If $f:X\to Y$ is an affine morphism of schemes, say with $Y$ irreducible, that is quasi-finite – all of the fibers, including the generic fiber, are finite – is it true that $f$ is finite?
If not, what is a good counterexample.
EDIT: I also am requiring that $f$ be closed.
Best Answer
Let $U = \mathbb{P}^1 - \{1\}$ and let $f : U \to \mathbb{P}^1$ be the squaring map $z\mapsto z^2$. Then $f$ is a counterexample. (Note that it is still surjective.)
The point is that "closed" is not sufficient. The property you want is "universally closed", that is, remains closed after any base change.
It is true that quasi-finite + affine + universally closed $\Rightarrow$ finite. (The precise statement is that finite = quasi-finite + proper.)