Let $X$ and $Y$ be projective varieties over a field and let $f:U\to V$ be a proper dominant separable map where $U\subset X$ and $V\subset Y$ are open subsets.
Question: Why is it true that $f$ is a finite map?
A proper dominant separable map.
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algebraic-geometry
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0Who told you so? There is no reason for $f$ to be finite with these assumptions. Maybe your opens are affine opens? – 2017-02-21
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0Can you give a precise reference (page number, line number) in the paper? There is obviously some context missing. (It doesn't matter if the open sets are affine or not, since as the question is written the dimensions need not even match.) – 2017-02-21
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0@Nefertiti I do not get your last point. Affine proper morphisms are finite: http://mathoverflow.net/questions/125740/why-is-a-proper-affine-morphism-finite If $U$ and $V$ would be affine opens with a proper morphism between them, then this would be automatically finite and we would get $\dim X = \dim U = \dim V = \dim Y$ (assuming the varieties are irreducible) a posteriori. Of course this would not necessarily mean that $f$ extends to a finite morphism $X \to Y$. – 2017-02-22
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0@MooS: right, sorry. In any case the OP gave (but then deleted) a reference for where this came from. The original paper did not make the claimed assertion. – 2017-02-22
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As MooS indicated this is obviously false : take for $f$ the unique morphism $$U=\mathbb P^1_\mathbb C\to V=\operatorname {Proj}\mathbb C[T]=\{\ast\}$$