## 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 $N_i$.