66.10 Schematic locus

In this section we prove that a decent algebraic space has a dense open subspace which is a scheme. We first prove this for reasonable algebraic spaces.

Proposition 66.10.1. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. If $X$ is reasonable, then there exists a dense open subspace of $X$ which is a scheme.

Proof. By Properties of Spaces, Lemma 64.13.1 the question is local on $X$. Hence we may assume there exists an affine scheme $U$ and a surjective étale morphism $U \to X$ (Properties of Spaces, Lemma 64.6.1). Let $n$ be an integer bounding the degrees of the fibres of $U \to X$ which exists as $X$ is reasonable, see Definition 66.6.1. We will argue by induction on $n$ that whenever

1. $U \to X$ is a surjective étale morphism whose fibres have degree $\leq n$, and

2. $U$ is isomorphic to a locally closed subscheme of an affine scheme

then the schematic locus is dense in $X$.

Let $X_ n \subset X$ be the open subspace which is the complement of the closed subspace $Z_{n - 1} \subset X$ constructed in Lemma 66.8.1 using the morphism $U \to X$. Let $U_ n \subset U$ be the inverse image of $X_ n$. Then $U_ n \to X_ n$ is finite locally free of degree $n$. Hence $X_ n$ is a scheme by Properties of Spaces, Proposition 64.14.1 (and the fact that any finite set of points of $U_ n$ is contained in an affine open of $U_ n$, see Properties, Lemma 28.29.5).

Let $X' \subset X$ be the open subspace such that $|X'|$ is the interior of $|Z_{n - 1}|$ in $|X|$ (see Topology, Definition 5.21.1). Let $U' \subset U$ be the inverse image. Then $U' \to X'$ is surjective étale and has degrees of fibres bounded by $n - 1$. By induction we see that the schematic locus of $X'$ is an open dense $X'' \subset X'$. By elementary topology we see that $X'' \cup X_ n \subset X$ is open and dense and we win. $\square$

Theorem 66.10.2 (David Rydh). Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. If $X$ is decent, then there exists a dense open subspace of $X$ which is a scheme.

Proof. Assume $X$ is a decent algebraic space for which the theorem is false. By Properties of Spaces, Lemma 64.13.1 there exists a largest open subspace $X' \subset X$ which is a scheme. Since $X'$ is not dense in $X$, there exists an open subspace $X'' \subset X$ such that $|X''| \cap |X'| = \emptyset$. Replacing $X$ by $X''$ we get a nonempty decent algebraic space $X$ which does not contain any open subspace which is a scheme.

Choose a nonempty affine scheme $U$ and an étale morphism $U \to X$. We may and do replace $X$ by the open subscheme corresponding to the image of $|U| \to |X|$. Consider the sequence of open subspaces

$X = X_0 \supset X_1 \supset X_2 \ldots$

constructed in Lemma 66.8.2 for the morphism $U \to X$. Note that $X_0 = X_1$ as $U \to X$ is surjective. Let $U = U_0 = U_1 \supset U_2 \ldots$ be the induced sequence of open subschemes of $U$.

Choose a nonempty open affine $V_1 \subset U_1$ (for example $V_1 = U_1$). By induction we will construct a sequence of nonempty affine opens $V_1 \supset V_2 \supset \ldots$ with $V_ n \subset U_ n$. Namely, having constructed $V_1, \ldots , V_{n - 1}$ we can always choose $V_ n$ unless $V_{n - 1} \cap U_ n = \emptyset$. But if $V_{n - 1} \cap U_ n = \emptyset$, then the open subspace $X' \subset X$ with $|X'| = \mathop{\mathrm{Im}}(|V_{n - 1}| \to |X|)$ is contained in $|X| \setminus |X_ n|$. Hence $V_{n - 1} \to X'$ is an étale morphism whose fibres have degree bounded by $n - 1$. In other words, $X'$ is reasonable (by definition), hence $X'$ contains a nonempty open subscheme by Proposition 66.10.1. This is a contradiction which shows that we can pick $V_ n$.

By Limits, Lemma 32.4.3 the limit $V_\infty = \mathop{\mathrm{lim}}\nolimits V_ n$ is a nonempty scheme. Pick a morphism $\mathop{\mathrm{Spec}}(k) \to V_\infty$. The composition $\mathop{\mathrm{Spec}}(k) \to V_\infty \to U \to X$ has image contained in all $X_ d$ by construction. In other words, the fibred $U \times _ X \mathop{\mathrm{Spec}}(k)$ has infinite degree which contradicts the definition of a decent space. This contradiction finishes the proof of the theorem. $\square$

Lemma 66.10.3. 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 65.30.6) 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 64.4.3). 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 64.14.1. 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 63.5.9). 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. $\square$

At this point we have several different ways for proving the following lemma.

Lemma 66.10.4. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. If there exists a finite, étale, surjective morphism $U \to X$ where $U$ is a scheme, then there exists a dense open subspace of $X$ which is a scheme.

