The structure theorem tells us that a finitely generated module $M$ over a principal ideal domain $R$ is isomorphic to a direct sum $ \bigoplus _{i}R/(q_{i}) $ where $ ( q_i ) ≠ R (q_{i})\neq R$ and the $(q_{i})$ are primary ideals. (In a PID, nonzero primary ideals are powers of primes, so $(q_{i})=(p_{i}^{{r_{i}}})=(p_{i})^{{r_{i}}}$).
I’m trying to see why each of the direct summands of the form $R/(q_{i})$ is an indecomposable module. The Wikipedia article seems to take that for granted and I haven’t found the reason myself.
If $q=p^n$, where $p$ is prime, then every proper submodule of $R/(q)$ is contained in $(p)/(q)$, so if $M,N \subsetneq R/(q)$ are two proper submodules, then $M,N \subseteq (p)/(q)$ and hence $M+N \subseteq (p)/(q)$, whence $M+N \subsetneq R/(q)$ is a proper submodule. Thus it is impossible that $R/(q)$ is a direct sum of two proper submodules.