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Completions of Regular Rings K. R. Goodearl Department of Mathematics, University of Utah, Salt Lake City, Utah 84112, USA

This paper is concerned with the structure of the completion of a (von Neumann) regular ring R with respect to a pseudo-rank function, and with the connections between the ring-theoretic structure of such a completion and the geometric structure of the compact convex set lP(R) of all pseudo-rank functions on R. Given a finite or infinite positive convex combination N = 2;~ RPRin 1P(R), it is shown that if the Pk lie in pairwise disjoint faces of IP(R), then the N-completion of R is naturally isomorphic to the direct product of the Pk-completions of R. Given N, Pc IP(R), it is shown that P extends to a pseudo-rank function on the N-completion of R if and only if P lies in the closure of the face generated by N in IP(R). Other results develop representations of the faces and extreme points in IP(R). This research was partially supported by Grant No. GP-43029 of the National Science Foundation (USA). All rings in this paper are associative with unit, and ring maps are assumed to preserve the unit. 1. Rank and Pseudo-Rank Functions

Definition. A pseudo-rank function [4] on a regular ring R is a map N:R--*[0, 1] such that (a) N(1)= 1. (b) N(xy) < U(x), N(y). (c) N(e + f ) = N(e) + N ( f ) for all orthogonal idempotents e, f 6 R. A (normalized) rank function [12, p. 231] on R is a pseudo-rank function N with the additional property (d) N(x)=0 if and only if x = 0 . Note as a consequence of (b) that if x, y~ R with xR ~ yR, then N(x) = N(y). Lemma 1.1. Let N be a pseudo-rank function on a regular ring R. (a) N ( x + y ) < N ( x ) + N ( y ) for all x, yER. (b) IN(x)- N(y)I < N ( x - y) for all x, ye R. (c) The set K = {xe RJN(x)=O} is a proper two-sided ideal of R. (d) N induces a rank junction N' on R/K accordin9 to the rule N'(.~ = N(x).

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K . R . Goodearl

Proof [12, Corollary, p. 23t] and [4, Lemma 5]. [] Definition. If N is a pseudo-rank function on a regular ring R, the ideal {x~ RI N(x)=0} is called the kernel of N, denoted ker(N). Definition. For any regular ring R, let 1P(R) denote the set of all pseudo-rank functions on R. Note that IP(R) is contained in the real vector space ~ R We consider IRR to be equipped with the product topology, under which it is a locally convex, Hausdorff, linear topological space. According to [4, Lemma 7], IP(R) is a compact subset of IRR. In addition, IP(R) is convex [4, p. 273]. Definition. Let K be a compact convex subset of a locally convex, Hausdorff, linear topological space. Given a countable sequence of vectors Vl, v2 .... e K and nonnegative real numbers ~1, ~2 .... such that Z~k = t, standard arguments show that the series XakV k must converge to some element of K. (For the case K = IP(R), this is very direct: see [4, p. 273].) We refer to ZC~kVk as a a-convex combination of the Vk. (Of course, finite convex combinations may also be regarded as a-convex combinations.) A convex or a-convex combination SekV k is said to be positive provided the coefficients ek are all positive. Definition. If R is a regular ring with a pseudo-rank function N, then the rule 6 ( x , y ) = N ( x - y ) defines a pseudo-metric 6 on R as in [12, pp. 231, 232]. In the interest of brevity, we refer to 6 as the N-metric on R, even though 6 is not a metric unless N is a rank function. By passing to the rank function N' induced by N on R/ker(N), we obtain an actual metric 6', which we sometimes refer to as the N-metric on R/ker(N). As shown in [12, p. 232], the ring operations and the map N are all uniformly continuous with respect to the N-metric 6. Thus the (Hausdorff) completion of R with respect to 6 is again a ring, which we call the N-completion of R. Note that the kernel of the natural map from R into the N-completion is just ker(N). Theorem 1.2 [8, Theorem 3.7]. Let R be a regular ring, let N e IP(R), and let denote the N-completion of R, Then R is a regular ring, N extends continuously to a rank function t~1 on R, and R is complete in the N-metric. [] Definition. Under the conditions of Theorem 1.2, we refer to b7 as the natural extension of N to P(/~). Because of continuity, N is unique. Theorem 1.3 [4, Corollary 15]. Let R be a regular ring and let N e F(R). The N-completion of R is a right and left self-injective ring. [] Definition. Let K be a convex subset of a real vector space. An extreme point of K is any point of K which cannot be expressed as a positive convex combination of two distinct points of K. Theorem 1.4 [4, Theorem 19]. Let R be a regular ring, let N ~ 1P(R), and let k be a positive integer. Then the N-completion of R is a direct product of k simple rings if and only if N is a positive convex combination of k distinct extreme points in 1P(R). [] Theorem 1.5 [4, Theorem 22]. Let R be a regular ring, and let N e IP(R). Then the N-completion of R is an infinite direct product of simple rings if and only if N is a positive a-convex combination of infinitely many distinct extreme points in

IP(R). []

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231

2. The Decomposition Problem

Let R be a regular ring, let N6 IP(R), and let /~ denote the N-completion of R. It is not hard to see that any ring decomposition of/~ induces a "decomposition" of N as a positive convex combination in IP(R). For simplicity, consider a finite direct product decomposition R = R ~ × . . . × / ~ k , with each /~=0. Since the natural extension ~r of N to P(/~) is a rank function, we obtain N(ei)=4=O, where el denotes the identity element of/~i. The rule Ni(x)=IV(eix)/N(ei)defines a pseudorank function ~7i~ IP(/~), and N = N(el)b71 +.-. + bT(ek)bTk. Pulling everything back to R, we obtain a positive convex combination N=N(eONI+...+N(ek)Nk in P(R). Further, it can be shown that/~ is naturally isomorphic to the N~-completion of R. We investigate the converse of this situation: If N is a positive convex (or a-convex) combination of some Pi in IP(R), when is the N-completion of R isomorphic to the direct product of the Pccompletions of R? This does hold, for instance, when the P~ are distinct extreme points of IP(R). (This may be deduced from the proofs of Theorems 1.4 and 1.5.) On the other hand, it does not hold in general, as the following example shows. Let R be the case t = 3 of [4, Example B], so that IP(R) is a triangle, as in Diagram I. Using Theorem 1.4, we see that the N-completion/~ is a direct product of three simple rings, while each of the Pccompletions/~ is a direct product of two simple rings. Thus R~R~ ×R2. Q3

\°2

Diagram I

There are three simple rings Si involved in this example, namely the Qicompletions of R. As indicated by the previous discussion, we have/~1 ~$1 x S), R2----$2 ×$3, and R~S1 ×$2 ×$3. Thus /~ fails to be isomorphic to /~ ×R2 because/~! and R2 each contain a factor isomorphic to $3, only one of which occurs in R. Geometrically, this corresponds to the fact that P1 and P2 are midpoints of edges which have a common vertex Q3. Thinking of vertices as zerodimensional faces, edges as one-dimensional faces, etc., we see that P1 and P2 do not lie in disjoint faces of IP(R). We may interpret this as a kind of geometric dependence between P~ and P2. This example suggests the following problem. Let R be a regular ring, and let N~ IP(R) be a positive a-convex combination of some Pie 1P(R). If the Pi lie in pairwise disjoint faces of ~'(R), show that the N-completion of R is isomorphic to the direct product of the Pccompletions of R. There is a natural map from the N-completion of R into the product of the Pi-completions, which we develop in

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the present section. The next section is devoted to a study of faces in IP(R), after which we can give the solution to this decomposition problem. Definition. Let R be a regular ring, and let N, Pc IP(R). We say that P is (uniformly) continuous with respect to N provided P is a (uniformly) continuous map from R, equipped with the pseudo-metric induced by N, to [0, 1], with the usual metric. Note in particular that ifP is continuous with respect to N, then ker(N)C__ker(P). Lemma 2.1. Let R be a regular ring. For N, Pc IP(R), the following conditions are equivalent: (a) P is uniformly continuous with respect to N. (b) P is continuous with respect to N. (c) For each ~ > 0 there exists 6 > 0 such that for all x ~ R , N(x)<3 implies

