Let $R$ be PID and $M$ a $R$-Module. Suppose ${\rm Ann}(M)=(a)$. Let $a=p_1^{e_1}p_2^{e_2}...p_k^{e_k}$ be a factorization of $a$ into primes. Define $M_i=\{m \in M|p_i^{e_i}m=0\}.$ Show $M=\bigoplus M_i$ ($i$ runs 1 to $k$).
So I did the most obiovs thing which was to use C.R.T. to get
$M \cong M\diagup(a)M=M \diagup \prod p_i^{e_i}M\cong M \diagup p_1^{e_1}M \times ...M \diagup p_k^{e_k}M$
After this I'm stuck. If I can show $M_i \cong \faktor{M}{p_i^{e_i}M}$ I think I will be close (I think then I will need only show the uniqueness of the sum), but I have no intuitive feeling for why this should be true.
We proceed by induction on $k$. For the case when $k=2$, write $a=p_1^{\alpha}p_2^{\alpha_2}$ where $p_1,p_2$ are distinct primes in $R$. This means $(p_1)$ and $(p_2)$ are comaximal and so (exercise) $(p_1^{\alpha_1})$ and $(p_2^{\alpha_2})$ are comaximal. Then there exists $x,y\in R$ such that $p_1^{\alpha_1}x+p_2^{\alpha_2}y=1$. Since $a$ annihilates $M$, then $am=0$ for all $m\in M$. This means that \begin{equation*} m=p_1^{\alpha_1}xm+p_2^{\alpha_2}ym \end{equation*} and multiplying throughout by $p_1^{\alpha_1}$, we see that $p_1^{\alpha_1}\left(m-p_1^{\alpha_1}xm\right)=p_1^{\alpha_1}p_2^{\alpha_2}ym=0$. This means that $p_2^{\alpha}ym\in M_{2}$, and by a similar argument, $p^{\alpha}xm\in M_{1}$. This shows that $M=M_{1}+M_{2}$. Now we show the general case. Write $a=p_{1}^{\alpha_{1}}\cdots p_{k}^{\alpha_{k}}$. Let $\gamma=p_{1}^{\alpha_{1}}\cdots p_{k-1}^{\alpha_{k-1}}$, so that $\gamma$ and $p_{k}^{\alpha_{k}}$ are pairwise coprime and hence the principal ideals generated by them are comaximal. Repeating the argument above, we get the desired result. Thus, we have shown that $M=M_{1}+\dots+M_{k}$.
We now show that the sum is indeed direct. We keep the notation as above. Since $\gamma$ and $p_{1}^{\alpha_{1}}$ are coprime in $R$, then we can write $1=\gamma s+p_{1}^{\alpha_{1}}t$ for some $s,t\in R$. Let \begin{equation*} n\in M_{i}\cap\sum_{\substack{j=1\\ j\neq i}}^{k}M_{j} \end{equation*} so that $\gamma n=0$ and $p_{1}^{\alpha_{1}}n=0$. Using $1=\gamma s+p_{1}^{\alpha_{1}}t$, then $n=\gamma s n+p_{1}^{\alpha_{1}}tn=0$, and hence the sum is direct.