The Stacks project


Theorem 80.14.9. Let $S$ be a scheme. Let $B$ be a quasi-compact and quasi-separated algebraic space over $S$. Let $X \to B$ be a separated, finite type morphism. Then $X$ has a compactification over $B$.

Proof. We first reduce to the Noetherian case. We strongly urge the reader to skip this paragraph. First, we may replace $S$ by $\mathop{\mathrm{Spec}}(\mathbf{Z})$. See Spaces, Section 64.16 and Properties of Spaces, Definition 65.3.1. There exists a closed immersion $X \to X'$ with $X' \to B$ of finite presentation and separated. See Limits of Spaces, Proposition 69.11.7. If we find a compactification of $X'$ over $B$, then taking the scheme theoretic closure of $X$ in this will give a compactification of $X$ over $B$. Thus we may assume $X \to B$ is separated and of finite presentation. We may write $B = \mathop{\mathrm{lim}}\nolimits B_ i$ as a directed limit of a system of Noetherian algebraic spaces of finite type over $\mathop{\mathrm{Spec}}(\mathbf{Z})$ with affine transition morphisms. See Limits of Spaces, Proposition 69.8.1. We can choose an $i$ and a morphism $X_ i \to B_ i$ of finite presentation whose base change to $B$ is $X \to B$, see Limits of Spaces, Lemma 69.7.1. After increasing $i$ we may assume $X_ i \to B_ i$ is separated, see Limits of Spaces, Lemma 69.6.9. If we can find a compactification of $X_ i$ over $B_ i$, then the base change of this to $B$ will be a compactification of $X$ over $B$. This reduces us to the case discussed in the next paragraph.

Assume $B$ is of finite type over $\mathbf{Z}$ in addition to being quasi-compact and quasi-separated. Let $U \to X$ be an ├ętale morphism of algebraic spaces such that $U$ has a compactification $Y$ over $\mathop{\mathrm{Spec}}(\mathbf{Z})$. The morphism

\[ U \longrightarrow B \times _{\mathop{\mathrm{Spec}}(\mathbf{Z})} Y \]

is separated and quasi-finite by Morphisms of Spaces, Lemma 66.27.10 (the displayed morphism factors into an immersion hence is a monomorphism). Hence by Zariski's main theorem (More on Morphisms of Spaces, Lemma 75.34.3) there is an open immersion of $U$ into an algebraic space $Y'$ finite over $B \times _{\mathop{\mathrm{Spec}}(\mathbf{Z})} Y$. Then $Y' \to B$ is proper as the composition $Y' \to B \times _{\mathop{\mathrm{Spec}}(\mathbf{Z})} Y \to B$ of two proper morphisms (use Morphisms of Spaces, Lemmas 66.45.9, 66.40.4, and 66.40.3). We conclude that $U$ has a compactification over $B$.

There is a dense open subspace $U \subset X$ which is a scheme. (Properties of Spaces, Proposition 65.13.3). In fact, we may choose $U$ to be an affine scheme (Properties, Lemmas 28.5.7 and 28.29.1). Thus $U$ has a compactification over $\mathop{\mathrm{Spec}}(\mathbf{Z})$; this is easily shown directly but also follows from the theorem for schemes, see More on Flatness, Theorem 38.33.8. By the previous paragraph $U$ has a compactification over $B$. By Noetherian induction we can find a maximal dense open subspace $U \subset X$ which has a compactification over $B$. We will show that the assumption that $U \not= X$ leads to a contradiction. Namely, by Lemma 80.14.8 we can find a strictly larger open $U \subset W \subset X$ and a distinguished square $(U \subset W, f : V \to W)$ with $V$ affine and $U \times _ W V$ dense image in $U$. Since $V$ is affine, as before it has a compactification over $B$. Hence Lemma 80.14.7 applies to show that $W$ has a compactification over $B$ which is the desired contradiction. $\square$

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