordering on C(X) by f 2 0 if f(x) > 0 for all x E X, and operations pointwise, then C(X) is a commutative semiprime fring with bounded inversion and identity. We will use the following notational conventions. Q(X) = Q(C(X)), q(X) = q(C(X)), S(X) = SK(C(X)), C*(X) = {f E C(X) : f < n for some n E N}. Lemma 3.3.1 Let X be a Tychonoff space. Then C(6X) is isomorphic to C*(X) and S(#X) is fisomorphic to S(X). PROOF That C(3X) and C*(X) are isomorphic as rings is Theorem 4.6(i) in [27] and is given by f if fO from C*(X) to C(3X). That this map preserves order follows from the density of X in /3X. The second eisomorphism follows from the first and the observation that S(C*(X)) = S(X). QED In subsequent arguments involving C(PX) and S(P3X), we will use the above eisomorphisms extensively; that is, to prove a result for C(PX), we will prove the result for C*(X). The next theorem is an improvement of Theorem 3.2.3 for the case A = C(X) Theorem 3.3.1 Let X be a Tychonoff space. The following are equivalent. 1. X is a specker space. 2. For every 0 $ f E C(X) there is a K C X clopen and a 0 ~ c E R such that fIK=C. 3. 3X is a specker space. 4. S(X) = Q(X). PROOF Suppose that X is a specker space. Let 0 = f E C(X). Since X is a specker space, there is an s E S(X) such that 0 6 sf E S(X). We can write s = rIXK, + +rnXKn where we may assume that the ri are distinct, nonzero and the K, are disjoint, clopen. Since 0 f sf E S(X), we can write sf = q1XKI + .. + qmXK' where again, the qi are distinct, nonzero, and the Kf are disjoint, clopen. Then for any 1 < i < m, sf K, = qi 0. In particular, sIK, 0. Then K' C coz(s) so that K'I n Kj 5 0 for some 1 < j < n. Let T = K' n Kj. Then T is clopen and sf T = rjfIT = qi, so that fIT = qi/rj = 0. Suppose now that (2) holds. Let 0 7 f E C(X). Then there is a 0 = K C X clopen and a 0 $ c E R with fK = c. Let s = XK. Then s E S(X) and 0 0 sf E S(X). Therefore X is a specker space. We will next show that (2) and (3) are equivalent. Suppose that (2) holds. It suffices to show that C*(X) is a speckertype ring. Let 0 7 f E C*(X). Since C*(X) C C(X), there is a K C X clopen and a 0 7 c E R such that fIK = c. Let s = XK. Then s E S(C*(X)) = S(X) and 0 $ sf E S(X). Now suppose that #X is a specker space. Let 0 : f E C(X). We will first show that we may assume that 0 < f. If 0 4 f then either f < 0 or coz(f V 0) 5 0. If f < 0 then f > 0 and f IK = c / 0 implies fIK = c # 0. If coz(f V 0) # 0 then there is a 0 5 T C X with T C coz(f V 0). Then 0 < g = f IT C(X). If there is a 0 # K C X clopen and a c 5 0 with gIK = c, then glK = fITngK = c with 0 : T n K clopen. We will show that if C*(X) is a speckertype ring then (2) holds. If f < 1, then f E C*(X) and we are done. Assume then that f X 1. Then (f V 1) > 1 and (f V 1) $ 1. Since C(X) has bounded inversion, (f V 1)1 E C(X) and 0 < 1 (f V 1)1 < 1 with (f V 1)1 # 0,1. In particular, 1 (f V 1)1 E C*(X). Then there is a 0 K C C X clopen, and a c 0 0 with 1 (f V 1)1IK = c, where 0 < c < 1. Now, 0 < (f V 1)1K = 1 c < 1 so that (f V 1)IK = 1 > 1. Therefore, fl > 1IK so that fIK = (f V 1)K = C 0. The equivalence of (1) and (4) follows directly from Theorem 3.2.3, taking A = C(X). QED Since a subset of X is clopen if and only if its characteristic function is continuous, it is straightforward to show that if X is a (quasi)specker space and Y C X is clopen, then Y is a (quasi)specker space. As immediate corollaries of Theorem 3.3.1, we have the following results. Corollary 3.3.1 If X contains a dense set of isolated points then X is a specker space. PROOF If 0 $ f E C(X), then by density, there is an isolated point x with f(x) : 0. Take K = {x} and apply (2) in Theorem 3.3.1. QED Definition 3.3.1 A point p in X is called a ppoint if p E Z(f) implies that p E int(Z(f)). Corollary 3.3.2 If X is zerodimensional and contains a dense set of ppoints, then X is a speaker space. PROOF We will again appeal to (2) in Theorem 3.3.1. Suppose that 0 # f E C(X). By density, there is a ppoint p such that f(p) = c 7 0. Since p is a ppoint, there is an open set O with p E O such that flo = c. Since X is zerodimensional, there is a K clopen with p E K C O. Then f K = c and therefore X is a specker space. QED Definition 3.3.2 A collection B of open sets of X is called a rbase if for every non empty open set 0 there is a 0 # U E B with U C O. We have the following near analogue of Theorem 3.3.1 for quasispecker spaces. Theorem 3.3.2 Let X be a Tychonoff space. 1. X is a quasispecker space. 2. /3X is a quasispecker space. 3. X has a clopen ibase. Then (1) and (2) are equivalent; (1) implies (3), and if X is compact then (3) implies (1). PROOF Suppose that X is a quasispecker space. Then S(X) is order large in C(X). Since S(X) C C*(X) C C(X) as frings, S(X) is order large in C*(X). Suppose now that PX is a quasispecker space. Let 0 <_ f C(X). Then 0 < (f A 1) e C*(X), so there is an s E S(X) and an n E N such that 0 < s < n(fA 1). But n(fA1) = nf An < nf so that X is a quasispecker space. To show that (1) implies (3), let 0 $ O C X be open. Since X is Tychonoff, there is an f E C(X) with 0 : coz(f) C O. By the hypothesis, there is an 0 # s E S(X) and an n E N such that 0 < s < nf. Say s = r1XK1 + X+ rXK where as before the rid's are distinct, nonzero and the Ki's are nonempty disjoint clopen subsets of X. Then for any i, Ki C coz(nf) = coz(f) C O. Therefore X has a clopen 7rbase. Suppose now that X is compact with a clopen 7base. Let 0 < f E C(X) with 0 f f. Then there is a 0 5 K clopen with K C coz(f). Since X is compact and K is clopen, K is compact. Therefore there is an 0 < a E R such that a = min{f(x) : x E K}. Choose n E N so that n > 1. If we take s = XK E S(X), then nf > if > s and X is a quasispecker space. QED We have seen in the previous section that for frings (respectively falgebras) if A is a speckertype ring (algebra) then A is a quasispecker ring, and if A is a local global ring with zero Jacobson radical, then A is a speckertype ring (algebra). In the present context these results translate as follows. First, by taking A = C(X) in Theorem 3.2.4, we get the following corollary. Corollary 3.3.3 If X is a speaker space then X is a quasispecker space. Corollary 3.3.4 If X is strongly zerodimensional then X is a quasispecker space. PROOF If X is strongly zerodimensional then fX is compact and zerodimensional. C(fX) is then localglobal, and as C(Y) is archimedean for any space Y, #X is a quasispecker space, by Theorem 3.2.5. Therefore, by Theorem 3.3.2, X is a quasi specker space. QED The following examples show that the converse of the above corollaries do not hold and that being a specker space is independent of strong zerodimensionality. Example Of a space X which is a specker space and hence a quasispecker space, but which