I'm trying to answer the following and would appreciate a hint:
Let $R$ be a PID and $M$ an $R$-module. Let $p, q \in R$ be distinct primes. Define $$M_p=\{m \in M \mid p^\alpha m=0 \mbox{ for some } \alpha \geq 0\}$$ and $$M_q=\{m \in M \mid q^\beta m=0 \mbox{ for some } \beta \geq 0\}.$$ Suppose $p^2q^2M=0$. Prove that $M = M_p \oplus M_q$.
In particular, I'm having trouble seeing why if $m\in M$, there some reason that it should break into a parts which are annihilated by powers of $p$ and $q$, respectively.
Without any other assumptions on $M$, I'm not sure how to proceed. My first thought was to try to define a map $M_p \oplus M_q \rightarrow M$ by $(m_1 ,m_2) \mapsto m_1+m_2$, but it's not clear to me that that's even surjective.
By Bezout's identity (or by applying the PID property), there are $a,b\in R$ such that $ap+bq=1$. Then, take an appropriate power of it: $$a^3\,p^3+3a^2b\,p^2q+3ab^2\,pq^2+b^3\,q^3=1$$
For any $m\in M$, we have $p^3m,\,p^2q\,m\in M_q$ and $pq^2m,q^3m\in M_p$, so the above sum will yield a decomposition of $m\in M_q+M_p$.
It remains to prove that $M_p\cap M_q=\{0\}$.
Suppose $m\in M_p\cap M_q$, that is, $p^\alpha m=q^\beta m=0$ for some $\alpha,\beta\ge 0$.
Then the above trick can be applied again, by taking an appropriate power ($\alpha+\beta$) of the identity $ap+bq=1$, so that each term on the left should be a multiple of either $p^\alpha$ or $q^\beta$. That will give $m=0$.