Let $A$ be an integral domain and $S \subset A$ a saturated subset (multiplicative subset s.t. if $ab \in S$ then $a \in S$ and $b \in S$). Would you please supply a hint, how to prove that $A \setminus S$ is a union of prime ideals?

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    $\begingroup$ For $a\notin S$ the ideal $aA$ is disjoint from $S$. Take an ideal that is maximal among those containinig $aA$ and disjoint from $S$ (why does such a maximal ideal exist? what sort of ideal is it?). $\endgroup$ – Georges Elencwajg Aug 15 '11 at 21:22
  • $\begingroup$ You can also note that if $a \notin S$, then $S^{-1}(a)$ is a proper ideal of $S^{-1}A$. $\endgroup$ – Dylan Moreland Aug 15 '11 at 21:32
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    $\begingroup$ There isn't in general such a thing as "the" saturated subset, so rather than "its saturated subset" that should be " a saturated subset". $\endgroup$ – Arturo Magidin Aug 16 '11 at 3:26
  • $\begingroup$ Thanks to all. Solved this (as far as I understand). @Georges I guess it would be better to trnslate your comment to answer. It's exactly what I wanted. $\endgroup$ – Artem Pelenitsyn Aug 16 '11 at 5:50
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    $\begingroup$ @Arturo: you are right, of course, but I'm pretty sure that Artem expressed himself the way he did, not because of a mathematical misconception, but because in his native Russian there are no articles.He wrote "its saturated subset" and very probably meant "a saturated subset of it". $\endgroup$ – Georges Elencwajg Aug 16 '11 at 10:33

At Artem's request I'm transforming my comment into an answer.

Take any $a\notin S$. The principal ideal $aA$ it generates is disjoint from $S$, that is $aA\cap S=\emptyset$ (why?). Take a maximal element $I$ in the set of ideals of $A$ containing $aA$ and disjoint from $S$ (why does such a maximal ideal exist?). This ideal has a property that allows you to conclude (in case of difficulty, look for inspiration at the proof of Proposition 1.8 page 5 in Atiyah-Macdonald's Introduction to Commutative Algebra).


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