Notation and conventions: I will use $\oplus$ for the "external direct sums" only. All the modules will be over a commutative ring with unity, $A$.
Problem: In the proof of one of the theorems that our professor presented in a lecture, he essentially used the following. Given a module $M$ and a submodule $N$ such that $M\cong N\oplus M$, it was claimed that $N = 0$. But why? Note that it's "$\cong$" and not "$=$". So, I attempted to craft a relevant lemma. This is what I came up with:
Let $\{N_i\}$ be modules of $M$ such that $\sum_i N_i\cong\oplus_i N_i$. Then $N_i$'s are independent.
(This would then imply that $N$ and $M$ are independent, and hence $N = 0$, for $M$ is the whole module.)
However, I am having trouble proving this. (I know that this can be proven because while surfing, I came across an uncannily similar problem in Dummit and Foote.)
Any help?
This is false. For instance, let $X$ be any nonzero module and let $M$ be an infinite direct sum of copies of $X$. If $N$ is one of the direct summands of $M$, then $M\cong M\oplus N$ (since if you remove one of the infinitely many copies of $X$ you still have the same number of them), but $N\neq 0$.
The correct statement is that if $M$ is the internal direct sum of $M$ and $N$, then $N=0$. That is, if the specific map $M\oplus N\to M$ given by $(m,n)\mapsto m+n$ is an isomorphism, then $N=0$.