Let $A,B$ be Noetherian local rings, and let $A \to B$ be a ring homomorphism such that the induced map $\operatorname{Spec} B \to \operatorname{Spec} A$ is surjective and quasifinite (of finite type and has finite fibers). Is it necessarily true that $B$ is finite over $A$?
What I know so far:
- The usual counterexample to finiteness of quasifinite morphisms is given by taking $A = k[[x]]$, $B = k((x))$, and $A \to B$ to be the obvious inclusion map, but this counterexample fails to satisfy the surjectivity requirement.
- One approach is to try to show that the map $\operatorname{Spec} B \to \operatorname{Spec} A$ is universally closed, because universally closed affine maps are integral. By https://stacks.math.columbia.edu/tag/0205, it suffices to show that the pulled-back map $\mathbb{A}_B^n \to \mathbb{A}_A^n$ is closed for every $n$. The surjectivity assumption implies that this is true for $n = 0$, but I'm not sure how to prove closedness for $n > 0$.
Consider $k[t]\subset k[u]$, $t\mapsto u^2$. Then, $t-1\mapsto u^2-1=(u-1)(u+1)$. Let $A$ be the localization of $k[t]$ at the maximal ideal $(t-1)$ and $B$ be the localization of $k[u]$ at the maximal ideal $(u-1)$. Then, $A\to B$ satisfies all your requirements, but not finite.