MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $R/\langle p^{s-s_1}\rangle\oplus\cdots\oplus R/\langle p^{s-s_r}\rangle$.

share|cite|improve this question
This is better suited for MSE. You could start with $r=1$, with $M=R/p^sR$, $N=R/p^{s_1}R$. – Dietrich Burde Jun 26 '13 at 8:23
up vote 3 down vote accepted

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.

share|cite|improve this answer
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? – Binzhou Xia Jul 13 '13 at 13:17

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.