Generalization of the Primary Decomposition Theorem (10.3.22)

Dummit and Foote Abstract Algebra, section 10.3, exercise 22:

Let $M=Tor(M)$ be an $R$-module for $R$ a PID and let $p∈R$ be prime. The $p$-primary component of $M$ is the set of all elements of $M$ that are annihilated by some positive power of $p$.

(a) Prove the $p$-primary component of $M$ is a submodule of $M$.
(b) Prove this definition of $p$-primary components of $M$ is consistent with the one given in 10.3.18 when $Ann(M)≠0$.
(c) Letting $M_p$ be the $p$-primary component of $M$, prove $$M=\oplus_{p\text{prime}}M_p$$ Proof: (a) Let $x,y∈M_p$ with $p^ax=p^by=0$. We have $p^{a+b}(x+ry)=p^b(p^ax)+p^ar(p^by)=0$ so that $x+ry∈M_p$.

(b) For $Ann(M)=(a)$ for $a=p_1^{\alpha_1}...p_k^{\alpha_k}$ with the previous definition of $M^*_{p_i}=\{m~|~p_i^{\alpha_i}m=0\}$ we clearly have $M^*_{p_i} \subseteq M_{p_i}$. As well, for any $m∈M_{p_i}$ we have $am=0$, so that for $Ann(m)=(d)$ we have $d \mid a$. As well, by $m∈M_{p_i}$ we have $d \mid p_i^a$ for some positive $a$. This implies $d \mid (a,p_i^a)=p_i^b$ for $b ≤ \alpha_i$ so that now $p_i^{\alpha_i}m=p^c(dm)=0$ and $M_{p_i} = M^*_{p_i}$.

(c) Choose arbitrary $x∈M$. Letting $X=Rx$ we have $(a) ⊆ Ann(x) = (b)$ so that $b | a = p_1^{\alpha_1}...p_k^{\alpha_k}$. Say $b = q_1^{β_1}...q_k^{β_k}$ where $β_i ≤ α_i$. By 10.3.18 we have $X=X_{q_1} \oplus ... \oplus X_{q_k} ⊆ M_{p_1} ⊕ ... ⊕ M_{p_k}$ so that $x$ is writable as a sum while in $M_{p_1} + ... + M_{p_k}$ and now $M=\sum_{p prime}M_p$.

Proceed in the fashion of 10.3.22(ii) $⇒$ (i). Let $m∈M_{p_1} \cap (M_{p_2} + ... + M_{p_k})$ for any selection such that $p_j$ are distinct primes. Write $m=n_1=n_2+...+n_k$ in accordance with the first and second halves, with $p_i^{\alpha_i}n_i=0$ for any $i$. For $Ann(m)=(d)$ have $d \mid p_1^{\alpha_1}$ for some positive $a$, and by observing the result of $(\prod_{2 ≤ i ≤ k}p_k^{\alpha_k})m$, we notice $d \mid \prod_{2 ≤ i ≤ k}p_k^{\alpha_k}$ and now $d=1$ and $1m=m=0$.$~\square$