First proof. The morphism $U \to X$ is finite locally free. Hence there is a decomposition of $X$ into open and closed subspaces $X_ d \subset X$ such that $U \times _ X X_ d \to X_ d$ is finite locally free of degree $d$. Thus we may assume $U \to X$ is finite locally free of degree $d$. In this case, let $U_ i \subset U$, $i \in I$ be the set of affine opens. For each $i$ the morphism $U_ i \to X$ is étale and has universally bounded fibres (namely, bounded by $d$). In other words, $X$ is reasonable and the result follows from Proposition 66.10.1. $\square$

Second proof. The question is local on $X$ (Properties of Spaces, Lemma 64.13.1), hence may assume $X$ is quasi-compact. Then $U$ is quasi-compact. Then there exists a dense open subscheme $W \subset U$ which is separated (Properties, Lemma 28.29.3). Set $Z = U \setminus W$. Let $R = U \times _ X U$ and $s, t : R \to U$ the projections. Then $t^{-1}(Z)$ is nowhere dense in $R$ (Topology, Lemma 5.21.6) and hence $\Delta = s(t^{-1}(Z))$ is an $R$-invariant closed nowhere dense subset of $U$ (Morphisms, Lemma 29.47.7). Let $u \in U \setminus \Delta$ be a generic point of an irreducible component. Since these points are dense in $U \setminus \Delta$ and since $\Delta$ is nowhere dense, it suffices to show that the image $x \in X$ of $u$ is in the schematic locus of $X$. Observe that $t(s^{-1}(\{ u\} )) \subset W$ is a finite set of generic points of irreducible components of $W$ (compare with Properties of Spaces, Lemma 64.11.1). By Properties, Lemma 28.29.1 we can find an affine open $V \subset W$ such that $t(s^{-1}(\{ u\} )) \subset V$. Since $t(s^{-1}(\{ u\} ))$ is the fibre of $|U| \to |X|$ over $x$, we conclude by Lemma 66.10.3. $\square$

Third proof. (This proof is essentially the same as the second proof, but uses fewer references.) Assume $X$ is an algebraic space, $U$ a scheme, and $U \to X$ is a finite étale surjective morphism. Write $R = U \times _ X U$ and denote $s, t : R \to U$ the projections as usual. Note that $s, t$ are surjective, finite and étale. Claim: The union of the $R$-invariant affine opens of $U$ is topologically dense in $U$.

Proof of the claim. Let $W \subset U$ be an affine open. Set $W' = t(s^{-1}(W)) \subset U$. Since $s^{-1}(W)$ is affine (hence quasi-compact) we see that $W' \subset U$ is a quasi-compact open. By Properties, Lemma 28.29.3 there exists a dense open $W'' \subset W'$ which is a separated scheme. Set $\Delta ' = W' \setminus W''$. This is a nowhere dense closed subset of $W''$. Since $t|_{s^{-1}(W)} : s^{-1}(W) \to W'$ is open (because it is étale) we see that the inverse image $(t|_{s^{-1}(W)})^{-1}(\Delta ') \subset s^{-1}(W)$ is a nowhere dense closed subset (see Topology, Lemma 5.21.6). Hence, by Morphisms, Lemma 29.47.7 we see that

$\Delta = s\left((t|_{s^{-1}(W)})^{-1}(\Delta ')\right)$

is a nowhere dense closed subset of $W$. Pick any point $\eta \in W$, $\eta \not\in \Delta$ which is a generic point of an irreducible component of $W$ (and hence of $U$). By our choices above the finite set $t(s^{-1}(\{ \eta \} )) = \{ \eta _1, \ldots , \eta _ n\}$ is contained in the separated scheme $W''$. Note that the fibres of $s$ is are finite discrete spaces, and that generalizations lift along the étale morphism $t$, see Morphisms, Lemmas 29.35.12 and 29.25.9. In this way we see that each $\eta _ i$ is a generic point of an irreducible component of $W''$. Thus, by Properties, Lemma 28.29.1 we can find an affine open $V \subset W''$ such that $\{ \eta _1, \ldots , \eta _ n\} \subset V$. By Groupoids, Lemma 39.24.1 this implies that $\eta$ is contained in an $R$-invariant affine open subscheme of $U$. The claim follows as $W$ was chosen as an arbitrary affine open of $U$ and because the set of generic points of irreducible components of $W \setminus \Delta$ is dense in $W$.

Using the claim we can finish the proof. Namely, if $W \subset U$ is an $R$-invariant affine open, then the restriction $R_ W$ of $R$ to $W$ equals $R_ W = s^{-1}(W) = t^{-1}(W)$ (see Groupoids, Definition 39.19.1 and discussion following it). In particular the maps $R_ W \to W$ are finite étale also. It follows in particular that $R_ W$ is affine. Thus we see that $W/R_ W$ is a scheme, by Groupoids, Proposition 39.23.9. On the other hand, $W/R_ W$ is an open subspace of $X$ by Spaces, Lemma 63.10.2. Hence having a dense collection of points contained in $R$-invariant affine open of $U$ certainly implies that the schematic locus of $X$ (see Properties of Spaces, Lemma 64.13.1) is open dense in $X$. $\square$

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