The Stacks project

Proof. Assume $\mathcal{X}$ is well-nigh affine. Choose an affine scheme $U$ and a surjective, flat, finite, and finitely presented morphism $U \to \mathcal{X}$. Set $R = U \times _\mathcal {X} U$. Then we obtain a groupoid $(U, R, s, t, c)$ in algebraic spaces and an isomorphism $[U/R] \to \mathcal{X}$, see Algebraic Stacks, Lemma 94.16.1 and Remark 94.16.3. Since $s, t : R \to U$ are flat, finite, and finitely presented morphisms (as base changes of $U \to \mathcal{X})$ we see that $s, t$ are finite locally free (Morphisms, Lemma 29.48.2). This implies that $R$ is affine (as finite morphisms are affine) and hence (2) holds.

Suppose that we have a groupoid scheme $(U, R, s, t, c)$ with $U$ and $R$ are affine and $s, t : R \to U$ finite locally free. Set $\mathcal{X} = [U/R]$. Then $\mathcal{X}$ is an algebraic stack by Criteria for Representability, Theorem 97.17.2 (strictly speaking we don't need this here, but it can't be stressed enough that this is true). The morphism $U \to \mathcal{X}$ is surjective, flat, and locally of finite presentation by Criteria for Representability, Lemma 97.17.1. Thus we can check whether $U \to \mathcal{X}$ is finite by checking whether the projection $U \times _\mathcal {X} U \to U$ has this property, see Properties of Stacks, Lemma 100.3.3. Since $U \times _\mathcal {X} U = R$ by Groupoids in Spaces, Lemma 78.22.2 we see that this is true. Thus $\mathcal{X}$ is well-nigh affine.

Proof of the final statement. We see that $\mathcal{X}$ is quasi-compact by Properties of Stacks, Lemma 100.6.2. We see that $\mathcal{X} = [U/R]$ is quasi-DM and separated by Morphisms of Stacks, Lemma 101.20.1. $\square$

Proof. Part (1) follows from immediately from Limits of Spaces, Lemma 70.15.1. However, this is overkill, since (1) also follows from Lemma 106.13.2 combined with Groupoids, Proposition 39.23.9.

To prove (2) we choose an affine scheme $U$ and a surjective, flat, finite, and finitely presented morphism $U \to \mathcal{X}$. Then $U' = \mathcal{X}' \times _\mathcal {X} U$ admits an affine morphism to $U$ (Morphisms of Stacks, Lemma 101.9.2). Therefore $U'$ is an affine scheme. Of course $U' \to \mathcal{X}'$ is surjective, flat, finite, and finitely presented as a base change of $U \to \mathcal{X}$. $\square$

Proof. Write $\mathcal{X} = [U/R]$ with $(U, R, s, t, c)$ as in Lemma 106.13.2. Let $C$ be the ring of $R$-invariant functions on $U$, see Groupoids, Section 39.23. We set $M = \mathop{\mathrm{Spec}}(C)$. The $R$-invariant morphism $U \to M$ corresponds to a morphism $f : \mathcal{X} \to M$ by Lemma 106.12.2. The characterization of morphisms into affine schemes given in Schemes, Lemma 26.6.4 immediately guarantees that $\phi : U \to M$ is a categorical quotient in the category of affine schemes. Hence $f$ is a categorical moduli space in the category of affine schemes (Lemma 106.12.3).

Since $\mathcal{X}$ is separated by Lemma 106.13.2 we find that $f$ is separated by Morphisms of Stacks, Lemma 101.4.12.

Since $U \to \mathcal{X}$ is surjective and since $U \to M$ is quasi-compact, we see that $f$ is quasi-compact by Morphisms of Stacks, Lemma 101.7.6.

By Groupoids, Lemma 39.23.4 the composition

is an integral morphism of affine schemes. In particular, it is universally closed (Morphisms, Lemma 29.44.7). Since $U \to \mathcal{X}$ is surjective, it follows that $\mathcal{X} \to M$ is universally closed (Morphisms of Stacks, Lemma 101.37.6). To conclude that $\mathcal{X} \to M$ is a universal homeomorphism, it is enough to show that it is universally bijective, i.e., surjective and universally injective.

