Radical Endomorphisms of Decomposable Modules

Transcription

1 Radical Endomorphisms of Decomposable Modules Julius M. Zelmanowitz University of California, Oakland, CA USA Abstract An element of the Jacobson radical of the endomorphism ring of a decomposable module is characterized in terms of its action on the components of the decomposition. This extends to arbitrary decomposable modules a result previously known only for the special case of free modules. Key words: AMS Classification: 16N20, 16S50 1 Introduction The historical motivation for this study may be said to begin with a question raised by N. Jacobson in Structure of Rings, first published in 1956 [4]. On page 23 of that seminal text, Jacobson asked for a characterization of the elements in the radical of the ring M I (R) of I I row-finite matrices over an arbitrary ring R. In general, it is not true that a matrix whose entries lie in the radical of R is in the radical of M I (R), as will be demonstrated below. For a ring R we let J(R) denote the Jacobson radical of R, and we consider the free R-module of infinite rank consisting of all row vectors (r 1, r 2,...) such that each r i R and r i = 0 for almost all i. Let a 1, a 2,... be a sequence of elements from J(R) and consider the row-finite matrix The author wishes to acknowledge with gratitude the assistance of the Department of Physics at the University of California, Santa Barbara, with the preparation of this manuscript. address: julius.zelmanowitz@ucop.edu (Julius M. Zelmanowitz). Preprint submitted to Elsevier Science 1 April 2004

2 0 a a 3 0 α = a If 1 α has a left inverse, then there will be a row vector (b 1, b 2,...) with (b 1, b 2,...)(1 α) = (a 1, 0, 0,...). This leads to the system of equations: b 1 = a 1 b i b i 1 a i = 0, for all i 2. Solving recursively, we get b i = a 1 a 2... a i for all i 1. Since b n = 0 for some n 1 we conclude that a 1 a 2... a n Thus, a necessary condition for J(M I (R)) = M I (J(R)) to hold for I an infinite set is that J(R) be left T-nilpotent. In 1961, E.M. Paterson showed conversely that left T-nilpotence of J(R) is sufficient for J(M I (R)) = M I (J(R)) to hold [6]. In particular then, any local integral domain which is not a field is an example of a ring for which J(M I (R)) M I (J(R)) whenever I =. Jacobson s question was settled in 1969 by N. Sexauer and J. Warnock who employed a daunting calculation to establish the following characterization of the radical elements in a ring of row-finite matrices. Theorem. (N. Sexauer & J. Warnock [7]) For an arbitrary ring R and a matrix α M I (R), α J(M I (R)) if and only if α M I (J(R)) and the column left ideals of α are right vanishing. We explain the preceding terminology. The column left ideals of α = (a ij ) i,j I M I (R) are the left ideals of the form { } A j (α) = r i a ij r i R and r i = 0, for almost all i I i I for each j I. An arbitrary family of left ideals {A j j I} of R is called right vanishing if for every sequence of elements a ik A ik with i 1, i 2,... a sequence of distinct elements of I, there exists an integer n 1 with a i1 a i2... a in A ring of row-finite matrices over a ring with identity element is isomorphic to the endomorphism ring of a free module. Accordingly, this theorem was generalized to the endomorphism ring of an arbitrary projective module in [8], where a more conceptual proof of the Sexauer & Warnock theorem was also presented. The key ideas in [8] were the use of the characterization of the radical of a ring as the sum of the small (superfluous) one-sided ideals as well as an often-mimicked calculation contained in the work of H. Bass [1]. 2

