Find all non-trivial submodules of a direct sum of two non-isomorphic simple modules 
Let $R$ be a ring with $1$. Let $M_1$ and $M_2$ be two non-isomorphic simple (nonzero) $R$-modules. Find all non-trivial submodules of $M_1 \bigoplus M_2$.

Solution: $M_1 \bigoplus M_2 \cong M_1 \times M_2$.  The submodules of $M_1 \times M_2$ are $M_1 \times \{0\}$, $\{0\} \times M_2$, $M_1 \times M_2$ and $\{0\} \times \{0\}$
 A: Here's how you prove there aren't any submodules besides the obvious ones listed in the question.  Suppose $N\subseteq M_1\oplus M_2$ is a submodule.  Consider the projection $p:N\to M_1$.  If the image of $p$ is $0$, then $N$ is contained in $0\oplus M_2\cong M_2$, so since $M_2$ is simple the only possibilities for $N$ are then $0\oplus 0$ and $0\oplus M_2$.
Since $M_1$ is simple, the only other possibility is that $p$ is surjective.  In that case, consider the kernel of $p$, which is contained in $0\oplus M_2\cong M_2$.  If the kernel of $p$ is $0$, then $p$ is an isomorphism.  Now consider the projection $q:N\to M_2$.  Since $N\cong M_1$ and $M_1$ and $M_2$ are non-isomorphic simple modules, $q=0$.  The argument of the previous paragraph now shows that $N\cong 0\oplus 0$ or $M_1\oplus 0$ (actually it must be the latter since we are assuming $p$ is surjective).
The final case is that $p$ is surjective and the kernel of $p$ is $0\oplus M_2$.  In that case $0\oplus M_2\subseteq N$ but $p$ is surjective on $N$, which implies $N$ is all of $M_1\oplus M_2$ (surjectivity means elements of $N$ can have any first coordinate, and then you can change the second coordinate arbitrarily).
A: Consider the composition series $$0\subset M_1\subset M_1\oplus M_2$$
If $N$ is a non-trivial submodule, then the series $$0\subset N\subset M_1\oplus M_2$$ can be refined to a composition series.
Since composition series have the same length, that means $0\subset N\subset M_1\oplus M_2$ is already a composition series,  which is equivalent to the first by Jordan-Holder Theorem.
That is, $N\cong M_1$ or $M_2$.
Currently, that is what I can think of. I would be interested in a simpler solution to this question.
