Primary Decomposition Theorem (10.3.18)

Dummit and Foote Abstract Algebra, section 10.3, exercise 18:

Let $M$ be an $R$-module for $R$ a PID, and let $Ann(M)=(a)≠0$. Let $a=p_1^{\alpha_1}...p_k^{\alpha_k}$ be its decomposition into prime factors. Let $M_i=Ann(p_i^{\alpha_i})$, i.e. the $p_i$-primary component of $M$. Prove$$M=M_1 \oplus~...~\oplus M_k$$ Proof: Let $n_i=a/p_i^{\alpha_i}$. We can see that no prime factor divides all of the $n_i$, so that their greatest common divisor is $1$ and there exist $r_i$ such that $1=r_1n_1+...+r_kn_k$. For any $m∈M$ we have $m=r_1n_1m+...+r_kn_km$, and since $p_i^{\alpha_i}(r_in_im)=r_i(am)=0$, we have $r_in_im∈M_i$ and now $M=M_1+...+M_k$. To observe uniqueness of sums, let$$m∈M_j∩(M_1+...+M_{j-1}+M_{j+1}+...+M_k)$$with $Ann(m)=(d)$. We have $d \mid p_j^{\alpha_j}$ from the first half as well as $d \mid n_j$ from the second so that $d=1$ and now $1m=m=0$.$~\square$