Let $\mathbb{k}$ be a field and $A$ be a commutative $\mathbb{k}$-algebra. Then the algebra of formal power series $A[[x]]$ can be viewed as a $\mathbb{k}[[x]]$-module in a natural way. My question is whether $A[[x]]$ is a free $\mathbb{k}[[x]]$-module.
Note that $\mathbb{k}[[x]]$ is a P.I.D. and $A[[x]]$ is a torsion-free $\mathbb{k}[[x]]$-module, so $A[[x]]$ must be flat as a $\mathbb{k}[[x]]$-module. Since $A[[x]]$ may not be finitely generated as a $\mathbb{k}[[x]]$-module, the structure theorem for finitely generated modules over a P.I.D. is invalid here.
Any help would be appreciated!
If $A$ is finite-dimensional as a $k$-vector space then trivially $A[[x]]$ is a free $k[[x]]$-module (any basis for $A$ over $k$ is also a basis for $A[[x]]$ over $k[[x]]$). On the other hand, if $A$ is infinite-dimensional, then $A[[x]]$ is never a free $k[[x]]$-module. Since a submodule of a free $k[[x]]$-module is free, it suffices to prove this in the case that $A$ is countable-dimensional. In that case, note that as shown here, $A[[x]]$ is not countably generated as a $k[[x]]$-module, so if it were free, its rank would have to be uncountable. However, $A[[x]]/xA[[x]]\cong A$ is countably generated as a $k[[x]]/(x)\cong k$-module, so $A[[x]]$ cannot be free of uncountable rank.