TITLE: Does the regularity of $A$ imply the regularity of $A[X]$?
QUESTION [5 upvotes]: Let $A$ be a commutative Noetherian ring. We say it is regular if its localization at every prime ideal is a regular local ring. 
If this is the case, is it true that $A[X]$ is regular?

REPLY [9 votes]: Yes.  Let $\mathfrak q$ be a prime of $A[X]$.  Then $\mathfrak q$ lies over
a prime $\mathfrak p$ of $A$, and the formation of $(A[X])_{\mathfrak q}$ can
be broken up into two steps: first localize at $\mathfrak p$, and then
localize this at the prime ideal generated by $\mathfrak q$.
The first step gives a ring which is canonically isomorphic to $A_{\mathfrak p}[X]$, and so the second step involves localizing at a prime of this ring
which contains the maximal ideal of $A_{\mathfrak p}.$
Since $A$ was assumed regular, we have reduced to the following situation:
let $A$ be a regular local ring with maximal ideal $\mathfrak m$, and
let $\mathfrak q$ be a prime ideal of $A[X]$ which contains $\mathfrak m.$
We have to show that $A[X]_{\mathfrak q}$ is regular.  
Can you see how to handle this special case?  (Hint: think first about
the case when $A$ is actually a field, and then try to reduce to this case.)