Let $A$ be a commutative ring and $f\in A,I\vartriangleleft A$. We have obvious submonoids $f^\mathbb{N},1+I\subset A$.
- The submonoid $f^\mathbb{N}\subset A$ is saturated, and its complement is the union of primes not containing $f$.
- The complement of the saturation of the submonoid $1+I\subset A$ is the union of maximal ideals containing $I$.
Question. Which primes of $A$ (do not) intersect the submonoid $f^\mathbb{N}(1+fA)$? (here $fA=(f)\vartriangleleft A$.)
Well, a prime $P$ intersects $f^\mathbb{N}(1+fA)$ iff it intersects either $f^\mathbb{N}$ or $1+fA$. This happens iff either $f\in P$ or $f$ is a unit mod $P$.
So, $P$ does not intersect $f^\mathbb{N}(1+fA)$ iff $f$ is nonzero and not a unit mod $P$. Equivalently, this means that $P$ can be extended to both prime ideals that contain $f$ and prime ideals that don't. Algebro-geometrically, if $\operatorname{Spec}A$ is a variety, then such primes $P$ are the generic points of subvarieties which intersect but are not contained in the vanishing set of $f$.