# When is a map of local rings finite?

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.
• 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 $(1,y)$. – Ashvin Swaminathan Jan 4 at 15:39
• 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 $t=1$. – Mohan Jan 4 at 17:04
• 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? – Ashvin Swaminathan Jan 4 at 17:53
• 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. – Mohan Jan 4 at 17:59