## 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 $M = \oplus_{i=1}^n \mathsf{Ann}_M(p_i^{k_i})$.