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Suppose $M$ is a finitely generated torsion $Z_p[[T]]$-module; the torsion comes from the $\mu$-invariant and the $\lambda$-invariant. Consider $M/(p)$ and $M[p]$ ($p$-torsion of $M$) which are $F_p[[T]]$-modules, so they have an $F_p[[T]]$-rank and an $F_p[[T]]$-torsion. How to understand the $\lambda$-invariant and $\mu$-invariant of $M$ in terms of these $?$

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  • $\begingroup$ I thought one could find everything there is to say in the references I gave to your last question, mathoverflow.net/a/150455/27465 $\endgroup$ Commented Dec 5, 2013 at 13:04
  • $\begingroup$ Indeed, I see now that this is a duplicate question. $\endgroup$ Commented Dec 5, 2013 at 13:06

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Both $\mu$ and $\lambda$ are about the $\Lambda$-torsion part of $M$. Neither of both invariants is visible in either $M/pM$ or $M[p]$ over $\Omega=\mathbb{F}_p[\![T]\!]$.

The $\Lambda$-rank of $M$ is the difference between the $\Omega$-ranks of $M/pM$ and the one of $M[p]$. Now suppose $M$ is torsion. If $M/pM$ (or equivalently) $M[p]$ is $\Omega$-torsion, then $\mu=0$, otherwise $\mu>0$, but you won't be able to say what it is. So suppose $M$ is torsion and $\mu=0$, then $M/pM$ and $M[p]$ are both finite dimensional $\mathbb{F}_p$-vector spaces and $\lambda$ is the difference between the dimensions of $M/pM$ and the one of $M[p]$. I guess that is all one can ever say.

Reason: Simply test what they are in the four cases $M=\Lambda$, $M=\Lambda/p^\mu$, $M= \Lambda/f$ for an irreducible distinguished polynomial $f$ and $M$ a finite $\Lambda$-module.

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