# Direct sum over an arbitrary index

In section 10.3 of Abstract Algebra by Dummit and Foote, exercise 21 asks the reading to prove that these 4 statements are equal: If I is a nonempty index set and $$N_i$$ is a submodule of $$M$$ $$\forall i\in I$$

1: $$\sum_{i\in I}N_i \cong \bigoplus_{i \in I}N_i$$

2: If $$I' \subseteq I$$ is a finite subset of $$I$$ and $$i_1 \in I'$$, $$N_{i_1}\cap \sum_{i\in I'-{i_1}}N_i=0$$

3: $$\sum_{i\in I'}N_i \cong \bigoplus_{i \in I'}N_i$$

4: $$\forall x\in \sum_{i\in I}N_i$$, $$x$$ can be expressed uniquely as a finite sum of nonzero elements in each $$N_i$$

(1)->(2) seems pretty clear since if there was a nontrivial element in the intersection then 0 would have 2 unique representations so we wouldn't be able to construct a bijection between the modules in (1) so the isomorphism in (1) necessarily forces this condition.

(2)->(3) the criteria described in the pages before the exercises makes this clear (Proposition 5)

(3)->(4) This one I'm not sure how to go about. Im pretty sure I have to show that if $$x\in \sum_{i\in I}N_i$$ then $$x \in \sum_{i\in I'}N_i$$ for some finite subset $$I'$$ of $$I$$. I've been trying to find a way to show this for a while and haven't been able to come up with anything. Would this be the correct approach? If so, how can I show this?

(4)->(1) assuming (4), all elements of the submodule generated by all the $$N_i$$ are expressible as a finite sum of $$a_i \in N_i$$ so mapping each $$a_i$$ to the $$i$$'th cooridnate in $$\bigoplus_{i \in I}N_i$$ is clearly bijective (surjective since every element has such an expression and well-defined/injective because this expression is unique for each $$x$$).

Any help regarding (3)->(4) would be appreciated.

• Is 3 also quantified over all finite subset $I’$ of $I$? Jun 19, 2020 at 21:15
• @ArturoMagidin There was no classification of $I'$ other than it being finite so I'm fairly certain were supposed to assume its any finite subset of $I$ Jun 19, 2020 at 21:23
• My query isn’t whether it is any finite subset, but whether one is expected to assume it is similar to the subset mentioned in 2. In a sense, the description of $I’$ that you find in item 2 “expires” at the period ending item 2. Jun 19, 2020 at 21:25
• @ArturoMagidin I guess you can set it to be the same subset in (2) because $I'$ was chosen arbitrarily so the result is the same either way Jun 19, 2020 at 21:35
• I’m not sure off the top of my head how to prove that 3 implies 4, but it seems fairly easy to prove the equivalence of 2 and 3, and an implication from 2 to 4, which would also establish that the four statements are equivalent. Jun 19, 2020 at 21:56

Let $$x\in\sum_{i\in I}N_i$$. Then by the definition of $$\sum_{i\in I}N_i$$ (or by one of several equivalent definitions), there exist nonzero $$x_{i_1}\in N_{i_1},\ldots,x_{i_n}\in N_{i_n}$$ with $$x=\sum_{k=1}^nx_{i_k}$$ for some distinct $$i_1,\ldots,i_n\in I$$. To show this expression is unique, suppose there exist nonzero $$y_{j_1}\in N_{j_1},\ldots,y_{j_m}\in N_{j_m}$$ such that $$x=\sum_{k=1}^my_{j_k}$$ for some distinct $$j_1,\ldots,j_m\in I$$. Let $$I'=\{i_1,\ldots,i_n,j_1,\ldots,j_m\}$$ and use the assumption that the canonical map $$\oplus_{i\in I'}N_i\to\sum_{i\in I'}N_i$$ is an isomorphism to show that there exists a bijection $$\sigma:\{1,\ldots,n\}\to\{1,\ldots,m\}$$ such that $$y_{j_k}=x_{i_{\sigma(k)}}$$ (and consequently, $$n=m$$ and $$j_k=i_{\sigma(k)}$$).