Let $A$ be a finitely generated torsion $\mathbb{Z}_p[[X]]$module, $B$ = { $x \in A$ such that $px=0$ } and $C=A/B$ where $\mathbb{Z}_p$ denotes the $p$adic integers. Given $ 0 \rightarrow B/pB \rightarrow A/pA \rightarrow C/pC \rightarrow 0$ is a short exact sequence of abelian groups. Now $A/pA, B/pB, C/pC$ are finitely generated $\mathbb{F}_p[[X]]$modules where $\mathbb{F}_p$ denotes the finite field. How to prove $A/pA \cong B/pB \oplus C/pC$ as $\mathbb{F}_p[[X]]$modules $?$

$\begingroup$ Please don't post on MO and MSE at the same time. (math.stackexchange.com/questions/645535/…) $\endgroup$ – Asaf Karagila Jan 20 '14 at 22:26

$\begingroup$ Sorry. Deleted from MSE. $\endgroup$ – Robert Jan 20 '14 at 22:27

1$\begingroup$ You're asking how to prove that every short exact sequence of f.g. $\mathbb{F}_p[[X]]$modules splits. It clearly splits mod $X$, since then they're just $\mathbb{F}_p$ vector spaces. My first thought would be to assume it splits mod $(X^n)$ and try to lift to a splitting mod $(X^{n+1})$. Have you tried that? You'll end up with some Ext$^1$ group that you'll need to check is trivial. $\endgroup$ – Joe Silverman Jan 20 '14 at 23:17

1$\begingroup$ Since $R = \mathbb{F}_p[[X]]$ is a DVR, $C/pC$ is isomorphic to a finite direct sum of $k$ $R$modules of the form $R/(X^{n_i})$, where $n_1 \leq n_2 \leq \cdots \leq n_k$ and $(X^{n_k}) = \operatorname{Ann}(C/pC)$. The exponents $n_i$ are all zero iff $C/pC$ is torsionfree, iff it is free, iff it projective, in which case the sequence splits. But in your case $C/pC$ is torsion, hence not projective. The sequence may still split, but not every sequence splits over $R$ since in that case $R$ would have to be artinian. mathoverflow.net/questions/62464/… $\endgroup$ – Jesse Elliott Jan 22 '14 at 10:16
Better not to prove it, as it is wrong. Writing $M[p] = \lbrace m \in M: pm = 0 \rbrace$ for the exact $p$torsion of a module $M$, I understand your question as follows: If the exact sequence $$0 \rightarrow A[p] \rightarrow A \rightarrow A/A[p] \rightarrow 0$$ remains exact modulo $p$ (which is not always the case: it is equivalent to $A[p] \cap pA = 0$), then this exact reduced sequence $$0 \rightarrow (A[p])/p \rightarrow A/p \rightarrow (A/A[p])/p \rightarrow 0$$ splits. A counterexample is $A = \mathbb{Z}_p[\![X]\!] /(pX)$ where the reduced sequence $$ 0 \rightarrow X\mathbb{F}_p[\![X]\!] \rightarrow \mathbb{F}_p[\![X]\!] \rightarrow \mathbb{F}_p \rightarrow 0$$ is exact, but does not split, nor is there even any chance to write the middle term as direct sum of the outer ones (as $\mathbb{F}_p[\![X]\!]$module).
Maybe the illusion that the statement might be true comes from the fact that it is true for $p$torsion modules (in particular, pseudonull modules) and the standard modules $\bigoplus_i\mathbb{Z}_p[\![X]\!]/(p^{\mu_i})\oplus\bigoplus_j\mathbb{Z}_p[\![X]\!]/(f_j^{m_j})$, as a direct check shows.
Computing a pseudoisomorphism from the above $A$ to the standard module $\mathbb{Z}_p[\![X]\!]/(p)\oplus\mathbb{Z}_p[\![X]\!]/(X)$ and seeing where things go wrong might be a good exercise in Iwasawa theory, compatible to questions like Dual of a module , Iwasawa algebra , Rank of a $ \mathbb{Z}_{p}[[T]] $ module or Iwasawa invariants.

$\begingroup$ @Tosten Please give some references where I can have the proof of the splitting for ptorsion modules (in particular, pseudonull modules) and the standard modules. $\endgroup$ – Robert Jan 23 '14 at 10:43

$\begingroup$ For a $p$torsion module, $A[p] \cap pA = 0$ implies $A=A[p]$; this also settles one summand of the standard module. You should really do the rest yourself. Maybe I am unfair, but I feel uncomfortable about these seemingly related, nottoohard questions about Iwasawa theory from several accounts. $\endgroup$ – Torsten Schoeneberg Jan 23 '14 at 12:58