Is $D^{-1}R$ regular?

Question

If $D$ is a multiplicative subset of a regular ring $R$, then is $D^{-1}R$ also regular?

I've been trying to think of an easy counterexample, but I'm stuck on figuring out how to either think of a counterexample or confirm my conjecture.

Answer 1

Recall prime ideals of $D^{-1}R$ correspond to prime ideals of $R$ which do not intersect $D$. If $\mathfrak p$ is such a prime ideal of $R$, corresponding to the prime ideal $D^{-1}\mathfrak p\subseteq D^{-1}R$, then it's a straightforward check that $(D^{-1}R)_{D^{-1}\mathfrak p}\cong R_\mathfrak p$.

To show $D^{-1}R$ is noetherian, you need to take an ideal, lift it to an ideal of $R$, take a finite set of generators there, and prove that the images of these generators under $R\to D^{-1}R$ generate the original ideal.