Completions of regular rings

...

79 downloads 906 Views 2MB Size

Math. Annalen 220, 229--252 (1976) - - © by Springer-Verlag 1976

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).

230

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). []

Completions of Regular Rings

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

232

K.R. Goodearl

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

234

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. []

Completions of Regular Rings

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)}].

236

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 llR@... 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 } ] ) =O}. The cone K* induces a partial order < * on E, where x < *y if and only if y - xe K*. If the subspace of E spanned by K is a lattice with respect to the partial order <*, then K is called a simplex. If, in addition, E is locally convex Hausdorff and K is compact, then K is called a Choquet

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 1. Then the h, are orthogonal central idempotents in R such that ~ h , R = u e , R=uJ,,. In view of L e m m a 3.7, Zh, converges to a central idempotent he R, and we claim that h = 1. Suppose not. Fix n for a moment. Since (1 - h ) R ~ J , =0, each nonzero submodule of(1 - h ) R must contain a nonzero element from the set X={xeRJP(x)>nQ(x)}. Thus (1 - h)R must have an essential submodule which is a direct sum of cyclic submodules with generators from X. Inasmuch as R has a rank function, this direct sum must be countable, hence we obtain a direct sum x i R O x z R G . . , which is essential in ( 1 - h ) R , and with each xkeX. Since R is right self-injective (Theorem 1.3), there exist orthogonal idempotents g~,g2 .... 6R such that gkR=xkR. According to L e m m a 3.7, 229k converges to a central idempotent which generates ( 1 - h ) R . Thus Xgk--, 1--h, and consequently P ( 1 - h) > nQ(1 - h). This inequality holds for all n = 1, 2,..., whence Q ( 1 - h ) = 0 . Since k e r ( Q ) < ker(P), we obtain P(1 - h) = 0, which contradicts the inequality P(1 - h) > Q(1 - h). Therefore h = t as claimed, i.e., 2Jh,--, 1. Set I = {n>OIP(h,)+-O}. For nel, we may define P , e IP(R) by the rule P , ( x ) = P(h,x)/P(h,). N o w h,R < e,R = J,, hence P,(x) = P(h,x)/ P(h,) < nQ(h,x)/ P(h,,) ~ nQ(x)/P(h,) for all xeR, i.e., P,<=[n/P(h,,)]Q. According to Corollary 3.3, P. lies in the face generated by Q. Given xeR, we have F,h,x-,x and so P(x)=XP(h,x). Since P(h.x)
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)0. If Q'(x) < 6, then Qk(X)< ~; ' Q'(x) < 6k for k = 1..... n, whence

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,.

Completions of Regular Rings

239

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.

240

K.R. Goodearl

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) 1. (c) Pl(e,1), Pz(e,2) ez2R > .... and that el 1 + e21 + . . . -t- enl = e - e,2 for all n. According to Lemma 3.7, the series Se, l converges (in the N-metric), whence the sequence {en2} converges as well, to some xe R. Since enzek2 = ek2 for all k > n, we find that enzX=X for all n. Thus P2(x)< P2(e,2)< P l(e)/2 "+ 1 for all n, whence Pz(X)=0, Since k e r ( P 2 ) < ( 1 - e ) R while x e e l z R < e R , it follows that x = 0 . Therefore e,2~0, so that Ze,1--*e. Inasmuch as P1 is continuous with respect to N, we obtain Pt(e) = ~ Pl(e,1) < n=l

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)
Completions of Regular Rings

241

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

Completions of regular rings

Recommend Documents