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Jan 23, 2014 at 10:19 vote accept Robert
Jan 22, 2014 at 15:39 answer added Torsten Schoeneberg timeline score: 3
Jan 22, 2014 at 10:16 comment added Jesse Elliott 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 torsion-free, 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/…
Jan 20, 2014 at 23:17 comment added Joe Silverman 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.
Jan 20, 2014 at 22:37 review First posts
Jan 20, 2014 at 22:42
Jan 20, 2014 at 22:27 comment added Robert Sorry. Deleted from MSE.
Jan 20, 2014 at 22:26 comment added Asaf Karagila Please don't post on MO and MSE at the same time. (math.stackexchange.com/questions/645535/…)
Jan 20, 2014 at 22:21 history asked Robert CC BY-SA 3.0