3 Another direction for generalization is to determine the elements of the radical of the endomorphism ring of a decomposable module M = i I M i for some suitable family of modules {M i i I} in terms of their action on the components of the decomposition; the test criterion for suitability being that the case where each M i = R is subsumed and the Sexauer & Warnock theorem captured as a special case. This can lead to consideration of a number of possible extensions: to the case when all M i are isomorphic; to the case when all M i are cyclic; to the case when all M i are finitely generated, and so on. Before we describe our principal result, it is worth considering motivation for this line of research coming from another source. In a successful search for extensions of the classical refinement theorems for direct sum decompositions of modules, Crawley and Jónsson introduced the concept of an exchange property for a module. An R-module M is said to have the (finite) exchange property if whenever M occurs as a direct summand of a (finite) direct sum N = i I N i, then N = M ( i I N i) for some submodules N i N i [2]. For modules that are direct sums of indecomposable modules, the (finite) exchange property completely characterizes when such decompositions complement direct summands. So does a local vanishing condition on endomorphisms of the module, as we now explain. First, a definition: a family of modules {M i i I} is called locally semi-tnilpotent if for each sequence M i1 Mi2 Mi3... of non-isomorphisms f 1 f 2 f 3 with pairwise distinct indices i k I, and for each x M i1, there exists an integer n 1 such that xf 1 f 2... f n Theorem. Suppose that M = i I M i where each M i is indecomposable and let S = End R M. Then the following conditions are equivalent. (1) M has the exchange property. (2) M has the finite exchange property. (3) The decomposition M = i I M i complements direct summands. (4) Each M i has a local endomorphism ring and the family {M i i I} is locally semi-t-nilpotent. (5) Each M i has a local endomorphism ring, S/J(S) is von Neumann regular and idempotents lift modulo J(S). The equivalence of (1), (2) and (4) is due to Zimmerman-Huisgen and Zimmerman [10], and the equivalence of the last three conditions to Harada and collaborators (see [3] for more complete references). From our perspective, condition (5) indicates that knowledge of the structure of J(S) determines exchange properties for a completely decomposable module, and condition (4) suggests a link with the Sexauer & Warnock theorem since the criterion in (4) is expressed by a vanishing condition. Motivated by these observations, we develop a characterization of the elements in the Jacobson radical of the 3

