Etale map has image whose complement is the vanishing locus of a finitely generated ideal

Question

While working through a proof of this 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. 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 $A$?

Answer 1

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 $I = (f_1, \ldots, f_n)$.