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