4 endomorphism ring of an arbitrary decomposable module. Somewhat surprisingly, the characterization in Theorem 1 does not require any restriction on the summands of the decomposition. In Theorem 2 we also describe the radical elements in the important subring of those endomorphisms which are locally of finite rank. 2 Radical elements of End R ( i I M i ). We begin by introducing some convenient notation. Let R be an arbitrary ring (not necessarily containing an identity element), let M = i I M i be a decomposition of left R-modules, and set S = End R M acting as right operators on M. For i I, we let e i S denote the projection homomorphism of M onto M i across j I\{i} M j, and we set S i = End R M i. Similarly, for any subset F I, set M F = i F M i, S F = End R (M F ), and let e F denote the projection homomorphism of M onto M F across j I\F M j. We will consistently regard S i and S F, respectively, as subrings of S by extending homomorphisms trivially across the complementary summands j I\{i} M j and i I\F M i, respectively; that is, we identify S i = e i Se i and S F = e F Se F. For any α S, i, j I, and F, G I, we abbreviate α ij = e i αe j, α F G = e F αe G, and α F = α F F. (This notation is consistent with the definitions of e i and e F as projection endomorphisms if we take e = 1 S, the identity endomorphism in S.) Finally, observe that for an infinite family β j S, j J, the sum β = j J β j defines an element of S provided that for every x M, xβ j = 0 for almost all j J. The principal result of this paper is the following characterization of the radical elements in S = End R M. Theorem 1. For M = i I M i with I an infinite index set and α S = End R M, either one of the following conditions are necessary and sufficient for α J(S) to hold. (1) For every γ S and every i I, (γα) ii J(S i ) and, for every sequence i 1, i 2,... of distinct elements of I and every x M, there exists a positive integer n with x(γα) i1 i 2 (γα) i2 i 3... (γα) ini n+1 (2) For every i I, (Sα) ii J(S i ) and, for every sequence i 1, i 2,... of distinct elements of I, every sequence γ 1, γ 2,... S, and every x M, there exists a positive integer n with x(γα) i1 i 2 (γα) i2 i 3... (γα) ini n+1 The proof that condition (2) is sufficient to imply that α J(S) relies on a lemma from [9], which is itself an adaptation of an argument presented in [5] in the special case when the summands of the decomposition have local endomorphism rings. Its statement is as follows. 4

5 Lemma 1. Suppose that M = i I M i with I an infinite set and that β S = End R M is such that β F = e F βe F has a left inverse in S F = End R M F for every finite subset F I. Then for any x M there exists a sequence of finite subsets = F 0 F 1 F 2... of I and endomorphisms µ 1, µ 2,... S, δ 1, δ 2,... S such that: (i) for each n 1, xµ n β = x(1 + δ 1 δ 2... δ n ); and (ii) each δ k S Fk β Fk,F k+1 \F k. This lemma will be applied to establish the invertibility of β = 1 + α when condition (2) of the theorem holds. For, under those circumstances, condition (ii) allows one to conclude that the product δ 1 δ 2... δ n is eventually zero, and condition (i) then implies that β = 1 + α is an epimorphism. We include a brief proof of the lemma in order to keep this exposition self-contained. Proof of the lemma. We proceed by induction on n. For F I, set F = I\F. Choose F 1 a finite subset of I with x M F1. Set β 1 = β F1 = e F1 βe F1 S F1, β 1 = e F 1 βe F 1, and β 1 = e F 1 β; then β = β 1 + β 1 + β 1. By hypothesis, there exists µ 1 S F1 with x = xµ 1 β 1 = xµ 1 (β β 1 β 1 ) = xµ 1 (β β 1) because M F1 β 1 Hence xµ 1β = x(1 + µ 1 β 1 ). Choose F 2 to be a finite subset of I which properly contains F 1 and with xµ 1 β M F2. Then xµ 1 β 1 = xµ 1β x M F2 M F 1 = M F2 \F 1 so, taking δ 1 = µ 1 β 1e F2 \F 1 S F1 β F1,F 2 \F 1, we have xµ 1 β = x(1 + µ 1 β 1 ) = x(1 + µ 1β 1 e F 2 \F 1 ) = x(1 + δ 1 ), which establishes the case n = 1. Now