# Modules over ring of formal power series

I am facing the following question: Let $K$ be a commutative unital ring (can be taken to be $\mathbb{R}$ or $\mathbb{C}$), and let $K[[\lambda]]$ be the associated ring of formal power series. I know that if $M$ is a $K$-module, then $M[[\lambda]]$ inherits a natural $K[[\lambda]]$-module structure. The question is,

what condition ensures that a given $K[[\lambda]]$-module, say ${\cal M}$, to be of the form $M_0[[\lambda]],$ for some $K$-module $M_0$?

I found the answer in the paper "Algebraic cohomology and deformation theory" of M. Gerstenhaber and D. Schack, published in the book "Deformation theory of algebras and structures and application". There it is just stated, without proof, the following:

A $K[[\lambda]]$-module ${\cal M}$ is of the form $M_0[[\lambda]]$ if an only if ${\cal M}$ is $\lambda$-adically complete and $\lambda$-torsion free, in wich case $M_0 = {\cal M}/\lambda{\cal M}.$

Trying to prove the "if" part, I consider the following map $$M_0[[\lambda]]\rightarrow {\cal M};\ \ \sum_{j=0}^\infty\lambda^j[x]_j\mapsto\sum_{j=0}^\infty\lambda^jx_j.$$

Then, the series in ${\cal M}$ converges, due to the $\lambda$-adic completeness of ${\cal M},$ for any choice of the representant $x_j$ of $[x]_j,$ yet, I can't prove that the limit is independent of the representant. I suppose that can be shown using the $\lambda$-torsion free property, but then I realised that I don't even have a clear idea of what exactly does that property mean.

So, now my question is:

What is the definition of a $\lambda$-torsion free $K[[\lambda]]$-module?

Also, the map proposed above is the "natural one" for me. But I am not sure that is the right one to solve the problem. Perhaps, one I get the right definition of a $\lambda$-torsion free module, I can decide if the map is the needed one.

Thanks in advance for any help.

• If $r\in R$ and $M$ is an $R$-module, the $r$-torsion in $M$ is the set of nonzero elements such that $rm=0$, and $M$ is $r$-torsion free if there is no $r$ torsion. Phrased differently, $M$ is $r$-torsion free if the map $r:M\to M$ given by multiplication by $r$ is an injection. – Aaron Nov 10 '16 at 20:26
• That seems a reasonable definition, may you tell me where did you take it from? Also, adapting that definition when $I\subset R$ is an ideal could be that: $0\neq x\in M$ is an $I$-torsion element if $Ix = 0$? And thus the Module $M$ would be $I$-torsion free if it has no $I$-torsion elements. Anyway, if that is the right definition, it seems that the map I proposed to prove the assertion I am interested in does not work. – Inocencio Nov 11 '16 at 18:48
• Not where I got the definition, but you should look here. In particular, they define $S$-torsion for $S$ a multiplicatively closed set (not necessarily an ideal), and it's not that all of $S$ annihilates $m$, but rather that some $s\in S$ does. This definition seems useful when talking about localization. – Aaron Nov 11 '16 at 19:10