Q:

Part of Wedderburn's theorem

In Dummit, Foote the theorem is divided into several exercises with hints. I am stuck proving the following result.
Suppose $R$ is a ring with $1$ which is isomorphic as R-module to $$R \cong L_1 \oplus L_2 \ldots \oplus L_n,$$
where $L_i \cong Re_i, \; e_ie_j = 0, i \neq j, e_i^2 =e_i, \; \sum e_i = 1.$
Then any R-module $N$ is completely reducible. The hint suggest to prove first that $N$ contains simple modules.
Are modules of the form $Re_ix, \; x \in R$ what I am looking for? 
Then the hint says I need to apply Zorn's lemma to the set of all the direct sums of simple submodules of $N$. I guess it works since the set of such submodules is not empty and the union of the elements of any chain is of this form (...is it?). So there exist a maximal submodule $M \subset N$ with respect to this property. 
Finally, if $M \neq N$ consider $\pi: N \to N/M$ and some simple submodule $A \subset N/M$ (why does it exist?). Define $M_1: \pi^{-1}(A)$. It contains $M$ contradicting maximality of $M$.
Thus,  using the hint I wrote a sketch of the proof. It does't convince me at all so I need help to work out the details (mostly questions in the brackets). 
Also it doesn't seem to essentially rely on the assumption. Is it true that we used the assumption (on the structure of the ring) just once to show that there are in fact simple submodules? So is that the key part? 
Edit: it is exercise 3 after 18.2 (Wedderburn's theorem and some consequences). Also I forgot to mention that all of $L_i$ are simple.  

A:

As @runway44 points out, this statement is false in general if the $L_i$ are not simple, so I will assume that they are supposed simple. Now, if I am correct Dummit and Foote, define "completely reducible" to mean decomposable as a direct sum of simple submodules, so I will operate on that definition. First, two lemmas. (Feel free to skip the first if you have already seen it; the main point is that a simple submodule will intersect any other submodule in either $\{0\}$ or itself.)

Lemma 1: a module $M$ is completely reducible if and only if it a sum of simple submodules.
Proof: For the non-trivial direction, suppose we have a decomposition $M=\sum_{i\in I} M_i$ of $M$ as a sum of simple submodules $\{M_i\}_{i\in I}$. We will apply Zorn's lemma to the set $S=\{J\subseteq I:\sum_{j\in J}M_j=\bigoplus_{j\in J}M_j\}$, partially ordered by inclusion.
Note, we have $\emptyset\in S$, so $S$ is indeed non-empty. Hence let $\{J_k\}_{k\in K}$ be a chain in $S$; we claim $J:=\bigcup_{k\in K}J_k$ is an upper bound. Indeed, suppose for contradiction that $J\notin S$, ie that the sum of all $M_j, j\in J$ is not a direct sum. Then in particular there is $x\in\sum_{j\in J}M_j$ such that $x$ can be decomposed as a sum of elements of the $M_j$ in two distinct ways. But any such decomposition has only finitely many have non-zero terms (say $x=a_1+\dots+a_m$ and $x=b_1+\dots+b_n$), and so in particular there is some $k\in K$ such that each $a_i$ (resp. each $b_i$) lies in some $M_j,\space j\in J_k$, contradicting that $J_k\in S$. Hence indeed $J\in S$, so we may apply Zorn's lemma to $S$ to obtain some $J_0\subseteq I$ maximal in $S$. We claim $M=\bigoplus_{j\in J_0} M_j$.
Indeed, suppose not, and let $\tilde M=\bigoplus_{j\in J_0} M_j$. Then since $M=\sum_{i\in I} M_i$ there must be some $i\in I$ such that $M_i\nsubseteq\tilde M$, ie such that $M_i\cap\tilde M\neq M_i$. But $M_i$ is simple and so we must have $M_i\cap\tilde M=\{0\}$, whence $J_0\cup\{i\}\in S$ and so contradicts maximality of $J_0$. Thus the lemma is proved.

Lemma 2: If $R$ is as in your problem statement (but where each $L_i$ is simple) and $N$ is a non-zero $R$-module, then $N$ contains a simple submodule.
Proof: Let $x\neq 0\in N$. We claim $M_i:=R\cdot(e_i\cdot x)$ is a simple $R$-submodule of $N$ for each $i$; indeed, recall that $M_i\cong R/\text{ker}(\varphi_{i})$, where $\varphi_i:R\rightarrow N$ is the $R$-module homomorphism given by $r\mapsto r\cdot(e_i\cdot x)$. Then for any $r=a_1 e_1+\dots+a_n e_n\in R$, by the central idempotency of the $e_i$ we have that $r\cdot(e_i\cdot x)=a_i\cdot(e_i\cdot x)=a_i e_i\cdot x$, whence $\text{ker}\varphi_{i}=\sum_{j\neq i} L_j$. Since $L_i$ is simple, evidently $\sum_{j\neq i} L_j$ is a maximal ideal of $R$, and so $M_i\cong R/\text{ker}\varphi_{i}$ is a simple $R$-submodule of $N$, as desired.

Now let $R$, $N$ be as in your problem statement, and let $S$ be the set of all completely reducible submodules of $N$, partially ordered by inclusion. In particular, $\{0\}\in S$, and so $S$ is non-empty.
Now let $\{M_i\}_{i\in I}$ be a chain in $S$, and let $M=\sum_{i\in I} M_i$; certainly $M$ is a submodule of $N$ and contains each $M_i$. We claim also that $M\in S$; indeed, by definition of $S$, each $M_i$ is a sum of simple submodules, and so $M$ is also a sum of simple submodules. But now by lemma 1 $M$ is completely reducible, and so indeed $M\in S$, as desired. Thus we may apply Zorn's Lemma to $S$; hence let $M\in S$ be maximal.
Now suppose for contradiction that $M\neq N$, and let $x\in N\setminus M$. Now $N/M$ is a non-zero $R$-module, and so by lemma 2 $\bar M_i:=R\cdot (e_i\cdot x+M)$ is a simple submodule of $N/M$ for each $i$. Clearly the preimage of $\bar M_i$ under the projection map $N\rightarrow N/M$ is just $M_i:=M+R\cdot(e_i\cdot x)$. But now note, again by the argument of lemma 2, $R\cdot(e_i\cdot x)$ is itself a simple submodule of $N$. Since $M$ is completely reducible and hence a sum of simple submodules, this means that $M_i$ is also a sum of simple submodules, and so again by lemma 1 is itself completely reducible. This contradicts maximality of $M$, and so we indeed must have that $M=N$, and hence that $N$ is completely reducible, as desired.