P(x)<~. Proof (a)~(b)=>(c) is clear. (c)~(a): With e, 6 as in (c), we see from Lemma 1.1 that N ( x - y ) < 6 implies l P ( x ) - P ( y ) l < P ( x - y ) < e . [] For finite, countably additive measures, the criterion given in Lemma 2.1(c) (viewing R as the Boolean algebra of sets on which the measures act) is equivalent to absolute continuity [7, Theorem B, p. 125]. The basic source of continuous pseudo-rank functions is a positive a-convex combination N XO~kPk. For Pk ~ ~k ~N, whence it is automatic from Lemma 2.1 that Pk is continuous with respect to N. Lemma 2.2 [4, Lemma 16]. Let R be a regular ring, let N6 IP(R), and let R be the N-completion of R. Let Pc IP(R) be continuous with respect to N, and let P be a positive a-convex combination XO~kPk of some Pk E IP(R). (a) P extends uniquely to a continuous P6 IP(R). (b) Each Pk extends uniquely to a continuous Pk ~ IP(R). =

(c) P=S~kPk. [] Lemma 2.3 [4, Proposition 18]. Let R be a regular ring, let N~ 1P(R), and let be the N-completion of R. Let Pc 1P(R) be continuous with respect to N, and let P be the continuous extension of P to IP(R). Then ker(P) is generated by a central idempotent in R. [] Lemma 2.4. Let R be a regular ring, let N, Pc IP(R), and assume that P is continuous with respect to N. Let R and R1 denote the N-completion and the P-completion of R. (a) The natural map R/ker(N)~R/ker(P) extends uniquely to a uniformly continuous function qS:R ~ R 1. Moreover, c~ is a ring morphism. (b) If P is the continuous extension of P to IP(R) and P1 is the natural extension of P to IP(RO, then P=P~cL (c) ker(q~) = ker(/3). Proof Ca) The existence and uniqueness of ~bare standard properties of completions. Since the ring operations in /~ and /~1 are continuous, we see that 4) is a ring map. (b) Diagram II shows the relationship of the functions in question, and we observe that the rectangle and the outer trapezoid commute. Thus P and P~ q5 agree on R/ker(N). Inasmuch as R/ker(N) is dense in R, we conclude from continuity that P = Pt ~.

Completions of Regular Rings c =

R/ker(N)

R/ker(P)

- -

=

,/~

R~"

233 p

~ [0, 13

Diagram Il

(c) follows from (b), because k e r ( P 0 = 0 . [] Definition. In the situation of Lemma 2.4, we refer to ~b as the natural map from R ~ R ~. Suppose we are given a regular ring R and a positive a-convex combination N = Z a k P k in IP(R). Let /~ denote the N-completion of R, and let /~k denote the Pk-completion of R. Each Pk is continuous with respect to N, hence we obtain the natural map Ok:R-~Rk. These ~bk induce a map ~ : R ~ I I R k, which we of course call the natural map. If N is the natural extension of N to IP(/~) and Pk is the continuous extension of Pk to IP(/~), then ]V=Y, ekPk by Lemma 2.2. As a result, r~ ker(Pk)= ker(N)=0. According to Lemma 2.4, ker(~k)= ker(Pk) for each k, whence c~ ker(q~k)=0. Therefore 0 is injective. Proposition 2.5. Let R be a regular ring, and let N = XekPk be a positive aconvex combination in IP(R). Let R denote the N-completion of R, and let Rk denote the Pk-completion of R. For each k, let Pk be the continuous extension of Pk to ~'(R). Then the natural map c~:R ~ II R k is an isomorphism if and only if ker(Pi)+ ker(fii)= for all i:~ j. Proof There is no loss of generality in assuming that ker(N)=0. Note that the natural map c~:R-~R~ is the composition of q~with the projection I1Rk--*R i. If q~ is an isomorphism, then clearly ker(~bi)+ker(~bi)=/~ for all i#:j, whence Lemma 2.4 shows that ker(Pi)+ ker(Pj.)~ R. Conversely, assume that ker(Pi)+ ker(PA = R for all i#:j. By_kemma 2.3, there exist central idempotents ekeR such that (1--ek)R=ker(Pk). Inasmuch as (1 -ei)R + (1 - e j ) R = R for all i+j, we see that the ekare pairwise orthogonal. If is the natural extension of N to IP(/~), then N = £~kPg by Lemma 2.2, and consequently C~(1--ek)/~=C~ ker(Pk)=ker(/f/)=0. Thus the annihilator of the ideal ®ekR is zero. Inasmuch as R is right self-injective (Theorem 1.3), it now follows as in [2, Theorem 18] that the natural map ~:R-~llekR is an isomorphism. For each k, ker(~bk)= ker(Pk) by Lemma 2,4, hence ~bk induces a map 0~: e f l ~ t~/ker(Pk)~R k. These Ok induce a map 0: HekI~-~HRk, and we observe that 0~p= ~b. Thus ~ is an isomorphism if and only if 0 is an isomorphism, hence it suffices to show that each 0~ is an isomorphism. Fix an index k, and note that Okis injective. Let P* denote the natural extension of P, to IP(Rk). Observing that the natural map p,: R ~ R R is the composition of Ok with the map R--*R-~e,I~, we see that P,(x)=P~Pk(X)--P~O~(e;O for all x e R . By continuity, it follows that P~(x)= P*O,(ekx) for all x~ R. Given any x~/~,, there is a sequence {x,}__ZRsuch that p,(X.)=O~(ekx,)~X in the P]'-metric. Inasmuch as P,(ey,c,.- e~.) = Vt(Ok(e~x,,)- Ok(ekX,)) for all m, n, we find that the sequence {e~x,,}C_R is Cauchy in the Pk-metric. For all j:~k, ekR <= (I-ei)/~=ker(P~), whence i~=~tkP~ on e,/~. Thus {e~,} is also Cauchy in the /q-metric, hence by completeness there must be some y~/~ such that ekx.~e~v

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K . R . Goodearl

in the/q-metric. Now P*(Ok(ekX,)-- Ok(eky))= Pk(ekX, -- eky) = ~k 1l~(ekX" _ edY) for all n, and consequently Ok(ekX,)--*Ok(eky) in the P*-metric. Thus Ok(eky)=x, whence 0 k is surjective. Therefore each Ok is an isomorphism, so that 0 and ~b are isomorphisms. [] Corollary 2.6. Let R be a regular ring, and let N = ZctkPt be a positive a-convex combination in IF(R). Let R denote the N-completion of R, and let Rk denote the Pk-completion of R. If ker(Pi) + ker(P) = R for all i 4:j, then R ~ H R k. [] In Corollary 2.6, the condition "ker(P 3 + ker(P) = R for all i4:f' is not necessary for R ~ I l R k . For example, let R be the case t = 2 of [4, Example B], so that R is a simple regular ring and IF(R) has two distinct extreme points P~, Pz. Note that ker(PO+ker(Pz)=O. However, if N=(1/2)Pl+(1/2)P2, then the natural map R ~ R ~ x/~2 is an isomorphism, as we shall see from Corollary 4.4. 3. Faces Definition. Let K be a convex subset of a real vector space. A face of K is a convex subset F (possibly empty) with the following property: If x, ye K and 0 < • < 1 with e x + ( 1 - ~ ) y ~ F , then x e F (and y e F , by symmetry). Note that a point x e K is an extreme point of K if and only if {x} is a face of K. Clearly any intersection of faces of K is again a face. Thus for any X__cK there is a smallest face of K which contains X, called the face generated by X (in K). Lemma 3.1 [1, (1.9)]. Let K be a convex subset of a real vector space, and let X c=K. A point x e K belongs to the face generated by X if and only if there exist ye K and 0 < ~t< 1 such that o~x + (1 -ct)y ties in the convex hull of X. [] It turns out that faces in IF(R) are naturally related to the partial order on IRR I f < g if and only i f f ( x ) < g ( x ) for all xe R]. This relationship is most conveniently studied within the following set. Definition. Let R be a regular ring. The cone (in IRR) with base IP(R) is the set of all nonnegative real scalar multiples of members of IF(R). Proposition 3.2. Let R be a regular ring such that IF(R) is nonempty, and let X be the cone with base IP(R). (a) A map A : R - , [ 0 , ~ ) belongs to X if and only if A(x)=A(y)+A(z) for all x, y, ze R such that x R ~ - y R O z R . (b) For A, Be X, A <=B if and only if B - Ae X. Proof (a) The given condition is clearly satisfied by any Pc IF(R), hence by any A e X . Conversely, consider any A : R ~ [ 0 , ~ ) which satisfies the condition. Clearly A(e+ f ) = A(e) + A ( f ) for all orthogonal idempotents e, f e R. Given x, ye R, we have xR = x y R ~ zR for some ze R, whence A(xy)= A ( x ) - A ( z ) < A(x). In addition, y R ~ x y R 0 wR for some we R, whence A(x y) ~ A(y). If A(1)=0, then A(x)<_A(1)=O for all x e R , in which case A =OeX. If A(1)>0, then A/A(1)e IF(R), and again Ae X. (b) If B - A e X , then B-A>__O and so A < B . If A ~ B , then B - A maps R-~[0, Go). Since A and B both satisfy the condition in (a), so does B - A , whence B - A ~ X. []

