The Stacks project

Lemma 68.10.3. Let $S$ be a scheme. Let $X \to Y$ be a surjective finite locally free morphism of algebraic spaces over $S$. For $y \in |Y|$ the following are equivalent

  1. $y$ is in the schematic locus of $Y$, and

  2. there exists an affine open $U \subset X$ containing the preimage of $y$.

Proof. If $y \in Y$ is in the schematic locus, then it has an affine open neighbourhood $V \subset Y$ and the inverse image $U$ of $V$ in $X$ is an open finite over $V$, hence affine. Thus (1) implies (2).

Conversely, assume that $U \subset X$ as in (2) is given. Set $R = X \times _ Y X$ and denote the projections $s, t : R \to X$. Consider $Z = R \setminus s^{-1}(U) \cap t^{-1}(U)$. This is a closed subset of $R$. The image $t(Z)$ is a closed subset of $X$ which can loosely be described as the set of points of $X$ which are $R$-equivalent to a point of $X \setminus U$. Hence $U' = X \setminus t(Z)$ is an $R$-invariant, open subspace of $X$ contained in $U$ which contains the fibre of $X \to Y$ over $y$. Since $X \to Y$ is open (Morphisms of Spaces, Lemma 67.30.6) the image of $U'$ is an open subspace $V' \subset Y$. Since $U'$ is $R$-invariant and $R = X \times _ Y X$, we see that $U'$ is the inverse image of $V'$ (use Properties of Spaces, Lemma 66.4.3). After replacing $Y$ by $V'$ and $X$ by $U'$ we see that we may assume $X$ is a scheme isomorphic to an open subscheme of an affine scheme.

Assume $X$ is a scheme isomorphic to an open subscheme of an affine scheme. In this case the fppf quotient sheaf $X/R$ is a scheme, see Properties of Spaces, Proposition 66.14.1. Since $Y$ is a sheaf in the fppf topology, obtain a canonical map $X/R \to Y$ factoring $X \to Y$. Since $X \to Y$ is surjective finite locally free, it is surjective as a map of sheaves (Spaces, Lemma 65.5.9). We conclude that $X/R \to Y$ is surjective as a map of sheaves. On the other hand, since $R = X \times _ Y X$ as sheaves we conclude that $X/R \to Y$ is injective as a map of sheaves. Hence $X/R \to Y$ is an isomorphism and we see that $Y$ is representable. $\square$

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