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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?

LCL
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  • Consider the max ideal $M$ in $S^{-1}R$. The "numerators" from $M$ form an ideal $I$ of $R$. What might maximality (and hence primality) of $M$ say about $I$? – walkar Aug 25 '15 at 16:43

1 Answers1

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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.

rschwieb
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    Conversely, it is easy to see that the elements of $R$ that become units in $S^{-1}R$ are exactly the divisors of elements of $S$. So if $S$ is saturated and $S^{-1}R$ is local, then $S$ is the complement of a prime ideal. – Eric Wofsey Aug 25 '15 at 17:16