Module of power series and change of rings Let $R$ be a commutative ring with unit an $M$ an $R$-module. Define the formal power series over $M$ as $$M[[X]]=\{\sum_{n\ge 0}m_nX^n: m_n\in M\}$$ and make this an $R[[X]]$-module by $(aX^i)(mX^j)=amX^{i+j}$.
In general, $M[[X]]$ is not isomorphic to $R[[X]]\otimes_R M$. (An example to support this is welcome.) I've read in this paper at page 20 that such an isomorphism holds if $R$ is Noetherian and $M$ is a finite free module, but there is no proof. 
Question. What are the most general conditions which ensure $M[[X]]\cong R[[X]]\otimes_RM$? 
Proofs are welcome, but references are equally appreciated.
 A: The definition of the $R[[X]]$-module structure on $M[[X]]$ is incomplete since the monomials don't generate the ring of power series (this only holds as topological rings). You can write down the module structure explicitly:
$$\sum_i r_i X^i \cdot \sum_i m_i X^i := \sum_i \left(\sum_{p+q=i} r_p m_q\right) X^i$$
This describes a bilinear map $R[[X]] \times M[[X]] \to M[[X]]$. By restriction we get a bilinear map $R[[X]] \times M \to M[[X]]$, hence a linear map $R[[X]] \otimes_R M \to M[[X]]$. Let $\mathcal{S}$ be the class of all $R$-modules for which this is an isomorphism. We have the following obvious properties: 


*

*$0 \in \mathcal{S}$ and $R \in \mathcal{S}$.

*$\mathcal{S}$ is closed under finite direct sums.

*$\mathcal{S}$ is closed under direct summands.

*$\mathcal{S}$ is closed under $\aleph_1$-directed colimits (i.e. $M = \varinjlim_i M_i$ where the index set has the property that every countable set has an upper bound; notice that then $M[[X]] = \varinjlim M_i[[X]]$).


The first three properties imply that every finitely generated projective module is contained in $\mathcal{S}$.
For the other direction, assume that $R$ is noetherian and that $M \in \mathcal{S}$. If $m_0,m_1,\dotsc$ is a sequence of elements in $M$, then $\sum_i m_i X^i \in M[[X]]$ has a preimage in $R[[X]] \otimes_R M$, say $\sum_{j=1}^{n} p_j \otimes n_j$ with finitely many (!) $p_j \in R[[X]]$ and $n_j \in M$. It follows that $\langle \{m_i : i \in \mathbb{N}\} \rangle \subseteq \langle \{n_j : 1 \leq j \leq n\} \rangle$ is finitely generated (using that $R$ is noetherian). Hence, every countably-generated submodule of $M$ is finitely generated.
In particular, when $R$ is a non-zero noetherian ring, we see that any free module of rank $\geq \aleph_0$ does not lie in $\mathcal{S}$.
