Let $f:X \rightarrow Y$ be a morphism of algebraic varieties over an algebraically closed field. If all fibers $f^{-1}(y)$ with $y$ closed point in $Y$ are finite, can one conclude that an arbitrary fiber (i.e. with $y$ not necessarily closed point) is finite?

Edit: By a fiber being finite I just mean it to consist of a finite number of points.

  • $\begingroup$ You should first think what you mean by a fiber being finite if it is not closed; I can see no obvious meaning. Think of $X=Y$ the affine line over your field and $f$ the identity; would you call the fiber over the generic point of $Y$ finite? $\endgroup$ May 17, 2012 at 9:14
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    $\begingroup$ Dear @Marc, a fiber is finite if it contains a finite number of points: it is the naïve set-theoretic meaning. So in your example, yes, the fiber over the generic point is finite. And Cyril's question has an affirmative answer: see my post below. $\endgroup$ May 17, 2012 at 9:23
  • $\begingroup$ @GeorgesElencwajg: Thanks for brushing up my algebraic geometry a bit. $\endgroup$ May 17, 2012 at 9:36
  • $\begingroup$ Dear @Marc: you are welcome. Actually it has happened to me too that, since scheme theory is so sophisticated, I couldn't believe that some concept or result was just what one naïvely would have guessed! $\endgroup$ May 17, 2012 at 10:15
  • $\begingroup$ sorry for not immediately clarifying that point... $\endgroup$
    – Cyril
    May 17, 2012 at 14:37

1 Answer 1


Yes, all fibers of $f$ are finite ($f$ is quasi-finite).

Let $Z\subset Y$ the set of points with finite fibers.The key point is that $Z$ is also the set of points where $\dim f^{-1}(y)=0$.
Thus $Z$ is a constructible set and since it contains all the closed points of $Y$, it is equal to $Y$.
You will find details in the book by Görtz-Wedhorn, Remark 12.16


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