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235

Corollary 3.3. Let R be a regular ring, Pe IP(R), Xc= IP(R). Then P lies in the face generated by X if and only if P < a Q for some a > 0 and some Q in the convex hull of X. Proof. If P lies in the face generated by X, then by L e m m a 3.1 we must have a P + ( 1 - a ) P ' = Q for some 0 < ~ < 1 , some P ' e IP(R), and some Q in the convex hull of X. As a result, P < c~- 1Q. Conversely, assume that P<=~Q for some a > 0 and some Q in the convex hull of X. In view of Proposition 3.2, we see that ~ Q - P = f l P ' for some f l > 0 and some P'e 1P(R). Evaluating this equation at the element l e R, we find that a - 1 =fl, whence a - 1 + a - 1fl = 1. Thus ? = e - 1 is a real number for which 0 < 7 < 1 and ? P + ( I - ? ) P ' = Q , from which we conclude that P lies in the face generated by X. [] Corollary 3.4. Let R be a regular ring, and let X be the cone with base IP(R). Let Q1, Q2e IP(R), and let F i denote the face generated by Qi. Then F 1 and F2 are disjoint if and only if Q 1 A Q2 = 0 in x . Proof If there exists P e F1 ~ F 2 , then by Corollary 3.3 we must have P < cq Q1 and P < ct2Q2 for some el, e2 > 0. Setting a = min {c~ l, e~ 1}, we obtain an element

a P e X such that Q1, Q2 > a P > 0 . Conversely, if Q 1 , Q 2 > a P > 0 for some e > 0 and some PeP(R), then by Corollary 3.3 we find that P e F lc~F2. [] Proposition 3.5. Let R be a regular ring such that IP(R) is nonempty, and let X be the cone with base IP(R). Then X is a lattice. For any A, Be X and any xe R, we have (a) (A /x B)(x)= inf{Z inf {A(xi), B(xi)}[xR "~ x l R O . . . O x , R } = inf{[inf {A(xl), B(x 0}] + [inf{A(x2), n(x2)}]lxR = x l R O x 2 R } . (b) (h v B)(x) = sup {Z sup {A(xi), B(xi) }lxR'~ x lR 0 . . . • x,R } = sup {[sup {A(xl), B(xt)}] + [sup {A(x2), B(x2)}]IxR =x1ROxzR}. Proof (a) For all x e R, set

C(x) = inf{Z inf {A(x~), B(xl) }lx R ~ x lR 0 . . . • xnR } C'(x)--inf{[inf{A(x0, B(x0}] + [inf{A(x2), B ( x 2 ) } ] l x R = x l R O x 2 R } . Obviously C(x)

C'(x) < [inf{A(y0, B(y0}] + [inf{A(y2), B(y/)}] = [A(xl) + . . . + A(xk)] + [B(Xk+ 1) + . . - + B(x,)]

= Z inf{A(xi), B(xi)}. Therefore C'(x)= C(x). Consider any x,y, z e R such that xR"~-yROzR. If y R ~ - y ~ R O . . . O y , R and zR"~-zlRO...OzkR, then xR~-y~ R O . . . O y , R O z l R O . . . O z k R and so

C(x) < [ ~ inf{A(y~), B(y~)}] + [Z inf{A(zj), B(zj)}].

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K.R. Goodearl

As a result, C(x)< CO')+ C(z). On the other hand, if x R ~ x l R O . . . O x , R, then by [6, Lemma 3.8] there exist decompositions x iR = x i i R O x i z R such that x [email protected] O x , I R ~ y R and x 1 2 R O . . . O x n 2 R ~ z R . Thus

C(y) + C(z)<=[X inf{A(xiO, B(xi0}] + [X inf{A(xi2 ), B(xi2)}] = X([inf{A(xil), B(xi 1)}] + [inf{A(xi:), B ( x j } ] ) =

simplex. Corollary 3.6. If R is a regular ring such that IP(R) is nonempty, then IP(R) is a Choquet simplex. Proof We know that IP(R) is a convex subset of the real vector space IRR, and we note that IP(R) is contained in the hyperplane {felRRlf(1)= 1 } which misses the origin. Let K* be the cone in IRn with base IP(R), and let =<* be the partial order on IRR induced by K*. As far as elements of K* are concerned, Proposition 3.2 shows that < * coincides with the natural partial order, whence Proposition 3.5 says that (K*, _<_*) is a lattice. According to [10, p. 60], it follows that IP(R) is a Choquet simplex. [] Definition. Let K be a compact convex subset of a locally convex, Hausdorff, linear topological space. Note that the intersection of any family of a-convex faces of K is again a a-convex face. Thus for any X __cK there is a smallest a-convex face of K containing X, called the a-convex face generated by X (in K). In general, we do not know whether the a-convex face generated by X coincides with the a-convex hull of the face generated by X, but we shall prove this is the case in ~(R). Lemma 3.'7. Let R be a regular ring with a rank function N, and assume that R is complete in the N-metric. Let J be a two-sided ideal of R which is closed in the N-metric, and let e 1, e2 .... be orthogonal idempotents in J. (a) Se, converges to an idempotent e~ J. (b) If @enR is essential in J, then e R = J and e is central. (c) J is generated by a central idempotent. Proof (a) The real numbers N(eO + , , , + N(e,) form a nondecreasing sequence, and N(el)+... + N(e,)= N(e~+.,, + e , ) < 1 for all n, hence the series SN(e,) must converge. As a result, given any e > 0 there must exist a positive integer K such that N(e,+... + e,) = N(e 0 +... + N(e,) < e whenever n >_k_> K. Thus the partial sums

Completions of Regular Rings

237

of the series 22e, form a Cauchy sequence with respect to the N-metric. By completeness, Xe, converges to some ee R, and of course e is an idempotent. Since e~ + . . . + e , lies in the closed set J for all n, we also have eeJ. (b) Inasmuch as (el + ... + e k ) e , = e , for all k>n, we obtain e e , = e , for all n, and consequently G e , R < e R < J. As a result, eR is essential in J, from which we infer that eR = J. Since eR is now a two-sided ideal in the semiprime ring R, it follows that e must be central in R. (c) According to [4, L e m m a 13], there exist orthogonal idempotents f t , f~ .... ~ R such that O f , R is essential in J. [] Proposition 3.8. Let R be a regular ring with a rank function N, and assume that R is complete in the N-metric. Let P, Qe 1P(R) be continuous with respect to N. If ker(Q)< ker(P), then P lies in the a-convex hull of the face generated by Q.

Proof For n = 1, 2 ..... set J,,= {x~RIP

P(x)-e such that h R f < k(xR)f for some f e F . Note that e f eF, and consider any 9eF such that 9 ~ e f Then n(xR)o<.,~mR9 and hR9 < k(xR)#, whence Lemma 6.6 says that nNo(gx)

nriiI

n~l

238

K . R . Goodearl

P(1)= 1. Therefore P = ~ P(h,)P, is a a-convex combination of these P,, and consequently P lies in the a-convex hull of the face generated by Q. [] Proposition 3.8 may fail if R is not complete. For example, let R be the case t = 2 of [4, Example B]. Then R is a simple regular ring, and 1P(R) has two distinct extreme points P, Q. Note that P and Q are each continuous with respect to N=(1/2)P+(1/2)Q. Since R is simple, we have ker(P)= ker(Q)=0. However, the a-convex hull of the face generated by Q is {Q}, which does not contain P. Likewise, Q does not lie in the a-convex hull of the face generated byP. Theorem 3.9. Let R be a regular ring, Pc IP(R), X c=IP(R). Then the following conditions are equivalent: (a) P lies in the a-convex face generated by X. (b) P lies in the a-convex hull of the face generated by X. (c) P is continuous with respect to some Q in the a-convex hull of X. Proof (b)=>(a) is clear. (a)=>(c): Let Y denote the set of those members of IP(R) which are continuous with respect to some member of the a-convex hull of X. We first show that Y is a a-convex face of IP(R). Suppose we are given a a-convex combination P'=ZCtkP k with each Pk~ Y. There is no loss of generality in assuming that all ~k>0. There exist Q,, Q2,.-. in the a-convex hull of X such that each Pk is continuous with respect to Qk. Then Q'=2,~kQ k lies in the a-convex hull of X, and we claim that P' is continuous with respect to Q'. Given e > 0 , choose n such that

