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Let $f\colon X \to Y$ be a finite morphism of schemes. If $U$ is a quasi-compact open subset of $X$, is $f\colon U \to Y$ a quasi-finite morphism? Obviously it has finite fibers, but I cannot see why it is a quasi-compact morphism. Should we add the assumption that the open immersion $i\colon U \to X$ is quasi-compact? Thanks!

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up vote 3 down vote accepted

A quasi-finite morphism of schemes is a morphism $f: X \to Y$ such that:

  1. $f$ is of finite type.
  2. $f$ has finite fibers.

The composite of two quasi-finite morphisms is quasi-finite (this is easy to see because each of the two above properties is preserved under composition). So if the open embedding $U \to X$ is quasi-compact, then it is of finite type, and thus quasi-finite (since the fibers are either a point or empty) and the composite $U \to X \to Y$ is thus quasi-finite.

If the inclusion $U \to X$ is not quasi-compact, then the composite need not be quasi-finite. For instance, take $f$ to be the identity map. Then the open immersion $U \to X$ itself is not quasi-finite.

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Thank you very much, your simple example is helpful. – Yubin Sep 7 '11 at 5:17

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