Arbitrary internal direct sums of modules

Let R be a ring and let M be a left R-module. Let \{N_i\}_I be a family of submodules of M indexed by a nonempty set I. Prove that the following are equivalent:

  1. The mapping \sigma : \bigoplus_I N_i \rightarrow M given by \sigma(a_i) = \sum_{a_i \neq 0} a_i is an injective R-module homomorphism whose image is \sum_I N_i.
  2. If K \subseteq I is a (nonempty) finite subset and k \in K, then N_k \cap (\sum_{K \setminus k} N_i) = 0.
  3. If K \subseteq I is a (nonempty) finite subset, then \sum_K N_k = \bigoplus_K N_k. (Where this \bigoplus denotes a finite internal direct product.)
  4. For every element x \in R(\bigcup_I N_i), there exist unique elements a_i \in N_i such that all but finitely many a_i are zero and x = \sum_{a_i \neq 0} a_i.

We will use the following strategy: (1) \Rightarrow (2) \Rightarrow (3) \Rightarrow (4) \Rightarrow (1).

(1) \Rightarrow (2): Suppose \sigma is injective and that its image is the submodule generated by the N_i. Now let K \subseteq I be a finite nonempty subset and let k \in K. Suppose now that x \in N_k \cap (\sum_{K \setminus k} N_t). Say x = a_k and x = \sum_{K \setminus k} a_t, where a_t \in N_t. Let x_i = a_k if i = k and 0 otherwise, and let y_i = a_i if i \in K \setminus k and 0 otherwise. Certainly (x_i), (y_i) \in \bigoplus_I N_i, and moreover \sigma(x_i) = \sigma(y_i). So (x_i) = (y_i). Comparing entries, we see that a_i = 0 for all i, and hence x = 0. Thus N_k \cap (\sum_{K \setminus k} N_t) = 0 as desired.

(2) \Rightarrow (3) Suppose K \subseteq I is a finite nonempty subset. Note that (2) is one of the equivalent conditions that define what it means for the internal sum \sum N_k to be direct; thus \sum N_k = \bigoplus N_k.

(3) \Rightarrow (4) Let x \in R(\bigcup_I N_i). By definition, there exist a_i \in N_i such that finitely many a_i are nonzero and x = \sum_{a_i \neq 0} a_i. Suppose now that there exist b_i \in N_i such that finitely many b_i are nonzero and x = \sum_{b_i \neq 0} b_i. Let K be the set of all indices k such that either a_k or b_k is nonzero. Now x \in \sum_K N_k = \bigoplus_K N_k; thus we have a_k = b_k for all k, so that the expansion x = \sum_{a_k \neq 0} a_k is unique.

(4) \Rightarrow (1) Define \sigma : \bigoplus_I N_i \rightarrow M by \sigma(a_i) = \sum_{a_i \neq 0} a_i. \sigma is clearly an R-module homomorphism. Now if \sigma(a_i) = \sigma(b_i), then \sum_{a_i \neq 0\ \mathsf{or}\ b_i \neq 0} a_i = \sum_{a_i \neq 0 \mathsf{or}\ b_i \neq 0} b_i. Since the sum over the (finite number of) indices k such that a_k \neq 0 or b_k \neq 0 is direct, we have a_k = b_k for all k. Thus (a_i) = (b_i), and so \sigma is injective. It is clear that the image of \sigma is \sum_I N_i.

Extending our notion of (finite) internal direct sums, if \{N_i\}_I is a family of submodules satisfying any of these equivalent conditions we will say that \sum_I N_i = \bigoplus_I N_i is the internal direct sum of the N_i.

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