Take the 2-minute tour ×  Can anyone give me a module $M$ over a
ring $R$, such that $M$ is indecomposable, but $M$ has a
submodule $N$ such that $N$ is decomposable?  In general, what
further assumptions can we put on the ring $R$ or $M$ to
guarantee that $M$ has no such decomposable submodule $N$?
share|improve this question add comment  2 Answers  up vote 1
down vote accepted  If you take $R=\begin{pmatrix}k & 0 &
0\\k&k&0\\k&0&k\end{pmatrix}$, where $k$ is a field and the
multiplication is given by matrix multiplication, then the
direct summand of $R$ given by
$\begin{pmatrix}k&0&0\\k&0&0\\k&0&0\end{pmatrix}$ is
indecomposable, but has the submodule
$\begin{pmatrix}0&0&0\\k&0&0\\k&0&0\end{pmatrix}$ decomposes
into two one-dimensional modules.  I did not try for a rigorous
proof, but an (at least sufficient) assumption I know in the
context of finite dimensional $k$-algebras is that of left (or
right, depending if you work with left or right modules)
uniseriality as an assumption on $R$, that is the set o