We have $|\mathcal{X}| = |U|/|R|$ by Morphisms of Stacks, Lemma 101.20.2. Thus $|f|$ is surjective and even bijective by Groupoids, Lemma 39.23.6.

Let $C \to C'$ be a ring map. Let $(U', R', s', t', c')$ be the base change of $(U, R, s, t, c)$ by $M' = \mathop{\mathrm{Spec}}(C') \to M$. Setting $\mathcal{X}' = [U'/R']$, we observe that $M' \times _ M \mathcal{X} = \mathcal{X}'$ by Quotients of Groupoids, Lemma 83.3.6. Let $C^1$ be the ring of $R'$-invariant functions on $U'$. Set $M^1 = \mathop{\mathrm{Spec}}(C^1)$ and consider the diagram

By Groupoids, Lemma 39.23.5 and Algebra, Lemma 10.46.11 the morphism $M^1 \to M'$ is a homeomorphism. On the other hand, the previous paragraph applied to $(U', R', s', t', c')$ shows that $|f'|$ is bijective. We conclude that $f$ induces a bijection on points after any base change by an affine scheme. Thus $f$ is universally injective by Morphisms of Stacks, Lemma 101.14.7.

Finally, we still have to show that $f$ is a uniform moduli space in the category of affine schemes. This follows from the discussion above and the fact that if the ring map $C \to C'$ is flat, then $C' \to C^1$ is an isomorphism by Groupoids, Lemma 39.23.5. $\square$

Proof. Observe that $h$ is representable by algebraic spaces by Morphisms of Stacks, Lemmas 101.45.3 and 101.45.1. Choose an affine scheme $U$ and a surjective, flat, finite, and finitely presented morphism $U \to \mathcal{X}$. Then $U' = \mathcal{X}' \times _\mathcal {X} U$ is an algebraic space with a finite (in particular affine) morphism $U' \to \mathcal{X}'$. By Lemma 106.13.3 we conclude that $U'$ is affine. Setting $R = U \times _\mathcal {X} U$ and $R' = U' \times _{\mathcal{X}'} U'$ we obtain groupoids $(U, R, s, t, c)$ and $(U', R', s', t', c')$ such that $\mathcal{X} = [U/R]$ and $\mathcal{X}' = [U'/R']$, see proof of Lemma 106.13.2. we see that the diagrams

are cartesian where $G$ and $G'$ are the stabilizer group schemes. This follows for the first two by transitivity of fibre products and for the last one this follows because it is the pullback of the isomorphism $\mathcal{I}_{\mathcal{X}'} \to \mathcal{X}' \times _\mathcal {X} \mathcal{I}_\mathcal {X}$ (by the already used Morphisms of Stacks, Lemma 101.45.3). Recall that $M$, resp. $M'$ was constructed in Lemma 106.13.4 as the spectrum of the ring of $R$-invariant functions on $U$, resp. the ring of $R'$-invariant functions on $U'$. Thus $M' \to M$ is étale and $U' = M' \times _ M U$ by Groupoids, Lemma 39.23.7. It follows that $R' = M' \times _ M U$, in other words the groupoid $(U', R', s', t', c')$ is the base change of $(U, R, s, t, c)$ by $M' \to M$. This implies that the diagram in the lemma is cartesian by Quotients of Groupoids, Lemma 83.3.6. $\square$

Proof. We already know that $M$ is a uniform categorical moduli space in the category of affine schemes. By Lemma 106.12.4 it suffices to show that the base change $f' : M' \times _ M \mathcal{X} \to M'$ is a categorical moduli space for any flat morphism $M' \to M$ of affine schemes. Observe that $\mathcal{X}' = M' \times _ M \mathcal{X}$ is well-nigh affine by Lemma 106.13.3. This after replacing $\mathcal{X}$ by $\mathcal{X}'$ and $M$ by $M'$, we reduce to proving $f$ is a categorical moduli space.

Let $g : \mathcal{X} \to Y$ be a morphism where $Y$ is an algebraic space. We have to show that $g = h \circ f$ for a unique morphism $h : M \to Y$.

