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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.

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A counterexample is $A=0$. – Martin Brandenburg Jan 4 '13 at 22:36
up vote 2 down vote accepted

Try inverting $n$ in $\mathbb{Z}$. Then the primes of this are literally the primes of $\mathbb{Z}$ coprime to $n$.

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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?

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+1 This answer is hilarious!! – user38268 Jan 17 '13 at 0:26

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.

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Thank you very much, you have been very kind, clear and useful. – Corra Jan 4 '13 at 20:10

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.

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Welcome to Math.SE! This is a nice way to contrast the assumption and purported conclusion in the question. – user53153 Jan 5 '13 at 8:14

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.)

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