is not strongly zerodimensional. Let N be the natural numbers with the discrete topology and let aN denote the one point compactification of N where we denote by oo the point adjoined. Let X = [0, 1] x aN where the topology is a refinement of the product topology such that points of the form {(r,n)} for n E N are isolated. The idea is that the isolated points of aN remain isolated. Suppose now that 0 I f E C(X). Then f(r,n) = c 0 for some (r,n) E X \ {[0, 1] x {oo}}. Since (r,n) is isolated, s = X{(r,n)} E S(X) and 0 5 cx{(r,n)} = sf E S(X). Therefore X is a specker space. Since X contains [0, 1] as a connected subset, it is not (strongly) zerodimensional. Example Of a space X which is strongly zerodimensional and hence a quasi specker space, but is not a specker space. Let X = Q as a subspace of R with the topology generated by the open intervals. Then X is strongly zerodimensional; intervals of the form (r, s) with r, s E IR\ Q are a clopen base. Let f E C(X) be the identity map. Suppose there is an s E S(X) such that 0 7 sf E S(X). Then write sf = rlXKi +' + rnXKn where the rid's are distinct nonzero and the Ki's are nonempty disjoint clopen and s = qlXK' + + qmXK, similarly. Then there are i,j such that T = KinKK 5 0. T is clopen and f T = ri/qj. This is a contradiction since points are not clopen. These examples show that the following hold. 1. That "X is a specker space" is independent of whether or not X is strongly zerodimensional. 2. That X is a quasispecker space does not imply that X is a specker space. 3. That X is a quasispecker space does not imply that X is strongly zero dimensional. 3.4 The Absolute of a Hausdorff Space In what follows we will consider one construction of the absolute of a Hausdorff space and some of its properties. The following is the construction presented in "Extensions and Absolutes of Hausdorff Spaces" [27], Chapter 6.6 of the Gleason Absolute of a compact Hausdorff space first done by A.M. Gleason [17]. We will need the following definitions. Definition 3.4.1 Let X and Y be topological spaces and let f : X Y be a function. Then f is said to be perfect if f is closed and for each y E Y, fl(y) is a compact subset of X. In the definition of perfect maps there is no assumption about continuity, but in the case that f is continuous and X is compact, f is perfect. Definition 3.4.2 Let X and Y be topological spaces and let f : X Y be a function. Then f is said to be irreducible if f is closed, onto, and if K is a proper closed subset of Y, then f(K) 5 Y. Irreducible maps have many interesting and useful properties. For more on irre ducible maps, see Chapter 6.5 [27]. In particular, irreducible maps have the following properties. Lemma 3.4.1 Let X, Y, and Z be spaces, f : X  Y an irreducible map. Then, 1. If g : Y  Z is an irreducible map then so is the composite g o f : X + Z. 2. If0 $ U c X is open, then int(f(U)) 0. 3. If W C Y is dense, then f'(W) is dense and ff,(w) f'(W) W is irreducible. Example Let aN denote the one point compactification of the naturals. The map from : ON * aN given by 0(n) = n for n E N and (P/N \ N) = oo is irreducible. Moreover, if N is any compactification of N then the Stone extension iZ : ON  N of the embedding of N in N is irreducible. Definition 3.4.3 A space X is said to be extremally disconnected if the closure of each open set is open. If we restrict our attention to Tychonoff spaces we have the following, which is a compilation of results from [27], 6.2 Theorems (b) and (c), and [15] problem 3N4. Theorem 3.4.1 Let X be a Tychonoff space. The following are equivalent. 1. X is extremally disconnected. 2. Disjoint open sets in X have disjoint closures. 3. Each dense (open) subset of X is extremally disconnected. 4. Each dense (open) subset of X is C* embedded. 5. fX is extremally disconnected. 6. C(X) is Dedekind complete. Example Clearly any discrete space D is extremally disconnected. Theorem 3.4.1 tells us then that PD is also extremally disconnected. Both D and fD contain a dense set of isolated points. To find an extremally disconnected space without isolated points, begin with a complete atomless boolean algebra B. Then St(B) is a compact extremally disconnected space without isolated points. Definition 3.4.4 For a space X, a subset K of X is called a regular closed subset if K = cl(int(K)) Now let X be a compact Hausdorff space and let RI(X) denote the regular closed subsets of X. We define the following operations on R(X). Definition 3.4.5 For A, B E R(X), B C 7(X), define 1. A binary meet by A A B = cl(int(A n B)). 2. A binary join by AVB = AU B. 3. Complementation by A' = X \ int(A). and an infinitary meet and join by 4. AB = cl(int(nB)) and VB = cl(int(UB)) By Proposition 2.2(c) and Example 3.1(e)(4) in [27], (R(X), A, V,') is a complete Boolean algebra. Let EX denote the Stone dual of R(X). Then EX is a compact, Hausdorff extremally disconnected space. For emphasis we will recall here that the points of EX are RZ(X)ultrafilters and that the basic open sets of EX are of the form: for K e R(X), OK = {a EX : K al. By Theorem 6.6(e) in [27] the map e: EX X, defined by e(a) = na, is a continuous irreducible surjection. The above construction generalizes to the noncompact case by taking EX to be the subspace of the Stone dual of R7(X) whose points are the fixed ultrafilters. In this case, EX is an extremally disconnected zerodimensional dense subspace of the Stone dual of RI(X) and e : EX  X as defined above is a continuous perfect irreducible surjection. We will call e the canonical surjection. The space EX as defined above is called the Gleason Absolute of X, in the case that X is compact, and the Illiadis Absolute for X noncompact. We will agree to refer to EX as the absolute of X in either case. In addition to the above mentioned properties of the absolute, EX is unique in the following sense. Theorem 3.4.2 Let X be a Hausdorff space and let (Y, ) be such that Y is a zero dimensional, extremally disconnected space and i : Y * X is a perfect irreducible function. Then there is a homeomorphism 0 : EX + Y such that f o 0 = e. Of particular interest will be the relationship between olarge embeddings of f rings and irreducible maps of topological spaces. Recall that if A is a commutative semiprime fring with identity and bounded inversion and if B is an fsubring of A with bounded inversion and 1 E B, then there is an induced map : Max(A) + Max(B) that is a continuous surjection. We translate part of the exposition contained in Chapter 11 of "Rings of Quotients of Rings of Functions" [12] into a language suitable for our purposes. The discussion that follows was originally done for annihilator ideals of semiprime rings and the Boolean algebra of regular open sets. Let P e P(A) and consider Kp = {M