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Let $\Lambda$ be a commutative integral ring with an automorphism $\sigma$ (I have in mind $\mathbb Z_p[[t]]$ and $\sigma(t) = (1+t)^\alpha - 1$ with $\alpha \in \Lambda^\times$) and $R = \Lambda\{F\}$ with $F\lambda = \sigma(\lambda)F$ for $\lambda \in \Lambda$.

Is there a classification of finitely generated modules over $R$ that are free and finite as modules over over $\Lambda$? I allow faithfully flat base changes of $\Lambda$ so that we can assume it's fraction field is algebraically closed (among other things).

Ultimately, I am only interested in the eigenvalues of F, if that makes sense.

When $\Lambda$ is a field, there is a classification similar to the standard one over PID's in chapter three of "The theory of rings" by Nathan Jacobson.

What about the general case or at least my specific example? Or even when $\Lambda$ is a PID? Ideally, I would want any finitely generated module to be isomorphic to a direct sum of modules generated by one element, perhaps up to finite kernel and cokernel.

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  • $\begingroup$ Do you mean that $R$ is a skew-polynomial ring in the sense of en.wikipedia.org/wiki/…, so that typical elements are of the form $\sum_{i=0}^n \lambda_i F^i$ with $\lambda_i\in \Lambda$? I think you must but wanted to check. $\endgroup$ Commented Apr 12, 2020 at 13:35
  • $\begingroup$ Yes, that's exactly right. I didn't know the name for it. $\endgroup$
    – Asvin
    Commented Apr 12, 2020 at 13:36
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    $\begingroup$ I don't know about your specific example, but for PIDs in general: for $\Lambda=\mathbb{Z}$ and $\sigma=\text{id}$, so you're looking at $\mathbb{Z}[t]$- modules, I believe it's known that the classification of these (even if you restrict to those that are finitely generated and free over $\mathbb{Z}$) is a wild problem. $\endgroup$ Commented Apr 12, 2020 at 14:49
  • $\begingroup$ Really? At least for similar examples to that, I thought I had an atgument where you tensor with Q, use the classification over PIDs and finally use the freeness over Z to classify such modules. I must have made a mistake somewhere. Do you have a reference for modules over Z[t] that are free as Z modules? $\endgroup$
    – Asvin
    Commented Apr 12, 2020 at 15:02
  • $\begingroup$ @Jeremy Rickard I am not sure why it is wild. If you restrict to finitely-generated free it seems to be covered by Latimer-MacDuffee: en.wikipedia.org/wiki/Latimer%E2%80%93MacDuffee_theorem $\endgroup$
    – Bugs Bunny
    Commented Apr 12, 2020 at 15:24

1 Answer 1

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I haven't thought about base changes, but the original problem for $\alpha=1$ (so $\sigma$ is the identity map and $R=\mathbb{Z}_p[[t]][F]$ is just a polynomial ring over $\mathbb{Z}_p[[t]]$, and classifying $R$-modules that are finitely generated and free over $\mathbb{Z}_p[[t]]$ up to isomorphism is equivalent to classifying square matrices over $\mathbb{Z}_p[[t]]$ up to conjugacy) is a "wild" problem (i.e., if you could classify these then you could classify pairs of matrices over some field up to simultaneous conjugacy), and so is probably intractable.

In fact, Theorem 2 of

Gudivok, P. M.; Oros, V. M.; Rojter, A. V., Representations of finite $p$-groups over the ring of formal power series with integral $p$-adic coefficients, Ukr. Math. J. 44, No. 6, 678-689 (1992); translation from Ukr. Mat. Zh. 44, No. 6, 753-765 (1992). ZBL0787.20006.

shows that the classification of representations of the cyclic group $C_{p^2}$ over $\mathbb{Z}_p[[t]]$ is a wild problem, and this is the subproblem of classifying those matrices whose $p^2$th power is the identity.

For $\alpha\neq1$, I think it should still be a wild problem, as the problem of classifying $R$-modules that are finitely generated and free over $\mathbb{Z}_p[[t]]$ should be at least as hard as classifying representations of finite cyclic groups over $\mathbb{Z}_p$, and this is a wild problem for $G=C_{p^3}$ ($p$ odd) and $C_{16}$ ($p=2$) (see the main theorem of

Dieterich, Ernst, Group rings of wild representation type, Math. Ann. 266, 1-22 (1983). ZBL0506.16021.)

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