On reflexivity of direct sums

Necessary and sufficient conditions are presented to insure that the direct sum of two reflexive representations of a finite dimensional algebra is reflexive, and it is shown that for each such algebra, there is an integer k such that the direct sum of k copies of each of its representations is reflexive. Given a ring i\, our results are actually presented in the more general setting of A\-representations of a ring R. Over the last thirty years, a significant amount of attention has been given to the problem of determining when an algebra A of operators on a vector space V (often a Hilbert space) over a field K is reflexive in the sense that no larger algebra of operators on that space has the same lattice of invariant subspaces. (See, for example, [2], [3], [6], [7], [12], [13], [14].) Of course, A is an algebra of operators on a vector space V if and only if V is a faithful left A-module. Thus, to attack this problem from a more module theoretic point of view, based on notation of Halmos [14], the following notions were presented in [9]: For any R A-bimodule M= RMA one defines alglat(M) {a E End(MA\) I aL C L for all RL alglat(MA) denote the canonical ring homomorphism, M is called a reflexive bimodule (or A-representation of R) if A is surjective. Thus the K-algebra of operators A on V is reflexive if and only if A VK is a reflexive bimodule (and then we simply say that the A-module V is reflexive). The problems we shall deal with here have their roots in the papers [5] and [6] of Deddens and Fillmore and [2] of Azoff. Stated in module theoretic terms, in the first pair of papers the question was posed and answered (affirmatively) of whether a direct sum of two finitely generated reflexive modules over an algebra generated by a single complex matrix is again reflexive; and Azoff showed that for each finite dimensional module over a C-algebra, there is a positive integer k, depending on its dimension, such that the direct sum of k copies of that module is reflexive. Other results in this vein can be found in [4] where Brenner and Butler showed that the direct sum of two copies of the regular representation of a finite dimensional algebra is reflexive, and in Habibi and Gustafson's [12] from which the same result follows for any faithful representation of a split serial algebra. (See [9], [10] and [11] for related and more general results.) Our main objectives are to provide, in Theorem 1.2, a necessary and sufficient condition for a direct sum of two reflexive bimodules to be reflexive; and to show, in Received by the editors September 10, 1998 and, in revised form, November 10, 1998. 1991 Mathematics Subject Classification. Primary 16D20, 16G99, 16P10; Secondary 47A15. ?)2000 American Mathematical Society 2855 This content downloaded from 157.55.39.191 on Tue, 11 Oct 2016 05:10:44 UTC All use subject to http://about.jstor.org/terms 2856 V. P. CAMILLO AND K. R. FULLER Theorem 3.2, that whenever R is a left artinian ring with composition length c(RR), there is a positive integer k (am, fn). Thus if -y E alglat(M e N), then a = (-YIM, -YIN). A common subquotient of a pair of left modules RM and RN is a module RX that is isomorphic to a subquotient of both RM and RN. This notion together with the following lemma allows us to determine just when the direct sum of a pair of reflexive modules is reflexive. Lemma 1.1. If r, s E R, the ordered pair (r, s) E alglat(M D N) if and only if (r s)X = 0, for every (equivalently, every cyclic) common subquotient of RM and RN. Proof. Note that a E R annihilates every common subquotient of M and N if and only it annihilates each of their cyclic common subquotients. According to Goursats' Lemma, W is an R-submodule of M ED N, if and only if there are submodules M2 N,N2 such that W {(ml, ni) E Ml @ N, f(ml + M2) ni + N2}. (Given W and the orthogonal projections 7rM and 7rN for M ED N, one checks that M1 = rM(W), M2 = M n W, N1 = 7rN(W), and N2 = N n W.) (=>) If (ml,ni) E W and (rmi,sni) E W, then f(rmi + M2) = sn1 + N2 = sf(ml + M2) so, since f is an R-isomorphism, (r s)(Ml/M2)= 0. (^=) Suppose (r s)(Ml/M2) = 0, and f: M1/M2 -* N1/N2 is an isomorphism and (m1,ri) E W. Then f(rmi + M2) = rf(ml + M2) = rnm + N2 = sn, + N2 so (rm1,sn1) E W. Now we are able to provide