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?
Complement of saturated set
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commutative-algebra
maximal-and-prime-ideals
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3For $a\notin S$ 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?). – 2011-08-15
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0You can also note that if $a \notin S$, then $S^{-1}(a)$ is a proper ideal of $S^{-1}A$. – 2011-08-15
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1There isn't in general such a thing as "the" saturated subset, so rather than "its saturated subset" that should be " *a* saturated subset". – 2011-08-16
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0Thanks 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. – 2011-08-16
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1@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". – 2011-08-16
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0@Arturo: Georges is pretty right about articles. – 2011-08-20
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0@Artem: My apologies if it came across badly. – 2011-08-20
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0@Arturo That was perfectly well. This artiles' related stuff is quite interesting and I'll try to take your comment into account in the future. Thank you! – 2011-08-20
1 Answers
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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).