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

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