Pre-order set homomorphism from R to Closed set in Spec(R)/from I(R) to open set in Spec(R)

Let $R$$R$ be a commutative ring.

Prop 1

$$\begin{array}{}\text{(1)}& a|b⟹(a)\supseteq (b)\end{array}$$

Proof. Follows from $b=ar$$b=ar$ directly.

The closed sets in Zariski topology of a commutative ring map each ideal $I$$I$ to the upset of prime ideal.

i.e.

$$\begin{array}{}\text{(2)}& \mathcal{Z}[I]:=\{\mathcal{p}\in \mathrm{Spec}(R)|I\subseteq \mathcal{p}\}\end{array}$$

Prop 2

$$\begin{array}{}\text{(3)}& (a)\supseteq (b)⟹\mathcal{Z}[(a)]\subseteq \mathcal{Z}[(b)]\end{array}$$

Proof. By definition of $\mathcal{Z}[I]$$\mathcal Z[I]$.

Hence

$$\begin{array}{}\text{(4)}& a|b⟹\mathcal{Z}[(a)]\subseteq \mathcal{Z}[(b)]\end{array}$$

It give a pre-order set homomorphism, or a functor.

Define $\mathcal{O}[I]:=\mathcal{Z}[I{]}^c$$\mathcal O[I]:=\mathcal{Z}[I]^c$ be the open set in the Zariski Topology, then

$$\begin{array}{}\text{(5)}& \mathcal{O}[I]\supseteq \mathcal{O}[J]⟺\mathcal{Z}[I]\subseteq \mathcal{Z}[J]\end{array}$$

Hence

$$\begin{array}{}\text{(6)}& I\subseteq J⟹\mathcal{O}[I]\subseteq \mathcal{O}[J]\end{array}$$

 

Posetalization Functor

Let $(P,⪯)$$(P,\preceq)$ be a pre-order set, $\pi (a)=\pi (b)⟺a⪯b\wedge b⪯a$$\pi(a)=\pi(b)\iff a\preceq b\wedge b\preceq a$.

Then $\pi (a)\le b⟺a⪯{\pi }^{-1}(b)$$\pi(a)\le b\iff a\preceq \pi^{-1}(b)$, hence it is an adjoint functor.

Moreover, let $F$$F$ be the posetalization functor and i be the inclusion functor. Then we have the following adjoint.

$$\begin{array}{}\text{(1)}& {\mathrm{Hom}}_{\mathrm{Pos}}(F(P),Q)\cong {\mathrm{Hom}}_{\mathrm{Pre}}(P,i(Q))\end{array}$$

The proof follows from the commutative diagram directly. Since $Q$$Q$ is a partial order set, so any pre-order homomorphism must factor through $\pi$$\pi$.

i.e. $a\le b\wedge b\le a\Rightarrow f(a)\le f(b)\wedge f(b)\le f(a)\Rightarrow f(a)=f(b)$$a\le b\wedge b\le a\Rightarrow f(a)\le f(b)\wedge f(b)\le f(a)\Rightarrow f(a)=f(b)$