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.

  • $\begingroup$ Thanks for the helpful response, but I did your example and got that it was finite. Your claim seems to be that, for example, the projection of a parabola $x=y^2-y$ onto the $x$-axis fails to be finite if one localizes the parabola at the origin $x=y=0$. But clearly $k[[x,y]]/(x-y^2+y)$ is finite over $k[[x]]$ with basis $(1,y)$. $\endgroup$ – Ashvin Swaminathan Jan 4 at 15:39
  • $\begingroup$ No, it is not finite. The integral closure of $A$ in the fraction field of $B$ is semilocal and not local, since there are two points above $t=1$. $\endgroup$ – Mohan Jan 4 at 17:04
  • $\begingroup$ Thanks, just to confirm, are you saying that in my comment, $k[[x,y]]/(x-y^2+y)$ is not finite over $k[[x]]$, or are you saying my comment is actually different from your counterexample? $\endgroup$ – Ashvin Swaminathan Jan 4 at 17:53
  • 1
    $\begingroup$ No, your case is different from mine. First, an application of Weirstrass preparation theorem will tell you that if $A$ is complete (for example $A=k[[x]]$), then quasifinite is finite. If you took your example over $k[x]$ localized at say $(x-1)$ (not powerseries) and localize $k[x,y]/(x-y^2+y)$ at one of the roots of $y^2-y-1$, you will have a quasifinite non-finite extension as you want. $\endgroup$ – Mohan Jan 4 at 17:59

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