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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?

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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
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.

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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
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
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).

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