~k

~ k~n+

1

6k>0 such that Qk(X)<6 k implies Pk(X)

k=n+

~ k=n+

i

~k

Therefore P' is continuous with respect to Q', and so P'~ Y. Thus Y is a-convex. Now consider P , , P2 e IP(R) and 0 < c~< 1 such that c~P, + (1 - c~)P2 = P'e Y Then P' is continuous with respect to some Q' in the a-convex hull of X. Inasmuch as P 1 N e - ' P ' , we infer that P1 is continuous with respect to Q', whence P , e Y Thus Y is a face of IP(R). Clearly X _ Y, hence Y must contain the a-convex face generated by X. Therefore P e Y. (c)=~(b): Let/~ denote the Q-completion of R, and let ~b:R--./] be the natural map. Let (~ be the natural extension of Q to IP(/~), and let P be the continuous extension of P to IP(/~). There is a a-convex combination Q=2cckQk for suitable Qke X. According to Lemma 2.2, each Qk has a continuous extension (~ke IP(/~), and Q = ~kQ,.

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By Lemma 2.3, there exist central idempotents e~, e 2 .... e/~ such that (1 - ek)/~ = ker((~k). Note that

ann(XekR) = c~(1 -- ek)/~ = C~ker(Qk) = ker((~)=0. Set f l = e l and fk=ek(1 --e0(1 --e2)...(1 --e k_ i) for all k > 1. The fi, are orthogonal central idempotents in /~, hence Lamina 3.7 says that Sfk converges to some central idempotent fE/~. Observing that eke f i r +... + fkR < f R for all k, we see that ann(fI~)=O, whence f = 1. Therefore Sfk--' 1. Now set I={k>OIP(fk)#O}. For kel, we may define PkelP(R) by the rule Pk(x) = P(fkx)/P(fk). As in the proof of Proposition 3.8, P is a a-convex combination of these/5. Setting Pk = Pkd?e 1P(R)for all ke I, we see that P is a a-convex combination of these Pk" TO get P in the a-convex hull of the face generated by X, it thus suffices to show that each Pk lies in the a-convex hull of the face generated by X. _ Given ke I, we have Pk<[1/P(fk)]P. Since P is continuous with respect to Q, it follows that fir is continuous with respect to (~. In addition, we know that Ok is continuous with respect to Q. Inasmuch as fk ~ ~ ekR, we have ker((~k) = (1 -- ek)/~ =<(1 -- f~)/~ _<--ker(i~k), whence Proposition 3.8 shows that Pk lies in the a-convex hull of the face generated by (~k- Utilizing Corollary 3.3, we infer that Pk lies in the a-convex hull of the face generated by Qk, hence in the a-convex hull of the lace generated by X. [] Theorem 3.9 may not be strengthened to say that P lies in the face generated by X. In fact, it is possible to have P, QE IP(R) such that P and Q are continuous with respect to each other, yet neither lies in the face generated by the other. For example, let F 1, F z .... be fields, and set R =IIF,. Let ek denote the identity o f f k, and let Pk be the unique pseudo-rank function in ~'(R) with kernel (1 - ek)R. For n = 1, 2..... set ~2. = f12.- 1 = 1/(1 + 2") and Ctz._ ~= f12. = 1/(2" + 22n), and note that e2.- 1 +cta.=fl2.- 1+f12. = I/2". Thus P=SekP k and Q=ZflkP k are rank functions on R. Each Pk lies in the face generated by Q, hence P lies in the aconvex face generated by Q. According to Theorem 3.9, P is continuous with respect to Q, and likewise Q is continuous with respect to P. For all n, P(e2.)=c~2.= 2"fl2.= 2"Q(e2.), from which we see that P;~o~Q for all ct>0. By Corollary 3.3, P does not lie in the face generated by Q, and likewise Q does not lie in the face generated by P.

4. Decomposition of Completions Definition. Let K be a convex subset of a real vector space, let {x~}_ K, and for each i let F i be the face generated by x i in K. If the F~ are pairwise disjoint, we shall say that the xi are facially independent (in K). If not, we shall say that the xl are facially dependent. Lamina 4.1. Let R be a regular ring with a rank .function N, and assume that R is complete in the N-metric. Let P1, P2 e IP(R) be continuous with respect to N. If k e r ( P 0 + ker(P2)4:R, then P1 and P2 are facially dependent. Proof. According to Lemma 2.3, each ker(P~) is generated by a central idempotent. Thus there is a central idempotent eaR such that (1-e)R=ker(P1)+ ker(P2), and e#:0 by hypothesis. Note that P1(e), P2(e)>0.

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Suppose that P1 and P2 are facially independent. Then by Corollary 3.4, P1 A P2 = 0 in the cone with base IP(R). Now (Pa A Pz)(e)=0, so by Proposition 3.5 there exists a decomposition

eR= x l R ® x z R such that [inf{Pl(Xl), P2(x0}] + [inf{P1(x2), P2(x2)}] < rain {P~(e)/4, P2(e)}. We cannot have each inf{Pl(xi), Pz(xi)} = Ps(xi) for the same j, because Ps(xl)+ Pj(x2) > rain {Pl(e), Pz(e)}. Thus Xl, x2 4=0 and we may renumber so that Pl(xl)< P2(x 0 and Pz(x2)

Pl(et 1) + Pz(el 2) = [inf{Pl(x 1), P2(x 1)}] + [inf{PI(x2), P2(x2)}] < Pl(e)/4, hence Pl(el 1), P2(el 2) < Pl(e)/4. Since e12 ~ 0 and (P1/x Pz)(e12)=0, we may proceed as in the last paragraph to find nonzero orthogonal idempotents ezl.e22~R such that ezl+ezz=e12 and Pl(ezl),P2(e22)

f

Pl(e)/2 "+1, which is impossible. Therefore P1 and P2 must be facially de-

n=l

pendent.

[]

Proposition 4.2. Let R be a regular ring, let N e IP(R), and let R be the N-completion of R. Let P1, P2e 1P(R) be continuous with respect to N, and let P~ denote the continuous extension of Pi to IP(/~). Then k e r ( P 0 + ker(/52)=/~ if and only if PI and P2 are facially independent in IP(R). Proof If ker(P0+ker(Pz)+/~, then by Lemma 4.1 P1 and /52 are facially dependent in IP(iff), from which we infer that P1 and P2 are facially dependent in IP(R). Conversely, assume that Pa and P2 are facially dependent in IP(R). In view of Corollary 3.3, there exist P~IP(R) and ~ > 0 such that P<~PI,~P2. Since Pt is continuous with respect to N, it follows that P is continuous with respect to N. Thus P has a continuous extension/56 IP(R), and by continuity we see that P < ~P~, ~/52. Therefore ker(Pl)+ ker(P2)=< ker(/5)

