Morphisms of Algebraic Stacks

Introduction

In this chapter we introduce some types of morphisms of algebraic stacks. A reference in the case of quasi-separated algebraic stacks with representable diagonal is [LM-B].

The goal is to extend the definition of each of the types of morphisms of algebraic spaces to morphisms of algebraic stacks. Each case is slightly different and it seems best to treat them all separately.

For morphisms of algebraic stacks which are representable by algebraic spaces we have already defined a large number of types of morphisms, see Properties of Stacks, Section 04XB. For each corresponding case in this chapter we have to make sure the definition in the general case is compatible with the definition given there.

Conventions and abuse of language

We continue to use the conventions and the abuse of language introduced in Properties of Stacks, Section 04XA.

Properties of diagonals

The diagonal of an algebraic stack is closely related to the \(\mathit{Isom}\)-sheaves, see Algebraic Stacks, Lemma 045G. By the second defining property of an algebraic stack these \(\mathit{Isom}\)-sheaves are always algebraic spaces.

Lemma

Let \(\mathcal{X}\) be an algebraic stack. Let \(T\) be a scheme and let \(x, y\) be objects of the fibre category of \(\mathcal{X}\) over \(T\). Then the morphism \(\mathit{Isom}_\mathcal{X}(x, y) \to T\) is locally of finite type.

Proof

By Algebraic Stacks, Lemma 04T5 we may assume that \(\mathcal{X} = [U/R]\) for some smooth groupoid in algebraic spaces. By Descent on Spaces, Lemma 041S it suffices to check the property fppf locally on \(T\). Thus we may assume that \(x, y\) come from morphisms \(x', y' : T \to U\). By Groupoids in Spaces, Lemma 044V we see that in this case \(\mathit{Isom}_\mathcal{X}(x, y) = T \times_{(y', x'), U \times_S U} R\). Hence it suffices to prove that \(R \to U \times_S U\) is locally of finite type. This follows from the fact that the composition \(s : R \to U \times_S U \to U\) is smooth (hence locally of finite type, see Morphisms of Spaces, Lemmas 04AJ and 0464) and Morphisms of Spaces, Lemma 0462.

Lemma

Let \(\mathcal{X}\) be an algebraic stack. Let \(T\) be a scheme and let \(x, y\) be objects of the fibre category of \(\mathcal{X}\) over \(T\). Then

1. \(\mathit{Isom}_\mathcal{X}(y, y)\) is a group algebraic space over \(T\), and

2. \(\mathit{Isom}_\mathcal{X}(x, y)\) is a pseudo torsor for \(\mathit{Isom}_\mathcal{X}(y, y)\) over \(T\).

Proof

See Groupoids in Spaces, Definitions 043H and 04TW. The lemma follows immediately from the fact that \(\mathcal{X}\) is a stack in groupoids.

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. The diagonal of \(f\) is the morphism \[\Delta_f : \mathcal{X} \longrightarrow \mathcal{X} \times_\mathcal{Y} \mathcal{X}\] Here are two properties that every diagonal morphism has.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. Then

1. \(\Delta_f\) is representable by algebraic spaces, and

2. \(\Delta_f\) is locally of finite type.

Proof

Let \(T\) be a scheme and let \(a : T \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}\) be a morphism. By definition of the fibre product and the \(2\)-Yoneda lemma the morphism \(a\) is given by a triple \(a = (x, x', \alpha)\) where \(x, x'\) are objects of \(\mathcal{X}\) over \(T\), and \(\alpha : f(x) \to f(x')\) is a morphism in the fibre category of \(\mathcal{Y}\) over \(T\). By definition of an algebraic stack the sheaves \(\mathit{Isom}_\mathcal{X}(x, x')\) and \(\mathit{Isom}_\mathcal{Y}(f(x), f(x'))\) are algebraic spaces over \(T\). In this language \(\alpha\) defines a section of the morphism \(\mathit{Isom}_\mathcal{Y}(f(x), f(x')) \to T\). A \(T'\)-valued point of \(\mathcal{X} \times_{\mathcal{X} \times_\mathcal{Y} \mathcal{X}, a} T\) for \(T' \to T\) a scheme over \(T\) is the same thing as an isomorphism \(x|_{T'} \to x'|_{T'}\) whose image under \(f\) is \(\alpha|_{T'}\). Thus we see that [04XT]\[\begin{equation} \vcenter{ \xymatrix{ \mathcal{X} \times_{\mathcal{X} \times_\mathcal{Y} \mathcal{X}, a} T \ar[d] \ar[r] & \mathit{Isom}_\mathcal{X}(x, x') \ar[d] \\ T\ar[r]^-\alpha & \mathit{Isom}_\mathcal{Y}(f(x), f(x')) } } \end{equation}\] is a fibre square of sheaves over \(T\). In particular we see that \(\mathcal{X} \times_{\mathcal{X} \times_\mathcal{Y} \mathcal{X}, a} T\) is an algebraic space which proves part (1) of the lemma.

To prove the second statement we have to show that the left vertical arrow of Diagram (04XT) is locally of finite type. By Lemma 04XR the algebraic space \(\mathit{Isom}_\mathcal{X}(x, x')\) is locally of finite type over \(T\). Hence the right vertical arrow of Diagram (04XT) is locally of finite type, see Morphisms of Spaces, Lemma 0462. We conclude by Morphisms of Spaces, Lemma 03XH.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks which is representable by algebraic spaces. Then

1. \(\Delta_f\) is representable (by schemes),

2. \(\Delta_f\) is locally of finite type,

3. \(\Delta_f\) is a monomorphism,

4. \(\Delta_f\) is separated, and

5. \(\Delta_f\) is locally quasi-finite.

Proof

We have already seen in Lemma 04XS that \(\Delta_f\) is representable by algebraic spaces. Hence the statements (2) – (5) make sense, see Properties of Stacks, Section 04XB. Also Lemma 04XS guarantees (2) holds. Let \(T \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}\) be a morphism and contemplate Diagram (04XT). By Algebraic Stacks, Lemma 02ZY the right vertical arrow is injective as a map of sheaves, i.e., a monomorphism of algebraic spaces. Hence also the morphism \(T \times_{\mathcal{X} \times_\mathcal{Y} \mathcal{X}} \mathcal{X} \to T\) is a monomorphism. Thus (3) holds. We already know that \(T \times_{\mathcal{X} \times_\mathcal{Y} \mathcal{X}} \mathcal{X} \to T\) is locally of finite type. Thus Morphisms of Spaces, Lemma 0463 allows us to conclude that \(T \times_{\mathcal{X} \times_\mathcal{Y} \mathcal{X}} \mathcal{X} \to T\) is locally quasi-finite and separated. This proves (4) and (5). Finally, Morphisms of Spaces, Proposition 03XX implies that \(T \times_{\mathcal{X} \times_\mathcal{Y} \mathcal{X}} \mathcal{X}\) is a scheme which proves (1).

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks representable by algebraic spaces. Then the following are equivalent

1. \(f\) is separated,

2. \(\Delta_f\) is a closed immersion,

3. \(\Delta_f\) is proper, or

4. \(\Delta_f\) is universally closed.

Proof

The statements “\(f\) is separated”, “\(\Delta_f\) is a closed immersion”, “\(\Delta_f\) is universally closed”, and “\(\Delta_f\) is proper” refer to the notions defined in Properties of Stacks, Section 04XB. Choose a scheme \(V\) and a surjective smooth morphism \(V \to \mathcal{Y}\). Set \(U = \mathcal{X} \times_\mathcal{Y} V\) which is an algebraic space by assumption, and the morphism \(U \to \mathcal{X}\) is surjective and smooth. By Categories, Lemma 04YR and Properties of Stacks, Lemma 04XD we see that for any property \(P\) (as in that lemma) we have: \(\Delta_f\) has \(P\) if and only if \(\Delta_{U/V} : U \to U \times_V U\) has \(P\). Hence the equivalence of (2), (3) and (4) follows from Morphisms of Spaces, Lemma 04Y0 applied to \(U \to V\). Moreover, if (1) holds, then \(U \to V\) is separated and we see that \(\Delta_{U/V}\) is a closed immersion, i.e., (2) holds. Finally, assume (2) holds. Let \(T\) be a scheme, and \(a : T \to \mathcal{Y}\) a morphism. Set \(T' = \mathcal{X} \times_\mathcal{Y} T\). To prove (1) we have to show that the morphism of algebraic spaces \(T' \to T\) is separated. Using Categories, Lemma 04YR once more we see that \(\Delta_{T'/T}\) is the base change of \(\Delta_f\). Hence our assumption (2) implies that \(\Delta_{T'/T}\) is a closed immersion, hence \(T' \to T\) is separated as desired.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks representable by algebraic spaces. Then the following are equivalent

1. \(f\) is quasi-separated,

2. \(\Delta_f\) is quasi-compact, or

3. \(\Delta_f\) is of finite type.

Proof

The statements “\(f\) is quasi-separated”, “\(\Delta_f\) is quasi-compact”, and “\(\Delta_f\) is of finite type” refer to the notions defined in Properties of Stacks, Section 04XB. Note that (2) and (3) are equivalent in view of the fact that \(\Delta_f\) is locally of finite type by Lemma 04YQ (and Algebraic Stacks, Lemma 045F). Choose a scheme \(V\) and a surjective smooth morphism \(V \to \mathcal{Y}\). Set \(U = \mathcal{X} \times_\mathcal{Y} V\) which is an algebraic space by assumption, and the morphism \(U \to \mathcal{X}\) is surjective and smooth. By Categories, Lemma 04YR and Properties of Stacks, Lemma 04XD we see that we have: \(\Delta_f\) is quasi-compact if and only if \(\Delta_{U/V} : U \to U \times_V U\) is quasi-compact. If (1) holds, then \(U \to V\) is quasi-separated and we see that \(\Delta_{U/V}\) is quasi-compact, i.e., (2) holds. Assume (2) holds. Let \(T\) be a scheme, and \(a : T \to \mathcal{Y}\) a morphism. Set \(T' = \mathcal{X} \times_\mathcal{Y} T\). To prove (1) we have to show that the morphism of algebraic spaces \(T' \to T\) is quasi-separated. Using Categories, Lemma 04YR once more we see that \(\Delta_{T'/T}\) is the base change of \(\Delta_f\). Hence our assumption (2) implies that \(\Delta_{T'/T}\) is quasi-compact, hence \(T' \to T\) is quasi-separated as desired.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks representable by algebraic spaces. Then the following are equivalent

1. \(f\) is locally separated, and

2. \(\Delta_f\) is an immersion.

Proof

The statements “\(f\) is locally separated”, and “\(\Delta_f\) is an immersion” refer to the notions defined in Properties of Stacks, Section 04XB. Proof omitted. Hint: Argue as in the proofs of Lemmas 04YS and 04YT.

Separation axioms

Let \(\mathcal{X} = [U/R]\) be a presentation of an algebraic stack. Then the properties of the diagonal of \(\mathcal{X}\) over \(S\), are the properties of the morphism \(j : R \to U \times_S U\). For example, if \(\mathcal{X} = [S/G]\) for some smooth group \(G\) in algebraic spaces over \(S\) then \(j\) is the structure morphism \(G \to S\). Hence the diagonal is not automatically separated itself (contrary to what happens in the case of schemes and algebraic spaces). To say that \([S/G]\) is quasi-separated over \(S\) should certainly imply that \(G \to S\) is quasi-compact, but we hesitate to say that \([S/G]\) is quasi-separated over \(S\) without also requiring the morphism \(G \to S\) to be quasi-separated. In other words, requiring the diagonal morphism to be quasi-compact does not really agree with our intuition for a “quasi-separated algebraic stack”, and we should also require the diagonal itself to be quasi-separated.

What about “separated algebraic stacks”? We have seen in Morphisms of Spaces, Lemma 04Y0 that an algebraic space is separated if and only if the diagonal is proper. This is the condition that is usually used to define separated algebraic stacks too. In the example \([S/G] \to S\) above this means that \(G \to S\) is a proper group scheme. This means algebraic stacks of the form \([\Spec(k)/E]\) are proper over \(k\) where \(E\) is an elliptic curve over \(k\) (insert future reference here). In certain situations it may be more natural to assume the diagonal is finite.

Definition

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks.

1. We say \(f\) is DM if \(\Delta_f\) is unramified¹.

2. We say \(f\) is quasi-DM if \(\Delta_f\) is locally quasi-finite².

3. We say \(f\) is separated if \(\Delta_f\) is proper.

4. We say \(f\) is quasi-separated if \(\Delta_f\) is quasi-compact and quasi-separated.

In this definition we are using that \(\Delta_f\) is representable by algebraic spaces and we are using Properties of Stacks, Section 04XB to make sense out of imposing conditions on \(\Delta_f\). We note that these definitions do not conflict with the already existing notions if \(f\) is representable by algebraic spaces, see Lemmas 04YT and 04YS. There is an interesting way to characterize these conditions by looking at higher diagonals, see Lemma 04Z0.

Definition

Let \(\mathcal{X}\) be an algebraic stack over the base scheme \(S\). Denote \(p : \mathcal{X} \to S\) the structure morphism.

1. We say \(\mathcal{X}\) is DM over \(S\) if \(p : \mathcal{X} \to S\) is DM.

2. We say \(\mathcal{X}\) is quasi-DM over \(S\) if \(p : \mathcal{X} \to S\) is quasi-DM.

3. We say \(\mathcal{X}\) is separated over \(S\) if \(p : \mathcal{X} \to S\) is separated.

4. We say \(\mathcal{X}\) is quasi-separated over \(S\) if \(p : \mathcal{X} \to S\) is quasi-separated.

5. We say \(\mathcal{X}\) is DM if \(\mathcal{X}\) is DM³ over \(\Spec(\mathbf{Z})\).

6. We say \(\mathcal{X}\) is quasi-DM if \(\mathcal{X}\) is quasi-DM over \(\Spec(\mathbf{Z})\).

7. We say \(\mathcal{X}\) is separated if \(\mathcal{X}\) is separated over \(\Spec(\mathbf{Z})\).

8. We say \(\mathcal{X}\) is quasi-separated if \(\mathcal{X}\) is quasi-separated over \(\Spec(\mathbf{Z})\).

In the last 4 definitions we view \(\mathcal{X}\) as an algebraic stack over \(\Spec(\mathbf{Z})\) via Algebraic Stacks, Definition 04X6.

Thus in each case we have an absolute notion and a notion relative to our given base scheme (mention of which is usually suppressed by our abuse of notation introduced in Properties of Stacks, Section 04XA). We will see that (1) \(\Leftrightarrow\) (5) and (2) \(\Leftrightarrow\) (6) in Lemma 050N. We spend some time proving some standard results on these notions.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks.

1. If \(f\) is separated, then \(f\) is quasi-separated.

2. If \(f\) is DM, then \(f\) is quasi-DM.

3. If \(f\) is representable by algebraic spaces, then \(f\) is DM.

Proof

To see (1) note that a proper morphism of algebraic spaces is quasi-compact and quasi-separated, see Morphisms of Spaces, Definition 03ZM. To see (2) note that an unramified morphism of algebraic spaces is locally quasi-finite, see Morphisms of Spaces, Lemma 05W0. Finally (3) follows from Lemma 04YQ.

Lemma

All of the separation axioms listed in Definition 04YW are stable under base change.

Proof

Let \(f : \mathcal{X} \to \mathcal{Y}\) and \(\mathcal{Y}' \to \mathcal{Y}\) be morphisms of algebraic stacks. Let \(f' : \mathcal{Y}' \times_\mathcal{Y} \mathcal{X} \to \mathcal{Y}'\) be the base change of \(f\) by \(\mathcal{Y}' \to \mathcal{Y}\). Then \(\Delta_{f'}\) is the base change of \(\Delta_f\) by the morphism \(\mathcal{X}' \times_{\mathcal{Y}'} \mathcal{X}' \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}\), see Categories, Lemma 04YR. By the results of Properties of Stacks, Section 04XB each of the properties of the diagonal used in Definition 04YW is stable under base change. Hence the lemma is true.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. Let \(W \to \mathcal{Y}\) be a surjective, flat, and locally of finite presentation where \(W\) is an algebraic space. If the base change \(W \times_\mathcal{Y} \mathcal{X} \to W\) has one of the separation properties of Definition 04YW then so does \(f\).

Proof

Denote \(g : W \times_\mathcal{Y} \mathcal{X} \to W\) the base change. Then \(\Delta_g\) is the base change of \(\Delta_f\) by the morphism \(q : W \times_\mathcal{Y} (\mathcal{X} \times_\mathcal{Y} \mathcal{X}) \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}\). Since \(q\) is the base change of \(W \to \mathcal{Y}\) we see that \(q\) is representable by algebraic spaces, surjective, flat, and locally of finite presentation. Hence the result follows from Properties of Stacks, Lemma 06TY.

Lemma

Let \(S\) be a scheme. The property of being quasi-DM over \(S\), quasi-separated over \(S\), or separated over \(S\) (see Definition 050D) is stable under change of base scheme, see Algebraic Stacks, Definition 04X7.

Proof

Follows immediately from Lemma 050F.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Z}\), \(g : \mathcal{Y} \to \mathcal{Z}\) and \(\mathcal{Z} \to \mathcal{T}\) be morphisms of algebraic stacks. Consider the induced morphism \(i : \mathcal{X} \times_\mathcal{Z} \mathcal{Y} \to \mathcal{X} \times_\mathcal{T} \mathcal{Y}\). Then

1. \(i\) is representable by algebraic spaces and locally of finite type,

2. if \(\Delta_{\mathcal{Z}/\mathcal{T}}\) is quasi-separated, then \(i\) is quasi-separated,

3. if \(\Delta_{\mathcal{Z}/\mathcal{T}}\) is separated, then \(i\) is separated,

4. if \(\mathcal{Z} \to \mathcal{T}\) is DM, then \(i\) is unramified,

5. if \(\mathcal{Z} \to \mathcal{T}\) is quasi-DM, then \(i\) is locally quasi-finite,

6. if \(\mathcal{Z} \to \mathcal{T}\) is separated, then \(i\) is proper, and

7. if \(\mathcal{Z} \to \mathcal{T}\) is quasi-separated, then \(i\) is quasi-compact and quasi-separated.

Proof

The following diagram \[\xymatrix{ \mathcal{X} \times_\mathcal{Z} \mathcal{Y} \ar[r]_i \ar[d] & \mathcal{X} \times_\mathcal{T} \mathcal{Y} \ar[d] \\ \mathcal{Z} \ar[r]^-{\Delta_{\mathcal{Z}/\mathcal{T}}} \ar[r] & \mathcal{Z} \times_\mathcal{T} \mathcal{Z} }\] is a \(2\)-fibre product diagram, see Categories, Lemma 04Z1. Hence \(i\) is the base change of the diagonal morphism \(\Delta_{\mathcal{Z}/\mathcal{T}}\). Thus the lemma follows from Lemma 04XS, and the material in Properties of Stacks, Section 04XB.

Lemma

Let \(\mathcal{T}\) be an algebraic stack. Let \(g : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks over \(\mathcal{T}\). Consider the graph \(i : \mathcal{X} \to \mathcal{X} \times_\mathcal{T} \mathcal{Y}\) of \(g\). Then

1. \(i\) is representable by algebraic spaces and locally of finite type,

2. if \(\mathcal{Y} \to \mathcal{T}\) is DM, then \(i\) is unramified,

3. if \(\mathcal{Y} \to \mathcal{T}\) is quasi-DM, then \(i\) is locally quasi-finite,

4. if \(\mathcal{Y} \to \mathcal{T}\) is separated, then \(i\) is proper, and

5. if \(\mathcal{Y} \to \mathcal{T}\) is quasi-separated, then \(i\) is quasi-compact and quasi-separated.

Proof

This is a special case of Lemma 050H applied to the morphism \(\mathcal{X} = \mathcal{X} \times_\mathcal{Y} \mathcal{Y} \to \mathcal{X} \times_\mathcal{T} \mathcal{Y}\).

Lemma

Let \(f : \mathcal{X} \to \mathcal{T}\) be a morphism of algebraic stacks. Let \(s : \mathcal{T} \to \mathcal{X}\) be a morphism such that \(f \circ s\) is \(2\)-isomorphic to \(\text{id}_\mathcal{T}\). Then

1. \(s\) is representable by algebraic spaces and locally of finite type,

2. if \(f\) is DM, then \(s\) is unramified,

3. if \(f\) is quasi-DM, then \(s\) is locally quasi-finite,

4. if \(f\) is separated, then \(s\) is proper, and

5. if \(f\) is quasi-separated, then \(s\) is quasi-compact and quasi-separated.

Proof

This is a special case of Lemma 050I applied to \(g = s\) and \(\mathcal{Y} = \mathcal{T}\) in which case \(i : \mathcal{T} \to \mathcal{T} \times_\mathcal{T} \mathcal{X}\) is \(2\)-isomorphic to \(s\).

Lemma

All of the separation axioms listed in Definition 04YW are stable under composition of morphisms.

Proof

Let \(f : \mathcal{X} \to \mathcal{Y}\) and \(g : \mathcal{Y} \to \mathcal{Z}\) be morphisms of algebraic stacks to which the axiom in question applies. The diagonal \(\Delta_{\mathcal{X}/\mathcal{Z}}\) is the composition \[\mathcal{X} \longrightarrow \mathcal{X} \times_\mathcal{Y} \mathcal{X} \longrightarrow \mathcal{X} \times_\mathcal{Z} \mathcal{X}.\] Our separation axiom is defined by requiring the diagonal to have some property \(\mathcal{P}\). By Lemma 050H above we see that the second arrow also has this property. Hence the lemma follows since the composition of morphisms which are representable by algebraic spaces with property \(\mathcal{P}\) also is a morphism with property \(\mathcal{P}\), see our general discussion in Properties of Stacks, Section 04XB and Morphisms of Spaces, Lemmas 03ZI, 03XK, 04XY, 03HG, and 03KQ.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks over the base scheme \(S\).

1. If \(\mathcal{Y}\) is DM over \(S\) and \(f\) is DM, then \(\mathcal{X}\) is DM over \(S\).

2. If \(\mathcal{Y}\) is quasi-DM over \(S\) and \(f\) is quasi-DM, then \(\mathcal{X}\) is quasi-DM over \(S\).

3. If \(\mathcal{Y}\) is separated over \(S\) and \(f\) is separated, then \(\mathcal{X}\) is separated over \(S\).

4. If \(\mathcal{Y}\) is quasi-separated over \(S\) and \(f\) is quasi-separated, then \(\mathcal{X}\) is quasi-separated over \(S\).

5. If \(\mathcal{Y}\) is DM and \(f\) is DM, then \(\mathcal{X}\) is DM.

6. If \(\mathcal{Y}\) is quasi-DM and \(f\) is quasi-DM, then \(\mathcal{X}\) is quasi-DM.

7. If \(\mathcal{Y}\) is separated and \(f\) is separated, then \(\mathcal{X}\) is separated.

8. If \(\mathcal{Y}\) is quasi-separated and \(f\) is quasi-separated, then \(\mathcal{X}\) is quasi-separated.

Proof

Parts (1), (2), (3), and (4) follow immediately from Lemma 050K and Definition 050D. For (5), (6), (7), and (8) think of \(\mathcal{X}\) and \(\mathcal{Y}\) as algebraic stacks over \(\Spec(\mathbf{Z})\) and apply Lemma 050K. Details omitted.

The following lemma is a bit different to the analogue for algebraic spaces. To compare take a look at Morphisms of Spaces, Lemma 03KR.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) and \(g : \mathcal{Y} \to \mathcal{Z}\) be morphisms of algebraic stacks.

1. If \(g \circ f\) is DM then so is \(f\).

2. If \(g \circ f\) is quasi-DM then so is \(f\).

3. If \(g \circ f\) is separated and \(\Delta_g\) is separated, then \(f\) is separated.

4. If \(g \circ f\) is quasi-separated and \(\Delta_g\) is quasi-separated, then \(f\) is quasi-separated.

Proof

Consider the factorization \[\mathcal{X} \to \mathcal{X} \times_\mathcal{Y} \mathcal{X} \to \mathcal{X} \times_\mathcal{Z} \mathcal{X}\] of the diagonal morphism of \(g \circ f\). Both morphisms are representable by algebraic spaces, see Lemmas 04XS and 050H. Hence for any scheme \(T\) and morphism \(T \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}\) we get morphisms of algebraic spaces \[A = \mathcal{X} \times_{(\mathcal{X} \times_\mathcal{Z} \mathcal{X})} T \longrightarrow B = (\mathcal{X} \times_\mathcal{Y} \mathcal{X}) \times_{(\mathcal{X} \times_\mathcal{Z} \mathcal{X})} T \longrightarrow T\] If \(g \circ f\) is DM (resp. quasi-DM), then the composition \(A \to T\) is unramified (resp. locally quasi-finite). Hence \(A \to B\) is unramified (resp. locally quasi-finite) by Morphisms of Spaces, Lemma 06G6 (resp. Morphisms of Spaces, Lemma 03XN). Now consider the diagram \[\xymatrix{ A \times_B T \ar[r] \ar[d] & T \ar[d] \\ A \ar[r] \ar[d] & B \ar[r] \ar[d] & T \ar[d] \\ \mathcal{X} \ar[r] & \mathcal{X} \times_\mathcal{Y} \mathcal{X} \ar[r] & \mathcal{X} \times_\mathcal{Z} \mathcal{X} }\] with all squares \(2\)-cartesian where the arrow \(T \to B\) comes from the arrow \(T \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}\) we started with. The arrow \(A \times_B T \to T\) is unramified (resp. locally quasi-finite) by Morphisms of Spaces, Lemma 03ZJ (resp. Morphisms of Spaces, Lemma 03XL). Noting that \(A \times_B T\) is equal to \(\mathcal{X} \times_{\mathcal{X} \times_\mathcal{Y} \mathcal{X}} T\) we conclude that \(f\) is DM (resp. quasi-DM). This proves (1) and (2).

Proof of (4). Assume \(g \circ f\) is quasi-separated and \(\Delta_g\) is quasi-separated. Consider the factorization \[\mathcal{X} \to \mathcal{X} \times_\mathcal{Y} \mathcal{X} \to \mathcal{X} \times_\mathcal{Z} \mathcal{X}\] of the diagonal morphism of \(g \circ f\). Both morphisms are representable by algebraic spaces and the second one is quasi-separated, see Lemmas 04XS and 050H. Hence for any scheme \(T\) and morphism \(T \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}\) we get morphisms of algebraic spaces \[A = \mathcal{X} \times_{(\mathcal{X} \times_\mathcal{Z} \mathcal{X})} T \longrightarrow B = (\mathcal{X} \times_\mathcal{Y} \mathcal{X}) \times_{(\mathcal{X} \times_\mathcal{Z} \mathcal{X})} T \longrightarrow T\] such that \(B \to T\) is quasi-separated. The composition \(A \to T\) is quasi-compact and quasi-separated as we have assumed that \(g \circ f\) is quasi-separated. Hence \(A \to B\) is quasi-separated by Morphisms of Spaces, Lemma 03KR. And \(A \to B\) is quasi-compact by Morphisms of Spaces, Lemma 03KS. Thus \(f\) is quasi-separated.

Proof of (3). Assume \(g \circ f\) is separated and \(\Delta_g\) is separated. Consider the factorization \[\mathcal{X} \to \mathcal{X} \times_\mathcal{Y} \mathcal{X} \to \mathcal{X} \times_\mathcal{Z} \mathcal{X}\] of the diagonal morphism of \(g \circ f\). Both morphisms are representable by algebraic spaces and the second one is separated, see Lemmas 04XS and 050H. Hence for any scheme \(T\) and morphism \(T \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}\) we get morphisms of algebraic spaces \[A = \mathcal{X} \times_{(\mathcal{X} \times_\mathcal{Z} \mathcal{X})} T \longrightarrow B = (\mathcal{X} \times_\mathcal{Y} \mathcal{X}) \times_{(\mathcal{X} \times_\mathcal{Z} \mathcal{X})} T \longrightarrow T\] such that \(B \to T\) is separated. The composition \(A \to T\) is proper as we have assumed that \(g \circ f\) is quasi-separated. Hence \(A \to B\) is proper by Morphisms of Spaces, Lemma 04NX which means that \(f\) is separated.

Lemma

Let \(\mathcal{X}\) be an algebraic stack over the base scheme \(S\).

1. \(\mathcal{X}\) is DM \(\Leftrightarrow\) \(\mathcal{X}\) is DM over \(S\).

2. \(\mathcal{X}\) is quasi-DM \(\Leftrightarrow\) \(\mathcal{X}\) is quasi-DM over \(S\).

3. If \(\mathcal{X}\) is separated, then \(\mathcal{X}\) is separated over \(S\).

4. If \(\mathcal{X}\) is quasi-separated, then \(\mathcal{X}\) is quasi-separated over \(S\).

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks over the base scheme \(S\).

1. If \(\mathcal{X}\) is DM over \(S\), then \(f\) is DM.

2. If \(\mathcal{X}\) is quasi-DM over \(S\), then \(f\) is quasi-DM.

3. If \(\mathcal{X}\) is separated over \(S\) and \(\Delta_{\mathcal{Y}/S}\) is separated, then \(f\) is separated.

4. If \(\mathcal{X}\) is quasi-separated over \(S\) and \(\Delta_{\mathcal{Y}/S}\) is quasi-separated, then \(f\) is quasi-separated.

Proof

Parts (5), (6), (7), and (8) follow immediately from Lemma 050M and Spaces, Definition 02X5. To prove (3) and (4) think of \(X\) and \(Y\) as algebraic stacks over \(\Spec(\mathbf{Z})\) and apply Lemma 050M. Similarly, to prove (1) and (2), think of \(\mathcal{X}\) as an algebraic stack over \(\Spec(\mathbf{Z})\) consider the morphisms \[\mathcal{X} \longrightarrow \mathcal{X} \times_S \mathcal{X} \longrightarrow \mathcal{X} \times_{\Spec(\mathbf{Z})} \mathcal{X}\] Both arrows are representable by algebraic spaces. The second arrow is unramified and locally quasi-finite as the base change of the immersion \(\Delta_{S/\mathbf{Z}}\). Hence the composition is unramified (resp. locally quasi-finite) if and only if the first arrow is unramified (resp. locally quasi-finite), see Morphisms of Spaces, Lemmas 03ZI and 06G6 (resp. Morphisms of Spaces, Lemmas 03XK and 03XN).

Lemma

Let \(\mathcal{X}\) be an algebraic stack. Let \(W\) be an algebraic space, and let \(f : W \to \mathcal{X}\) be a surjective, flat, locally finitely presented morphism.

1. If \(f\) is unramified (i.e., étale, i.e., \(\mathcal{X}\) is Deligne-Mumford), then \(\mathcal{X}\) is DM.

2. If \(f\) is locally quasi-finite, then \(\mathcal{X}\) is quasi-DM.

Proof

Note that if \(f\) is unramified, then it is étale by Morphisms of Spaces, Lemma 06LU. This explains the parenthetical remark in (1). Assume \(f\) is unramified (resp. locally quasi-finite). We have to show that \(\Delta_\mathcal{X} : \mathcal{X} \to \mathcal{X} \times \mathcal{X}\) is unramified (resp. locally quasi-finite). Note that \(W \times W \to \mathcal{X} \times \mathcal{X}\) is also surjective, flat, and locally of finite presentation. Hence it suffices to show that \[W \times_{\mathcal{X} \times \mathcal{X}, \Delta_\mathcal{X}} \mathcal{X} = W \times_\mathcal{X} W \longrightarrow W \times W\] is unramified (resp. locally quasi-finite), see Properties of Stacks, Lemma 04XD. By assumption the morphism \(\text{pr}_i : W \times_\mathcal{X} W \to W\) is unramified (resp. locally quasi-finite). Hence the displayed arrow is unramified (resp. locally quasi-finite) by Morphisms of Spaces, Lemma 06G6 (resp. Morphisms of Spaces, Lemma 03XN).

Lemma

A monomorphism of algebraic stacks is separated and DM. The same is true for immersions of algebraic stacks.

Proof

If \(f : \mathcal{X} \to \mathcal{Y}\) is a monomorphism of algebraic stacks, then \(\Delta_f\) is an isomorphism, see Properties of Stacks, Lemma 04ZZ. Since an isomorphism of algebraic spaces is proper and unramified we see that \(f\) is separated and DM. The second assertion follows from the first as an immersion is a monomorphism, see Properties of Stacks, Lemma 0504.

Lemma

Let \(\mathcal{X}\) be an algebraic stack. Let \(x \in |\mathcal{X}|\). Assume the residual gerbe \(\mathcal{Z}_x\) of \(\mathcal{X}\) at \(x\) exists. If \(\mathcal{X}\) is DM, resp. quasi-DM, resp. separated, resp. quasi-separated, then so is \(\mathcal{Z}_x\).

Proof

This is true because \(\mathcal{Z}_x \to \mathcal{X}\) is a monomorphism hence DM and separated by Lemma 06MY. Apply Lemma 050L to conclude.

Inertia stacks

The (relative) inertia stack of a stack in groupoids is defined in Stacks, Section 036X. The actual construction, in the setting of fibred categories, and some of its properties is in Categories, Section 04Z2.

Lemma

Let \(\mathcal{X}\) be an algebraic stack. Then the inertia stack \(\mathcal{I}_\mathcal{X}\) is an algebraic stack as well. The morphism \[\mathcal{I}_\mathcal{X} \longrightarrow \mathcal{X}\] is representable by algebraic spaces and locally of finite type. More generally, let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. Then the relative inertia \(\mathcal{I}_{\mathcal{X}/\mathcal{Y}}\) is an algebraic stack and the morphism \[\mathcal{I}_{\mathcal{X}/\mathcal{Y}} \longrightarrow \mathcal{X}\] is representable by algebraic spaces and locally of finite type.

