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Suppose $X\to Y$ is a quasi-finite morphism , it is closed on underlying topological spaces, is it a finite morphism? Is there a counterexample? If there is, will it be true if $X$ is a curve?

  • Often the definition of quasifinite will include a finite type hypothesis. It seems like you aren't requiring this. – Hoot Oct 24 '15 at 03:51
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    As long as you include PROPER. In fact, finite is the same thing as quasi-finite plus proper. So, for example, a map between projective objects which is quasi-finite is finite. – Alex Youcis Oct 24 '15 at 03:55
  • @Hoot Ah..you are right, I should use the definition that it is finite type and each point $x$ is open in $f^{-1}(f(x))$. –  Oct 24 '15 at 03:59

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Let $f:C\to D$ be a finite map of curves, and suppose $c_0\in C$ is a (closed) point that isn't the entire fiber over a point in $D$. Let $g$ be the restriction of $f$ to $C\backslash\{c_0\}$. Then $g$ is quasi-finite and closed, but not finite.

For comparison, if $C$ and $D$ are defined over $\mathbb{C}$, then $g$ will not be closed for the analytic topology.

Julian Rosen
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The map $\operatorname{Spec}\overline{ \Bbb{Q}} \to \operatorname{Spec} \Bbb{Q}$ is quasi-finite and closed because it is a map between one point spaces. However it is not a finite morphism.

David Benjamin Lim
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