Uniqueness. Suppose we have two morphisms $h_ i : M \to Y$ with $g = h_1 \circ f = h_2 \circ f$. Let $M' \subset M$ be the equalizer of $h_1$ and $h_2$. Then $M' \to M$ is a monomorphism and $f : \mathcal{X} \to M$ factors through $M'$. Thus $M' \to M$ is a universal homeomorphism. We conclude $M'$ is affine (Morphisms, Lemma 29.45.5). But then since $f : \mathcal{X} \to M$ is a categorical moduli space in the category of affine schemes, we see $M' = M$.

Existence. Below we will show that for every $p \in M$ there exists a cartesian square

with $M' \to M$ an étale morphism of affines and $p$ in the image such that the composition $\mathcal{X}' \to \mathcal{X} \to Y$ factors through $M'$. This means we can construct the map $h : M \to Y$ étale locally on $M$. Since $Y$ is a sheaf for the étale topology and by the uniqueness shown above, this is enough (small detail omitted).

Let $y \in |Y|$ be the image of $p$. Let $(V, v) \to (Y, y)$ be an étale morphism with $V$ affine. Consider $\mathcal{X}' = V \times _ Y \mathcal{X}$. Observe that $\mathcal{X}' \to \mathcal{X}$ is separated and étale as the base change of $V \to Y$. Moreover, $\mathcal{X}' \to \mathcal{X}$ induces isomorphisms on automorphism groups (Morphisms of Stacks, Remark 101.19.5) as this is true for $V \to Y$, see Morphisms of Stacks, Lemma 101.45.5. Choose a presentation $\mathcal{X} = [U/R]$ as in Lemma 106.13.2. Set $U' = \mathcal{X}' \times _\mathcal {X} U = V \times _ Y U$ and choose $u' \in U'$ mapping to $p$ and $v$ (possible by Properties of Spaces, Lemma 66.4.3). Since $U' \to U$ is separated and étale we see that every finite set of points of $U'$ is contained in an affine open, see More on Morphisms, Lemma 37.45.1. On the other hand, the morphism $U' \to \mathcal{X}'$ is surjective, finite, flat, and locally of finite presentation. Setting $R' = U' \times _{\mathcal{X}'} U'$ we see that $s', t' : R' \to U'$ are finite locally free. By Groupoids, Lemma 39.24.1 there exists an $R'$-invariant affine open subscheme $U'' \subset U'$ containing $u'$. Let $\mathcal{X}'' \subset \mathcal{X}'$ be the corresponding open substack. Then $\mathcal{X}''$ is well-nigh affine. By Lemma 106.13.5 we obtain a cartesian square

with $M'' \to M$ étale. Since $\mathcal{X}'' \to M''$ is a categorical moduli space in the category of affine schemes we obtain a morphism $M'' \to V$ such that the composition $\mathcal{X}'' \to \mathcal{X}' \to V$ is equal to the composition $\mathcal{X}'' \to M'' \to V$. This proves our claim and finishes the proof. $\square$

Proof. Choose an affine scheme $U$ and a surjective, flat, finite, and locally finitely presented morphism $U \to \mathcal{X}$. Since $h$ is representable by algebraic spaces (Morphisms of Stacks, Lemmas 101.45.3 and 101.45.1) we see that $U' = \mathcal{X}' \times _\mathcal {X} U$ is an algebraic space. Since $U' \to U$ is separated and étale, we see that $U'$ is a scheme and that every finite set of points of $U'$ is contained in an affine open, see Morphisms of Spaces, Lemma 67.51.1 and More on Morphisms, Lemma 37.45.1. Setting $R' = U' \times _{\mathcal{X}'} U'$ we see that $s', t' : R' \to U'$ are finite locally free. By Groupoids, Lemma 39.24.1 there exists an open covering $U' = \bigcup U'_ i$ by $R'$-invariant affine open subschemes $U'_ i \subset U'$. Let $\mathcal{X}'_ i \subset \mathcal{X}'$ be the corresponding open substacks. These are well-nigh affine as $U'_ i \to \mathcal{X}'_ i$ is surjective, flat, finite and of finite presentation. By Lemma 106.13.5 we obtain cartesian diagrams