Proof

By Categories, Lemma 034H there are equivalences \[\mathcal{I}_\mathcal{X} \to \mathcal{X} \times_{\Delta, \mathcal{X} \times_S \mathcal{X}, \Delta} \mathcal{X} \quad\text{and}\quad \mathcal{I}_{\mathcal{X}/\mathcal{Y}} \to \mathcal{X} \times_{\Delta, \mathcal{X} \times_\mathcal{Y} \mathcal{X}, \Delta} \mathcal{X}\] which shows that the inertia stacks are algebraic stacks. Let \(T \to \mathcal{X}\) be a morphism given by the object \(x\) of the fibre category of \(\mathcal{X}\) over \(T\). Then we get a \(2\)-fibre product square \[\xymatrix{ \mathit{Isom}_\mathcal{X}(x, x) \ar[d] \ar[r] & \mathcal{I}_\mathcal{X} \ar[d] \\ T \ar[r]^x & \mathcal{X} }\] This follows immediately from the definition of \(\mathcal{I}_\mathcal{X}\). Since \(\mathit{Isom}_\mathcal{X}(x, x)\) is always an algebraic space locally of finite type over \(T\) (see Lemma 04XR) we conclude that \(\mathcal{I}_\mathcal{X} \to \mathcal{X}\) is representable by algebraic spaces and locally of finite type. Finally, for the relative inertia we get \[\vcenter{ \xymatrix{ \mathit{Isom}_\mathcal{X}(x, x) \ar[d] & K \ar[l] \ar[d] \ar[r] & \mathcal{I}_{\mathcal{X}/\mathcal{Y}} \ar[d] \\ \mathit{Isom}_\mathcal{Y}(f(x), f(x)) & T \ar[l]_-e \ar[r]^x & \mathcal{X} } }\] with both squares \(2\)-fibre products. This follows from Categories, Lemma 04Z6. The left vertical arrow is a morphism of algebraic spaces locally of finite type over \(T\), and hence is locally of finite type, see Morphisms of Spaces, Lemma 0462. Thus \(K\) is an algebraic space and \(K \to T\) is locally of finite type. This proves the assertion on the relative inertia.

Remark

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. In Properties of Stacks, Remark 04ZQ we have seen that the \(2\)-category of morphisms \(\mathcal{Z} \to \mathcal{X}\) representable by algebraic spaces with target \(\mathcal{X}\) forms a category. In this category the inertia stack of \(\mathcal{X}/\mathcal{Y}\) is a group object. Recall that an object of \(\mathcal{I}_{\mathcal{X}/\mathcal{Y}}\) is just a pair \((x, \alpha)\) where \(x\) is an object of \(\mathcal{X}\) and \(\alpha\) is an automorphism of \(x\) in the fibre category of \(\mathcal{X}\) that \(x\) lives in with \(f(\alpha) = \text{id}\). The composition \[c : \mathcal{I}_{\mathcal{X}/\mathcal{Y}} \times_\mathcal{X} \mathcal{I}_{\mathcal{X}/\mathcal{Y}} \longrightarrow \mathcal{I}_{\mathcal{X}/\mathcal{Y}}\] is given by the rule on objects \[((x, \alpha), (x', \alpha'), \beta) \mapsto (x, \alpha \circ \beta^{-1} \circ \alpha' \circ \beta)\] which makes sense as \(\beta : x \to x'\) is an isomorphism in the fibre category by our definition of fibre products. The neutral element \(e : \mathcal{X} \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}\) is given by the functor \(x \mapsto (x, \text{id}_x)\). We omit the proof that the axioms of a group object hold.

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks and let \(\mathcal{I}_{\mathcal{X}/\mathcal{Y}}\) be its inertia stack. Let \(T\) be a scheme and let \(x\) be an object of \(\mathcal{X}\) over \(T\). Set \(y = f(x)\). We have seen in the proof of Lemma 050Q that for any scheme \(T\) and object \(x\) of \(\mathcal{X}\) over \(T\) there is an exact sequence of sheaves of groups [0CPJ]\[\begin{equation} 0 \to \mathit{Isom}_{\mathcal{X}/\mathcal{Y}}(x, x) \to \mathit{Isom}_\mathcal{X}(x, x) \to \mathit{Isom}_\mathcal{Y}(y, y) \end{equation}\] The group structure on the second and third term is the one defined in Lemma 04YP and the sequence gives a meaning to the first term. Also, there is a canonical cartesian square \[\xymatrix{ \mathit{Isom}_{\mathcal{X}/\mathcal{Y}}(x, x) \ar[d] \ar[r] & \mathcal{I}_{\mathcal{X}/\mathcal{Y}} \ar[d] \\ T \ar[r]^x & \mathcal{X} }\] In fact, the group structure on \(\mathcal{I}_{\mathcal{X}/\mathcal{Y}}\) discussed in Remark 050R induces the group structure on \(\mathit{Isom}_{\mathcal{X}/\mathcal{Y}}(x, x)\). This allows us to define the sheaf \(\mathit{Isom}_{\mathcal{X}/\mathcal{Y}}(x, x)\) also for morphisms from algebraic spaces to \(\mathcal{X}\). We formalize this in the following definition.

Definition

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. Let \(Z\) be an algebraic space.

1. Let \(x : Z \to \mathcal{X}\) be a morphism. We set \[\mathit{Isom}_{\mathcal{X}/\mathcal{Y}}(x, x) = Z \times_{x, \mathcal{X}} \mathcal{I}_{\mathcal{X}/\mathcal{Y}}\] We endow it with the structure of a group algebraic space over \(Z\) by pulling back the composition law discussed in Remark 050R. We will sometimes refer to \(\mathit{Isom}_{\mathcal{X}/\mathcal{Y}}(x, x)\) as the relative sheaf of automorphisms of \(x\).

2. Let \(x_1, x_2 : Z \to \mathcal{X}\) be morphisms. Set \(y_i = f \circ x_i\). Let \(\alpha : y_1 \to y_2\) be a \(2\)-morphism. Then \(\alpha\) determines a morphism \(\Delta^\alpha : Z \to Z \times_{y_1, \mathcal{Y}, y_2} Z\) and we set \[\mathit{Isom}_{\mathcal{X}/\mathcal{Y}}^\alpha(x_1, x_2) = (Z \times_{x_1, \mathcal{X}, x_2} Z) \times_{Z \times_{y_1, \mathcal{Y}, y_2} Z, \Delta^\alpha} Z.\] We will sometimes refer to \(\mathit{Isom}_{\mathcal{X}/\mathcal{Y}}^\alpha(x_1, x_2)\) as the relative sheaf of isomorphisms from \(x_1\) to \(x_2\).

If \(\mathcal{Y} = \Spec(\mathbf{Z})\) or more generally when \(\mathcal{Y}\) is an algebraic space, then we use the notation \(\mathit{Isom}_\mathcal{X}(x, x)\) and \(\mathit{Isom}_\mathcal{X}(x_1, x_2)\) and we use the terminology sheaf of automorphisms of \(x\) and sheaf of isomorphisms from \(x_1\) to \(x_2\).

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. Let \(Z\) be an algebraic space and let \(x_i : Z \to \mathcal{X}\), \(i = 1, 2\) be morphisms. Then

1. \(\mathit{Isom}_{\mathcal{X}/\mathcal{Y}}(x_2, x_2)\) is a group algebraic space over \(Z\),

2. there is an exact sequence of groups \[0 \to \mathit{Isom}_{\mathcal{X}/\mathcal{Y}}(x_2, x_2) \to \mathit{Isom}_\mathcal{X}(x_2, x_2) \to \mathit{Isom}_\mathcal{Y}(f \circ x_2, f \circ x_2)\]

3. there is a map of algebraic spaces \(\mathit{Isom}_\mathcal{X}(x_1, x_2) \to \mathit{Isom}_\mathcal{Y}(f \circ x_1, f \circ x_2)\) such that for any \(2\)-morphism \(\alpha : f \circ x_1 \to f \circ x_2\) we obtain a cartesian diagram \[\xymatrix{ \mathit{Isom}_{\mathcal{X}/\mathcal{Y}}^\alpha(x_1, x_2) \ar[d] \ar[r] & Z \ar[d]^\alpha \\ \mathit{Isom}_\mathcal{X}(x_1, x_2) \ar[r] & \mathit{Isom}_\mathcal{Y}(f \circ x_1, f \circ x_2) }\]

4. for any \(2\)-morphism \(\alpha : f \circ x_1 \to f \circ x_2\) the algebraic space \(\mathit{Isom}_{\mathcal{X}/\mathcal{Y}}^\alpha(x_1, x_2)\) is a pseudo torsor for \(\mathit{Isom}_{\mathcal{X}/\mathcal{Y}}(x_2, x_2)\) over \(Z\).

Proof

Part (1) follows from Definition 06PP. Part (2) comes from the exact sequence (0CPJ) étale locally on \(Z\). Part (3) can be seen by unwinding the definitions. Locally on \(Z\) in the étale topology part (4) reduces to part (2) of Lemma 04YP.

Lemma

Let \(\pi : \mathcal{X} \to \mathcal{Y}\) and \(f : \mathcal{Y}' \to \mathcal{Y}\) be morphisms of algebraic stacks. Set \(\mathcal{X}' = \mathcal{X} \times_\mathcal{Y} \mathcal{Y}'\). Then both squares in the diagram \[\xymatrix{ \mathcal{I}_{\mathcal{X}'/\mathcal{Y}'} \ar[r] \ar[d]_{ \text{Categories, Equation}\ (\href{categories.html#categories-equation-functorial}{04Z4}) } & \mathcal{X}' \ar[r]_{\pi'} \ar[d] & \mathcal{Y}' \ar[d]^f \\ \mathcal{I}_{\mathcal{X}/\mathcal{Y}} \ar[r] & \mathcal{X} \ar[r]^\pi & \mathcal{Y} }\] are fibre product squares.

Proof

The inertia stack \(\mathcal{I}_{\mathcal{X}'/\mathcal{Y}'}\) is defined as the category of pairs \((x', \alpha')\) where \(x'\) is an object of \(\mathcal{X}'\) and \(\alpha'\) is an automorphism of \(x'\) with \(\pi'(\alpha') = \text{id}\), see Categories, Section 04Z2. Suppose that \(x'\) lies over the scheme \(U\) and maps to the object \(x\) of \(\mathcal{X}\). By the construction of the \(2\)-fibre product in Categories, Lemma 0040 we see that \(x' = (U, x, y', \beta)\) where \(y'\) is an object of \(\mathcal{Y}'\) over \(U\) and \(\beta\) is an isomorphism \(\beta : \pi(x) \to f(y')\) in the fibre category of \(\mathcal{Y}\) over \(U\). By the very construction of the \(2\)-fibre product the automorphism \(\alpha'\) is a pair \((\alpha, \gamma)\) where \(\alpha\) is an automorphism of \(x\) over \(U\) and \(\gamma\) is an automorphism of \(y'\) over \(U\) such that \(\alpha\) and \(\gamma\) are compatible via \(\beta\). The condition \(\pi'(\alpha') = \text{id}\) signifies that \(\gamma = \text{id}\) whereupon the condition that \(\alpha, \beta, \gamma\) are compatible is exactly the condition \(\pi(\alpha) = \text{id}\), i.e., means exactly that \((x, \alpha)\) is an object of \(\mathcal{I}_{\mathcal{X}/\mathcal{Y}}\). In this way we see that the left square is a fibre product square (some details omitted).

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a monomorphism of algebraic stacks. Then the diagram \[\xymatrix{ \mathcal{I}_\mathcal{X} \ar[r] \ar[d] & \mathcal{X} \ar[d] \\ \mathcal{I}_\mathcal{Y} \ar[r] & \mathcal{Y} }\] is a fibre product square.

Proof

This follows immediately from the fact that \(f\) is fully faithful (see Properties of Stacks, Lemma 04ZZ) and the definition of the inertia in Categories, Section 04Z2. Namely, an object of \(\mathcal{I}_\mathcal{X}\) over a scheme \(T\) is the same thing as a pair \((x, \alpha)\) consisting of an object \(x\) of \(\mathcal{X}\) over \(T\) and a morphism \(\alpha : x \to x\) in the fibre category of \(\mathcal{X}\) over \(T\). As \(f\) is fully faithful we see that \(\alpha\) is the same thing as a morphism \(\beta : f(x) \to f(x)\) in the fibre category of \(\mathcal{Y}\) over \(T\). Hence we can think of objects of \(\mathcal{I}_\mathcal{X}\) over \(T\) as triples \(((y, \beta), x, \gamma)\) where \(y\) is an object of \(\mathcal{Y}\) over \(T\), \(\beta : y \to y\) in \(\mathcal{Y}_T\) and \(\gamma : y \to f(x)\) is an isomorphism over \(T\), i.e., an object of \(\mathcal{I}_\mathcal{Y} \times_\mathcal{Y} \mathcal{X}\) over \(T\).

Lemma

Let \(\mathcal{X}\) be an algebraic stack. Let \([U/R] \to \mathcal{X}\) be a presentation. Let \(G/U\) be the stabilizer group algebraic space associated to the groupoid \((U, R, s, t, c)\). Then \[\xymatrix{ G \ar[d] \ar[r] & U \ar[d] \\ \mathcal{I}_\mathcal{X} \ar[r] & \mathcal{X} }\] is a fibre product diagram.

Proof

Immediate from Groupoids in Spaces, Lemma 06PC.

Higher diagonals

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. In this situation it makes sense to consider not only the diagonal \[\Delta_f : \mathcal{X} \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}\] but also the diagonal of the diagonal, i.e., the morphism \[\Delta_{\Delta_f} : \mathcal{X} \longrightarrow \mathcal{X} \times_{(\mathcal{X} \times_\mathcal{Y} \mathcal{X})} \mathcal{X}\] Because of this we sometimes use the following terminology. We denote \(\Delta_{f, 0} = f\) the zeroth diagonal, we denote \(\Delta_{f, 1} = \Delta_f\) the first diagonal, and we denote \(\Delta_{f, 2} = \Delta_{\Delta_f}\) the second diagonal. Note that \(\Delta_{f, 1}\) is representable by algebraic spaces and locally of finite type, see Lemma 04XS. Hence \(\Delta_{f, 2}\) is representable, a monomorphism, locally of finite type, separated, and locally quasi-finite, see Lemma 04YQ.

We can describe the second diagonal using the relative inertia stack. Namely, the fibre product \(\mathcal{X} \times_{(\mathcal{X} \times_\mathcal{Y} \mathcal{X})} \mathcal{X}\) is equivalent to the relative inertia stack \(\mathcal{I}_{\mathcal{X}/\mathcal{Y}}\) by Categories, Lemma 034H. Moreover, via this identification the second diagonal becomes the neutral section \[\Delta_{f, 2} = e : \mathcal{X} \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}\] of the relative inertia stack. By analogy with what happens for groupoids in algebraic spaces (Groupoids in Spaces, Lemma 0455) we have the following equivalences.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks.

1. The following are equivalent

   1. \(\mathcal{I}_{\mathcal{X}/\mathcal{Y}} \to \mathcal{X}\) is separated,

   2. \(\Delta_{f, 1} = \Delta_f : \mathcal{X} \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}\) is separated, and

   3. \(\Delta_{f, 2} = e : \mathcal{X} \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}\) is a closed immersion.

2. The following are equivalent

   1. \(\mathcal{I}_{\mathcal{X}/\mathcal{Y}} \to \mathcal{X}\) is quasi-separated,

   2. \(\Delta_{f, 1} = \Delta_f : \mathcal{X} \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}\) is quasi-separated, and

   3. \(\Delta_{f, 2} = e : \mathcal{X} \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}\) is quasi-compact.

3. The following are equivalent

   1. \(\mathcal{I}_{\mathcal{X}/\mathcal{Y}} \to \mathcal{X}\) is locally separated,

   2. \(\Delta_{f, 1} = \Delta_f : \mathcal{X} \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}\) is locally separated, and

   3. \(\Delta_{f, 2} = e : \mathcal{X} \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}\) is an immersion.

4. The following are equivalent

   1. \(\mathcal{I}_{\mathcal{X}/\mathcal{Y}} \to \mathcal{X}\) is unramified,

   2. \(f\) is DM.

5. The following are equivalent

   1. \(\mathcal{I}_{\mathcal{X}/\mathcal{Y}} \to \mathcal{X}\) is locally quasi-finite,

   2. \(f\) is quasi-DM.

Proof

Proof of (1), (2), and (3). Choose an algebraic space \(U\) and a surjective smooth morphism \(U \to \mathcal{X}\). Then \(G = U \times_\mathcal{X} \mathcal{I}_{\mathcal{X}/\mathcal{Y}}\) is an algebraic space over \(U\) (Lemma 050Q). In fact, \(G\) is a group algebraic space over \(U\) by the group law on relative inertia constructed in Remark 050R. Moreover, \(G \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}\) is surjective and smooth as a base change of \(U \to \mathcal{X}\). Finally, the base change of \(e : \mathcal{X} \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}\) by \(G \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}\) is the identity \(U \to G\) of \(G/U\). Thus the equivalence of (a) and (c) follows from Groupoids in Spaces, Lemma 06P6. Since \(\Delta_{f, 2}\) is the diagonal of \(\Delta_f\) we have (b) \(\Leftrightarrow\) (c) by definition.

Proof of (4) and (5). Recall that (4)(b) means \(\Delta_f\) is unramified and (5)(b) means that \(\Delta_f\) is locally quasi-finite. Choose a scheme \(Z\) and a morphism \(a : Z \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}\). Then \(a = (x_1, x_2, \alpha)\) where \(x_i : Z \to \mathcal{X}\) and \(\alpha : f \circ x_1 \to f \circ x_2\) is a \(2\)-morphism. Recall that \[\vcenter{ \xymatrix{ \mathit{Isom}_{\mathcal{X}/\mathcal{Y}}^\alpha(x_1, x_2) \ar[d] \ar[r] & Z \ar[d] \\ \mathcal{X} \ar[r]^{\Delta_f} & \mathcal{X} \times_\mathcal{Y} \mathcal{X} } } \quad\text{and}\quad \vcenter{ \xymatrix{ \mathit{Isom}_{\mathcal{X}/\mathcal{Y}}(x_2, x_2) \ar[d] \ar[r] & Z \ar[d]^{x_2} \\ \mathcal{I}_{\mathcal{X}/\mathcal{Y}} \ar[r] & \mathcal{X} } }\] are cartesian squares. By Lemma 0CPK the algebraic space \(\mathit{Isom}_{\mathcal{X}/\mathcal{Y}}^\alpha(x_1, x_2)\) is a pseudo torsor for \(\mathit{Isom}_{\mathcal{X}/\mathcal{Y}}(x_2, x_2)\) over \(Z\). Thus the equivalences in (4) and (5) follow from Groupoids in Spaces, Lemma 0DSK.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. The following are equivalent:

1. the morphism \(f\) is representable by algebraic spaces,

2. the second diagonal of \(f\) is an isomorphism,

3. the group stack \(\mathcal{I}_{\mathcal{X}/\mathcal{Y}}\) is trivial over \(\mathcal X\), and

4. for a scheme \(T\) and a morphism \(x : T \to \mathcal{X}\) the kernel of \(\mathit{Isom}_\mathcal{X}(x, x) \to \mathit{Isom}_\mathcal{Y}(f(x), f(x))\) is trivial.

Proof

We first prove the equivalence of (1) and (2). Namely, \(f\) is representable by algebraic spaces if and only if \(f\) is faithful, see Algebraic Stacks, Lemma 04Y5. On the other hand, \(f\) is faithful if and only if for every object \(x\) of \(\mathcal{X}\) over a scheme \(T\) the functor \(f\) induces an injection \(\mathit{Isom}_\mathcal{X}(x, x) \to \mathit{Isom}_\mathcal{Y}(f(x), f(x))\), which happens if and only if the kernel \(K\) is trivial, which happens if and only if \(e : T \to K\) is an isomorphism for every \(x : T \to \mathcal{X}\). Since \(K = T \times_{x, \mathcal{X}} \mathcal{I}_{\mathcal{X}/\mathcal{Y}}\) as discussed above, this proves the equivalence of (1) and (2). To prove the equivalence of (2) and (3), by the discussion above, it suffices to note that a group stack is trivial if and only if its identity section is an isomorphism. Finally, the equivalence of (3) and (4) follows from the definitions: in the proof of Lemma 050Q we have seen that the kernel in (4) corresponds to the fibre product \(T \times_{x, \mathcal{X}} \mathcal{I}_{\mathcal{X}/\mathcal{Y}}\) over \(T\).

This lemma leads to the following hierarchy for morphisms of algebraic stacks.

Lemma

A morphism \(f : \mathcal{X} \to \mathcal{Y}\) of algebraic stacks is

1. a monomorphism if and only if \(\Delta_{f, 1}\) is an isomorphism, and

2. representable by algebraic spaces if and only if \(\Delta_{f, 1}\) is a monomorphism.

Moreover, the second diagonal \(\Delta_{f, 2}\) is always a monomorphism.

Proof

Recall from Properties of Stacks, Lemma 04ZZ that a morphism of algebraic stacks is a monomorphism if and only if its diagonal is an isomorphism of stacks. Thus Lemma 04YY can be rephrased as saying that a morphism is representable by algebraic spaces if the diagonal is a monomorphism. In particular, it shows that condition (3) of Lemma 04YQ is actually an if and only if, i.e., a morphism of algebraic stacks is representable by algebraic spaces if and only if its diagonal is a monomorphism.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. Then

1. \(\Delta_{f, 1}\) separated \(\Leftrightarrow\) \(\Delta_{f, 2}\) closed immersion \(\Leftrightarrow\) \(\Delta_{f, 2}\) proper \(\Leftrightarrow\) \(\Delta_{f, 2}\) universally closed,

2. \(\Delta_{f, 1}\) quasi-separated \(\Leftrightarrow\) \(\Delta_{f, 2}\) finite type \(\Leftrightarrow\) \(\Delta_{f, 2}\) quasi-compact, and

3. \(\Delta_{f, 1}\) locally separated \(\Leftrightarrow\) \(\Delta_{f, 2}\) immersion.

Proof

Follows from Lemmas 04YS, 04YT, and 04YU applied to \(\Delta_{f, 1}\).

The following lemma is kind of cute and it may suggest a generalization of these conditions to higher algebraic stacks.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. Then

1. \(f\) is separated if and only if \(\Delta_{f, 1}\) and \(\Delta_{f, 2}\) are universally closed, and

2. \(f\) is quasi-separated if and only if \(\Delta_{f, 1}\) and \(\Delta_{f, 2}\) are quasi-compact.

3. \(f\) is quasi-DM if and only if \(\Delta_{f, 1}\) and \(\Delta_{f, 2}\) are locally quasi-finite.

4. \(f\) is DM if and only if \(\Delta_{f, 1}\) and \(\Delta_{f, 2}\) are unramified.

Proof

Proof of (1). Assume that \(\Delta_{f, 2}\) and \(\Delta_{f, 1}\) are universally closed. Then \(\Delta_{f, 1}\) is separated and universally closed by Lemma 04YZ. By Morphisms of Spaces, Lemma 04XW and Algebraic Stacks, Lemma 045F we see that \(\Delta_{f, 1}\) is quasi-compact. Hence it is quasi-compact, separated, universally closed and locally of finite type (by Lemma 04XS) so proper. This proves “\(\Leftarrow\)” of (1). The proof of the implication in the other direction is omitted.

Proof of (2). This follows immediately from Lemma 04YZ.

Proof of (3). This follows from the fact that \(\Delta_{f, 2}\) is always locally quasi-finite by Lemma 04YQ applied to \(\Delta_f = \Delta_{f, 1}\).

Proof of (4). This follows from the fact that \(\Delta_{f, 2}\) is always unramified as Lemma 04YQ applied to \(\Delta_f = \Delta_{f, 1}\) shows that \(\Delta_{f, 2}\) is locally of finite type and a monomorphism. See More on Morphisms of Spaces, Lemma 05W6.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a separated (resp. quasi-separated, resp. quasi-DM, resp. DM) morphism of algebraic stacks. Then

1. given algebraic spaces \(T_i\), \(i = 1, 2\) and morphisms \(x_i : T_i \to \mathcal{X}\), with \(y_i = f \circ x_i\) the morphism \[T_1 \times_{x_1, \mathcal{X}, x_2} T_2 \longrightarrow T_1 \times_{y_1, \mathcal{Y}, y_2} T_2\] is proper (resp. quasi-compact and quasi-separated, resp. locally quasi-finite, resp. unramified),

2. given an algebraic space \(T\) and morphisms \(x_i : T \to \mathcal{X}\), \(i = 1, 2\), with \(y_i = f \circ x_i\) the morphism \[\mathit{Isom}_\mathcal{X}(x_1, x_2) \longrightarrow \mathit{Isom}_\mathcal{Y}(y_1, y_2)\] is proper (resp. quasi-compact and quasi-separated, resp. locally quasi-finite, resp. unramified).

Proof

Proof of (1). Observe that the diagram \[\xymatrix{ T_1 \times_{x_1, \mathcal{X}, x_2} T_2 \ar[d] \ar[r] & T_1 \times_{y_1, \mathcal{Y}, y_2} T_2 \ar[d] \\ \mathcal{X} \ar[r] & \mathcal{X} \times_\mathcal{Y} \mathcal{X} }\] is cartesian. Hence this follows from the fact that \(f\) is separated (resp. quasi-separated, resp. quasi-DM, resp. DM) if and only if the diagonal is proper (resp. quasi-compact and quasi-separated, resp. locally quasi-finite, resp. unramified).

Proof of (2). This is true because \[\mathit{Isom}_\mathcal{X}(x_1, x_2) = (T \times_{x_1, \mathcal{X}, x_2} T) \times_{T \times T, \Delta_T} T\] hence the morphism in (2) is a base change of the morphism in (1).

Quasi-compact morphisms

Let \(f\) be a morphism of algebraic stacks which is representable by algebraic spaces. In Properties of Stacks, Section 04XB we have defined what it means for \(f\) to be quasi-compact. Here is another characterization.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks which is representable by algebraic spaces. The following are equivalent:

1. \(f\) is quasi-compact (as in Properties of Stacks, Section 04XB), and

2. for every quasi-compact algebraic stack \(\mathcal{Z}\) and any morphism \(\mathcal{Z} \to \mathcal{Y}\) the algebraic stack \(\mathcal{Z} \times_\mathcal{Y} \mathcal{X}\) is quasi-compact.

Proof

Assume (1), and let \(\mathcal{Z} \to \mathcal{Y}\) be a morphism of algebraic stacks with \(\mathcal{Z}\) quasi-compact. By Properties of Stacks, Lemma 04YC there exists a quasi-compact scheme \(U\) and a surjective smooth morphism \(U \to \mathcal{Z}\). Since \(f\) is representable by algebraic spaces and quasi-compact we see by definition that \(U \times_\mathcal{Y} \mathcal{X}\) is an algebraic space, and that \(U \times_\mathcal{Y} \mathcal{X} \to U\) is quasi-compact. Hence \(U \times_\mathcal{Y} \mathcal{X}\) is a quasi-compact algebraic space. The morphism \(U \times_\mathcal{Y} \mathcal{X} \to \mathcal{Z} \times_\mathcal{Y} \mathcal{X}\) is smooth and surjective (as the base change of the smooth and surjective morphism \(U \to \mathcal{Z}\)). Hence \(\mathcal{Z} \times_\mathcal{Y} \mathcal{X}\) is quasi-compact by another application of Properties of Stacks, Lemma 04YC

Assume (2). Let \(Z \to \mathcal{Y}\) be a morphism, where \(Z\) is a scheme. We have to show that the morphism of algebraic spaces \(p : Z \times_\mathcal{Y} \mathcal{X} \to Z\) is quasi-compact. Let \(U \subset Z\) be affine open. Then \(p^{-1}(U) = U \times_\mathcal{Y} Z\) and the algebraic space \(U \times_\mathcal{Y} Z\) is quasi-compact by assumption (2). Hence \(p\) is quasi-compact, see Morphisms of Spaces, Lemma 03KG.

This motivates the following definition.

Definition

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. We say \(f\) is quasi-compact if for every quasi-compact algebraic stack \(\mathcal{Z}\) and morphism \(\mathcal{Z} \to \mathcal{Y}\) the fibre product \(\mathcal{Z} \times_\mathcal{Y} \mathcal{X}\) is quasi-compact.

By Lemma 050T above this agrees with the already existing notion for morphisms of algebraic stacks representable by algebraic spaces. In particular this notion agrees with the notions already defined for morphisms between algebraic stacks and schemes.

Lemma

The base change of a quasi-compact morphism of algebraic stacks by any morphism of algebraic stacks is quasi-compact.

Proof

Omitted.

Lemma

The composition of a pair of quasi-compact morphisms of algebraic stacks is quasi-compact.

Proof

Omitted.

Lemma

A closed immersion of algebraic stacks is quasi-compact.

Proof

This follows from the fact that immersions are always representable and the corresponding fact for closed immersion of algebraic spaces.

Lemma

Let \[\xymatrix{ \mathcal{X} \ar[rr]_f \ar[rd]_p & & \mathcal{Y} \ar[dl]^q \\ & \mathcal{Z} }\] be a \(2\)-commutative diagram of morphisms of algebraic stacks. If \(f\) is surjective and \(p\) is quasi-compact, then \(q\) is quasi-compact.

Proof

Let \(\mathcal{T}\) be a quasi-compact algebraic stack, and let \(\mathcal{T} \to \mathcal{Z}\) be a morphism. By Properties of Stacks, Lemma 04ZU the morphism \(\mathcal{T} \times_\mathcal{Z} \mathcal{X} \to \mathcal{T} \times_\mathcal{Z} \mathcal{Y}\) is surjective and by assumption \(\mathcal{T} \times_\mathcal{Z} \mathcal{X}\) is quasi-compact. Hence \(\mathcal{T} \times_\mathcal{Z} \mathcal{Y}\) is quasi-compact by Properties of Stacks, Lemma 04YC.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) and \(g : \mathcal{Y} \to \mathcal{Z}\) be morphisms of algebraic stacks. If \(g \circ f\) is quasi-compact and \(g\) is quasi-separated then \(f\) is quasi-compact.

Proof

This is true because \(f\) equals the composition \((1, f) : \mathcal{X} \to \mathcal{X} \times_\mathcal{Z} \mathcal{Y} \to \mathcal{Y}\). The first map is quasi-compact by Lemma 050J because it is a section of the quasi-separated morphism \(\mathcal{X} \times_\mathcal{Z} \mathcal{Y} \to \mathcal{X}\) (a base change of \(g\), see Lemma 050F). The second map is quasi-compact as it is the base change of \(f\), see Lemma 050V. And compositions of quasi-compact morphisms are quasi-compact, see Lemma 050W.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks.

1. If \(\mathcal{X}\) is quasi-compact and \(\mathcal{Y}\) is quasi-separated, then \(f\) is quasi-compact.

2. If \(\mathcal{X}\) is quasi-compact and quasi-separated and \(\mathcal{Y}\) is quasi-separated, then \(f\) is quasi-compact and quasi-separated.

3. A fibre product of quasi-compact and quasi-separated algebraic stacks is quasi-compact and quasi-separated.

Proof

Part (1) follows from Lemma 050Y. Part (2) follows from (1) and Lemma 050M. For (3) let \(\mathcal{X} \to \mathcal{Y}\) and \(\mathcal{Z} \to \mathcal{Y}\) be morphisms of quasi-compact and quasi-separated algebraic stacks. Then \(\mathcal{X} \times_\mathcal{Y} \mathcal{Z} \to \mathcal{Z}\) is quasi-compact and quasi-separated as a base change of \(\mathcal{X} \to \mathcal{Y}\) using (2) and Lemmas 050V and 050F. Hence \(\mathcal{X} \times_\mathcal{Y} \mathcal{Z}\) is quasi-compact and quasi-separated as an algebraic stack quasi-compact and quasi-separated over \(\mathcal{Z}\), see Lemmas 050L and 050W.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a quasi-compact morphism of algebraic stacks. Let \(y \in |\mathcal{Y}|\) be a point in the closure of the image of \(|f|\). There exists a valuation ring \(A\) with fraction field \(K\) and a commutative diagram \[\xymatrix{ \Spec(K) \ar[r] \ar[d] & \mathcal{X} \ar[d] \\ \Spec(A) \ar[r] & \mathcal{Y} }\] such that the closed point of \(\Spec(A)\) maps to \(y\).

Proof

Choose an affine scheme \(V\) and a point \(v \in V\) and a smooth morphism \(V \to \mathcal{Y}\) sending \(v\) to \(y\). Consider the base change diagram \[\xymatrix{ V \times_\mathcal{Y} \mathcal{X} \ar[r] \ar[d]_g & \mathcal{X} \ar[d]^f \\ V \ar[r] & \mathcal{Y} }\] Recall that \(|V \times_\mathcal{Y} \mathcal{X}| \to |V| \times_{|\mathcal{Y}|} |\mathcal{X}|\) is surjective (Properties of Stacks, Lemma 04XH). Because \(|V| \to |\mathcal{Y}|\) is open (Properties of Stacks, Lemma 04XL) we conclude that \(v\) is in the closure of the image of \(|g|\). Thus it suffices to prove the lemma for the quasi-compact morphism \(g\) (Lemma 050V) which we do in the next paragraph.

Assume \(\mathcal{Y} = Y\) is an affine scheme. Then \(\mathcal{X}\) is quasi-compact as \(f\) is quasi-compact (Definition 050U). Choose an affine scheme \(W\) and a surjective smooth morphism \(W \to \mathcal{X}\). Then the image of \(|f|\) is the image of \(W \to Y\). By Morphisms, Lemma 02JQ we can choose a diagram \[\xymatrix{ \Spec(K) \ar[r] \ar[d] & W \ar[d] \ar[r] & \mathcal{X} \ar[d] \\ \Spec(A) \ar[r] & Y \ar[r] & Y }\] such that the closed point of \(\Spec(A)\) maps to \(y\). Composing with \(W \to \mathcal{X}\) we obtain a solution.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. Let \(W \to \mathcal{Y}\) be surjective, flat, and locally of finite presentation where \(W\) is an algebraic space. If the base change \(W \times_\mathcal{Y} \mathcal{X} \to W\) is quasi-compact, then \(f\) is quasi-compact.

