Indecomposable or not? Maybe I am just being silly here. But any hints would be appreciated. 
My question is: If a module has two non- isomophic simple submodules, can it be indecomposable? My guts feelings are that it could still be but I can't come up with an argument. 
Thanks very much! 
 A: Here's an example.  Let $k$ be a field and let $R=k\langle x,y,z\rangle$, the free $k$-algebra on three generators.  Consider the $R$-module $M=k^3$ with $x,y,$ and $z$ acting by the following matrices:
$$x: \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1\end{pmatrix}$$
$$y: \begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix}$$
$$z: \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0\end{pmatrix}$$
Writing $e_1,e_2,e_3$ for the standard basis vectors, then the span $V$ of $e_2$ and the span $W$ of $e_3$ are non-isomorphic simple submodules of $M$ (since $x$ acts nontrivially on $e_3$ but not on $e_2$).  However, $M$ is indecomposable.  Indeed, note that any vector whose first coordinate is nonzero generates all of $M$ (by applying $y$ and $z$ it generates $V$ and $W$, and then together with the original vector they span everything).  So, any proper submodule is contained in $V+W$, and thus $M$ cannot be the sum of two proper submodules.
(Of course, you could get an example where $R$ is nice and small by replacing it with the subring of $M_3(k)$ generated by these three matrices.)
