Reading a book I saw the following assertion:
Let $R$ be a ring (not necessarilly commutative) and $\varepsilon$ be the poset (w.r.t. inclusion) of all internal direct factors of an $R$-module $M$. If $M_1$ is a maximal element of $\varepsilon$, then its complement $N_1$ is a minimal element of $\varepsilon$ and hence $N_1$ is indecomposable.
My question: But I don't see how that fact that $M_1$ is maximal implies $N_1$ minimal in the poset $\varepsilon$. Any hint on how to show this ?
I tried to show that $N_1$ is indecomposable as follows:
Suppose that there exists two submodules of $N_2\neq 0$ and $N_3\neq 0$ such that $N_{1}=N_2\oplus N_3$. Then, $$M=M_{1}\oplus N_1=M_{1}\oplus(N_2\oplus N_3)=(M_1\oplus N_2)\oplus N_3.$$ but $M_{1}\subsetneqq M_{1}\oplus N_{2}$ as $N_2\neq 0$ and this contradicts the maximality of $M_{1}$. So, $N_{1}$ is indecomposable.
Answer from comments:
Your proof also shows that $N_1$ is minimal: if $N_1$ were not minimal, it would contain a proper submodule which was also a direct factor of $M$, and thus $N_1$ would be decomposable. – KReiser 14 hours ago