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Representation of regular rings of finite index
JOURNAL
OF ALGEBRA
39,
512-526
Representation
(1976)
of Regular ANDREW B.
Mathematics
Department,
of
University
Communicated Received
I.
Rings of Finite
Index
CARSON*
Saskatchewan,
Saskatoon,
S7N
OWO,
Canada
by I. W. Herstein July
20, 1973
INTRODUCTION
In this paper, all rings have unity, all ring homomorphisms preserve the unity, and all modules are unital, except when otherwise specified. By a monomorphism (epimorphism) we mean a homomorphism that is into (onto). We shall use freely certain standard terminology, notation, and results from model theory, the construction of complete quotient rings, and sheaf theory. More specifically, the definitions of, and some basic facts about, an elementary property, an elementary extension, a universal extension, and an ultrapower, occur in [2], If the ring S is an elementary extension of the ring R, we write R @ S. If it is a universal extension we write R Cv S. The concepts of an essential extension of a module, an injective module, and a complete (left) (right) quotient ring, are discussed in [9]. The definitions of a clopen set, a Boolean space, and a sheaf k of rings over a Boolean space X are given in [l 11. In particular, if U C X, F( U, k) denotes the ring of all sections of k over U. DEFINITION
(i)
1.1. A ring R is: Regular, if it satisfies the sentence (Vr)(L)(rsr
= r);
(ii) Biregular, if each principal two-sided ideal in R is generated by a central idempotent; (iii) Quasi-biregular, if each nonzero two sided ideal in R contains a nonzero central idempotent; (iv) Of index n, where n 3 1 is an integer, if for all m 3 n, it satisfies the sentence (VT) (rm = 0 -+ P = 0); * The author was supported grant while conducting research
by a National Research for this paper.
512 Copyright AU rights
0 1976 by Academic Press, Inc. of reproduction in any form reserved.
Council
of Canada
operating
REGULAR
RINGS
OF FINITE
513
INDEX
(v)
Of finite index if it is of index n, for some integer n 3 1;
(vi)
Reduced if it is of index 1 (i.e., if it satisfies the sentence (VY)
(vii)
(Y2
= 0+
Y
= 0);
Strongly regular, if it satisfies the sentence (VT)
(3s)
(Y2S
=
Y);
(viii) Homogeneous of degree n if, for each primitive R/M is an n x n matrix ring over some division ring.
ideal M in R,
It is known (cf. [7, Lemma 3 and Theorem 31 and [I, Theorem 3.21) that a regular ring is reduced iff it is strongly regular, and a strongly regular ring is regular, biregular, and homogeneous of degree 1. Clearly, any commutative regular ring is both strongly regular and reduced. Some structure results for certain reduced regular rings are [ 1, Theorems 5.1-5.3, 6.1,6.2,&l, and 8.2; 8, Theorem 5.1; 11, especially Theorem 10.1; 3, Theorem 3.3; 4, Theorems 4.6 and 4.9; 5, Theorems 1.4 and 2.11. Although stating this only for the commutative case, Pierce shows ([ 11, Theorem 10.11) that any reduced (or equivalently strongly regular) ring R can be represented as the ring r(X, k), for some sheaf K of division rings over some Boolean space X. Results from [l, 3, 8, 11, Sect. 241 show that certain of such rings can be represented as suitable rings of continuous functions or (see [4]) as elementary extensions of such function rings. Results of Kaplansky, [8], specialize to yield the following, where rings are not assumed to have a unity: THEOREM 1.2. Let R be a regular ring of index n. Whenever 0 < m < n, let I, = fl (M : M is a primitive ideal in R such that RIM is an Y x Y matrix ring over some division ring, for some Y < m}. Then:
OY
(i) R either contains injinitely many orthogonal central idempotents, is Artinian.
(ii) For each primitive ideal M, RIM division ring, for some Y < n.
is a
Y x
Y
matrix ring over a
(iii) 0 = I, CInP1 C *a- C I1 C I, = R and I,-JI, is homogeneous of degree m, whenever 0 < m < n. Consequently, In+ is homogeneous of degree n. Also each I, is regular. (iv) Z(J) C Z(R), for any ideal Jpf R (where Z(S) denotes the center of S). (v) The natural map (R/I,) -+ 17(RIM : RIM is a Y x Y matrix ring over a division ring, for some Y < m} represents R/I,,, as a subdirect product. Hence, R/I,,, has index m.
514
ANDREW
B. CARSON
(vi) If R is homogeneous of degree n and either is countable or has an identity, then R is an n x n matrix ring over some strongly regular ring, and hence, is biregular. Kaplansky’s results do not indicate how the homogeneousparts I, of R “fit together.” Assuming that rings have unity, in this paper we prove: A. Let Q be a regular
THEOREM
complete left quotient
ring of index n.
Then:
I.
Q is biregular,
and
II. There exist strongly regular rings D, ,..., D, , and distinct integers n(1) < n,..., n(m) < n, for some m < n, such that
Q zx
Wnw
positive
0 ... 0 VLJn(n)
expresses Q as a direct sum of matrix rings. DEFINITION
1.3. For any ring R let B(R) = (e ER: e2 = e and e is central in R}.
View B(R) as either the Boolean algebra (B(R), +‘, *) or the Boolean ring (B(R), +“, .>, where e+‘f
=e+f-e-f
THEOREM B. Let R be a regular as a Boolean algebra. Then:
and
e+“f
=e+f-2e.f.
ring of index n such that B(R) is complete
(*) Rex WC 4, f or some sheaf k over a Boolean any stalk k, of k has the form
for some division rings Fi,, andpositive dependent upon x.
space X, such that
integers m < n and n(i) < n (1 < i < m)
THEOREM C. Let R be any regular ring of index n. Then there is an elementary extension S of R such that (*) from Theorem B holds, with S in place of R.
Our main tool, developed in Pierce’s memoir [ll], allows any ring R to be representedas I’(X, k), for somesheaf k of rings, where X is the Stone spaceof the Booleanalgebra B(R). The main results we need are now summarized.
REGULAR
DEFINITION
RINGS
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515
INDEX
1.4. Let R be a ring.
(i) Let X(R) denote the Stone space of R. Let k(R) be the sheaf of rings over X(R) such that k,(R) = R/m ( w h ere m is the ideal in R generated by the ideal M in B(R)) and the map r + r of R into P(X(R), k(R)) such that for any r E R and ME X(R), is an isomorphism. r(M)=(r+WEh, (That the sheaf k(R) with these properties exists, is established in [ll, Theorem 4.41. It is reduced in the sense that for any e E B(R) and x E X(R), e(x) = 0, or e(x) = 1.) . (ii) For any a E P(X(R), k(R)) let Supp(a) = {X E X(R) : u(x) # 0} and Zer(a) = {X E X(R) : u(x)‘=
01.
(It is a standard fact that Zer(u) is open and Supp(u) is closed. However, if R is biregular, then Supp(u) and 2 er ( u) are clopen. (cf. [l 1, pp. 37 and 451). By methods from the proof of [l 1, 23.11 we obtain: THEOREM
(i) U'
1.5. Let R be a ring. Suppose that R is self-injective. Then:
(*) For any open subset U of X(R) E P(X(R), k(R)) such that u’ lo = u.
and u E P( U, k(R)),
there exists
(ii) Suppose instead that R is strongly regular. Then R is left (or right) self-injective if(*) holds. Thus, R is left self-injective iff it is right self-injective. All results in this paper hinge upon the following Kaplansky’s work:
consequence
of
THEOREM 1.6. In this theorem, rings need not have a unity. Let R be a regular ring of finite index. Then R is quasi-biregular.