with $M'_ i \to M$ an étale morphism of affine schemes and vertical arrows as in Lemma 106.13.4. Observe that $\mathcal{X}'_{ij} = \mathcal{X}'_ i \cap \mathcal{X}'_ j$ is an open subspace of $\mathcal{X}'_ i$ and $\mathcal{X}'_ j$. Hence we get corresponding open subschemes $V_{ij} \subset M'_ i$ and $V_{ji} \subset M'_ j$. By the result of Lemma 106.13.6 we see that both $\mathcal{X}'_{ij} \to V_{ij}$ and $\mathcal{X}'_{ji} \to V_{ji}$ are categorical moduli spaces! Thus we get a unique isomorphism $\varphi _{ij} : V_{ij} \to V_{ji}$ such that

is commutative. These isomorphisms satisfy the cocyclce condition of Schemes, Section 26.14 by a computation (and another application of the previous lemma) which we omit. Thus we can glue the affine schemes in to scheme $M'$, see Schemes, Lemma 26.14.1. Let us identify the $M'_ i$ with their image in $M'$. We claim there is a morphism $\mathcal{X}' \to M'$ fitting into cartesian diagrams

This is clear from the description of the morphisms into the glued scheme $M'$ in Schemes, Lemma 26.14.1 and the fact that to give a morphism $\mathcal{X}' \to M'$ is the same thing as given a morphism $T \to M'$ for any morphism $T \to \mathcal{X}'$. Similarly, there is a morphism $M' \to M$ restricting to the given morphisms $M'_ i \to M$ on $M'_ i$. The morphism $M' \to M$ is étale (being étale on the members of an étale covering) and the fibre product property holds as it can be checked on members of the (affine) open covering $M' = \bigcup M'_ i$. Finally, $M' \to M$ is separated because the composition $U' \to \mathcal{X}' \to M'$ is surjective and universally closed and we can apply Morphisms, Lemma 29.41.11. $\square$

Proof. For any $x \in |\mathcal{X}|$ we can choose $g : \mathcal{U} \to \mathcal{X}$, $\mathcal{U} = [U/R]$, and $u$ as in Morphisms of Stacks, Lemma 101.32.4. Then by Morphisms of Stacks, Lemma 101.45.4 we see that there exists an open substack $\mathcal{U}' \subset \mathcal{U}$ containing $u$ such that $\mathcal{I}_{\mathcal{U}'} \to \mathcal{U}' \times _\mathcal {X} \mathcal{I}_\mathcal {X}$ is an isomorphism. Let $U' \subset U$ be the $R$-invariant open corresponding to the open substack $\mathcal{U}'$. Let $u' \in U'$ be a point of $U'$ mapping to $u$. Observe that $t(s^{-1}(\{ u'\} ))$ is finite as $s : R \to U$ is finite. By Properties, Lemma 28.29.5 and Groupoids, Lemma 39.24.1 we can find an $R$-invariant affine open $U'' \subset U'$ containing $u'$. Let $R''$ be the restriction of $R$ to $U''$. Then $\mathcal{U}'' = [U''/R'']$ is an open substack of $\mathcal{U}'$ containing $u$, is well-nigh affine, $\mathcal{I}_{\mathcal{U}''} \to \mathcal{U}'' \times _\mathcal {X} \mathcal{I}_\mathcal {X}$ is an isomorphism, and $\mathcal{U}'' \to \mathcal{X}$ and is representable by algebraic spaces and étale. Finally, $\mathcal{U}'' \to \mathcal{X}$ is separated as $\mathcal{U}''$ is separated (Lemma 106.13.2) the diagonal of $\mathcal{X}$ is separated (Morphisms of Stacks, Lemma 101.6.1) and separatedness follows from Morphisms of Stacks, Lemma 101.4.12. Since the point $x \in |\mathcal{X}|$ is arbitrary the proof is complete. $\square$

Proof. We choose a set $I$¹ and for $i \in I$ a morphism of algebraic stacks $g_ i : \mathcal{X}_ i \to \mathcal{X}$ as in Lemma 106.13.8; we will use all of the properties listed in this lemma without further mention. Let

be as in Lemma 106.13.4. Consider the stacks

for $i, j \in I$. The projections $\mathcal{X}_{ij} \to \mathcal{X}_ i$ and $\mathcal{X}_{ij} \to \mathcal{X}_ j$ are separated by Morphisms of Stacks, Lemma 101.4.4, étale by Morphisms of Stacks, Lemma 101.35.3, and induce isomorphisms on automorphism groups (as in Morphisms of Stacks, Remark 101.19.5) by Morphisms of Stacks, Lemma 101.45.5. Thus we may apply Lemma 106.13.7 to find a commutative diagram

with cartesian squares where $M_{ij} \to M_ i$ and $M_{ij} \to M_ j$ are separated étale morphisms of schemes; here we also use that $f_ i$ is a uniform categorical quotient by Lemma 106.13.6. Claim:

is an étale equivalence relation.

