power series ring is faithfully flat but not free Let $A$ be a commutative ring. 
Question 1: Why is the power series ring $A[[x]]$ not free over $A$ in general?
Question 2: Why is $A[[x]]$ faithfully flat over $A$?
 A: If $A =\mathbb Z$, then as a $\mathbb Z$-module, there is an isomorphism
$$\mathbb Z [[x]] \cong \prod_{i = 0}^{\infty} \mathbb Z.$$
The fact that this abelian group is not free is a standard fact, discussed e.g. here (indirectly; if it were free,
its $\mathbb Z$-dual would be huge, rather than just $\bigoplus_i \mathbb Z$; see here)
and here.
Note that if e.g. $A$ is a field then $A[[x]]$ is free over $A$, so this depends on the ring $A$.
One way to see flatness when $A$ is Noetherian is that $A[x]$ is obviously flat over $A$ (being free), and completions of Noetherian rings are flat over the original ring; hence $A[[x]]$ is flat over $A[x]$, and so over $A$. For faithful flatness, one can then use that there is a surjection $A[[x]] \to A$ given by mapping $x \to 0.$
In general (i.e. the non-Noetherian case), one has that for a finitely presented $A$-module, there is a natural isomorphism $M\otimes_A A[[x]] \cong 
M[[x]]$, so that $\text{--}\otimes_A A[[x]]$ is exact on finitely presented $A$-modules.
If $A$ is furthermore coherent (e.g. Noetherian!) then any finitely generated submodule of a finitely presented module is itself finitely presented, and so we may write an injection of arbitrary
$A$-modules as the direct limit of an injection of finitely presented $A$-modules. Since  tensor product commutes with direct limits, we conclude (when $A$ is coherent) that $\text{--}\otimes_A A[[x]]$ is exact on arbitrary $A$-modules, and hence that $A[[x]]$ is $A$-flat.  Faithful flatness again follows from the existence of the map $A[[x]] \to A$.
There are rings that are coherent but not Noetherian (see e.g. here, so this is slightly more general.
Actually, as is explained here, if an arbitrary  product of flat $A$-modules is again flat, then $A$ is necessarily coherent.  I'm not sure what the story is for the countable product $A[[x]]$
of copies of $A$ (i.e. exactly what condition is necessary for this particular $A$-module to be flat), but as noted in the link, if $A = k[t_1,t_2,\ldots]/(t_i t_j)_{i,j = 1,\ldots,\infty},$ then $A[[x]]$ is not flat over $A$.
(Thanks to user26857 for their comment below, which led me to correct a blunder in an earlier version of this answer, and drew my attention to the above link.)
A: For question 2: you can use a general criterion for faithful flatness of ring maps:
$\DeclareMathOperator{Spec}{Spec}$
Proposition: Let $\varphi : R \to S$ be a ring homomorphism. Then $S$ is faithfully flat over $R$ iff $S$ is flat over $R$ and the induced map on spectra ${}^a\varphi : \Spec S \to \Spec R$ is surjective.
For question 1: no counterexample comes to mind at the moment, but let me try to convince you why it's not obviously free:
As an $A$-module, $A[[x]]$ is isomorphic to a countable direct product of copies of $A$, $\prod_{i \in \mathbb{N}} A$. In general, there's no reason to expect an infinite direct product to be isomorphic to a direct sum. For cardinality reasons alone, an uncountable direct sum would be needed. 
However, finding a counterexample is made more difficult by the fact that if $A$ is Artinian local, then $A[[x]]$ is indeed a free $A$-module. In particular, it holds for fields, which are the most common ground rings for power series. 
An interesting question that I do not know the answer to is whether $A[[x]]$ is projective over $A$. Of course, if $A$ is local, this is equivalent to being free. One should thus consider $A$ nonlocal, and presumably not a domain, in searching for counterexamples.
