## Every torsion module over a PID is the internal direct sum of its p-primary components

Let $R$ be a principal ideal domain and let $M$ be a torsion unital left $R$-module. If $p$ is a prime in $R$, the $p$-primary component of $M$ is the set $M_p = \{m \in M \ |\ p^k \cdot m = 0\ \mathrm{for\ some}\ k \geq 1 \}$.

1. Prove that $M_p$ is a submodule of $M$.
2. Prove that this definition of $p$-primary component agrees with the definition given in this previous exercise if $M$ has a nonzero annihilator.
3. If $P \subseteq R$ is the set of all primes in $R$ (up to associates), prove that $R = \bigoplus_{p \in P} M_p$. (Where $\bigoplus$ denotes an internal direct sum.)

1. It is clear that $M_p = \bigcup_{k \in \mathbb{N}^+} \mathsf{Ann}_M(p^k)$. Note also that if $k \leq \ell$ and $p^k \cdot m = 0$, then $p^\ell \cdot m = 0$. Hence $\mathsf{Ann}_M(p^k) \subseteq \mathsf{Ann}_M(p^\ell)$. That is, $\{\mathsf{Ann}_M(p^k)\}_{k \in \mathbb{N}^+}$ is a chain of submodules of $M$. Thus $M_p$ is a submodule of $M$.
2. Suppose $\mathsf{Ann}_R(M) = (a)$ is nonzero; say $a = \prod p_i^{k_i}$. In this previous exercise we defined the $p_i$-primary component of $M$ to be $\mathsf{Ann}_M(p_i^{k_i})$. Certainly $\mathsf{Ann}_M(p_i^{k_i}) \subseteq M_{p_i}$. Now suppose $m \in M_{p_i}$; say $p_i^k \cdot m = 0$. Since $a \cdot m = 0$ and $R$ is a PID, $(a, p_i^k) = (p_i^t)$ is in $\mathsf{Ann}_R(m)$ where t \leq k_i\$. So $m \in \mathsf{Ann}_M(p_i^{k_i})$, and so $M_p = \mathsf{Ann}_M(p_i^{k_i})$. Thus if $M$ has a nonzero annihilator, the two notions of $p$-primary component agree.
3. Now let $K \subseteq P$ be a finite set of primes in $R$, and fix some $p \in K$. Suppose $m \in M_p \cap (\sum_{P \setminus p} M_q)$. Say $m = m_p = \sum m_q$, where $m_p \in M_p$ and $m_q \in M_q$ for each $q$. Since $M_p = \bigcup_{k \in \mathbb{Z}^+} \mathsf{Ann}_M(p^k)$, we have $e_p$ and $e_q$ (for each $q$ such that $p^{e_p} \cdot m_p = 0$ and $q^{e_q} \cdot m_q = 0$. Note that $(\prod q^{e_q}) \cdot m = \sum (\prod q^{e_q}) \cdot m_q = 0$ and $p^{e_p} \cdot m = p^{e_p} \cdot m_p = 0$. Now $(p^{e_p}, \prod q^{e_q}) \subseteq \mathsf{Ann}_R(m)$. Since $R$ is a PID and $p^{e_p}$ and $\prod q^{e_q}$ are relatively prime (as we know their factorizations), $\mathsf{Ann}_R(m) = R$. In particular, $m = 1 \cdot m = 0$. Thus $M_p \cap (\sum M_q) = 0$. Moreover, note that if $m \in M$, then $a \cdot m = 0$ for some $a \in R$ since $M$ is torsion. Say $a = \prod_T p_i^{k_i}$. Let $q_t = \prod_{i \neq t} p_i^{k_i}$. Note that $(q_t \ |\ t \in T) = R$, so that $1 = \sum x_tq_t$ for some $x_t \in R$. Finally, $p_i^{k_i} \cdot (x_tq_t \cdot m) = 0$, so that $x_tq_t \cdot m \in M_{p_i}$. Finally, $m = 1 \dot m = \sum x_tq_t \cdot m \in \sum_T M_{p_i}$. Thus $M = \bigoplus_P M_p$, by this characterization of arbitrary internal direct sums.