There is a theorem claiming that every finitely generated torsion module $M \ne \{0\}$ over a principal ideal domain $R$ such that the annihilator of $M$ is $(p^e), p \in R, p \text{-prime}, e \in \mathbb N$ can be written as a direct sum of cyclic sub-modules $S_i$ of $M$, whose annihilators has the form $(p^{e_i}), 1 \le e_i \le e$, i.e. $$M=\oplus_{i=1}^{n}S_i$$
In my lecture I also see a claim that $$pM=\oplus_{i=1}^{n}pS_i$$ where $pM=\{pm: m \in M \}$, $pS_i=\{ps: s \in S_i \}$.
I easily see why $pM=\sum_{i=1}^{n}pS_i$, but not why $\forall i, pS_i \bigcap\sum_{j=1, j \ne i}^{n}pS_j = \{0\}$.
My attempt
$ps_i \in pS_i$ and $ps_i = p(\sum_{j=1, j \ne i}^{n}s_j), s_j \in S_j$, then $p(s_i - \sum_{j=1, j \ne i}^{n}s_j) = 0$, and it follows that $s_i - \sum_{j=1, j \ne i}^{n}s_j \in\ker\phi$, where $\phi: M \to M, m \mapsto pm$. It would be nice to show $\ker\phi = \{0\}$ but I do not see how.
Let $x$ be an element that's both in $pS_i$ and in $\sum_{j \neq i} pS_j$. Then you can write $$x = p s_i = p \cdot \left( \sum_{j \neq i} s_j \right)$$ where $s_k \in S_k$. But all the $S_i$ are submodules, from which it follows that $x$ is both in $S_i$ and in $\sum_{j \neq i} S_j$ (if $s_i \in S_i$, then $ps_i = s_i + \dots + s_i \in S_i$ too). Since these submodules were in direct sum, it follows that $x = 0$.