Proof. It suffices to show that, where 0 # r E R, (Rr R) n B(R) # 0. First assume that R is homogeneous. Find, since R is semisimple, a primitive ideal M such that r 4 M and, by [8, Theorem 4.11, 0 # e E B(R) such that e $ M. Then 0 # R er R is an ideal in the homogeneous ring with identity Re, hence, by Theorem 1.2(vi), 0 # f E R er R C R r R, for some f E B(Re) C B(R). The general proof proceeds by induction on n. If n = 1, then R is homogeneous, so that the result holds. Assume the result for all regular rings of index n and suppose that R has index n+ 1. Let I,, where 0 < m < n + 1, be as in Theorem 1.2. As
ANDREW B. CARSON
516
(R Y R) n I, is an ideal in the homogeneousregular ring 1, and Z(1,) _CE(R), (R Y R) n B(R) # 0 follows if (R Y R) n I,, # 0. Now supposethat (R Y R) n I,, = 0. Since R is regular, so is the ideal R Y R. Let h: R -+ l7(R/M : M is primitive in R) be the standard subdirect product representation of R. For any primitive ideal M of R, if R YR $ M, then I, C M, since (R Y R) n I,, = 0. Hence, by Theorem 1.2, the homomorphic image R/M of R/I,, has index n. Thus, h (R r R) and R Y R haye index 12.The result for R now follows from Theorem 1.2(iv) and the induction hypothesis applied to R Y R. [ We shall repeatedly combine the above with the following: PROPOSITION 1.7. Suppose that 0 # e E (R Y R) n B(R), where f elementof the ring R. Then:
is
an
(i) er # 0; (ii) ;f Y EB(R), then er = e. Proof.
Find rr ,..., r, , si ,..., s, ER such that e = C rirsi . Then:
(i) 0 # e = e2 = C riresi , so that re # 0; (ii)
if Y E B(R), then er = 1 rir2si = C rirsi = e. 1
The following is an easyconsequenceof Theorem 1.6 and Proposition 1.7: COROLLARY 1.8. Let R be a regular ring of $nite index and 0 # Y E
r(X(R),
k(R)). Then Supp(r) hasa nonemptyinterior. 2. REGULAR QUOTIENT RINGS OF FINITE INDEX
In this section, we prove Theorem A. To do this we require the following consequenceof Theorem 1.2(i): PROPOSITION 2.1. Any indecomposable (and hence, any simple) regular ring of index n is a full m x m matrix ring over somedivision ring, for some m
(i) For each x EX(R), there exists a divzkion ring F, and a positive integer h(x) < n suchthat k,(R) = (F&c) .
REGULAR
RINGS
OF FINITE
INDEX
517
(ii) For eachx EX(R) there existsa neighborhoodN of x suchthat h(x) dividesh(z), for all x EN.
Proof. (i) Since R is regular, so are the stalks of k(R). They also have index n for if some K, does not, we can find u E r(X(R), k(R)) (see Definition 1.4) and m such that (*)
c44P
= 0,
yet (44)n
f- 0,
when z = x. However (*) also holds for all z in some clopen neighborhood W of x. Thus, 7m = 0 yet 7% # 0, where r E I”(X(R), k(R)) E R is defined by T(Z) = U(Z) when .s E W and T(Z) = 0 otherwise. This would contradict the hypothesis that R has index n. These stalks are simple rings, by [ 11, Theorem 11 .l], since R is biregular. The result now follows from Proposition 2.1. (ii) First note that, for any division ring F and positive integers a and b, if there is an embedding (F)a -+ (F), , then a divides b. This holds since a minimal left ideal in (F)b has dimension b as a vector space over F yet, as a (F),-module, it is a direct sum of simple modules (cf. [9, Sect. 3.3]), and these have dimension a over F. Identify R = r(X(R), k(R)). Let x E X(R) and, when i, j E{l,..., X(x)}, let eij(x) E(F)~(~ = k,(R), where eij E R, be the matrix with entry 1 in the i, jth place and entries zero elsewhere. Then the finite set of formulas: (CZ) eii(z) # e&z),
unless i = K and j = p,
(B)
e&4
ek,(4
= %&I, = 0
(r)
C2?
e&4
= 1,
if if
j=K j # k, and
holds when z = x. Since R is biregular, all of its elements have clopen support in X(R), so that (a), (p), and (y) hold for all x in some neighborhood N of x. Thus, for any z EN, the subring of k,(R) = (F,),(,) generated by F, and the eii(z) is isomorphic to (F&d . Hence, by our opening remarks, h(x) divides A(z). 1 THEOREM 2.3. Let Q be any regular completeleft quotient ring. Then the stalksof k(Q) are all regular indecomposable rings.
Proof. Let Y = X(Q), K = k(Q), and identify Q = r(Y, K). Suppose that some K, is not indecomposable. Thus there exists e2 = e E Q such that e(y) is central in K, yet 0 # e(y) # 1. Since the sheaf K is reduced (cf. Definition 1.4), for each neighborhood N of y, there exists z E N and s E Q such that s(x) e(x) - e(x) s(x) # 0, (1) holds when x = z.
518
ANDREW
B.
CARSON
In fact, we may choose s such that Supp(s) = Supp(se - es).
(2)
To see this, let sr = (ese - se), sa = (ese - es), and note that es, = 0, sic = s, , es, = sa , and sac = 0. Hence, (2) holds with either sr or sa in place of s. Further, z E Supp(s,) u Supp(s,) for otherwise we would have (ese - se)(z) = 0 = (ese - es)(z), hence, (1) fails for z = X, a contradiction. We replace s with an si satisfying x E Supp(sJ. By Zorn’s lemma, find a family {tol} of elements of Q maximal with respect to the properties: (4
SuppkJ
= %-v&e
- et,) f
0,
for each (Y, and (b)
Supp(t,) n Supp(&) = ia, when 01# ,6.
Let C, = Supp(t,), for each t, . In view of (a), the sentences involving (1) and (2), and the maximality of the family, we have: YE cl u (Cm) * ( a 1
(3)
For each Mlet Qol= (4 EQ : Supp(q) C C,}. Each QDlis a two sided ideal in Q. By (b) the sum Ca Qolis direct. Since Q is a complete left quotient ring that is regular, Q is injective (cf. [9, Sect. 4.3 and 4.5]), sothat the left injective hull of CNQUis contained in Q. Further, &Q, is essentialover z:u Qa since the regularity of Q ensuresthat QQII # 0, for each 01.Therefore
CQ.C~Q~CQ, as left Q-modules. Thus 4 * kJ = (4 * !?cJ>
(4)
for any q E Q and (q=} E IIQ,, . We now show that GIJ - {d = {% * 91
(5)
alsoholds. Fix q EQ and define the left Q-module morphismf: DQa+ Q by f({q=}) = {q=} . q - {qa * q}. Clearly, C QaC ker( f),and (5) would follow if nQ,,C ker( f), Thus, if (5) fails, there exists {q,} EnQ,,, and e EQ such that 0 # e = f{(qJ). Since Q is regular we may assume,without loss of generality, that e2 = e. Thus, using (4), 0 # e = e2 = ef({qa}) =f({eq,}), so that eqB # 0, for some /3. Let t = eq, and choose t’ EQ such that tt’t = t. Replace t’ with
REGULAR
RINGS
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519
INDEX
t’tt’, zo that t’ E Qe also holds. Then t’ . {Q~} = {t’ * qol} = t’ * q8 E Q0 , hence t’ * e = f(t’ * {qa}) = f(t’ . qs) = 0, contradicting (eq,) t’(eq& = t-q8 # 0. Thus, (5) holds. Finally, {t&4
44
f
44k&>,
when x = y. To see this, suppose that borhood N of y, such that (6) fails to be such that Y(z) = 1 when x E N and Ye{&} so that, by (4) and (5), (ta Ye) = for suitable /I, and conclude that x $ Supp(t,e
(6)
it is false, and obtain a clopen neighhold for all z E N. Let YE r( Y, K) Y(z) = 0 otherwise. Then {tm}Ye = {Yet,}. By (3), find some z E C, n N,
- et& = C, ,
a contradiction. Hence, each K, is indecomposable. they are epimorphic images of Q. 1 Theorem
2.3 also holds if Q is a regular
Proof of TheoremA.
Let Y = X(Q),
The stalks are regular
complete
right
quotient
as
ring.
K = k(Q), and identify Q = r( Y, K).
(I) By [ 11, Theorem 11. l] Q is biregular iff the stalks of K are simple rings. However, they have index 71,and hence, by Proposition 2.1 and Theorem 2.3, are simple. (II) For each y E Y there exists, by (I) and Proposition 2.2(i), a division ring F, and a positive integer h(y) < n, such that K, = (Fy)A(y). To begin, we show that the function X is locally constant. Suppose that X is constant in no neighborhood of y, for some y E Y. Then, since Proposition 2.2(ii) holds, y E cl( I’), where V = (Ja (Supp(q,)), p > I\(y), and {qa} is a subset of Q maximal with respect to these properties: (a)
(q2
(b)
Supp(q,J
= 0 yet SUPP[(P~-~]
n Supp(~,J
= Supp(q,)
= m, whenever
# O,
for
any a,
and
a # /I.
