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?
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).