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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.

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    sorry for not immediately clarifying that point...2012-05-17

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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