Proof

Assume \(W \times_\mathcal{Y} \mathcal{X} \to W\) is quasi-compact. Let \(\mathcal{Z} \to \mathcal{Y}\) be a morphism with \(\mathcal{Z}\) a quasi-compact algebraic stack. Choose a scheme \(U\) and a surjective smooth morphism \(U \to W \times_\mathcal{Y} \mathcal{Z}\). Since \(U \to \mathcal{Z}\) is flat, surjective, and locally of finite presentation and \(\mathcal{Z}\) is quasi-compact, we can find a quasi-compact open subscheme \(U' \subset U\) such that \(U' \to \mathcal{Z}\) is surjective. Then \(U' \times_\mathcal{Y} \mathcal{X} = U' \times_W (W \times_\mathcal{Y} \mathcal{X})\) is quasi-compact by assumption and surjects onto \(\mathcal{Z} \times_\mathcal{Y} \mathcal{X}\). Hence \(\mathcal{Z} \times_\mathcal{Y} \mathcal{X}\) is quasi-compact as desired.

Noetherian algebraic stacks

We have already defined locally Noetherian algebraic stacks in Properties of Stacks, Section 04YE.

Definition

Let \(\mathcal{X}\) be an algebraic stack. We say \(\mathcal{X}\) is Noetherian if \(\mathcal{X}\) is quasi-compact, quasi-separated and locally Noetherian.

Note that a Noetherian algebraic stack \(\mathcal{X}\) is not just quasi-compact and locally Noetherian, but also quasi-separated. In the language of Section 04YX if we denote \(p : \mathcal{X} \to \Spec(\mathbf{Z})\) the “absolute” structure morphism (i.e., the structure morphism of \(\mathcal{X}\) viewed as an algebraic stack over \(\mathbf{Z}\)), then \[\mathcal{X}\text{ Noetherian} \Leftrightarrow \mathcal{X}\text{ locally Noetherian and } \Delta_{p, 0}, \Delta_{p, 1}, \Delta_{p, 2} \text{ quasi-compact}.\] This will later mean that an algebraic stack of finite type over a Noetherian algebraic stack is not automatically Noetherian.

Lemma

Let \(j : \mathcal{X} \to \mathcal{Y}\) be an immersion of algebraic stacks.

1. If \(\mathcal{Y}\) is locally Noetherian, then \(\mathcal{X}\) is locally Noetherian and \(j\) is quasi-compact.

2. If \(\mathcal{Y}\) is Noetherian, then \(\mathcal{X}\) is Noetherian.

Proof

Choose a scheme \(V\) and a surjective smooth morphism \(V \to \mathcal{Y}\). Then \(U = \mathcal{X} \times_\mathcal{Y} V\) is a scheme and \(V \to U\) is an immersion, see Properties of Stacks, Definition 04YL. Recall that \(\mathcal{Y}\) is locally Noetherian if and only if \(V\) is locally Noetherian. In this case \(U\) is locally Noetherian too (Morphisms, Lemmas 01T5 and 01T6) and \(U \to V\) is quasi-compact (Properties, Lemma 01OX). This shows that \(j\) is quasi-compact (Lemma 0DTL) and that \(\mathcal{X}\) is locally Noetherian. Finally, if \(\mathcal{Y}\) is Noetherian, then we see from the above that \(\mathcal{X}\) is quasi-compact and locally Noetherian. To finish the proof observe that \(j\) is separated and hence \(\mathcal{X}\) is quasi-separated because \(\mathcal{Y}\) is so by Lemma 050L.

Lemma

Let \(\mathcal{X}\) be an algebraic stack.

1. If \(\mathcal{X}\) is locally Noetherian then \(|\mathcal{X}|\) is a locally Noetherian topological space.

2. If \(\mathcal{X}\) is quasi-compact and locally Noetherian, then \(|\mathcal{X}|\) is a Noetherian topological space.

Proof

Assume \(\mathcal{X}\) is locally Noetherian. Choose a scheme \(U\) and a surjective smooth morphism \(U \to \mathcal{X}\). As \(\mathcal{X}\) is locally Noetherian we see that \(U\) is locally Noetherian. By Properties, Lemma 01OZ this means that \(|U|\) is a locally Noetherian topological space. Since \(|U| \to |\mathcal{X}|\) is open and surjective we conclude that \(|\mathcal{X}|\) is locally Noetherian by Topology, Lemma 04Z8. This proves (1). If \(\mathcal{X}\) is quasi-compact and locally Noetherian, then \(|\mathcal{X}|\) is quasi-compact and locally Noetherian. Hence \(|\mathcal{X}|\) is Noetherian by Topology, Lemma 04ZB.

Lemma

Let \(\mathcal{X}\) be a locally Noetherian algebraic stack. Then \(|\mathcal{X}|\) is quasi-sober (Topology, Definition 004X).

Proof

We have to prove that every irreducible closed subset \(T \subset |\mathcal{X}|\) has a generic point. Choose an affine scheme \(U\) and a smooth morphism \(f : U \to \mathcal{X}\) such that \(f^{-1}(T) \subset |U|\) is nonempty. Since \(U\) is Noetherian, the closed subset \(f^{-1}(T)\) has finitely many irreducible components (Topology, Lemma 0052). Say \(f^{-1}(T) = Z_1 \cup \ldots \cup Z_n\) is the decomposition into irreducible components. As \(f\) is open, the image of \(f|_{f^{-1}(T)} : f^{-1}(T) \to T\) contains a nonempty open subset of \(T\). Since \(T\) is irreducible, this means that \(f(f^{-1}(T))\) is dense. Since \(T\) is irreducible, it follows that \(f(Z_i)\) is dense for some \(i\). Then if \(\xi_i \in Z_i\) is the generic point we see that \(f(\xi_i)\) is a generic point of \(T\).

Affine morphisms

Affine morphisms of algebraic stacks are defined as follows.

Definition

A morphism of algebraic stacks is said to be affine if it is representable and affine in the sense of Properties of Stacks, Section 04XB.

For us it is a little bit more convenient to think of an affine morphism of algebraic stacks as a morphism of algebraic stacks which is representable by algebraic spaces and affine in the sense of Properties of Stacks, Section 04XB. (Recall that the default for “representable” in the Stacks project is representable by schemes.) Since this is clearly equivalent to the notion just defined we shall use this characterization without further mention. We prove a few simple lemmas about this notion.

Lemma

Let \(\mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. Let \(\mathcal{Z} \to \mathcal{Y}\) be an affine morphism of algebraic stacks. Then \(\mathcal{Z} \times_\mathcal{Y} \mathcal{X} \to \mathcal{X}\) is an affine morphism of algebraic stacks.

Proof

This follows from the discussion in Properties of Stacks, Section 04XB.

Lemma

Compositions of affine morphisms of algebraic stacks are affine.

Proof

This follows from the discussion in Properties of Stacks, Section 04XB and Morphisms of Spaces, Lemma 03WH.

Lemma

Let \[\xymatrix{ \mathcal{X} \ar[rr]_f \ar[rd]_a & & \mathcal{Y} \ar[dl]^b \\ & \mathcal{Z} }\] be a commutative diagram of morphisms of algebraic stacks. If \(a\) is affine and \(\Delta_b\) is affine, then \(f\) is affine.

Proof

The base change \(\text{pr}_2 : \mathcal{X} \times_\mathcal{Z} \mathcal{Y} \to \mathcal{Y}\) of \(a\) is affine by Lemma 0CHR. The morphism \((1, f) : \mathcal{X} \to \mathcal{X} \times_\mathcal{Z} \mathcal{Y}\) is the base change of \(\Delta_b : \mathcal{Y} \to \mathcal{Y} \times_\mathcal{Z} \mathcal{Y}\) by the morphism \(\mathcal{X} \times_\mathcal{Z} \mathcal{Y} \to \mathcal{Y} \times_\mathcal{Z} \mathcal{Y}\) (see material in Categories, Section 003O). Hence it is affine by Lemma 0CHR. The composition \(f = \text{pr}_2 \circ (1, f)\) of affine morphisms is affine by Lemma 0CHS and the proof is done.

Integral and finite morphisms

Integral and finite morphisms of algebraic stacks are defined as follows.

Definition

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks.

1. We say \(f\) is integral if \(f\) is representable and integral in the sense of Properties of Stacks, Section 04XB.

2. We say \(f\) is finite if \(f\) is representable and finite in the sense of Properties of Stacks, Section 04XB.

For us it is a little bit more convenient to think of an integral, resp. finite morphism of algebraic stacks as a morphism of algebraic stacks which is representable by algebraic spaces and integral, resp. finite in the sense of Properties of Stacks, Section 04XB. (Recall that the default for “representable” in the Stacks project is representable by schemes.) Since this is clearly equivalent to the notion just defined we shall use this characterization without further mention. We prove a few simple lemmas about this notion.

Lemma

Let \(\mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. Let \(\mathcal{Z} \to \mathcal{Y}\) be an integral (or finite) morphism of algebraic stacks. Then \(\mathcal{Z} \times_\mathcal{Y} \mathcal{X} \to \mathcal{X}\) is an integral (or finite) morphism of algebraic stacks.

Proof

This follows from the discussion in Properties of Stacks, Section 04XB.

Lemma

Compositions of integral, resp. finite morphisms of algebraic stacks are integral, resp. finite.

Proof

This follows from the discussion in Properties of Stacks, Section 04XB and Morphisms of Spaces, Lemma 03ZR.

Open morphisms

Let \(f\) be a morphism of algebraic stacks which is representable by algebraic spaces. In Properties of Stacks, Section 04XB we have defined what it means for \(f\) to be universally open. Here is another characterization.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks which is representable by algebraic spaces. The following are equivalent

1. \(f\) is universally open (as in Properties of Stacks, Section 04XB), and

2. for every morphism of algebraic stacks \(\mathcal{Z} \to \mathcal{Y}\) the morphism of topological spaces \(|\mathcal{Z} \times_\mathcal{Y} \mathcal{X}| \to |\mathcal{Z}|\) is open.

Proof

Assume (1), and let \(\mathcal{Z} \to \mathcal{Y}\) be as in (2). Choose a scheme \(V\) and a surjective smooth morphism \(V \to \mathcal{Z}\). By assumption the morphism \(V \times_\mathcal{Y} \mathcal{X} \to V\) of algebraic spaces is universally open, in particular the map \(|V \times_\mathcal{Y} \mathcal{X}| \to |V|\) is open. By Properties of Stacks, Section 04XE in the commutative diagram \[\xymatrix{ |V \times_\mathcal{Y} \mathcal{X}| \ar[r] \ar[d] & |\mathcal{Z} \times_\mathcal{Y} \mathcal{X}| \ar[d] \\ |V| \ar[r] & |\mathcal{Z}| }\] the horizontal arrows are open and surjective, and moreover \[|V \times_\mathcal{Y} \mathcal{X}| \longrightarrow |V| \times_{|\mathcal{Z}|} |\mathcal{Z} \times_\mathcal{Y} \mathcal{X}|\] is surjective. Hence as the left vertical arrow is open it follows that the right vertical arrow is open. This proves (2). The implication (2) \(\Rightarrow\) (1) follows from the definitions.

Thus we may use the following natural definition.

Definition

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks.

1. We say \(f\) is open if the map of topological spaces \(|\mathcal{X}| \to |\mathcal{Y}|\) is open.

2. We say \(f\) is universally open if for every morphism of algebraic stacks \(\mathcal{Z} \to \mathcal{Y}\) the morphism of topological spaces \[|\mathcal{Z} \times_\mathcal{Y} \mathcal{X}| \to |\mathcal{Z}|\] is open, i.e., the base change \(\mathcal{Z} \times_\mathcal{Y} \mathcal{X} \to \mathcal{Z}\) is open.

Lemma

The base change of a universally open morphism of algebraic stacks by any morphism of algebraic stacks is universally open.

Proof

This is immediate from the definition.

Lemma

The composition of a pair of (universally) open morphisms of algebraic stacks is (universally) open.

Proof

Omitted.

Submersive morphisms

Let \(f\) be a morphism of algebraic stacks which is representable by algebraic spaces. In Properties of Stacks, Section 04XB we have defined what it means for \(f\) to be universally submersive. Here is another characterization.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks which is representable by algebraic spaces. The following are equivalent

1. \(f\) is universally submersive (as in Properties of Stacks, Section 04XB), and

2. for every morphism of algebraic stacks \(\mathcal{Z} \to \mathcal{Y}\) the morphism of topological spaces \(|\mathcal{Z} \times_\mathcal{Y} \mathcal{X}| \to |\mathcal{Z}|\) is submersive.

Proof

Assume (1), and let \(\mathcal{Z} \to \mathcal{Y}\) be as in (2). Choose a scheme \(V\) and a surjective smooth morphism \(V \to \mathcal{Z}\). By assumption the morphism \(V \times_\mathcal{Y} \mathcal{X} \to V\) of algebraic spaces is universally submersive, in particular the map \(|V \times_\mathcal{Y} \mathcal{X}| \to |V|\) is submersive. By Properties of Stacks, Section 04XE in the commutative diagram \[\xymatrix{ |V \times_\mathcal{Y} \mathcal{X}| \ar[r] \ar[d] & |\mathcal{Z} \times_\mathcal{Y} \mathcal{X}| \ar[d] \\ |V| \ar[r] & |\mathcal{Z}| }\] the horizontal arrows are open and surjective, and moreover \[|V \times_\mathcal{Y} \mathcal{X}| \longrightarrow |V| \times_{|\mathcal{Z}|} |\mathcal{Z} \times_\mathcal{Y} \mathcal{X}|\] is surjective. Hence as the left vertical arrow is submersive it follows that the right vertical arrow is submersive. This proves (2). The implication (2) \(\Rightarrow\) (1) follows from the definitions.

Thus we may use the following natural definition.

Definition

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks.

1. We say \(f\) is submersive⁴ if the continuous map \(|\mathcal{X}| \to |\mathcal{Y}|\) is submersive, see Topology, Definition 0406.

2. We say \(f\) is universally submersive if for every morphism of algebraic stacks \(\mathcal{Y}' \to \mathcal{Y}\) the base change \(\mathcal{Y}' \times_\mathcal{Y} \mathcal{X} \to \mathcal{Y}'\) is submersive.

We note that a submersive morphism is in particular surjective.

Lemma

The base change of a universally submersive morphism of algebraic stacks by any morphism of algebraic stacks is universally submersive.

Proof

This is immediate from the definition.

Lemma

The composition of a pair of (universally) submersive morphisms of algebraic stacks is (universally) submersive.

Proof

Omitted.

Universally closed morphisms

Let \(f\) be a morphism of algebraic stacks which is representable by algebraic spaces. In Properties of Stacks, Section 04XB we have defined what it means for \(f\) to be universally closed. Here is another characterization.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks which is representable by algebraic spaces. The following are equivalent

1. \(f\) is universally closed (as in Properties of Stacks, Section 04XB), and

2. for every morphism of algebraic stacks \(\mathcal{Z} \to \mathcal{Y}\) the morphism of topological spaces \(|\mathcal{Z} \times_\mathcal{Y} \mathcal{X}| \to |\mathcal{Z}|\) is closed.

Proof

Assume (1), and let \(\mathcal{Z} \to \mathcal{Y}\) be as in (2). Choose a scheme \(V\) and a surjective smooth morphism \(V \to \mathcal{Z}\). By assumption the morphism \(V \times_\mathcal{Y} \mathcal{X} \to V\) of algebraic spaces is universally closed, in particular the map \(|V \times_\mathcal{Y} \mathcal{X}| \to |V|\) is closed. By Properties of Stacks, Section 04XE in the commutative diagram \[\xymatrix{ |V \times_\mathcal{Y} \mathcal{X}| \ar[r] \ar[d] & |\mathcal{Z} \times_\mathcal{Y} \mathcal{X}| \ar[d] \\ |V| \ar[r] & |\mathcal{Z}| }\] the horizontal arrows are open and surjective, and moreover \[|V \times_\mathcal{Y} \mathcal{X}| \longrightarrow |V| \times_{|\mathcal{Z}|} |\mathcal{Z} \times_\mathcal{Y} \mathcal{X}|\] is surjective. Hence as the left vertical arrow is closed it follows that the right vertical arrow is closed. This proves (2). The implication (2) \(\Rightarrow\) (1) follows from the definitions.

Thus we may use the following natural definition.

Definition

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks.

1. We say \(f\) is closed if the map of topological spaces \(|\mathcal{X}| \to |\mathcal{Y}|\) is closed.

2. We say \(f\) is universally closed if for every morphism of algebraic stacks \(\mathcal{Z} \to \mathcal{Y}\) the morphism of topological spaces \[|\mathcal{Z} \times_\mathcal{Y} \mathcal{X}| \to |\mathcal{Z}|\] is closed, i.e., the base change \(\mathcal{Z} \times_\mathcal{Y} \mathcal{X} \to \mathcal{Z}\) is closed.

Lemma

The base change of a universally closed morphism of algebraic stacks by any morphism of algebraic stacks is universally closed.

Proof

This is immediate from the definition.

Lemma

The composition of a pair of (universally) closed morphisms of algebraic stacks is (universally) closed.

Proof

Omitted.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. The following are equivalent

1. \(f\) is universally closed,

2. for every scheme \(Z\) and every morphism \(Z \to \mathcal{Y}\) the projection \(|Z \times_\mathcal{Y} \mathcal{X}| \to |Z|\) is closed,

3. for every affine scheme \(Z\) and every morphism \(Z \to \mathcal{Y}\) the projection \(|Z \times_\mathcal{Y} \mathcal{X}| \to |Z|\) is closed, and

4. there exists an algebraic space \(V\) and a surjective smooth morphism \(V \to \mathcal{Y}\) such that \(V \times_\mathcal{Y} \mathcal{X} \to V\) is a universally closed morphism of algebraic stacks.

Proof

We omit the proof that (1) implies (2), and that (2) implies (3).

Assume (3). Choose a surjective smooth morphism \(V \to \mathcal{Y}\). We are going to show that \(V \times_\mathcal{Y} \mathcal{X} \to V\) is a universally closed morphism of algebraic stacks. Let \(\mathcal{Z} \to V\) be a morphism from an algebraic stack to \(V\). Let \(W \to \mathcal{Z}\) be a surjective smooth morphism where \(W = \coprod W_i\) is a disjoint union of affine schemes. Then we have the following commutative diagram \[\xymatrix{ \coprod_i |W_i \times_\mathcal{Y} \mathcal{X}| \ar@{=}[r] \ar[d] & |W \times_\mathcal{Y} \mathcal{X}| \ar[r] \ar[d] & |\mathcal{Z} \times_\mathcal{Y} \mathcal{X}| \ar[d] \ar@{=}[r] & |\mathcal{Z} \times_V (V \times_\mathcal{Y} \mathcal{X})| \ar[ld] \\ \coprod |W_i| \ar@{=}[r] & |W| \ar[r] & |\mathcal{Z}| }\] We have to show the south-east arrow is closed. The middle horizontal arrows are surjective and open (Properties of Stacks, Lemma 04XL). By assumption (3), and the fact that \(W_i\) is affine we see that the left vertical arrows are closed. Hence it follows that the right vertical arrow is closed.

Assume (4). We will show that \(f\) is universally closed. Let \(\mathcal{Z} \to \mathcal{Y}\) be a morphism of algebraic stacks. Consider the diagram \[\xymatrix{ |(V \times_\mathcal{Y} \mathcal{Z}) \times_V (V \times_\mathcal{Y} \mathcal{X})| \ar@{=}[r] \ar[rd] & |V \times_\mathcal{Y} \mathcal{X}| \ar[r] \ar[d] & |Z \times_\mathcal{Y} \mathcal{X}| \ar[d] \\ & |V \times_\mathcal{Y} \mathcal{Z}| \ar[r] & |\mathcal{Z}| }\] The south-west arrow is closed by assumption. The horizontal arrows are surjective and open because the corresponding morphisms of algebraic stacks are surjective and smooth (see reference above). It follows that the right vertical arrow is closed.

Universally injective morphisms

Let \(f\) be a morphism of algebraic stacks which is representable by algebraic spaces. In Properties of Stacks, Section 04XB we have defined what it means for \(f\) to be universally injective. Here is another characterization.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks which is representable by algebraic spaces. The following are equivalent

1. \(f\) is universally injective (as in Properties of Stacks, Section 04XB), and

2. for every morphism of algebraic stacks \(\mathcal{Z} \to \mathcal{Y}\) the map \(|\mathcal{Z} \times_\mathcal{Y} \mathcal{X}| \to |\mathcal{Z}|\) is injective.

Proof

Assume (1), and let \(\mathcal{Z} \to \mathcal{Y}\) be as in (2). Choose a scheme \(V\) and a surjective smooth morphism \(V \to \mathcal{Z}\). By assumption the morphism \(V \times_\mathcal{Y} \mathcal{X} \to V\) of algebraic spaces is universally injective, in particular the map \(|V \times_\mathcal{Y} \mathcal{X}| \to |V|\) is injective. By Properties of Stacks, Section 04XE in the commutative diagram \[\xymatrix{ |V \times_\mathcal{Y} \mathcal{X}| \ar[r] \ar[d] & |\mathcal{Z} \times_\mathcal{Y} \mathcal{X}| \ar[d] \\ |V| \ar[r] & |\mathcal{Z}| }\] the horizontal arrows are open and surjective, and moreover \[|V \times_\mathcal{Y} \mathcal{X}| \longrightarrow |V| \times_{|\mathcal{Z}|} |\mathcal{Z} \times_\mathcal{Y} \mathcal{X}|\] is surjective. Hence as the left vertical arrow is injective it follows that the right vertical arrow is injective. This proves (2). The implication (2) \(\Rightarrow\) (1) follows from the definitions.

Thus we may use the following natural definition.

Definition

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. We say \(f\) is universally injective if for every morphism of algebraic stacks \(\mathcal{Z} \to \mathcal{Y}\) the map \[|\mathcal{Z} \times_\mathcal{Y} \mathcal{X}| \to |\mathcal{Z}|\] is injective.

Lemma

The base change of a universally injective morphism of algebraic stacks by any morphism of algebraic stacks is universally injective.

Proof

This is immediate from the definition.

Lemma

The composition of a pair of universally injective morphisms of algebraic stacks is universally injective.

Proof

Omitted.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. The following are equivalent

1. \(f\) is universally injective,

2. \(\Delta : \mathcal{X} \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}\) is surjective, and

3. for an algebraically closed field, for \(x_1, x_2 : \Spec(k) \to \mathcal{X}\), and for a \(2\)-arrow \(\beta : f \circ x_1 \to f \circ x_2\) there is a \(2\)-arrow \(\alpha : x_1 \to x_2\) with \(\beta = \text{id}_f \star \alpha\).

Proof

(1) \(\Rightarrow\) (2). If \(f\) is universally injective, then the first projection \(|\mathcal{X} \times_\mathcal{Y} \mathcal{X}| \to |\mathcal{X}|\) is injective, which implies that \(|\Delta|\) is surjective.

(2) \(\Rightarrow\) (1). Assume \(\Delta\) is surjective. Then any base change of \(\Delta\) is surjective (see Properties of Stacks, Section 04ZR). Since the diagonal of a base change of \(f\) is a base change of \(\Delta\), we see that it suffices to show that \(|\mathcal{X}| \to |\mathcal{Y}|\) is injective. If not, then by Properties of Stacks, Lemma 04XH we find that the first projection \(|\mathcal{X} \times_\mathcal{Y} \mathcal{X}| \to |\mathcal{X}|\) is not injective. Of course this means that \(|\Delta|\) is not surjective.

(3) \(\Rightarrow\) (2). Let \(t \in |\mathcal{X} \times_\mathcal{Y} \mathcal{X}|\). Then we can represent \(t\) by a morphism \(t : \Spec(k) \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}\) with \(k\) an algebraically closed field. By our construction of \(2\)-fibre products we can represent \(t\) by \((x_1, x_2, \beta)\) where \(x_1, x_2 : \Spec(k) \to \mathcal{X}\) and \(\beta : f \circ x_1 \to f \circ x_2\) is a \(2\)-morphism. Then (3) implies that there is a \(2\)-morphism \(\alpha : x_1 \to x_2\) mapping to \(\beta\). This exactly means that \(\Delta(x_1) = (x_1, x_1, \text{id})\) is isomorphic to \(t\). Hence (2) holds.

(2) \(\Rightarrow\) (3). Let \(x_1, x_2 : \Spec(k) \to \mathcal{X}\) be morphisms with \(k\) an algebraically closed field. Let \(\beta : f \circ x_1 \to f \circ x_2\) be a \(2\)-morphism. As in the previous paragraph, we obtain a morphism \(t = (x_1, x_2, \beta) : \Spec(k) \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}\). By Lemma 04XS \[T = \mathcal{X} \times_{\Delta, \mathcal{X} \times_\mathcal{Y} \mathcal{X}, t} \Spec(k)\] is an algebraic space locally of finite type over \(\Spec(k)\). Condition (2) implies that \(T\) is nonempty. Then since \(k\) is algebraically closed, there is a \(k\)-point in \(T\). Unwinding the definitions this means there is a morphism \(\alpha : x_1 \to x_2\) in \(\Mor(\Spec(k), \mathcal{X})\) such that \(\beta = \text{id}_f \star \alpha\).

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a universally injective morphism of algebraic stacks. Let \(y : \Spec(k) \to \mathcal{Y}\) be a morphism where \(k\) is an algebraically closed field. If \(y\) is in the image of \(|\mathcal{X}| \to |\mathcal{Y}|\), then there is a morphism \(x : \Spec(k) \to \mathcal{X}\) with \(y = f \circ x\).

Proof

We first remark this lemma is not a triviality, because the assumption that \(y\) is in the image of \(|f|\) means only that we can lift \(y\) to a morphism into \(\mathcal{X}\) after possibly replacing \(k\) by an extension field. To prove the lemma we may base change \(f\) by \(y\), hence we may assume we have a nonempty algebraic stack \(\mathcal{X}\) and a universally injective morphism \(\mathcal{X} \to \Spec(k)\) and we want to find a \(k\)-valued point of \(\mathcal{X}\). We may replace \(\mathcal{X}\) by its reduction. We may choose a field \(k'\) and a surjective, flat, locally finite type morphism \(\Spec(k') \to \mathcal{X}\), see Properties of Stacks, Lemma 06MN. Since \(\mathcal{X} \to \Spec(k)\) is universally injective, we find that \[\Spec(k') \times_\mathcal{X} \Spec(k') \to \Spec(k' \otimes_k k')\] is surjective as the base change of the surjective morphism \(\Delta : \mathcal{X} \to \mathcal{X} \times_{\Spec(k)} \mathcal{X}\) (Lemma 0CPN). Since \(k\) is algebraically closed \(k' \otimes_k k'\) is a domain (Algebra, Lemma 09P9). Let \(\xi \in \Spec(k') \times_\mathcal{X} \Spec(k')\) be a point mapping to the generic point of \(\Spec(k' \otimes_k k')\). Let \(U\) be the reduced induced closed subscheme structure on the connected component of \(\Spec(k') \times_\mathcal{X} \Spec(k')\) containing \(\xi\). Then the two projections \(U \to \Spec(k')\) are locally of finite type, as this was true for the projections \(\Spec(k') \times_\mathcal{X} \Spec(k') \to \Spec(k')\) as base changes of the morphism \(\Spec(k') \to \mathcal{X}\). Applying Varieties, Proposition 04MK we find that the integral closures of the two images of \(k'\) in \(\Gamma(U, \mathcal{O}_U)\) are equal. Looking in \(\kappa(\xi)\) means that any element of the form \(\lambda \otimes 1\) is algebraically dependent on the subfield \[1 \otimes k' \subset (\text{fraction field of }k' \otimes_k k') \subset \kappa(\xi).\] Since \(k\) is algebraically closed, this is only possible if \(k' = k\) and the proof is complete.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. The following are equivalent:

1. \(f\) is universally injective,

2. for every affine scheme \(Z\) and any morphism \(Z \to \mathcal{Y}\) the morphism \(Z \times_\mathcal{Y} \mathcal{X} \to Z\) is universally injective, and

3. add more here.

Proof

The implication (1) \(\Rightarrow\) (2) is immediate. Assume (2) holds. We will show that \(\Delta_f : \mathcal{X} \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}\) is surjective, which implies (1) by Lemma 0CPN. Consider an affine scheme \(V\) and a smooth morphism \(V \to \mathcal{Y}\). Since \(g : V \times_\mathcal{Y} \mathcal{X} \to V\) is universally injective by (2), we see that \(\Delta_g\) is surjective. However, \(\Delta_g\) is the base change of \(\Delta_f\) by the smooth morphism \(V \to \mathcal{Y}\). Since the collection of these morphisms \(V \to \mathcal{Y}\) are jointly surjective, we conclude \(\Delta_f\) is surjective.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. Let \(W \to \mathcal{Y}\) be surjective, flat, and locally of finite presentation where \(W\) is an algebraic space. If the base change \(W \times_\mathcal{Y} \mathcal{X} \to W\) is universally injective, then \(f\) is universally injective.

Proof

Observe that the diagonal \(\Delta_g\) of the morphism \(g : W \times_\mathcal{Y} \mathcal{X} \to W\) is the base change of \(\Delta_f\) by \(W \to \mathcal{Y}\). Hence if \(\Delta_g\) is surjective, then so is \(\Delta_f\) by Properties of Stacks, Lemma 04XD. Thus the lemma follows from the characterization (2) in Lemma 0CPN.

Universal homeomorphisms

Let \(f\) be a morphism of algebraic stacks which is representable by algebraic spaces. In Properties of Stacks, Section 04XB we have defined what it means for \(f\) to be a universal homeomorphism. Here is another characterization.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks which is representable by algebraic spaces. The following are equivalent

1. \(f\) is a universal homeomorphism (Properties of Stacks, Section 04XB), and

2. for every morphism of algebraic stacks \(\mathcal{Z} \to \mathcal{Y}\) the map of topological spaces \(|\mathcal{Z} \times_\mathcal{Y} \mathcal{X}| \to |\mathcal{Z}|\) is a homeomorphism.

Proof

Assume (1), and let \(\mathcal{Z} \to \mathcal{Y}\) be as in (2). Choose a scheme \(V\) and a surjective smooth morphism \(V \to \mathcal{Z}\). By assumption the morphism \(V \times_\mathcal{Y} \mathcal{X} \to V\) of algebraic spaces is a universal homeomorphism, in particular the map \(|V \times_\mathcal{Y} \mathcal{X}| \to |V|\) is a homeomorphism. By Properties of Stacks, Section 04XE in the commutative diagram \[\xymatrix{ |V \times_\mathcal{Y} \mathcal{X}| \ar[r] \ar[d] & |\mathcal{Z} \times_\mathcal{Y} \mathcal{X}| \ar[d] \\ |V| \ar[r] & |\mathcal{Z}| }\] the horizontal arrows are open and surjective, and moreover \[|V \times_\mathcal{Y} \mathcal{X}| \longrightarrow |V| \times_{|\mathcal{Z}|} |\mathcal{Z} \times_\mathcal{Y} \mathcal{X}|\] is surjective. Hence as the left vertical arrow is a homeomorphism it follows that the right vertical arrow is a homeomorphism. This proves (2). The implication (2) \(\Rightarrow\) (1) follows from the definitions.

Thus we may use the following natural definition.

Definition

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. We say \(f\) is a universal homeomorphism if for every morphism of algebraic stacks \(\mathcal{Z} \to \mathcal{Y}\) the map of topological spaces \[|\mathcal{Z} \times_\mathcal{Y} \mathcal{X}| \to |\mathcal{Z}|\] is a homeomorphism.

Lemma

The base change of a universal homeomorphism of algebraic stacks by any morphism of algebraic stacks is a universal homeomorphism.

Proof

This is immediate from the definition.

Lemma

The composition of a pair of universal homeomorphisms of algebraic stacks is a universal homeomorphism.

Proof

Omitted.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. Let \(W \to \mathcal{Y}\) be surjective, flat, and locally of finite presentation where \(W\) is an algebraic space. If the base change \(W \times_\mathcal{Y} \mathcal{X} \to W\) is a universal homeomorphism, then \(f\) is a universal homeomorphism.

Proof

Assume \(g : W \times_\mathcal{Y} \mathcal{X} \to W\) is a universal homeomorphism. Then \(g\) is universally injective, hence \(f\) is universally injective by Lemma 0DTP. On the other hand, let \(\mathcal{Z} \to \mathcal{Y}\) be a morphism with \(\mathcal{Z}\) an algebraic stack. Choose a scheme \(U\) and a surjective smooth morphism \(U \to W \times_\mathcal{Y} \mathcal{Z}\). Consider the diagram \[\xymatrix{ W \times_\mathcal{Y} \mathcal{X} \ar[d]^g & U \times_\mathcal{Y} \mathcal{X} \ar[d] \ar[l] \ar[r] & \mathcal{Z} \times_\mathcal{Y} \mathcal{X} \ar[d] \\ W & U \ar[l] \ar[r] & \mathcal{Z} }\] The middle vertical arrow induces a homeomorphism on topological space by assumption on \(g\). The morphism \(U \to \mathcal{Z}\) and \(U \times_\mathcal{Y} \mathcal{X} \to \mathcal{Z} \times_\mathcal{Y} \mathcal{X}\) are surjective, flat, and locally of finite presentation hence induce open maps on topological spaces. We conclude that \(|\mathcal{Z} \times_\mathcal{Y} \mathcal{X}| \to |\mathcal{Z}|\) is open. Surjectivity is easy to prove; we omit the proof.

Types of morphisms smooth local on source-and-target

Given a property of morphisms of algebraic spaces which is smooth local on the source-and-target, see Descent on Spaces, Definition 06F8 we may use it to define a corresponding property of morphisms of algebraic stacks, namely by imposing either of the equivalent conditions of the lemma below.

Lemma

Let \(\mathcal{P}\) be a property of morphisms of algebraic spaces which is smooth local on the source-and-target. Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. Consider commutative diagrams \[\xymatrix{ U \ar[d]_a \ar[r]_h & V \ar[d]^b \\ \mathcal{X} \ar[r]^f & \mathcal{Y} }\] where \(U\) and \(V\) are algebraic spaces and the vertical arrows are smooth. The following are equivalent

1. for any diagram as above such that in addition \(U \to \mathcal{X} \times_\mathcal{Y} V\) is smooth the morphism \(h\) has property \(\mathcal{P}\), and

2. for some diagram as above with \(a : U \to \mathcal{X}\) surjective the morphism \(h\) has property \(\mathcal{P}\).

If \(\mathcal{X}\) and \(\mathcal{Y}\) are representable by algebraic spaces, then this is also equivalent to \(f\) (as a morphism of algebraic spaces) having property \(\mathcal{P}\). If \(\mathcal{P}\) is also preserved under any base change, and fppf local on the base, then for morphisms \(f\) which are representable by algebraic spaces this is also equivalent to \(f\) having property \(\mathcal{P}\) in the sense of Properties of Stacks, Section 04XB.

