0
$\begingroup$

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 $?$

$\endgroup$
2
  • $\begingroup$ I thought one could find everything there is to say in the references I gave to your last question, mathoverflow.net/a/150455/27465 $\endgroup$ Dec 5, 2013 at 13:04
  • $\begingroup$ Indeed, I see now that this is a duplicate question. $\endgroup$ Dec 5, 2013 at 13:06

1 Answer 1

4
$\begingroup$

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.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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