Let $A \in M_n(\mathbb{Z}_p)$ be a nonsingular matrix which is nilpotent mod $p$, so $A^r \in pM_n(\mathbb{Z}_p)$ for some $r$. Then $\mathbb{Z}_p[[T]]$ acts on $\mathbb{Z}_p^n$ with $T$ acting by $A$. If I take $M/pM$, or equivalently tensor with $\mathbb{F}_p[[T]]$, then since $\mathbb{F}_p[[T]]$ is a local ring I get a decomposition of the form $\bigoplus_i \mathbb{F}_p[[T]]/(T^{n_i})$. However, the structure theorem for modules over the Iwasawa algebra tells us that there is a quasi-isomorphism (i.e. a $\mathbb{Z}_p[[T]]$-module map with finite kernel and cokernel) $$M \to \bigoplus_{i} \mathbb{Z}_p[[T]]/(f_i(T)^{n_i})$$ where the $f_i$ are either distinguished polynomials (there is no free part and no $p$-torsion do to the initial assumptions on $M$). If I take the module $\bigoplus_{i} \mathbb{Z}_p[[T]]/(f_i(T)^{n_i})$ and reduce mod $p$ to get an $\mathbb{F}_p[[T]]$-module, then it's already in the form $\bigoplus_i \mathbb{F}_p[[T]]/(\bar{f}_i(T)^{n_i})$ (where the $\bar{f}_i$'s are just powers of $T$ since the $f_i$ were distinguished polynomials).

My question is, is the $\mathbb{F}_p[[T]]$-module I get by reducing $\bigoplus_{i} \mathbb{Z}_p[[T]]/(f_i(T)^{n_i})$ mod $p$ the same as the one I got by reducing $M$ mod $p$, and then finding its decomposition? So in some sense I'm asking whether the operations "decompose my module" and "reduce mod $p$" commute. Note that this decomposition holds more information than the $\lambda$-invariant of $M$.

As Chris Wuthrich points out in a comment responding to an earlier version of this question where I only assumed $M$ is a finitely-generated torsion module which is free when restricted to a $\mathbb{Z}_p$-module, the answer is "no" under these conditions. However, the setup above gives some additional conditions, and hopefully these are enough to make the above true.

Note: I'm aware that the quasi-isomorphism from $M$ to the decomposed module will in general NOT become an isomorphism after tensoring the two modules with $\mathbb{F}_p[[T]]$, but I'm asking whether they will nonetheless be isomorphic by a different map.

  • $\begingroup$ The PID $\mathbf F_p[[T]]$ is still a local ring, so in fact the structure theorem gives you that $M$ is isomorphic to a direct sum of modules like $\mathbf F_p[T]/ T^n$. In other words, your irreducible polynomials $f_i$ in the first paragraph are actually just $f_i(T) = T$. $\endgroup$
    – user94041
    Mar 29, 2018 at 2:31
  • $\begingroup$ Yes, I didn’t write that down but thanks for pointing it out $\endgroup$ Mar 29, 2018 at 2:50
  • $\begingroup$ I updated the question as well. $\endgroup$ Mar 29, 2018 at 3:39
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    $\begingroup$ What about $M$ being the quotient of the maximal ideal of $\Lambda$ by $T^2$? I think both quotients by $p$ will be of dimension $2$, but $T$ does not act the same on them... $\endgroup$ Mar 29, 2018 at 10:17
  • $\begingroup$ @ChrisWuthrich You're right that this is a counterexample to my question as stated. There are a few extra conditions that apply in the particular setting I'm interested in, and in particular rule out your example, so perhaps a salvage is possible under the condition below (I've added this to the question text as well). For what I'm interested in, $M$ is $\mathbb{Z}_p^n$ with $T$ acting some $A \in M_n(\mathbb{Z}_p)$ which is nilpotent mod $p$ (this condition is necessary for power series in $A$ to converge $p$-adically) but nonsingular. Do you think my claim is true with these hypotheses? $\endgroup$ Mar 30, 2018 at 3:33


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