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Theorem 4.3. Let R be a regular ring, let N~ IP(R), and let R denote the Ncompletion of R. Assume that N is a positive a-convex combination of some P~_6IP(R), and let Rk denote the Pk-completion of R. Then the natural map R ~ I 1 R k is an isomorphism if and only if the Pk are facially independent in ~(R). Proof Propositions 2.5 and 4.2. [] Corollary 4.4. Let R be a regular ring, let N61P(R), and let R denote the Ncompletion of R. Assume that N is a positive a-convex combination of distinct extreme points Pk~ IP(R), and let Rk denote the Pk-completion of R. Then R ~ IIR k. Proof The face generated by each Pk is {Pk}, whence the Pk are facially independent. [] In general, while the completion of a regular ring R with respect to a a-convex combination L'~kPk in IP(R) need not be isomorphic to the direct product of the Pk-completions of R, it is at least isomorphic to a subdirect product of the Pkcompletions, which follows from the next theorem. Theorem 4.5. Let R be a regular ring, let N~ IP(R), and let R denote the Ncompletion of R. Let P~ IP(R) becontinuous with respect to N, let fi be the continuous extension of P to IP(R), and let R1 denote the P-completion of R. Then the natural map c~:R ~ R 1 is surjective, and ker(q~)= ker(P). Proof Without loss of generality, we may assume that ker(N)=0. We already have ker(~b)= ker(P), by Lemma 2.4. According to Lemma 2.3, there is a central idempotent ee/~ such that (1-e)/~ = ker(P). Let N be the natural extension of N to lP(/~). Observing that e=~0, we see that ]q(e) ~0. Thus we may define N*~ IP(/~) according to the rule N*(x)= 1V(ex)/IV(e). Since N* < [1//V(e)]N, N* is continuous with respect to/V. Inasmuch as ker(N*)= (1-e)/~=ker(/3), we now see by Proposition 3.8 and Theorem 3.9 that N* is continuous with respect to 15. Now let 151 denote the natural extension of P to IP(/~l), and recall from Lemma 2.4 that/514~ = P. Given x~/~ t, there exists a sequence {x,,} c__R such that (a(x,)~x in the/Sj-metric. Since (1-e)/~=ker(/5)=ker(qS), we see that c~(ex,,)---,x as well. Now P(ex,,-ex,)=Pl(dp(eXm)-(~(ex,)) for all m, n, hence the sequence {ex,} must be Cauchy in the P-metric. Inasmuch as N* is continuous with respect to/5, it follows that {ex,} is also Cauchy in the N*-metric. In addition, Iq(ex,,-ex.)= N(e)N*(exm-ex,) for all m, n, whence {ex.} is Cauchy in the bT-metric. By completeness, {ex,} must converge (in the N-metric) to some y~ iff.Since ~bis continuous, ~ ( e x , ) ~ ( y ) in the/51-metric, and consequently (p(y)=x, Therefore

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tinuous pseudo-rank functions to extend to the N-completion, as the following example shows. Choose a field F, set Fk=F for k = t, 2,..., and let/~ = H F k. Set J = ® F k, and let R be the F-subalgebra of/~ generated by 1 and J. There is a unique rank function N k on F k, and we define a rank function N on/~ according to the rule lq(X)=SNk(Xk)/2 k. Letting N denote the restriction of ~7 to R, we see that/~ is the N-completion of R, and that N is the natural extension of N to IP(/~). Since R/J~-F, there is a unique Pc 1P(R) for which ker(P)=J. If x,, denotes the sum of the identity elements of F~ ..... F,, we observe that x,--*l in the Nmetric. However, each P(x,)=0, hence P(x,) does not converge to P(1). Thus P is not continuous with respect to N. Now let M be any maximal ideal of/~ which contains J. There is a unique P'~ IP(/~) for which ker(P')= M. Observing that ker(P'lR)= M n R = J, we see from the uniqueness of P that P'[R= P. Therefore P extends to IP(/~). Thus we are faced with the following problem: Given a regular ring R and some N~ IP(R), which pseudo-rank functions on R extend to pseudo-rank functions on the N-completion of R? The natural map qS:R~/~ induces a continuous map ~b*:IP(/~)---,IP(R), and our problem is to determine the range of qS*. Note that qS*(IP(/~)) is a continuous image of a compact space and so is compact. Thus th*(IP(/~)) is closed in IP(R). It is clear that the face generated by N in IP(R) is contained in qS*(IP(/~)), and we shall prove that in fact ~*(IP(/~)) is the closure of the face generated by N. In order to prove this, we must develop some results about closures of faces. As with the decomposition problem, we first consider what happens in/~, which by Theorem 1.3 is self-injective. Since extreme points play a crucial role in this process, we devote the next section to studying the extreme points among the pseudo-rank functions on a regular self-injective ring.

6. Extreme Points

Theorem 6.1. Let R be a regular, right self-injective ring, and let K be a twosided ideal of R. If R/K contains no uncountable direct sums of nonzero right ideals, then R/K is right self-injective. Proof Let I be a nonzero right ideal of R/K, f ~ HomR/K(I, R/K). Since R/K has no uncountable direct sums of nonzero right ideals, we infer that R/K must have a countably generated right ideal d which is essential in I. Choose generators x~, x2 .... for d. According to [4, Lemma 12], there exist orthogonal idempotents el, e2 ~ R such that O)e,R=Zx,R. Then e 1, e2 .... are orthogonal idempotents in R/K, and O~,(R/K)=Z,Y.(R/K)=J is essential in I. For each n, f(~,)= f(E,)~,, hence f(~',)= )7 for some y~ Re,. There is a map g: O)e,R~R such that 9(e.)=y n for all n. Since RR is injective, it follows that there exists y~ R such that Yen = Y, for all n. As a result, yen= ; . = f(E.) for all n, hence left multiplication by ; agrees with f on O~n(R/K). Inasmuch as I/[GEn(R/K)] is a singular right (R/K)-module while (R/K)~IK is nonsingular, we conclude that f must be given by left multiplication by ~7. Therefore R/K is right self-injective. [] ....

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Corollary 6.2. Let R be a regular, right self-injective ring, and let Pe IP(R). Then R/ker( P) is right self-injective. Proof Since P induces a rank function on R/ker(P), we infer that R/ker(P) contains no uncountable direct sums of nonzero right ideals. [] Definition. For any ring R, we use B(R) to denote the Boolean algebra of central idempotents in R. The space of maximal ideals of B(R) we call the Boolean spectrum of R, denoted BS(R).

Definition. A ring R is directly finite provided all one-sided inverses in R are two-sided, i.e., xy = 1 implies yx = 1. As noted in [4, p. 268], any regular ring with a rank function must be directly finite. Theorem 6.3. Let R be a regular, right self-injective ring. For any Pc IP(R), the following conditions are equivalent: (a) P is an extreme point of IP(R). (b) There exists a unique rank function on R/ker(P). (c) ker(P) is a maximal two-sided ideal of R. (d) ker(P) is a prime ideal of R. (e) B(R)c~ker(P)~ BS(R).

Proof. Note that P induces a rank function P' on R/](er(P), so that R/ker(P) is directly finite. Also, R/ker(P) is right self-injective, by Corollary 6.2. (a)=~(d): Clearly P' is an extreme point of IP(R/ker(P)), from which we infer that R/ker(P) must be indecomposable (as a ring). Since R/ker(P) is also regular and right self-injective, [2, Proposition 3] now shows that R/ker(P) is a prime ring. (d)=~(c): Since R/ker(P) is a prime, regular, right self-injective ring which is directly finite, it must be a simple ring [6, Proposition 1.4]. (c)~(b): The ring R/ker(P) is simple, regular, and directly finite, as well as right self-injective, In view of [2, Lemma 5', p. 832], we see that for any x, ye R~ ker(P), either x[R/ker(P)] is isomorphic to a submodule of y[R/ker(P)], or vice versa. Thus R/ker(P) satisfies the "comparability axiom" of [6]. According to [6, Corollary 3.15], there is a unique rank function on R/ker(P). (b)=~(a): Given any Q~lP(R/ker(P)), we see that (1/2)P'+(1/2)Q is a rank function on R/ker(P), whence (1/2)P'+(1/2)Q=P', and consequently Q = P ' . Thus IP(R/ker(P)) = {P' }. Now suppose that P = ~P1 + (i - ~)P2 for some 0 < ~ < 1 and some P1, P2eP(R). Observing that ker(P)=ker(P1)c~ker(P2), we see that P1 and P2 induce P'I, P~ ¢ P(R/ker(P)). As a result, P'l = P'2 = P', whence P1 -----P2 = P. Therefore P is an extreme point of IP(R). (d)~(e) is clear. (e)~(d): Since M=B(R)c~ker(P) belongs to BS(R), [3, Theorem 2.3] shows that MR is a prime ideal of R. According to [2, Corollary 7], any proper two-sided ideal of R which contains MR must be prime, whence ker(P) is prime. [] It is not hard to see that the implication (a)~(d) in Theorem 6.3 holds for any regular ring R. For if ker(P) is not prime, we infer that the P-completion of R is not prime and thus not simple, whence Theorem 1.4 shows that P is not an extreme point of IP(R). On the other hand, the implications (a)~(e) and (e)~(a) do not hold in general, as the following examples show.

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First, let R be the case t = 2 of [4, Example B], so that R is a simple regular ring and IP(R) has distinct extreme points P1, P2. Then P = ( 1 / 2 ) P 1+(1/2)P 2 is not an extreme point of IP(R), yet ker(P)=0 is a maximal two-sided ideal of R. For the next example, choose a field F. For n = 1, 2,..., let S, be the ring of all

2"x2"matricesoverF, andmapS,~S,+lalongthediagonal,

i.e.,x~(;

~).