E Max(A) : M D P} and Op = {M E Max(A) : M P'}. Then Kp is closed, Op is open and since maximal ideals are prime, Op C Kp. It is always the case that cl(Op) = Kp and if J(A) = 0, then int(Kp) = Op. In particular if J(A) = 0, then Kp is a regular closed set. Moreover, when J(A) = 0 the map P H Kp is a Boolean isomorphism from P(A) onto TR(Max(A)) (Theorem 11.8 [12]). Therefore, if J(A) = 0, the regular closed sets of Max(A) are precisely those of the form Kp = {M E Max(A): M D P} for some P E P(A). Theorem 3.4.3 Let A, B and q be as in the preceding paragraphs. If B is olarge in A, then 0 is irreducible. If in addition A is archimedean then the converse holds. PROOF Since Max(A) is compact and Max(B) is Hausdorff, q is closed. Recall that is defined by O(M) = the unique maximal ideal of B containing M n B. Denote by A/. the basic open sets of Max(B) and by Ma the basic open sets of Max(A). Let K be a proper closed subset of Max(A). Then there is an 0 < a E A such that M, C Max(A) \ K. Since B is olarge, there is a b E B and a n E N such that 0 < b < na. Then Mb C Ma so that K C Max(A) \ Mb. If M E Max(A) \ Mb, then b E M and as b E B, b E Mn Max(B) C O(M). Then O(M) E Max(B) \ AF and consequently O(K) C Max(B) \ A/b Max(B). Therefore is irreducible. For the converse we will show a bit more. That if J(A) = 0 and q is irreducible then the contraction P 'P P n B is a Boolean isomorphism from P(A) onto P(B). Then by Theorem 3.1.3, in the case that A is archimedean, the result follows. Also by Theorem 3.1.3, suffices to show that if 0 f a E A then a1l n B 0 0. Suppose we know that this holds for B(1) C A(1). Let denote the polar operation in A(1). Let 0 : a E A. Since A(1) is order large in A we have that a1 nf A(1) 5 0. In fact, all n A(1) = (a A 1)**. Therefore (a A 1)** n B(1) # 0, so that a1 n B 5 0. Since Max(A) 2 Max(A(1)), Max(B) Max(B(1)) and J(A) = 0 implies J(A(1)) = 0, we may assume without loss of generality that A has strong unit. Suppose that A has a strong unit with J(A) = 0, B is an fsubring of A and S: Max(A) , Max(B) is irreducible. Since A has strong unit, O(M) = M n B. Let 0 $ a E A. Since J(A) = 0, there is an M E Max(A) with a V M, so that K = {M E Max(A): M D a} is a proper regular closed subset of Max(A). Since is irreducible O(K) is a proper regular closed subset of Max(B). Therefore there is a Q E P(B) such that b(K) = {N E Max(B): N D Q}. Each N E O(K) is of the form N = Mn B for some M E K so that O(K)) = {M n B :M BD Q}. Suppose now that a1 n B = 0 and Q # 0. Then there is a 0 < q E Q with q V a". Since q V a"1, there is a 0 < c E a1 with 0 < cq. Since J(A) = 0, there is an M E Max(A) with cq M. Then c V M and q V M. Since c M, a1 ( M and as M is prime, a1 C M. Then M E K so that O(M) = M n BD Q. This is a contradiction as q V M. Therefore, if aI n B = 0 then Q = 0. If Q = 0 then O(K) = Max(B), but is irreducible and K is a proper closed subset of Max(A) so that Q $ 0, and hence a11 n B = 0. We have then for 0 7 a E A that all nB $ 0 and so by Theorem 3.1.3, the map P P n B is a Boolean isomorphism from P(A) onto P(B). QED We can also argue from A archimedean to the order largeness directly by noting that if A is archimedean and 0 is irreducible, by Lemma 11.8 in [12], the map P F+ {M E Max(A) : M D P} is a Boolean isomorphism from P(A) onto R(Max(A)). Since B also is archimedean we have that P(B) is Boolean isomorphic to R(Max(B)). By the embedding theorem for archimedean 1groups, Be = Ae. Then B C A C B' and since B is olarge in Be, it is olarge in A. 3.5 Specker Spaces and Absolutes In this section we look at the "inheritability" of being a speckertype space in the context of Tychonoff spaces visavis their absolutes. Lemma 3.5.1 Suppose that A = S6(A) and AL = Ae. Then Max(A) is a specker space. PROOF Since A = S(A), A is hyperarchimedean and therefore, Max(A) Min(A) is zerodimensional. Since Max(A) is compact, by Corollary 2.3.4, C(Max(A)) is localglobal. Then by Theorem 3.2.5, C(Max(A)) is a quasispecker ring. That is S(Max(A)) is olarge in C(Max(A)), so that C(Max(A)) is an fsubring of S(Max(A))e. For any ring R, let Id(R) denote the idempotents of R. For a E Id(A), the map a 4 XMa extends linearly to an fring isomorphism from A onto S(Max(A)). There fore S(Max(A))e Ae = AL S(Max(A))L. By Lemma 3.2.1, S(Max(A))L = Q(S(Max(A))) therefore C(Max(A)) C Q(S(Max(A))). QED Lemma 3.5.2 If X is a quasispecker space and S(X)L is oessentially closed then X is a specker space. PROOF Since S(X) is a quasispecker, S(X) is olarge in C(X). By Theorems 3.1.3 and 3.1.5, S(X)e = C(X)e. By Lemma 3.2.1, S(X)L = Q(S(X)), therefore C(X) C Q(S(X)) and X is a specker space. QED In the preceding lemma we cannot omit the assumption that X is a quasispecker space, as the following example shows. Example Let X = [0, 1] with the open interval topology. Since X is connected, the only idempotents in C(X) are the constant functions 0 and 1. Then S(X) R and is essentially closed, as is S(X)L, but X has no proper clopen subsets and is therefore not a specker space. Lemma 3.5.3 If X is a compact quasispecker space and EX is a specker space, then X is a speaker space. PROOF Let 0 : f E C(X). Then 0 7 f oe E C(EX), where e is the canonical surjection. Since EX is a speaker space, there is a nonempty clopen K C EX and a c : 0 with fo e (K) = c. Since K is clopen hence regular closed and e is irreducible, e (K) is a nonempty regular closed subset of X. In particular, int(e (K)) $ 0. Since X is compact and quasispecker, there is a nonempty B C X clopen with B C int(e (K)). Then f(B) = c so that X is a specker space. QED The preceding result generalizes to the following. Theorem 3.5.1 If X is a quasispecker space and EX is a specker space, then X is a specker space. PROOF Since X is a quasispecker space, fX is a compact quasispecker space. Since EX is a specker space, 3(EX) is a specker space. Since X is Tychonoff, by Theorem 6.9(b)(4) in [27], /(EX) E(OX) so that E(/X) is a specker space. By Lemma 3.5.3, fX is a specker space and therefore X is a specker space. QED For extremally disconnected spaces the following characterization obtains. Theorem 3.5.2 If X is extremally disconnected then X is a specker space if and only if S(X)L is essentially closed. PROOF Suppose that X is a specker space. Since X is extremally disconnected, by The orem 3.4.1, C(X) is Dedekind complete. Since X is a specker space, C(X) is an olarge subring of S(X)L. Since X is Tychonoff and extremally disconnected, X is strongly zerodimensional and hence a quasispecker space. In particular, S(X) is olarge in C(X). Since C(X) is Dedekind complete, S(X)^ is an olarge subring of C(X) and hence of S(X)L. But then S(X)e = (S(X)^) C S(X) C S(X)e and S(X)L is essentially closed. Suppose now that S(X)L is essentially