Lemma 29.51.5. Let $X$, $Y$ be schemes. Let $f : X \to Y$ be locally of finite type. Let $X^0$, resp. $Y^0$ denote the set of generic points of irreducible components of $X$, resp. $Y$. Assume

1. $X^0$ and $Y^0$ are finite and $f^{-1}(Y^0) = X^0$,

2. either $f$ is quasi-compact or $f$ is separated.

Then there exists a dense open $V \subset Y$ such that $f^{-1}(V) \to V$ is finite.

Proof. Since $Y$ has finitely many irreducible components, we can find a dense open which is a disjoint union of its irreducible components. Thus we may assume $Y$ is irreducible affine with generic point $\eta$. Then the fibre over $\eta$ is finite as $X^0$ is finite.

Assume $f$ is separated and $Y$ irreducible affine. Choose $V \subset Y$ and $U \subset X$ as in Lemma 29.51.1 part (3). Since $f|_ U : U \to V$ is finite, we see that $U \subset f^{-1}(V)$ is closed as well as open (Lemmas 29.41.7 and 29.44.11). Thus $f^{-1}(V) = U \amalg W$ for some open subscheme $W$ of $X$. However, since $U$ contains all the generic points of $X$ we conclude that $W = \emptyset$ as desired.

Assume $f$ is quasi-compact and $Y$ irreducible affine. Then $X$ is quasi-compact, hence there exists a dense open subscheme $U \subset X$ which is separated (Properties, Lemma 28.29.3). Since the set of generic points $X^0$ is finite, we see that $X^0 \subset U$. Thus $\eta \not\in f(X \setminus U)$. Since $X \setminus U \to Y$ is quasi-compact, we conclude that there is a nonempty open $V \subset Y$ such that $f^{-1}(V) \subset U$, see Lemma 29.8.3. After replacing $X$ by $f^{-1}(V)$ and $Y$ by $V$ we reduce to the separated case which we dealt with in the preceding paragraph. $\square$

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