If S denotes the direct limit of the S,, then as seen at the end of [6, Section D] there is a unique rank function N on S. Let T be the N-completion of S. By [6, Theorems 4.5, 2.4], T is a simple, regular, right and left self-injective ring which is not artinian. Let N denote the natural extension of N to IP(T). Since T is not artinian, it must have a proper essential right ideal J. In view of [4, Lemma 12], we may assume that J = Oe, T for some sequence el, e2,.., of orthogonal idempotents in T Now set f,=el + ... +e, for all n, let R be the Fsubalgebra of T generated by 1 and u f , Tf,, and note that R is a non-simple regular ring. Thus ifP denotes the restriction of N to R, ker(P)=0 is not a maximal two-sided ideal of R. Inasmuch as T is complete in the N-metric, we infer from Lemma 3.7 that f , ~ l . Given any xET, we have f, xf,~R for all n, and f, xf,~x. Thus R is dense in T in the N-metric, from which it follows that T is the P-completion of R. Since T is simple, we conclude from Theorem 1.4 that P is an extreme point of IP(R).

Definition. A module A is subisomorphic to a module B, written A ~ B, provided A is isomorphic to a submodule of B. For any nonnegative integer n, we use nA to denote the direct sum of n copies of A. Theorem 6.4. Let R be a directly finite, regular, right self-injective ring, and let Me BS(R). (a) There is a unique PM~ IP(R)for which B(R)c~ker(PM)= M. (b) PM(x)=inf{m/nln(xR)e<~mRe for some e~B(R)-M} for all x~R. (c) PM(x)= sup {h/k[hRe <~k(xR)e for some e~ B(R)- M} for all x~ R. Proof. (a)(b) By [5, Lemma 12.3], the rule (b) defines a pseudo-rank function PM~IP(R). It is clear that M ~ ker(PM), whence B(R)c~ker(PM)=M. Now consider any Q~ 1P(R) such that B(R)c~ker(Q)= M. According to Theorem 6.3, ker(Q) and ker(PM) are maximal two-sided ideals of R. By [3, Theorem 2.3], MR is a prime ideal of R, whence [2, Theorem 6] shows that the two-sided ideals in R/MR are linearly ordered. Therefore R/MR has exactly one maximal twosided ideal, from which we obtain ker(Q)--ker(PM). Using Theorem 6.3 again, there is a unique rank function on R/ker(PM), hence we infer that Q = PM- Therefore PM is unique. (c) Let x e R, and set p = sup{h/kthRe < k(xR)e for some ee B(R)- M}. Suppose we have hRe <~k(xR)e for some es B ( R ) - m and n(xR)f <~mRf for some f e B ( R ) - M . Then e f C M and so e f 4:0. Also, hnRef

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245

k(xR)e < (h + 1)Re and (h + 1)R(1 - e) < k(xR) (1 - e). Inasmuch as (h + 1)/k> p, we must have 1 - e e M, whence e 6 M. As a result, PM(X)< (h + 1)/k < p + (1/k). Therefore Pu(x)=p.

[]

Theorem 6.5. Let R be a directly finite, regular, right self-injective ring. Let X be the set of all extreme points of IP(R), with the relative topology. Then the rule P~-~B(R)~ker(P) defines a homeomorphism of X onto BS(R). Proof. In view of Theorem 6.3, the rule ¢(P)=B(R)nker(P) defines a m a p ¢:X~BS(R). Theorem 6.4 shows that ¢ is injective, and in combination with Theorem 6.3, it also shows that ~b is surjective. There is a basis of BS(R) consisting of clopen sets, each of which has the form Y~={MeBS(R)[eeM} for eeB(R). Now c~-I(Y~)={PeXIP(e)4=O}, which is a relative open set in X. Thus ~b is continuous. There is a subbasis for X consisting of sets of the form {PeX]P(x)a}, for xeR, aelR. To show that ~b is an open map, it suffices to show that ¢ carries any such set to an open set in BS(R). First consider the set Y = {PeXIP(x)

W = {eeB(R)[n(xR)e

for some

rain

In view of Theorem 6.4, we see that for any PeX, P(x)a}) is open. Therefore ¢ is an open map, hence a homeomorphism. [] F o r a directly finite, regular, right self-injective ring R, Theorems 6.3 and 6.4 give a complete description of the extreme points of 1P(R). In turn, every pseudorank function on R can be described as an integral of these extreme points, for the proof of which we need the following lemmas. L e m m a 6 . 6 . Let R be a regular ring, PelP(R). Let x t ..... xn, Yl ..... ykeR. If x ~RO... • x,R <~Y t RO... Oyk R, then P(x 1) + . . . + P{x,) <=P{Yl) +... + P{Yk). Proof. We have x l R G . . . Gx,RGA_~ [email protected] OykR for some A. According to [6, L e m m a 3.8], there exist decompositions x i R = A , G...(~Ai~ (i= 1..... n) and A=A~+IAO...OA,+I. k such that A l j O . . . G A , + L j ~ y j R for j = l ..... k. F o r each i= 1..... n, there exist orthogonal idempotents ell ..... e~keR such that ( e , + •.. +eik)R=x~R and each e~jR= A~j. For each j= 1..... k, there exist orthogonal idempotents f l j ..... f ~ + l , f R such that (flj+...+f,+l,i)R=yjR and each f i j R - Aij. Thus

i=1 k

= Y, E P(I jl

i = l j =I

[since

j=li=l k n+l k E E P(fu)---- E P(fli+'"+f.+l,J) j = l i=1 j=l

=P(Y~)+... + P(Yk). []

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In particular, Lemma 6.6 shows that if R is a regular ring, P e IP(R), x, y e R, and n(xR)<~k(yR) for some positive integers n, k, then nP(x)

Proof Let x e R , and consider a positive integer n. For k = l , . . . , n, set ek= v{eeB(R)ln(xR)e

[]

Definition. Recall that a probability measure on a compact Hausdorff space X is a (nonnegative) regular Borel measure I~ such that /a(X)= 1. Lemma 6.8. Let X be a compact Hausdor[f totally disconnected space, and let la be a nonnegative finitely additive measure on the Boolean algebra of clopen subsets of X such that la(X)= 1. Then la extends uniquely to a probability measure on X. Proof Let Wdenote the collection of all compact subsets of X. For each Ae W, define 2(A) = inf{#(B)lA c B, B clopen }. This defines a monotone map 2: W--, [0, 1], and the following properties of 2 are routinely verified: (a))~(AwB)<__2(A)+)c(B) for all A, Be W. (b) 2(A uB) = 2(A) + 2(B) for all disjoint A, Be W. (c) 2(A) = inf{2(B)LA ~ interior(B), Be W } for all Ae W. These properties show that 2 is a regular content on X [7, pp. 231,237]. According to [7, Theorems E, A, pp. 234, 237], 2 induces a regular Borel measure/a' on X such t h a t / / ( A ) = 2(A) for all As W. It is clear that/~' is a probability measure on X which extends p. Uniqueness follows from the regularity of probability measures. [] For use in the next theorem, we must consider pseudo-rank functions on the Boolean algebra B(R), for certain regular rings R. This is allowable because B(R), when made into a ring in the usual manner, is regular. Care must be taken, however, because these ring operations may not coincide with those from R. The two multiplications always coincide, and the two additions at least coincide when applied to orthogonal idempotents. Since this is all that occurs in the definition of a pseudo-rank function, we see that the operations from R may be used when checking whether a map from B(R)~[0, 1] is a pseudo-rank function on B(R). As a result, we note that the restriction of any pseudo-rank function in IP(R) to B(R) yields a pseudo-rank function in IP(B(R)).