closed. Since X is extremally disconnected, X is a quasispecker space and the result follows from Lemma 3.5.2. QED We have shown that for quasispecker spaces, if the absolute is a specker space, then the space itself is a specker space. We would like to reverse this implication. So far we have only partial success. Theorem 3.5.3 Let X be a quasispecker space. If S(X)L is essentially closed then EX is a specker space. PROOF The continuous map e : EX  X induces an fring homomorphism C(e) : C(X)  C(EX) given by; for f E C(X), C(e)(f) = f o e. Since e is onto, C(e) is an embedding and since e is perfect and irreducible, this embedding is o large. Then since X is a quasispecker space, S(X) is olarge in C(X) and therefore S(X) is olarge in C(EX). The restriction of C(e) to S(X) is an embedding into S(EX), and as S(X) is olarge in C(EX), S(X) is olarge in S(EX). Therefore, S(X)L C S(EX)L. By the hypothesis, S(X)L = S(X)e. Since S(X) is olarge in C(EX), C(EX) C S(X){. Thus C(EX) C S(EX)L and EX is a specker space. QED Definition 3.5.1 A space X is said to have the countable chain condition if every collection of disjoint open sets is at most countable. The following conjecture is the motivation for much of what follows. Conjecture 3.5.1 Let X be a compact Tychonoff space with the countable chain con dition. If X is a specker space, then EX is a specker space. Recall that a collection of open subsets B of a space X is called a irbase for X if for every nonempty open 0 C X there is a nonempty U E B with U C O. As a special case of this conjecture we will consider the case for X a compact Tychonoff space with a countable rbase where we can get a positive result. We will then look at what progress has been made with regards to the conjecture itself. We need the following notion which is due to J. Martinez [23]. Definition 3.5.2 A Tychonoff space X is said to be a weakly specker space if whenever 0 0 f E C(X), then for each open set V on which f does not vanish, there exists an open U C V such that flu is nonzero and constant. Lemma 3.5.4 If X is weakly specker, then for any 0 = f E C(X), there is a collection of open sets with dense union such that f is constant on each open set. PROOF Let S = {O, : flo, = co 5 0} U {O : flo, = 0}. If S is not dense, then V = X \ cl(S) is a nonempty open set on which f does not vanish. Since X is specker, there is an open U C V such that flu is nonzero and constant. This is a contradiction. Therefore, S is dense. QED The following is an unpublished result due to W. W. Comfort. Theorem 3.5.4 If X is a metric space with no isolated points, then there is a contin uous f : X + [0,1] such that if U is a nonempty open subset of X, then If(U)I > 2. For our purposes, the importance of Comfort's Theorem is the following corollary. Corollary 3.5.1 If X is a weakly specker metric space, then X contains a dense set of isolated points. PROOF If X contains no isolated points, then the function guaranteed by Theorem 3.5.4 is not constant on any open set. Therefore a weakly specker metric space contains at least one isolated point. Let D be the set of isolated points of X. If cl(D) : X, then X\cl(D) is a nonempty open set without isolated points. Therefore K = cl(X\cl(D)) is a metric space that is a closed subspace of X that contains no isolated points. Let 0 7 f E C*(K) be the function guaranteed by Theorem 3.5.4. By Tietze's extension theorem, there is an f E C(X) such that f IK = g. Let {O,} be a collection of open sets such that f1, is constant. Since U{O,} is dense, there is an a such that X \ cl(D) n OQ = U 5 0. Then U is an open subset of K such that If(U)I = 1. This is a contradiction. Therefore cl(D) = X and hence D is dense. QED Suppose now that X is a specker space and that 0 : X  Y is a continuous perfect irreducible map. Let g E C(Y). Then g o 0 E C(X). Since X is a specker space, there is a K C X clopen and a 0 7 c E R such that g o 0(K) = c. But 0 is irreducible so that int(O(K)) 5 0 is open. This proves the following lemma. Lemma 3.5.5 If X is a speaker space and 0 : X  Y is a continuous perfect irre ducible map, then Y is a weakly specker space. For X, a compact Tychonoff space with a countable t7base, we will construct a quotient space that is an irreducible image of X and is also a metric space. This is the content of the following theorem. Theorem 3.5.5 If X is a compact Tychonoff space with a countable clopen itbase, then there is a metric space Y and an irreducible surjection p : X  Y. In particular, EX EY. PROOF Let II' = {B, : n E N} be a countable clopen 7rbase for X. Let I be the Boolean algebra generated by I'. Then II is a countable clopen 7rbase for X. Define a relation on X by x ~ y if x and y are in exactly the same sets of II. It is easily verified that this is an equivalence relation on X. Let Y be the quotient space of X modulo this equivalence relation and let p : X + Y be the projection map. Since p is continuous and X is compact, Y is compact. Let [x] denote the equivalence class of x in X. We claim that [x] = n{B E H : x e B}. For if y E [x] then y E B for every B with x E B so that y E n{B E II : x e B}. Conversely, if y n{B E II : x E B} then x E B implies y E B. Since II is a Boolean algebra, if there is a B e 1 with y E B and x ( B, then X \ B E II has y ( X \ B D n{B II : x e B}. This is a contradiction. Thus y e B implies x E B so that y E [x]. In particular, since each B II is clopen any such intersection is closed so that the points of Y are closed. We will next show that the projection map is closed. By Proposition 1.6, Chapter VI [10], it suffices to show that if K C X is closed, then p'(p(K)) is closed. Let K C X be closed, and let x E X \ pl(p(K)). Then p(x) ) p(K), and since H is a Boolean algebra, there is a B E II with x E B. Therefore, for every z E K there is a B E II with x E Bz and z V Bz. Then {X \B, : z E K} is an open cover of K. Since K is compact, there is a finite subcover say {X \Bi : 1 < i < n}. We then have that x E nf=l Bi C X \ K and since each Bi is clopen, nl=1 Bi is clopen. It remains to show that nf=l Bi C X \p1(p(K)). Suppose that z E (n=l Bi) n p1(p(K)). Then p(z) = p(y) for some y E K. But then z E n7=l Bi implies y E nf=l Bi C X \ K; a contradiction. Therefore nl=t Bi C X \p'(p(K)) and X \ p(p(K)) is open, so that pl(p(K)) is closed. Now let K be a proper closed subset of X and suppose that p(K) = Y. Since 0 X \ K is open and II is a irbase, there is a B E II with B C X \ K. Let x E B. Since p(K) = Y, there is a y E K with p(y) = p(x), but this implies that y E B. A contradiction. Therefore p is irreducible. We now have that p is a continuous irreducible map from a compact Tychonoff space X onto a compact T1 space Y. Then p is perfect and by Theorem 5.2, Chapter XI [10], Y is regular. We