The Stacks project

68.16 Finite cover by a scheme

As an application of the limit results of this chapter, we prove that given any quasi-compact and quasi-separated algebraic space $X$, there is a scheme $Y$ and a surjective, finite morphism $Y \to X$. We will rely on the already proven result that we can find a finite integral cover by a scheme, which was proved in Decent Spaces, Section 66.9.

Proposition 68.16.1. Let $S$ be a scheme. Let $X$ be a quasi-compact and quasi-separated algebraic space over $S$.

  1. There exists a surjective finite morphism $Y \to X$ of finite presentation where $Y$ is a scheme,

  2. given a surjective ├ętale morphism $U \to X$ we may choose $Y \to X$ such that for every $y \in Y$ there is an open neighbourhood $V \subset Y$ such that $V \to X$ factors through $U$.

Proof. Part (1) is the special case of (2) with $U = X$. Let $Y \to X$ be as in Decent Spaces, Lemma 66.9.2. Choose a finite affine open covering $Y = \bigcup V_ j$ such that $V_ j \to X$ factors through $U$. We can write $Y = \mathop{\mathrm{lim}}\nolimits Y_ i$ with $Y_ i \to X$ finite and of finite presentation, see Lemma 68.11.2. For large enough $i$ the algebraic space $Y_ i$ is a scheme, see Lemma 68.5.11. For large enough $i$ we can find affine opens $V_{i, j} \subset Y_ i$ whose inverse image in $Y$ recovers $V_ j$, see Lemma 68.5.7. For even larger $i$ the morphisms $V_ j \to U$ over $X$ come from morphisms $V_{i, j} \to U$ over $X$, see Proposition 68.3.8. This finishes the proof. $\square$

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