Lemma 37.29.4. Let f : X \to Y, s : Y \to X be as in Situation 37.29.1. If f is of finite presentation then X^0 is locally constructible in X.
Proof. Let x \in X. We have to show that there exists an open neighbourhood U of x such that X^0 \cap U is constructible in U. This reduces us to the case where Y is affine. Write Y = \mathop{\mathrm{Spec}}(A) and A = \mathop{\mathrm{colim}}\nolimits A_ i as a directed limit of finite type \mathbf{Z}-algebras. By Limits, Lemma 32.10.1 we can find an i and a morphism f_ i : X_ i \to \mathop{\mathrm{Spec}}(A_ i) of finite presentation, endowed with a section s_ i : \mathop{\mathrm{Spec}}(A_ i) \to X_ i whose base change to Y recovers f and the section s. By Lemma 37.29.2 it suffices to prove the lemma for f_ i, s_ i. Thus we reduce to the case where Y is the spectrum of a Noetherian ring.
Assume Y is a Noetherian affine scheme. Since f is of finite presentation, i.e., of finite type, we see that X is a Noetherian scheme too, see Morphisms, Lemma 29.15.6. In order to prove the lemma in this case it suffices to show that for every irreducible closed subset Z \subset X the intersection Z \cap X^0 either contains a nonempty open of Z or is not dense in Z, see Topology, Lemma 5.16.3. Let x \in Z be the generic point, and let y = f(x). By Lemma 37.29.3 there exists a nonempty open subset V \subset \overline{\{ y\} } such that X^0 \cap X_ V is open and closed in X_ V. Since f(Z) \subset \overline{\{ y\} } and f(x) = y \in V we see that W = f^{-1}(V) \cap Z is a nonempty open subset of Z. It follows that X^0 \cap W is open and closed in W. Since W is irreducible we see that X^0 \cap W is either empty or equal to W. This proves the lemma. \square
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