Let 4: I’-+ K be such that q(z) = qn(x) when z E Supp(q,J. Then V is open and q is continuous, since the qmhave clopen support. Since oQ is injective (cf. [8, Sect. 4.5]), Theorem 1.5 yields q* E r(Y, K) satisfying q* IV = q. Since y E cl(V) and elements of Q have clopen support, we have (q*)“-l(y) # 0 yet (q*)” (y) = 0. Th is contradicts h(y) < p and shows that A is locally constant. Let {n(l),..., n(m)} = {X(y): y E Y}, where the n(i) =$ n are distinct and m < n. For each i, 1 < i < m, let Vi = {y E Y : X(y) = n(r)}. The Ui # ia and, since X is locally constant, partition X into clopen sets. Clearly,
520
ANDREW
B. CARSON
(I) Qg @Cz,Qi, where Qi = r(Ui, K), for 1
Qi g (DJnti) , when 1 < i < m.
The theorem follows from (1) and (2). 1 Many structure results for strongly regular rings are already available in [l, 3-5, 8, 111. To conclude this section, we represent the complete (right) quotient ring of a regular ring R of finite index, in terms of the sheafk(R). For the rest of this section, let Q denote the complete right quotient ring of the regular ring R of finite index. For any subsetS of R let S be the two-sided ideal of R generatedby S. It is known [9, Sect. 4.3, Proposition 61that the complete right quotient ring of any ring T has the form inj lim[Hom#, T): I is a denseright ideal in R]. PROPOSITION 2.4. (1) Each denseright ideal of R contains a two-sided ideal that is denseas a right ideal of R.
(2) Q is alsothe completeleft quotient ring of R. (3) Q E inj lim[Hom,(D, , R,): D is a denseideal in B(R)]. Proof. We shall usethe standard fact (cf. [9, Sect. 4.51)that the essential and the denseright idealsin a regular ring coincide. Consider the following condition for a ring S: (1’) Any essentialright ideal in S contains a two-sided ideal that is essentialas a right ideal. Martindale ([lo, Theorem 41) has established(1’) and (2) for any semiprime polynomial identity ring S, using Rowen’sresult ([12, Theorem 11)that any nonzero two sided ideal of S contains a nonzero central element. We may use Theorem 1.6 and Martindale’s proof to establish(1) and (2), for R. Result (3) would follow from (1) and the remark preceding this theorem, if D were densein R as a right ideal, where D = In B(R) for sometwo sided ideal I that is dense as a right ideal. However, this is true since if O#aER,thenO#eE[(RaR)nI]nB(R) = Dn[RaR],(byTheorem 1.6)sothatOfaeEaD. 1 In future, the common left and right complete quotient rings of R will be simply called the complete quotient ring of R. COROLLARY
2.5. (1) Q z inj lim{r(U, k(R)): U is a denseofxn subset
of WW.
(2) Q is regular, biregular and has index n.
REGULAR RINGS OF FINITE INDEX
521
Proof. (1) Let D be a dense ideal in B(R). Identify R = r(X(R), k(R)), and let U = (J {Supp(e) : e ED}. Then, since k(R) is a reduced sheaf, U is a dense open subset of X(R) and D = {r E R : Supp(r) C U}. Much as in [l 1, Lemma 23.11, the map u -+ Y, satisfying y,,[e](x) = U(X). e(x), is an isomorphismr(U, k(R) -+ Hom,(DR , RR). The result now follows from Proposition 2.4.(3).
(2) This is immediate from (1) and Theorem A. The above is reminiscent of the following result, due to Fine, Gillman, and Lambek in [6]: (For any topological space Y and topological ring T, U(Y, T) denotesthe ring of all continuous functions from Y into T.) Let X denote a compact Hausdorff space,R the real number field in the usual topology, and Q the complete quotient ring of S = U(X, R). Then Q g inj lim{%( U, R): U is a denseopen subsetof X}. However, the two situations are different. For example, if X is also infinite and connected, then B(S) = (0, l} yet B(Q) is uncountable. This contrasts sharply with Proposition 3.1.
3. REGULAR RINGS OF FINITE INDEX WITH A COMPLETE BOOLEAN ALGEBRA OF CENTRAL TDEMPOTENTS In this section we prove Theorem B. PROPOSITION 3.1. Let R be a regular ring of jinite complete quotient ring of the ring S. Then
Q(W))
=
index, let Q(S)
be the
WQW
Proof. For any e E B(R), the left R-module morphism f: Q(R) ---f Q(R) defined by f(r) = re - er satisfies R C ker( f). Thus, by [9, Sect. 4.31, Q(R) C ker( f ). Hence, B(R) C B(Q(W
Any regular ring is a complete left quotient ring iff it is left self-injective. Thus, as in the commutative case,(using Corollary 1.8), cf. [ll, Proposition 24.1]), X(Q(R)) is extremally disconnected. Since X(Q(R)) = X(B(Q(R))) it is thus standard(cf. [3, Lemma 2.5(l)]) that B(Q(R)) is a complete quotient ring. Hence, the result follows if B(Q(R)) is essentialover B(R). Let 0 # e E B(Q(R)). Then, since RQ(R) is essentialover RR, there exists r E R such that 0 # re E R. Since R is quasi-biregular, find 0 f f E (RreR) n B(R). Then 0 # fe = f EB(R), so that B(Q(R)) is essentialover B(R). 1
ANDREW B. CARSON
522
Proof of TheoremB. Let Q denote the complete quotient ring of R. Since B(R) is complete (as a Boolean algebra), it is its own complete quotient ring (cf. [9, Sect. 2.41). Hence, by Proposition 3.1, B(R) = B(Q) so that X(R) = X(Q) and k(R) is a subsheaf of k(Q). Thus, by Theorem A, each stalk k,(R) is a regular subring of index n of k,(Q) = (F3c)I\(z), for some division ring F, and positive integer A(x) < n. The result now follows from Theorem 1.2(i). 1 Theorem B does not give complete information about R since the topology on k(R), as well as the structure of its stalks, is required to recover r(X(R), k(R)). However, elements of I’(X(R), k(R)) do have clopen support, since Q is biregular and k(R) is a subsheaf of k(Q). Equivalently k(R) is Hausdorff [ll, Corollary 10.61. Similarly, the topology of k(R) is determined by that of k(Q) and it is determined by the Di , where the Di are as in Theorem A. This is only a partial help, since most complete structure results for strongly regular rings require that they be commutative algebraic algebras over some field. However, we do have the following Theorem. (Notation: for a ring S and topological space Y, let V(Y, S) denote the ring of all continuous, functions from Y to S, where S has the discrete topology.) THEOREM 3.2. Let R be a regular ring of index n that is alsoan algebraic algebraover the real numberjield, and let F denotethe quaternions.Suppose that B(R) is complete.Then there exist distinct integersn(1) < n,..., n(m) < n, for somem < n, and a partition X, ,..., X, of X(R) into clopensetssuchthat k(R) Ix.t is a subsheafof the simplesheafXi x (F),ti) , whenever1 < i < m.
Proof. Let Q be the complete quotient ring of R. By Theorem A find distinct integers n(1) < n,..., B(m) < n (for some m < n) and strongly regular rings D, ,... , D, such that
For each i (1 < i < m) let Xi = X(DJ and Ei = (d EDi : f (d) = 0 for some polynomial f with real coefficients). Then Ei is a ring, and by [3, concluding remark C] Ed _C U(Xi , F), when 1 < i < m. Clearly, us, since R is algebraic over the reals Th KG 1F)lnw s VXi >Ph). R C V-&
3 Ph)
0 ... 0 Vhn,
, P)n(m)).
By [ 11, pp. 26-271, the Xi partition X(Q) = X(R) so that since k(W(Xi, (F)a(dj) is the simple sheaf Xi x (F),ci) , k(R) Ixi is a subsheaf of X, x (F),ti) , whenever 1 < i < m. The topology of k(R) is completely determined, when Theorem 3.2 applies, since the Xi are clopen in X(R). 1
523
REGULAR RINGS OF FINITE INDEX
4. ELEMENTARY PROPERTIES OF REGULAR RINGS OF FINITE INDEX The main result in this section is Theorem C. Proof of TheoremC. We shall construct a sequence (*)
R,-tS,-tR,~S,~...-tR,~S,~...
of rings and subring inclusions such that R = R, and (whenever i < i) 6) RtO 4, (ii) Qi is the complete quotient ring of Ri , (iii) (iv)
B(QJ C WC+J, Si is the subring of Ri+l generatedby Ri and B(Q,).