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247

Theorem 6.9. Let R be a directly finite, regular, right self-injective ring. (a) For each Me BS(R), there is a unique PMe IP(R)for which B(R)c~ker(P~) = M. (b) Given N~ 1P(B(R)), there exists a unique probability measure p on BS(R) such that p({MeBS(R)Ieq~M})=N(e) for all eeB(R). Moreover, the rule P(x)=

PM(x)dp defines a pseudo-rank function P~ IP(R) which extends N. M~BS(R)

(c) Every P~ IP(R) may be uniquely represented as in (b). Proof (a) Theorem 6.4. (b) Recall that the rule e~{M~BS(R)IeCM } defines an isomorphism of B(R) onto the Boolean algebra B' of clopen subsets of BS(R). Thus the rule p'({Me BS(R)ie¢M})=N(e ) defines a nonnegative finitely additive measure p' on B' such that p'(BS(R))= 1. By Lemma 6.8, p' extends uniquely to a probability measure # on BS(R). Given x e R, the evaluation map Q~+Q(x) is a continuous map of IP(R)--*[0, 1]. In view of (a) and Theorem 6.5, the rule M~-~PM defines a continuous map of BS(R)~IP(R), whence the composite map M~-~PM(x) is a continuous map of BS(R)~ [0, 1]. Since/t is a Borel measure, this map is measurable, hence we can form P(x) = S PM(x)dp. It is straightforward to check that this defines a pseudoMeBS(R)

rank function P c IP(R). For any eeB(R), the set X = {M~BS(R)LeCM } is a clopen subset of BS(R) such that p(X) = N(e). Observing that Pu(e) = 1 for all M e X and P~t(e)= 0 for all M 6 B S ( R ) - X , we conclude that P(e)= N(e). Therefore P extends N. (c) Let N be the restriction of P to IP(B(R)), and construct Q6 IP(R) from N as in (b). Then P and Q agree on B(R), whence Lemma 6.7 shows that P = Q . Uniqueness is clear. [] Corollary 6.10. Let R be a directly finite, regular, right self-injective ring. Then the restriction map ~a: IP(R)~IP(B(R)) is an affine homeomorphism.

Proof We already know that ~b is affine and continuous. Lemma 6.7 shows that ~b is injective, and Theorem 6.9 shows that ~b is surjective. Since ~(R) is compact, we conclude that ~b is a homeomorphism. []

7. Extending Pseudo-Rank Functions to Completions The main theorem of this section shows that for a regular ring R and N, P c IP(R), P extends to a pseudo-rank function on the N-completion of R if and only ifP lies in the closure of the face generated by N. By analogy with Theorem 3.9, we might expect this to be equivalent to P lying in the closed face generated by N (i.e., the smallest closed face containing N). This does hold when R is directly finite self-injective (Theorem 7.1), but not in general, as the following example shows. Choose a field F, and for n = l, 2 .... let R, be a copy of the ring of all 2 × 2 matrices over F. Set J = OR,, which is a two-sided ideal of HRn. Let C denote the subring of IIRn consisting of all constant sequences of diagonal matrices, so that C ~ F x F. Then R= C + J is a subring of I1Rn, and we observe that R is regular. Note also that R/J-~ F x F.

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For n = 1, 2 ..... let e. denote the identity matrix in R,, which is a central idempotent in R. Inasmuch as R/(1 -e.)R ~ R., which has a unique rank function [4, Proposition 2], there exists a unique pseudo-rank function P.~ IP(R) for which ker(P.)=(1-e.)R. Set N =

~ P./2", and note that N is a rank function on R. n=l

We shall show that the closure of the face generated by N in IP(R) is properly contained in the closed face generated by N. Let W, X, and Y denote the face, the a-convex face, and the closed face generated by N in ~'(R). According to Theorem 3.9, X equals the a-convex hull of W. As a result, X is contained in the closure W, and consequently X = VV.Thus it suffices to show that X is properly contained in E L e t f d e n o t e the constant sequence {(~

~),(;

;),...}, which is an idempotent

in R. Recalling that R/J~_F ×F, and noting that f and 1 - f map to nonzero idempotents in R/J, we see that R/(J + f R ) ~ F and R/(J + ( 1 - f ) R ) ~ F. Consequently, there exist unique pseudo-rank functions Q1, Q2c IP(R) for which ker(Q 1) = J + f R and ker(Qz) = J + (1 - f)R. Note that any P6 IP(R) for which P(J) = 0 must be a convex combination of Q1 and Q2. We also observe that the sequence {P,} converges to (Q1 +Q2)/2 in IP(R). Given any Pe IP(R), we claim that there exist nonnegative real numbers ~, fl such that

a+fl+ ~ P ( e . ) = l

and

n=l

P=~Qt+flQ2+ ~ P(e.)P.. n=l

Since this is clear when 2~P(e.)=0, consider the case when S P ( e . ) = 7 > 0 . In this case, we obtain Q = Z[P(e.)/y]P. in IP(R),and we note that 7Q < P. According to Proposition 3.2, P - 7 Q = y'Q' for some 7' > 0 and some Q'~ IP(R). Clearly Q'(J)=0, hence Q'=a'Q1 +fl'Q2 for some ~', fl'>0. Setting ~=y'~' and fl=7'fl', we obtain P = aQ 1+ flQ2 + ZP(e.)P. and a + fl +,FP(e.) = 1, as required. Now { P~IP(R)I .=1 ~ P ( e , ) = l } is a a-convex face of IP(R)which contains N, hence it must contain X. On the other hand, if P~ P(R) and ZP(e.) = 1, then we see from the results above that P = XP(e.)P., so that P is a a-convex combination of the P.. Since the P. all belong to X, so must any such P, Thus X = {PeP(R),.=1 ~ P(e.)=l}. For n = t , 2 ..... we see that e.fR~-e.(1-f)R, hence 2(e.fR)~e.R, and consequently 2P(e.f)= P(e.) for any PC IP(R). In particular, P.(e.f)= P.(e.)/2= 1/2. Given any P~X, we have seen that P=,FP(e.)P., from which we now obtain P ( f ) = t/2. Thus X is contained in the set {PE IP(R)IP(f)= t/2}. Since this set is closed in IP(R), it must contain .~ as well. Now consider any P~ IP(R) for which P ( f ) = 1/2. Writing P=~Q1 +flQz +XP(e.)P. as above, we obtain

or+ ~ P(e.)/Z=P(1- f)=l/Z=P(f)=fl+ ~ P(e.)/2, n=l

n=l

Completions of Regular Rings

249

hence a=fl. Thus P=a(Ql+Oz)+ZP(e,)P,. Inasmuch as PR~(Qt+Q2)/2 in 1P(R), it now follows that the pseudo-rank functions Tk=2aPk+~,P(e,)P, must converge to P. Since each Tke X, we find that P e X. Therefore )~ = {Pc IP(R)IP(f)= 1/2}. Since Y is a closed face of IP(R) which contains X, we have 1~ ~ Y. In particular, all P,~ Y. Observing that (Q1 +QE)/2~R, we have (Q1 +Q2)/2 in the face Y, and consequently Q1, Q2e Y. Now Y is closed and convex, hence a-convex. Inasmuch as every P e IP(R) is a a-convex combination of Q1, Q2, and the P,, we conclude that Y = IP(R). Therefore the closed face generated by N is IP(R), while the closure of the face generated by N is {P~ IP(R)[P(f)= 1/2}, which is a proper subset of IP(R). Theorem 7.1. Let R be a directly finite, regular, right self-injective ring. Let V c=IP(R), and set ker(V)= n{ker(N)[Ne V}. If W denotes the closure of the face generated by V in IP(R), then W = {Pc IP(R)I ker(V)c__ ker(P)} = {Pc IP(R)IB(R)n ker(V)~ ker(P)}.

In addition, W equals the closed face generated by V. Proof Set X = {Pc IP(R)I ker(V)__c ker(P)} and Y = {P~IP(R)IB(R)nker(V)~ ker(P)}. If V is empty, then k e r ( V ) = R and so X and Y are empty, in which case W = X = Y. Thus we may assume that V is nonempty. It is clear that X and Y are closed faces of IP(R). Obviously V____X, and consequently W~X. Note that X ~ Y as well. Since Y is closed in IP(R), it is compact, hence the Krein-Milman Theorem [-9, p. 131] says that Y is the closure of the convex hull of its extreme points. Now W is closed and convex, so to get Y ____W it suffices to show that any extreme point P of Y belongs to W. Since Y is a face of IP(R), P is also an extreme point of IP(R). Thus M = B(R)n ker(P) is a maximal ideal of B(R), by Theorem 6.3. According to Theorem 6.4,

P(x)= inf{m/nln(xR)e < mRe for some ee B(R)- M} = sup{h/klhRe

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As a result,

P(x) - a < h/k < N o(gx)/ g g(g) = Pg(x) < m/n < P(x) + ~ . Thus IPo(x)-P(x)j < e for all g e F such that g < e f Therefore Pe(x)~P(x) for all x e R , so that P ~ P in IP(R). Since W is closed, we obtain P e W, as desired. Now all extreme points of Y belong to W, and consequently Y __cW. Therefore W = X = Y. Finally, W = X is a closed face of P(R), hence W must also be the closed face generated by V. [] Theorem 7.1 fails for non-self-injective rings. For example, let R be the case t = 2 of [4, Example B]. Then R is a simple regular ring, and IP(R) has two distinct extreme points P1,P2. Since R is simple, {Pe lP(R)lker(POC__ker(P)}=IP(R). However, the closure of the face generated by P1 is just {P1 }. Corollary 7.2. Let R be a directly finite, regular, right self-injective ring. Then the lattice of closed faces of IP(R) is anti-isomorphic to the lattice of ideals of B(R).

