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Let $f : X\to Y$ be a dominant morphism of integral algebraic varieties over $\mathbb C$. Suppose $[K(X): K(Y)]=n$. Then there exists a dense open subset $U$ of $Y$ such that $f^{-1}(y)$ consists in $n$ points for all $y\in U$.

This is a property I quoted from answer Number of points in the fibre and the degree of field extension. To prove the property, the author reduced it to the affine case, i.e. $X=\mathrm{Spec}\,B,Y=\mathrm{Spec}\,A$ for some algebras $A,B$ over $\mathbb{C}$. Why is it possible? In my view, the author choosed some open affine subsets $V$ of $Y$, but how to make sure $f^{-1}(V)$ is also affine?

user498029
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  • You're getting ahead of yourself - you don't need to have $f^{-1}(V)$ affine when you make the reduction. – KReiser Jun 07 '21 at 00:24
  • @KReiser Do you mean we can choose some affine open $U$ contained in $f^{-1}(V)$? How to ensure the fiber is unchanged when restricts $f$ to $U\to V$? – user498029 Jun 07 '21 at 00:33
  • Yes. Look at the closed subset $f^{-1}(V)\setminus U$: its image under $f$ is contained in a lower-dimensional subvariety of $V$, so you can just throw that out to see that on a dense open subset of $V$, we have the same fibers. – KReiser Jun 07 '21 at 01:05
  • @KReiser Thank you for your comment. I see how it works. – user498029 Jun 07 '21 at 01:16

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Let me compile my comments in to an answer. One doesn't need to have $f^{-1}(V)$ affine in this reduction: the difference $f^{-1}(V)\setminus U$ is a proper closed subvariety, so its image under $f$ will be contained in a lower-dimensional subvariety of $V$ and therefore on a dense open subset of $V$ we have the same fibers.

KReiser
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