Proof of the claim. Set $R = \coprod M_{ij}$ and $U = \coprod M_ i$. We have already seen that $t : R \to U$ and $s : R \to U$ are étale. Let us construct a morphism $c : R \times _{s, U, t} R \to R$ compatible with $\text{pr}_{13} : U \times U \times U \to U \times U$. Namely, for $i, j, k \in I$ we consider

Arguing exactly as in the previous paragraph, we find that $M_{ijk} = M_{ij} \times _{M_ j} M_{jk}$ is a categorical moduli space for $\mathcal{X}_{ijk}$. In particular, there is a canonical morphism $M_{ijk} = M_{ij} \times _{M_ j} M_{jk} \to M_{ik}$ coming from the projection $\mathcal{X}_{ijk} \to \mathcal{X}_{ik}$. Putting these morphisms together we obtain the morphism $c$. In a similar fashion we construct a morphism $e : U \to R$ compatible with $\Delta : U \to U \times U$ and $i : R \to R$ compatible with the flip $U \times U \to U \times U$. Let $k$ be an algebraically closed field. Then

is bijective on isomorphism classes and the same remains true after any base change by a morphism $M' \to M$. This follows from our choice of $f_ i$ and Morphisms of Stacks, Lemmas 101.14.5 and 101.14.6. By construction of $2$-fibred products the diagram

is a fibre product of categories. By our choice of $g_ i$ the functors in this diagram induce bijections on automorphism groups. It follows that this diagram induces a fibre product diagram on sets of isomorphism classes! Thus we see that

where $|\mathop{\mathrm{Mor}}\nolimits (\mathop{\mathrm{Spec}}(k), \mathcal{X})|$ denotes the set of isomorphism classes. In particular, for any algebraically closed field $k$ the map on $k$-valued point is an equivalence relation. We conclude the claim holds by Groupoids, Lemma 39.3.5.

Let $M = U/R$ be the algebraic space which is the quotient of the above étale equivalence relation, see Spaces, Theorem 65.10.5. There is a canonical morphism $f : \mathcal{X} \to M$ fitting into commutative diagrams

Namely, such a morphism $f$ is given by a functor

for any scheme $T$ compatible with base change. Let $a : T \to \mathcal{X}$ be an object of the left hand side. We obtain an étale covering $\{ T_ i \to T\} $ with $T_ i = \mathcal{X}_ i \times _\mathcal {X} T$ and morphisms $a_ i : T_ i \to \mathcal{X}_ i$. Then we get $b_ i = f_ i \circ a_ i : T_ i \to M_ i$. Since $T_ i \times _ T T_ j = \mathcal{X}_{ij} \times _\mathcal {X} T$ we moreover get a morphism $a_{ij} : T_ i \times _ T T_ j \to \mathcal{X}_{ij}$. Setting $b_{ij} = f_{ij} \circ a_{ij}$ we find that $b_ i \times b_ j$ factors through the monomorphism $M_{ij} \to M_ i \times M_ j$. Hence the morphisms

agree on $T_ i \times _ T T_ j$. As $M$ is a sheaf for the étale topology, we see that these morphisms glue to a unique morphism $b = f(a) : T \to M$. We omit the verification that this construction is compatible with base change and we omit the verification that the diagrams (106.13.9.1) commute.