assume that n 2 and that the (n 1) st case has been established, so that xµ n 1 β = x(1 + δ 1... δ n 1 ) with each δ k S Fk β Fk,F k+1 \F k. As above, write β = β n + β n + β n where β n = β Fn = e Fn βe Fn S Fn, β n = e F n βe F n, and β n = e F n β. Then xδ 1... δ n 1 M Fn, so by hypothesis, there exists ν n S Fn with xδ 1... δ n 1 = xδ 1... δ n 1 ν n β n = xδ 1... δ n 1 ν n (β β n β n ) = xδ 1... δ n 1 ν n (β β n ) because M Fn β n From the induction hypothesis, so that xµ n 1 β = x + xδ 1... δ n 1 = x + xδ 1... δ n 1 ν n (β β n ), x(µ n 1 δ 1... δ n 1 ν n )β = x(1 δ 1... δ n 1 ν n β n ). (1) Set µ n = µ n 1 δ 1... δ n 1 ν n and choose F n+1 to be a finite subset of I 5

6 which properly contains F n and xµ n β. Then xδ 1... δ n 1 ν n β n = x xµ nβ M Fn+1 M F n = M Fn+1 \F n so, taking we have δ n = ν n β n e F n+1 \F n = ν n e Fn βe F n e Fn+1 \F n (2) = ν n β Fn,F n+1 \F n S Fn β Fn,F n+1 \F n, (3) xµ n β = x xδ 1... δ n 1 ν n β n = x xδ 1... δ n 1 ν n β n e F n+1 \F n = x(1+δ 1... δ n 1 δ n ). This establishes the lemma. Proof that condition (2) implies that α J(S). Since condition (2) is also satisfied by να for every ν S, it suffices to prove that β = 1 + α is a unit in S. Let F be any finite subset of I, and for each j F, set C j = { i F µ ij µ S and e j Sµ ij J(S j ) for every i F }. Then each C j is a left ideal of S F, in fact a quasi-regular left ideal of S F because, as is easily checked, µ jj J(S j ) and (e F i F µ ij ) 1 = (e j µ jj ) 1 + i F \{j} (e i + µ ij (e j µ jj ) 1 ), with (e j µ jj ) 1 the inverse of e j µ jj in S j. Hence µ F = i,j F µ ij j F C j J(S F ) for every finite subset F of I. In particular, since, by hypothesis, e j Sα ij (Sα) jj J(S j ) for every j F, α F J(S F ) and β F = (1 + α) F = e F + α F is a unit in S F for every finite subset F of I. We may therefore apply the Lemma to learn that for each x M there exists a sequence of finite subsets = F 0 F 1 F 2... of I and homomorphisms µ 1, µ 2,... S, δ 1, δ 2,... S such that: (i) for each n 1, xµ n β = x(1 + δ 1 δ 2... δ n ); and (ii) each δ k S Fk β Fk,F k+1 \F k. We first establish that xδ 1 δ 2... δ n = 0 for some n 1. By an application of the König Graph Theorem, it suffices to show that for every choice of i k F k \F k 1 with k 1, there exists m 1 (depending on the choice of the sequence {i k }) with x(e i1 δ 1 e i2 )(e i2 δ 2 e i3 )... (e im δ m e im+1 ) Using (ii), for each k 1 we may write each δ k = γ k β Fk,F k+1 \F k with γ k S Fk. Then for each k 1, e ik δ k e ik+1 = e ik γ k β Fk,F k+1 \F k e ik+1 = e ik γ k e Fk (1 + α)e Fk+1 \F k e ik+1 = e ik γ k αe ik+1 = (γ k α) ik i k+1. 6

7 Hence, from condition (2), it follows that there exists m 1 with x(e i1 δ 1 e i2 )(e i2 δ 2 e i3 )... (e im δ m e im+1 ) = x(γ 1 α) i1 i 2 (γ 2 α) i2 i 3... (γ m α) imi m+1 = 0, and thus xδ 1 δ 2... δ n = 0 for some n 1. From (i) we know that xµ n β = x and, since x M was arbitrary, this proves that β is an epimorphism. Since β F is a unit for each finite subset F of I, β is also a monomorphism, and this completes the proof that condition (2) is sufficient for α J(S) to hold. The fact that (1) implies (2) follows from the observation that for every sequence i 1, i 2,... of distinct elements of I and every sequence γ 1, γ 2,... S, (γ k α) ik i k+1 = e ik γ k αe ik+1 = e ik e ij γ j αe ik+1 j 1 = e ij γ j α j 1 where γ = j 1 e ij γ j S. i k i k+1 = (γα) ik i k+1 In order to prove the necessity of (1), let α J(S) be given. Since γα J(S) for every γ S, it suffices to show that α ii J(S i ) for every i I (which is clear since α ii = e i αe i e i J(S)e i = J(S i )), and that for every sequence i 1, i 2,... of distinct elements of I and every x M, there exists a positive integer n 1 with xα i1 i 2 α i2 i 3... α ini n+1 Without loss of generality, we may assume that {1, 2,...