TITLE: Converse of "localization at a prime is local"
QUESTION [5 upvotes]: Suppose $S^{-1}R$ is the localization of a ring R at a multiplicative subset S, and is local. Must S be the complement of a prime ideal?

REPLY [4 votes]: Hint: try $R=\Bbb R[[x]]$, and $S=\{1\}$ and you will find your answer.
Complements of prime ideals necessarily have the property of being "saturated" in the sense that divisors of elements in the complement are in the complement.

Much later
Also, you could even do this with $R=F_3$, the field of three elements. The above example is better if you think it's too restrictive.