When does the regularity of $A$ implies the regularity of $A[w]$? Let $A$ be a commutative noetherian ring (I do not mind to assume that $A$ is a UFD), and assume that $A$ is regular.
Recall that a commutative noetherian ring is called regular if all its localizations at maximal ideals are regular local rings.
Let $w$ be algebraic over $A$ (I do not mind to assume that $w$ is integral over $A$. I do NOT want to assume that $w$ is in the field of fractions of $A$), and assume that $A[w]$ is noetherian.
My question: When is $A[w]$ regular? 
(I suspect there exists a counterexample, even under the additional conditions: $A$ is a UFD, $w$ is integral over $A$).
Notice the following question, which deals with $w$ transcendental over $A$:
Does the regularity of $A$ imply the regularity of $A[X]$?
 A: The nonregular ring $R=k[x,y]/\langle x^2 - y^3\rangle$, which is the coordinate ring of the cusp (not regular at the maximal ideal $\langle x,y\rangle$), is of this form. Indeed, you can set $A:=k[t^2]\cong k[t]$ which is a polynomial ring and then, $R=A[t^3]$. Here, $t^3$ is integral over $A$, but it is not in the field of fractions for reasons of degree. 
Geometry: You are basically taking a smooth affine scheme $X$ and then you consider a closed subscheme $Z\subseteq \mathbb A^1_X$ whose projection to $X$ is finite. That closed subscheme may well be singular. However, this is not the case if that finite projection is actually étale, i.e. smooth of relative dimension zero - in this case $Z$ must be smooth. There are several equivalent definitions of being étale, one would be flat and unramified. These are now just properties of the ring extension $A\subseteq A[w]$.
Edit. When you can guarantee flatness and you want to use the above criterion, you have to check the following: For any prime ideal $\mathfrak q\subseteq A[w]$, the prime ideal $\mathfrak p := \mathfrak q\cap A$ (of $A$) generates the maximal ideal of the local ring $A[w]_{\mathfrak q}$. This is the condition for $A\hookrightarrow A[w]$ to be unramified.
A: If $A$ is regular, then $A[X]$ is regular. Now let $f$ be a monic polynomial in the square of a maximal ideal of $A[X]$. Then $A[X]/(f)=A[w]$ is not regular. (It's worthwhile to mention that $A[w]$ is however CM; see Bruns and Herzog, Proposition 2.2.11.)
Concrete example: $A=k[t]$, and $f(X)=X^2-t^3$. The maximal ideal $\mathfrak m=(t,X)$ has the property $f\in\mathfrak m^2$. (I've chosen this example on purpose, to show that the example given by the other answer is a particular case of the construction I proposed: just replace here $t$ by $t^2$.)
