Finitely generated module over $A[[x]]$ I have a question about module over power series ring.
Let $A$ is a local ring with maximal ideal $\mathfrak m$, I'm more interested in the case that $A=\mathbb Z_p/p^n\mathbb Z_p$
If $M\neq0$ is a finitely generated torsion module over $A[[x]]$, then is $M$ finitely generated over $A$?
I think it suffices to assume that $M$ is generated by one element $m$ and $m$ is killed by some series $f(x)=\sum_{i=0}a_ix^i$. But if $a_0\in\mathfrak m$, I can't deduce anything.
Thanks.
 A: $p$ is nilpotent. For any $f\in pA[[x]], g\in x^m A[[x]]^\times$,
$f-g$ is not a zero-divisor and any non-zero divisor is of this form.
let $$h=\prod_{m=0}^{n-1}(f^{2^m}+g^{2^m})$$
we have that $$(f-g) (h)=(f^{2^n}-g^{2^n})=(g^{2^n})=(g)^{2^n}=(x^m)^{2^n}=(x^{m 2^n})$$
When including non-zero divisor in the definition of torsion module you get that
$M$ is a finitely generated $A[[x]]$-module killed by such a non-zero divisor $f-g\in A[[x]]$, whence $M$ is killed by some $x^{m2^n}$,  it is a finitely generated $A[[x]]/(x^{m 2^n})$ module.
Therefore it is also a finitely generated $A$-module.
A: You say you know the structure theory for f.g. torsion modules over the Iwasawa algebra $\Lambda := \mathbb Z_p[[x]]$. But then the result follows easily.
Namely, the ring $(\mathbb Z_p/p^n \mathbb Z_p) [[x]]$ you look at is just $\Lambda / (p^n)$, and like over any ring, a $\Lambda / (p^n)$-module is the same as a $\Lambda$-module on which $p^n$ operates as zero. Further, such a module $M$ being torsion and f.g. as $\Lambda / (p^n)$-module implies that it is f.g. and torsion as $\Lambda$-module. So we can apply said structure theory and now there is a homomorphism with finite kernel and cokernel
$$ M \rightarrow \bigoplus_i\mathbb{Z}_p[\![x]\!]/(p^{\mu_i})\oplus\bigoplus_j\mathbf{Z}_p[\![x]\!]/(f_j^{m_j})$$
Since being f.g. over $\mathbb Z_p$ will not be changed by those finite kernel and cokernel, w.l.o.g. we can assume this is an isomorphism; then, to be torsion as $\Lambda/(p^n)$-module means that the first sum must vanish (because it's not torsion). But to have $p^n$ operate as $0$, the second sum must vanish as well. Meaning that up to something finite, $M =0$; meaning that $M$ is finite; meaning that it's finitely generated as module over whatever ring you want.
