While working through a proof of [this](https://arxiv.org/pdf/1711.04148.pdf) paper, at the end of page 46, the author seems to claim along the lines that the following is true:



> 
> Let $A\rightarrow B$ be an [etale map of rings](https://stacks.math.columbia.edu/tag/00U0). Then the underlying map
>  $$
> \text{Spec}(B)\rightarrow \text{Spec}(A)
> $$
>  is open and the complement of the image is the vanishing locus of a *finitely* generated ideal in $A$.
> 
> 
> 


The fact that the underlying map is open is well-known. Why does the part about the finite generation hold with no Noetherianity assumptions on $A$?



Etale map has image whose complement is the vanishing locus of a finitely generated ideal
Let $U$ be the image of $\operatorname{Spec} B$ in $\operatorname{Spec} A$. Since the map $\operatorname{Spec} B \to \operatorname{Spec} A$ is open, $U$ is open. And since $\operatorname{Spec} B$ is quasi-compact, $U$ is quasi-compact. Therefore, $U$ is the union of finitely many basic opens in $\operatorname{Spec} A$, i.e. there exist $f\_1, \ldots, f\_n \in A$ such that
$$ U = \bigcup\_{i = 1}^n\operatorname{Spec} A[{1}/{f\_i}]$$
as subsets of $\operatorname{Spec} A$. Then the complement of $U$ in $\operatorname{Spec} A$ is exactly the vanishing locus of the finitely generated ideal $I = (f\_1, \ldots, f\_n)$. 