} I and we replace each i j by j. To further simplify this presentation, we introduce some additional notation. Let 1 = k 1 < k 2 <... be an arbitrary increasing sequence of positive integers; later, we will specify a particular sequence for the purpose of this proof. Set x 1 = xe 1, and for each i 1, set x i+1 = x i α ki,k i +1α ki +1,k i α ki+1 1,k i+1 = x i α ki,k i +1β ki +1,k i+1, where β ki +1,k i+1 = α ki +1,k i α ki+1 1,k i+1 Hom R (M ki +1, M ki+1 ). Since x n+1 = xα 12 α α kn+1 1,k n+1 we have to show that x n+1 = 0 for some n 1. Set β = i 1 β ki +1,k i+1 ; β is a well-defined element of S, and the notations are compatible in the sense that for each i 1, e ki +1βe ki+1 = β ki +1,k i+1. Also, observe that e h βe l = 0 whenever (h, l) (k i +1, k i+1 ) for some i 1. Further, note that for each i 1, e ki αβ = j 1 α ki,k j +1β kj +1,k j+1, which we rewrite as e ki + i 1 j=1 α k i,k j +1β kj +1,k j+1 = (e ki j i α ki,k j +1β kj +1,k j+1 ) + e ki αβ. We introduce the abbreviations u i = e ki j i α ki,k j +1β kj +1,k j+1 and v i = 7

8 e ki + i 1 j=1 α k i,k j +1β kj +1,k j+1 for i 1 (by convention, v 1 = e k1 ), and we let U = {s S s = i 1 s i u i for some sequence s i S}. Then each u i, v i S, v i = u i + e ki αβ and U is a left ideal of S. Furthermore, U Hom R (M, M 0 ) where M 0 = i 1 M ki. Finally, put v = i 1 v i ; v is a well-defined element of S and v Hom R (M, M 0 ). We begin the proof proper by first showing that v M0 is an automorphism of M 0. To see this, note that v M0 is the ascending union of {v (m) m 1} where v (m) = m i=1 v i = m i=1 (e ki + i 1 j=1 α ki,k j +1β kj +1,k j+1 ) = m i=1 e ki + mi=1 i 1 j=1 e ki αe kj +1βe kj+1 End R (M k1... M km ). Furthermore, m i 1 i=1 j=1 e ki αe kj +1βe kj+1 J(End R (M k1... M km )) because α J(S). Since m i=1 e ki is the identity automorphism of M k1... M km, it follows that v (m) is an automorphism of M k1... M km for each m 1. Hence v M0 is an automorphism of M 0. Next, v = i 1 v i = i 1 u i + i 1 e ki αβ = ( i 1 u i ) + e0 αβ U + Sαβ, where e 0 = i 1 e ki is the identity element of M 0. Then e 0 = (v M0 ) 1 (v M0 ) = (v M0 ) 1 v U + Sαβ Hom R (M, M 0 ) = Se 0, and therefore U + Sαβ = Se 0. Since α J(S), Sαβ is a small submodule of S S, hence of Se 0, and therefore U = Se 0. In particular, e 1 = e k1 = e k1 e 0 Se 0 = U, so we may write e 1 = i 1 s i u i U for some choice of s i S. Since x 1 = xe 1 = i 1 xs i u i, there must exist an integer n 1 with xs k u k = 0 for all k > n. Hence n n x 1 = xs i u i = xs i (e ki α ki,k j +1β kj +1,k j+1 ) i=1 i=1 j i n i 1 = xs 1 e k1 + (xs i e ki xs j α kj,k i 1 +1β ki 1 +1,k i ) i=2 j=1 i 1 xs j α kj,k i 1 +1β ki 1 +1,k i. i>n j=1 Examining the first n+1 components of this equation in M 0 = i 1 M ki yields the following system of equations: x 1 = xs 1 e k1 xs i e ki = i 1 j=1 xs j α kj,k i 1 +1β ki 1 +1,k i for i = 2, 3,..., n nj=1 xs j α kj,k n+1β kn+1,k n+1 We now fix a particular choice for the sequence 1 = k 1 < k 2 <... recursively as follows: k 1 = 1 and k i having been chosen, k i+1 > k i is chosen so that x i α ki,h = 8

