Adjunctions capturing duality between ideals and saturated monoids in a commutative ring? Let $R$ be a commutative ring. A saturated monoid in $R$ is a multiplicative submonoid $S\subset R$ which is closed under divisors, i.e $xy\in S\implies x\in S$. This is the converse of the analogous axiom for ideals $x\in I\implies xy\in I$. So saturated monoids and ideals have a sort of duality, part of which interchanges addition and multiplication, and part of which reverses the direction of implication.
The only elements we are given suggest a pair of operations between ideals and saturated monoids
 $$\begin{gathered}\mathrm{Ideal}(R)\rightleftarrows\mathrm{SatMon}(R), \\ I\mapsto (1+I)_{\text{sat}}\text{ and } S \mapsto \left\langle S-1 \right\rangle. \end{gathered}$$
Taking the poset structures induced from the powerset of $R$, given by inclusions, both of the above operations are poset morphisms (take inclusions to inclusions). However, they are not adjoint functors:
$$\begin{gathered} \left\langle S-1 \right\rangle ⊂ I\iff S-1 \subset I\iff S\subset 1+I \\ S\supset(1+I)_\text{sat} \iff S\supset 1+I \end{gathered}$$
This is a strange situation of functors $\mathsf C\rightleftarrows \mathsf D$ say $F,G$ and yet bijections $$\mathsf D(FA,B)\cong \mathsf C(GB,A)= \mathsf C^\text{op}(A,GB).$$
Question 1. Do the operations above underlie some adjunction I'm missing?
If not, then:
Question 2. Is the formal duality between ideals and saturated monoids captured by some other adjunction?
If not, then:
Question 3. Is there more structure to the abstract setting I described above, with interesting category theory?

Added remark. Between prime ideals and prime saturated monoids (prime here means $0\notin S$ and $x+y\in S\implies x\in S\vee y\in S$), taking set-complements is actually a bijection. This perfect duality is really a consequence of the fact the added primeness assumptions make the definitions perfectly dual.
 A: I fail to see how the "strange situation" you describe actually occurs in your case: the sets $\mathrm{Ideal}(R)(\langle S-1 \rangle, I), \mathrm{SatMon}(R)((1+I)_{\mathrm{sat}}, S)$ will not be bijective for every pair $I, S$. If $I \subsetneq J$ are two ideals and $S=(1+J)_{\mathrm{sat}}$, then $\mathrm{SatMon}(R)((1+I)_{\mathrm{sat}}, S)=\{\subseteq\}$ while $\mathrm{Ideal}(R)(\langle S-1 \rangle, I)=\emptyset$ since $\langle S-1\rangle \supseteq J \supsetneq I.$
To describe what adjoint situations actually do occur here, I would suggest adding an intermediate step to make the situation a bit clearer:


*

*Consider first rather $\mathrm{Ideal}(R)\overset{F}{\underset{G}\leftrightarrows}\mathrm{Mon}(R)$ where $F: S \mapsto \langle S-1 \rangle$ and $G: I \mapsto 1+I$. Then your first line of iffs shows that $F \dashv G$.

*Now consider separately $\mathrm{Mon}(R)\overset{U}{\underset{\mathrm{sat}}\leftrightarrows}\mathrm{SatMon}(R)$ where $U$ is the forgetful functor. When you replace in your second line of iffs "$1+I$" by some general multiplicative monoid $S'$, the statement still holds. That is, given a monoid $S'$ and a saturated monoid $S,$ we have $S'_\mathrm{sat}\subseteq S$ iff $S' \subseteq S$. Thus, $\mathrm{sat}\dashv U$.
So the adjunction-like thing that you have can be described as the pair given by a composition of a left adjoint with a right adjoint, and the composition of their respective adjoints in the opposite order.
