The Stacks project

66.9 Integral cover by a scheme

Here we prove that given any quasi-compact and quasi-separated algebraic space $X$, there is a scheme $Y$ and a surjective, integral morphism $Y \to X$. After we develop some theory about limits of algebraic spaces, we will prove that one can do this with a finite morphism, see Limits of Spaces, Section 68.16.

Lemma 66.9.1. Let $S$ be a scheme. Let $j : V \to Y$ be a quasi-compact open immersion of algebraic spaces over $S$. Let $\pi : Z \to V$ be an integral morphism. Then there exists an integral morphism $\nu : Y' \to Y$ such that $Z$ is $V$-isomorphic to the inverse image of $V$ in $Y'$.

Proof. Since both $j$ and $\pi $ are quasi-compact and separated, so is $j \circ \pi $. Let $\nu : Y' \to Y$ be the normalization of $Y$ in $Z$, see Morphisms of Spaces, Section 65.48. Of course $\nu $ is integral, see Morphisms of Spaces, Lemma 65.48.5. The final statement follows formally from Morphisms of Spaces, Lemmas 65.48.4 and 65.48.10. $\square$

Lemma 66.9.2. Let $S$ be a scheme. Let $X$ be a quasi-compact and quasi-separated algebraic space over $S$.

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

  2. given a surjective étale 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 part (2) where $U = X$. Choose a surjective étale morphism $U' \to U$ where $U'$ is a scheme. It is clear that we may replace $U$ by $U'$ and hence we may assume $U$ is a scheme. Since $X$ is quasi-compact, there exist finitely many affine opens $U_ i \subset U$ such that $U' = \coprod U_ i \to X$ is surjective. After replacing $U$ by $U'$ again, we see that we may assume $U$ is affine. Since $X$ is quasi-separated, hence reasonable, there exists an integer $d$ bounding the degree of the geometric fibres of $U \to X$ (see Lemma 66.5.1). We will prove the lemma by induction on $d$ for all quasi-compact and separated schemes $U$ mapping surjective and étale onto $X$. If $d = 1$, then $U = X$ and the result holds with $Y = U$. Assume $d > 1$.

We apply Morphisms of Spaces, Lemma 65.52.2 and we obtain a factorization

\[ \xymatrix{ U \ar[rr]_ j \ar[rd] & & Y \ar[ld]^\pi \\ & X } \]

with $\pi $ integral and $j$ a quasi-compact open immersion. We may and do assume that $j(U)$ is scheme theoretically dense in $Y$. Note that

\[ U \times _ X Y = U \amalg W \]

where the first summand is the image of $U \to U \times _ X Y$ (which is closed by Morphisms of Spaces, Lemma 65.4.6 and open because it is étale as a morphism between algebraic spaces étale over $Y$) and the second summand is the (open and closed) complement. The image $V \subset Y$ of $W$ is an open subspace containing $Y \setminus U$.

The étale morphism $W \to Y$ has geometric fibres of cardinality $< d$. Namely, this is clear for geometric points of $U \subset Y$ by inspection. Since $|U| \subset |Y|$ is dense, it holds for all geometric points of $Y$ by Lemma 66.8.1 (the degree of the fibres of a quasi-compact étale morphism does not go up under specialization). Thus we may apply the induction hypothesis to $W \to V$ and find a surjective integral morphism $Z \to V$ with $Z$ a scheme, which Zariski locally factors through $W$. Choose a factorization $Z \to Z' \to Y$ with $Z' \to Y$ integral and $Z \to Z'$ open immersion (Lemma 66.9.1). After replacing $Z'$ by the scheme theoretic closure of $Z$ in $Z'$ we may assume that $Z$ is scheme theoretically dense in $Z'$. After doing this we have $Z' \times _ Y V = Z$. Finally, let $T \subset Y$ be the induced closed subspace structure on $Y \setminus V$. Consider the morphism

\[ Z' \amalg T \longrightarrow X \]

This is a surjective integral morphism by construction. Since $T \subset U$ it is clear that the morphism $T \to X$ factors through $U$. On the other hand, let $z \in Z'$ be a point. If $z \not\in Z$, then $z$ maps to a point of $Y \setminus V \subset U$ and we find a neighbourhood of $z$ on which the morphism factors through $U$. If $z \in Z$, then we have an open neighbourhood of $z$ in $Z$ (which is also an open neighbourhood of $z$ in $Z'$) which factors through $W \subset U \times _ X Y$ and hence through $U$. $\square$

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