Let $p$ be a prime and $M$ is a finitely generated $ \mathbb{Z}_{p}[[T]] $ module. Suppose $M[p]$ denotes the $p$-torsion of $M$. Then $M[p]$ and $M/(p)$ are both $ F_{p}$ vector spaces. So we can talk of their dimensions. Now what can we say about the rank of $M$ from looking at the dimensions of $M[p]$ and $M/(p) ?$

  • $\begingroup$ What's $\mathbb Z_p$? Not the $p$-adic integers? What do you mean by torsion of a $\mathbb Z_p[[T]]$-module? $\endgroup$ – Fernando Muro Dec 1 '13 at 8:34
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    $\begingroup$ Wouldn't you want to consider the $\mathbb{F}_p[[t]]$-module structure of the reduction, instead of the $\mathbb{F}_p$-module structure? $\endgroup$ – S. Carnahan Dec 1 '13 at 9:34
  • $\begingroup$ @FernandoMuro $ \mathbb{Z}_{p} $ is the p-adic integers. $\endgroup$ – Suman Dec 1 '13 at 18:12

Not much : take $M=\mathbb{Z}_p[[T]]^r$. Then $M[p]$ is zero, and $M/pM=\mathbb{F}_p[[T]]^r$ is infinite-dimensional...


It should be noted that the ring $\mathbb{Z}_p[[T]]$ is the Iwasawa algebra of the additive group $\mathbb{Z}_p$, and there is a nice structure theory for finitely generated modules over it. See the linked Wikipedia article, and the references therein, as well as e.g. chapters 7 and 13 of Washington's Introduction to Cyclotomic Fields or chapter 5 of Lang's Cyclotomic Fields (and for a much more general account, Bourbaki's Commutative Algebra ch. VII §4). However, this structure theory works "up to pseudo-isomorphism" i.e. up to so-called "pseudo-null" modules which in this case are exactly the ones with finite cardinality. All subquotients which are finite-dimensional $\mathbb{F}_p$-vector spaces are in this class; on the other hand, the rank is invariant under pseudo-isomorphism. So it is highly unlikely that the dimensions in your question will give you any non-trivial information about the rank of your module, as illustrated by abx's answer.

Edit: When you consider the $\mathbb{F}_p[[T]]$-module structure as you did between two edits, from this structure theory you get

$rank_{\mathbb{Z}_p[[T]]} M = rank_{\mathbb{F}_p[[T]]} (M/(p)) - rank_{\mathbb{F}_p[[T]]} (M[p])$

but not much more, as far as I can see.


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