# Complement of saturated set

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?

• 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?). – Georges Elencwajg Aug 15 '11 at 21:22
• You can also note that if $a \notin S$, then $S^{-1}(a)$ is a proper ideal of $S^{-1}A$. – Dylan Moreland Aug 15 '11 at 21:32
• There isn't in general such a thing as "the" saturated subset, so rather than "its saturated subset" that should be " a saturated subset". – Arturo Magidin Aug 16 '11 at 3:26
• 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. – Artem Pelenitsyn Aug 16 '11 at 5:50
• @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". – Georges Elencwajg Aug 16 '11 at 10:33

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).