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.
This is not true for arbitrary commutative rings $K$, and I would also not consider the suggested map "natural" when it depends on choices of representatives. We need to add a third condition to fix this.
Lemma. A $K[[\lambda]]$-module $M$ is isomorphic to $N[[\lambda]]$ for some $K$-module $N$ if and only if the following three conditions are satisfied:
Notice that the third condition is automatic when $K$ is a field, but not in general (see below). We only demand here a complement in the category of $K$-modules, not in the category of $K[[\lambda]]$-modules (which would be too strong).
The direction $\implies$ is easy. Notice that $N[[\lambda]] \lambda = \prod_{n>0} N \lambda^n$ has a canonical $K$-module complement inside $N[[\lambda]]$, namely $N = N \lambda^0$. By the way, I will work with right modules here, basically since we are used to writing coefficients of power series in front of the monomials.
The direction $\impliedby$ is more interesting. Let $M$ be a $K[[\lambda]]$-module that has no $\lambda$-torsion, is $\lambda$-adically complete, and such that there is a $K$-submodule $N \subseteq M$ with $M = M\lambda \oplus N$. Since $M$ is $\lambda$-adically complete, the map $$\alpha : N[[\lambda]] \to M,\quad \sum_{n=0}^{\infty} u_n \lambda^n \to \sum_{n=0}^{\infty} u_n \lambda^n$$ is well-defined. To explain the notation a bit, the power series on the left is formal, the one on the right is a convergent series in $M$.
$\alpha$ is injective. Assume that $\sum_{n=0}^{\infty} u_n \lambda^n$ lies in the kernel. Then $u_0 = -\sum_{n=0}^{\infty} u_{n+1} \lambda^{n+1}$ holds in $M$. This shows $u_0 \in N \cap M \lambda = 0$, hence $u_0 = 0$. We proceed by induction. Assume that we have already shown that $u_0=\dotsc=u_{k-1}=0$. Then $\sum_{n \geq k}^{\infty} u_n \lambda^n = 0$ holds in $M$. Since $\lambda^k : M \to M$ is injective, this implies $\sum_{n \geq k}^{\infty} u_n \lambda^{n-k} = 0$. So again we see $u_k \in N \cap M \lambda = 0$ and hence $u_k = 0$. This finishes the induction.
$\alpha$ is surjective. (My gut feeling is that there is a more elegant argument, but anyway let's go.) Let $m \in M$. Since $M = M \lambda + N$, we can write $m = m_0 \lambda + u_0$ with $m_0 \in M$ and $u_0 \in N$. We proceed with $m_0$ in the same way and find $m_1 \in M$, $u_1 \in N$ with $m_0 = m_1 \lambda + u_1$. Hence, $m = m_1 \lambda^2 + u_1 \lambda + u_0$. Recursively, we find $u_n \in N$ and $m_n \in M$ with $$m = m_n \lambda^{n+1} + u_n \lambda^n + \cdots + u_1 \lambda + u_0.$$ The limit $m' := \sum_{n=0}^{\infty} u_n \lambda^n$ exists in $M$. Then $m - m'$ is the limit of the sequence $m_n \lambda^{n+1}$, hence $0$. So, $m = m'$, and we see that $m \in \mathrm{im}(\alpha)$. $\quad \square$
Example. The third condition is really necessary. Consider the ring of $p$-adic integers $\mathbb{Z}_p$. It can be realized as $\mathbb{Z}[[X]]/\langle X-p \rangle$ and hence carries a natural $\mathbb{Z}[[X]]$-module structure. It has no $p$-torsion, and it is well-known to be $p$-adically complete. But it is not isomorphic to $N[[X]]$ for a subgroup $N \subseteq \mathbb{Z}$. Otherwise, it would be isomorphic to $\mathbb{Z}[[X]]$ as a $\mathbb{Z}[[X]]$-module, which is not true since it has torsion over $\mathbb{Z}[[X]]$.