The required ring S will be inj lim(&). The rings Ri and Qi satisfying (i) and (ii) are defined inductively. Let R = R,, and chooseQ,, satisfying (ii). Supposethat Ri and Qi satisfying (i) and (ii) have been defined for all i and j < somem. It is immediate from Proposition 3.1 that: (v) in
Each atom (i.e., minimal nonzero element) in B(R,) is an atom
BfQd.
By [5, Lemma 2.21, (V) ensures that B(R,) C vB(Q,J. [2, Chap. 9, Lemma 3.91):
Hence, (by
(vi) There is an embedding LY,: B(Q,) -+ (B(R,))I/F, into some ultrapower of B(R,), which extends the natural map B(R,) -+ (B(R,))I/F. Let R,+l = (R#/F and chooseQ,,,+r to satisfy (ii). By Los’s Theorem (cf. [2, Chap. 5, Th eorem 2.1]), (i) holds whenever i
524
ANDREW
B.
CARSON
Let 0 # s = r1 . e, + ... + ra . e, be an arbitrary element of Si where the r, E R and the e, E B(Q). We may (and do) select the rB and eBso that the e, are mutually orthogonal, and some r4 . e, # 0. Thus, r, * e, # 0 so that there is 0 # e E [(RYE * eoR) n R] n B(R). Thus 0 # e * (I~ * e,) = Y, * (e * eB) = ye * e E R. Since e . s = e . ye . e, = ye . e = Y, * e, this establishes(viii). As a consequencewe have, after suitable identifications, (ix)
S1C Qi .
Further Z(S) !EZ(Qi), since Qi is essentialover Si . Since B(Qi) C B(Si), this yields: (X)
B(Qi) = B(Si)*
Thus, X(S,) = X(Q,) so that, by [I 1, Proposition 6.81, k(Si) is a subsheaf of k(QJ and k,(Si) _Ck,(Qi), for all x E X($). In view of Theorem A and Proposition 2.2, this yields: (xi) k,(S,) has at most n nonzero orthogonal idempotents, for each XE X(Si). Let S = inj lim(&). Th en also S = inj lim(R,) so that by [2, Chap. 4, Theorem 2.11, Ri @ S, whenever i > 0. Identify S = r(X(S), k(S)). (xii) k,(S) has at most n nonzero orthogonal idempotents, for each x E X(S). To see (xii) supposethat O(x), el(x),..., e,+,(x), are distinct orthogonal idempotentsin k,(S), for somex EX(S). We assumewithout lossof generality that the ej are idempotents in S. Choose i such that e, ,..., e,,, E Si . Let s + s be the natural map Si z r(X(S,), k(S,)), and let 7: X(S) -+ X(S,) be the natural map induced by B(S,) C B(S) * (B(S,) _CB(S) since B(Si) = B(QJ C B(R,+J and Ri+l @ S.) As in [II, Lemma 6.31, there is a homomorphism @r:k,JSi) - k&Q such that @&s(rr(x))) = s(x), for all s E Si . This contradicts our choice of the ej since,by (xii), at leasttwo of O(rr(x)), e,(r(x)),..., e,+,(rr(x)), coincide. Thus, (xii) holds. Since R = R, @ S, S and its stalks are regular and of index n. The theorem now follows from (xii) and Theorem 1.2(i). An example follows to show that, in Theorem C, the structure result for the elementary extension S of R neednot apply to R itself. In fact, somestalk (s) of k(R) may even fail to be Artinian.
REGULAR
RINGS
OF FINITE
INDEX
525
EXAMPLE. Let X be the Boolean space w + 1, where w is the first infinite ordinal. Let F be a field, ki E (F)a when i < W, and k, = 9?(X, F), the ring of locally constant functions from X to F. Let k be the disjoint union {k, : x E X}. Let CL:w ---f w x w be a bijection and let p1and t+ be the induced maps pl: w-+w x w-w x {0} ++ w and ~a:w--+w x w--(O) x w t-$ w, respectively. Topologize k such that, when i < W, the points in ki are clopen in k and, for any neighborhood N of w in X andf E C(X, F).
is open in k. Let R = r(X, k). Then R is a regular ring of index 2. By construction, k is a reduced sheaf, so that X(R) = X and k(R) = k. The stalk k,,, is not Artinian. We conclude with someremarks: (a) In [l l] Pierce showedthat any ring R can be representedasI’(X(R), k(R)). If R is commutative, the stalks of k(R) are indecomposable,and if R is alsoregular, they are fields. Thus, many problems concerning commutative (regular) rings are reduced to simpler onesabout sheavesof indecomposable rings (fields). However, if R is noncommutative, the stalks of k(R) may be, to quote Pierce (cf. [l 1, Example 11.4]), “almost arbitrarily bad.” Theorem B shows that, for suitable noncommutative regular R, the stalks of k(R) are still manageable.In fact, for any regular complete left quotient ring T, the stalks of k(T) are indecomposable,although often of infinite index. Our present knowledge of such stalksis insufficient to produce deepresults for T. (b) Our results indicate that for some rings R, the Stone space(i.e., spectrum) of B(R) is more useful than the maximal spectrum of R, in achieving a representation of R as a ring of continuous functions (sections) with values in some nice ring (s). The spectrums of R and B(R) coincide if R is biregular or, in particular, if R is commutative and regular. (c) Although the structure result in Theorem C is for an elementary extension S of R, it doesyield somefacts about R itself. To illustrate this, we sketch a short proof of Fisher’s result that any regular ring R of finite index is zcnit reguZur. By definition, a ring is unit regular if it has the elementary PropeW (**)
(Vr)(Zls)(ilt)
[TST= Yand st = ts = 11.
Thus, it sufficesto show that the elementary extension S of R, described in Theorem C, is unit regular. Since the stalks of k(S) are given by (*) from Theorem B, it is standardthat they satisfy (* *). Thus, sincethe universal quantifier in (**) precedesthe existential quantifiers, [l I, Proposition 3.41 ensuresthat (**) also holds in S. 481/39/2-x3
526
ANDREW
B. CARSON
(d) If 2 is a unit in R, then each k,(S) is a finite direct sum of full matrix rings over division rings, none of which has Characteristic 2. Therefore, any element in k,(S) can be written as a sum of two units of k,(S), hence, the same result holds with R in place of k,(S). This result, for R, originated from Fisher.
ACKNOWLEDGMENT The author would like to thank the referee for detecting an error in the original proof of Theorem 1.6, and suggesting its present nonsheaf theoretic presentation.
REFERENCES 1. R. ARENS AND I. KAPLANSKY, Topological representation of algebras, Trans. Amer. Math. Sot. 63 (1948), 457-481. 2. J. L. BELL AND A. B. SLOMSON, “Models and Ultraproducts: An Introduction,” North-Holland, Amsterdam, 1969. 3. A. B. CARSON, Representation of semi-simple algebraic algebras, I. Algebra 24 (1973), 245-257. 4. A. B. CARSON, The model completion of the theory of commutative regular rings, J. Algebra 27 (1973), 136-146. 5. A. B. CARSON, Algebraically closed regular rings, Canad. J. Math. 26 (1974), 1036-1049. 6. N. J. FINE, L. GILLMAN, AND J. LAMBEK, “Rings of Quotients of Rings of Functions,” McGill Univ. Press, Montreal, 1965. 7. A. FORSYTHE AND N. H. MCCOY, The commutativity of certain rings, Bull. Amer. Math. Sot. 52 (1946), 523-526. 8. I. KAPLANSKY, Topological representation of algebras. II, Trans. Amer. Math. Sot. 68 (1950), 62-75. 9. J. LAMBEK, “Lectures on Rings and Modules,” Blaisdell, Waltham, Massachusetts, 1966. 10. W. MARTINDALE III, On semi-prime P.I. rings, Pvoc. Amer. Math. Sot. 40 (1973). 365-369. 11. R. S. PIERCE, Modules over commutative regular rings, Mem. Amer. Math. Sot. 70 (1967). 12. L. ROWEN, Some results on the centre of a ring with polynomial identity, Bull. Amer. Math. Sot. 79 (1973), 219-223. 13. L. LIPSHITZ AND D. SARACINO, The model companion of the theory of commutative rings without nilpotent elements, PYOC. Am. Math. Sot. 38 (1973). 381-387.
