Question about torsion submodules and Decomposition theorem

Question

Let $A$ be an principal ideal domain, and $M$ an $A$-module. If $p$ is irreducible in $A$, let's define $$\mathrm{Tor}_p(M):=\{m\in M\mid p^km=0\text{ for some }k\in\mathbb{N}\}.$$

I need to show that if $M$ is finitely generated, then $\mathrm{Tor}(M)$ can be written as a direct sum of the submodules $\mathrm{Tor}_p(M)$.

Through the Decomposition Theorem, I have that $M$ is a direct sum of a finite number of cyclic modules and a free module. Thus, the torsion submodule of M is the direct sum of the torsion submodules of each of these cyclic submodules (the torsion submodule of the free module is zero). But how can I show that $\mathrm{Tor}(C_{p_i^{\alpha_i}})=\mathrm{Tor}_p(M)$ or that $\mathrm{Tor}(C_{p_i^{\alpha_i}})$ is a direct sum of several $\mathrm{Tor}_p(M)$, for some irreducibles $p\in A$, where $p_i^{\alpha_i}$ is the order of the cyclic submodule? I can't guarantee that the generator of each cyclic submodule has a factorization in irreducibles, since I have no additional hypotheses over the ring $A$ except that it is a integral domain.

Any help would be appreciated.

Answer 1

You mentioned that you needed a tool to split the cyclic modules into "primary" pieces. Here's the lemma I think you're looking for:

Lemma: Let $Ax$ be a torsion module over a principal ideal domain $A$ such that $ann(x)=(pq)$ where $p$ and $q$ are coprime elements of $A$. Then $Ax=Ay\oplus Az$ where $ann(y)=(p)$ and $ann(z)=(q)$.

Proof: Set $y=qx, z=px$. Clearly $y,z\in Ax$. Notice $Ay+Az$ contains $x$, because we can find elements such that $1=ap+bq$ and then $x=(ap+bq)x=ay+bz$. Thus $Ax\subseteq Ay+Az$, proving $Ax=Ay+Az$. If $u\in Ay\cap Az$, then $u=u(ap+bq)=0$. So, the sum is direct.

Using this with induction, you can break composite cyclic modules into sums of prime power cyclic modules.

Good luck!