will next show that Y is metrizable by showing that Y is second countable and appealing to a result due to P.Urysohn [10] which says for second countable spaces, regularity is equivalent to metrizability. Let y E Y. Then y = p(x) for some x E X. For every B E II with x E B, y = p(x) E p(B) so that y E N{p(B) : B E n and x E B}. For B E H, B C p'(p(B)). If x E p1(p(B)), then p(x) E p(B) so that p(x) = p(z) for some z E B. But then x E B and therefore B = p1(p(B)). Since B is open and the open sets U of Y are precisely those for which p1(U) is open, we have that p(B) is open for every B II Since I is countable, {p(B) : B E H and x E B} is at most countable. Let y' E Y with y' = y. Then y' = p(x') for some x' E X. Since y' # y, there is a B E I with x E B, x' B. If p(x') E p(B) then there is a z E B with p(x') = p(z) and then x' E B. Therefore, y' = p(x') i p(B) and as x E B, p(B) D N{p(B) : B E II and x E B}. Therefore {y} = f{p(B) : B E H and x E B} so that every point of Y is a G6. In particular, by Theorem 3.5(1)(1) [27], since Y is compact and every point is a G6, Y is first countable. Now let 0 0 U C Y be open and let y E U. By the above, there is a countable subset of n say {Bi : i E N} such that {y} = n{p(Bi) : i E N}. We have already seen that each p(B) is open for B E H and, since p is a closed map, each p(B) is clopen. Then {Y \p(Bi) : i E N} is an open cover of Y \ U. Since Y \ U is closed and hence compact, there is a finite subset, {Y \ p(Bi) : 1 < i < n} suitably reindexed, with Y \U C U{Y \p(Bi) : 1 < i < n}. Then yE N{p(Bi) : 1 < i < n} C U and n{p(B.) : 1 < i < n} is open. Let Bo = n{Bi : 1 < i < n}. Clearly Bo E Ii and p(Bo) C n{p(B2) : 1 < i < n}. If z E n{p(Bi) : 1 < i < n}, then p1(z) C p'({p(Bi) : 1 < i < n}) = {p(p(Bi)) : 1 < i < n} = n{Bi: 1 < i < n} = Bo, so that z E Bo. Therefore p(Bo) = ({p(Bi) : 1 < i < n}. This gives us that {p(B) : B E II} is a countable basis for Y and therefore that Y is second countable. We now have a perfect irreducible map p from X onto a metric space Y. Then p o e : EX + Y is a perfect irreducible map from an extremally disconnected zero dimensional space onto Y. By the "uniqueness" of Theorem 3.4.2, EX 2 EY. QED As a corollary we have a partial converse of Theorem 3.5.1. Theorem 3.5.6 Let X be a compact Tychonoff space with a countable 7rbase. If X is a speaker space, then EX is a specker space. If this is the case, both X and EX contain a countable dense set of isolated points. PROOF Let II be a countable 7rbase for X and let U E I. Since X is Tychonoff, there is an f E C(X) with coz(f) C U. Since X is a specker space, there is a B clopen and a 0 5 c R such that f(B) = c. In particular, B C coz(f) C U. If we choose one such B for each U E I, then the resulting collection is a countable clopen r base for X. By Theorem 3.5.5, there is a metric space Y and an irreducible map p : X + Y. By Lemma 3.5.5, Y is a weakly specker metric space. By Corollary 3.5.1, Y contains a dense set of isolated points. Let D be this dense set of isolated points and let t : EX  Y be the irreducible map p o e as in the proof of Theorem 3.5.5. By Lemma 3.4.1, t'(D) is dense in EX. By Proposition 6.9(e) [27], the points of 1I(D) are all isolated. Therefore, EX contains a dense set of isolated points, and so EX is a specker space. Let I(EX) and I(X) denote the set of isolated points of EX and X respectively. Again, by Proposition 6.9(e) [27], eII(EX) is a bijection from I(EX) onto I(X). If H is a irbase for X, then II must contain I(X). Therefore if H is countable, then I(X) and hence I(EX) are countable. QED Recall that a space is locally compact if each point has a compact neighborhood. In Chapter 6 [15], it is shown that X is open in #X if and only if X is locally compact. This gives most of the following corollary. Corollary 3.5.2 Let X be a locally compact Tychonoff space with a countable rbase. If X is a specker space, then EX is a specker space. PROOF Let II be a countable rbase for X. Let 0 $ 0 C iX be open. Since X is locally compact, by the above observation, and since X is dense in #X, On X is a nonempty open subset of X. Then there is a B E H with B C O n X. Again, as X is open in 3X, B is open in 3X so that II is a countable irbase for fX. Then fX is a compact specker space with a countable rbase so that by Theorem 3.5.6, E(/3X) is a specker space. Since E(/X) /3(EX), EX is a specker space. QED To return to the standing of Conjecture 3.5.1, we begin with the following result which will allow us to restrict our attention to compact zerodimensional spaces. Lemma 3.5.6 If X is a compact quasispecker space then there is a compact zero dimensional space Y such that EX EY. PROOF Since X is a quasispecker space, S(X) is olarge in C(X). By Theorem 3.4.3, the map 0: Max(C(X))  Max(S(X)) isdn irreducible. Since X is compact, X  Max(C(X)) and since S(X) is hyperarchimedean, Max(S(X)) is zerodimensional. QED Now if X has the countable chain condition and 0 : X  Y is any continuous map, then Y has then countable chain condition. Therefore if X is a compact quasispecker space with the countable chain condition, then X is coabsolute with a compact zero dimensional space Y with the countable chain condition. The advantage of dealing with compact zerodimensional spaces is that in the context of the following theorem, due to A. Hager, we can translate our topological problem to one of Boolean algebras. Definition 3.5.3 Let A be a Boolean algebra. For C, D C A we say that C refines D, denoted by C < D, if for every c C there is a d E D such that c < d. Definition 3.5.4 Let A be a Boolean algebra. A quasicover of A is 0 C C A with VC = 1. A cover is a finite quasicover. A partition is a cover by pairwise disjoint elements. Definition 3.5.5 A Boolean algebra is called speaker if every sequence of covers has a common refinement by a quasicover. We will need the following lemma. Lemma 3.5.7 A zerodimensional space X is a specker space if and only if for every 0 : f E C(X) there is a C C B(X) such that UC is dense and for every C E C, fic is constant. PROOF Suppose that X is a specker space and let 0 : f E C(X). Then there is a clopen K C X such that f K is a nonzero constant. Let C = {K E B(X) : f K is constant). If UC is dense, we are done. Suppose that UC is not dense. Then int(X \ (UC)) 7 0 is open. Since X is zerodimensional, there is a 0 # T clopen with T C int(X\ (UC)). Let g = fXT. Then 0 # g E C(X), and g is not a nonzero constant on any clopen subset. This is a contradiction. Therefore UC is dense. The converse is clear. QED It should be pointed out that for the above result it is sufficient that X have a clopen 7rbase. With this observation the preceding lemma is actually a restatement of Proposition 0.5 [23]; X is a specker space if and only