Over a PID, every module with a nonzero annihilator is an internal direct sum of its p-primary components

Let R be a principal ideal domain. Let M be a left unital R-module, and suppose \mathsf{Ann}_R(M) \supseteq (a) where a is nonzero. Say a = \prod p_i^{k_i} is the irreducible factorization of a in R. Prove that M is the internal direct sum of its submodules of the form \mathsf{Ann}_M(p_i^{k_i}). (This \mathsf{Ann}_M(p_i^{k_i}) is called the p_i-primary component of M.)


For each t, let q_t = \prod_{i \neq t} p_i^{k_i}.

We claim that \mathsf{Ann}_M(p_t^{k_t}) = (q_t)M for all t. (\supseteq) Suppose q_t \cdot m \in (q_t)M. Note that p_t^{k_t} \cdot (q_t \cdot m) = a \cdot m = 0, so that q_t \cdot m \in \mathsf{Ann}_M(p_t^{k_t}). Thus (q_t)M \subseteq \mathsf{Ann}_M(p_t^{k_t}). (\subseteq) Suppose now that m \in \mathsf{Ann}_M(p_t^{k_t}). Since R is a principal ideal domain and p_t^{k_t} and q_t are relatively prime, we have 1 = q_tx + p_t^{k_t}y for some x,y \in R. Note then that m = 1 \cdot m = (q_tx + p_t^{k_t}y) \cdot m = q_tx \cdot m + p_t^{k_t}y \cdot m q_tx \cdot m \in (q_t)M. Thus \mathsf{Ann}_M(p_t^{k_t}) \subseteq (q_t)M, and we have \mathsf{Ann}_M(p_t^{k_t}) = (q_t)M.

Next, we claim that (q_t)M \cap \sum (q_s)M = 0 for all t. To that end, let m \in (q_t)M \cap \sum (q_s)M. Note that p_t^{k_t} and q_t are relatively prime (by definition). Now m = 1 \cdot m = xp_t^{k_t} \cdot m + yq_t \cdot m = 0, since p_t^{k_t} \cdot (q_t)M = 0 while q_t \cdot (\sum (q_s)M) = 0. Thus (q_t)M \cap (\sum (q_s)M) = 0 as desired.

Next, we claim that (q_i \ |\ 1 \leq i \leq n) = R. To see this, note that we have the irreducible factorization of each q_i, and that these have no irreducible factors in common.

Now let m \in M. Since (q_i \ |\ 1 \leq i \leq n) = R, we have 1 = \sum r_i q_i for some r_i \in R. Now m = 1 \cdot m = \sum r_iq_i \cdot m \in \sum (q_i)M.

Thus M = \sum_{i=1}^n (q_i)M, and in fact this sum is direct. So M = \oplus_{i=1}^n \mathsf{Ann}_M(p_i^{k_i}).

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