9 0 for every h > k i+1. This choice is possible because x i α ki,h = x i e ki αe h = 0 for almost all h. With this choice for the sequence, we can solve the preceding system of equations. Substituting the first equation in the second gives xs 2 e k2 = xs 1 α k1,k 1 +1β k1 +1,k 2 = xs 1 e k1 α k1,k 1 +1β k1 +1,k 2 = x 1 α k1,k 1 +1β k1 +1,k 2 = x 2. Substituting this into the third equation gives xs 3 e k3 = xs 1 α k1,k 2 +1β k2 +1,k 3 + xs 2 α k2,k 2 +1β k2 +1,k 3 = xs 1 e k1 α k1,k 2 +1β k2 +1,k 3 + xs 2 e k2 α k2,k 2 +1β k2 +1,k 3 = x 1 α k1,k 2 +1β k2 +1,k 3 + x 2 α k2,k 2 +1β k2 +1,k 3 = x 2 α k2,k 2 +1β k2 +1,k 3 = x 3 because x 1 α k1,h = 0 for all h > k 2. Continuing in this manner through the n th equation, we have that xs i e ki = x i for i = 2, 3,..., n. Inserting this into the final equation yields n j=1 x j α kj,k n+1β kn+1,k n+1 Again, since x j α kj,h = 0 for every h > k j+1, we conclude that x n α kn,k n+1β kn+1,k n+1 That is, x n+1 = 0, and with this the proof of the theorem is concluded. 3 Endomorphisms locally of finite rank. In this section, with all notation as in the previous section, we consider an important subring of S = End R M relative to the decomposition M = i I M i, namely S 0 = {α S for each i I, M i αe j = 0 for almost all j I}. Very loosely speaking, S 0 consists of the endomorphisms of M which are locally of finite rank relative to the decomposition M = i I M i. Other descriptions are S 0 = {α S for each i I, e i α = e i αe G for some finite subset G I}, and S 0 = {α S for each finite subset F I, e F α = e F αe G for some finite subset G I}. Observe that e G S 0 for every subset G I and that S F S 0 for every finite subset F I. A relatively straightforward adaptation of the proofs of Lemma 1 and Theorem 1 provides a characterization of the elements in the Jacobson radical of S 0. The description is a simplification of that given in Theorem 1 and proofs will therefore be omitted. We first recall some information about the Jacobson radical in R-Mod. For K, L R-Mod, J(Hom R (K, L)) = {f Hom R (K, L) for all g Hom R (L, K), 1 L +gf is an automorphism of L} = {f Hom R (K, L) for all h Hom R (L, K), 1 K + fh is an automorphism of K}. Also, J(Hom R (K, L)) is an ideal in the 9

10 category R-Mod; that is, it is closed under addition and for any X, Y R-Mod, Hom R (X, K)J(Hom R (K, L))Hom R (L, Y ) J(Hom R (X, Y )). Lemma 2. Suppose that M = i I M i with I an infinite set and that β S 0 End R M is such that β F = e F βe F has a left inverse in S F = End R M F for every finite subset F I. Then for any finite subset F 1 I there exists a sequence of finite subsets F 1 F 2 F 3... of I and endomorphisms µ 1, µ 2,... S 0, δ 1, δ 2,... S 0 such that: (i) for each n 1, µ n β = e F1 (1 + δ 1 δ 2... δ n ); and (ii) each δ k S Fk β Fk,F k+1 \F k. Theorem 2. For M = i I M i with I an infinite index set and α S 0 End R M, either one of the following conditions are necessary and sufficient for α J(S 0 ) to hold. (1) For every γ S 0 and every i I, (γα) ii J(S i ) and, for every sequence i 1, i 2,... of distinct elements of I, there exists a positive integer n with (γα) i1 i 2 (γα) i2 i 3... (γα) ini n+1 (2) For every i, j I, α ij J(Hom R (M i, M j )) and, for every sequence i 1, i 2,... of distinct elements of I, and every sequence γ 1, γ 2,... S 0, there exists a positive integer n with (γ 1 α) i1 i 2 (γ 2 α) i2 i 3... (γ n α) ini n+1 Since S = S 0 whenever