OF ALGEBRA
39,
512-526
Representation
(1976)
of Regular ANDREW B.
Mathematics
Department,
of
University
Communicated Received
I.
Rings of Finite
Index
CARSON*
Saskatchewan,
Saskatoon,
S7N
OWO,
Canada
by I. W. Herstein July
20, 1973
INTRODUCTION
In this paper, all rings have unity, all ring homomorphisms preserve the unity, and all modules are unital, except when otherwise specified. By a monomorphism (epimorphism) we mean a homomorphism that is into (onto). We shall use freely certain standard terminology, notation, and results from model theory, the construction of complete quotient rings, and sheaf theory. More specifically, the definitions of, and some basic facts about, an elementary property, an elementary extension, a universal extension, and an ultrapower, occur in [2], If the ring S is an elementary extension of the ring R, we write R @ S. If it is a universal extension we write R Cv S. The concepts of an essential extension of a module, an injective module, and a complete (left) (right) quotient ring, are discussed in [9]. The definitions of a clopen set, a Boolean space, and a sheaf k of rings over a Boolean space X are given in [l 11. In particular, if U C X, F( U, k) denotes the ring of all sections of k over U. DEFINITION
(i)
1.1. A ring R is: Regular, if it satisfies the sentence (Vr)(L)(rsr
= r);
(ii) Biregular, if each principal two-sided ideal in R is generated by a central idempotent; (iii) Quasi-biregular, if each nonzero two sided ideal in R contains a nonzero central idempotent; (iv) Of index n, where n 3 1 is an integer, if for all m 3 n, it satisfies the sentence (VT) (rm = 0 -+ P = 0); * The author was supported grant while conducting research
by a National Research for this paper.
512 Copyright AU rights
0 1976 by Academic Press, Inc. of reproduction in any form reserved.
Council
of Canada
operating
REGULAR
RINGS
OF FINITE
513
INDEX
(v)
Of finite index if it is of index n, for some integer n 3 1;
(vi)
Reduced if it is of index 1 (i.e., if it satisfies the sentence (VY)
(vii)
(Y2
= 0+
Y
= 0);
Strongly regular, if it satisfies the sentence (VT)
(3s)
(Y2S
=
Y);
(viii) Homogeneous of degree n if, for each primitive R/M is an n x n matrix ring over some division ring.
ideal M in R,
It is known (cf. [7, Lemma 3 and Theorem 31 and [I, Theorem 3.21) that a regular ring is reduced iff it is strongly regular, and a strongly regular ring is regular, biregular, and homogeneous of degree 1. Clearly, any commutative regular ring is both strongly regular and reduced. Some structure results for certain reduced regular rings are [ 1, Theorems 5.1-5.3, 6.1,6.2,&l, and 8.2; 8, Theorem 5.1; 11, especially Theorem 10.1; 3, Theorem 3.3; 4, Theorems 4.6 and 4.9; 5, Theorems 1.4 and 2.11. Although stating this only for the commutative case, Pierce shows ([ 11, Theorem 10.11) that any reduced (or equivalently strongly regular) ring R can be represented as the ring r(X, k), for some sheaf K of division rings over some Boolean space X. Results from [l, 3, 8, 11, Sect. 241 show that certain of such rings can be represented as suitable rings of continuous functions or (see [4]) as elementary extensions of such function rings. Results of Kaplansky, [8], specialize to yield the following, where rings are not assumed to have a unity: THEOREM 1.2. Let R be a regular ring of index n. Whenever 0 < m < n, let I, = fl (M : M is a primitive ideal in R such that RIM is an Y x Y matrix ring over some division ring, for some Y < m}. Then:
OY
(i) R either contains injinitely many orthogonal central idempotents, is Artinian.
(ii) For each primitive ideal M, RIM division ring, for some Y < n.
is a
Y x
Y
matrix ring over a
(iii) 0 = I, CInP1 C *a- C I1 C I, = R and I,-JI, is homogeneous of degree m, whenever 0 < m < n. Consequently, In+ is homogeneous of degree n. Also each I, is regular. (iv) Z(J) C Z(R), for any ideal Jpf R (where Z(S) denotes the center of S). (v) The natural map (R/I,) -+ 17(RIM : RIM is a Y x Y matrix ring over a division ring, for some Y < m} represents R/I,,, as a subdirect product. Hence, R/I,,, has index m.
514
ANDREW
B. CARSON
(vi) If R is homogeneous of degree n and either is countable or has an identity, then R is an n x n matrix ring over some strongly regular ring, and hence, is biregular. Kaplansky’s results do not indicate how the homogeneousparts I, of R “fit together.” Assuming that rings have unity, in this paper we prove: A. Let Q be a regular
THEOREM
complete left quotient
ring of index n.
Then:
I.
Q is biregular,
and
II. There exist strongly regular rings D, ,..., D, , and distinct integers n(1) < n,..., n(m) < n, for some m < n, such that
Q zx
Wnw
positive
0 ... 0 VLJn(n)
expresses Q as a direct sum of matrix rings. DEFINITION
1.3. For any ring R let B(R) = (e ER: e2 = e and e is central in R}.
View B(R) as either the Boolean algebra (B(R), +‘, *) or the Boolean ring (B(R), +“, .>, where e+‘f
=e+f-e-f
THEOREM B. Let R be a regular as a Boolean algebra. Then:
and
e+“f
=e+f-2e.f.
ring of index n such that B(R) is complete
(*) Rex WC 4, f or some sheaf k over a Boolean any stalk k, of k has the form
for some division rings Fi,, andpositive dependent upon x.
space X, such that
integers m < n and n(i) < n (1 < i < m)
THEOREM C. Let R be any regular ring of index n. Then there is an elementary extension S of R such that (*) from Theorem B holds, with S in place of R.
Our main tool, developed in Pierce’s memoir [ll], allows any ring R to be representedas I’(X, k), for somesheaf k of rings, where X is the Stone spaceof the Booleanalgebra B(R). The main results we need are now summarized.
REGULAR
DEFINITION
RINGS
OF FINITE
515
INDEX
1.4. Let R be a ring.
(i) Let X(R) denote the Stone space of R. Let k(R) be the sheaf of rings over X(R) such that k,(R) = R/m ( w h ere m is the ideal in R generated by the ideal M in B(R)) and the map r + r of R into P(X(R), k(R)) such that for any r E R and ME X(R), is an isomorphism. r(M)=(r+WEh, (That the sheaf k(R) with these properties exists, is established in [ll, Theorem 4.41. It is reduced in the sense that for any e E B(R) and x E X(R), e(x) = 0, or e(x) = 1.) . (ii) For any a E P(X(R), k(R)) let Supp(a) = {X E X(R) : u(x) # 0} and Zer(a) = {X E X(R) : u(x)‘=
01.
(It is a standard fact that Zer(u) is open and Supp(u) is closed. However, if R is biregular, then Supp(u) and 2 er ( u) are clopen. (cf. [l 1, pp. 37 and 451). By methods from the proof of [l 1, 23.11 we obtain: THEOREM
(i) U'
1.5. Let R be a ring. Suppose that R is self-injective. Then:
(*) For any open subset U of X(R) E P(X(R), k(R)) such that u’ lo = u.
and u E P( U, k(R)),
there exists
(ii) Suppose instead that R is strongly regular. Then R is left (or right) self-injective if(*) holds. Thus, R is left self-injective iff it is right self-injective. All results in this paper hinge upon the following Kaplansky’s work:
consequence
of
THEOREM 1.6. In this theorem, rings need not have a unity. Let R be a regular ring of finite index. Then R is quasi-biregular.
