I posed the question here, 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 $R/\langle p^{s-s_1}\rangle\oplus\cdots\oplus R/\langle p^{s-s_r}\rangle$.

  • $\begingroup$ This is better suited for MSE. You could start with $r=1$, with $M=R/p^sR$, $N=R/p^{s_1}R$. $\endgroup$ – Dietrich Burde Jun 26 '13 at 8:23

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.

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  • $\begingroup$ Following is the way I can think of to work for the case when some $s_i=0$. Get $m_i$'s in your argument for nonzero $s_i$'s, and prove that they are $R$-linearly independent by showing a correspinding minor of $A$ is invertible mod $p^s$. Then extend these $m_i$'s to an $R$-basis of $M$ as $M$ is homocyclic. Is there any simpler way? $\endgroup$ – Binzhou Xia Jul 13 '13 at 13:17

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