Proof

Let us prove the implication (1) \(\Rightarrow\) (2). Pick an algebraic space \(V\) and a surjective and smooth morphism \(V \to \mathcal{Y}\). Pick an algebraic space \(U\) and a surjective and smooth morphism \(U \to \mathcal{X} \times_\mathcal{Y} V\). Note that \(U \to \mathcal{X}\) is surjective and smooth as well, as a composition of the base change \(\mathcal{X} \times_\mathcal{Y} V \to \mathcal{X}\) and the chosen map \(U \to \mathcal{X} \times_\mathcal{Y} V\). Hence we obtain a diagram as in (1). Thus if (1) holds, then \(h : U \to V\) has property \(\mathcal{P}\), which means that (2) holds as \(U \to \mathcal{X}\) is surjective.

Conversely, assume (2) holds and let \(U, V, a, b, h\) be as in (2). Next, let \(U', V', a', b', h'\) be any diagram as in (1). Picture \[\xymatrix{ U \ar[d] \ar[r]_h & V \ar[d] \\ \mathcal{X} \ar[r]^f & \mathcal{Y} } \quad\quad \xymatrix{ U' \ar[d] \ar[r]_{h'} & V' \ar[d] \\ \mathcal{X} \ar[r]^f & \mathcal{Y} }\] To show that (2) implies (1) we have to prove that \(h'\) has \(\mathcal{P}\). To do this consider the commutative diagram \[\xymatrix{ U \ar[dd]^h & U \times_\mathcal{X} U' \ar[d] \ar[l] \ar@/^6ex/[dd]^{(h, h')} \ar[r] & U' \ar[dd]^{h'} \\ & U \times_\mathcal{Y} V' \ar[lu] \ar[d] & \\ V & V \times_\mathcal{Y} V' \ar[l] \ar[r] & V' }\] of algebraic spaces. Note that the horizontal arrows are smooth as base changes of the smooth morphisms \(V \to \mathcal{Y}\), \(V' \to \mathcal{Y}\), \(U \to \mathcal{X}\), and \(U' \to \mathcal{X}\). Note that \[\xymatrix{ U \times_\mathcal{X} U' \ar[d] \ar[r] & U' \ar[d] \\ U \times_\mathcal{Y} V' \ar[r] & \mathcal{X} \times_\mathcal{Y} V' }\] is cartesian, hence the left vertical arrow is smooth as \(U', V', a', b', h'\) is as in (1). Since \(\mathcal{P}\) is smooth local on the target by Descent on Spaces, Lemma 06F9 part (2) we see that the base change \(U \times_\mathcal{Y} V' \to V \times_\mathcal{Y} V'\) has \(\mathcal{P}\). Since \(\mathcal{P}\) is smooth local on the source by Descent on Spaces, Lemma 06F9 part (1) we can precompose by the smooth morphism \(U \times_\mathcal{X} U' \to U \times_\mathcal{Y} V'\) and conclude \((h, h')\) has \(\mathcal{P}\). Since \(V \times_\mathcal{Y} V' \to V'\) is smooth we conclude \(U \times_\mathcal{X} U' \to V'\) has \(\mathcal{P}\) by Descent on Spaces, Lemma 06F9 part (3). Finally, since \(U \times_X U' \to U'\) is surjective and smooth and \(\mathcal{P}\) is smooth local on the source (same lemma) we conclude that \(h'\) has \(\mathcal{P}\). This finishes the proof of the equivalence of (1) and (2).

If \(\mathcal{X}\) and \(\mathcal{Y}\) are representable, then Descent on Spaces, Lemma 06FA applies which shows that (1) and (2) are equivalent to \(f\) having \(\mathcal{P}\).

Finally, suppose \(f\) is representable, and \(U, V, a, b, h\) are as in part (2) of the lemma, and that \(\mathcal{P}\) is preserved under arbitrary base change. We have to show that for any scheme \(Z\) and morphism \(Z \to \mathcal{Y}\) the base change \(Z \times_\mathcal{Y} \mathcal{X} \to Z\) has property \(\mathcal{P}\). Consider the diagram \[\xymatrix{ Z \times_\mathcal{Y} U \ar[d] \ar[r] & Z \times_\mathcal{Y} V \ar[d] \\ Z \times_\mathcal{Y} \mathcal{X} \ar[r] & Z }\] Note that the top horizontal arrow is a base change of \(h\) and hence has property \(\mathcal{P}\). The left vertical arrow is smooth and surjective and the right vertical arrow is smooth. Thus Descent on Spaces, Lemma 06FA kicks in and shows that \(Z \times_\mathcal{Y} \mathcal{X} \to Z\) has property \(\mathcal{P}\).

Definition

Let \(\mathcal{P}\) be a property of morphisms of algebraic spaces which is smooth local on the source-and-target. We say a morphism \(f : \mathcal{X} \to \mathcal{Y}\) of algebraic stacks has property \(\mathcal{P}\) if the equivalent conditions of Lemma 06FM hold.

Remark

Let \(\mathcal{P}\) be a property of morphisms of algebraic spaces which is smooth local on the source-and-target and stable under composition. Then the property of morphisms of algebraic stacks defined in Definition 06FN is stable under composition. Namely, let \(f : \mathcal{X} \to \mathcal{Y}\) and \(g : \mathcal{Y} \to \mathcal{Z}\) be morphisms of algebraic stacks having property \(\mathcal{P}\). Choose an algebraic space \(W\) and a surjective smooth morphism \(W \to \mathcal{Z}\). Choose an algebraic space \(V\) and a surjective smooth morphism \(V \to \mathcal{Y} \times_\mathcal{Z} W\). Finally, choose an algebraic space \(U\) and a surjective and smooth morphism \(U \to \mathcal{X} \times_\mathcal{Y} V\). Then the morphisms \(V \to W\) and \(U \to V\) have property \(\mathcal{P}\) by definition. Whence \(U \to W\) has property \(\mathcal{P}\) as we assumed that \(\mathcal{P}\) is stable under composition. Thus, by definition again, we see that \(g \circ f : \mathcal{X} \to \mathcal{Z}\) has property \(\mathcal{P}\).

Remark

Let \(\mathcal{P}\) be a property of morphisms of algebraic spaces which is smooth local on the source-and-target and stable under base change. Then the property of morphisms of algebraic stacks defined in Definition 06FN is stable under base change. Namely, let \(f : \mathcal{X} \to \mathcal{Y}\) and \(g : \mathcal{Y}' \to \mathcal{Y}\) be morphisms of algebraic stacks and assume \(f\) has property \(\mathcal{P}\). Choose an algebraic space \(V\) and a surjective smooth morphism \(V \to \mathcal{Y}\). Choose an algebraic space \(U\) and a surjective smooth morphism \(U \to \mathcal{X} \times_\mathcal{Y} V\). Finally, choose an algebraic space \(V'\) and a surjective and smooth morphism \(V' \to \mathcal{Y}' \times_\mathcal{Y} V\). Then the morphism \(U \to V\) has property \(\mathcal{P}\) by definition. Whence \(V' \times_V U \to V'\) has property \(\mathcal{P}\) as we assumed that \(\mathcal{P}\) is stable under base change. Considering the diagram \[\xymatrix{ V' \times_V U \ar[r] \ar[d] & \mathcal{Y}' \times_\mathcal{Y} \mathcal{X} \ar[r] \ar[d] & \mathcal{X} \ar[d] \\ V' \ar[r] & \mathcal{Y}' \ar[r] & \mathcal{Y} }\] we see that the left top horizontal arrow is smooth and surjective, whence by definition we see that the projection \(\mathcal{Y}' \times_\mathcal{Y} \mathcal{X} \to \mathcal{Y}'\) has property \(\mathcal{P}\).

Remark

Let \(\mathcal{P}, \mathcal{P}'\) be properties of morphisms of algebraic spaces which are smooth local on the source-and-target. Suppose that we have \(\mathcal{P} \Rightarrow \mathcal{P}'\) for morphisms of algebraic spaces. Then we also have \(\mathcal{P} \Rightarrow \mathcal{P}'\) for the properties of morphisms of algebraic stacks defined in Definition 06FN using \(\mathcal{P}\) and \(\mathcal{P}'\). This is clear from the definition.

Morphisms of finite type

The property “locally of finite type” of morphisms of algebraic spaces is smooth local on the source-and-target, see Descent on Spaces, Remark 06FC. It is also stable under base change and fpqc local on the target, see Morphisms of Spaces, Lemma 03XH and Descent on Spaces, Lemma 041S. Hence, by Lemma 06FM above, we may define what it means for a morphism of algebraic spaces to be locally of finite type as follows and it agrees with the already existing notion defined in Properties of Stacks, Section 04XB when the morphism is representable by algebraic spaces.

Definition

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks.

1. We say \(f\) locally of finite type if the equivalent conditions of Lemma 06FM hold with \(\mathcal{P} = \text{locally of finite type}\).

2. We say \(f\) is of finite type if it is locally of finite type and quasi-compact.

Lemma

The composition of finite type morphisms is of finite type. The same holds for locally of finite type.

Proof

Combine Remark 06FP with Morphisms of Spaces, Lemma 03XG.

Lemma

A base change of a finite type morphism is finite type. The same holds for locally of finite type.

Proof

Combine Remark 06FQ with Morphisms of Spaces, Lemma 03XH.

Lemma

An immersion is locally of finite type.

Proof

Combine Remark 06PS with Morphisms of Spaces, Lemma 06ED.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. If \(f\) is locally of finite type and \(\mathcal{Y}\) is locally Noetherian, then \(\mathcal{X}\) is locally Noetherian.

Proof

Let \[\xymatrix{ U \ar[d] \ar[r] & V \ar[d] \\ \mathcal{X} \ar[r] & \mathcal{Y} }\] be a commutative diagram where \(U\), \(V\) are schemes, \(V \to \mathcal{Y}\) is surjective and smooth, and \(U \to V \times_\mathcal{Y} \mathcal{X}\) is surjective and smooth. Then \(U \to V\) is locally of finite type. If \(\mathcal{Y}\) is locally Noetherian, then \(V\) is locally Noetherian. By Morphisms, Lemma 01T6 we see that \(U\) is locally Noetherian, which means that \(\mathcal{X}\) is locally Noetherian.

The following two lemmas will be improved on later (after we have discussed morphisms of algebraic stacks which are locally of finite presentation).

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. Let \(W \to \mathcal{Y}\) be a surjective, flat, and locally of finite presentation where \(W\) is an algebraic space. If the base change \(W \times_\mathcal{Y} \mathcal{X} \to W\) is locally of finite type, then \(f\) is locally of finite type.

Proof

Choose an algebraic space \(V\) and a surjective smooth morphism \(V \to \mathcal{Y}\). Choose an algebraic space \(U\) and a surjective smooth morphism \(U \to V \times_\mathcal{Y} \mathcal{X}\). We have to show that \(U \to V\) is locally of finite presentation. Now we base change everything by \(W \to \mathcal{Y}\): Set \(U' = W \times_\mathcal{Y} U\), \(V' = W \times_\mathcal{Y} V\), \(\mathcal{X}' = W \times_\mathcal{Y} \mathcal{X}\), and \(\mathcal{Y}' = W \times_\mathcal{Y} \mathcal{Y} = W\). Then it is still true that \(U' \to V' \times_{\mathcal{Y}'} \mathcal{X}'\) is smooth by base change. Hence by our definition of locally finite type morphisms of algebraic stacks and the assumption that \(\mathcal{X}' \to \mathcal{Y}'\) is locally of finite type, we see that \(U' \to V'\) is locally of finite type. Then, since \(V' \to V\) is surjective, flat, and locally of finite presentation as a base change of \(W \to \mathcal{Y}\) we see that \(U \to V\) is locally of finite type by Descent on Spaces, Lemma 041S and we win.

Lemma

Let \(\mathcal{X} \to \mathcal{Y} \to \mathcal{Z}\) be morphisms of algebraic stacks. Assume \(\mathcal{X} \to \mathcal{Z}\) is locally of finite type and that \(\mathcal{X} \to \mathcal{Y}\) is representable by algebraic spaces, surjective, flat, and locally of finite presentation. Then \(\mathcal{Y} \to \mathcal{Z}\) is locally of finite type.

Proof

Choose an algebraic space \(W\) and a surjective smooth morphism \(W \to \mathcal{Z}\). Choose an algebraic space \(V\) and a surjective smooth morphism \(V \to W \times_\mathcal{Z} \mathcal{Y}\). Set \(U = V \times_\mathcal{Y} \mathcal{X}\) which is an algebraic space. We know that \(U \to V\) is surjective, flat, and locally of finite presentation and that \(U \to W\) is locally of finite type. Hence the lemma reduces to the case of morphisms of algebraic spaces. The case of morphisms of algebraic spaces is Descent on Spaces, Lemma 06EW.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\), \(g : \mathcal{Y} \to \mathcal{Z}\) be morphisms of algebraic stacks. If \(g \circ f : \mathcal{X} \to \mathcal{Z}\) is locally of finite type, then \(f : \mathcal{X} \to \mathcal{Y}\) is locally of finite type.

Proof

We can find a diagram \[\xymatrix{ U \ar[r] \ar[d] & V \ar[r] \ar[d] & W \ar[d] \\ \mathcal{X} \ar[r] & \mathcal{Y} \ar[r] & \mathcal{Z} }\] where \(U\), \(V\), \(W\) are schemes, the vertical arrow \(W \to \mathcal{Z}\) is surjective and smooth, the arrow \(V \to \mathcal{Y} \times_\mathcal{Z} W\) is surjective and smooth, and the arrow \(U \to \mathcal{X} \times_\mathcal{Y} V\) is surjective and smooth. Then also \(U \to \mathcal{X} \times_\mathcal{Z} V\) is surjective and smooth (as a composition of a surjective and smooth morphism with a base change of such). By definition we see that \(U \to W\) is locally of finite type. Hence \(U \to V\) is locally of finite type by Morphisms, Lemma 01T8 which in turn means (by definition) that \(\mathcal{X} \to \mathcal{Y}\) is locally of finite type.

Points of finite type

Let \(\mathcal{X}\) be an algebraic stack. A finite type point \(x \in |\mathcal{X}|\) is a point which can be represented by a morphism \(\Spec(k) \to \mathcal{X}\) which is locally of finite type. Finite type points are a suitable replacement of closed points for algebraic spaces and algebraic stacks. There are always “enough of them” for example.

Lemma

Let \(\mathcal{X}\) be an algebraic stack. Let \(x \in |\mathcal{X}|\). The following are equivalent:

1. There exists a morphism \(\Spec(k) \to \mathcal{X}\) which is locally of finite type and represents \(x\).

2. There exists a scheme \(U\), a closed point \(u \in U\), and a smooth morphism \(\varphi : U \to \mathcal{X}\) such that \(\varphi(u) = x\).

Proof

Let \(u \in U\) and \(U \to \mathcal{X}\) be as in (2). Then \(\Spec(\kappa(u)) \to U\) is of finite type, and \(U \to \mathcal{X}\) is representable and locally of finite type (by Morphisms of Spaces, Lemmas 0468 and 0464). Hence we see (1) holds by Lemma 06FT.

Conversely, assume \(\Spec(k) \to \mathcal{X}\) is locally of finite type and represents \(x\). Let \(U \to \mathcal{X}\) be a surjective smooth morphism where \(U\) is a scheme. By assumption \(U \times_\mathcal{X} \Spec(k) \to U\) is a morphism of algebraic spaces which is locally of finite type. Pick a finite type point \(v\) of \(U \times_\mathcal{X} \Spec(k)\) (there exists at least one, see Morphisms of Spaces, Lemma 06EH). By Morphisms of Spaces, Lemma 06EI the image \(u \in U\) of \(v\) is a finite type point of \(U\). Hence by Morphisms, Lemma 02J2 after shrinking \(U\) we may assume that \(u\) is a closed point of \(U\), i.e., (2) holds.

Definition

Let \(\mathcal{X}\) be an algebraic stack. We say a point \(x \in |\mathcal{X}|\) is a finite type point⁵ if the equivalent conditions of Lemma 06FX are satisfied. We denote \(\mathcal{X}_{\text{ft-pts}}\) the set of finite type points of \(\mathcal{X}\).

We can describe the set of finite type points as follows.

Lemma

Let \(\mathcal{X}\) be an algebraic stack. We have \[\mathcal{X}_{\text{ft-pts}} = \bigcup\nolimits_{\varphi : U \to \mathcal{X}\text{ smooth}} |\varphi|(U_0)\] where \(U_0\) is the set of closed points of \(U\). Here we may let \(U\) range over all schemes smooth over \(\mathcal{X}\) or over all affine schemes smooth over \(\mathcal{X}\).

Proof

Immediate from Lemma 06FX.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. If \(f\) is locally of finite type, then \(f(\mathcal{X}_{\text{ft-pts}}) \subset \mathcal{Y}_{\text{ft-pts}}\).

Proof

Take \(x \in \mathcal{X}_{\text{ft-pts}}\). Represent \(x\) by a locally finite type morphism \(x : \Spec(k) \to \mathcal{X}\). Then \(f \circ x\) is locally of finite type by Lemma 06FT. Hence \(f(x) \in \mathcal{Y}_{\text{ft-pts}}\).

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. If \(f\) is locally of finite type and surjective, then \(f(\mathcal{X}_{\text{ft-pts}}) = \mathcal{Y}_{\text{ft-pts}}\).

Proof

We have \(f(\mathcal{X}_{\text{ft-pts}}) \subset \mathcal{Y}_{\text{ft-pts}}\) by Lemma 06G0. Let \(y \in |\mathcal{Y}|\) be a finite type point. Represent \(y\) by a morphism \(\Spec(k) \to \mathcal{Y}\) which is locally of finite type. As \(f\) is surjective the algebraic stack \(\mathcal{X}_k = \Spec(k) \times_\mathcal{Y} \mathcal{X}\) is nonempty, therefore has a finite type point \(x \in |\mathcal{X}_k|\) by Lemma 06FZ. Now \(\mathcal{X}_k \to \mathcal{X}\) is a morphism which is locally of finite type as a base change of \(\Spec(k) \to \mathcal{Y}\) (Lemma 06FU). Hence the image of \(x\) in \(\mathcal{X}\) is a finite type point by Lemma 06G0 which maps to \(y\) by construction.

Lemma

Let \(\mathcal{X}\) be an algebraic stack. For any locally closed subset \(T \subset |\mathcal{X}|\) we have \[T \not = \emptyset \Rightarrow T \cap \mathcal{X}_{\text{ft-pts}} \not = \emptyset.\] In particular, for any closed subset \(T \subset |\mathcal{X}|\) we see that \(T \cap \mathcal{X}_{\text{ft-pts}}\) is dense in \(T\).

Proof

Let \(i : \mathcal{Z} \to \mathcal{X}\) be the reduced induced substack structure on \(T\), see Properties of Stacks, Remark 06FK. An immersion is locally of finite type, see Lemma 06FV. Hence by Lemma 06G0 we see \(\mathcal{Z}_{\text{ft-pts}} \subset \mathcal{X}_{\text{ft-pts}} \cap T\). Finally, any nonempty affine scheme \(U\) with a smooth morphism towards \(\mathcal{Z}\) has at least one closed point, hence \(\mathcal{Z}\) has at least one finite type point by Lemma 06FZ. The lemma follows.

Here is another, more technical, characterization of a finite type point on an algebraic stack. It tells us in particular that the residual gerbe of \(\mathcal{X}\) at \(x\) exists whenever \(x\) is a finite type point!

Lemma

Let \(\mathcal{X}\) be an algebraic stack. Let \(x \in |\mathcal{X}|\). The following are equivalent:

1. \(x\) is a finite type point,

2. there exists an algebraic stack \(\mathcal{Z}\) whose underlying topological space \(|\mathcal{Z}|\) is a singleton, and a morphism \(f : \mathcal{Z} \to \mathcal{X}\) which is locally of finite type such that \(\{x\} = |f|(|\mathcal{Z}|)\), and

3. the residual gerbe \(\mathcal{Z}_x\) of \(\mathcal{X}\) at \(x\) exists and the inclusion morphism \(\mathcal{Z}_x \to \mathcal{X}\) is locally of finite type.

Proof

(All of the morphisms occurring in this paragraph are representable by algebraic spaces, hence the conventions and results of Properties of Stacks, Section 04XB are applicable.) Assume \(x\) is a finite type point. Choose an affine scheme \(U\), a closed point \(u \in U\), and a smooth morphism \(\varphi : U \to \mathcal{X}\) with \(\varphi(u) = x\), see Lemma 06FZ. Set \(u = \Spec(\kappa(u))\) as usual. Set \(R = u \times_\mathcal{X} u\) so that we obtain a groupoid in algebraic spaces \((u, R, s, t, c)\), see Algebraic Stacks, Lemma 04T4. The projection morphisms \(R \to u\) are the compositions \[R = u \times_\mathcal{X} u \to u \times_\mathcal{X} U \to u \times_\mathcal{X} X = u\] where the first arrow is of finite type (a base change of the closed immersion of schemes \(u \to U\)) and the second arrow is smooth (a base change of the smooth morphism \(U \to \mathcal{X}\)). Hence \(s, t : R \to u\) are locally of finite type (as compositions, see Morphisms of Spaces, Lemma 03XG). Since \(u\) is the spectrum of a field, it follows that \(s, t\) are flat and locally of finite presentation (by Morphisms of Spaces, Lemma 06G4). We see that \(\mathcal{Z} = [u/R]\) is an algebraic stack by Criteria for Representability, Theorem 06FI. By Algebraic Stacks, Lemma 04T4 we obtain a canonical morphism \[f : \mathcal{Z} \longrightarrow \mathcal{X}\] which is fully faithful. Hence this morphism is representable by algebraic spaces, see Algebraic Stacks, Lemma 04Y5 and a monomorphism, see Properties of Stacks, Lemma 04ZZ. It follows that the residual gerbe \(\mathcal{Z}_x \subset \mathcal{X}\) of \(\mathcal{X}\) at \(x\) exists and that \(f\) factors through an equivalence \(\mathcal{Z} \to \mathcal{Z}_x\), see Properties of Stacks, Lemma 06MX. By construction the diagram \[\xymatrix{ u \ar[d] \ar[r] & U \ar[d] \\ \mathcal{Z} \ar[r]^f & \mathcal{X} }\] is commutative. By Criteria for Representability, Lemma 06FH the left vertical arrow is surjective, flat, and locally of finite presentation. Consider \[\xymatrix{ u \times_\mathcal{X} U \ar[d] \ar[r] & \mathcal{Z} \times_\mathcal{X} U \ar[r] \ar[d] & U \ar[d] \\ u \ar[r] & \mathcal{Z} \ar[r]^f & \mathcal{X} }\] As \(u \to \mathcal{X}\) is locally of finite type, we see that the base change \(u \times_\mathcal{X} U \to U\) is locally of finite type. Moreover, \(u \times_\mathcal{X} U \to \mathcal{Z} \times_\mathcal{X} U\) is surjective, flat, and locally of finite presentation as a base change of \(u \to \mathcal{Z}\). Thus \(\{u \times_\mathcal{X} U \to \mathcal{Z} \times_\mathcal{X} U\}\) is an fppf covering of algebraic spaces, and we conclude that \(\mathcal{Z} \times_\mathcal{X} U \to U\) is locally of finite type by Descent on Spaces, Lemma 06EV. By definition this means that \(f\) is locally of finite type (because the vertical arrow \(\mathcal{Z} \times_\mathcal{X} U \to \mathcal{Z}\) is smooth as a base change of \(U \to \mathcal{X}\) and surjective as \(\mathcal{Z}\) has only one point). Since \(\mathcal{Z} = \mathcal{Z}_x\) we see that (3) holds.

It is clear that (3) implies (2). If (2) holds then \(x\) is a finite type point of \(\mathcal{X}\) by Lemma 06G0 and Lemma 06G2 to see that \(\mathcal{Z}_{\text{ft-pts}}\) is nonempty, i.e., the unique point of \(\mathcal{Z}\) is a finite type point of \(\mathcal{Z}\).

Automorphism groups

Let \(\mathcal{X}\) be an algebraic stack. Let \(x \in |\mathcal{X}|\) correspond to \(x : \Spec(k) \to \mathcal{X}\). In this situation we often use the phrase “let \(G_x/k\) be the automorphism group algebraic space of \(x\)”. This just means that \[G_x = \mathit{Isom}_\mathcal{X}(x, x) = \Spec(k) \times_\mathcal{X} \mathcal{I}_\mathcal{X}\] is the group algebraic space of automorphism of \(x\). This is a group algebraic space over \(\Spec(k)\). If \(k'/k\) is an extension of fields then the automorphism group algebraic space of the induced morphism \(x' : \Spec(k') \to \mathcal{X}\) is the base change of \(G_x\) to \(\Spec(k')\).

Lemma

In the situation above \(G_x\) is a scheme if one of the following holds

1. \(\Delta : \mathcal{X} \to \mathcal{X} \times \mathcal{X}\) is quasi-separated

2. \(\Delta : \mathcal{X} \to \mathcal{X} \times \mathcal{X}\) is locally separated,

3. \(\mathcal{X}\) is quasi-DM,

4. \(\mathcal{I}_\mathcal{X} \to \mathcal{X}\) is quasi-separated,

5. \(\mathcal{I}_\mathcal{X} \to \mathcal{X}\) is locally separated, or

6. \(\mathcal{I}_\mathcal{X} \to \mathcal{X}\) is locally quasi-finite.

Proof

Observe that (1) \(\Rightarrow\) (4), (2) \(\Rightarrow\) (5), and (3) \(\Rightarrow\) (6) by Lemma 0CL0. In case (4) we see that \(G_x\) is a quasi-separated algebraic space and in case (5) we see that \(G_x\) is a locally separated algebraic space. In both cases \(G_x\) is a decent algebraic space (Decent Spaces, Section 03I7 and Lemma 088J). Then \(G_x\) is separated by More on Groupoids in Spaces, Lemma 08BH whereupon we conclude that \(G_x\) is a scheme by More on Groupoids in Spaces, Proposition 0B8G. In case (6) we see that \(G_x \to \Spec(k)\) is locally quasi-finite and hence \(G_x\) is a scheme by Spaces over Fields, Lemma 06S0.

Lemma

Let \(\mathcal{X}\) be an algebraic stack. Let \(x \in |\mathcal{X}|\) be a point. Let \(P\) be a property of algebraic spaces over fields which is invariant under ground field extensions; for example \(P(X/k) = X \to \Spec(k)\text{ is finite}\). The following are equivalent

1. for some morphism \(x : \Spec(k) \to \mathcal{X}\) in the class of \(x\) the automorphism group algebraic space \(G_x/k\) has \(P\), and

2. for any morphism \(x : \Spec(k) \to \mathcal{X}\) in the class of \(x\) the automorphism group algebraic space \(G_x/k\) has \(P\).

Proof

Omitted.

Remark

Let \(P\) be a property of algebraic spaces over fields which is invariant under ground field extensions. Given an algebraic stack \(\mathcal{X}\) and \(x \in |\mathcal{X}|\), we say the automorphism group of \(\mathcal{X}\) at \(x\) has \(P\) if the equivalent conditions of Lemma 0DTT are satisfied. For example, we say the automorphism group of \(\mathcal{X}\) at \(x\) is finite, if \(G_x \to \Spec(k)\) is finite whenever \(x : \Spec(k) \to \mathcal{X}\) is a representative of \(x\). Similarly for smooth, proper, etc. (There is clearly an abuse of language going on here, but we believe it will not cause confusion or imprecision.)

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. Let \(x \in |\mathcal{X}|\) be a point. The following are equivalent

1. for some morphism \(x : \Spec(k) \to \mathcal{X}\) in the class of \(x\) setting \(y = f \circ x\) the map \(G_x \to G_y\) of automorphism group algebraic spaces is an isomorphism, and

2. for any morphism \(x : \Spec(k) \to \mathcal{X}\) in the class of \(x\) setting \(y = f \circ x\) the map \(G_x \to G_y\) of automorphism group algebraic spaces is an isomorphism.

Proof

This comes down to the fact that being an isomorphism is fpqc local on the target, see Descent on Spaces, Lemma 041Y. Namely, suppose that \(k'/k\) is an extension of fields and denote \(x' : \Spec(k') \to \mathcal{X}\) the composition and set \(y' = f \circ x'\). Then the morphism \(G_{x'} \to G_{y'}\) is the base change of \(G_x \to G_y\) by \(\Spec(k') \to \Spec(k)\). Hence \(G_x \to G_y\) is an isomorphism if and only if \(G_{x'} \to G_{y'}\) is an isomorphism. Thus we see that the property propagates through the equivalence class if it holds for one.

Remark

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. Let \(x \in |\mathcal{X}|\) be a point. To indicate the equivalent conditions of Lemma 0DTV are satisfied for \(f\) and \(x\) in the literature the terminology \(f\) is stabilizer preserving at \(x\) or \(f\) is fixed-point reflecting at \(x\) is used. We prefer to say \(f\) induces an isomorphism between automorphism groups at \(x\) and \(f(x)\).

Presentations and properties of algebraic stacks

Let \((U, R, s, t, c)\) be a groupoid in algebraic spaces. If \(s, t : R \to U\) are flat and locally of finite presentation, then the quotient stack \([U/R]\) is an algebraic stack, see Criteria for Representability, Theorem 06FI. In this section we study what properties of \((U, R, s, t, c)\) imply for the algebraic stack \([U/R]\).

Lemma

Let \((U, R, s, t, c)\) be a groupoid in algebraic spaces such that \(s, t : R \to U\) are flat and locally of finite presentation. Consider the algebraic stack \(\mathcal{X} = [U/R]\) (see above).

1. If \(R \to U \times U\) is separated, then \(\Delta_\mathcal{X}\) is separated.

2. If \(U\), \(R\) are separated, then \(\Delta_\mathcal{X}\) is separated.

3. If \(R \to U \times U\) is locally quasi-finite, then \(\mathcal{X}\) is quasi-DM.

4. If \(s, t : R \to U\) are locally quasi-finite, then \(\mathcal{X}\) is quasi-DM.

5. If \(R \to U \times U\) is proper, then \(\mathcal{X}\) is separated.

6. If \(s, t : R \to U\) are proper and \(U\) is separated, then \(\mathcal{X}\) is separated.

7. Add more here.

Proof

Observe that the morphism \(U \to \mathcal{X}\) is surjective, flat, and locally of finite presentation by Criteria for Representability, Lemma 06FH. Hence the same is true for \(U \times U \to \mathcal{X} \times \mathcal{X}\). We have the cartesian diagram \[\xymatrix{ R = U \times_\mathcal{X} U \ar[r] \ar[d] & U \times U \ar[d] \\ \mathcal{X} \ar[r] & \mathcal{X} \times \mathcal{X} }\] (see Groupoids in Spaces, Lemma 04M9). Thus we see that \(\Delta_\mathcal{X}\) has one of the properties listed in Properties of Stacks, Section 04XB if and only if the morphism \(R \to U \times U\) does, see Properties of Stacks, Lemma 04XD. This explains why (1), (3), and (5) are true. The condition in (2) implies \(R \to U \times U\) is separated hence (2) follows from (1). The condition in (4) implies the condition in (3) hence (4) follows from (3). The condition in (6) implies the condition in (5) by Morphisms of Spaces, Lemma 04NX hence (6) follows from (5).

Lemma

Let \((U, R, s, t, c)\) be a groupoid in algebraic spaces such that \(s, t : R \to U\) are flat and locally of finite presentation. Consider the algebraic stack \(\mathcal{X} = [U/R]\) (see above). Then the image of \(|R| \to |U| \times |U|\) is an equivalence relation and \(|\mathcal{X}|\) is the quotient of \(|U|\) by this equivalence relation.

Proof

The induced morphism \(p : U \to \mathcal{X}\) is surjective, flat, and locally of finite presentation, see Criteria for Representability, Lemma 06FH. Hence \(|U| \to |\mathcal{X}|\) is surjective by Properties of Stacks, Lemma 04XI. Note that \(R = U \times_\mathcal{X} U\), see Groupoids in Spaces, Lemma 04M9. Hence Properties of Stacks, Lemma 04XH implies the map \[|R| \longrightarrow |U| \times_{|\mathcal{X}|} |U|\] is surjective. Hence the image of \(|R| \to |U| \times |U|\) is exactly the set of pairs \((u_1, u_2) \in |U| \times |U|\) such that \(u_1\) and \(u_2\) have the same image in \(|\mathcal{X}|\). Combining these two statements we get the result of the lemma.

Special presentations of algebraic stacks

In this section we prove two important theorems. The first is the characterization of quasi-DM stacks \(\mathcal{X}\) as the stacks of the form \(\mathcal{X} = [U/R]\) with \(s, t : R \to U\) locally quasi-finite (as well as flat and locally of finite presentation). The second is the statement that DM algebraic stacks are Deligne-Mumford.

The following lemma gives a criterion for when a “slice” of a presentation is still flat over the algebraic stack.

Lemma

Let \(\mathcal{X}\) be an algebraic stack. Consider a cartesian diagram \[\xymatrix{ U \ar[d] & F \ar[l]^p \ar[d] \\ \mathcal{X} & \Spec(k) \ar[l] }\] where \(U\) is an algebraic space, \(k\) is a field, and \(U \to \mathcal{X}\) is flat and locally of finite presentation. Let \(f_1, \ldots, f_r \in \Gamma(U, \mathcal{O}_U)\) and \(z \in |F|\) such that \(f_1, \ldots, f_r\) map to a regular sequence in the local ring \(\mathcal{O}_{F, \overline{z}}\). Then, after replacing \(U\) by an open subspace containing \(p(z)\), the morphism \[V(f_1, \ldots, f_r) \longrightarrow \mathcal{X}\] is flat and locally of finite presentation.

