Lemma 37.25.5. Let $f : X \to Y$ be a morphism which is quasi-compact and locally of finite presentation. Then the set

is locally constructible in $Y$.

Lemma 37.25.5. Let $f : X \to Y$ be a morphism which is quasi-compact and locally of finite presentation. Then the set

\[ E = \{ y \in Y \mid X_ y\text{ is geometrically reduced}\} \]

is locally constructible in $Y$.

**Proof.**
Let $y \in Y$. We have to show that there exists an open neighbourhood $V$ of $y$ in $Y$ such that $E \cap V$ is constructible in $V$. Thus we may assume that $Y$ is affine. Then $X$ is quasi-compact. Choose a finite affine open covering $X = U_1 \cup \ldots \cup U_ n$. Then the fibres of $U_ i \to Y$ at $y$ form an affine open covering of the fibre of $X \to Y$ at $y$. Hence we may assume $X$ is affine as well. Write $Y = \mathop{\mathrm{Spec}}(A)$. Write $A = \mathop{\mathrm{colim}}\nolimits A_ i$ 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 whose base change to $Y$ recovers $f$. By Lemma 37.25.2 it suffices to prove the lemma for $f_ i$. Thus we reduce to the case where $Y$ is the spectrum of a Noetherian ring.

We will use the criterion of Topology, Lemma 5.16.3 to prove that $E$ is constructible in case $Y$ is a Noetherian scheme. To see this let $Z \subset Y$ be an irreducible closed subscheme. We have to show that $E \cap Z$ either contains a nonempty open subset or is not dense in $Z$. If $X_\xi $ is geometrically reduced, then Lemma 37.25.4 (applied to the morphism $X_ Z \to Z$) implies that all fibres $X_ y$ are geometrically reduced for a nonempty open $V \subset Z$. If $X_\xi $ is not geometrically reduced, then Lemma 37.25.3 (applied to the morphism $X_ Z \to Z$) implies that all fibres $X_ y$ are geometrically reduced for a nonempty open $V \subset Z$. Thus we win. $\square$

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