Let $f:X \to Y$ be a finite surjective morphism of quasiprojective schemes over $\mathbb{C}$, $X$ is reduced and $Y$ is integral. Suppose that there exists an integer $n$ such that for every closed point $y \in Y$, the fiber $f^{1}(y)$ is reduced and consists of $n$ distinct closed points. Is it true that $f$ is flat?
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$\begingroup$ I suspect there is a ground field. Is it assumed algebraically closed? $\endgroup$– Laurent MoretBaillyMar 13, 2015 at 17:41

$\begingroup$ Yes. You can also assume characteristic zero. $\endgroup$– user46578Mar 13, 2015 at 17:55

2$\begingroup$ The normalization of a cuspidal curve is a bijective finite map which is not flat. $\endgroup$– Piotr AchingerMar 13, 2015 at 18:14

1$\begingroup$ Hartshorne also assumes $Y$ is regular. $\endgroup$– SashaMar 13, 2015 at 19:08

2$\begingroup$ The normalization of the cuspidal curve is actually bijective (in fact, it is a homeomorphism), but the fibre over the cusp is not reduced. $\endgroup$– Francesco PolizziMar 13, 2015 at 23:36

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1 Answer
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Apply Hartshorne, Exercise II.5.8 to $f_*\mathcal{O}_X$.

$\begingroup$ @Starr: Thank you very much for the answer. I have a small question. In this exercise Hartshorne considers dimesion of pullback of coherent sheaves to points on $Y$. Are these points necessarily closed? As one can see, for example in the proof of III. Theorem 9.9 of Hartshorne, we see that he does not necessarily restrict to closed points, when he says points. $\endgroup$ Mar 15, 2015 at 6:57

1$\begingroup$ @user46578. You should read the rest of the section, or maybe just the rest of the exercise. The rank function is upper semicontinuous. So if you know the rank is constant at all closed points, then you know the rank is constant at all points. $\endgroup$ Mar 15, 2015 at 8:58