## 70.9 Schematic locus

We have already proven a number of results on the schematic locus of an algebraic space. Here is a list of references:

Properties of Spaces, Sections 64.13 and 64.14,

Decent Spaces, Section 66.10,

Properties of Spaces, Lemma 64.15.3 $\Leftarrow $ Decent Spaces, Lemma 66.12.8 $\Leftarrow $ Decent Spaces, Lemma 66.14.2,

Limits of Spaces, Section 68.15, and

Limits of Spaces, Section 68.17.

There are some cases where certain types of morphisms of algebraic spaces are automatically representable, for example separated, locally quasi-finite morphisms (Morphisms of Spaces, Lemma 65.51.1), and flat monomorphisms (More on Morphisms of Spaces, Lemma 74.4.1) In Section 70.10 we will study what happens with the schematic locus under extension of base field.

Lemma 70.9.1. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. In each of the following cases $X$ is a scheme:

$X$ is quasi-compact and quasi-separated and $\dim (X) = 0$,

$X$ is locally of finite type over a field $k$ and $\dim (X) = 0$,

$X$ is Noetherian and $\dim (X) = 0$, and

add more here.

**Proof.**
Cases (2) and (3) follow immediately from case (1) but we will give a separate proofs of (2) and (3) as these proofs use significantly less theory.

Proof of (3). Let $U$ be an affine scheme and let $U \to X$ be an étale morphism. Set $R = U \times _ X U$. The two projection morphisms $s, t : R \to U$ are étale morphisms of schemes. By Properties of Spaces, Definition 64.9.2 we see that $\dim (U) = 0$ and $\dim (R) = 0$. Since $R$ is a locally Noetherian scheme of dimension $0$, we see that $R$ is a disjoint union of spectra of Artinian local rings (Properties, Lemma 28.10.5). Since we assumed that $X$ is Noetherian (so quasi-separated) we conclude that $R$ is quasi-compact. Hence $R$ is an affine scheme (use Schemes, Lemma 26.6.8). The étale morphisms $s, t : R \to U$ induce finite residue field extensions. Hence $s$ and $t$ are finite by Algebra, Lemma 10.53.4 (small detail omitted). Thus Groupoids, Proposition 39.23.9 shows that $X = U/R$ is an affine scheme.

Proof of (2) – almost identical to the proof of (4). Let $U$ be an affine scheme and let $U \to X$ be an étale morphism. Set $R = U \times _ X U$. The two projection morphisms $s, t : R \to U$ are étale morphisms of schemes. By Properties of Spaces, Definition 64.9.2 we see that $\dim (U) = 0$ and similarly $\dim (R) = 0$. On the other hand, the morphism $U \to \mathop{\mathrm{Spec}}(k)$ is locally of finite type as the composition of the étale morphism $U \to X$ and $X \to \mathop{\mathrm{Spec}}(k)$, see Morphisms of Spaces, Lemmas 65.23.2 and 65.39.9. Similarly, $R \to \mathop{\mathrm{Spec}}(k)$ is locally of finite type. Hence by Varieties, Lemma 33.20.2 we see that $U$ and $R$ are disjoint unions of spectra of local Artinian $k$-algebras finite over $k$. The same thing is therefore true of $U \times _{\mathop{\mathrm{Spec}}(k)} U$. As

\[ R = U \times _ X U \longrightarrow U \times _{\mathop{\mathrm{Spec}}(k)} U \]

is a monomorphism, we see that $R$ is a finite(!) union of spectra of finite $k$-algebras. It follows that $R$ is affine, see Schemes, Lemma 26.6.8. Applying Varieties, Lemma 33.20.2 once more we see that $R$ is finite over $k$. Hence $s, t$ are finite, see Morphisms, Lemma 29.43.14. Thus Groupoids, Proposition 39.23.9 shows that the open subspace $U/R$ of $X$ is an affine scheme. Since the schematic locus of $X$ is an open subspace (see Properties of Spaces, Lemma 64.13.1), and since $U \to X$ was an arbitrary étale morphism from an affine scheme we conclude that $X$ is a scheme.

Proof of (1). By Cohomology of Spaces, Lemma 67.10.1 we have vanishing of higher cohomology groups for all quasi-coherent sheaves $\mathcal{F}$ on $X$. Hence $X$ is affine (in particular a scheme) by Cohomology of Spaces, Proposition 67.16.7.
$\square$

The following lemma tells us that a quasi-separated algebraic space is a scheme away from codimension $1$.