Proof. It suffices to show that, where 0 # r E R, (Rr R) n B(R) # 0. First assume that R is homogeneous. Find, since R is semisimple, a primitive ideal M such that r 4 M and, by [8, Theorem 4.11, 0 # e E B(R) such that e $ M. Then 0 # R er R is an ideal in the homogeneous ring with identity Re, hence, by Theorem 1.2(vi), 0 # f E R er R C R r R, for some f E B(Re) C B(R). The general proof proceeds by induction on n. If n = 1, then R is homogeneous, so that the result holds. Assume the result for all regular rings of index n and suppose that R has index n+ 1. Let I,, where 0 < m < n + 1, be as in Theorem 1.2. As
ANDREW B. CARSON
516
(R Y R) n I, is an ideal in the homogeneousregular ring 1, and Z(1,) _CE(R), (R Y R) n B(R) # 0 follows if (R Y R) n I,, # 0. Now supposethat (R Y R) n I,, = 0. Since R is regular, so is the ideal R Y R. Let h: R -+ l7(R/M : M is primitive in R) be the standard subdirect product representation of R. For any primitive ideal M of R, if R YR $ M, then I, C M, since (R Y R) n I,, = 0. Hence, by Theorem 1.2, the homomorphic image R/M of R/I,, has index n. Thus, h (R r R) and R Y R haye index 12.The result for R now follows from Theorem 1.2(iv) and the induction hypothesis applied to R Y R. [ We shall repeatedly combine the above with the following: PROPOSITION 1.7. Suppose that 0 # e E (R Y R) n B(R), where f elementof the ring R. Then:
is
an
(i) er # 0; (ii) ;f Y EB(R), then er = e. Proof.
Find rr ,..., r, , si ,..., s, ER such that e = C rirsi . Then:
(i) 0 # e = e2 = C riresi , so that re # 0; (ii)
if Y E B(R), then er = 1 rir2si = C rirsi = e. 1
The following is an easyconsequenceof Theorem 1.6 and Proposition 1.7: COROLLARY 1.8. Let R be a regular ring of $nite index and 0 # Y E
r(X(R),
k(R)). Then Supp(r) hasa nonemptyinterior. 2. REGULAR QUOTIENT RINGS OF FINITE INDEX
In this section, we prove Theorem A. To do this we require the following consequenceof Theorem 1.2(i): PROPOSITION 2.1. Any indecomposable (and hence, any simple) regular ring of index n is a full m x m matrix ring over somedivision ring, for some m
(i) For each x EX(R), there exists a divzkion ring F, and a positive integer h(x) < n suchthat k,(R) = (F&c) .
REGULAR
RINGS
OF FINITE
INDEX
517
(ii) For eachx EX(R) there existsa neighborhoodN of x suchthat h(x) dividesh(z), for all x EN.
Proof. (i) Since R is regular, so are the stalks of k(R). They also have index n for if some K, does not, we can find u E r(X(R), k(R)) (see Definition 1.4) and m such that (*)
c44P
= 0,
yet (44)n
f- 0,
when z = x. However (*) also holds for all z in some clopen neighborhood W of x. Thus, 7m = 0 yet 7% # 0, where r E I”(X(R), k(R)) E R is defined by T(Z) = U(Z) when .s E W and T(Z) = 0 otherwise. This would contradict the hypothesis that R has index n. These stalks are simple rings, by [ 11, Theorem 11 .l], since R is biregular. The result now follows from Proposition 2.1. (ii) First note that, for any division ring F and positive integers a and b, if there is an embedding (F)a -+ (F), , then a divides b. This holds since a minimal left ideal in (F)b has dimension b as a vector space over F yet, as a (F),-module, it is a direct sum of simple modules (cf. [9, Sect. 3.3]), and these have dimension a over F. Identify R = r(X(R), k(R)). Let x E X(R) and, when i, j E{l,..., X(x)}, let eij(x) E(F)~(~ = k,(R), where eij E R, be the matrix with entry 1 in the i, jth place and entries zero elsewhere. Then the finite set of formulas: (CZ) eii(z) # e&z),
unless i = K and j = p,
(B)
e&4
ek,(4
= %&I, = 0
(r)
C2?
e&4
= 1,
if if
j=K j # k, and
holds when z = x. Since R is biregular, all of its elements have clopen support in X(R), so that (a), (p), and (y) hold for all x in some neighborhood N of x. Thus, for any z EN, the subring of k,(R) = (F,),(,) generated by F, and the eii(z) is isomorphic to (F&d . Hence, by our opening remarks, h(x) divides A(z). 1 THEOREM 2.3. Let Q be any regular completeleft quotient ring. Then the stalksof k(Q) are all regular indecomposable rings.
Proof. Let Y = X(Q), K = k(Q), and identify Q = r(Y, K). Suppose that some K, is not indecomposable. Thus there exists e2 = e E Q such that e(y) is central in K, yet 0 # e(y) # 1. Since the sheaf K is reduced (cf. Definition 1.4), for each neighborhood N of y, there exists z E N and s E Q such that s(x) e(x) - e(x) s(x) # 0, (1) holds when x = z.
518
ANDREW
B.
CARSON
In fact, we may choose s such that Supp(s) = Supp(se - es).
(2)
To see this, let sr = (ese - se), sa = (ese - es), and note that es, = 0, sic = s, , es, = sa , and sac = 0. Hence, (2) holds with either sr or sa in place of s. Further, z E Supp(s,) u Supp(s,) for otherwise we would have (ese - se)(z) = 0 = (ese - es)(z), hence, (1) fails for z = X, a contradiction. We replace s with an si satisfying x E Supp(sJ. By Zorn’s lemma, find a family {tol} of elements of Q maximal with respect to the properties: (4
SuppkJ
= %-v&e
- et,) f
0,
for each (Y, and (b)
Supp(t,) n Supp(&) = ia, when 01# ,6.
Let C, = Supp(t,), for each t, . In view of (a), the sentences involving (1) and (2), and the maximality of the family, we have: YE cl u (Cm) * ( a 1
(3)
For each Mlet Qol= (4 EQ : Supp(q) C C,}. Each QDlis a two sided ideal in Q. By (b) the sum Ca Qolis direct. Since Q is a complete left quotient ring that is regular, Q is injective (cf. [9, Sect. 4.3 and 4.5]), sothat the left injective hull of CNQUis contained in Q. Further, &Q, is essentialover z:u Qa since the regularity of Q ensuresthat QQII # 0, for each 01.Therefore
CQ.C~Q~CQ, as left Q-modules. Thus 4 * kJ = (4 * !?cJ>
(4)
for any q E Q and (q=} E IIQ,, . We now show that GIJ - {d = {% * 91
(5)
alsoholds. Fix q EQ and define the left Q-module morphismf: DQa+ Q by f({q=}) = {q=} . q - {qa * q}. Clearly, C QaC ker( f),and (5) would follow if nQ,,C ker( f), Thus, if (5) fails, there exists {q,} EnQ,,, and e EQ such that 0 # e = f{(qJ). Since Q is regular we may assume,without loss of generality, that e2 = e. Thus, using (4), 0 # e = e2 = ef({qa}) =f({eq,}), so that eqB # 0, for some /3. Let t = eq, and choose t’ EQ such that tt’t = t. Replace t’ with
REGULAR
RINGS
OF FINITE
519
INDEX
t’tt’, zo that t’ E Qe also holds. Then t’ . {Q~} = {t’ * qol} = t’ * q8 E Q0 , hence t’ * e = f(t’ * {qa}) = f(t’ . qs) = 0, contradicting (eq,) t’(eq& = t-q8 # 0. Thus, (5) holds. Finally, {t&4
44
f
44k&>,
when x = y. To see this, suppose that borhood N of y, such that (6) fails to be such that Y(z) = 1 when x E N and Ye{&} so that, by (4) and (5), (ta Ye) = for suitable /I, and conclude that x $ Supp(t,e
(6)
it is false, and obtain a clopen neighhold for all z E N. Let YE r( Y, K) Y(z) = 0 otherwise. Then {tm}Ye = {Yet,}. By (3), find some z E C, n N,
- et& = C, ,
a contradiction. Hence, each K, is indecomposable. they are epimorphic images of Q. 1 Theorem
2.3 also holds if Q is a regular
Proof of TheoremA.
Let Y = X(Q),
The stalks are regular
complete
right
quotient
as
ring.
K = k(Q), and identify Q = r( Y, K).
(I) By [ 11, Theorem 11. l] Q is biregular iff the stalks of K are simple rings. However, they have index 71,and hence, by Proposition 2.1 and Theorem 2.3, are simple. (II) For each y E Y there exists, by (I) and Proposition 2.2(i), a division ring F, and a positive integer h(y) < n, such that K, = (Fy)A(y). To begin, we show that the function X is locally constant. Suppose that X is constant in no neighborhood of y, for some y E Y. Then, since Proposition 2.2(ii) holds, y E cl( I’), where V = (Ja (Supp(q,)), p > I\(y), and {qa} is a subset of Q maximal with respect to these properties: (a)
(q2
(b)
Supp(q,J
= 0 yet SUPP[(P~-~]
n Supp(~,J
= Supp(q,)
= m, whenever
# O,
for
any a,
and
a # /I.
