Question
I am wondering whether or not it is true that if $A$ is a reduced ring, then
is it the case that the localization of $A$ at any of its prime ideals is an integral domain?
Discussion
Recall that $A$ is reduced if it contains no nonzero nilpotents, i.e. $Nil(A)=\{0\}$.
I have already shown the following two facts:
- A ring $A$ being reduced is a local property, i.e. $A$ is reduced if and only if the localization of $A$ at any prime ideal is reduced.
- $A$ reduced does not imply $A$ is an integral domain (in general).
The counter example I used to prove $2$ was $\mathbb{Z}/6\mathbb{Z}$. Unfortunately, in this case all the localizations at prime ideals are integral domains, and they seem to be in every example I can think of. More generally, one could ask the question (since localizations are local rings), when is a reduced local ring an integral domain?
If anyone had a good idea for a counterexample to the original question (or a proof if it is true-although I doubt this since being an integral domain seems much stronger than being reduced), that would be much appreciated.
This is one of 12 parts to a question which I had on my midterm a few weeks ago, and the only part I have not figured out of that question
Best Answer
This is not true. The ring $k[x,y]/(xy)$ is reduced but the localization at $(x,y)$ is not an integral domain. For instance, because $xy = 0$ in the localization but $x \neq 0$ and $y \neq 0$.