if it is a weakly specker space with a clopen rbase. Suppose that {Ci : i E N} is a sequence of covers of A. Suppose also that for each i E N, Ci = {cj : 1 < j < ni}. Let di = cl and, for 1 < j < ni, let dj = cj \ V{ck : 1 < k < j 1}. Let 2Di = {dj : 1 < j < ni}. Then Di is a partition of A and AI < Ci. If D is a common refinement of the sequence {Di : i E N} by a quasicover, then Z> is a common refinement of the sequence {Ci : i E N} by a quasi cover. Define 'D inductively by VDo = Do and VYn+ = {a A b: a E VYn and b E ++ }. Then {12 : i E N} is a sequence of partitions with D < Di such that D' > D' > . If D is a common refinement by a quasicover of the sequence {DfI : i E N}, then D is a common refinement by a quasicover of the sequence {Di : i E N}. Call a sequence of covers {D, : i E N} with D1 > D2 > a decreasing sequence of covers. We have proved the following lemma. Lemma 3.5.8 A Boolean algebra A is specker if and only if every decreasing sequence of partitions has a common refinement by a quasicover. We are now ready to state and sketch the proof of the following theorem which is due to A. Hager. Theorem 3.5.7 A Boolean algebra A is specker if and only if its Stone dual St(A) is a specker space. PROOF Suppose A is is specker. Let 0 7 f E C(St(A)). Since St(A) is compact and zerodimensional, there is a finite cover of St(A) by clopen sets such that f varies less than 2" on each set of this cover. For each n, let U4 be this cover. Then in B(St(A)) A, {1U : n E N} is a sequence of covers. By the hypothesis, there is a common refinement by a quasicover U. Then VU = cl(UU) = St(A) so that UU is dense. Since U < U4 for all n, if K E U then f varies less than 2" on K for all n. Therefore is f is constant on K. Suppose now that St(A) is a specker space. Let {Ci} be a sequence of covers of A. By Lemma 3.5.7, we may assume without loss of generality that {Ci} is a decreasing sequence of partitions. Via the isomorphism A B(St(A)) we may view these as a sequence of finite clopen partitions of St(A). We then construct a Cantor type function f E C(St(A)) as the uniform limit of simple functions f, defined on C1 having the property that if Ci E Ci has C1 D C2 D .. then there exists an r E R such that nCi = fl(r) and if r E f(St(A)) then f'(r) is of this form. By the hypothesis and Lemma 3.5.6, there exists a C C St(A) with UC dense such that for every C E C, f\C is constant. Viewed in A, C is a quasicover and a common refinement of the sequence {Ci}. QED For our purposes, we have that a compact zerodimensional space X is a specker space if and only if B(X) is a specker boolean algebra. This is where we can make a translation of our original problem. All of the results used in this paragraph can be found in [27] Chapters 3 and 6. Suppose that X is a compact zerodimensional specker space with the countable chain condition. Then B(X) is a specker boolean algebra with the countable chain condition. The irreducible map e : EX  X induces an olarge embedding of B(X) in B(EX), and a boolean isomorphism of R(X) and R(EX). Since EX is extremally disconnected, R(EX) = B(EX). Identifying isomorphic algebras, we have an olarge embedding of B(X) in 7R(X). Since R(X) is the completion of B(X), our original question translates to the following. Is the completion of a specker boolean algebra with the countable chain condition a specker boolean algebra? What follows is part of an attempt to answer this question. Definition 3.5.6 Let {C : i E N} with C1 > C2 > be a decreasing sequence of covers of A. C1 > C2 > .. is said to be a binary sequence of covers if for every i E N and for every c E CC, I{b C,+i : b < c}1 < 2. We will show that specker boolean algebras can be characterized in terms of binary sequences of covers. We first need some terminology and notation. For a boolean algebra A, C C A is called a chain if C is a linearly ordered subset of A. For a decreasing sequence of covers C1 > C2 > *, a chain cl > c2 > is a representative chain if c, E Ci for all i E N. A chain cl 2 c2 > .. is called a proper chain if there exists an i = j such that c; = cj, provided that {Ci} is not a trivial sequence. For a e A, let [0,a] = {b E A : 0 < b < a}. For r E R, let [rJ denote the largest integer less than or equal to r and let Fr] denote the least integer greater than or equal to r. We first need the following technical lemma. Lemma 3.5.9 Let A be a boolean algebra. If for every binary sequence of covers B1 > B2 > there is a proper representative chain { bi} with a nonzero lower bound, then for every sequence of covers C1 > C2 > ... there is a proper representative chain {ci} with a nonzero lower bound PROOF By way of contradiction, suppose that Ci > C2 > ... is a sequence of covers such that every proper representative chain {ci} has A{ci : i E N} = 0. We will construct a binary sequence of covers with the same property. The idea of this proof is to construct, for every C > Ci+1, a binary sequence of covers between" Ci and Ci+l. We do this pointwise for each element a of Ci by taking pairwise joins of the elements of Ci+1 that are below a. Let i E N and for a E Ci, let Ci+i(a) = {c E Ci+1 : c < a}. Then Ci+i(a) is a binary sequence of covers of [0,a]. Suppose that we have constructed Bl(a) > B'_1 (a) > >. 3z(a) = Ci+l(a) a binary sequence of covers of [0, a]. Construct B' as follows. First there is an indexing of the elements of Bk(a) = {b, : 1 < n < mk}. For 1 < n < ["Lj, let an = b2n1 V b2n and if mk is odd, let arJi = b,,. Now take S= {a 1 < n < [ 1 }. For some finite k E N, 8 = {a}. Let B = B'n for 1 < n < k. We have constructed {a} = Bi(a) > B2(a) > > Bk(a) = Ci+l(a) a binary sequence of covers of [0, a]. Now repeat this process for each a E Ci. Suppose that Ci = {an : 1 < m < mi}. Let kn be such that Bkn(an) = Ci+i(an) for 1 < m < mi. Let k = max{k,n : 1 < n < mi}. We need to do some manipulation of the indices. For each 1 < n < mi and for each j E N with k, < j < k, let Bj(a,) = Bkn(a,). Now for 1 < 1 < k, let BI = Bi(al) U B2(al) U ** U Bi(am,). We have constructed Ci = B1 > B2 > *. > k = Ci+l a binary sequence of covers of A. Repeat this process for every pair C, > Ci+1 of the original sequence of covers and reindex the resulting binary sequence so that C1 = B3 > B2 > .. Then for each Ci there is an ij such that Ci = Bi,. Now let {a,} be a proper representative chain of the sequence B,. By construction there is a chain {aj } with {ai,} E C, which is cofinal in {as}. Since {a,} is a proper chain and {ai, } is cofinal, {ai, } is a proper representative chain of the sequence {Ci}. By the hypothesis, A{a,,} = 0. As any lower bound for {an} is a lower bound for {a, }, we have that A{a,} = 0 QED We then have the following theorem which allows us to characterize specker boolean algebras in terms