Let 4: I’-+ K be such that q(z) = qn(x) when z E Supp(q,J. Then V is open and q is continuous, since the qmhave clopen support. Since oQ is injective (cf. [8, Sect. 4.5]), Theorem 1.5 yields q* E r(Y, K) satisfying q* IV = q. Since y E cl(V) and elements of Q have clopen support, we have (q*)“-l(y) # 0 yet (q*)” (y) = 0. Th is contradicts h(y) < p and shows that A is locally constant. Let {n(l),..., n(m)} = {X(y): y E Y}, where the n(i) =$ n are distinct and m < n. For each i, 1 < i < m, let Vi = {y E Y : X(y) = n(r)}. The Ui # ia and, since X is locally constant, partition X into clopen sets. Clearly,
520
ANDREW
B. CARSON
(I) Qg @Cz,Qi, where Qi = r(Ui, K), for 1
Qi g (DJnti) , when 1 < i < m.
The theorem follows from (1) and (2). 1 Many structure results for strongly regular rings are already available in [l, 3-5, 8, 111. To conclude this section, we represent the complete (right) quotient ring of a regular ring R of finite index, in terms of the sheafk(R). For the rest of this section, let Q denote the complete right quotient ring of the regular ring R of finite index. For any subsetS of R let S be the two-sided ideal of R generatedby S. It is known [9, Sect. 4.3, Proposition 61that the complete right quotient ring of any ring T has the form inj lim[Hom#, T): I is a denseright ideal in R]. PROPOSITION 2.4. (1) Each denseright ideal of R contains a two-sided ideal that is denseas a right ideal of R.
(2) Q is alsothe completeleft quotient ring of R. (3) Q E inj lim[Hom,(D, , R,): D is a denseideal in B(R)]. Proof. We shall usethe standard fact (cf. [9, Sect. 4.51)that the essential and the denseright idealsin a regular ring coincide. Consider the following condition for a ring S: (1’) Any essentialright ideal in S contains a two-sided ideal that is essentialas a right ideal. Martindale ([lo, Theorem 41) has established(1’) and (2) for any semiprime polynomial identity ring S, using Rowen’sresult ([12, Theorem 11)that any nonzero two sided ideal of S contains a nonzero central element. We may use Theorem 1.6 and Martindale’s proof to establish(1) and (2), for R. Result (3) would follow from (1) and the remark preceding this theorem, if D were densein R as a right ideal, where D = In B(R) for sometwo sided ideal I that is dense as a right ideal. However, this is true since if O#aER,thenO#eE[(RaR)nI]nB(R) = Dn[RaR],(byTheorem 1.6)sothatOfaeEaD. 1 In future, the common left and right complete quotient rings of R will be simply called the complete quotient ring of R. COROLLARY
2.5. (1) Q z inj lim{r(U, k(R)): U is a denseofxn subset
of WW.
(2) Q is regular, biregular and has index n.
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Proof. (1) Let D be a dense ideal in B(R). Identify R = r(X(R), k(R)), and let U = (J {Supp(e) : e ED}. Then, since k(R) is a reduced sheaf, U is a dense open subset of X(R) and D = {r E R : Supp(r) C U}. Much as in [l 1, Lemma 23.11, the map u -+ Y, satisfying y,,[e](x) = U(X). e(x), is an isomorphismr(U, k(R) -+ Hom,(DR , RR). The result now follows from Proposition 2.4.(3).
(2) This is immediate from (1) and Theorem A. The above is reminiscent of the following result, due to Fine, Gillman, and Lambek in [6]: (For any topological space Y and topological ring T, U(Y, T) denotesthe ring of all continuous functions from Y into T.) Let X denote a compact Hausdorff space,R the real number field in the usual topology, and Q the complete quotient ring of S = U(X, R). Then Q g inj lim{%( U, R): U is a denseopen subsetof X}. However, the two situations are different. For example, if X is also infinite and connected, then B(S) = (0, l} yet B(Q) is uncountable. This contrasts sharply with Proposition 3.1.
3. REGULAR RINGS OF FINITE INDEX WITH A COMPLETE BOOLEAN ALGEBRA OF CENTRAL TDEMPOTENTS In this section we prove Theorem B. PROPOSITION 3.1. Let R be a regular ring of jinite complete quotient ring of the ring S. Then
Q(W))
=
index, let Q(S)
be the
WQW
Proof. For any e E B(R), the left R-module morphism f: Q(R) ---f Q(R) defined by f(r) = re - er satisfies R C ker( f). Thus, by [9, Sect. 4.31, Q(R) C ker( f ). Hence, B(R) C B(Q(W
Any regular ring is a complete left quotient ring iff it is left self-injective. Thus, as in the commutative case,(using Corollary 1.8), cf. [ll, Proposition 24.1]), X(Q(R)) is extremally disconnected. Since X(Q(R)) = X(B(Q(R))) it is thus standard(cf. [3, Lemma 2.5(l)]) that B(Q(R)) is a complete quotient ring. Hence, the result follows if B(Q(R)) is essentialover B(R). Let 0 # e E B(Q(R)). Then, since RQ(R) is essentialover RR, there exists r E R such that 0 # re E R. Since R is quasi-biregular, find 0 f f E (RreR) n B(R). Then 0 # fe = f EB(R), so that B(Q(R)) is essentialover B(R). 1
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Proof of TheoremB. Let Q denote the complete quotient ring of R. Since B(R) is complete (as a Boolean algebra), it is its own complete quotient ring (cf. [9, Sect. 2.41). Hence, by Proposition 3.1, B(R) = B(Q) so that X(R) = X(Q) and k(R) is a subsheaf of k(Q). Thus, by Theorem A, each stalk k,(R) is a regular subring of index n of k,(Q) = (F3c)I\(z), for some division ring F, and positive integer A(x) < n. The result now follows from Theorem 1.2(i). 1 Theorem B does not give complete information about R since the topology on k(R), as well as the structure of its stalks, is required to recover r(X(R), k(R)). However, elements of I’(X(R), k(R)) do have clopen support, since Q is biregular and k(R) is a subsheaf of k(Q). Equivalently k(R) is Hausdorff [ll, Corollary 10.61. Similarly, the topology of k(R) is determined by that of k(Q) and it is determined by the Di , where the Di are as in Theorem A. This is only a partial help, since most complete structure results for strongly regular rings require that they be commutative algebraic algebras over some field. However, we do have the following Theorem. (Notation: for a ring S and topological space Y, let V(Y, S) denote the ring of all continuous, functions from Y to S, where S has the discrete topology.) THEOREM 3.2. Let R be a regular ring of index n that is alsoan algebraic algebraover the real numberjield, and let F denotethe quaternions.Suppose that B(R) is complete.Then there exist distinct integersn(1) < n,..., n(m) < n, for somem < n, and a partition X, ,..., X, of X(R) into clopensetssuchthat k(R) Ix.t is a subsheafof the simplesheafXi x (F),ti) , whenever1 < i < m.
Proof. Let Q be the complete quotient ring of R. By Theorem A find distinct integers n(1) < n,..., B(m) < n (for some m < n) and strongly regular rings D, ,... , D, such that
For each i (1 < i < m) let Xi = X(DJ and Ei = (d EDi : f (d) = 0 for some polynomial f with real coefficients). Then Ei is a ring, and by [3, concluding remark C] Ed _C U(Xi , F), when 1 < i < m. Clearly, us, since R is algebraic over the reals Th KG 1F)lnw s VXi >Ph). R C V-&
3 Ph)
0 ... 0 Vhn,
, P)n(m)).
By [ 11, pp. 26-271, the Xi partition X(Q) = X(R) so that since k(W(Xi, (F)a(dj) is the simple sheaf Xi x (F),ci) , k(R) Ixi is a subsheaf of X, x (F),ti) , whenever 1 < i < m. The topology of k(R) is completely determined, when Theorem 3.2 applies, since the Xi are clopen in X(R). 1
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4. ELEMENTARY PROPERTIES OF REGULAR RINGS OF FINITE INDEX The main result in this section is Theorem C. Proof of TheoremC. We shall construct a sequence (*)
R,-tS,-tR,~S,~...-tR,~S,~...
of rings and subring inclusions such that R = R, and (whenever i < i) 6) RtO 4, (ii) Qi is the complete quotient ring of Ri , (iii) (iv)
B(QJ C WC+J, Si is the subring of Ri+l generatedby Ri and B(Q,).
