In Dummit and Foote's Abstract Algebra, the proof of the Wedderburn-Artin theorem is sketched out in a series of exercises for reader to solve. I'm currently working on the final exercise in the series. Here it is:
Suppose that $R$ is a ring with $1$ (not necessarily commutative). Suppose that $R$, when considered as a left $R$-module, is a direct sum $$ R = L_1 \oplus L_2 \oplus \dots \oplus L_n,$$ where each $L_i$ is a simple (i.e. irreducible) left $R$-module with $L_i = Re_i$ for some $e_i \in R$ with
- $e_ie_j = 0$ if $i \neq j$
- $e_i^2 = e_i$ for all $i$
- $\sum_{i=1}^n e_i = 1$
Show that every left $R$-module is completely reducible.
In case you're wondering what this question has to do with the title, I should probably mention that a previous exercise in the book showed that if $R$ is a direct product of matrix rings over division rings, then $R$ must have the above structure. The current exercise then shows that all $R$-modules are completely reducible.
I've managed to solve this exercise (or at least, I think I have). But what bothers me is that I didn't follow the hint given in the book, and can't see how the hint can possibly be made to work. This makes me suspect that I've made a mistake or missed something crucial.
Here is the hint:
Let $N$ be a nonzero $R$-module. First show $N$ contains a simple submodule by considering a cyclic submodule. Then use Zorn's lemma applied to the set of direct sums of simple submodules (appropriately ordered) to show that $N$ contains a maximal completely reducible submodule $M$. If $M \neq N$, let $M_1$ be the complete preimage in $N$ of a simple module in $N / M$ and contract the maximality of $M$.
So let's get started. Suppose $N$ is a nonzero $R$-module. My first observation is that if $n$ is any non-zero element of $N$, then it must be the case that $e_i n \neq 0$ for some $i$. This is because $0 \neq n = 1.n = \sum_{i=1}^n e_i n$.
$Re_i n$ is clearly an $R$-module. In fact, $Re_i n$ is a simple $R$-module. To see that $Re_i n$ is simple, consider the $R$-module homomorphism $\varphi : L_i \to Re_i n$ defined by $\varphi(l) = ln$. The kernel $\text{ker }\varphi$ is an $R$-submodule of $L_i$ and $L_i$ is a simple $R$-module by assumption, so either $\text{ker }\varphi = 0$ or $\text{ker }\varphi = L_i$. The latter would imply that $e_i n = 0$, a contradiction. So $\varphi$ is injective. Since $\varphi$ is obviously surjective, we find that $Re_i n$ is isomorphic as an $R$-module to the simple $R$-module $L_i$, and hence is simple.
So we've proven that $N$ must contain at least one simple submodule.
Using Zorn's lemma, it follows that there exists a collection of simple $R$-submodules $\{ S_i \}_{i \in I}$ of $N$ such that $S_i \cap (\sum_{i' \in I \setminus \{ i \} } S_{i'}) = 0$ for all $i \in I$, and which is maximal subject to this condition. (The condition $S_i \cap (\sum_{i' \in I \setminus \{ i \} } S_{i'}) = 0$ for $i \in I$ is just a scary-looking way of saying that the sum $\sum_{i \in I} S_i$ is in fact a direct sum $\oplus_{i \in I} S_i$. Apologies for my awful notation.)
From here, my approach is as follows. Let $M = \oplus_{i \in I} S_i$, and suppose for contradiction that $M \neq N$. Then there exists some $n' \in N \setminus M$. Arguing as before, we can show that $e_i n' \notin M$ for some $e_i$, and that $Re_i n'$ is a simple $R$-module. Furthermore, $M \cap Re_i n'$ is a submodule of $Re_i n'$, and since $Re_i n'$ is simple, we either have $M \cap Re_i n' = 0$ or $M \cap Re_i n' = M$. The latter is impossible because $e_i n' \notin M$. So $M \cap Re_i n' = 0$. Thus, if we add the simple submodule $Re_i n'$ to the collection $\{ S_i \}_{i \in I }$, we preserve condition that the sum of the modules in the collection is a direct sum. This contradicts the maximality of the collection $\{ S_i \}_{i \in I}$.
But the approach I took in the final paragraph is not what the hint told me to do!
Let's see what happens if we try to follow the hint. For sure, if $M \neq N$, then $N / M$ must contain a simple $R$-submodule. If $M_1$ is its complete preimage in $N$, then $M_1$ is an $R$-module, which contains $M$ and is not simple. What's the use of this? This line of thought seems to lead nowhere.
Am I missing something? Is there a way to get the hint to work? Did I make a mistake somewhere?
Minor aside: There's one other thing that makes me suspect I've made a mistake. Nowhere in my argument did I explicitly use the first two bullet points which tell us that $e_i e_j = 0$ if $i \neq j$ and $e_i^2 = e_i$. for all $i$.
That said, these two bullet points contain no additional information - they are implied by the fact that $\sum_{i = 1}^n e_i = 1$ and by the direct sum decomposition $R = L_1 \oplus L_2 \oplus \dots \oplus L_n$. Just multiply $\sum_{i = 1}^n e_i = 1$ on the left by one of the $e_j$'s to see this.