Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?


share|cite|improve this question

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.

share|cite|improve this answer
Dear Matt: I'm impressed! – Georges Elencwajg Feb 15 '12 at 23:28
@Georges: Dear Georges, Thank you! Best wishes, – Matt E Feb 16 '12 at 1:38
Dear Matt, thank you so much for your explanations. This helped me a lot! – eileendavid82 Feb 16 '12 at 2:51

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.