Saturated multiplicative subset ($ab\in S⟹a,b\in S$)

A multiplicative subset is saturated if every factor of a product in the subset also lies in it. We include $1$ in a multiplicative subset. A saturated subset containing $0$ is the whole ring. The units form a saturated subset in a commutative ring.

In $\mathbb{Z}$, the saturated multiplicative subset $\{-1,1\}$ has complement equal to the union of $(p)$ over all positive primes. No finite subcollection of prime ideals covers this complement: a prime outside the finite list is a nonunit not in any listed ideal.

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.