Proof

Choose a scheme \(W\) and a surjective smooth morphism \(W \to \mathcal{X}\). Choose an extension of fields \(k'/k\) and a morphism \(w : \Spec(k') \to W\) such that \(\Spec(k') \to W \to \mathcal{X}\) is \(2\)-isomorphic to \(\Spec(k') \to \Spec(k) \to \mathcal{X}\). This is possible as \(W \to \mathcal{X}\) is surjective. Consider the commutative diagram \[\xymatrix{ U \ar[d] & U \times_\mathcal{X} W \ar[l]^-{\text{pr}_0} \ar[d] & F' \ar[l]^-{p'} \ar[d] \\ \mathcal{X} & W \ar[l] & \Spec(k') \ar[l] }\] both of whose squares are cartesian. By our choice of \(w\) we see that \(F' = F \times_{\Spec(k)} \Spec(k')\). Thus \(F' \to F\) is surjective and we can choose a point \(z' \in |F'|\) mapping to \(z\). Since \(F' \to F\) is flat we see that \(\mathcal{O}_{F, \overline{z}} \to \mathcal{O}_{F', \overline{z}'}\) is flat, see Morphisms of Spaces, Lemma 04NG. Hence \(f_1, \ldots, f_r\) map to a regular sequence in \(\mathcal{O}_{F', \overline{z}'}\), see Algebra, Lemma 00LM. Note that \(U \times_\mathcal{X} W \to W\) is a morphism of algebraic spaces which is flat and locally of finite presentation. Hence by More on Morphisms of Spaces, Lemma 06LW we see that there exists an open subspace \(U'\) of \(U \times_\mathcal{X} W\) containing \(p'(z')\) such that the intersection \(U' \cap (V(f_1, \ldots, f_r) \times_\mathcal{X} W)\) is flat and locally of finite presentation over \(W\). Note that \(\text{pr}_0(U')\) is an open subspace of \(U\) containing \(p(z)\) as \(\text{pr}_0\) is smooth hence open. Now we see that \(U' \cap (V(f_1, \ldots, f_r) \times_\mathcal{X} W) \to \mathcal{X}\) is flat and locally of finite presentation as the composition \[U' \cap (V(f_1, \ldots, f_r) \times_\mathcal{X} W) \to W \to \mathcal{X}.\] Hence Properties of Stacks, Lemma 06M2 implies \(\text{pr}_0(U') \cap V(f_1, \ldots, f_r) \to \mathcal{X}\) is flat and locally of finite presentation as desired.

Lemma

Let \(\mathcal{X}\) be an algebraic stack. Consider a cartesian diagram \[\xymatrix{ U \ar[d] & F \ar[l]^p \ar[d] \\ \mathcal{X} & \Spec(k) \ar[l] }\] where \(U\) is an algebraic space, \(k\) is a field, and \(U \to \mathcal{X}\) is locally of finite type. Let \(z \in |F|\) be such that \(\dim_z(F) = 0\). Then, after replacing \(U\) by an open subspace containing \(p(z)\), the morphism \[U \longrightarrow \mathcal{X}\] is locally quasi-finite.

Proof

Since \(f : U \to \mathcal{X}\) is locally of finite type there exists a maximal open \(W(f) \subset U\) such that the restriction \(f|_{W(f)} : W(f) \to \mathcal{X}\) is locally quasi-finite, see Properties of Stacks, Remark 06M5 (06M7). Hence all we need to do is prove that \(p(z)\) is a point of \(W(f)\). Moreover, the remark referenced above also shows the formation of \(W(f)\) commutes with arbitrary base change by a morphism which is representable by algebraic spaces. Hence it suffices to show that the morphism \(F \to \Spec(k)\) is locally quasi-finite at \(z\). This follows immediately from Morphisms of Spaces, Lemma 04NV.

A quasi-DM stack has a locally quasi-finite “covering” by a scheme.

Theorem

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 06MB. 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 06FX. Denote \(u = \Spec(\kappa(u))\) 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 06G3 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 0448). 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 06FF we have \(\dim(R) = 0\).

Let \(e : u \to R\) be the identity of the groupoid. The composition \(u \to R \to u\) is 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 \(F\); call it \(z\). Consider the maps of étale local rings \[\mathcal{O}_{U, u} \xrightarrow{p^\sharp} \mathcal{O}_{F, \overline{z}} \longrightarrow \mathcal{O}_{R, \overline{e}}\] 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 \(p^\sharp(\mathfrak m_u)\mathcal{O}_{F, \overline{z}}\) as \(R\) is cut out in \(F\) by \(\mathfrak m_u\). Thus we see that \[\mathfrak m_{\overline{z}} = \sqrt{p^\sharp(\mathfrak m_u)\mathcal{O}_{F, \overline{z}}}\] 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 06M1). Hence \(\mathcal{O}_{F, \overline{z}}\) is a regular local ring (Properties of Spaces, Lemma 06LQ). Note that a regular local ring is Cohen-Macaulay (Algebra, Lemma 00NQ). Let \(d = \dim(\mathcal{O}_{F, \overline{z}})\). By Algebra, Lemma 06LC 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 06MD 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 \(z\) is a point of this closed subspace, and that \[\dim(\mathcal{O}_{F_Z, \overline{z}}) = 0.\] By Morphisms of Spaces, Lemma 04NQ it follows that \(\dim_z(F_Z) = 0\) because the transcendence degree of \(z\) relative to \(u\) is zero. Hence it follows from Lemma 06ME 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 04XL. By construction the image contains all finite type points of \(\mathcal{X}\), hence \(f\) is surjective by Lemma 06G2 (and Properties of Stacks, Lemma 04XI).

Lemma

Let \(\mathcal{Z}\) be a DM, locally Noetherian, reduced algebraic stack with \(|\mathcal{Z}|\) a singleton. Then there exists a field \(k\) and a surjective étale morphism \(\Spec(k) \to \mathcal{Z}\).

Proof

By Properties of Stacks, Lemma 06MP there exists a field \(k\) and a surjective, flat, locally finitely presented morphism \(\Spec(k) \to \mathcal{Z}\). Set \(U = \Spec(k)\) and \(R = U \times_\mathcal{Z} U\) so we obtain a groupoid in algebraic spaces \((U, R, s, t, c)\), see Algebraic Stacks, Lemma 02ZY. Note that by Algebraic Stacks, Remark 04WY we have an equivalence \[f_{can} : [U/R] \longrightarrow \mathcal{Z}\] The projections \(s, t : R \to U\) are locally of finite presentation. As \(\mathcal{Z}\) is DM we see that the stabilizer group algebraic space \[G = U \times_{U \times U} R = U \times_{U \times U} (U \times_\mathcal{Z} U) = U \times_{\mathcal{Z} \times \mathcal{Z}, \Delta_\mathcal{Z}} \mathcal{Z}\] is unramified over \(U\). In particular \(\dim(G) = 0\) and by More on Groupoids in Spaces, Lemma 06FF we have \(\dim(R) = 0\). This implies that \(R\) is a scheme, see Spaces over Fields, Lemma 06LZ. By Varieties, Lemma 06LH we see that \(R\) (and also \(G\)) is the disjoint union of spectra of Artinian local rings finite over \(k\) via either \(s\) or \(t\). Let \(P = \Spec(A) \subset R\) be the open and closed subscheme whose underlying point is the identity \(e\) of the groupoid scheme \((U, R, s, t, c)\). As \(s \circ e = t \circ e = \text{id}_{\Spec(k)}\) we see that \(A\) is an Artinian local ring whose residue field is identified with \(k\) via either \(s^\sharp : k \to A\) or \(t^\sharp : k \to A\). Note that \(s, t : \Spec(A) \to \Spec(k)\) are finite (by the lemma referenced above). Since \(G \to \Spec(k)\) is unramified we see that \[G \cap P = P \times_{U \times U} U = \Spec(A \otimes_{k \otimes k} k)\] is unramified over \(k\). On the other hand \(A \otimes_{k \otimes k} k\) is local as a quotient of \(A\) and surjects onto \(k\). We conclude that \(A \otimes_{k \otimes k} k = k\). It follows that \(P \to U \times U\) is universally injective (as \(P\) has only one point with residue field \(k\)), unramified (by the computation of the fibre over the unique image point above), and of finite type (because \(s, t\) are) hence a monomorphism (see Étale Morphisms, Lemma 05VH). Thus \(s|_P, t|_P : P \to U\) define a finite flat equivalence relation. Thus we may apply Groupoids, Proposition 03BM to conclude that \(U/P\) exists and is a scheme \(\overline{U}\). Moreover, \(U \to \overline{U}\) is finite locally free and \(P = U \times_{\overline{U}} U\). In fact \(\overline{U} = \Spec(k_0)\) where \(k_0 \subset k\) is the ring of \(R\)-invariant functions. As \(k\) is a field it follows from the definition Groupoids, Equation (03BF) that \(k_0\) is a field.

We claim that [06N1]\[\begin{equation} \Spec(k_0) = \overline{U} = U/P \to [U/R] = \mathcal{Z} \end{equation}\] is the desired surjective étale morphism. It follows from Properties of Stacks, Lemma 06MM that this morphism is surjective. Thus it suffices to show that (06N1) is étale⁶. Instead of proving the étaleness directly we first apply Bootstrap, Lemma 04S4 to see that there exists a groupoid scheme \((\overline{U}, \overline{R}, \overline{s}, \overline{t}, \overline{c})\) such that \((U, R, s, t, c)\) is the restriction of \((\overline{U}, \overline{R}, \overline{s}, \overline{t}, \overline{c})\) via the quotient morphism \(U \to \overline{U}\). (We verified all the hypothesis of the lemma above except for the assertion that \(j : R \to U \times U\) is separated and locally quasi-finite which follows from the fact that \(R\) is a separated scheme locally quasi-finite over \(k\).) Since \(U \to \overline{U}\) is finite locally free we see that \([U/R] \to [\overline{U}/\overline{R}]\) is an equivalence, see Groupoids in Spaces, Lemma 046T.

Note that \(s, t\) are the base changes of the morphisms \(\overline{s}, \overline{t}\) by \(U \to \overline{U}\). As \(\{U \to \overline{U}\}\) is an fppf covering we conclude \(\overline{s}, \overline{t}\) are flat, locally of finite presentation, and locally quasi-finite, see Descent, Lemmas 02L2, 02KY, and 02VI. Consider the commutative diagram \[\xymatrix{ U \times_{\overline{U}} U \ar@{=}[r] \ar[rd] & P \ar[r] \ar[d] & R \ar[d] \\ & \overline{U} \ar[r]^{\overline{e}} & \overline{R} }\] It is a general fact about restrictions that the outer four corners form a cartesian diagram. By the equality we see the inner square is cartesian. Since \(P\) is open in \(R\) we conclude that \(\overline{e}\) is an open immersion by Descent, Lemma 02L3.

But of course, if \(\overline{e}\) is an open immersion and \(\overline{s}, \overline{t}\) are flat and locally of finite presentation then the morphisms \(\overline{t}, \overline{s}\) are étale. For example you can see this by applying More on Groupoids, Lemma 04R9 which shows that \(\Omega_{\overline{R}/\overline{U}} = 0\) implies that \(\overline{s}, \overline{t} : \overline{R} \to \overline{U}\) is unramified (see Morphisms, Lemma 02G5), which in turn implies that \(\overline{s}, \overline{t}\) are étale (see Morphisms, Lemma 02GV). Hence \(\mathcal{Z} = [\overline{U}/\overline{R}]\) is an étale presentation of the algebraic stack \(\mathcal{Z}\) and we conclude that \(\overline{U} \to \mathcal{Z}\) is étale by Properties of Stacks, Lemma 04XD.

Lemma

Let \(\mathcal{X}\) be an algebraic stack. Consider a cartesian diagram \[\xymatrix{ U \ar[d] & F \ar[l]^p \ar[d] \\ \mathcal{X} & \Spec(k) \ar[l] }\] where \(U\) is an algebraic space, \(k\) is a field, and \(U \to \mathcal{X}\) is flat and locally of finite presentation. Let \(z \in |F|\) be such that \(F \to \Spec(k)\) is unramified at \(z\). Then, after replacing \(U\) by an open subspace containing \(p(z)\), the morphism \[U \longrightarrow \mathcal{X}\] is étale.

Proof

Since \(f : U \to \mathcal{X}\) is flat and locally of finite presentation there exists a maximal open \(W(f) \subset U\) such that the restriction \(f|_{W(f)} : W(f) \to \mathcal{X}\) is étale, see Properties of Stacks, Remark 06M5 (06MA). Hence all we need to do is prove that \(p(z)\) is a point of \(W(f)\). Moreover, the remark referenced above also shows the formation of \(W(f)\) commutes with arbitrary base change by a morphism which is representable by algebraic spaces. Hence it suffices to show that the morphism \(F \to \Spec(k)\) is étale at \(z\). Since it is flat and locally of finite presentation as a base change of \(U \to \mathcal{X}\) and since \(F \to \Spec(k)\) is unramified at \(z\) by assumption, this follows from Morphisms of Spaces, Lemma 06LU.

A DM stack is a Deligne-Mumford stack.

Theorem

Let \(\mathcal{X}\) be an algebraic stack. The following are equivalent

1. \(\mathcal{X}\) is DM,

2. \(\mathcal{X}\) is Deligne-Mumford, and

3. there exists a scheme \(W\) and a surjective étale morphism \(W \to \mathcal{X}\).

Proof

Recall that (3) is the definition of (2), see Algebraic Stacks, Definition 03YO. The implication (3) \(\Rightarrow\) (1) is Lemma 06MB. 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.

By Lemma 06G3 the residual gerbe \(\mathcal{Z}_x\) of \(\mathcal{X}\) at \(x\) exists and \(\mathcal{Z}_x \to \mathcal{X}\) is locally of finite type. By Lemma 06MZ the algebraic stack \(\mathcal{Z}_x\) is DM. By Lemma 06N0 there exists a field \(k\) and a surjective étale morphism \(z : \Spec(k) \to \mathcal{Z}_x\). In particular the composition \(x : \Spec(k) \to \mathcal{X}\) is locally of finite type (by Morphisms of Spaces, Lemmas 03XG and 06LT).

Pick a scheme \(U\) and a smooth morphism \(U \to \mathcal{X}\) such that \(x\) is in the image of \(|U| \to |\mathcal{X}|\). Consider the following fibre square \[\xymatrix{ U \ar[d] & F \ar[l] \ar[d] \\ \mathcal{X} & \Spec(k) \ar[l]_-x }\] in other words \(F = U \times_{\mathcal{X}, x} \Spec(k)\). By Properties of Stacks, Lemma 04XH we see that \(F\) is nonempty. As \(\mathcal{Z}_x \to \mathcal{X}\) is a monomorphism we have \[\Spec(k) \times_{z, \mathcal{Z}_x, z} \Spec(k) = \Spec(k) \times_{x, \mathcal{X}, x} \Spec(k)\] with étale projection maps to \(\Spec(k)\) by construction of \(z\). Since \[F \times_U F = (\Spec(k) \times_\mathcal{X} \Spec(k)) \times_{\Spec(k)} F\] we see that the projections maps \(F \times_U F \to F\) are étale as well. It follows that \(\Delta_{F/U} : F \to F \times_U F\) is étale (see Morphisms of Spaces, Lemma 05W3). By Morphisms of Spaces, Lemma 05W5 this implies that \(\Delta_{F/U}\) is an open immersion, which finally implies by Morphisms of Spaces, Lemma 05W1 that \(F \to U\) is unramified.

Pick a nonempty affine scheme \(V\) and an étale morphism \(V \to F\). (This could be avoided by working directly with \(F\), but it seems easier to explain what’s going on by doing so.) Picture \[\xymatrix{ U \ar[d] & F \ar[l] \ar[d] & V \ar[l] \ar[ld] \\ \mathcal{X} & \Spec(k) \ar[l]_-x }\] Then \(V \to \Spec(k)\) is a smooth morphism of schemes and \(V \to U\) is an unramified morphism of schemes (see Morphisms of Spaces, Lemmas 03ZD and 03ZI). Pick a closed point \(v \in V\) with \(k \subset \kappa(v)\) finite separable, see Varieties, Lemma 056U. Let \(u \in U\) be the image point. The local ring \(\mathcal{O}_{V, v}\) is regular (see Varieties, Lemma 056S) and the local ring homomorphism \[\varphi : \mathcal{O}_{U, u} \longrightarrow \mathcal{O}_{V, v}\] coming from the morphism \(V \to U\) is such that \(\varphi(\mathfrak m_u)\mathcal{O}_{V, v} = \mathfrak m_v\), see Morphisms, Lemma 02GF. Hence we can find \(f_1, \ldots, f_d \in \mathcal{O}_{U, u}\) such that the images \(\varphi(f_1), \ldots, \varphi(f_d)\) form a basis for \(\mathfrak m_v/\mathfrak m_v^2\) over \(\kappa(v)\). Since \(\mathcal{O}_{V, v}\) is a regular local ring this implies that \(\varphi(f_1), \ldots, \varphi(f_d)\) form a regular sequence in \(\mathcal{O}_{V, v}\) (see Algebra, Lemma 00NQ). After replacing \(U\) by an open neighbourhood of \(u\) we may assume \(f_1, \ldots, f_d \in \Gamma(U, \mathcal{O}_U)\). After replacing \(U\) by a possibly even smaller open neighbourhood of \(u\) we may assume that \(V(f_1, \ldots, f_d) \to \mathcal{X}\) is flat and locally of finite presentation, see Lemma 06MD. By construction \[V(f_1, \ldots, f_d) \times_\mathcal{X} \Spec(k) \longleftarrow V(f_1, \ldots, f_d) \times_U V\] is étale and \(V(f_1, \ldots, f_d) \times_U V\) is the closed subscheme \(T \subset V\) cut out by \(f_1|_V, \ldots, f_d|_V\). Hence by construction \(v \in T\) and \[\mathcal{O}_{T, v} = \mathcal{O}_{V, v}/(\varphi(f_1), \ldots, \varphi(f_d)) = \kappa(v)\] a finite separable extension of \(k\). It follows that \(T \to \Spec(k)\) is unramified at \(v\), see Morphisms, Lemma 02GF. By definition of an unramified morphism of algebraic spaces this means that \(V(f_1, \ldots, f_d) \times_\mathcal{X} \Spec(k) \to \Spec(k)\) is unramified at the image of \(v\) in \(V(f_1, \ldots, f_d) \times_\mathcal{X} \Spec(k)\). Applying Lemma 06N2 we see that on shrinking \(U\) to yet another open neighbourhood of \(u\) the morphism \(V(f_1, \ldots, f_d) \to \mathcal{X}\) is étale.

We conclude that for every finite type point \(x\) of \(\mathcal{X}\) there exists an étale morphism \(f_x : W_x \to \mathcal{X}\) with \(x\) in the image of \(|f_x|\). Set \(W = \coprod_x W_x\) and \(f = \coprod f_x\). Then \(f\) is étale. In particular the image of \(|f|\) is open, see Properties of Stacks, Lemma 04XL. By construction the image contains all finite type points of \(\mathcal{X}\), hence \(f\) is surjective by Lemma 06G2 (and Properties of Stacks, Lemma 04XI).

Here is a useful corollary which tells us that the “fibres” of a DM morphism of algebraic stacks are Deligne-Mumford.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a DM morphism of algebraic stacks. Then

1. For every DM algebraic stack \(\mathcal{Z}\) and morphism \(\mathcal{Z} \to \mathcal{Y}\) there exists a scheme and a surjective étale morphism \(U \to \mathcal{X} \times_\mathcal{Y} \mathcal{Z}\).

2. For every algebraic space \(Z\) and morphism \(Z \to \mathcal{Y}\) there exists a scheme and a surjective étale morphism \(U \to \mathcal{X} \times_\mathcal{Y} Z\).

Proof

Proof of (1). As \(f\) is DM we see that the base change \(\mathcal{X} \times_\mathcal{Y} \mathcal{Z} \to \mathcal{Z}\) is DM by Lemma 050F. Since \(\mathcal{Z}\) is DM this implies that \(\mathcal{X} \times_\mathcal{Y} \mathcal{Z}\) is DM by Lemma 050L. Hence there exists a scheme \(U\) and a surjective étale morphism \(U \to \mathcal{X} \times_\mathcal{Y} \mathcal{Z}\), see Theorem 06N3. Part (2) is a special case of (1) since an algebraic space (when viewed as an algebraic stack) is DM by Lemma 050E.

The Deligne-Mumford locus

Every algebraic stack has a largest open substack which is a Deligne-Mumford stack; this is more or less clear but we also write out the proof below. Of course this substack may be empty, for example if \(X = [\Spec(\mathbf{Z})/\mathbf{G}_{m, \mathbf{Z}}]\). Below we will characterize the points of the DM locus.

Lemma

Let \(\mathcal{X}\) be an algebraic stack. There exist open substacks \[\mathcal{X}'' \subset \mathcal{X}' \subset \mathcal{X}\] such that \(\mathcal{X}''\) is DM, \(\mathcal{X}'\) is quasi-DM, and such that these are the largest open substacks with these properties.

Proof

All we are really saying here is that if \(\mathcal{U} \subset \mathcal{X}\) and \(\mathcal{V} \subset \mathcal{X}\) are open substacks which are DM, then the open substack \(\mathcal{W} \subset \mathcal{X}\) with \(|\mathcal{W}| = |\mathcal{U}| \cup |\mathcal{V}|\) is DM as well. (Similarly for quasi-DM.) Although this is a cheat, let us use Theorem 06N3 to prove this. By that theorem we can choose schemes \(U\) and \(V\) and surjective étale morphisms \(U \to \mathcal{U}\) and \(V \to \mathcal{V}\). Then of course \(U \amalg V \to \mathcal{W}\) is surjective and étale. The quasi-DM case is proven by exactly the same method using Theorem 06MF.

Lemma

Let \(\mathcal{X}\) be an algebraic stack. Let \(x \in |\mathcal{X}|\) correspond to \(x : \Spec(k) \to \mathcal{X}\). Let \(G_x/k\) be the automorphism group algebraic space of \(x\). Then

1. \(x\) is in the DM locus of \(\mathcal{X}\) if and only if \(G_x \to \Spec(k)\) is unramified, and

2. \(x\) is in the quasi-DM locus of \(\mathcal{X}\) if and only if \(G_x \to \Spec(k)\) is locally quasi-finite.

Proof

Proof of (2). Choose a scheme \(U\) and a surjective smooth morphism \(U \to \mathcal{X}\). Consider the fibre product \[\xymatrix{ G \ar[r] \ar[d] & \mathcal{I}_\mathcal{X} \ar[d] \\ U \ar[r] & \mathcal{X} }\] Recall that \(G\) is the automorphism group algebraic space of \(U \to \mathcal{X}\). By Groupoids in Spaces, Lemma 0DSJ there is a maximal open subscheme \(U' \subset U\) such that \(G_{U'} \to U'\) is locally quasi-finite. Moreover, formation of \(U'\) commutes with arbitrary base change. In particular the two inverse images of \(U'\) in \(R = U \times_\mathcal{X} U\) are the same open subspace of \(R\) (since after all the two maps \(R \to \mathcal{X}\) are isomorphic and hence have isomorphic automorphism group spaces). Hence \(U'\) is the inverse image of an open substack \(\mathcal{X}' \subset \mathcal{X}\) by Properties of Stacks, Lemma 0507 and we have a cartesian diagram \[\xymatrix{ G_{U'} \ar[r] \ar[d] & \mathcal{I}_{\mathcal{X}'} \ar[d] \\ U' \ar[r] & \mathcal{X}' }\] Thus the morphism \(\mathcal{I}_{\mathcal{X}'} \to \mathcal{X}'\) is locally quasi-finite and we conclude that \(\mathcal{X}'\) is quasi-DM by Lemma 0CL0 part (5). On the other hand, if \(\mathcal{W} \subset \mathcal{X}\) is an open substack which is quasi-DM, then the inverse image \(W \subset U\) of \(\mathcal{W}\) must be contained in \(U'\) by our construction of \(U'\) since \(\mathcal{I}_\mathcal{W} = \mathcal{W} \times_\mathcal{X} \mathcal{I}_\mathcal{X}\) is locally quasi-finite over \(\mathcal{W}\). Thus \(\mathcal{X}'\) is the quasi-DM locus. Finally, choose a field extension \(K/k\) and a \(2\)-commutative diagram \[\xymatrix{ \Spec(K) \ar[r] \ar[d] & \Spec(k) \ar[d]^x \\ U \ar[r] & \mathcal{X} }\] Then we find an isomorphism \(G_x \times_{\Spec(k)} \Spec(K) \cong G \times_U \Spec(K)\) of group algebraic spaces over \(K\). Hence \(G_x\) is locally quasi-finite over \(k\) if and only if \(\Spec(K) \to U\) maps into \(U'\) (use the commutation of formation of \(U'\) and Groupoids in Spaces, Lemma 0DSJ applied to \(\Spec(K) \to \Spec(k)\) and \(G_x\) to see this). This finishes the proof of (2). The proof of (1) is exactly the same.

Locally quasi-finite morphisms

The property “locally quasi-finite” of morphisms of algebraic spaces is not smooth local on the source-and-target so we cannot use the material in Section 06FL to define locally quasi-finite morphisms of algebraic stacks. We do already know what it means for a morphism of algebraic stacks representable by algebraic spaces to be locally quasi-finite, see Properties of Stacks, Section 04XB. To find a condition suitable for general morphisms we make the following observation.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. Assume \(f\) is representable by algebraic spaces. The following are equivalent

1. \(f\) is locally quasi-finite (as in Properties of Stacks, Section 04XB), and

2. \(f\) is locally of finite type and for every morphism \(\Spec(k) \to \mathcal{Y}\) where \(k\) is a field the space \(|\Spec(k) \times_\mathcal{Y} \mathcal{X}|\) is discrete.

Proof

Assume (1). In this case the morphism of algebraic spaces \(\mathcal{X}_k \to \Spec(k)\) is locally quasi-finite as a base change of \(f\). Hence \(|\mathcal{X}_k|\) is discrete by Morphisms of Spaces, Lemma 06RW. Conversely, assume (2). Pick a surjective smooth morphism \(V \to \mathcal{Y}\) where \(V\) is a scheme. It suffices to show that the morphism of algebraic spaces \(V \times_\mathcal{Y} \mathcal{X} \to V\) is locally quasi-finite, see Properties of Stacks, Lemma 04XD. The morphism \(V \times_\mathcal{Y} \mathcal{X} \to V\) is locally of finite type by assumption. For any morphism \(\Spec(k) \to V\) where \(k\) is a field \[\Spec(k) \times_V (V \times_\mathcal{Y} \mathcal{X}) = \Spec(k) \times_\mathcal{Y} \mathcal{X}\] has a discrete space of points by assumption. Hence we conclude that \(V \times_\mathcal{Y} \mathcal{X} \to V\) is locally quasi-finite by Morphisms of Spaces, Lemma 06RW.

A morphism of algebraic stacks which is representable by algebraic spaces is quasi-DM, see Lemma 050E. Combined with the lemma above we see that the following definition does not conflict with the already existing notion in the case of morphisms representable by algebraic spaces.

Definition

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. We say \(f\) is locally quasi-finite if \(f\) is quasi-DM, locally of finite type, and for every morphism \(\Spec(k) \to \mathcal{Y}\) where \(k\) is a field the space \(|\mathcal{X}_k|\) is discrete.

The condition that \(f\) be quasi-DM is natural. For example, let \(k\) be a field and consider the morphism \(\pi : [\Spec(k)/\mathbf{G}_m] \to \Spec(k)\) which has singleton fibres and is locally of finite type. As we will see later this morphism is smooth of relative dimension \(-1\), and we’d like our locally quasi-finite morphisms to have relative dimension \(0\). Also, note that the section \(\Spec(k) \to [\Spec(k)/\mathbf{G}_m]\) does not have discrete fibres, hence is not locally quasi-finite, and we’d like to have the following permanence property for locally quasi-finite morphisms: If \(f : \mathcal{X} \to \mathcal{X}'\) is a morphism of algebraic stacks locally quasi-finite over the algebraic stack \(\mathcal{Y}\), then \(f\) is locally quasi-finite (in fact something a bit stronger holds, see Lemma 06UG).

Another justification for the definition above is Lemma 06UF below which characterizes being locally quasi-finite in terms of the existence of suitable “presentations” or “coverings” of \(\mathcal{X}\) and \(\mathcal{Y}\).

Lemma

A base change of a locally quasi-finite morphism is locally quasi-finite.

Proof

We have seen this for quasi-DM morphisms in Lemma 050F and for locally finite type morphisms in Lemma 06FU. It is immediate that the condition on fibres is inherited by a base change.

Lemma

Let \(\mathcal{X} \to \Spec(k)\) be a locally quasi-finite morphism where \(\mathcal{X}\) is an algebraic stack and \(k\) is a field. Let \(f : V \to \mathcal{X}\) be a locally quasi-finite morphism where \(V\) is a scheme. Then \(V \to \Spec(k)\) is locally quasi-finite.

Proof

By Lemma 06FT we see that \(V \to \Spec(k)\) is locally of finite type. Assume, to get a contradiction, that \(V \to \Spec(k)\) is not locally quasi-finite. Then there exists a nontrivial specialization \(v \leadsto v'\) of points of \(V\), see Morphisms, Lemma 01TH. In particular \(\text{trdeg}_k(\kappa(v)) > \text{trdeg}_k(\kappa(v'))\), see Morphisms, Lemma 06RU. Because \(|\mathcal{X}|\) is discrete we see that \(|f|(v) = |f|(v')\). Consider \(R = V \times_\mathcal{X} V\). Then \(R\) is an algebraic space and the projections \(s, t : R \to V\) are locally quasi-finite as base changes of \(V \to \mathcal{X}\) (which is representable by algebraic spaces so this follows from the discussion in Properties of Stacks, Section 04XB). By Properties of Stacks, Lemma 04XH we see that there exists an \(r \in |R|\) such that \(s(r) = v\) and \(t(r) = v'\). By Morphisms of Spaces, Lemma 06RX we see that the transcendence degree of \(v/k\) is equal to the transcendence degree of \(r/k\) is equal to the transcendence degree of \(v'/k\). This contradiction proves the lemma.

Lemma

A composition of a locally quasi-finite morphisms is locally quasi-finite.

Proof

We have seen this for quasi-DM morphisms in Lemma 050K and for locally finite type morphisms in Lemma 06FT. Let \(\mathcal{X} \to \mathcal{Y}\) and \(\mathcal{Y} \to \mathcal{Z}\) be locally quasi-finite. Let \(k\) be a field and let \(\Spec(k) \to \mathcal{Z}\) be a morphism. It suffices to show that \(|\mathcal{X}_k|\) is discrete. By Lemma 06UB the morphisms \(\mathcal{X}_k \to \mathcal{Y}_k\) and \(\mathcal{Y}_k \to \Spec(k)\) are locally quasi-finite. In particular we see that \(\mathcal{Y}_k\) is a quasi-DM algebraic stack, see Lemma 050N. By Theorem 06MF we can find a scheme \(V\) and a surjective, flat, locally finitely presented, locally quasi-finite morphism \(V \to \mathcal{Y}_k\). By Lemma 06UC we see that \(V\) is locally quasi-finite over \(k\), in particular \(|V|\) is discrete. The morphism \(V \times_{\mathcal{Y}_k} \mathcal{X}_k \to \mathcal{X}_k\) is surjective, flat, and locally of finite presentation hence \(|V \times_{\mathcal{Y}_k} \mathcal{X}_k| \to |\mathcal{X}_k|\) is surjective and open. Thus it suffices to show that \(|V \times_{\mathcal{Y}_k} \mathcal{X}_k|\) is discrete. Note that \(V\) is a disjoint union of spectra of Artinian local \(k\)-algebras \(A_i\) with residue fields \(k_i\), see Varieties, Lemma 06LH. Thus it suffices to show that each \[|\Spec(A_i) \times_{\mathcal{Y}_k} \mathcal{X}_k| = |\Spec(k_i) \times_{\mathcal{Y}_k} \mathcal{X}_k| = |\Spec(k_i) \times_\mathcal{Y} \mathcal{X}|\] is discrete, which follows from the assumption that \(\mathcal{X} \to \mathcal{Y}\) is locally quasi-finite.

Before we characterize locally quasi-finite morphisms in terms of coverings we do it for quasi-DM morphisms.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. The following are equivalent

1. \(f\) is quasi-DM,

2. for any morphism \(V \to \mathcal{Y}\) with \(V\) an algebraic space there exists a surjective, flat, locally finitely presented, locally quasi-finite morphism \(U \to \mathcal{X} \times_\mathcal{Y} V\) where \(U\) is an algebraic space, and

3. there exist algebraic spaces \(U\), \(V\) and a morphism \(V \to \mathcal{Y}\) which is surjective, flat, and locally of finite presentation, and a morphism \(U \to \mathcal{X} \times_\mathcal{Y} V\) which is surjective, flat, locally of finite presentation, and locally quasi-finite.

Proof

The implication (2) \(\Rightarrow\) (3) is immediate.

Assume (1) and let \(V \to \mathcal{Y}\) be as in (2). Then \(\mathcal{X} \times_\mathcal{Y} V \to V\) is quasi-DM, see Lemma 050F. By Lemma 050E the algebraic space \(V\) is DM, hence quasi-DM. Thus \(\mathcal{X} \times_\mathcal{Y} V\) is quasi-DM by Lemma 050L. Hence we may apply Theorem 06MF to get the morphism \(U \to \mathcal{X} \times_\mathcal{Y} V\) as in (2).

Assume (3). Let \(V \to \mathcal{Y}\) and \(U \to \mathcal{X} \times_\mathcal{Y} V\) be as in (3). To prove that \(f\) is quasi-DM it suffices to show that \(\mathcal{X} \times_\mathcal{Y} V \to V\) is quasi-DM, see Lemma 06TZ. By Lemma 06MB we see that \(\mathcal{X} \times_\mathcal{Y} V\) is quasi-DM. Hence \(\mathcal{X} \times_\mathcal{Y} V \to V\) is quasi-DM by Lemma 050N and (1) holds. This finishes the proof of the lemma.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. The following are equivalent

1. \(f\) is locally quasi-finite,

2. \(f\) is quasi-DM and for any morphism \(V \to \mathcal{Y}\) with \(V\) an algebraic space and any locally quasi-finite morphism \(U \to \mathcal{X} \times_\mathcal{Y} V\) where \(U\) is an algebraic space the morphism \(U \to V\) is locally quasi-finite,

3. for any morphism \(V \to \mathcal{Y}\) from an algebraic space \(V\) there exists a surjective, flat, locally finitely presented, and locally quasi-finite morphism \(U \to \mathcal{X} \times_\mathcal{Y} V\) where \(U\) is an algebraic space such that \(U \to V\) is locally quasi-finite,

4. there exists algebraic spaces \(U\), \(V\), a surjective, flat, and locally of finite presentation morphism \(V \to \mathcal{Y}\), and a morphism \(U \to \mathcal{X} \times_\mathcal{Y} V\) which is surjective, flat, locally of finite presentation, and locally quasi-finite such that \(U \to V\) is locally quasi-finite.

