TITLE: Faithfully Flat Ring Homomorphism of Power Series
QUESTION [11 upvotes]: Let $R$ be a one-dimensional local ring and let $f:R[[x]][y] \rightarrow R[y][[x]]$ be the inclusion map. 
How can I show that $f$ is a faithfully flat ring homomorphism? 
Or can you give me a reference?
Thanks.

REPLY [11 votes]: The usual way to show that a map of this kind is faithfully flat is to use the
Artin--Rees Lemma and its corollaries, which show that adic completions of Noetherian rings are flat, and faithfully flat under appropriate hypotheses.
For example, the completion of a Noetherian ring at any ideal contained in
its Jacobson radical is faithfully flat.
In your particular case, you can check that the target of your inclusion $f$
is the $x$-adic completion of the source, and so the map is flat.  
However, $x$ is not in the Jacobson radical of $R[[x]][y]$ (although it
is in the Jacobson radical of $R[[x]]$, and also in the Jacobson radical
of $R[y][[x]]$).  
E.g. let $\mathfrak m_R$ be the maximal ideal of $R$, and consider the
ideal $\mathfrak m := (\mathfrak m_R, xy -1)$.  The quotient $R[[x]][y]/\mathfrak m$ is equal to $(R/\mathfrak m_R)[[x]][1/x]$, which is
the field of Laurent series in $x$ over the residue field $R/\mathfrak m_R$.
Thus $\mathfrak m$ is maximal, but does not contain $x$.
By Artin--Rees, the tensor product of $R[y][[x]]$ with $R[[x]][y]/\mathfrak m$ 
over $R[[x]][y]$ is equal to the $x$-adic completion of $R[[x]][y]/\mathfrak m$, which
vanishes.  Thus $R[y][[x]]$ is not faithfully flat over $R[[x]][y]$.
In fact, we could have just looked at the ideal $I = (xy - 1)$; i.e. the tensor product $R[y][[x]]\otimes_{R[[x]][y]} (R[[x]][y]/I)$ already vanishes, because
$1 - xy$ is a unit in $R[y][[x]]$.  The reason I introduced $\mathfrak m$ at all is just to show explicitly that $x$ is not in the Jacobson radical of $R[[x]][y]$.
The basic intuition is that in $R[[x]][y]$ you are allowed to specialize $y$
to be $1/x$.  But in $R[y][[x]]$ you have elements of the form
$ 1 + x y + x^2 y^2 + \cdots + x^n y^n + \cdots$  (this is precisely the inverse of $1 - xy$) and you cannot substitute
$y = 1/x$ into such an element in a meaningful way.