Show that for a principal ideal domain $R$ with identity, $Q=R/(p^n)$ is an indecomposable module over $R$ where $p$ is a prime element in $R$ and $n$ is a positive integer.
Actually, I was trying to show that we can never decompose $Q$ into non-trivial submodules, but I am getting stuck.
A small hint is very much required at this moment. Thanks in advance.
If $R$ is a PID and $p \in R$ is a non-zero element generating a prime ideal $I:=(p)$ it follows the quotient $R^*:=R/(p^n)$ is a local ring: If $J^* \subseteq R/(p^n)$ is a prime ideal, it follows we get an inclusion $(p^n) \subseteq J \subseteq R$ and hence $(p) \subseteq J$. This implies that $(p)=J$ since $(p)$ is a maximal ideal. Hence $J^*$ is the only maximal ideal in $R/(p^n)$ and $R^* \rightarrow R^*/J^*$ is the only simple quotient of $R^*$. If $R^* \cong R_1 \oplus R_2$ is a non-trivial direct sum decomposition as $R^*$-modules, we may choose simple quotients $S_i:=R_i/m_i$ and get a surjection
$$\phi:R^* \rightarrow S_1 \oplus S_2$$
and since $R^*/J^*$ is the only simple quotient of $R^*$ we get a contradiction: The inverse image $m_i^*$ of $m_i$ in $R^*$ is an ideal, and since $R^*/J^*$ is the only simple quotient of $R^*$ we must have $m^*_i=J^*$. Since the map $\phi$ is surjective we must have $m^*_1$ and $m^*_2$ to be relatively prime - a contradiction.