Why does the set of submodules of a module that are direct sums of irreducible submodules have a maximal element? I was reading the proof that every R-module M of an artinian semisimple ring R is the direct sum of all irreducible R-submodules M. In the proof they stated that the set
$$\mathcal{F}:=\{N\subset M: N \text{ is the direct sum of its irreducible submodules}\}$$
has a maximal element. I was trying to see why such a maximal element exists.
My attempt: Let $\{N_i\}_{i\in I}\subset \mathcal{F}$ be a total ordered subset of $\mathcal{F}$ ($I$ is any indexation set). I want to prove that $N=\cup_{i\in I} N_i$ belongs to $\mathcal{F}$. We know that for each $i\in I$ there's an indexation set $J_i$ such that $N_i=\bigoplus_{k_i\in J_i} M_{k_i}$ with $M_{k_i}$ irreducible submodules of $N_i$. Define $$\Lambda :=\cup_{i\in I}\{k_i\in J_i:M_{k_i}=M_{k_{io}} \text{ for some }i_o\in I\setminus\{i\} \text{ and } k_{i_o}\in J_{i_o}\}$$
Obviously $\sum_{k_i\in (\cup_{i\in I} J_i)\setminus \Lambda} M_{k_i}$ is direct sum, but I got stuck for adding only one time such irreducible submodules $M_{i_k}$ associated to $\Lambda$ to asure the direct sum.
 A: It appears that the proof you are reading is incorrect.  There is no straightforward way to show that $\mathcal{F}$ is closed under unions of chains; knowing that each $N_i$ is a direct sum of simple submodules is not particularly helpful for proving that $\bigcup N_i$ is a direct sum of simple submodules.
For a correct proof, you want to use a different poset which forces the direct sum decompositions in such a chain to be "compatible".  Here's one way to formulate it.  Instead of using your poset $\mathcal{F}$, use the poset $$\mathcal{G}=
\left\{(N,S):N\subseteq M\text{ is a submodule, $S$ is a set of simple submodules of $N$, and }N=\bigoplus_{A\in S}A\right\}$$ ordered by saying $(N,S)\leq (N',S')$ if $N\subseteq N'$ and $S\subseteq S'$.
Now suppose you have a chain $\{(N_i,S_i):i\in I\}\subseteq\mathcal{G}$; I claim that $(N,S)\in \mathcal{G}$ where $N=\bigcup N_i$ (or $N=\{0\}$ if the chain is empty) and $S=\bigcup S_i$, and so $(N,S)$ is an upper bound for the chain.  It is clear that $N$ is a submodule of $M$, $S$ is a set of simple submodules of $N$, and $N$ is the sum of the elements of $S$.  It remains to show that this sum is direct.  If the sum were not direct, this would mean there are finitely many elements $A_1,\dots,A_n\in S$ and elements $a_j,b_j\in A_j$ such that $\sum a_j=\sum b_j$ but $a_j\neq b_j$ for some $j$.  But there is some $i\in I$ such that $A_1,\dots,A_n$ are all in $S_i$, and so this would contradict the fact that $N_i$ is the direct sum of the elements of $S_i$.
A: In fact, the set $\mathcal{F}$ is closed under the union of any chain. Let $\{N_i\}_{i\in I}\subset \mathcal{F}$ a total ordered subset of $\mathcal{F}$. Clearly $M=\cup_i N_i$ is a sum of irreducible $R$-modules. Write $M=\sum_{j\in J} T_j$ with $T_j$ irreducible modules. Let $\mathcal{S}\subset \mathcal{P}(J)$  such that for all $S_o\in \mathcal{S}$, $\sum_{s\in S_o}T_s=\bigoplus_{s\in S_o}T_s$. $S$ is not empty because for each $j\in J$, $\{j\}\in \mathcal{S}$. Let's see that $\mathcal{S}$ has a maximal. Let $\{S_r\}$ be a total ordered subset of $\mathcal{S}$ and let $\hat{S}=\cup_r S_r$. Pick any $s_o\in\hat{S}$ and  $x\in T_{s_o}\cap\sum_{s\in\hat{S}\setminus\{s_o\}} T_s$. Then $x=x_{s_1}+\cdots+x_{s_{n}}$ with $x_{s_l}\in T_{s_l}$. Since $\{S_r\}$ is a chain, there's $n_o\in\{1,\ldots,n\}$ such that $x_{s_l}\in T_{s_{n_o}}$ for all $l\in\{1,..,n\}$. Then $x\in T_{s_o}\cap T_{s_{n_o}}$. Since each $T$ is irreducible, $x=0$. Conclution $\mathcal{S}$ has a maximal, call it $S_o$.
Finally, let's prove that $M=\bigoplus_{s\in S_o} T_s$. Clearly $\bigoplus_{s\in S_o} T_s\subset M$. Suppose that there's $x\in M\setminus\bigoplus_{s\in S_o} T_s$ with $x\neq 0$. Then, $x\in T_p$ for some $p\in J$ with $p\notin S_o$. It's easy to check that $\sum_{s\in S_o\cup\{p\}}T_s=\bigoplus_{s\in S_o\cup\{p\}}T_s$. This contradicts the maximality of $S_o$. Hence $M$ is a direct sum of irreducible modules, i.e., it's the direct sum of all its irreducible submodules.