The required ring S will be inj lim(&). The rings Ri and Qi satisfying (i) and (ii) are defined inductively. Let R = R,, and chooseQ,, satisfying (ii). Supposethat Ri and Qi satisfying (i) and (ii) have been defined for all i and j < somem. It is immediate from Proposition 3.1 that: (v) in
Each atom (i.e., minimal nonzero element) in B(R,) is an atom
BfQd.
By [5, Lemma 2.21, (V) ensures that B(R,) C vB(Q,J. [2, Chap. 9, Lemma 3.91):
Hence, (by
(vi) There is an embedding LY,: B(Q,) -+ (B(R,))I/F, into some ultrapower of B(R,), which extends the natural map B(R,) -+ (B(R,))I/F. Let R,+l = (R#/F and chooseQ,,,+r to satisfy (ii). By Los’s Theorem (cf. [2, Chap. 5, Th eorem 2.1]), (i) holds whenever i
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Let 0 # s = r1 . e, + ... + ra . e, be an arbitrary element of Si where the r, E R and the e, E B(Q). We may (and do) select the rB and eBso that the e, are mutually orthogonal, and some r4 . e, # 0. Thus, r, * e, # 0 so that there is 0 # e E [(RYE * eoR) n R] n B(R). Thus 0 # e * (I~ * e,) = Y, * (e * eB) = ye * e E R. Since e . s = e . ye . e, = ye . e = Y, * e, this establishes(viii). As a consequencewe have, after suitable identifications, (ix)
S1C Qi .
Further Z(S) !EZ(Qi), since Qi is essentialover Si . Since B(Qi) C B(Si), this yields: (X)
B(Qi) = B(Si)*
Thus, X(S,) = X(Q,) so that, by [I 1, Proposition 6.81, k(Si) is a subsheaf of k(QJ and k,(Si) _Ck,(Qi), for all x E X($). In view of Theorem A and Proposition 2.2, this yields: (xi) k,(S,) has at most n nonzero orthogonal idempotents, for each XE X(Si). Let S = inj lim(&). Th en also S = inj lim(R,) so that by [2, Chap. 4, Theorem 2.11, Ri @ S, whenever i > 0. Identify S = r(X(S), k(S)). (xii) k,(S) has at most n nonzero orthogonal idempotents, for each x E X(S). To see (xii) supposethat O(x), el(x),..., e,+,(x), are distinct orthogonal idempotentsin k,(S), for somex EX(S). We assumewithout lossof generality that the ej are idempotents in S. Choose i such that e, ,..., e,,, E Si . Let s + s be the natural map Si z r(X(S,), k(S,)), and let 7: X(S) -+ X(S,) be the natural map induced by B(S,) C B(S) * (B(S,) _CB(S) since B(Si) = B(QJ C B(R,+J and Ri+l @ S.) As in [II, Lemma 6.31, there is a homomorphism @r:k,JSi) - k&Q such that @&s(rr(x))) = s(x), for all s E Si . This contradicts our choice of the ej since,by (xii), at leasttwo of O(rr(x)), e,(r(x)),..., e,+,(rr(x)), coincide. Thus, (xii) holds. Since R = R, @ S, S and its stalks are regular and of index n. The theorem now follows from (xii) and Theorem 1.2(i). An example follows to show that, in Theorem C, the structure result for the elementary extension S of R neednot apply to R itself. In fact, somestalk (s) of k(R) may even fail to be Artinian.
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EXAMPLE. Let X be the Boolean space w + 1, where w is the first infinite ordinal. Let F be a field, ki E (F)a when i < W, and k, = 9?(X, F), the ring of locally constant functions from X to F. Let k be the disjoint union {k, : x E X}. Let CL:w ---f w x w be a bijection and let p1and t+ be the induced maps pl: w-+w x w-w x {0} ++ w and ~a:w--+w x w--(O) x w t-$ w, respectively. Topologize k such that, when i < W, the points in ki are clopen in k and, for any neighborhood N of w in X andf E C(X, F).
is open in k. Let R = r(X, k). Then R is a regular ring of index 2. By construction, k is a reduced sheaf, so that X(R) = X and k(R) = k. The stalk k,,, is not Artinian. We conclude with someremarks: (a) In [l l] Pierce showedthat any ring R can be representedasI’(X(R), k(R)). If R is commutative, the stalks of k(R) are indecomposable,and if R is alsoregular, they are fields. Thus, many problems concerning commutative (regular) rings are reduced to simpler onesabout sheavesof indecomposable rings (fields). However, if R is noncommutative, the stalks of k(R) may be, to quote Pierce (cf. [l 1, Example 11.4]), “almost arbitrarily bad.” Theorem B shows that, for suitable noncommutative regular R, the stalks of k(R) are still manageable.In fact, for any regular complete left quotient ring T, the stalks of k(T) are indecomposable,although often of infinite index. Our present knowledge of such stalksis insufficient to produce deepresults for T. (b) Our results indicate that for some rings R, the Stone space(i.e., spectrum) of B(R) is more useful than the maximal spectrum of R, in achieving a representation of R as a ring of continuous functions (sections) with values in some nice ring (s). The spectrums of R and B(R) coincide if R is biregular or, in particular, if R is commutative and regular. (c) Although the structure result in Theorem C is for an elementary extension S of R, it doesyield somefacts about R itself. To illustrate this, we sketch a short proof of Fisher’s result that any regular ring R of finite index is zcnit reguZur. By definition, a ring is unit regular if it has the elementary PropeW (**)
(Vr)(Zls)(ilt)
[TST= Yand st = ts = 11.
Thus, it sufficesto show that the elementary extension S of R, described in Theorem C, is unit regular. Since the stalks of k(S) are given by (*) from Theorem B, it is standardthat they satisfy (* *). Thus, sincethe universal quantifier in (**) precedesthe existential quantifiers, [l I, Proposition 3.41 ensuresthat (**) also holds in S. 481/39/2-x3
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(d) If 2 is a unit in R, then each k,(S) is a finite direct sum of full matrix rings over division rings, none of which has Characteristic 2. Therefore, any element in k,(S) can be written as a sum of two units of k,(S), hence, the same result holds with R in place of k,(S). This result, for R, originated from Fisher.
ACKNOWLEDGMENT The author would like to thank the referee for detecting an error in the original proof of Theorem 1.6, and suggesting its present nonsheaf theoretic presentation.
REFERENCES 1. R. ARENS AND I. KAPLANSKY, Topological representation of algebras, Trans. Amer. Math. Sot. 63 (1948), 457-481. 2. J. L. BELL AND A. B. SLOMSON, “Models and Ultraproducts: An Introduction,” North-Holland, Amsterdam, 1969. 3. A. B. CARSON, Representation of semi-simple algebraic algebras, I. Algebra 24 (1973), 245-257. 4. A. B. CARSON, The model completion of the theory of commutative regular rings, J. Algebra 27 (1973), 136-146. 5. A. B. CARSON, Algebraically closed regular rings, Canad. J. Math. 26 (1974), 1036-1049. 6. N. J. FINE, L. GILLMAN, AND J. LAMBEK, “Rings of Quotients of Rings of Functions,” McGill Univ. Press, Montreal, 1965. 7. A. FORSYTHE AND N. H. MCCOY, The commutativity of certain rings, Bull. Amer. Math. Sot. 52 (1946), 523-526. 8. I. KAPLANSKY, Topological representation of algebras. II, Trans. Amer. Math. Sot. 68 (1950), 62-75. 9. J. LAMBEK, “Lectures on Rings and Modules,” Blaisdell, Waltham, Massachusetts, 1966. 10. W. MARTINDALE III, On semi-prime P.I. rings, Pvoc. Amer. Math. Sot. 40 (1973). 365-369. 11. R. S. PIERCE, Modules over commutative regular rings, Mem. Amer. Math. Sot. 70 (1967). 12. L. ROWEN, Some results on the centre of a ring with polynomial identity, Bull. Amer. Math. Sot. 79 (1973), 219-223. 13. L. LIPSHITZ AND D. SARACINO, The model companion of the theory of commutative rings without nilpotent elements, PYOC. Am. Math. Sot. 38 (1973). 381-387.