TITLE: When is a local, reduced, (commutative) ring an integral domain?
QUESTION [12 upvotes]: 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

REPLY [4 votes]: There is not a whole lot to say here. You are given that $\{0\}$ is semiprime, and you are asking when it is prime. You could say that $R$ is a domain iff $\{0\}$ is primary. fpqc's example demonstrates this nicely since $(xy)$ is a semiprime but not prime ideal of $k[x,y]$ and also not in the localization at $(x)$ or $(y)$.
Locality never really comes into play. 
The reason you did not find any counterexamples using $\Bbb Z/(n)$ is that this ring is only reduced when $n$ is squarefree, and for each such $n$ the ring is von Neumann regular, and the localizations at prime ideals are all fields.