While working through a proof of [this](https://arxiv.org/pdf/1711.04148.pdf) paper, at the beginning of page 42, the author seems to claim the following is true:



> 
> Let $R\subset S$ be rings, where $R$ is a finite type algebra over $\mathbb F\_p$. Consider the associated map of the prime spectra
>  $$
> \varphi:\text{Spec}(S)\rightarrow \text{Spec}(R).
> $$
>  Suppose that $K\subset \text{Spec}(R)$ is a constructible subset such that $\varphi^{-1}(K)=\varnothing$. Prove that there exists an $R\subset R^{'}\subset S$, such that $R'$ is a finite type $R-$algebra and that if 
>  $$
> \psi:\text{Spec}(R')\rightarrow \text{Spec}(R).
> $$
>  is the associated map of spectra, then $\psi^{-1}(K)=\varnothing$. 
> 
> 
> 


I believe that I have an argument for the case when $K$ is a finite subset. One could think of $S$ as a direct limit of its finitely generated $R$-subalgebras and therefore $Spec(S)$ should equal an inverse limit of the spectra of the finitely generated $R$-subalgebras. For each prime in $K$, choose a finitely generated $R$-subalgebra where it does not have a preimage, and the rest is clear. However, I don't know what to do for the general case. 



Preimage of a constructible set in spectrum of a subring
See for example [Tag 05F4](https://stacks.math.columbia.edu/tag/05F4). If you click through you find the following argument: write $K$ as the image of a map $\text{Spec}(A) \to \text{Spec}(R)$ for some ring map $R \to A$; this is where you use that $K$ is constructible. Then $\varphi^{-1}(K) = \emptyset$ means that $A \otimes\_R S$ is the zero ring, in other words, $1 = 0$ in $A \otimes\_R S$. Since $S = \text{colim} R'$ we find that $1 = 0$ already in some $A \otimes\_R R'$.


