A question on local rings 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.
 A: If $A$ is an arbitrary (commutative) ring and $f=1$, then  $S=\lbrace 1\rbrace$and  the canonical morphism $A\to S^{-1}A:a\mapsto \frac {a}{1}$ is an isomorphism of rings.
So your question is equivalent to "Is every ring $A$ local ?"  What do you think?
A: 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.
A: Try inverting $n$ in $\mathbb{Z}$. Then the primes of this are literally the primes of $\mathbb{Z}$ coprime to $n$. 
A: Just elaborating a little on uncookedfalcon's answer, it is a standard result that if $S \subset A$ is a multiplicative subset, then the prime ideals of $S^{-1}A$ are precisely those of the form $pS^{-1}A$, where $p$ is a prime ideal of $A$ such that $p \cap S = \emptyset$. Now try to deduce the example that uncookedalcon gave from this fact.
Also note that since your $f$ is not required to be non-zero or non-nilpotent, you can just localize at $0$ (or a nilpotent element), giving you the zero ring which has no prime ideals, hence is not local.
A: For $F$ a field, if you localize at $\{1,x,x^2,\dots\}$ for the ring $F[x]$, you get the domain of Laurent polynomials inside of the field $F(x)$.
This new ring isn't local because its units look like $fx^i$ for integers $i$ and $f\in F$, and so the nonunits are not closed under addition. (For example, $(x+2)-(x+1)$ is a unit.)
