Lemma 114.11.5. Let $f : X \to Y$ be a morphism schemes. Assume

1. $X$ and $Y$ are integral schemes,

2. $f$ is locally of finite type and dominant,

3. $f$ is either quasi-compact or separated,

4. $f$ is generically finite, i.e., one of (1) – (5) of Morphisms, Lemma 29.51.7 holds.

Then there is a nonempty open $V \subset Y$ such that $f^{-1}(V) \to V$ is finite locally free of degree $\deg (X/Y)$. In particular, the degrees of the fibres of $f^{-1}(V) \to V$ are bounded by $\deg (X/Y)$.

Proof. We may choose $V$ such that $f^{-1}(V) \to V$ is finite. Then we may shrink $V$ and assume that $f^{-1}(V) \to V$ is flat and of finite presentation by generic flatness (Morphisms, Proposition 29.27.1). Then the morphism is finite locally free by Morphisms, Lemma 29.48.2. Since $V$ is irreducible the morphism has a fixed degree. The final statement follows from this and Morphisms, Lemma 29.56.3. $\square$

Comment #5549 by Hao on

Form the proof of 0CC3, I guess we don't need $f$ to be dominant when considering (3).

Comment #5551 by Laurent Moret-Bailly on

Condition (3) refers to Lemma 02NX, but there $X$ and $Y$ are assumed integral.

Comment #5734 by on

Yes, thanks to both of you! The condition that $X$ and $Y$ should be integral and that $f$ is dominant were missing. Oops! This lemma is kind of obsolete and I have moved it to the obsoiete chapter; this doesn't mean it is wrong (I mean it was but now it is fixed). Feel free to tell me if it should be moved back to this chapter. See this commit.

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