Torsion module over PID

Question

Suppose $p$ is irreducible and $M$ is a tosion module over a PID $R$ that can be written as a direct sum of cyclic submodules with annihilators of the form $p^{a_1} | \cdots | p^{a_s}$ and $p^{a_i}|p^{a_i+1}$. Let now $N$ be a submodule of $M$. How can i prove that $N$ can be written a direct sum of cyclic modules with annihilators of the form $p^{b_1} | \cdots | p^{b_t}, t\leq s$ and $\ p^{b_i}| p^ {a_(s-t+i)}$?

I've already shown that $t\leq s$ considering the epimorphism from a free module to $M$ and from its submodule to $N$.

Answer 1

Here's an ugly argument that I know of (which works for abelian groups, and so I assume it works over arbitrary PID). You need to make three observations:

1. Maximal cyclic $p$-submodules of a $p$-module $M$ are the direct summands of $M$.
2. Every cyclic $p$-submodule of $M$ is contained in a maximal cyclic $p$-submodule.
3. If $N_1\subset M_1$ and $N_2\subset M_2$ are cyclic $p$-submodules of $M$, then $M_1$ and $M_2$ are disjoint if and only if $N_1$ and $N_2$ (this follows from observing that the submodules of a cyclic $p$-module are totally ordered under inclusion, and considering where $M_1\cap M_2$ is relative to $N_1$ and $N_2$ as a submodule of $M_1$ or $M_2$; you may have to use the fact that cyclic $p$-modules are indecomposable).

Once you have proven those, take a direct sum decomposition of $N$ into cylic $p$-submodules $N_1\oplus N_2\oplus\dots\oplus N_i$ (which exists by structure theorem of finitely generated modules over a PID), and use 1., 2. and 3. to obtain a decomposition $M=M_1\oplus\dots M_n\oplus M'$, where $M_i$ is the maximal cyclic $p$-submodule that contains the cyclic $p$-submodule $N_i$.

*by $p$-module I mean a module annihilated by some power of the prime $p$ in the PID $R$ (it's the primality of $p$ that matters, not irreducibility, for the total ordering of submodules, though the two coincide over PIDs, which is where you need to be anyway for the structure theorem to work)

Answer 2

See Lemma 9 here.

Answer 3

just look at the size of the subgroups annihilated by p^k for various k.

Answer 4

Here's a proof :

The result can be easily shown by strong induction to $\sum_{i=1}^{s}{a_i}$ .Suppose that the result is true if the sum above is less or equal to k.Suppose now that the sum is equal to k+1.Then for the induction step we use the invariant factors of pM for which the sum is less than k.