How to check if a subset is open in Zariski

Question

I'm having troubles determining if a given subset of $\operatorname{Spec}A$ is open or not. The contest is not trivial. I have to consider a morphism of finitely generated $k$-algebras $A\rightarrow B$, which are also integral domains. We assume that the map induced on the fraction fields defines a finite field extension. The task is to show that $$ U:=\{q\in \operatorname{Spec}A \mid B\otimes_A A_q\text{ is finite as $A_q$-modules}\} $$

I tried to consider $I=\displaystyle\bigcap_{q\in U^c}q$ and I claimed that $V(I)=U^c$, but i'm stack.

Can anybody give me a suggestion? No solutions, please, just a hint. Thank you!!

Answer 1

Here's a suggestion.

If you can prove that for any $q\in U$, there is some $f\in A\setminus q$ such that $B\otimes_A A_f$ is finite as an $A_f$-module, then you are done. Do you see why this would mean that you are done?

This is proving that every point in $U$ has a standard neighborhood, $q\in D(f)\subseteq U$. To see why $B\otimes_A A_f$ finite over $A_f$ implies $D(f)\subseteq U$, note that $B\otimes_A A_p = B\otimes_A A_f\otimes_{A_f} A_p$, where $p\in D(f)$, so if $B\otimes A_f$ is finite over $A_f$, then $B\otimes A_p$ is finite over $A_p$.