The Stacks project

Theorem 99.21.3. Let $\mathcal{X}$ be an algebraic stack. The following are equivalent

  1. $\mathcal{X}$ is quasi-DM, and

  2. there exists a scheme $W$ and a surjective, flat, locally finitely presented, locally quasi-finite morphism $W \to \mathcal{X}$.

Proof. The implication (2) $\Rightarrow $ (1) is Lemma 99.4.14. Assume (1). Let $x \in |\mathcal{X}|$ be a finite type point. We will produce a scheme over $\mathcal{X}$ which “works” in a neighbourhood of $x$. At the end of the proof we will take the disjoint union of all of these to conclude.

Let $U$ be an affine scheme, $U \to \mathcal{X}$ a smooth morphism, and $u \in U$ a closed point which maps to $x$, see Lemma 99.18.1. Denote $u = \mathop{\mathrm{Spec}}(\kappa (u))$ as usual. Consider the following commutative diagram

\[ \xymatrix{ u \ar[d] & R \ar[l] \ar[d] \\ U \ar[d] & F \ar[d] \ar[l]^ p \\ \mathcal{X} & u \ar[l] } \]

with both squares fibre product squares, in particular $R = u \times _\mathcal {X} u$. In the proof of Lemma 99.18.7 we have seen that $(u, R, s, t, c)$ is a groupoid in algebraic spaces with $s, t$ locally of finite type. Let $G \to u$ be the stabilizer group algebraic space (see Groupoids in Spaces, Definition 76.15.2). Note that

\[ G = R \times _{(u \times u)} u = (u \times _\mathcal {X} u) \times _{(u \times u)} u = \mathcal{X} \times _{\mathcal{X} \times \mathcal{X}} u. \]

As $\mathcal{X}$ is quasi-DM we see that $G$ is locally quasi-finite over $u$. By More on Groupoids in Spaces, Lemma 77.9.11 we have $\dim (R) = 0$.

Let $e : u \to R$ be the identity of the groupoid. Thus both compositions $u \to R \to u$ are equal to the identity morphism of $u$. Note that $R \subset F$ is a closed subspace as $u \subset U$ is a closed subscheme. Hence we can also think of $e$ as a point of $F$. Consider the maps of étale local rings

\[ \mathcal{O}_{U, u} \xrightarrow {p^\sharp } \mathcal{O}_{F, \overline{e}} \longrightarrow \mathcal{O}_{R, \overline{e}} \]

Note that $\mathcal{O}_{R, \overline{e}}$ has dimension $0$ by the result of the first paragraph. On the other hand, the kernel of the second arrow is $p^\sharp (\mathfrak m_ u)\mathcal{O}_{F, \overline{e}}$ as $R$ is cut out in $F$ by $\mathfrak m_ u$. Thus we see that

\[ \mathfrak m_{\overline{z}} = \sqrt{p^\sharp (\mathfrak m_ u)\mathcal{O}_{F, \overline{e}}} \]

On the other hand, as the morphism $U \to \mathcal{X}$ is smooth we see that $F \to u$ is a smooth morphism of algebraic spaces. This means that $F$ is a regular algebraic space (Spaces over Fields, Lemma 70.16.1). Hence $\mathcal{O}_{F, \overline{e}}$ is a regular local ring (Properties of Spaces, Lemma 64.25.1). Note that a regular local ring is Cohen-Macaulay (Algebra, Lemma 10.105.3). Let $d = \dim (\mathcal{O}_{F, \overline{e}})$. By Algebra, Lemma 10.103.10 we can find $f_1, \ldots , f_ d \in \mathcal{O}_{U, u}$ whose images $\varphi (f_1), \ldots , \varphi (f_ d)$ form a regular sequence in $\mathcal{O}_{F, \overline{z}}$. By Lemma 99.21.1 after shrinking $U$ we may assume that $Z = V(f_1, \ldots , f_ d) \to \mathcal{X}$ is flat and locally of finite presentation. Note that by construction $F_ Z = Z \times _\mathcal {X} u$ is a closed subspace of $F = U \times _\mathcal {X} u$, that $e$ is a point of this closed subspace, and that

\[ \dim (\mathcal{O}_{F_ Z, \overline{e}}) = 0. \]

By Morphisms of Spaces, Lemma 65.34.1 it follows that $\dim _ e(F_ Z) = 0$ because the transcendence degree of $e$ relative to $u$ is zero. Hence it follows from Lemma 99.21.2 that after possibly shrinking $U$ the morphism $Z \to \mathcal{X}$ is locally quasi-finite.

We conclude that for every finite type point $x$ of $\mathcal{X}$ there exists a locally quasi-finite, flat, locally finitely presented morphism $f_ x : Z_ x \to \mathcal{X}$ with $x$ in the image of $|f_ x|$. Set $W = \coprod _ x Z_ x$ and $f = \coprod f_ x$. Then $f$ is flat, locally of finite presentation, and locally quasi-finite. In particular the image of $|f|$ is open, see Properties of Stacks, Lemma 98.4.7. By construction the image contains all finite type points of $\mathcal{X}$, hence $f$ is surjective by Lemma 99.18.6 (and Properties of Stacks, Lemma 98.4.4). $\square$

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