Lemma 70.9.2. Let $S$ be a scheme. Let $X$ be a quasi-separated algebraic space over $S$. Let $x \in |X|$. The following are equivalent

$x$ is a point of codimension $0$ on $X$,

the local ring of $X$ at $x$ has dimension $0$, and

$x$ is a generic point of an irreducible component of $|X|$.

If true, then there exists an open subspace of $X$ containing $x$ which is a scheme.

**Proof.**
The equivalence of (1), (2), and (3) follows from Decent Spaces, Lemma 66.20.1 and the fact that a quasi-separated algebraic space is decent (Decent Spaces, Section 66.6). However in the next paragraph we will give a more elementary proof of the equivalence.

Note that (1) and (2) are equivalent by definition (Properties of Spaces, Definition 64.10.2). To prove the equivalence of (1) and (3) we may assume $X$ is quasi-compact. Choose

\[ \emptyset = U_{n + 1} \subset U_ n \subset U_{n - 1} \subset \ldots \subset U_1 = X \]

and $f_ i : V_ i \to U_ i$ as in Decent Spaces, Lemma 66.8.6. Say $x \in U_ i$, $x \not\in U_{i + 1}$. Then $x = f_ i(y)$ for a unique $y \in V_ i$. If (1) holds, then $y$ is a generic point of an irreducible component of $V_ i$ (Properties of Spaces, Lemma 64.11.1). Since $f_ i^{-1}(U_{i + 1})$ is a quasi-compact open of $V_ i$ not containing $y$, there is an open neighbourhood $W \subset V_ i$ of $y$ disjoint from $f_ i^{-1}(V_ i)$ (see Properties, Lemma 28.2.2 or more simply Algebra, Lemma 10.25.4). Then $f_ i|_ W : W \to X$ is an isomorphism onto its image and hence $x = f_ i(y)$ is a generic point of $|X|$. Conversely, assume (3) holds. Then $f_ i$ maps $\overline{\{ y\} }$ onto the irreducible component $\overline{\{ x\} }$ of $|U_ i|$. Since $|f_ i|$ is bijective over $\overline{\{ x\} }$, it follows that $\overline{\{ y\} }$ is an irreducible component of $U_ i$. Thus $x$ is a point of codimension $0$.

The final statement of the lemma is Properties of Spaces, Proposition 64.13.3.
$\square$

The following lemma says that a separated locally Noetherian algebraic space is a scheme in codimension $1$, i.e., away from codimension $2$.

slogan
Lemma 70.9.3. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $x \in |X|$. If $X$ is separated, locally Noetherian, and the dimension of the local ring of $X$ at $x$ is $\leq 1$ (Properties of Spaces, Definition 64.10.2), then there exists an open subspace of $X$ containing $x$ which is a scheme.

**Proof.**
(Please see the remark below for a different approach avoiding the material on finite groupoids.) We can replace $X$ by an quasi-compact neighbourhood of $x$, hence we may assume $X$ is quasi-compact, separated, and Noetherian. There exists a scheme $U$ and a finite surjective morphism $U \to X$, see Limits of Spaces, Proposition 68.16.1. Let $R = U \times _ X U$. Then $j : R \to U \times _ S U$ is an equivalence relation and we obtain a groupoid scheme $(U, R, s, t, c)$ over $S$ with $s, t$ finite and $U$ Noetherian and separated. Let $\{ u_1, \ldots , u_ n\} \subset U$ be the set of points mapping to $x$. Then $\dim (\mathcal{O}_{U, u_ i}) \leq 1$ by Decent Spaces, Lemma 66.12.6.

By More on Groupoids, Lemma 40.14.10 there exists an $R$-invariant affine open $W \subset U$ containing the orbit $\{ u_1, \ldots , u_ n\} $. Since $U \to X$ is finite surjective the continuous map $|U| \to |X|$ is closed surjective, hence submersive by Topology, Lemma 5.6.5. Thus $f(W)$ is open and there is an open subspace $X' \subset X$ with $f : W \to X'$ a surjective finite morphism. Then $X'$ is an affine scheme by Cohomology of Spaces, Lemma 67.17.1 and the proof is finished.
$\square$

## Comments (1)

Comment #5375 by Matthieu Romagny on