TITLE: A question on local rings
QUESTION [5 upvotes]: I was trying to get a counterexample of this fact: given a ring $A$, $f\in A$ and $S=\{1,f,f^2,...\}$, is $S^{-1}A$ always a local ring? Could you help me please? Thank you.

REPLY [2 votes]: In fact, in general if $S$ is a multiplicative set, then $S^{-1}A$ is local if and only if the saturation of $S$ is the complement of a prime ideal.  In general, a saturated multiplicatively closed set will always be the complement of a union of prime ideals -- but any such complement is a saturated multiplicatively closed set.  So, for a multiplicative set to give you a local ring when you invert it is a kind of maximality condition.  Multiplicative sets of the form you mention are in a sense minimal (since for any multiplicative set $S$ and any $f \in S$, all the powers of $f$ also have to be in $S$).  So one expects that such a multiplicative set will fail, in general, to yield a local ring.