Every ideal is contained in a prime ideal that is disjoint from a given multiplicative set Let $R$ be a ring $I\subset R$ an ideal and $S\subset R$ be a set for which holds: 
$1)$ $1\in S$
2) $a,b \in S\Rightarrow a\cdot b\in S$
Show that there exists a prime ideal $P$ in $R$ containing $I$ with $P\cap S =\emptyset$
Any ideas?

It's a standard argument in commutative algebra. Consider the set of all ideals that contain $I$ and avoid $S$, ordered by inclusion. By Zorn's lemma there are maximal elements (since unions are upper bounds) and you can prove that an ideal maximal via inclusion among those avoiding $S$ and containing $I$ turns out to be prime.
