Let $R$ be a noetherian regular domain. Suppose that $a, b \in R$, with $b \neq 0$, and consider the ring $S:=R[\frac{a}{b}]=R[X]/(bXa)$. Is $S$ regular? If this is not the case are there some conditions on $a$ and $b$ (or on $R$) that imply regularity? For example $a=1$ is enough.

1$\begingroup$ For example, if $R/(a,b)$ is regular, you should be okay. $\endgroup$ – Mohan May 11 '11 at 16:31

$\begingroup$ If $a=b\not\in R^{\times}$, then $S$ is not regular. $\endgroup$ – Laurent MoretBailly May 11 '11 at 17:38
If $R = k[x,y]$ and we throw in $\frac {x^2}y$, we get $k[x,y,z]/(zyx^2)$, not regular. This shows that it's not enough to assume $a,b$ form a regular sequence. The only sufficient condition I can come up with is that $a$ be outside the square of (every/the) maximal ideal

$\begingroup$ Thanks! Actually I'm interested in the case where $R= \mathbb Z_p[Y]$, with $a=Y$ and $b=p^n$ for some integer $n$. I think that in this case $S$ is always regular, do you agree with me? $\endgroup$ – Ricky May 12 '11 at 9:01

$\begingroup$ Yes, $\mathbb{Z}_p[Y,Z]/(Zp^nY)$ is regular. $\endgroup$ – Graham Leuschke May 12 '11 at 10:44