Saturated complement as a union of prime ideals
= Saturated complement as a union of prime ideals
{title2=$R\setminus S=\bigcup_{P\cap S=\varnothing}P$}
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.