Proof

Assume (1). Then \(f\) is quasi-DM by assumption. Let \(V \to \mathcal{Y}\) and \(U \to \mathcal{X} \times_\mathcal{Y} V\) be as in (2). By Lemma 06UD the composition \(U \to \mathcal{X} \times_\mathcal{Y} V \to V\) is locally quasi-finite. Thus (1) implies (2).

Assume (2). Let \(V \to \mathcal{Y}\) be as in (3). By Lemma 06UE we can find an algebraic space \(U\) and a surjective, flat, locally finitely presented, locally quasi-finite morphism \(U \to \mathcal{X} \times_\mathcal{Y} V\). By (2) the composition \(U \to V\) is locally quasi-finite. Thus (2) implies (3).

It is immediate that (3) implies (4).

Assume (4). We will prove (1) holds, which finishes the proof. By Lemma 06UE we see that \(f\) is quasi-DM. To prove that \(f\) is locally of finite type it suffices to prove that \(g : \mathcal{X} \times_\mathcal{Y} V \to V\) is locally of finite type, see Lemma 06U7. Then it suffices to check that \(g\) precomposed with \(h : U \to \mathcal{X} \times_\mathcal{Y} V\) is locally of finite type, see Lemma 06U8. Since \(g \circ h : U \to V\) was assumed to be locally quasi-finite this holds, hence \(f\) is locally of finite type. Finally, let \(k\) be a field and let \(\Spec(k) \to \mathcal{Y}\) be a morphism. Then \(V \times_\mathcal{Y} \Spec(k)\) is a nonempty algebraic space which is locally of finite presentation over \(k\). Hence we can find a finite extension \(k'/k\) and a morphism \(\Spec(k') \to V\) such that \[\xymatrix{ \Spec(k') \ar[r] \ar[d] & V \ar[d] \\ \Spec(k) \ar[r] & \mathcal{Y} }\] commutes (details omitted). Then \(\mathcal{X}_{k'} \to \mathcal{X}_k\) is representable (by schemes), surjective, and finite locally free. In particular \(|\mathcal{X}_{k'}| \to |\mathcal{X}_k|\) is surjective and open. Thus it suffices to prove that \(|\mathcal{X}_{k'}|\) is discrete. Since \[U \times_V \Spec(k') = U \times_{\mathcal{X} \times_\mathcal{Y} V} \mathcal{X}_{k'}\] we see that \(U \times_V \Spec(k') \to \mathcal{X}_{k'}\) is surjective, flat, and locally of finite presentation (as a base change of \(U \to \mathcal{X} \times_\mathcal{Y} V\)). Hence \(|U \times_V \Spec(k')| \to |\mathcal{X}_{k'}|\) is surjective and open. Thus it suffices to show that \(|U \times_V \Spec(k')|\) is discrete. This follows from the fact that \(U \to V\) is locally quasi-finite (either by our definition above or from the original definition for morphisms of algebraic spaces, via Morphisms of Spaces, Lemma 06RW).

Lemma

Let \(\mathcal{X} \to \mathcal{Y} \to \mathcal{Z}\) be morphisms of algebraic stacks. Assume that \(\mathcal{X} \to \mathcal{Z}\) is locally quasi-finite and \(\mathcal{Y} \to \mathcal{Z}\) is quasi-DM. Then \(\mathcal{X} \to \mathcal{Y}\) is locally quasi-finite.

Proof

Write \(\mathcal{X} \to \mathcal{Y}\) as the composition \[\mathcal{X} \longrightarrow \mathcal{X} \times_\mathcal{Z} \mathcal{Y} \longrightarrow \mathcal{Y}\] The second arrow is locally quasi-finite as a base change of \(\mathcal{X} \to \mathcal{Z}\), see Lemma 06UB. The first arrow is locally quasi-finite by Lemma 050I as \(\mathcal{Y} \to \mathcal{Z}\) is quasi-DM. Hence \(\mathcal{X} \to \mathcal{Y}\) is locally quasi-finite by Lemma 06UD.

Quasi-finite morphisms

We have defined “locally quasi-finite” morphisms of algebraic stacks in Section 06PT and “quasi-compact” morphisms of algebraic stacks in Section 050S. Since a morphism of algebraic spaces is by definition quasi-finite if and only if it is both locally quasi-finite and quasi-compact (Morphisms of Spaces, Definition 03XJ), we may define what it means for a morphism of algebraic stacks to be quasi-finite as follows and it agrees with the already existing notion defined in Properties of Stacks, Section 04XB when the morphism is representable by algebraic spaces.

Definition

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. We say \(f\) is quasi-finite if \(f\) is locally quasi-finite (Definition 06PU) and quasi-compact (Definition 050U).

Lemma

The composition of quasi-finite morphisms is quasi-finite.

Proof

Combine Lemmas 06UD and 050W.

Lemma

A base change of a quasi-finite morphism is quasi-finite.

Proof

Combine Lemmas 06UB and 050V.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) and \(g : \mathcal{Y} \to \mathcal{Z}\) be morphisms of algebraic stacks. If \(g \circ f\) is quasi-finite and \(g\) is quasi-separated and quasi-DM then \(f\) is quasi-finite.

Proof

Combine Lemmas 06UG and 050Y.

Flat morphisms

The property “being flat” of morphisms of algebraic spaces is smooth local on the source-and-target, see Descent on Spaces, Remark 06FC. It is also stable under base change and fpqc local on the target, see Morphisms of Spaces, Lemma 03MO and Descent on Spaces, Lemma 041W. Hence, by Lemma 06FM above, we may define what it means for a morphism of algebraic spaces to be flat as follows and it agrees with the already existing notion defined in Properties of Stacks, Section 04XB when the morphism is representable by algebraic spaces.

Definition

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. We say \(f\) is flat if the equivalent conditions of Lemma 06FM hold with \(\mathcal{P} = \text{flat}\).

Lemma

The composition of flat morphisms is flat.

Proof

Combine Remark 06FP with Morphisms of Spaces, Lemma 03MN.

Lemma

A base change of a flat morphism is flat.

Proof

Combine Remark 06FQ with Morphisms of Spaces, Lemma 03MO.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. Let \(\mathcal{Z} \to \mathcal{Y}\) be a surjective flat morphism of algebraic stacks. If the base change \(\mathcal{Z} \times_\mathcal{Y} \mathcal{X} \to \mathcal{Z}\) is flat, then \(f\) is flat.

Proof

Choose an algebraic space \(W\) and a surjective smooth morphism \(W \to \mathcal{Z}\). Then \(W \to \mathcal{Z}\) is surjective and flat (Morphisms of Spaces, Lemma 04TA) hence \(W \to \mathcal{Y}\) is surjective and flat (by Properties of Stacks, Lemma 04ZT and Lemma 06PX). Since the base change of \(\mathcal{Z} \times_\mathcal{Y} \mathcal{X} \to \mathcal{Z}\) by \(W \to \mathcal{Z}\) is a flat morphism (Lemma 06PY) we may replace \(\mathcal{Z}\) by \(W\).

Choose an algebraic space \(V\) and a surjective smooth morphism \(V \to \mathcal{Y}\). Choose an algebraic space \(U\) and a surjective smooth morphism \(U \to V \times_\mathcal{Y} \mathcal{X}\). We have to show that \(U \to V\) is flat. Now we base change everything by \(W \to \mathcal{Y}\): Set \(U' = W \times_\mathcal{Y} U\), \(V' = W \times_\mathcal{Y} V\), \(\mathcal{X}' = W \times_\mathcal{Y} \mathcal{X}\), and \(\mathcal{Y}' = W \times_\mathcal{Y} \mathcal{Y} = W\). Then it is still true that \(U' \to V' \times_{\mathcal{Y}'} \mathcal{X}'\) is smooth by base change. Hence by our definition of flat morphisms of algebraic stacks and the assumption that \(\mathcal{X}' \to \mathcal{Y}'\) is flat, we see that \(U' \to V'\) is flat. Then, since \(V' \to V\) is surjective as a base change of \(W \to \mathcal{Y}\) we see that \(U \to V\) is flat by Morphisms of Spaces, Lemma 05VW (2) and we win.

Lemma

Let \(\mathcal{X} \to \mathcal{Y} \to \mathcal{Z}\) be morphisms of algebraic stacks. If \(\mathcal{X} \to \mathcal{Z}\) is flat and \(\mathcal{X} \to \mathcal{Y}\) is surjective and flat, then \(\mathcal{Y} \to \mathcal{Z}\) is flat.

Proof

Choose an algebraic space \(W\) and a surjective smooth morphism \(W \to \mathcal{Z}\). Choose an algebraic space \(V\) and a surjective smooth morphism \(V \to W \times_\mathcal{Z} \mathcal{Y}\). Choose an algebraic space \(U\) and a surjective smooth morphism \(U \to V \times_\mathcal{Y} \mathcal{X}\). We know that \(U \to V\) is flat and that \(U \to W\) is flat. Also, as \(\mathcal{X} \to \mathcal{Y}\) is surjective we see that \(U \to V\) is surjective (as a composition of surjective morphisms). Hence the lemma reduces to the case of morphisms of algebraic spaces. The case of morphisms of algebraic spaces is Morphisms of Spaces, Lemma 05VY.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a flat morphism of algebraic stacks. Let \(\Spec(A) \to \mathcal{Y}\) be a morphism where \(A\) is a valuation ring. If the closed point of \(\Spec(A)\) maps to a point of \(|\mathcal{Y}|\) in the image of \(|\mathcal{X|} \to |\mathcal{Y}|\), then there exists a commutative diagram \[\xymatrix{ \Spec(A') \ar[r] \ar[d] & \mathcal{X} \ar[d] \\ \Spec(A) \ar[r] & \mathcal{Y} }\] where \(A \to A'\) is an extension of valuation rings (More on Algebra, Definition 0ASG).

Proof

The base change \(\mathcal{X}_A \to \Spec(A)\) is flat (Lemma 06PY) and the closed point of \(\Spec(A)\) is in the image of \(|\mathcal{X}_A| \to |\Spec(A)|\) (Properties of Stacks, Lemma 04XH). Thus we may assume \(\mathcal{Y} = \Spec(A)\). Let \(U \to \mathcal{X}\) be a surjective smooth morphism where \(U\) is a scheme. Then we can apply Morphisms of Spaces, Lemma 089F to the morphism \(U \to \Spec(A)\) to conclude.

Flat at a point

We still have to develop the general machinery needed to say what it means for a morphism of algebraic stacks to have a given property at a point. For the moment the following lemma is sufficient.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. Let \(x \in |\mathcal{X}|\). Consider commutative diagrams \[\vcenter{ \xymatrix{ U \ar[d]_a \ar[r]_h & V \ar[d]^b \\ \mathcal{X} \ar[r]^f & \mathcal{Y} } } \quad\text{with points} \vcenter{ \xymatrix{ u \in |U| \ar[d] \\ x \in |\mathcal{X}| } }\] where \(U\) and \(V\) are algebraic spaces, \(b\) is flat, and \((a, h) : U \to \mathcal{X} \times_\mathcal{Y} V\) is flat. The following are equivalent

1. \(h\) is flat at \(u\) for one diagram as above,

2. \(h\) is flat at \(u\) for every diagram as above.

Proof

Suppose we are given a second diagram \(U', V', u', a', b', h'\) as in the lemma. Then we can consider \[\xymatrix{ U \ar[d] & U \times_\mathcal{X} U' \ar[l] \ar[d] \ar[r] & U' \ar[d] \\ V & V \times_\mathcal{Y} V' \ar[l] \ar[r] & V' }\] By Properties of Stacks, Lemma 04XH there is a point \(u'' \in |U \times_\mathcal{X} U'|\) mapping to \(u\) and \(u'\). If \(h\) is flat at \(u\), then the base change \(U \times_V (V \times_\mathcal{Y} V') \to V \times_\mathcal{Y} V'\) is flat at any point over \(u\), see Morphisms of Spaces, Lemma 05VW. On the other hand, the morphism \[U \times_\mathcal{X} U' \to U \times_\mathcal{X} (\mathcal{X} \times_\mathcal{Y} V') = U \times_\mathcal{Y} V' = U \times_V (V \times_\mathcal{Y} V')\] is flat as a base change of \((a', h')\), see Lemma 06PY. Composing and using Morphisms of Spaces, Lemma 05VX we conclude that \(U \times_\mathcal{X} U' \to V \times_\mathcal{Y} V'\) is flat at \(u''\). Then we can use composition by the flat map \(V \times_\mathcal{Y} V' \to V'\) to conclude that \(U \times_\mathcal{X} U' \to V'\) is flat at \(u''\). Finally, since \(U \times_\mathcal{X} U' \to U'\) is flat at \(u''\) and \(u''\) maps to \(u'\) we conclude that \(U' \to V'\) is flat at \(u'\) by Morphisms of Spaces, Lemma 05VY.

Definition

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. Let \(x \in |\mathcal{X}|\). We say \(f\) is flat at \(x\) if the equivalent conditions of Lemma 0CIC hold.

Morphisms of finite presentation

The property “locally of finite presentation” of morphisms of algebraic spaces is smooth local on the source-and-target, see Descent on Spaces, Remark 06FC. It is also stable under base change and fpqc local on the target, see Morphisms of Spaces, Lemma 03XR and Descent on Spaces, Lemma 041T. Hence, by Lemma 06FM above, we may define what it means for a morphism of algebraic stacks to be locally of finite presentation as follows and it agrees with the already existing notion defined in Properties of Stacks, Section 04XB when the morphism is representable by algebraic spaces.

Definition

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks.

1. We say \(f\) locally of finite presentation if the equivalent conditions of Lemma 06FM hold with \(\mathcal{P} = \text{locally of finite presentation}\).

2. We say \(f\) is of finite presentation if it is locally of finite presentation, quasi-compact, and quasi-separated.

Note that a morphism of finite presentation is not just a quasi-compact morphism which is locally of finite presentation.

Lemma

The composition of finitely presented morphisms is of finite presentation. The same holds for morphisms which are locally of finite presentation.

Proof

Combine Remark 06FP with Morphisms of Spaces, Lemma 03XQ.

Lemma

A base change of a finitely presented morphism is of finite presentation. The same holds for morphisms which are locally of finite presentation.

Proof

Combine Remark 06FQ with Morphisms of Spaces, Lemma 03XR.

Lemma

A morphism which is locally of finite presentation is locally of finite type. A morphism of finite presentation is of finite type.

Proof

Combine Remark 06PS with Morphisms of Spaces, Lemma 0464.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks.

1. If \(\mathcal{Y}\) is locally Noetherian and \(f\) locally of finite type then \(f\) is locally of finite presentation.

2. If \(\mathcal{Y}\) is locally Noetherian and \(f\) of finite type and quasi-separated then \(f\) is of finite presentation.

Proof

Assume \(f : \mathcal{X} \to \mathcal{Y}\) locally of finite type and \(\mathcal{Y}\) locally Noetherian. This means there exists a diagram as in Lemma 06FM with \(h\) locally of finite type and surjective vertical arrow \(a\). By Morphisms of Spaces, Lemma 06G4 \(h\) is locally of finite presentation. Hence \(\mathcal{X} \to \mathcal{Y}\) is locally of finite presentation by definition. This proves (1). If \(f\) is of finite type and quasi-separated then it is also quasi-compact and quasi-separated and (2) follows immediately.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) and \(g : \mathcal{Y} \to \mathcal{Z}\) be morphisms of algebraic stacks If \(g \circ f\) is locally of finite presentation and \(g\) is locally of finite type, then \(f\) is locally of finite presentation.

Proof

Choose an algebraic space \(W\) and a surjective smooth morphism \(W \to \mathcal{Z}\). Choose an algebraic space \(V\) and a surjective smooth morphism \(V \to \mathcal{Y} \times_\mathcal{Z} W\). Choose an algebraic space \(U\) and a surjective smooth morphism \(U \to \mathcal{X} \times_\mathcal{Y} V\). The lemma follows upon applying Morphisms of Spaces, Lemma 05WT to the morphisms \(U \to V \to W\).

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks with diagonal \(\Delta : \mathcal{X} \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}\). If \(f\) is locally of finite type then \(\Delta\) is locally of finite presentation. If \(f\) is quasi-separated and locally of finite type, then \(\Delta\) is of finite presentation.

Proof

Note that \(\Delta\) is a morphism over \(\mathcal{X}\) (via the second projection). If \(f\) is locally of finite type, then \(\mathcal{X}\) is of finite presentation over \(\mathcal{X}\) and \(\text{pr}_2 : \mathcal{X} \times_\mathcal{Y} \mathcal{X} \to \mathcal{X}\) is locally of finite type by Lemma 06FU. Thus the first statement holds by Lemma 06Q6. The second statement follows from the first and the definitions (because \(f\) being quasi-separated means by definition that \(\Delta_f\) is quasi-compact and quasi-separated).

Lemma

An open immersion is locally of finite presentation.

Proof

In view of Properties of Stacks, Definition 04YL this follows from Morphisms of Spaces, Lemma 06CN.

Lemma

Let \(P\) be a property of morphisms of algebraic spaces which is fppf local on the target and preserved by arbitrary base change. Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks representable by algebraic spaces. Let \(\mathcal{Z} \to \mathcal{Y}\) be a morphism of algebraic stacks which is surjective, flat, and locally of finite presentation. Set \(\mathcal{W} = \mathcal{Z} \times_\mathcal{Y} \mathcal{X}\). Then \[(f\text{ has }P) \Leftrightarrow (\text{the projection }\mathcal{W} \to \mathcal{Z}\text{ has }P).\] For the meaning of this statement see Properties of Stacks, Section 04XB.

Proof

Choose an algebraic space \(W\) and a morphism \(W \to \mathcal{Z}\) which is surjective, flat, and locally of finite presentation. By Properties of Stacks, Lemma 04ZT and Lemmas 06PX and 06Q3 the composition \(W \to \mathcal{Y}\) is also surjective, flat, and locally of finite presentation. Denote \(V = W \times_\mathcal{Z} \mathcal{W} = V \times_\mathcal{Y} \mathcal{X}\). By Properties of Stacks, Lemma 04XD we see that \(f\) has \(\mathcal{P}\) if and only if \(V \to W\) does and that \(\mathcal{W} \to \mathcal{Z}\) has \(\mathcal{P}\) if and only if \(V \to W\) does. The lemma follows.

Lemma

Let \(\mathcal{P}\) be a property of morphisms of algebraic spaces which is smooth local on the source-and-target and fppf local on the target. Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. Let \(\mathcal{Z} \to \mathcal{Y}\) be a surjective, flat, locally finitely presented morphism of algebraic stacks. If the base change \(\mathcal{Z} \times_\mathcal{Y} \mathcal{X} \to \mathcal{Z}\) has \(\mathcal{P}\), then \(f\) has \(\mathcal{P}\).

Proof

Assume \(\mathcal{Z} \times_\mathcal{Y} \mathcal{X} \to \mathcal{Z}\) has \(\mathcal{P}\). Choose an algebraic space \(W\) and a surjective smooth morphism \(W \to \mathcal{Z}\). Observe that \(W \times_\mathcal{Z} \mathcal{Z} \times_\mathcal{Y} \mathcal{X} = W \times_\mathcal{Y} \mathcal{X}\). Thus by the very definition of what it means for \(\mathcal{Z} \times_\mathcal{Y} \mathcal{X} \to \mathcal{Z}\) to have \(\mathcal{P}\) (see Definition 06FN and Lemma 06FM) we see that \(W \times_\mathcal{Y} \mathcal{X} \to W\) has \(\mathcal{P}\). On the other hand, \(W \to \mathcal{Z}\) is surjective, flat, and locally of finite presentation (Morphisms of Spaces, Lemmas 04TA and 04AJ) hence \(W \to \mathcal{Y}\) is surjective, flat, and locally of finite presentation (by Properties of Stacks, Lemma 04ZT and Lemmas 06PX and 06Q3). Thus we may replace \(\mathcal{Z}\) by \(W\).

Choose an algebraic space \(V\) and a surjective smooth morphism \(V \to \mathcal{Y}\). Choose an algebraic space \(U\) and a surjective smooth morphism \(U \to V \times_\mathcal{Y} \mathcal{X}\). We have to show that \(U \to V\) has \(\mathcal{P}\). Now we base change everything by \(W \to \mathcal{Y}\): Set \(U' = W \times_\mathcal{Y} U\), \(V' = W \times_\mathcal{Y} V\), \(\mathcal{X}' = W \times_\mathcal{Y} \mathcal{X}\), and \(\mathcal{Y}' = W \times_\mathcal{Y} \mathcal{Y} = W\). Then it is still true that \(U' \to V' \times_{\mathcal{Y}'} \mathcal{X}'\) is smooth by base change. Hence by Lemma 06FM used in the definition of \(\mathcal{X}' \to \mathcal{Y}' = W\) having \(\mathcal{P}\) we see that \(U' \to V'\) has \(\mathcal{P}\). Then, since \(V' \to V\) is surjective, flat, and locally of finite presentation as a base change of \(W \to \mathcal{Y}\) we see that \(U \to V\) has \(\mathcal{P}\) as \(\mathcal{P}\) is local in the fppf topology on the target.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. Let \(\mathcal{Z} \to \mathcal{Y}\) be a surjective, flat, locally finitely presented morphism of algebraic stacks. If the base change \(\mathcal{Z} \times_\mathcal{Y} \mathcal{X} \to \mathcal{Z}\) is locally of finite presentation, then \(f\) is locally of finite presentation.

Proof

The property “locally of finite presentation” satisfies the conditions of Lemma 0DN6. Smooth local on the source-and-target we have seen in the introduction to this section and fppf local on the target is Descent on Spaces, Lemma 041T.

Lemma

Let \(\mathcal{X} \to \mathcal{Y} \to \mathcal{Z}\) be morphisms of algebraic stacks. If \(\mathcal{X} \to \mathcal{Z}\) is locally of finite presentation and \(\mathcal{X} \to \mathcal{Y}\) is surjective, flat, and locally of finite presentation, then \(\mathcal{Y} \to \mathcal{Z}\) is locally of finite presentation.

Proof

Choose an algebraic space \(W\) and a surjective smooth morphism \(W \to \mathcal{Z}\). Choose an algebraic space \(V\) and a surjective smooth morphism \(V \to W \times_\mathcal{Z} \mathcal{Y}\). Choose an algebraic space \(U\) and a surjective smooth morphism \(U \to V \times_\mathcal{Y} \mathcal{X}\). We know that \(U \to V\) is flat and locally of finite presentation and that \(U \to W\) is locally of finite presentation. Also, as \(\mathcal{X} \to \mathcal{Y}\) is surjective we see that \(U \to V\) is surjective (as a composition of surjective morphisms). Hence the lemma reduces to the case of morphisms of algebraic spaces. The case of morphisms of algebraic spaces is Descent on Spaces, Lemma 06EV.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks which is surjective, flat, and locally of finite presentation. Then for every scheme \(U\) and object \(y\) of \(\mathcal{Y}\) over \(U\) there exists an fppf covering \(\{U_i \to U\}\) and objects \(x_i\) of \(\mathcal{X}\) over \(U_i\) such that \(f(x_i) \cong y|_{U_i}\) in \(\mathcal{Y}_{U_i}\).

Proof

We may think of \(y\) as a morphism \(U \to \mathcal{Y}\). By Properties of Stacks, Lemma 04ZU and Lemmas 06Q4 and 06PY we see that \(\mathcal{X} \times_\mathcal{Y} U \to U\) is surjective, flat, and locally of finite presentation. Let \(V\) be a scheme and let \(V \to \mathcal{X} \times_\mathcal{Y} U\) smooth and surjective. Then \(V \to \mathcal{X} \times_\mathcal{Y} U\) is also surjective, flat, and locally of finite presentation (see Morphisms of Spaces, Lemmas 04TA and 04AJ). Hence also \(V \to U\) is surjective, flat, and locally of finite presentation, see Properties of Stacks, Lemma 04ZT and Lemmas 06Q3, and 06PX. Hence \(\{V \to U\}\) is the desired fppf covering and \(x : V \to \mathcal{X}\) is the desired object.

Lemma

Let \(f_j : \mathcal{X}_j \to \mathcal{X}\), \(j \in J\) be a family of morphisms of algebraic stacks which are each flat and locally of finite presentation and which are jointly surjective, i.e., \(|\mathcal{X}| = \bigcup |f_j|(|\mathcal{X}_j|)\). Then for every scheme \(U\) and object \(x\) of \(\mathcal{X}\) over \(U\) there exists an fppf covering \(\{U_i \to U\}_{i \in I}\), a map \(a : I \to J\), and objects \(x_i\) of \(\mathcal{X}_{a(i)}\) over \(U_i\) such that \(f_{a(i)}(x_i) \cong y|_{U_i}\) in \(\mathcal{X}_{U_i}\).

Proof

Apply Lemma 06QA to the morphism \(\coprod_{j \in J} \mathcal{X}_j \to \mathcal{X}\). (There is a slight set theoretic issue here – due to our setup of things – which we ignore.) To finish, note that a morphism \(x_i : U_i \to \coprod_{j \in J} \mathcal{X}_j\) is given by a disjoint union decomposition \(U_i = \coprod U_{i, j}\) and morphisms \(U_{i, j} \to \mathcal{X}_j\). Then the fppf covering \(\{U_{i, j} \to U\}\) and the morphisms \(U_{i, j} \to \mathcal{X}_j\) do the job.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be flat and locally of finite presentation. Then \(|f| : |\mathcal{X}| \to |\mathcal{Y}|\) is open.

Proof

Choose a scheme \(V\) and a smooth surjective morphism \(V \to \mathcal{Y}\). Choose a scheme \(U\) and a smooth surjective morphism \(U \to V \times_\mathcal{Y} \mathcal{X}\). By assumption the morphism of schemes \(U \to V\) is flat and locally of finite presentation. Hence \(U \to V\) is open by Morphisms, Lemma 01UA. By construction of the topology on \(|\mathcal{Y}|\) the map \(|V| \to |\mathcal{Y}|\) is open. The map \(|U| \to |\mathcal{X}|\) is surjective. The result follows from these facts by elementary topology.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. Let \(\mathcal{Z} \to \mathcal{Y}\) be a surjective, flat, locally finitely presented morphism of algebraic stacks. If the base change \(\mathcal{Z} \times_\mathcal{Y} \mathcal{X} \to \mathcal{Z}\) is quasi-compact, then \(f\) is quasi-compact.

Proof

We have to show that given \(\mathcal{Y}' \to \mathcal{Y}\) with \(\mathcal{Y}'\) quasi-compact, we have \(\mathcal{Y}' \times_\mathcal{Y} \mathcal{X}\) is quasi-compact. Denote \(\mathcal{Z}' = \mathcal{Z} \times_\mathcal{Y} \mathcal{Y}'\). Then \(|\mathcal{Z}'| \to |\mathcal{Y}'|\) is open, see Lemma 06R7. Hence we can find a quasi-compact open substack \(\mathcal{W} \subset \mathcal{Z}'\) mapping onto \(\mathcal{Y}'\). Because \(\mathcal{Z} \times_\mathcal{Y} \mathcal{X} \to \mathcal{Z}\) is quasi-compact, we know that \[\mathcal{W} \times_\mathcal{Z} \mathcal{Z} \times_\mathcal{Y} \mathcal{X} = \mathcal{W} \times_\mathcal{Y} \mathcal{X}\] is quasi-compact. And the map \(\mathcal{W} \times_\mathcal{Y} \mathcal{X} \to \mathcal{Y}' \times_\mathcal{Y} \mathcal{X}\) is surjective, hence we win. Some details omitted.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\), \(g : \mathcal{Y} \to \mathcal{Z}\) be composable morphisms of algebraic stacks with composition \(h = g \circ f : \mathcal{X} \to \mathcal{Z}\). If \(f\) is surjective, flat, locally of finite presentation, and universally injective and if \(h\) is separated, then \(g\) is separated.

Proof

Consider the diagram \[\xymatrix{ \mathcal{X} \ar[r]_\Delta \ar[rd] & \mathcal{X} \times_\mathcal{Y} \mathcal{X} \ar[r] \ar[d] & \mathcal{X} \times_\mathcal{Z} \mathcal{X} \ar[d] \\ & \mathcal{Y} \ar[r] & \mathcal{Y} \times_\mathcal{Z} \mathcal{Y} }\] The square is cartesian. We have to show the bottom horizontal arrow is proper. We already know that it is representable by algebraic spaces and locally of finite type (Lemma 04XS). Since the right vertical arrow is surjective, flat, and locally of finite presentation it suffices to show the top right horizontal arrow is proper (Lemma 0CPP). Since \(h\) is separated, the composition of the top horizontal arrows is proper.

Since \(f\) is universally injective \(\Delta\) is surjective (Lemma 0CPN). Since the composition of \(\Delta\) with the projection \(\mathcal{X} \times_\mathcal{Y} \mathcal{X} \to \mathcal{X}\) is the identity, we see that \(\Delta\) is universally closed. By Morphisms of Spaces, Lemma 05Z2 we conclude that \(\mathcal{X} \times_\mathcal{Y} \mathcal{X} \to \mathcal{X} \times_\mathcal{Z} \mathcal{X}\) is separated as \(\mathcal{X} \to \mathcal{X} \times_\mathcal{Z} \mathcal{X}\) is separated. Here we use that implications between properties of morphisms of algebraic spaces can be transferred to the same implications between properties of morphisms of algebraic stacks representable by algebraic spaces; this is discussed in Properties of Stacks, Section 04XB. Finally, we use the same principle to conlude that \(\mathcal{X} \times_\mathcal{Y} \mathcal{X} \to \mathcal{X} \times_\mathcal{Z} \mathcal{X}\) is proper from Morphisms of Spaces, Lemma 08AJ.

Gerbes

An important type of algebraic stack are the stacks of the form \([B/G]\) where \(B\) is an algebraic space and \(G\) is a flat and locally finitely presented group algebraic space over \(B\) (acting trivially on \(B\)), see Criteria for Representability, Lemma 06PL. It turns out that an algebraic stack is a gerbe when it locally in the fppf topology is of this form, see Lemma 06QH. In this section we briefly discuss this notion and the corresponding relative notion.

Definition

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. We say \(\mathcal{X}\) is a gerbe over \(\mathcal{Y}\) if \(\mathcal{X}\) is a gerbe over \(\mathcal{Y}\) as stacks in groupoids over \((\Sch/S)_{fppf}\), see Stacks, Definition 06P2. We say an algebraic stack \(\mathcal{X}\) is a gerbe if there exists a morphism \(\mathcal{X} \to X\) where \(X\) is an algebraic space which turns \(\mathcal{X}\) into a gerbe over \(X\).

The condition that \(\mathcal{X}\) be a gerbe over \(\mathcal{Y}\) is defined purely in terms of the topology and category theory underlying the given algebraic stacks; but as we will see later this condition has geometric consequences. For example it implies that \(\mathcal{X} \to \mathcal{Y}\) is surjective, flat, and locally of finite presentation, see Lemma 06QI. The absolute notion is trickier to parse, because it may not be at first clear that \(X\) is well determined. Actually, it is.

Lemma

Let \(\mathcal{X}\) be an algebraic stack. If \(\mathcal{X}\) is a gerbe, then the sheafification of the presheaf \[(\Sch/S)_{fppf}^{opp} \to \textit{Sets}, \quad U \mapsto \Ob(\mathcal{X}_U)/\!\!\cong\] is an algebraic space and \(\mathcal{X}\) is a gerbe over it.

Proof

(In this proof the abuse of language introduced in Section 04XP really pays off.) Choose a morphism \(\pi : \mathcal{X} \to X\) where \(X\) is an algebraic space which turns \(\mathcal{X}\) into a gerbe over \(X\). It suffices to prove that \(X\) is the sheafification of the presheaf \(\mathcal{F}\) displayed in the lemma. It is clear that there is a map \(c : \mathcal{F} \to X\). We will use Stacks, Lemma 06P1 properties (2)(a) and (2)(b) to see that the map \(c^\# : \mathcal{F}^\# \to X\) is surjective and injective, hence an isomorphism, see Sites, Lemma 00WN. Surjective: Let \(T\) be a scheme and let \(f : T \to X\). By property (2)(a) there exists an fppf covering \(\{h_i : T_i \to T\}\) and morphisms \(x_i : T_i \to \mathcal{X}\) such that \(f \circ h_i\) corresponds to \(\pi \circ x_i\). Hence we see that \(f|_{T_i}\) is in the image of \(c\). Injective: Let \(T\) be a scheme and let \(x, y : T \to \mathcal{X}\) be morphisms such that \(c \circ x = c \circ y\). By (2)(b) we can find a covering \(\{T_i \to T\}\) and morphisms \(x|_{T_i} \to y|_{T_i}\) in the fibre category \(\mathcal{X}_{T_i}\). Hence the restrictions \(x|_{T_i}, y|_{T_i}\) are equal in \(\mathcal{F}(T_i)\). This proves that \(x, y\) give the same section of \(\mathcal{F}^\#\) over \(T\) as desired.

Lemma

Let \[\xymatrix{ \mathcal{X}' \ar[r] \ar[d] & \mathcal{X} \ar[d] \\ \mathcal{Y}' \ar[r] & \mathcal{Y} }\] be a fibre product of algebraic stacks. If \(\mathcal{X}\) is a gerbe over \(\mathcal{Y}\), then \(\mathcal{X}'\) is a gerbe over \(\mathcal{Y}'\).

Proof

Immediate from the definitions and Stacks, Lemma 06P3.

Lemma

Let \(\mathcal{X} \to \mathcal{Y}\) and \(\mathcal{Y} \to \mathcal{Z}\) be morphisms of algebraic stacks. If \(\mathcal{X}\) is a gerbe over \(\mathcal{Y}\) and \(\mathcal{Y}\) is a gerbe over \(\mathcal{Z}\), then \(\mathcal{X}\) is a gerbe over \(\mathcal{Z}\).