of binary sequences of covers. Theorem 3.5.8 Let A be a boolean algebra. A is specker if and only if for every binary sequence of partitions B3 > B2 > '* there is a proper representative chain {bi} such that {b,} has a nonzero lower bound. PROOF Suppose that A is specker. Let B1 > B2 > be a binary sequence of partitions. Let B be a common refinement by a quasicover. Since VB = 1, there is a 0 / b E B. Since B is a common refinement, for each Bi there is a bi E Bi with b < b,. Since each Bi is a partition, {bi} is a representative chain. If B1 > 2 > has no representative chains {b,} with bi / bj for some i 5 j, then we can take B = B1 and the sequence is trivial. Suppose that C1 > C2 > is a decreasing sequence of partitions of A. By Lemma 3.5.6, it suffices to show that this sequence has a common refinement by a quasicover. By the hypothesis and Lemma 3.5.7, there is a proper representative chain {c~ : i E N} such that {c, : i E N} has a nonzero lower bound. Let L be the set of all nonzero lower bounds for all proper representative chains of C1 > C2 > ... Then L is a common refinement of the Ci's. It remains to show that L is a quasicover. If VL = 1 we are done. Suppose that VL # 1. Then there is an a E A with 0 < a < 1 such that a is an upper bound for L. Let b = a' and let C,[b] = {cA b: b E C;}. Then Ci[b] is a decreasing sequence of partitions of [0, b]. Now, Cl[b]U {a} > C2[b]U {a} > ** is a decreasing sequence of partitions of A. Therefore, by the hypothesis and Lemma 3.5.9, there is a proper representative chain {bi} with a nonzero lower bound. Since a A b = 0, bi E Ci[b] for all i. For each i, bi = ci A b for some ci E Ci. Since {bi} is a chain and each Ci is a partition, {ci} is a chain. Since {bi} is a proper chain, so is {c;}. If d is a nonzero lower bound for {bi} then it is also a lower bound for {c;} and therefore d < a. This is a contradiction since d < b and a A b = 0. Therefore, V L = 1 and the proof is complete. QED The current goal with these results is to show that if X is a compact specker space with the countable chain condition, then X contains a dense set of isolated points. We begin by assuming that X has no isolated points. Then B(X) is an atomless specker boolean algebra. It may be possible to then construct an uncountable anti chain, violating the countable chain condition. This will show that X contains one isolated point. It should then be possible to argue that X in fact contains a dense set of isolated points, and therefore that EX is a specker space. CHAPTER 4 CONCLUSION The focus of this dissertation is an examination of frings which are rich in idem potents. We consider three different types of frings, reflecting three different degrees of "richness". In Chapter 2, we consider localglobal frings. For bounded rings, we obtain the equivalence of the following conditions. 1. A is a localglobal ring. 2. Every primitive quadratic polynomial with nonnegative coefficients in A rep resents a multiplicative unit. 3. Max(A), the maximal spectrum, is zerodimensional. As one measure of the richness of idempotents for localglobal frings, recall that in the verification of the localglobal condition, we began with a primitive polyno mial f E A[t] and constructed an element s E A such that f(s) is a multiplicative unit. This element s we constructed is, in fact, a linear combination of idempotents. The zerodimensionality of Max(A) also gives another indication of the richness of idempotents in localglobal rings. Here we need to recall that the clopen subsets of Max(A) are the basic open sets determined by an idempotent of A and that the zerodimensionality of Max(A) implies that the clopen sets are a base for the open sets of Max(A). Considering the case A = C(X) for a Tychonoff space X, we obtain a nice addition to the "AlgebraTopology Dictionary". Namely, that a space X is strongly zerodimensional if and only if every primitive quadratic polynomial with 87 nonnegative coefficients in C(X) represents a unit. This characterization is of addi tional interest because it provides a first order algebraic characterization of strongly zerodimensional spaces. For this reason, this condition merits further investigation. Chapter 2 also left a glaring open problem. That is, whether or not all of the above conditions are equivalent in the unbounded case. As indicated at the end of the chapter, the problem is one of "cutting down" primitive polynomials to the bounded subring A(1) of A and maintaining primitivity. We give sufficient conditions for this to occur, namely, that the ring be a Bezout ring. The working conjecture is that the general result does hold, but the proof may require some sort of set theoretical forcing argument. In Chapter 3, we consider two related measures of the richness of idempotents. Recall that an fring A is said to be of speckertype if A is an essential extension of S(A), the subalgebra generated by the idempotents of A, and A is said to be a quasispecker ring if A is an oessential extension of S(A). Recall also that for a ring A, Q(A) denotes the complete ring of quotients of A. We first obtain the equivalence of the following conditions. 1. A is a speckertype ring. 2. A is an fsubring of Q(S(A)). 3. S(A)L = Q(A). In particular, condition (3) gives some indication of the richness of idempotents in speckertype frings. It says, in effect, that every element of the complete ring of quotients of A can be written as a (possibly infinite) linear combination of idem potents. For quasispecker frings, the richness of idempotents is indicated by the defining condition that every nonzero element is larger than some scalar multiple of a nonzero idempotent. Results regarding speckertype and quasispecker rings become particularly interesting for the case A = C(X). Recall that a space X is said to be a specker space (quasispecker space) if C(X) is a speckertype (quasispecker) ring. For specker spaces, we have in addition to the previous conditions, suitably translated, the following. 