# Fibres in a power series ring versus fibres in a polynomial ring (a simple question)

Let $A$ be a commutative ring and $p \in \operatorname{Spec} A$.

In Matsumura's Commutative Ring Theory p. 118 it is mentioned that even though $A[x] \otimes \kappa(p) = \kappa(p)[x]$, it is not true in general that $A[[ x ]] \otimes \kappa(p) = \kappa(p)[[x]]$.

I tried to see why the equality fails for the power series ring and it seems to me that the problem is that when we have an element of $\kappa(p) [[x]]$, then in general we have infinitely many denominators and we can not write that element in the form "some element of $A/p[[x]]$ divided by some element of $A-p$".

Am i right, is that the reason? Also, can this fact be expressed more rigorously than i am expressing it?

• – user26857
Commented Jul 17, 2013 at 15:17
• More rigorously: $A[x]$ is $A$-flat, while $A[[x]]$ is not.
– user26857
Commented Jul 17, 2013 at 15:21
• @YACP: Both of your comments are very useful. Question 1: $A[x]$ is $A$-free, hence $A$-flat, OK. But how do we see that $A[[x]]$ is not flat? Question 2: How is flatness used to answer my question? As one comment mentions in your answer to the link you provided, this must have to do with the fact that $A[[x]] \otimes \kappa(p)$ has only "bounded denominators". This i also point out in my question. How does flatness relate to that? Commented Jul 17, 2013 at 15:24
• Well, maybe a better approach from your point of view is the following: is $A[[x]]\otimes A/I\simeq(A/I)[[x]]$ (canonically)? This is obviously true when $IA[[x]]=I[[x]]$, and now take a look here.
– user26857
Commented Jul 17, 2013 at 15:47
• @YACP: i see your point. Please make this an answer, i am covered. Also, i have posted a comment question in your answer to the first link you provide. Commented Jul 17, 2013 at 16:05

Well, maybe a better approach from your point of view is the following: is $A[[x]]\otimes A/I$ (canonically) isomorphic to $(A/I)[[x]]$? This holds when $IA[[x]]=I[[x]]$ and now can take look here.