Proof

Immediate from Stacks, Lemma 06R3.

Lemma

Let \[\xymatrix{ \mathcal{X}' \ar[r] \ar[d] & \mathcal{X} \ar[d] \\ \mathcal{Y}' \ar[r] & \mathcal{Y} }\] be a fibre product of algebraic stacks. If \(\mathcal{Y}' \to \mathcal{Y}\) is surjective, flat, and locally of finite presentation and \(\mathcal{X}'\) is a gerbe over \(\mathcal{Y}'\), then \(\mathcal{X}\) is a gerbe over \(\mathcal{Y}\).

Proof

Follows immediately from Lemma 06QA and Stacks, Lemma 06P4.

Lemma

Let \(\pi : \mathcal{X} \to U\) be a morphism from an algebraic stack to an algebraic space and let \(x : U \to \mathcal{X}\) be a section of \(\pi\). Set \(G = \mathit{Isom}_\mathcal{X}(x, x)\), see Definition 06PP. If \(\mathcal{X}\) is a gerbe over \(U\), then

1. there is a canonical equivalence of stacks in groupoids \[x_{can} : [U/G] \longrightarrow \mathcal{X}.\] where \([U/G]\) is the quotient stack for the trivial action of \(G\) on \(U\),

2. \(G \to U\) is flat and locally of finite presentation, and

3. \(U \to \mathcal{X}\) is surjective, flat, and locally of finite presentation.

Proof

Set \(R = U \times_{x, \mathcal{X}, x} U\). The morphism \(R \to U \times U\) factors through the diagonal \(\Delta_U : U \to U \times U\) as it factors through \(U \times_U U = U\). Hence \(R = G\) because \[\begin{align*} G & = \mathit{Isom}_\mathcal{X}(x, x) \\ & = U \times_{x, \mathcal{X}} \mathcal{I}_\mathcal{X} \\ & = U \times_{x, \mathcal{X}} (\mathcal{X} \times_{\Delta, \mathcal{X} \times_S \mathcal{X}, \Delta} \mathcal{X}) \\ & = (U \times_{x, \mathcal{X}, x} U) \times_{U \times U, \Delta_U} U \\ & = R \times_{U \times U, \Delta_U} U \\ & = R \end{align*}\] for the fourth equality use Categories, Lemma 02XF. Let \(t, s : R \to U\) be the projections. The composition law \(c : R \times_{s, U, t} R \to R\) constructed on \(R\) in Algebraic Stacks, Lemma 04T4 agrees with the group law on \(G\) (proof omitted). Thus Algebraic Stacks, Lemma 04T4 shows we obtain a canonical fully faithful \(1\)-morphism \[x_{can} : [U/G] \longrightarrow \mathcal{X}\] of stacks in groupoids over \((\Sch/S)_{fppf}\). To see that it is an equivalence it suffices to show that it is essentially surjective. To do this it suffices to show that any object of \(\mathcal{X}\) over a scheme \(T\) comes fppf locally from \(x\) via a morphism \(T \to U\), see Stacks, Lemma 046N. However, this follows the condition that \(\pi\) turns \(\mathcal{X}\) into a gerbe over \(U\), see property (2)(a) of Stacks, Lemma 06P1.

By Criteria for Representability, Lemma 06PL we conclude that \(G \to U\) is flat and locally of finite presentation. Finally, \(U \to \mathcal{X}\) is surjective, flat, and locally of finite presentation by Criteria for Representability, Lemma 06FH.

Lemma

Let \(\pi : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. The following are equivalent

1. \(\mathcal{X}\) is a gerbe over \(\mathcal{Y}\), and

2. there exists an algebraic space \(U\), a group algebraic space \(G\) flat and locally of finite presentation over \(U\), and a surjective, flat, and locally finitely presented morphism \(U \to \mathcal{Y}\) such that \(\mathcal{X} \times_\mathcal{Y} U \cong [U/G]\) over \(U\).

Proof

Assume (2). By Lemma 06QF to prove (1) it suffices to show that \([U/G]\) is a gerbe over \(U\). This is immediate from Groupoids in Spaces, Lemma 06PF.

Assume (1). Any base change of \(\pi\) is a gerbe, see Lemma 06QE. As a first step we choose a scheme \(V\) and a surjective smooth morphism \(V \to \mathcal{Y}\). Thus we may assume that \(\pi : \mathcal{X} \to V\) is a gerbe over a scheme. This means that there exists an fppf covering \(\{V_i \to V\}\) such that the fibre category \(\mathcal{X}_{V_i}\) is nonempty, see Stacks, Lemma 06P1 (2)(a). Note that \(U = \coprod V_i \to V\) is surjective, flat, and locally of finite presentation. Hence we may replace \(V\) by \(U\) and assume that \(\pi : \mathcal{X} \to U\) is a gerbe over a scheme \(U\) and that there exists an object \(x\) of \(\mathcal{X}\) over \(U\). By Lemma 06QG we see that \(\mathcal{X} = [U/G]\) over \(U\) for some flat and locally finitely presented group algebraic space \(G\) over \(U\).

Lemma

Let \(\pi : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. If \(\mathcal{X}\) is a gerbe over \(\mathcal{Y}\), then \(\pi\) is surjective, flat, and locally of finite presentation.

Proof

By Properties of Stacks, Lemma 06PM and Lemmas 06PZ and 06Q8 it suffices to prove to the lemma after replacing \(\pi\) by a base change with a surjective, flat, locally finitely presented morphism \(\mathcal{Y}' \to \mathcal{Y}\). By Lemma 06QH we may assume \(\mathcal{Y} = U\) is an algebraic space and \(\mathcal{X} = [U/G]\) over \(U\). Then \(U \to [U/G]\) is surjective, flat, and locally of finite presentation, see Lemma 06QG. This implies that \(\pi\) is surjective, flat, and locally of finite presentation by Properties of Stacks, Lemma 06PN and Lemmas 06Q0 and 06Q9.

Proposition

Let \(\mathcal{X}\) be an algebraic stack. The following are equivalent

1. \(\mathcal{X}\) is a gerbe, and

2. \(\mathcal{I}_\mathcal{X} \to \mathcal{X}\) is flat and locally of finite presentation.

Proof

Assume (1). Choose a morphism \(\mathcal{X} \to X\) into an algebraic space \(X\) which turns \(\mathcal{X}\) into a gerbe over \(X\). Let \(X' \to X\) be a surjective, flat, locally finitely presented morphism and set \(\mathcal{X}' = X' \times_X \mathcal{X}\). Note that \(\mathcal{X}'\) is a gerbe over \(X'\) by Lemma 06QE. Then both squares in \[\xymatrix{ \mathcal{I}_{\mathcal{X}'} \ar[r] \ar[d] & \mathcal{X}' \ar[r] \ar[d] & X' \ar[d] \\ \mathcal{I}_\mathcal{X} \ar[r] & \mathcal{X} \ar[r] & X }\] are fibre product squares, see Lemma 06PQ. Hence to prove \(\mathcal{I}_\mathcal{X} \to \mathcal{X}\) is flat and locally of finite presentation it suffices to do so after such a base change by Lemmas 06PZ and 06Q8. Thus we can apply Lemma 06QH to assume that \(\mathcal{X} = [U/G]\). By Lemma 06QG we see \(G\) is flat and locally of finite presentation over \(U\) and that \(x : U \to [U/G]\) is surjective, flat, and locally of finite presentation. Moreover, the pullback of \(\mathcal{I}_\mathcal{X}\) by \(x\) is \(G\) and we conclude that (2) holds by descent again, i.e., by Lemmas 06PZ and 06Q8.

Conversely, assume (2). Choose a smooth presentation \(\mathcal{X} = [U/R]\), see Algebraic Stacks, Section 04T3. Denote \(G \to U\) the stabilizer group algebraic space of the groupoid \((U, R, s, t, c, e, i)\), see Groupoids in Spaces, Definition 0448. By Lemma 06PR we see that \(G \to U\) is flat and locally of finite presentation as a base change of \(\mathcal{I}_\mathcal{X} \to \mathcal{X}\), see Lemmas 06PY and 06Q4. Consider the following action \[a : G \times_{U, t} R \to R, \quad (g, r) \mapsto c(g, r)\] of \(G\) on \(R\). This action is free on \(T\)-valued points for any scheme \(T\) as \(R\) is a groupoid. Hence \(R' = R/G\) is an algebraic space and the quotient morphism \(\pi : R \to R'\) is surjective, flat, and locally of finite presentation by Bootstrap, Lemma 06PH. The projections \(s, t : R \to U\) are \(G\)-invariant, hence we obtain morphisms \(s' , t' : R' \to U\) such that \(s = s' \circ \pi\) and \(t = t' \circ \pi\). Since \(s, t : R \to U\) are flat and locally of finite presentation we conclude that \(s', t'\) are flat and locally of finite presentation, see Morphisms of Spaces, Lemmas 05VY and Descent on Spaces, Lemma 06EV. Consider the morphism \[j' = (t', s') : R' \longrightarrow U \times U.\] We claim this is a monomorphism. Namely, suppose that \(T\) is a scheme and that \(a, b : T \to R'\) are morphisms which have the same image in \(U \times U\). By definition of the quotient \(R' = R/G\) there exists an fppf covering \(\{h_j : T_j \to T\}\) such that \(a \circ h_j = \pi \circ a_j\) and \(b \circ h_j = \pi \circ b_j\) for some morphisms \(a_j, b_j : T_j \to R\). Since \(a_j, b_j\) have the same image in \(U \times U\) we see that \(g_j = c(a_j, i(b_j))\) is a \(T_j\)-valued point of \(G\) such that \(c(g_j, b_j) = a_j\). In other words, \(a_j\) and \(b_j\) have the same image in \(R'\) and the claim is proved. Since \(j : R \to U \times U\) is a pre-equivalence relation (see Groupoids in Spaces, Lemma 043X) and \(R \to R'\) is surjective (as a map of sheaves) we see that \(j' : R' \to U \times U\) is an equivalence relation. Hence Bootstrap, Theorem 04S6 shows that \(X = U/R'\) is an algebraic space. Finally, we claim that the morphism \[\mathcal{X} = [U/R] \longrightarrow X = U/R'\] turns \(\mathcal{X}\) into a gerbe over \(X\). This follows from Groupoids in Spaces, Lemma 06PE as \(R \to R'\) is surjective, flat, and locally of finite presentation (if needed use Bootstrap, Lemma 04S1 to see this implies the required hypothesis).

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks which makes \(\mathcal{X}\) a gerbe over \(\mathcal{Y}\). Then

1. \(\mathcal{I}_{\mathcal{X}/\mathcal{Y}} \to \mathcal{X}\) is flat and locally of finite presentation,

2. \(\mathcal{X} \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}\) is surjective, flat, and locally of finite presentation,

3. given algebraic spaces \(T_i\), \(i = 1, 2\) and morphisms \(x_i : T_i \to \mathcal{X}\), with \(y_i = f \circ x_i\) the morphism \[T_1 \times_{x_1, \mathcal{X}, x_2} T_2 \longrightarrow T_1 \times_{y_1, \mathcal{Y}, y_2} T_2\] is surjective, flat, and locally of finite presentation,

4. given an algebraic space \(T\) and morphisms \(x_i : T \to \mathcal{X}\), \(i = 1, 2\), with \(y_i = f \circ x_i\) the morphism \[\mathit{Isom}_\mathcal{X}(x_1, x_2) \longrightarrow \mathit{Isom}_\mathcal{Y}(y_1, y_2)\] is surjective, flat, and locally of finite presentation.

Proof

Proof of (1). Choose a scheme \(Y\) and a surjective smooth morphism \(Y \to \mathcal{Y}\). Set \(\mathcal{X}' = \mathcal{X} \times_\mathcal{Y} Y\). By Lemma 06PQ we obtain cartesian squares \[\xymatrix{ \mathcal{I}_{\mathcal{X}'} \ar[r] \ar[d] & \mathcal{X}' \ar[r] \ar[d] & Y \ar[d] \\ \mathcal{I}_{\mathcal{X}/\mathcal{Y}} \ar[r] & \mathcal{X} \ar[r] & \mathcal{Y} }\] By Lemmas 06PZ and 06Q8 it suffices to prove that \(\mathcal{I}_{\mathcal{X}'} \to \mathcal{X}'\) is flat and locally of finite presentation. This follows from Proposition 06QJ (because \(\mathcal{X}'\) is a gerbe over \(Y\) by Lemma 06QE).

Proof of (2). With notation as above, note that we may assume that \(\mathcal{X}' = [Y/G]\) for some group algebraic space \(G\) flat and locally of finite presentation over \(Y\), see Lemma 06QH. The base change of the morphism \(\Delta : \mathcal{X} \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}\) over \(\mathcal{Y}\) by the morphism \(Y \to \mathcal{Y}\) is the morphism \(\Delta' : \mathcal{X}' \to \mathcal{X}' \times_Y \mathcal{X}'\). Hence it suffices to show that \(\Delta'\) is surjective, flat, and locally of finite presentation (see Lemmas 06PZ and 06Q8). In other words, we have to show that \[[Y/G] \longrightarrow [Y/G \times_Y G]\] is surjective, flat, and locally of finite presentation. This is true because the base change by the surjective, flat, locally finitely presented morphism \(Y \to [Y/G \times_Y G]\) is the morphism \(G \to Y\).

Proof of (3). Observe that the diagram \[\xymatrix{ T_1 \times_{x_1, \mathcal{X}, x_2} T_2 \ar[d] \ar[r] & T_1 \times_{y_1, \mathcal{Y}, y_2} T_2 \ar[d] \\ \mathcal{X} \ar[r] & \mathcal{X} \times_\mathcal{Y} \mathcal{X} }\] is cartesian. Hence (3) follows from (2).

Proof of (4). This is true because \[\mathit{Isom}_\mathcal{X}(x_1, x_2) = (T \times_{x_1, \mathcal{X}, x_2} T) \times_{T \times T, \Delta_T} T\] hence the morphism in (4) is a base change of the morphism in (3).

Proposition

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. The following are equivalent

1. \(\mathcal{X}\) is a gerbe over \(\mathcal{Y}\), and

2. \(f : \mathcal{X} \to \mathcal{Y}\) and \(\Delta : \mathcal{X} \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}\) are surjective, flat, and locally of finite presentation.

Proof

The implication (1) \(\Rightarrow\) (2) follows from Lemmas 06QI and 0CPR.

Assume (2). It suffices to prove (1) for the base change of \(f\) by a surjective, flat, and locally finitely presented morphism \(\mathcal{Y}' \to \mathcal{Y}\), see Lemma 06QF (note that the base change of the diagonal of \(f\) is the diagonal of the base change). Thus we may assume \(\mathcal{Y}\) is a scheme \(Y\). In this case \(\mathcal{I}_\mathcal{X} \to \mathcal{X}\) is a base change of \(\Delta\) and we conclude that \(\mathcal{X}\) is a gerbe by Proposition 06QJ. We still have to show that \(\mathcal{X}\) is a gerbe over \(Y\). Let \(\mathcal{X} \to X\) be the morphism of Lemma 06QD turning \(\mathcal{X}\) into a gerbe over the algebraic space \(X\) classifying isomorphism classes of objects of \(\mathcal{X}\). It is clear that \(f : \mathcal{X} \to Y\) factors as \(\mathcal{X} \to X \to Y\). Since \(f\) is surjective, flat, and locally of finite presentation, we conclude that \(X \to Y\) is surjective as a map of fppf sheaves (for example use Lemma 06QA). On the other hand, \(X \to Y\) is injective too: for any scheme \(T\) and any two \(T\)-valued points \(x_1, x_2\) of \(X\) which map to the same point of \(Y\), we can first fppf locally on \(T\) lift \(x_1, x_2\) to objects \(\xi_1, \xi_2\) of \(\mathcal{X}\) over \(T\) and second deduce that \(\xi_1\) and \(\xi_2\) are fppf locally isomorphic by our assumption that \(\Delta : \mathcal{X} \to \mathcal{X} \times_Y \mathcal{X}\) is surjective, flat, and locally of finite presentation. Whence \(x_1 = x_2\) by construction of \(X\). Thus \(X = Y\) and the proof is complete.

At this point we have developed enough machinery to prove that residual gerbes (when they exist) are gerbes.

Lemma

Let \(\mathcal{Z}\) be a reduced, locally Noetherian algebraic stack such that \(|\mathcal{Z}|\) is a singleton. Then \(\mathcal{Z}\) is a gerbe over a reduced, locally Noetherian algebraic space \(Z\) with \(|Z|\) a singleton.

Proof

By Properties of Stacks, Lemma 06MP there exists a surjective, flat, locally finitely presented morphism \(\Spec(k) \to \mathcal{Z}\) where \(k\) is a field. Then \(\mathcal{I}_Z \times_\mathcal{Z} \Spec(k) \to \Spec(k)\) is representable by algebraic spaces and locally of finite type (as a base change of \(\mathcal{I}_\mathcal{Z} \to \mathcal{Z}\), see Lemmas 050Q and 06FU). Therefore it is locally of finite presentation, see Morphisms of Spaces, Lemma 06G4. Of course it is also flat as \(k\) is a field. Hence we may apply Lemmas 06PZ and 06Q8 to see that \(\mathcal{I}_\mathcal{Z} \to \mathcal{Z}\) is flat and locally of finite presentation. We conclude that \(\mathcal{Z}\) is a gerbe by Proposition 06QJ. Let \(\pi : \mathcal{Z} \to Z\) be a morphism to an algebraic space such that \(\mathcal{Z}\) is a gerbe over \(Z\). Then \(\pi\) is surjective, flat, and locally of finite presentation by Lemma 06QI. Hence \(\Spec(k) \to Z\) is surjective, flat, and locally of finite presentation as a composition, see Properties of Stacks, Lemma 04ZT and Lemmas 06PX and 06Q3. Hence by Properties of Stacks, Lemma 06MP we see that \(|Z|\) is a singleton and that \(Z\) is locally Noetherian and reduced.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. If \(\mathcal{X}\) is a gerbe over \(\mathcal{Y}\) then \(f\) is a universal homeomorphism.

Proof

By Lemma 06QE the assumption on \(f\) is preserved under base change. Hence it suffices to show that the map \(|\mathcal{X}| \to |\mathcal{Y}|\) is a homeomorphism of topological spaces. Let \(k\) be a field and let \(y\) be an object of \(\mathcal{Y}\) over \(\Spec(k)\). By Stacks, Lemma 06P1 property (2)(a) there exists an fppf covering \(\{T_i \to \Spec(k)\}\) and objects \(x_i\) of \(\mathcal{X}\) over \(T_i\) with \(f(x_i) \cong y|_{T_i}\). Choose an \(i\) such that \(T_i \not = \emptyset\). Choose a morphism \(\Spec(K) \to T_i\) for some field \(K\). Then \(k \subset K\) and \(x_i|_K\) is an object of \(\mathcal{X}\) lying over \(y|_K\). Thus we see that \(|\mathcal{Y}| \to |\mathcal{X}|\). is surjective. The map \(|\mathcal{Y}| \to |\mathcal{X}|\) is also injective. Namely, if \(x, x'\) are objects of \(\mathcal{X}\) over \(\Spec(k)\) whose images \(f(x), f(x')\) become isomorphic (over an extension) in \(\mathcal{Y}\), then Stacks, Lemma 06P1 property (2)(b) guarantees the existence of an extension of \(k\) over which \(x\) and \(x'\) become isomorphic (details omitted). Hence \(|\mathcal{X}| \to |\mathcal{Y}|\) is continuous and bijective and it suffices to show that it is also open. This follows from Lemmas 06QI and 06R7.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks such that \(\mathcal{X}\) is a gerbe over \(\mathcal{Y}\). If \(\Delta_\mathcal{X}\) is quasi-compact, so is \(\Delta_\mathcal{Y}\).

Proof

Consider the diagram \[\xymatrix{ \mathcal{X} \ar[r] & \mathcal{X} \times_\mathcal{Y} \mathcal{X} \ar[r] \ar[d] & \mathcal{X} \times \mathcal{X} \ar[d] \\ & \mathcal{Y} \ar[r] & \mathcal{Y} \times \mathcal{Y} }\] By Proposition 0CPS we find that the arrow on the top left is surjective. Since the composition of the top horizontal arrows is quasi-compact, we conclude that the top right arrow is quasi-compact by Lemma 050X. The square is cartesian and the right vertical arrow is surjective, flat, and locally of finite presentation. Thus we conclude by Lemma 0DQK.

The following lemma tells us that residual gerbes exist for all points on any algebraic stack which is a gerbe.

Lemma

Let \(\mathcal{X}\) be an algebraic stack. If \(\mathcal{X}\) is a gerbe then for every \(x \in |\mathcal{X}|\) the residual gerbe of \(\mathcal{X}\) at \(x\) exists.

Proof

Let \(\pi : \mathcal{X} \to X\) be a morphism from \(\mathcal{X}\) into an algebraic space \(X\) which turns \(\mathcal{X}\) into a gerbe over \(X\). Let \(Z_x \to X\) be the residual space of \(X\) at \(x\), see Decent Spaces, Definition 06R0. Let \(\mathcal{Z} = \mathcal{X} \times_X Z_x\). By Lemma 06QE the algebraic stack \(\mathcal{Z}\) is a gerbe over \(Z_x\). Hence \(|\mathcal{Z}| = |Z_x|\) (Lemma 06R9) is a singleton. Since \(\mathcal{Z} \to Z_x\) is locally of finite presentation as a base change of \(\pi\) (see Lemmas 06QI and 06Q4) we see that \(\mathcal{Z}\) is locally Noetherian, see Lemma 06R6. Thus the residual gerbe \(\mathcal{Z}_x\) of \(\mathcal{X}\) at \(x\) exists and is equal to \(\mathcal{Z}_x = \mathcal{Z}_{red}\) the reduction of the algebraic stack \(\mathcal{Z}\). Namely, we have seen above that \(|\mathcal{Z}_{red}|\) is a singleton mapping to \(x \in |\mathcal{X}|\), it is reduced by construction, and it is locally Noetherian (as the reduction of a locally Noetherian algebraic stack is locally Noetherian, details omitted).

Stratification by gerbes

The goal of this section is to show that many algebraic stacks \(\mathcal{X}\) have a “stratification” by locally closed substacks \(\mathcal{X}_i \subset \mathcal{X}\) such that each \(\mathcal{X}_i\) is a gerbe. This shows that in some sense gerbes are the building blocks out of which any algebraic stack is constructed. Note that by stratification we only mean that \[|\mathcal{X}| = \bigcup\nolimits_i |\mathcal{X}_i|\] is a stratification of the topological space associated to \(\mathcal{X}\) and nothing more (in this section). Hence it is harmless to replace \(\mathcal{X}\) by its reduction (see Properties of Stacks, Section 0508) in order to study this stratification.

The following proposition tells us there is (almost always) a dense open substack of the reduction of \(\mathcal{X}\)

Proposition

Let \(\mathcal{X}\) be a reduced algebraic stack such that \(\mathcal{I}_\mathcal{X} \to \mathcal{X}\) is quasi-compact. Then there exists a dense open substack \(\mathcal{U} \subset \mathcal{X}\) which is a gerbe.

Proof

According to Proposition 06QJ it is enough to find a dense open substack \(\mathcal{U}\) such that \(\mathcal{I}_\mathcal{U} \to \mathcal{U}\) is flat and locally of finite presentation. Note that \(\mathcal{I}_\mathcal{U} = \mathcal{I}_\mathcal{X} \times_\mathcal{X} \mathcal{U}\), see Lemma 06PQ.

Choose a presentation \(\mathcal{X} = [U/R]\). Let \(G \to U\) be the stabilizer group algebraic space of the groupoid \(R\). By Lemma 06PR we see that \(G \to U\) is the base change of \(\mathcal{I}_\mathcal{X} \to \mathcal{X}\) hence quasi-compact (by assumption) and locally of finite type (by Lemma 050Q). Let \(W \subset U\) be the largest open (possibly empty) subscheme such that the restriction \(G_W \to W\) is flat and locally of finite presentation (we omit the proof that \(W\) exists; hint: use that the properties are local). By Morphisms of Spaces, Proposition 06QS we see that \(W \subset U\) is dense. Note that \(W \subset U\) is \(R\)-invariant by More on Groupoids in Spaces, Lemma 06R4. Hence \(W\) corresponds to an open substack \(\mathcal{U} \subset \mathcal{X}\) by Properties of Stacks, Lemma 0507. Since \(|U| \to |\mathcal{X}|\) is open and \(|W| \subset |U|\) is dense we conclude that \(\mathcal{U}\) is dense in \(\mathcal{X}\). Finally, the morphism \(\mathcal{I}_\mathcal{U} \to \mathcal{U}\) is flat and locally of finite presentation because the base change by the surjective smooth morphism \(W \to \mathcal{U}\) is the morphism \(G_W \to W\) which is flat and locally of finite presentation by construction. See Lemmas 06PZ and 06Q8.

The above proposition immediately implies that any point has a residual gerbe on an algebraic stack with quasi-compact inertia, as we will show in Lemma 06RD. It turns out that there doesn’t always exist a finite stratification by gerbes. Here is an example.

Example

Let \(k\) be a field. Take \(U = \Spec(k[x_0, x_1, x_2, \ldots])\) and let \(\mathbf{G}_m\) act by \(t(x_0, x_1, x_2, \ldots) = (tx_0, t^p x_1, t^{p^2} x_2, \ldots)\) where \(p\) is a prime number. Let \(\mathcal{X} = [U/\mathbf{G}_m]\). This is an algebraic stack. There is a stratification of \(\mathcal{X}\) by strata

1. \(\mathcal{X}_0\) is where \(x_0\) is not zero,

2. \(\mathcal{X}_1\) is where \(x_0\) is zero but \(x_1\) is not zero,

3. \(\mathcal{X}_2\) is where \(x_0, x_1\) are zero, but \(x_2\) is not zero,

4. and so on, and

5. \(\mathcal{X}_{\infty}\) is where all the \(x_i\) are zero.

Each stratum is a gerbe over a scheme with group \(\mu_{p^i}\) for \(\mathcal{X}_i\) and \(\mathbf{G}_m\) for \(\mathcal{X}_{\infty}\). The strata are reduced locally closed substacks. There is no coarser stratification with the same properties.

Nonetheless, using transfinite induction we can use Proposition 06RC find possibly infinite stratifications by gerbes...!

Lemma

Let \(\mathcal{X}\) be an algebraic stack such that \(\mathcal{I}_\mathcal{X} \to \mathcal{X}\) is quasi-compact. Then there exists a well-ordered index set \(I\) and for every \(i \in I\) a reduced locally closed substack \(\mathcal{U}_i \subset \mathcal{X}\) such that

1. each \(\mathcal{U}_i\) is a gerbe,

2. we have \(|\mathcal{X}| = \bigcup_{i \in I} |\mathcal{U}_i|\),

3. \(T_i = |\mathcal{X}| \setminus \bigcup_{i' < i} |\mathcal{U}_{i'}|\) is closed in \(|\mathcal{X}|\) for all \(i \in I\), and

4. \(|\mathcal{U}_i|\) is open in \(T_i\).

We can moreover arrange it so that either (a) \(|\mathcal{U}_i| \subset T_i\) is dense, or (b) \(\mathcal{U}_i\) is quasi-compact. In case (a), if we choose \(\mathcal{U}_i\) as large as possible (see proof for details), then the stratification is canonical.

Proof

Let \(T \subset |\mathcal{X}|\) be a nonempty closed subset. We are going to find (resp. choose) for every such \(T\) a reduced locally closed substack \(\mathcal{U}(T) \subset \mathcal{X}\) with \(|\mathcal{U}(T)| \subset T\) open dense (resp. nonempty quasi-compact). Namely, by Properties of Stacks, Lemma 0509 there exists a unique reduced closed substack \(\mathcal{X}' \subset \mathcal{X}\) such that \(T = |\mathcal{X}'|\). Note that \(\mathcal{I}_{\mathcal{X}'} = \mathcal{I}_\mathcal{X} \times_\mathcal{X} \mathcal{X}'\) by Lemma 06R5. Hence \(\mathcal{I}_{\mathcal{X}'} \to \mathcal{X}'\) is quasi-compact as a base change, see Lemma 050V. Therefore Proposition 06RC implies there exists a dense maximal (see proof proposition) open substack \(\mathcal{U} \subset \mathcal{X}'\) which is a gerbe. In case (a) we set \(\mathcal{U}(T) = \mathcal{U}\) (this is canonical) and in case (b) we simply choose a nonempty quasi-compact open \(\mathcal{U}(T) \subset \mathcal{U}\), see Properties of Stacks, Lemma 04Y9 (we can do this for all \(T\) simultaneously by the axiom of choice).

Using transfinite recursion we construct for every ordinal \(\alpha\) a closed subset \(T_\alpha \subset |\mathcal{X}|\). For \(\alpha = 0\) we set \(T_0 = |\mathcal{X}|\). Given \(T_\alpha\) set \[T_{\alpha + 1} = T_\alpha \setminus |\mathcal{U}(T_\alpha)|.\] If \(\beta\) is a limit ordinal we set \[T_\beta = \bigcap\nolimits_{\alpha < \beta} T_\alpha.\] We claim that \(T_\alpha = \emptyset\) for all \(\alpha\) large enough. Namely, assume that \(T_\alpha \not = \emptyset\) for all \(\alpha\). Then we obtain an injective map from the class of ordinals into the set of subsets of \(|\mathcal{X}|\) which is a contradiction.

The claim implies the lemma. Namely, let \[I = \{\alpha \mid \mathcal{U}_\alpha \not = \emptyset \}.\] This is a well-ordered set by the claim. For \(i = \alpha \in I\) we set \(\mathcal{U}_i = \mathcal{U}_\alpha\). So \(\mathcal{U}_i\) is a reduced locally closed substack and a gerbe, i.e., (1) holds. By construction \(T_i = T\alpha\) if \(i = \alpha \in I\), hence (3) holds. Also, (4) and (a) or (b) hold by our choice of \(\mathcal{U}(T)\) as well. Finally, to see (2) let \(x \in |\mathcal{X}|\). There exists a smallest ordinal \(\beta\) with \(x \not \in T_\beta\) (because the ordinals are well-ordered). In this case \(\beta\) has to be a successor ordinal by the definition of \(T_\beta\) for limit ordinals. Hence \(\beta = \alpha + 1\) and \(x \in |\mathcal{U}(T_\alpha)|\) and we win.

Remark

We can wonder about the order type of the canonical stratifications which occur as output of the stratifications of type (a) constructed in Lemma 06RF. A natural guess is that the well-ordered set \(I\) has cardinality at most \(\aleph_0\). We have no idea if this is true or false. If you do please email stacks.project@gmail.com.

The topological space of an algebraic stack

In this section we apply the previous results to the topological space \(|\mathcal{X}|\) associated to an algebraic stack.

Lemma

Let \(\mathcal{X}\) be a quasi-compact algebraic stack whose diagonal \(\Delta\) is quasi-compact. Then \(|\mathcal{X}|\) is a spectral topological space.

Proof

Choose an affine scheme \(U\) and a surjective smooth morphism \(U \to \mathcal{X}\), see Properties of Stacks, Lemma 04YC. Then \(|U| \to |\mathcal{X}|\) is continuous, open, and surjective, see Properties of Stacks, Lemma 04XL. Hence the quasi-compact opens of \(|\mathcal{X}|\) form a basis for the topology. For \(W_1, W_2 \subset |\mathcal{X}|\) quasi-compact open, we may choose a quasi-compact opens \(V_1, V_2\) of \(U\) mapping to \(W_1\) and \(W_2\). Since \(\Delta\) is quasi-compact, we see that \[V_1 \times_\mathcal{X} V_2 = (V_1 \times V_2) \times_{\mathcal{X} \times \mathcal{X}, \Delta} \mathcal{X}\] is quasi-compact. Then image of \(|V_1 \times_\mathcal{X} V_2|\) in \(|\mathcal{X}|\) is \(W_1 \cap W_2\) by Properties of Stacks, Lemma 04XH. Thus \(W_1 \cap W_2\) is quasi-compact. To finish the proof, it suffices to show that \(|\mathcal{X}|\) is sober, see Topology, Definition 08YG.

Let \(T \subset |\mathcal{X}|\) be an irreducible closed subset. We have to show \(T\) has a unique generic point. Let \(\mathcal{Z} \subset \mathcal{X}\) be the reduced induced closed substack corresponding to \(T\), see Properties of Stacks, Definition 050C. Since \(\mathcal{Z} \to \mathcal{X}\) is a closed immersion, we see that \(\Delta_\mathcal{Z}\) is quasi-compact: first show that \(\mathcal{Z} \to \mathcal{X} \times \mathcal{X}\) is quasi-compact as the composition of \(\mathcal{Z} \to \mathcal{X}\) with \(\Delta\), then write \(\mathcal{Z} \to \mathcal{X} \times \mathcal{X}\) as the composition of \(\Delta_\mathcal{Z}\) and \(\mathcal{Z} \times \mathcal{Z} \to \mathcal{X} \times \mathcal{X}\) and use Lemma 050Y and the fact that \(\mathcal{Z} \times \mathcal{Z} \to \mathcal{X} \times \mathcal{X}\) is separated. Thus we reduce to the case discussed in the next paragraph.

Assume \(\mathcal{X}\) is quasi-compact, \(\Delta\) is quasi-compact, \(\mathcal{X}\) is reduced, and \(|\mathcal{X}|\) irreducible. We have to show \(|\mathcal{X}|\) has a unique generic point. Since \(\mathcal{I}_\mathcal{X} \to \mathcal{X}\) is a base change of \(\Delta\), we see that \(\mathcal{I}_\mathcal{X} \to \mathcal{X}\) is quasi-compact (Lemma 050V). Thus there exists a dense open substack \(\mathcal{U} \subset \mathcal{X}\) which is a gerbe by Proposition 06RC. In other words, \(|\mathcal{U}| \subset |\mathcal{X}|\) is open dense. Thus we may assume that \(\mathcal{X}\) is a gerbe. Say \(\mathcal{X} \to X\) turns \(\mathcal{X}\) into a gerbe over the algebraic space \(X\). Then \(|\mathcal{X}| \cong |X|\) by Lemma 06R9. In particular, \(X\) is quasi-compact. By Lemma 0DQL we see that \(X\) has quasi-compact diagonal, i.e., \(X\) is a quasi-separated algebraic space. Then \(|X|\) is spectral by Properties of Spaces, Lemma 0A4G which implies what we want is true.

