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.

  • $\begingroup$ Is 3 also quantified over all finite subset $I’$ of $I$? $\endgroup$ Jun 19, 2020 at 21:15
  • $\begingroup$ @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 $I$ $\endgroup$
    – uhhhhidk
    Jun 19, 2020 at 21:23
  • $\begingroup$ 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. $\endgroup$ Jun 19, 2020 at 21:25
  • $\begingroup$ @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 $\endgroup$
    – uhhhhidk
    Jun 19, 2020 at 21:35
  • $\begingroup$ 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. $\endgroup$ Jun 19, 2020 at 21:56

1 Answer 1


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 $j_k=i_{\sigma(k)}$).


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