The Stacks project

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:

  1. Properties of Spaces, Sections 64.13 and 64.14,

  2. Decent Spaces, Section 66.10,

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

  4. Limits of Spaces, Section 68.15, and

  5. 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:

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

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

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

  4. 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

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

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

  3. $x$ 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. $\square$

Remark 70.9.4. Here is a sketch of a proof of Lemma 70.9.3 which avoids using More on Groupoids, Lemma 40.14.10.

Step 1. We may assume $X$ is a reduced Noetherian separated algebraic space (for example by Cohomology of Spaces, Lemma 67.17.1 or by Limits of Spaces, Lemma 68.15.3) and we may choose a finite surjective morphism $Y \to X$ where $Y$ is a Noetherian scheme (by Limits of Spaces, Proposition 68.16.1).

Step 2. After replacing $X$ by an open neighbourhood of $x$, there exists a birational finite morphism $X' \to X$ and a closed subscheme $Y' \subset X' \times _ X Y$ such that $Y' \to X'$ is surjective finite locally free. Namely, because $X$ is reduced there is a dense open subspace $U \subset X$ over which $Y$ is flat (Morphisms of Spaces, Proposition 65.32.1). Then we can choose a $U$-admissible blowup $b : \tilde X \to X$ such that the strict transform $\tilde Y$ of $Y$ is flat over $\tilde X$, see More on Morphisms of Spaces, Lemma 74.39.1. (An alternative is to use Hilbert schemes if one wants to avoid using the result on blowups). Then we let $X' \subset \tilde X$ be the scheme theoretic closure of $b^{-1}(U)$ and $Y' = X' \times _{\tilde X} \tilde Y$. Since $x$ is a codimension $1$ point, we see that $X' \to X$ is finite over a neighbourhood of $x$ (Lemma 70.3.2).

Step 3. After shrinking $X$ to a smaller neighbourhood of $x$ we get that $X'$ is a scheme. This holds because $Y'$ is a scheme and $Y' \to X'$ being finite locally free and because every finite set of codimension $1$ points of $Y'$ is contained in an affine open. Use Properties of Spaces, Proposition 64.14.1 and Varieties, Proposition 33.41.7.

Step 4. There exists an affine open $W' \subset X'$ containing all points lying over $x$ which is the inverse image of an open subspace of $X$. To prove this let $Z \subset X$ be the closure of the set of points where $X' \to X$ is not an isomorphism. We may assume $x \in Z$ otherwise we are already done. Then $x$ is a generic point of an irreducible component of $Z$ and after shrinking $X$ we may assume $Z$ is an affine scheme (Lemma 70.9.2). Then the inverse image $Z' \subset X'$ is an affine scheme as well. Say $x_1, \ldots , x_ n \in Z'$ are the points mapping to $x$. Then we can find an affine open $W'$ in $X'$ whose intersection with $Z'$ is the inverse image of a principal open of $Z$ containing $x$. Namely, we first pick an affine open $W' \subset X'$ containing $x_1, \ldots , x_ n$ using Varieties, Proposition 33.41.7. Then we pick a principal open $D(f) \subset Z$ containing $x$ whose inverse image $D(f|_{Z'})$ is contained in $W' \cap Z'$. Then we pick $f' \in \Gamma (W', \mathcal{O}_{W'})$ restricting to $f|_{Z'}$ and we replace $W'$ by $D(f') \subset W'$. Since $X' \to X$ is an isomorphism away from $Z' \to Z$ the choice of $W'$ guarantees that the image $W \subset X$ of $W'$ is open with inverse image $W'$ in $X'$.

Step 5. Then $W' \to W$ is a finite surjective morphism and $W$ is a scheme by Cohomology of Spaces, Lemma 67.17.1 and the proof is complete.


Comments (1)

Comment #5375 by Matthieu Romagny on

add final point at the end of the sentence "... and flat monomorphisms (More on Morphisms of Spaces, Lemma 0B8A)"


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