Showing quasicompactness is affine-local on the target.

Question

$\newcommand{\Spec}{\operatorname{Spec}}$Let $\pi:X\to Y$ be a morphism of schemes. I want to show that the property of $\pi$ being quasicompact is affine-local on the target. That is, if $\{U_i\}$ is an affine open cover of $Y$, and $\pi^{-1}(U_i)$ is quasicompact for each $i$, then $\pi$ is quasicompact.

I understand that by the Affine Communication Lemma, I just need to show

$(1)$ If $\Spec A\subset Y$ is affine open such that $\pi^{-1}(\Spec A)$ is quasicompact, then $f\in A$ implies $\pi^{-1}(\Spec A_f)$ is quasicompact.

$(2)$ If $(f_1,\dots,f_n)=A$ and $\pi^{-1}(\Spec A_{f_i})$ is quasicompact for each $i$, then $\pi^{-1}(\Spec A)$ is quasicompact.

The latter is obvious enough because $\Spec A=\cup_i\Spec A_{f_i}$, so

$$\pi^{-1}(\Spec A)=\cup_{i=1}^n\pi^{-1}(\Spec A_{f_i})$$

which is a finite union of quasicompact sets, hence quasicompact.

I'm having more trouble showing $(1)$ holds; if $f\in A$, then $\Spec A_{f_i}$ is open in $\Spec A$, hence open in $Y$, so $\pi^{-1}(\Spec A_{f_i})$ is open in $\pi^{-1}(\Spec A)$, the latter of which is quasicompact, but I don't know how to use this to show $\pi^{-1}(\Spec A_{f_i})$ is quasicompact. It would be true certainly if $\pi^{-1}(\Spec A)$ were Noetherian, but this isn't the case here.

Can anybody provide any insight as to what I'm missing to prove $(1)$?

Answer 1

The trick for your question is as follows. $\pi^{-1} \text{Spec}\,A_f$ is the complement of $Z(\pi^\sharp f)$ in $\pi^{-1} \text{Spec}\,A$. The latter might not be affine, but since it is quasicompact, it has a finite cover by affines, and taking the complement of $Z(\pi^\sharp f)$ in each of them will keep them affine.