Saturated complement as a union of prime ideals ($R\setminus S={\bigcup }_{P\cap S=\varnothing }P$)

The complement of a multiplicative subset is a union of prime ideals exactly when the subset is saturated. If $a\notin S$, saturation ensures $(a)\cap S=\varnothing$; extend this to a prime ideal avoiding a multiplicative subset. Conversely, a product in $S$ cannot have a factor in a prime ideal contained in the complement.