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.
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