Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $f\colon X \rightarrow Y$ be a separated morphism of finite type of schemes. Suppose $f^{-1}(y)$ is proper over $Spec(k(y))$ for every $y \in Y$, where $k(y)$ is the residue field of $y$. Is $f$ proper? If not, what conditions (in addition to the above one) should $Y$ (or $X$) satisfy to make $f$ proper?

share|improve this question
    
Deleted stupid comment. Of course generic-like points on $X$ don't have to live on a fiber. –  Matt Nov 30 '12 at 17:46
    
Dear @Matt: but yes, every point $x\in X$ (generic or not) lives in its fiber $f^{-1}(f(x))$ ! –  Georges Elencwajg Nov 30 '12 at 19:47

2 Answers 2

up vote 6 down vote accepted

No, the morphism $f$ needn't be proper.

For a counterexample, take $ X=\mathbb A^1_\mathbb C \setminus \lbrace 0\rbrace , Y=\mathbb A^1_\mathbb C$ and let $f:\mathbb A^1_\mathbb C \setminus \lbrace 0\rbrace \hookrightarrow \mathbb A^1_\mathbb C $ be the inclusion.
All fibers $f^{-1}(y)$ are proper over $Spec(k(y))$ but $f$ is not proper since it is not closed.

share|improve this answer
    
Any open immersion $f:X\to Y$ of schemes which isn't closed will do, right? –  Harry Nov 30 '12 at 7:32
    
@Harry: yes, that's right. –  Georges Elencwajg Nov 30 '12 at 8:24
1  
Dear Georges, I wonder what if $f$ is a closed morphism in the first place. –  Makoto Kato Nov 30 '12 at 19:54
    
Dear Makoto: I have no counterexample to (nor proof of) properness if you add that interesting hypothesis. –  Georges Elencwajg Nov 30 '12 at 20:16
1  
Keep the same $X, Y$, but consider $z\mapsto z^2-z$. It is surjective, quasi-finite but not finite (hence not proper). But a morphism from a curve to a curve is always closed (there are very few closed subsets). –  user18119 Apr 13 '13 at 20:42

I think, it is instructive to consider the case that $f$ has only finitely many points in each fiber, i.e. it is quasi-finite. It is not so difficult to show that every finite morphism is both proper and quasi-finite. But the converse is also true: At least if $X$ and $Y$ are noetherian, every proper quasi-finite morphism is automatically finite (Lemma 3.1.4 in Jonathan Wang's notes on the Zariski Main Theorem). Thus, in this case you would be asking whether every quasi-finite morphism is also finite. As in Georges example, open immersion are usually counter-examples to this.

The surprising thing is that these are more or less the only counter-examples. More precisely, a version of Zariski's main theorem states that under mild assumptions every quasi-finite morphism can be factored into an open embedding and a finite morphism. These mild assumptions could, e.g., be that $Y$ is quasi-compact and (quasi-)separated and $f$ is separated and of finite presentation (see Section 4 in Wang's notes).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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