Direct Sum Equivalences (10.3.21)

Dummit and Foote Abstract Algebra, section 10.3, exercise 21:

Let $I$ be an indexing set for submodules $N_i$ of $M$. Prove $$\sum_{i∈I}N_i ≅ \oplus_{i∈I}N_i ⇔$$ $$(\{i_1,...,i_k\} \subseteq I ⇒ N_{i_1} ∩ (N_{i_2}+...+N_{i_k}) = 0) ⇔$$ $$(\{i_1,...,i_k\} \subseteq I ⇒ N_{i_1}+...+N_{i_k} = N_{i_1} \oplus ... \oplus N_{i_k}) ⇔$$ $$(x∈\sum_{i∈I}N_i ⇒ ∃! \prod a_i ∈ \prod N_i~s.t.~(|\{y~|~0≠y∈\prod a_i\}|< \infty \land x=\sum y))$$ Proof: (i) $⇒$ (ii) Let $n∈N_{i_1} ∩ (N_{i_2}+...+N_{i_k}))$. By the first half, we can simply write $n$ as a sum, and by the second half we can write a sum only using elements of $N_{i_2},...,N_{i_k}$, which would necessarily be distinct sums when $n≠0$.

(ii) $⇒$ (iii) This is simply proposition 5 applied to $N_{i_1}+...+N_{i_k}$.

(iii) $⇒$ (iv) Since $x∈\sum_{i∈I}N_i$, we must necessarily have it written as a finite sum of nonzero elements, say $x=\sum_{k∈K} b_k$ for some finite $K \subseteq I$ where $b_k≠0$ for all $k$. Assume further that $x=\sum_{j∈J} c_j$ for some finite $J \subseteq I$ where $b_j≠0$ for all $j$, is another sum. We thus have $x∈\sum_{l∈J \cup K} N_l = \oplus_{l∈J \cup K} N_l$ is writable as two sums, so that these two sums are in fact the same.

(iv) $⇒$ (i) Define $φ : \sum_{i∈I}N_i → \oplus_{i∈I}N_i$ by $x = \sum n_i \mapsto \prod n_i$ where $\sum n_i$ is the unique sum representation of $x$. We easily see this is a well-defined isomorphism of $R$-modules.$~\square$