0
$\begingroup$

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 $n > 0$.
$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.