the promised characterization in terms of annihilators of subquotients of M and N. The left annihilator of a module M is ?R(M) {r E RI rM= O}. This content downloaded from 157.55.39.191 on Tue, 11 Oct 2016 05:10:44 UTC All use subject to http://about.jstor.org/terms REFLEXIVITY OF DIRECT SUMS 2857 Theorem 1.2. Let M and N be reflexive, and let {Xi I i E I} represent one copy of each of the (cyclic) common subquotients of RM and RN. Then M ED N is reflexive if and only if ?R(M) + tR(N) = iR(D xi) iEl Proof. ( ) Let r E iR(GiE' Xi). Then by Lemma 1.1, (r, 0) E alglat(M ED N). Thus, assuming that M ED N is reflexive, (r,0) = (s,s) and r = (r-s) + s EE R(M) + eR(N). Since always ?R(M) + ?R(N) C iR(@iEI Xi), the proof is complete. F We note that the proof (=>) above shows that if M e N is reflexive (regardless of reflexivity of M and N), then ?R(M) + ?R(N) = ?R(@iCI Xi). From the inclusions 0 C iR(M) + eR(N) C ?R(ED Xi) C iR(Xi) C R iCI we easily obtain the following two corollaries: Corollary 1.3. If M and N are reflexive and have a common faithful subquotient, then M D N is reflexive. Corollary 1.4. If RZ is semiperfect and M and N are reflexive and have no common composition factor, then M ffl N is reflexive. Proof. Suppose that e1, ..., em, em?1, ..., en is a complete set of primitive idempotents such that Rei/Jej is not a composition factor of M for i = 1, ..., m, and Rei/Jej is not a composition factor of N for i = m+ 1, ..., n. Then ?R(M) + iR( N) R. (See [1, Section 27].) Regarding the Deddens-Fillmore result, the algebra generated by a matrix over a field K, being isomorphic to a proper factor of the polynomial ring K[x], is an example of a split commutative uniserial algebra, i.e., a direct product of local uniserial rings. Since a module over a direct product of rings is reflexive precisely when its corresponding components are reflexive, to show that direct sums of reflexive modules are reflexive over such an algebra R, we may assume that R is a local uniserial algebra, so that the ideals of R are linearly ordered and every R-module is a direct sum of factors of R (see [1, Section 32]). In this case, R/IR(M) embeds in M and every subquotient of RM is a factor of R/lR(M). In the presence of these facts Theorem 1.2 yields the following corollary almost at once. We note, however, that [10, Theorem 1] is more general. This content downloaded from 157.55.39.191 on Tue, 11 Oct 2016 05:10:44 UTC All use subject to http://about.jstor.org/terms 2858 V. P. CAMILLO AND K. R. FULLER Corollary 1.5. The direct sum of a finite number of reflexive modules over a split uniserial algebra is reflexive. Proof. Since we may assume that the ideals of R are linearly ordered, given modules M and N, we may also assume that ?R(M) C ?R(N) C ?R(X) for every common subquotient X of M and N. So since R/lR(N) embeds in N, we see that both sides of the desired equality are equal to ?R(N). Next, as further applications of this theorem, we shall see that it yields particularly nice proofs of some key results in [10]. In [10], in order to show that direct sums of reflexive modules may fail to be reflexive over a split K-algebra R with radical J whose quiver contains a triple arrow, a pair of reflexive modules was constructed with diagrams (as in [8]) u V w M: al b\ cl and N: a, c x y z with a, b, c linearly independent elements of J \ J2. (The diagram indicates that au = x, aw = cw = z, etc.) Here ?R(M) + ?R(N) = ?R(M) + K(a c) + Kb, but the only common subquotients of M and N are simple modules, so ?R(G X =) =R(M) + J R ?R(M) + tR(N). iEI Thus the theorem shows that M D N is not reflexive. In the positive vein we shall employ the following lemma and Theorem 1.2 to obtain a simple proof of a key part of [10, Proposition 4]. Lemma 1.6. Let R be a split local K-algebra of dim(KR) < 3. If M is a faithful left R-module, then every proper cyclic module is isomorphic to a subquotient of M. Proof. (This proof is essentially contained in the proof of [10, Proposition 4].) If R is uniserial, then RR embeds in M, so we may assume that J2 = 0 and that dim(KJ) = 2. Let u E M\SocM and suppose that WR(u) : 0. Then, since u ? SocM, there is a b C J with ?R(u) = Kb. But then there is a v E M with bv z 0, and there is an a E J with Ka =R(V). Thus it follows that J = Ka DKb.