each module M i in the decomposition M = i I M i is finitely generated, we have the following immediate consequence of Theorem 2. Corollary 1. Suppose that M = i I M i with each M i a finitely generated R- module and let α S = End R M. Then α J(S) if and only if the following two conditions hold: (i) for every i, j I, α ij J(Hom R (M i, M j )); and (ii) for every sequence γ 1, γ 2,... S, and for every sequence i 1, i 2,... of distinct elements of I, there exists a positive integer n with (γ 1 α) i1 i 2 (γ 2 α) i2 i 3... (γ n α) ini n+1 Corollary 2. (N. Sexauer & J. Warnock [7]) For an arbitrary ring R and a matrix α M I (R), α J(M I (R)) if and only if α M I (J(R)) and the column left ideals of α are right vanishing. Proof. For a ring R with identity element, this follows immediately from the preceding corollary. For an arbitrary ring R, we can regard R as an ideal of an overring R 1 which contains an identity element. The result then follows immediately from the fact that M I (R) is an ideal of M I (R 1 ) and J(M I (R)) = J(M I (R 1 )) M I (R). 10

11 In certain circumstances the Jacobson radical of S 0 has a particularly simple structure. For example, consider E.M. Patterson s result, cited in the introduction to this paper, that J(M I (R)) = M I (J(R)) for I an infinite set if and only if J(R) is left T-nilpotent. This is extended to arbitrary module decompositions in Corollary 3 below. First, a definition: a family of modules {M i i I} is called semi-t-nilpotent if f 1 f 2 f 3 for each sequence M i1 Mi2 Mi3 with each fk J(Hom R (M ik, M ik+1 )) and with pairwise distinct indices i k I, there exists an integer n 1 such that f 1 f 2... f n (This definition is compatible with the earlier concept of locally semi-t-nilpotent families for modules with local endomorphism rings because the requirement that f k J(Hom R (M ik, M ik+1 )) is equivalent to f k being a non-isomorphism in that case.) Corollary 3. For M = i I M i with I an infinite index set, J(S 0 ) = {α S 0 α ij J(Hom R (M i, M j )) for all i, j I} if and only if {M i i I} is semi-t-nilpotent. Proof. Suppose that {M i i I} is semi-t-nilpotent. It is always true that J(S 0 ) {α S 0 α ij J(Hom R (M i, M j )) for all i, j I} so suppose that α S 0 is such that α ij J(Hom R (M i, M j )) for all i, j I. Then given distinct elements i 1, i 2,... of I and a sequence γ 1, γ 2,... S, there exists a positive integer n with (γ 1 α) i1 i 2 (γ 2 α) i2 i 3... (γ n α) ini n+1 = 0 because {M i i I} is semi-t-nilpotent. Hence, by Theorem 2(1), we have that α J(S 0 ). Conversely, suppose that J(S 0 ) = {α S 0 α ij J(Hom R (M i, M j )) for all f 1 f 2 f 3 i, j I} and let the sequence M i1 Mi2 Mi3 be given with each f k J(Hom R (M ik, M ik+1 )) and with pairwise distinct indices i k I. Set α = k 1 e ik f k e ik+1. Then α S 0 and 0, if (i, j) (i k, i k+1 ) for any k 1 α ij = f k, if (i, j) = (i k, i k+1 ) for some k 1, so α ij J(Hom R (M i, M j )) for each i, j I, and therefore α J(S 0 ). Hence from Theorem 2, there exists a positive integer n with α i1 i 2 α i2 i 3... α ini n+1 = 0; that is, f 1 f 2... f n = 0, proving that {M i i I} is semi-t-nilpotent. Question. Is an analogous result true for S = End R M, when M = i I M i? That is, is it true that J(S) = {α S α ij J(Hom R (M i, M j )) for all i, j I} if and only if {M i i I} is locally semi-t-nilpotent (with respect to sequences of radical homomorphisms)? An affirmative answer would shed additional light on the structure of completely decomposable exchange modules. As another application, in [7] it was shown that when J(R) is a prime ring then J(M I (R)) = {column-bounded matrices in M I (J(R))} = {(a ij ) i,j I a ij 11

