I posed the question [here](https://math.stackexchange.com/questions/428982/homocyclic-primary-module-over-pid), but get no answers yet.


Let $R$ be a PID, $M$ be an $R$-module. If $M$ is isomorphic to $r$ copies of cyclic primary module $R/\langle p^s\rangle$ where $p$ is a prime element of $R$, then does $M$ possess the following property?


Given any submodule $N$ of $M$ isomorphic to $R/\langle p^{s\_1}\rangle\oplus\cdots\oplus R/\langle p^{s\_r}\rangle$, $M/N$ is isomorphic to $R/\langle p^{s-s\_1}\rangle\oplus\cdots\oplus R/\langle p^{s-s\_r}\rangle$.



Homocyclic primary module over PID
Choose generators $n\_1,\dotsc,n\_r$ for $N$ with $p^{s\_i}n\_i=0$. It is not hard to see that the annihilator of $p^{s\_i}$ on $M$ is $p^{s-s\_i}M$, so we can choose $m\_i$ with $p^{s-s\_i}m\_i=n\_i$. If we can prove that the elements $m\_i$ form a basis for $M$ over the ring $R/p^s$, then everything else is clear.


The given assumptions on $N$ imply that the elements $p^{s-1}m\_i=p^{s\_i-1}n\_i$ are linearly independent over $R/p$ in the space $M[p]=\{m\in M:pm=0\}$, and by counting dimensions they must form a basis. Multiplication by $p^{s-1}$ gives an isomorphism $M/pM\to M[p]$, so the elements $m\_i$ form a basis for $M/pM$ over $R/p$. 


We can certainly write $m\_i=\sum\_ja\_{ij}e\_j$ for some matrix $A=(a\_{ij})$ over $R$, where $e\_1,\dotsc,e\_r$ is the standard basis for $M$. The above shows that $\det(A)$ is invertible mod $p$. It follows easily that it is invertible mod $p^s$ as well, which proves the claim.


My argument refers to $p^{s\_i-1}$ and so does not immediately work if $s\_i=0$ for some $i$, but that can be cured with a few more steps.


