Radical Endomorphisms of Decomposable Modules

Radical Endomorphisms of Decomposable Modules Julius M. Zelmanowitz University of California, Oakland, CA 94607 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. Email address: julius.zelmanowitz@ucop.edu (Julius M. Zelmanowitz). Preprint submitted to Elsevier Science 1 April 2004

0 a 2 0 0 0 0 a 3 0 α = 0 0 0 a 4........ 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

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

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

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

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

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 +2... 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 +2... α ki+1 1,k i+1 Hom R (M ki +1, M ki+1 ). Since x n+1 = xα 12 α 23... 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

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

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

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

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

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) 23 0. 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) 23... (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), 466-488. [2] P. Crawley and B. Jónsson, Refinements for infinite direct decompositions of algebraic systems, Pacific J. Math 14 (1964), 797-855. [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, 1983. [4] N. Jacobson, Structure of Rings, Amer. Math. Soc. Colloq. Publ. v. 37, Providence, 1956. 12

[5] F. Kasch, Moduln mit LE-Zerlegung und Harada-Moduln, Lecture Notes, Ludwig-Maximilians Univ., Munich, 1982. [6] E.M. Patterson, On the radicals of rings of row-finite matrices, Proc. Roy. Soc. Edinburgh Sect. A 66 (1961/62), 42-46. [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), 287-295. [8] R. Ware and J. Zelmanowitz, The Jacobson radical of the endomorphism ring of a projective module, Proc. Amer. Math. Soc. 26 (1970), 15-20. [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, 317-322. [10] B. Zimmerman-Huisgen and W. Zimmerman, Classes of modules with the exchange property, J. Algebra 88 (1984), 416-434. 13