12 J(R), and a ij = 0 for all j / K, K a finite subset of I}. We conclude by exhibiting a module-theoretic generalization of this result. First, with M = i I M i and notation as above, set C J ( i I M i ) = {f S 0 f ij J(Hom R (M i, M j )) for all i, j I, and fe j = 0 for almost all j I}. When each M i = R, a ring with identity element, then C J ( i I M i ) can be identified with the ring of column-bounded matrices in M I (R). From Theorem 2, we know that C J ( i I M i ) J(S 0 ). The following definition provides a sufficient condition for equality to hold: call the decomposition M = i I M i radical-prime if given 0 f J(Hom R (M i, M j )) and 0 g J(Hom R (M k, M l )) with j l there exists h Hom R (M j, M k ) with fhg 0. Corollary 4. If the decomposition M = i I M i is radical-prime then J(S 0 ) = C J ( i I M i ). In particular, if each M i = N for some module N with J(EndR N) a prime ring, then J(S 0 ) = {column-bounded matrices in M I (J(End R N))}. Proof. It suffices to show that if f S 0 \C J ( i I M i ) then f / J(S 0 ). We can assume that each f ij J(Hom R (M i, M j )); otherwise, from Theorem 2, f / J(S 0 ). Since f / C J ( i I M i ) we can also assume without loss of generality that {1, 2,...} I and that fe i 0 for all i 1. For each i 1, choose k i I with e ki fe i 0, and note that e ki fe i J(Hom R (M ki, M i )) for each i I. Since the decomposition is radical-prime there exists γ 1 Hom R (M 1, M k2 ) with f k1 1γ 1 f k2 2 = e k1 fe 1 γ 1 e k2 fe 2 0. Set f 1 = e 1 γ 1 e k2 fe 2 = (γ 1 f) 12 0 and note that f 1 J(Hom R (M 1, M 2 )). Next use the radical-prime property to choose γ 2 Hom R (M 2, M k3 ) with f 1 γ 2 f k3 3 = f 1 e 2 γ 2 e k3 fe 3 0 and set f 2 = e 2 γ 2 e k3 fe 3 = (γ 2 f) Then (γ 1 f) 12 (γ 2 f) 23 = f 1 f 2 0. Continuing in this manner for each n 1 we can find elements γ n S 0 with (γ 1 f) 12 (γ 2 f) (γ n f) n,n+1 0. Hence, from Theorem 2, f / J(S 0 ). References [1] H. Bass, Finitistic dimension and a homological generalization of semi-primary rings, Trans. Amer. Math. Soc. 95 (1960), [2] P. Crawley and B. Jónsson, Refinements for infinite direct decompositions of algebraic systems, Pacific J. Math 14 (1964), [3] M. Harada, Factor Categories with Applications to Direct Decompositions of Modules, Lecture Notes in Pure and Appl. Math. v. 88, Marcel Dekker, New York, [4] N. Jacobson, Structure of Rings, Amer. Math. Soc. Colloq. Publ. v. 37, Providence,

13 [5] F. Kasch, Moduln mit LE-Zerlegung und Harada-Moduln, Lecture Notes, Ludwig-Maximilians Univ., Munich, [6] E.M. Patterson, On the radicals of rings of row-finite matrices, Proc. Roy. Soc. Edinburgh Sect. A 66 (1961/62), [7] N.E. Sexauer and J.E. Warnock, The radical of the row-finite matrices over an arbitrary ring, Trans. Amer. Math. Soc. 139 (1969), [8] R. Ware and J. Zelmanowitz, The Jacobson radical of the endomorphism ring of a projective module, Proc. Amer. Math. Soc. 26 (1970), [9] J.M. Zelmanowitz, On the endomorphism ring of a discrete module: a theorem of F. Kasch, Advances in Ring Theory, S.K. Jain & S.T. Rizvi, eds., Birkhäuser, Boston, 1997, [10] B. Zimmerman-Huisgen and W. Zimmerman, Classes of modules with the exchange property, J. Algebra 88 (1984),