Claim: the diagrams (106.13.9.1) are cartesian. To see this we study the induced morphism

This is a morphism of stacks étale over $\mathcal{X}$ and hence $h_ i$ is étale (Morphisms of Stacks, Lemma 101.35.6). Since $g_ i$ is separated, we see $h_ i$ is separated (use Morphisms of Stacks, Lemma 101.4.12 and the fact seen above that the diagonal of $\mathcal{X}$ is separated). The morphism $h_ i$ induces isomorphisms on automorphism groups (Morphisms of Stacks, Remark 101.19.5) as this is true for $g_ i$. For an algebraically closed field $k$ the diagram

is a catesian diagram of categories and the top arrow induces bijections on automorphism groups. On the other hand, we have

by what we said above. Thus the right vertical arrow in the cartesian diagram above is a bijection on isomorphism classes. We conclude that $|\mathop{\mathrm{Mor}}\nolimits (\mathop{\mathrm{Spec}}(k), M_ i \times _ M \mathcal{X})| \to M_ i(k)$ is bijective. Review: $h_ i$ is a separated, étale, induces isomorphisms on automorphism groups (as in Morphisms of Stacks, Remark 101.19.5), and induces an equivalence on fibre categories over algebraically closed fields. Hence it is an isomorphism by Morphisms of Stacks, Lemma 101.45.7.

From the claim we get in particular the following: we have a surjective étale morphism $U \to M$ such that the base change of $f$ is separated, quasi-compact, and a universal homeomorphism. It follows that $f$ is separated, quasi-compact, and a universal homeomorphism. See Morphisms of Stacks, Lemma 101.4.5, 101.7.10, and 101.15.5

To finish the proof we have to show that $f : \mathcal{X} \to M$ is a uniform categorical moduli space. To prove this it suffices to show that given a flat morphism $M' \to M$ of algebraic spaces, the base change

is a categorical moduli space. Thus we consider a morphism

where $E$ is an algebraic space. For each $i$ we know that $f_ i$ is a uniform categorical moduli space. Hence we obtain

Since $\{ M' \times _ M M_ i \to M'\} $ is an étale covering, to obtain the desired morphism $\psi : M' \to E$ it suffices to show that $\psi _ i$ and $\psi _ j$ agree over $M' \times _ M M_ i \times _ M M_ j = M' \times _ M M_{ij}$. This follows easily from the fact that $f_{ij} : \mathcal{X}_{ij} = \mathcal{X}_ i \times _\mathcal {X} \mathcal{X}_ j \to M_{ij}$ is a uniform categorical quotient; details omitted. Then finally one shows that $\psi $ fits into the commutative diagram

because “$\{ M' \times _ M \mathcal{X}_ i \to M' \times _ M \mathcal{X}\} $ is an étale covering” and the morphisms $\psi _ i$ fit into the corresponding commutative diagrams by construction. This finishes the proof of the Keel-Mori theorem. $\square$

Proof. By Morphisms of Stacks, Lemma 101.45.3 we see that $\mathcal{I}_{\mathcal{X}'} \to \mathcal{X}' \times _\mathcal {X} \mathcal{I}_\mathcal {X}$ is an isomorphism. Hence $\mathcal{I}_{\mathcal{X}'} \to \mathcal{X}'$ is finite as a base change of $\mathcal{I}_\mathcal {X} \to \mathcal{X}$. Let $f' : \mathcal{X}' \to M'$ and $f : \mathcal{X} \to M$ be as in Theorem 106.13.9. We obtain a commutative diagram as in the lemma because $f'$ is categorical moduli space. Choose $I$ and $g'_ i : \mathcal{X}'_ i \to \mathcal{X}'$ as in Lemma 106.13.8. Observe that $g_ i = h \circ g'_ i$ is étale, separated, and induces isomorphisms on automorphism groups (Morphisms of Stacks, Remark 101.19.5). Let $f'_ i : \mathcal{X}'_ i \to M'_ i$ be as in Lemma 106.13.4. In the proof of Theorem 106.13.9 we have seen that the diagrams

are cartesian and that $M'_ i \to M'$ and $M'_ i \to M$ are étale (this also follows directly from the properties of the morphisms $g'_ i, g_ i, f', f'_ i, f$ listed sofar by arguing in exactly the same way). This first implies that $M' \to M$ is étale and second that the diagram in the lemma is cartesian. We still need to show that $M' \to M$ is separated. To do this we contemplate the diagram

The top horizontal arrow is universally closed as $\mathcal{X}' \to \mathcal{X}$ is separated. The vertical arrows are as in Theorem 106.13.9 (as flat base changes of $\mathcal{X} \to M$) hence universal homeomorphisms. Thus the lower horizontal arrow is universally closed. This (combined with it being an étale monomorphism of algebraic spaces) proves it is a closed immersion as desired. $\square$