4. For every nonzero f E C(X), there is a clopen set on which f is nonzero and constant. In the case that X is compact we are able to improve this characterization to, 5. For every f E C(X) there is a collection of clopen sets whose union is dense in X and such that f is constant on each of these clopen sets. Again, confusing the clopen sets of X with the idempotents of C(X), we get some sense of the richness of idempotents for specker spaces. For compact quasispecker spaces, the richness of idempotents of C(X) is reflected in the condition which char acterizes these spaces as those which have a countable clopen 7rbase. Motivated by the "tight" containment of C(X) in Q(X), we consider the specker condition for spaces visavis their absolutes. We are able to show that if a space X is a quasispecker space and EX, the absolute of X, is a specker space, then X is a specker space. As a partial converse, we have the conjecture that if X is compact with the countable chain condition and X is a specker, then EX is a specker space. As a special case of this conjecture we are able to show that if X compact, with a countable rbase, then X is a specker space implies that EX is a specker space. Finally, via A. Hager's result on specker boolean algebras and specker spaces, we are able to translate the problem of compact specker spaces to one of specker boolean algebras. Here, using the characterization obtained for specker algebras in terms of 90 refinements of binary sequences of covers, we hope to prove the following. If X is a compact specker space with the countable chain condition, then X contains a dense set of isolated points. Then EX will also contain a dense set of isolated points and therefore be a specker space. REFERENCES [1] F. Anderson. Latticeordered rings of quotients. Canad. J. Math., 17:434448, 1965. [2] M. Anderson and T. Feil. LatticeOrdered Groups. Reidel Texts in the Mathe matical Sciences. D. Reidel Publishing Company, Dordrecht, Holland, 1988. [3] B. Banaschewski. Maximal rings of quotients of semisimple commutative rings. Archiv Math., 16:414420, 1965. [4] S. Bernau. Unique representation of archimedean lattice groups and archimedean lattice rings. Proc. London Math. Soc., 15:599631, 1965. [5] A. Bigard, K. Keimel, and S. Wolfenstein. Groupes et Anneaux Reticules. Lec ture Notes in Mathematics. SpringerVerlag, New York, 1977. [6] G. Birkhoff. Lattice ordered groups. Ann. Math., 43:228331, 1942. [7] P. Conrad. The essential closure of an archimedean latticeordered group. Duke Math. J., 38:151160, 1971. [8] P. Conrad. Epiarchimedean groups. Czech. Math. J., 24(99):192218, 1974. [9] P. Conrad and J. Martinez. Complemented lattice ordered groups. Indag. Math., 3:281297, 1990. [10] J. Dugundji. Topology. Allyn and Bacon Series in Advanced Mathematics. Allyn and Bacon, Boston, 1966. [11] D. Estes and R. Guralnick. Module equivalences: Local to global when primitive polynomials represent units. J. Algebra, 77:138157, 1982. [12] N. Fine, L. Gillman, and J. Lambek. Rings of Quotients of Rings of Functions. McGill University Press, Montreal, 1965. [13] I. Gelfand and A. Kolmogoroff. On rings of continuous functions on topological spaces. Dokl. Akad. Nauk SSSR, 22:1115, 1939. [14] L. Gillman and M. Henriksen. Rings of continuous functions in which every finitely generated ideal is principal. Trans. Amer. Math. Soc., 82:366391, 1956. [15] L. Gillman and M. Jerison. Rings of Continuous Functions. Graduate Texts in Mathematics. SpringerVerlag, New York, 1976. [16] R. Gilmer. Multiplicative Ideal Theory. Pure and Applied Mathematics. Marcel Dekker, Inc., New York, 1972. [17] A.M. Gleason. Projective topological spaces. Ill. J. Math. 2:482489, 1958. [18] V. Glivenko. Sur quelques points de la logique de M. Brouwer. Bull. Acad. des Sci. des Belgiques, 15:183188, 1929. [19] A. Hager and L. Robertson. Representing and ringifying a Riesz space. Symposia Math., 21:411431, 1977. [20] M. Henriksen, J. Isbell, and D. Johnson. Residue class fields of lattice ordered algebras. Fund. Math., 50:107117, 1961. [21] O. Holder. Die axiome der quantitit und die lehre vom maB. Ber. Verh. Sichs. Ges. W1iss. Leipzig, Math.Phys. Cl., pages 164, 1901. [22] J. Lambek. Lectures on Rings and Modulfs. Blaisdell Publishing Company, Waltham, Massachusetts, 1966. [23] J. Martinez. Algebras and spaces of dense constancy. In preparation, 1992. [24] J. Martinez. The maximal ring of quotients of an fring. In preparation, 1992. [25] J. Martinez and S. Woodward. Bezout and priifer frings. Comm. Algebra, 1992. To appear. [26] R. Pierce. Modules over commutative regular rings, volume 70. Mem. Amer. Math. Soc., 1967. [27] J. Porter and R. Woods. Extensions and Absolutes of Hausdorff Spaces. SpringerVerlag, New York, 1988. [28] R. Sikorski. Boolean Algebras, volume 25 of Ergebnisse der M.althrmalik und Ihrer Grc n:g bhif I. SpringerVerlag, Berlin, 1964. [29] M. Stone. Application of the theory of boolean rings to general topology. Trans. Amer. Math. Soc., 41:375481, 1937. [30] Y. Utumi. On quotient rings. Osaka Math. J., 8:118, 1956. [31] F. Sik. Zur theorie der halbgeordneten gruppen. Czech. Math. J., 10:400424, 1960. [32] B. Vulich. Introduction to the Theory of Partially Ordered Spaces. Wolters Noordhoff, Groningen, 1967. [33] R. Walker. The StoneCech Compactification. Ergebnisse der Mathematik und ihrer Grenzgebiete. SpringerVerlag, New York, 1974. [34] K. Yosida. On the representation of a vector lattice. Proc. Imp. Acad. Tokyo, 18:339343, 1942. BIOGRAPHICAL SKETCH Scott David Woodward was born in Phoenix, Arizona, in 1955. Before returning to school in 1980, he was a journeyman mason working out of Orlando, Florida. He received his B.S. and M.S. degrees in mathematics from the University of Florida in 1983 and 1987 respectively. He is married to Rita WendtWoodward and has two children, Christopher and Michael. I certify that I have read this study and that in my opinion it conforms to accept able standards of scholarly presentation and is fully adequate, in scope and quality, as a dissertation for the degree of Doctor of Philosophy. Jorge Martinez, Chairman Professor of Mathematics I certify that I have read this study and that in my opinion it conforms to accept able standards of scholarly presentation and is fully adequate, in scope and quality, as a dissertation for the degree of Doctor of Philosophy. Krishnaswami Alladi Professor of Mathematics I certify that I have read this study and that in my opinion it conforms to accept able standards of scholarly presentation and is fully adequate, in scope and quality, as a dissertation for the degree of Doctor of Philosophy. Douglas 4. Center Professor of Mathematics  uJi~'~~ I certify that I have read this study and that in my opinion it conforms to accept able standards of scholarly presentation and is fully adequate, in scope and quality, as a dissertation for the degree of Doctor of Philosophy. Christopher W. Stark Associate Professor of Mathematics I certify that I have read this study and that in my opinion it conforms to accept able standards of scholarly presentation and is fully adequate, in scope and quality, as a dissertation for the degree of Doctor of Philosophy. El by J. olduC Jr. Professor of Instruction and Curri ulum This dissertation was submitted to the Graduate Faculty of the Department of Mathematics in the College of Liberal Arts and Sciences and to the Graduate School and was accepted as partial fulfillment of the requirements for the degree of Doctor of Philosophy. August 1992 Dean, Graduate School 