Proof Let X denote the lattice of closed faces of IP(R), and let Y denote the lattice of ideals of B(R). Given F e X , set ¢p(F)={eeB(R)IN(e)=O for all NEE}. Since q~(F) is the intersection of B(R) with the two-sided ideal c~{ker(N)lNeF} of R, we see that ~b(F) is an ideal of B(R). Thus we get a map q~:X~ Y, and it is clear that ~b reverses _. Given I e Y, set ~c,(I)= {Ne 1P(R)tN(e)=O for all ee I}, which is a closed face of IP(R). This gives us a map ~: Y ~ X , which also reverses __c Consider any F e X. Since F is the closure of the face generated by F, Theorem 7.1 shows that F = {Pc IP(R)IqS(F)C=ker(P)}, i.e., F=q~b(F). Now consider any l e Y, and note that I c__0~P(I). There is a collection W c=BS(R) such that c~W=I. For each M e W, Theorem 6.4 gives us a unique PMe IP(R) for which B(R)c~ ker(PM)= M. Since each such PMe~(I), we obtain q~tp(I)_C_c~{B(R)c~ ker(PM)lMe W } = c~W = I , so that qSq~(I)=1. Therefore q~ and ~p are inverse order anti-isomorphisms, hence lattice antiisomorphisms. [] Lemma 7.3. Let R be a regular rin 9, N, Q1 ..... Q,e IP(R). For each i, let h i denote the Qi-completion of R, and let c~i:R-~R i be the natural map. Let dp:R~ R1 x ... x R, be the map induced by the ~i. If N is continuous with respect to some positive convex combination of the Q~, then there exists N*e IP(R~ x ... x R,) such that N'q5 = N. Proof We are given that N is continuous with respect to some positive convex combination Q=a~Q~ +... +a,Q,. L e t / ~ denote the Q-completion of R, and let ~p:R~/~ be the natural map. Since N is continuous with respect to Q, it extends continuously to N e IP(/~) such that N ~ = N . In addition, each Q~ is continuous with respect to Q and so has a continuous extension Q~e tP(/~). For each i, we have the natural map O~:R-~R~ such that 0~tp= 4~. According to Theorem 4.5, 0i is surjective, and ker(0~)= ker((~). Thus by Lemma 2.3, there

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exists a central idempotent eic/~ such that (1 - el)~2= ker(O 3. Let O:R ~ R 1 x ... x/2, be the map induced by the 0 i, and note that 0~ = ~b. Set f l = e l and f i = e i ( 1 - e O ( 1 - e 2 ) . . . ( 1 - e i _ O for all i>1, so that fl ..... f, are orthogonal central idempotents in /2. Observing that el ..... e, c f i R + . . . + f,/2 = (j] +... + f,)/2, we see that

ann( f 1 +... + f ,) ~ c~(1 - e l)-R= ~ ker(0i) = c~ ker((~/)= ker((~)=0, whence f t + . . . + L = 1. Set / = {i137(ji)4:0}. For icI, we may' define NiclP(R ) by the rule Ni(x)= iq(fx)/N(fi). Inasmuch as f c el~2, we see that Ni(1-el) =0. Thus ker(0i)< ker(Ni), whence N i induces N*clP(/2i) such that N*Oi=Ni. Since ~ ~7(Jl)=/q(j~)+... + ieI

N ( f , ) = 1, the rule N*(x 1..... x , ) = ~ ,q(fi)N*(xi) defines a pseudo-rank function iel

N*c

IP(/21x ... x/2,).

For all xc/2,

N*O(x)=N*(O~(x) ..... * Oitx) ' • , 0,1x))= ~ N(fi)Ni i~I

= y~ ~(f,)Ni(x)= y, ~ 7 ( f , x ) = ~ 7 ( x ) . iel

iel

Therefore N*~b = N*O,p = N ~ = N, as required.

[]

Theorem 7.4. Let R be a regular ring, Pc IP(R), X = {Qi}---IP(R). For each i, let /2i denote the Qi-completion of R, and let $i:R'-+/2i be the natural map. Set S=H/2g, and let $:R--,S be the map induced by the ~ . Then P lies in the closure of the face generated by X if and only if P=P'd~ for some P'clP(S).

Proof Each/2i is a regular, right and left self-injective ring by Theorem 1.3, hence the same is true of S. According to [-11, Theorems 4.7, 5.1], it follows that S is directly finite. First assume that P=P'd) for some P'c IP(S). For each i, let p~:S-~/2~ be the projection, let Q~ be the natural extension of Q~ to IP(/23, and set Q'~--(2~p~. Note that Q'~b= Q~p~b= (~:b~ = Qi. Since Q~ is a rank function on/2,, we see that ker(Q'3 = ker(pi), whence c~ker(Q'3=OC=ker(P'), By Theorem 7.1, P' must lie in the closure of the face generated by the Q'~ in IP(S). As a result, we infer that P = P'ek must lie in the closure of the face generated by {Q}~b}- X in IP(R). Conversely, assume that P lies in the closure of the face generated by X, and set Y-- {P'ck]P'c IP(S)}_c_1P(R). Inasmuch as IP(S) is compact and the map P'~P'~b of IP(S)-~IP(R) is continuous, Y must be compact, hence closed in IP(R). Thus to get P c Y, it suffices to show that Y contains the face generated by X. Given any N in the face generated by X, we see from Corollary 3.3 that N <~zQ for some ~ > 0 and some Q in the convex hull of X. We may write Q as a positive convex combination of some Qt¢,)..... Q,,)._Let p : S - , R , t ) x ... ×/2i(,) be the projection, and note that p~b:R ~ R i ( I ) x . . . × Ri~,) is the map induced by ~ , , ) ..... ~b,,). Inasmuch as N is continuous with respect to Q, Lemma 7.3 shows that N*pc~=N for some N*elP(/2,,)× ... ×/2,,)). Setting N'=N*pclP(S), we have N'ck=N and so N c Y

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T h e r e f o r e Y c o n t a i n s the face g e n e r a t e d by X, h e n c e also the c l o s u r e o f t h a t face. T h u s P c g [] C o r o l l a r y 7.5. Let R be a regular tin 9, P, Q6 IP(R). Then P extends to a pseudorank function on the Q-completion of R if and only if P lies in the closure of the face generated by Q. []

References 1. Alfsen, E. M. : On the geometry of Choquet simplexes. Math. Scand. 15, 97--110 (1964) 2. Goodearl, K. R. : Prime ideals in regular self-injective rings. Canadian J. Math. 25, 829--839 (1973) 3. Goodearl, K. R.: Prime ideals in regular self-injective rings. II. J. Pure Appl. Algebra 3, 357--373 (1973) 4. Goodearl, K.R.: Simple regular rings and rank functions. Math. Annalen 214, 267--287 (1975) 5. Goodearl, K.R., Boyle, A. K. : Dimension theory for nonsingular injective modules. (to appear) 6. Goodearl, K.R., Handelman, D.: Simple self-injective rings. Communications in Algebra 3, 7 9 7 ~ 3 4 (1975) 7. Halmos, P. R. : Measure Theory. Princeton : Van Nostrand (t950) 8. Hatperin, I. : Regular rank rings. Canadian J. Math. 17, 709---719 (t965) 9. Keltey, J.L., Namioka, I.: Linear Topological Spaces. Princeton: Van Nostrand 1963 10. Phelps, R.R.: Lectures on Choquet's Theorem. Princeton: Van Nostrand 1966 1t. Utumi, Y. : On continuous rings and self-injective rings. Trans. American Math. Soc. 118, 158--173 (1965) 12. Von Neumann, J.: Continuous Geometry. Princeton: Princeton Univ. Press 1960

(Received August 20, 1976)