The Stacks project

Lemma 76.36.9. Let $S$ be a scheme. Let $f : X \to Y$ be a proper morphism of algebraic spaces over $S$. Let $X \to Y' \to Y$ be the Stein factorization of $f$ (Theorem 76.36.5). If $f$ is of finite presentation, flat, with geometrically reduced fibres (Definition 76.29.2), then $Y' \to Y$ is finite étale.

Proof. Formation of the Stein factorization commutes with flat base change, see Lemma 76.36.1. Thus we may work étale locally on $Y$ and we may assume $Y$ is an affine scheme. Then $Y'$ is an affine scheme and $Y' \to Y$ is integral.

Let $y \in Y$. Set $n$ be the number of connected components of the geometric fibre $X_{\overline{y}}$. Note that $n < \infty $ as the geometric fibre of $X \to Y$ at $y$ is a proper algebraic space over a field, hence Noetherian, hence has a finite number of connected components. By Lemma 76.36.2 there are finitely many points $y'_1, \ldots , y'_ m \in Y'$ lying over $y$ and for each $i$ we can pick a finite type point $x_ i \in |X_ y|$ mapping to $y'_ i$ the extension $\kappa (y'_ i)/\kappa (y)$ is finite. Thus More on Morphisms, Lemma 37.42.1 tells us that after replacing $Y$ by an étale neighbourhood of $y$ we may assume $Y' = V_1 \amalg \ldots \amalg V_ m$ as a scheme with $y'_ i \in V_ i$ and $\kappa (y'_ i)/\kappa (y)$ purely inseparable. In this case the algebraic spaces $X_{y_ i'}$ are geometrically connected over $\kappa (y)$, hence $m = n$. The algebraic spaces $X_ i = (f')^{-1}(V_ i)$, $i = 1, \ldots , n$ are proper, flat, of finite presentation, with geometrically reduced fibres over $Y$. It suffices to prove the lemma for each of the morphisms $X_ i \to Y$. This reduces us to the case where $X_{\overline{y}}$ is connected.

Assume that $X_{\overline{y}}$ is connected. By Lemma 76.36.8 we see that $X \to Y$ has geometrically connected fibres in a neighbourhood of $y$. Thus we may assume the fibres of $X \to Y$ are geometrically connected. Then $f_*\mathcal{O}_ X = \mathcal{O}_ Y$ by Derived Categories of Spaces, Lemma 75.26.8 which finishes the proof. $\square$


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