Lemma

Let \(\mathcal{X}\) be a quasi-compact and quasi-separated algebraic stack. Then \(|\mathcal{X}|\) is a spectral topological space.

Proof

This is a special case of Lemma 0DQN.

Lemma

Let \(\mathcal{X}\) be an algebraic stack whose diagonal is quasi-compact (for example if \(\mathcal{X}\) is quasi-separated). Then there is an open covering \(|\mathcal{X}| = \bigcup U_i\) with \(U_i\) spectral. In particular \(|\mathcal{X}|\) is a sober topological space.

Proof

Immediate consequence of Lemma 0DQN.

Existence of residual gerbes

The definition of a residual gerbe of a point on an algebraic stack is Properties of Stacks, Definition 06MU. We have already shown that residual gerbes exist for finite type points (Lemma 06G3) and for any point of a gerbe (Lemma 06RA). In this section we prove that residual gerbes exist on many algebraic stacks. First, here is the promised application of Proposition 06RC.

Lemma

Let \(\mathcal{X}\) be an algebraic stack such that \(\mathcal{I}_\mathcal{X} \to \mathcal{X}\) is quasi-compact. Then the residual gerbe of \(\mathcal{X}\) at \(x\) exists for every \(x \in |\mathcal{X}|\).

Proof

Let \(T = \overline{\{x\}} \subset |\mathcal{X}|\) be the closure of \(x\). By Properties of Stacks, Lemma 0509 there exists a reduced closed substack \(\mathcal{X}' \subset \mathcal{X}\) such that \(T = |\mathcal{X}'|\). Note that \(\mathcal{I}_{\mathcal{X}'} = \mathcal{I}_\mathcal{X} \times_\mathcal{X} \mathcal{X}'\) by Lemma 06R5. Hence \(\mathcal{I}_{\mathcal{X}'} \to \mathcal{X}'\) is quasi-compact as a base change, see Lemma 050V. Therefore Proposition 06RC implies there exists a dense open substack \(\mathcal{U} \subset \mathcal{X}'\) which is a gerbe. Note that \(x \in |\mathcal{U}|\) because \(\{x\} \subset T\) is a dense subset too. Hence a residual gerbe \(\mathcal{Z}_x \subset \mathcal{U}\) of \(\mathcal{U}\) at \(x\) exists by Lemma 06RA. It is immediate from the definitions that \(\mathcal{Z}_x \to \mathcal{X}\) is a residual gerbe of \(\mathcal{X}\) at \(x\).

If the stack is quasi-DM then residual gerbes exist too. In particular, residual gerbes always exist for Deligne-Mumford stacks.

Lemma

Let \(\mathcal{X}\) be a quasi-DM algebraic stack. Then the residual gerbe of \(\mathcal{X}\) at \(x\) exists for every \(x \in |\mathcal{X}|\).

Proof

Choose a scheme \(U\) and a surjective, flat, locally finite presented, and locally quasi-finite morphism \(U \to \mathcal{X}\), see Theorem 06MF. Set \(R = U \times_\mathcal{X} U\). The projections \(s, t : R \to U\) are surjective, flat, locally of finite presentation, and locally quasi-finite as base changes of the morphism \(U \to \mathcal{X}\). There is a canonical morphism \([U/R] \to \mathcal{X}\) (see Algebraic Stacks, Lemma 04T4) which is an equivalence because \(U \to \mathcal{X}\) is surjective, flat, and locally of finite presentation, see Algebraic Stacks, Remark 04WY. Thus we may assume that \(\mathcal{X} = [U/R]\) where \((U, R, s, t, c)\) is a groupoid in algebraic spaces such that \(s, t : R \to U\) are surjective, flat, locally of finite presentation, and locally quasi-finite. Set \[U' = \coprod\nolimits_{u \in U\text{ lying over }x} \Spec(\kappa(u)).\] The canonical morphism \(U' \to U\) is a monomorphism. Let \[R' = U' \times_\mathcal{X} U' = R \times_{(U \times U)} (U' \times U')\] Because \(U' \to U\) is a monomorphism we see that both projections \(s', t' : R' \to U'\) factor as a monomorphism followed by a locally quasi-finite morphism. Hence, as \(U'\) is a disjoint union of spectra of fields, using Spaces over Fields, Lemma 06S1 we conclude that the morphisms \(s', t' : R' \to U'\) are locally quasi-finite. Again since \(U'\) is a disjoint union of spectra of fields, the morphisms \(s', t'\) are also flat. Finally, \(s', t'\) locally quasi-finite implies \(s', t'\) locally of finite type, hence \(s', t'\) locally of finite presentation (because \(U'\) is a disjoint union of spectra of fields in particular locally Noetherian, so that Morphisms of Spaces, Lemma 06G4 applies). Hence \(\mathcal{Z} = [U'/R']\) is an algebraic stack by Criteria for Representability, Theorem 06FI. As \(R'\) is the restriction of \(R\) by \(U' \to U\) we see \(\mathcal{Z} \to \mathcal{X}\) is a monomorphism by Groupoids in Spaces, Lemma 046S and Properties of Stacks, Lemma 04ZZ. Since \(\mathcal{Z} \to \mathcal{X}\) is a monomorphism we see that \(|\mathcal{Z}| \to |\mathcal{X}|\) is injective, see Properties of Stacks, Lemma 0500. By Properties of Stacks, Lemma 04XH we see that \[|U'| = |\mathcal{Z} \times_\mathcal{X} U'| \longrightarrow |\mathcal{Z}| \times_{|\mathcal{X}|} |U'|\] is surjective which implies (by our choice of \(U'\)) that \(|\mathcal{Z}| \to |\mathcal{X}|\) has image \(\{x\}\). We conclude that \(|\mathcal{Z}|\) is a singleton. Finally, by construction \(U'\) is locally Noetherian and reduced, i.e., \(\mathcal{Z}\) is reduced and locally Noetherian. This means that the essential image of \(\mathcal{Z} \to \mathcal{X}\) is the residual gerbe of \(\mathcal{X}\) at \(x\), see Properties of Stacks, Lemma 06MX.

Lemma

Let \(\mathcal{X}\) be a locally Noetherian algebraic stack. Then the residual gerbe of \(\mathcal{X}\) at \(x\) exists for every \(x \in |\mathcal{X}|\).

Proof

Choose an affine scheme \(U\) and a smooth morphism \(U \to \mathcal{X}\) such that \(x\) is in the image of the open continuous map \(|U| \to |\mathcal{X}|\). We may and do replace \(\mathcal{X}\) with the open substack corresponding to the image of \(|U| \to |\mathcal{X}|\), see Properties of Stacks, Lemma 06FJ. Thus we may assume \(\mathcal{X} = [U/R]\) for a smooth groupoid \((U, R, s, t, c)\) in algebraic spaces where \(U\) is a Noetherian affine scheme, see Algebraic Stacks, Section 04T3.

Let \(E \subset |U|\) be the inverse image of \(\{x\} \subset |\mathcal{X}|\). Of course \(E \not = \emptyset\). Since \(|U|\) is a Noetherian topological space, we can choose an element \(u \in E\) such that \(\overline{\{u\}} \cap E = \{u\}\). As usual, we think of \(u = \Spec(\kappa(u))\) as the spectrum of its residue field. Let us write \[F = u \times_{U, t} R = u \times_\mathcal{X} U \quad\text{and}\quad R' = (u \times u) \times_{(U \times U), (t, s)} R = u \times_\mathcal{X} u\] Furthermore, denote \(Z = \overline{\{u\}} \subset U\) with the reduced induced scheme structure. Denote \(p : F \to U\) the morphism induced by the second projection (using \(s : R \to U\) in the first fibre product description of \(F\)). Then \(E\) is the set theoretic image of \(p\). The morphism \(R' \to F\) is a monomorphism which factors through the inverse image \(p^{-1}(Z)\) of \(Z\). This inverse image \(p^{-1}(Z) \subset F\) is a closed subscheme and the restriction \(p|_{p^{-1}(Z)} : p^{-1}(Z) \to Z\) has image set theoretically contained in \(\{u\} \subset Z\) by our careful choice of \(u \in E\) above. Since \(u = \lim W\) where the limit is over the nonempty affine open subschemes of the irreducible reduced scheme \(Z\), we conclude that the morphism \(p|_{p^{-1}(Z)} : p^{-1}(Z) \to Z\) factors through the morphism \(u \to Z\). Clearly this implies that \(R' = p^{-1}(Z)\). In particular the morphism \(t' : R' \to u\) is locally of finite presentation as the composition of the closed immersion \(p^{-1}(Z) \to F\) of locally Noetherian algebraic spaces with the smooth morphism \(\text{pr}_1 : F \to u\); use Morphisms of Spaces, Lemmas 04ZK, 084Q, and 03XQ. Hence the restriction \((u, R', s', t', c')\) of \((U, R, s, t, c)\) by \(u \to U\) is a groupoid in algebraic spaces where \(s'\) and \(t'\) are flat and locally of finite presentation. Therefore \(\mathcal{Z} = [u/R']\) is an algebraic stack by Criteria for Representability, Theorem 06FI. As \(R'\) is the restriction of \(R\) by \(u \to U\) we see \(\mathcal{Z} \to \mathcal{X}\) is a monomorphism by Groupoids in Spaces, Lemma 046S and Properties of Stacks, Lemma 04ZZ. Then \(\mathcal{Z}\) is (isomorphic to) the residual gerbe by the material in Properties of Stacks, Section 06ML.

Étale local structure

In this section we start discussing what we can say about the étale local structure of an algebraic stack.

Lemma

Let \(Y\) be an algebraic space. Let \((U, R, s, t, c)\) be a groupoid in algebraic spaces over \(Y\). Assume \(U \to Y\) is flat and locally of finite presentation and \(R \to U \times_Y U\) an open immersion. Then \(X = [U/R] = U/R\) is an algebraic space and \(X \to Y\) is étale.

Proof

The quotient stack \([U/R]\) is an algebraic stacks by Criteria for Representability, Theorem 06FI. On the other hand, since \(R \to U \times U\) is a monomorphism, it is an algebraic space (by our abuse of language and Algebraic Stacks, Proposition 04SZ) and of course it is equal to the algebraic space \(U/R\) (of Bootstrap, Theorem 04S6). Since \(U \to X\) is surjective, flat, and locally of finite presentation (Bootstrap, Lemma 06PG) we conclude that \(X \to Y\) is flat and locally of finite presentation by Morphisms of Spaces, Lemma 05VY and Descent on Spaces, Lemma 0AHC. Finally, consider the cartesian diagram \[\xymatrix{ R \ar[d] \ar[r] & U \times_Y U \ar[d] \\ X \ar[r] & X \times_Y X }\] Since the right vertical arrow is surjective, flat, and locally of finite presentation (small detail omitted), we find that \(X \to X \times_Y X\) is an open immersion as the top horizontal arrow has this property by assumption (use Properties of Stacks, Lemma 04XD). Thus \(X \to Y\) is unramified by Morphisms of Spaces, Lemma 05W1. We conclude by Morphisms of Spaces, Lemma 06LU.

Lemma

Let \(S\) be a scheme. Let \((U, R, s, t, c)\) be a groupoid in algebraic spaces over \(S\). Assume \(s, t\) are flat and locally of finite presentation. Let \(P \subset R\) be an open subspace such that \((U, P, s|_P, t|_P, c|_{P \times_{s, U, t} P})\) is a groupoid in algebraic spaces over \(S\). Then \[[U/P] \longrightarrow [U/R]\] is a morphism of algebraic stacks which is representable by algebraic spaces, surjective, and étale.

Proof

Since \(P \subset R\) is open, we see that \(s|_P\) and \(t|_P\) are flat and locally of finite presentation. Thus \([U/R]\) and \([U/P]\) are algebraic stacks by Criteria for Representability, Theorem 06FI. To see that the morphism is representable by algebraic spaces, it suffices to show that \([U/P] \to [U/R]\) is faithful on fibre categories, see Algebraic Stacks, Lemma 04Y5. This follows immediately from the fact that \(P \to R\) is a monomorphism and the explicit description of quotient stacks given in Groupoids in Spaces, Lemma 044X. Having said this, we know what it means for \([U/P] \to [U/R]\) to be surjective and étale by Algebraic Stacks, Definition 03YK. Surjectivity follows for example from Criteria for Representability, Lemma 06D7 and the description of objects of quotient stacks (see lemma cited above) over spectra of fields. It remains to prove that our morphism is étale.

To do this it suffices to show that \(U \times_{[U/R]} [U/P] \to U\) is étale, see Properties of Stacks, Lemma 04XD. By Groupoids in Spaces, Lemma 04Y4 the fibre product is equal to \([R/P \times_{s, U, t} R]\) with morphism to \(U\) induced by \(s : R \to U\). The maps \(s', t' : P \times_{s, U, t} R \to R\) are given by \(s' : (p, r) \mapsto r\) and \(t' : (p, r) \mapsto c(p, r)\). Since \(P \subset R\) is open we conclude that \((t', s') : P \times_{s, U, t} R \to R \times_{s, U, s} R\) is an open immersion. Thus we may apply Lemma 0DU1 to conclude.

Lemma

Let \(\mathcal{X}\) be an algebraic stack. Assume \(\mathcal{X}\) is quasi-DM with separated diagonal (equivalently \(\mathcal{I}_\mathcal{X} \to \mathcal{X}\) is locally quasi-finite and separated). Let \(x \in |\mathcal{X}|\). Then there exists a morphism of algebraic stacks \[\mathcal{U} \longrightarrow \mathcal{X}\] with the following properties

1. there exists a point \(u \in |\mathcal{U}|\) mapping to \(x\),

2. \(\mathcal{U} \to \mathcal{X}\) is representable by algebraic spaces and étale,

3. \(\mathcal{U} = [U/R]\) where \((U, R, s, t, c)\) is a groupoid scheme with \(U\), \(R\) affine, and \(s, t\) finite, flat, and locally of finite presentation.

Proof

(The parenthetical statement follows from the equivalences in Lemma 0CL0). Choose an affine scheme \(U\) and a flat, locally finitely presented, locally quasi-finite morphism \(U \to \mathcal{X}\) such that \(x\) is the image of some point \(u \in U\). This is possible by Theorem 06MF and the assumption that \(\mathcal{X}\) is quasi-DM. Let \((U, R, s, t, c)\) be the groupoid in algebraic spaces we obtain by setting \(R = U \times_\mathcal{X} U\), see Algebraic Stacks, Lemma 04T4. Let \(\mathcal{X}' \subset \mathcal{X}\) be the open substack corresponding to the open image of \(|U| \to |\mathcal{X}|\) (Properties of Stacks, Lemmas 04XL and 06FJ). Clearly, we may replace \(\mathcal{X}\) by the open substack \(\mathcal{X}'\). Thus we may assume \(U \to \mathcal{X}\) is surjective and then Algebraic Stacks, Remark 04WY gives \(\mathcal{X} = [U/R]\). Observe that \(s, t : R \to U\) are flat, locally of finite presentation, and locally quasi-finite. Since \(R = U \times U \times_{(\mathcal{X} \times \mathcal{X})} \mathcal{X}\) and since the diagonal of \(\mathcal{X}\) is separated, we find that \(R\) is separated. Hence \(s, t : R \to U\) are separated. It follows that \(R\) is a scheme by Morphisms of Spaces, Proposition 03XX applied to \(s : R \to U\).

Above we have verified all the assumptions of More on Groupoids in Spaces, Lemma 04S0 are satisfied for \((U, R, s, t, c)\) and \(u\). Hence we can find an elementary étale neighbourhood \((U', u') \to (U, u)\) such that the restriction \(R'\) of \(R\) to \(U'\) is quasi-split over \(u\). Note that \(R' = U' \times_\mathcal{X} U'\) (small detail omitted; hint: transitivity of fibre products). Replacing \((U, R, s, t, c)\) by \((U', R', s', t', c')\) and shrinking \(\mathcal{X}\) as above, we may assume that \((U, R, s, t, c)\) has a quasi-splitting over \(u\) (the point \(u\) is irrelevant from now on as can be seen from the footnote in More on Groupoids in Spaces, Definition 04RK). Let \(P \subset R\) be a quasi-splitting of \(R\) over \(u\). Apply Lemma 0DU2 to see that \[\mathcal{U} = [U/P] \longrightarrow [U/R] = \mathcal{X}\] has all the desired properties.

Lemma

Let \(\mathcal{X}\) be an algebraic stack. Assume \(\mathcal{X}\) is quasi-DM with separated diagonal (equivalently \(\mathcal{I}_\mathcal{X} \to \mathcal{X}\) is locally quasi-finite and separated). Let \(x \in |\mathcal{X}|\). Assume the automorphism group of \(\mathcal{X}\) at \(x\) is finite (Remark 0DTU). Then there exists a morphism of algebraic stacks \[g : \mathcal{U} \longrightarrow \mathcal{X}\] with the following properties

1. there exists a point \(u \in |\mathcal{U}|\) mapping to \(x\) and \(g\) induces an isomorphism between automorphism groups at \(u\) and \(x\) (Remark 0DTW),

2. \(\mathcal{U} \to \mathcal{X}\) is representable by algebraic spaces and étale,

3. \(\mathcal{U} = [U/R]\) where \((U, R, s, t, c)\) is a groupoid scheme with \(U\), \(R\) affine, and \(s, t\) finite, flat, and locally of finite presentation.

Proof

Observe that \(G_x\) is a group scheme by Lemma 0DTS. The first part of the proof is exactly the same as the first part of the proof of Lemma 0DU3. Thus we may assume \(\mathcal{X} = [U/R]\) where \((U, R, s, t, c)\) and \(u \in U\) mapping to \(x\) satisfy all the assumptions of More on Groupoids in Spaces, Lemma 04S0. Our assumption on \(G_x\) implies that \(G_u\) is finite over \(u\). Hence all the assumptions of More on Groupoids in Spaces, Lemma 0DTE are satisfied. Hence we can find an elementary étale neighbourhood \((U', u') \to (U, u)\) such that the restriction \(R'\) of \(R\) to \(U'\) is split over \(u\). Note that \(R' = U' \times_\mathcal{X} U'\) (small detail omitted; hint: transitivity of fibre products). Replacing \((U, R, s, t, c)\) by \((U', R', s', t', c')\) and shrinking \(\mathcal{X}\) as above, we may assume that \((U, R, s, t, c)\) has a splitting over \(u\). Let \(P \subset R\) be a splitting of \(R\) over \(u\). Apply Lemma 0DU2 to see that \[\mathcal{U} = [U/P] \longrightarrow [U/R] = \mathcal{X}\] is representable by algebraic spaces and étale. By construction \(G_u\) is contained in \(P\), hence this morphism defines an isomorphism on automorphism groups at \(u\) as desired.

Lemma

Let \(\mathcal{X}\) be an algebraic stack. Assume \(\mathcal{X}\) is quasi-DM with separated diagonal (equivalently \(\mathcal{I}_\mathcal{X} \to \mathcal{X}\) is locally quasi-finite and separated). Let \(x \in |\mathcal{X}|\). Assume \(x\) can be represented by a quasi-compact morphism \(\Spec(k) \to \mathcal{X}\). Then there exists a morphism of algebraic stacks \[g : \mathcal{U} \longrightarrow \mathcal{X}\] with the following properties

1. there exists a point \(u \in |\mathcal{U}|\) mapping to \(x\) and \(g\) induces an isomorphism between the residual gerbes at \(u\) and \(x\),

2. \(\mathcal{U} \to \mathcal{X}\) is representable by algebraic spaces and étale,

3. \(\mathcal{U} = [U/R]\) where \((U, R, s, t, c)\) is a groupoid scheme with \(U\), \(R\) affine, and \(s, t\) finite, flat, and locally of finite presentation.

Proof

The first part of the proof is exactly the same as the first part of the proof of Lemma 0DU3. Thus we may assume \(\mathcal{X} = [U/R]\) where \((U, R, s, t, c)\) and \(u \in U\) mapping to \(x\) satisfy all the assumptions of More on Groupoids in Spaces, Lemma 04S0. Observe that \(u = \Spec(\kappa(u)) \to \mathcal{X}\) is quasi-compact, see Properties of Stacks, Lemma 0DTK. Consider the cartesian diagram \[\xymatrix{ F \ar[d] \ar[r] & U \ar[d] \\ u \ar[r]^u & \mathcal{X} }\] Since \(U\) is an affine scheme and \(F \to U\) is quasi-compact, we see that \(F\) is quasi-compact. Since \(U \to \mathcal{X}\) is locally quasi-finite, we see that \(F \to u\) is locally quasi-finite. Hence \(F \to u\) is quasi-finite and \(F\) is an affine scheme whose underlying topological space is finite discrete (Spaces over Fields, Lemma 06S0). Observe that we have a monomorphism \(u \times_\mathcal{X} u \to F\). In particular the set \(\{r \in R : s(r) = u, t(r) = u\}\) which is the image of \(|u \times_\mathcal{X} u| \to |R|\) is finite. we conclude that all the assumptions of More on Groupoids in Spaces, Lemma 04RZ hold.

Thus we can find an elementary étale neighbourhood \((U', u') \to (U, u)\) such that the restriction \(R'\) of \(R\) to \(U'\) is strongly split over \(u'\). Note that \(R' = U' \times_\mathcal{X} U'\) (small detail omitted; hint: transitivity of fibre products). Replacing \((U, R, s, t, c)\) by \((U', R', s', t', c')\) and shrinking \(\mathcal{X}\) as above, we may assume that \((U, R, s, t, c)\) has a strong splitting over \(u\). Let \(P \subset R\) be a strong splitting of \(R\) over \(u\). Apply Lemma 0DU2 to see that \[\mathcal{U} = [U/P] \longrightarrow [U/R] = \mathcal{X}\] is representable by algebraic spaces and étale. Since \(P \subset R\) is open and contains \(\{r \in R : s(r) = u, t(r) = u\}\) by construction we see that \(u \times_\mathcal{U} u \to u \times_\mathcal{X} u\) is an isomorphism. The statement on residual gerbes then follows from Properties of Stacks, Lemma 0DTI (we observe that the residual gerbes in question exist by Lemma 06UI).

Smooth morphisms

The property “being smooth” of morphisms of algebraic spaces is smooth local on the source-and-target, see Descent on Spaces, Remark 06FC. It is also stable under base change and fpqc local on the target, see Morphisms of Spaces, Lemma 03ZE and Descent on Spaces, Lemma 0429. Hence, by Lemma 06FM above, we may define what it means for a morphism of algebraic spaces to be smooth as follows and it agrees with the already existing notion defined in Properties of Stacks, Section 04XB when the morphism is representable by algebraic spaces.

Definition

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. We say \(f\) is smooth if the equivalent conditions of Lemma 06FM hold with \(\mathcal{P} = \text{smooth}\).

Lemma

The composition of smooth morphisms is smooth.

Proof

Combine Remark 06FP with Morphisms of Spaces, Lemma 03ZD.

Lemma

A base change of a smooth morphism is smooth.

Proof

Combine Remark 06FQ with Morphisms of Spaces, Lemma 03ZE.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. Let \(\mathcal{Z} \to \mathcal{Y}\) be a surjective, flat, locally finitely presented morphism of algebraic stacks. If the base change \(\mathcal{Z} \times_\mathcal{Y} \mathcal{X} \to \mathcal{Z}\) is smooth, then \(f\) is smooth.

Proof

The property “smooth” satisfies the conditions of Lemma 0DN6. Smooth local on the source-and-target we have seen in the introduction to this section and fppf local on the target is Descent on Spaces, Lemma 0429.

Lemma

A smooth morphism of algebraic stacks is locally of finite presentation.

Proof

Omitted.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. There is a largest open substack \(\mathcal{U} \subset \mathcal{X}\) such that \(f|_\mathcal{U} : \mathcal{U} \to \mathcal{Y}\) is smooth. Moreover, formation of this open commutes with

1. precomposing by smooth morphisms,

2. base change by morphisms which are flat and locally of finite presentation,

3. base change by flat morphisms provided \(f\) is locally of finite presentation.

Proof

Choose a commutative diagram \[\xymatrix{ U \ar[d]_a \ar[r]_h & V \ar[d]^b \\ \mathcal{X} \ar[r]^f & \mathcal{Y} }\] where \(U\) and \(V\) are algebraic spaces, the vertical arrows are smooth, and \(a : U \to \mathcal{X}\) surjective. There is a maximal open subspace \(U' \subset U\) such that \(h_{U'} : U' \to V\) is smooth, see Morphisms of Spaces, Lemma 0DZI. Let \(\mathcal{U} \subset \mathcal{X}\) be the open substack corresponding to the image of \(|U'| \to |\mathcal{X}|\) (Properties of Stacks, Lemmas 04XL and 06FJ). By the equivalence in Lemma 06FM we find that \(f|_\mathcal{U} : \mathcal{U} \to \mathcal{Y}\) is smooth and that \(\mathcal{U}\) is the largest open substack with this property.

Assertion (1) follows from the fact that being smooth is smooth local on the source (this property was used to even define smooth morphisms of algebraic stacks). Assertions (2) and (3) follow from the case of algebraic spaces, see Morphisms of Spaces, Lemma 0DZI.

Lemma

Let \(X \to Y\) be a smooth morphism of algebraic spaces. Let \(G\) be a group algebraic space over \(Y\) which is flat and locally of finite presentation over \(Y\). Let \(G\) act on \(X\) over \(Y\). Then the quotient stack \([X/G]\) is smooth over \(Y\).

This holds even if \(G\) is not smooth over \(Y\)!

Proof

The quotient \([X/G]\) is an algebraic stack by Criteria for Representability, Theorem 06FI. The smoothness of \([X/G]\) over \(Y\) follows from the fact that smoothness descends under fppf coverings: Choose a surjective smooth morphism \(U \to [X/G]\) where \(U\) is a scheme. Smoothness of \([X/G]\) over \(Y\) is equivalent to smoothness of \(U\) over \(Y\). Observe that \(U \times_{[X/G]} X\) is smooth over \(X\) and hence smooth over \(Y\) (because compositions of smooth morphisms are smooth). On the other hand, \(U \times_{[X/G]} X \to U\) is locally of finite presentation, flat, and surjective (because it is the base change of \(X \to [X/G]\) which has those properties for example by Criteria for Representability, Lemma 06FH). Therefore we may apply Descent on Spaces, Lemma 0AHE.

Lemma

Let \(\pi : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. If \(\mathcal{X}\) is a gerbe over \(\mathcal{Y}\), then \(\pi\) is surjective and smooth.

Proof

We have seen surjectivity in Lemma 06QI. By Lemma 0DN7 it suffices to prove to the lemma after replacing \(\pi\) by a base change with a surjective, flat, locally finitely presented morphism \(\mathcal{Y}' \to \mathcal{Y}\). By Lemma 06QH we may assume \(\mathcal{Y} = U\) is an algebraic space and \(\mathcal{X} = [U/G]\) over \(U\) with \(G \to U\) flat and locally of finite presentation. Then we win by Lemma 0DLS.

Types of morphisms étale-smooth local on source-and-target

Given a property of morphisms of algebraic spaces which is étale-smooth local on the source-and-target, see Descent on Spaces, Definition 0CFZ we may use it to define a corresponding property of DM morphisms of algebraic stacks, namely by imposing either of the equivalent conditions of the lemma below.

Lemma

Let \(\mathcal{P}\) be a property of morphisms of algebraic spaces which is étale-smooth local on the source-and-target. Let \(f : \mathcal{X} \to \mathcal{Y}\) be a DM morphism of algebraic stacks. Consider commutative diagrams \[\xymatrix{ U \ar[d]_a \ar[r]_h & V \ar[d]^b \\ \mathcal{X} \ar[r]^f & \mathcal{Y} }\] where \(U\) and \(V\) are algebraic spaces, \(V \to \mathcal{Y}\) is smooth, and \(U \to \mathcal{X} \times_\mathcal{Y} V\) is étale. The following are equivalent

1. for any diagram as above the morphism \(h\) has property \(\mathcal{P}\), and

2. for some diagram as above with \(a : U \to \mathcal{X}\) surjective the morphism \(h\) has property \(\mathcal{P}\).

If \(\mathcal{X}\) and \(\mathcal{Y}\) are representable by algebraic spaces, then this is also equivalent to \(f\) (as a morphism of algebraic spaces) having property \(\mathcal{P}\). If \(\mathcal{P}\) is also preserved under any base change, and fppf local on the base, then for morphisms \(f\) which are representable by algebraic spaces this is also equivalent to \(f\) having property \(\mathcal{P}\) in the sense of Properties of Stacks, Section 04XB.

Proof

Let us prove the implication (1) \(\Rightarrow\) (2). Pick an algebraic space \(V\) and a surjective and smooth morphism \(V \to \mathcal{Y}\). As \(f\) is DM there exists a scheme \(U\) and a surjective étale morphism \(U \to V \times_\mathcal{Y} \mathcal{X}\), see Lemma 0CIA. Thus we see that (2) holds. Note that \(U \to \mathcal{X}\) is surjective and smooth as well, as a composition of the base change \(\mathcal{X} \times_\mathcal{Y} V \to \mathcal{X}\) and the chosen map \(U \to \mathcal{X} \times_\mathcal{Y} V\). Hence we obtain a diagram as in (1). Thus if (1) holds, then \(h : U \to V\) has property \(\mathcal{P}\), which means that (2) holds as \(U \to \mathcal{X}\) is surjective.

Conversely, assume (2) holds and let \(U, V, a, b, h\) be as in (2). Next, let \(U', V', a', b', h'\) be any diagram as in (1). Picture \[\xymatrix{ U \ar[d] \ar[r]_h & V \ar[d] \\ \mathcal{X} \ar[r]^f & \mathcal{Y} } \quad\quad \xymatrix{ U' \ar[d] \ar[r]_{h'} & V' \ar[d] \\ \mathcal{X} \ar[r]^f & \mathcal{Y} }\] To show that (2) implies (1) we have to prove that \(h'\) has \(\mathcal{P}\). To do this consider the commutative diagram \[\xymatrix{ U \ar[d]^h & U \times_\mathcal{X} U' \ar[l] \ar[d]^{(h, h')} \ar[r] & U' \ar[d]^{h'} \\ V & V \times_\mathcal{Y} V' \ar[l] \ar[r] & V' }\] of algebraic spaces. Note that the horizontal arrows are smooth as base changes of the smooth morphisms \(V \to \mathcal{Y}\), \(V' \to \mathcal{Y}\), \(U \to \mathcal{X}\), and \(U' \to \mathcal{X}\). Note that the squares \[\xymatrix{ U \ar[d] & U \times_\mathcal{X} U' \ar[l] \ar[d] & U \times_\mathcal{X} U' \ar[d] \ar[r] & U' \ar[d] \\ V \times_\mathcal{Y} \mathcal{X} & V \times_\mathcal{Y} U' \ar[l] & U \times_\mathcal{Y} V' \ar[r] & \mathcal{X} \times_\mathcal{Y} V' }\] are cartesian, hence the vertical arrows are étale by our assumptions on \(U', V', a', b', h'\) and \(U, V, a, b, h\). Since \(\mathcal{P}\) is smooth local on the target by Descent on Spaces, Lemma 0CG0 part (2) we see that the base change \(t : U \times_\mathcal{Y} V' \to V \times_\mathcal{Y} V'\) of \(h\) has \(\mathcal{P}\). Since \(\mathcal{P}\) is étale local on the source by Descent on Spaces, Lemma 0CG0 part (1) and \(s : U \times_\mathcal{X} U' \to U \times_\mathcal{Y} V'\) is étale, we see the morphism \((h, h') = t \circ s\) has \(\mathcal{P}\). Consider the diagram \[\xymatrix{ U \times_\mathcal{X} U' \ar[r]_{(h, h')} \ar[d] & V \times_\mathcal{Y} V' \ar[d] \\ U' \ar[r]^{h'} & V' }\] The left vertical arrow is surjective, the right vertical arrow is smooth, and the induced morphism \[U \times_\mathcal{X} U' \longrightarrow U' \times_{V'} (V \times_\mathcal{Y} V') = V \times_\mathcal{Y} U'\] is étale as seen above. Hence by Descent on Spaces, Definition 0CFZ part (3) we conclude that \(h'\) has \(\mathcal{P}\). This finishes the proof of the equivalence of (1) and (2).

If \(\mathcal{X}\) and \(\mathcal{Y}\) are representable, then Descent on Spaces, Lemma 0CG1 applies which shows that (1) and (2) are equivalent to \(f\) having \(\mathcal{P}\).

Finally, suppose \(f\) is representable, and \(U, V, a, b, h\) are as in part (2) of the lemma, and that \(\mathcal{P}\) is preserved under arbitrary base change. We have to show that for any scheme \(Z\) and morphism \(Z \to \mathcal{X}\) the base change \(Z \times_\mathcal{Y} \mathcal{X} \to Z\) has property \(\mathcal{P}\). Consider the diagram \[\xymatrix{ Z \times_\mathcal{Y} U \ar[d] \ar[r] & Z \times_\mathcal{Y} V \ar[d] \\ Z \times_\mathcal{Y} \mathcal{X} \ar[r] & Z }\] Note that the top horizontal arrow is a base change of \(h\) and hence has property \(\mathcal{P}\). The left vertical arrow is surjective, the induced morphism \[Z \times_\mathcal{Y} U \longrightarrow (Z \times_\mathcal{Y} \mathcal{X}) \times_Z (Z \times_\mathcal{Y} V)\] is étale, and the right vertical arrow is smooth. Thus Descent on Spaces, Lemma 0CG1 kicks in and shows that \(Z \times_\mathcal{Y} \mathcal{X} \to Z\) has property \(\mathcal{P}\).

Definition

Let \(\mathcal{P}\) be a property of morphisms of algebraic spaces which is étale-smooth local on the source-and-target. We say a DM morphism \(f : \mathcal{X} \to \mathcal{Y}\) of algebraic stacks has property \(\mathcal{P}\) if the equivalent conditions of Lemma 06FM hold.

Remark

Let \(\mathcal{P}\) be a property of morphisms of algebraic spaces which is étale-smooth local on the source-and-target and stable under composition. Then the property of DM morphisms of algebraic stacks defined in Definition 0CIG is stable under composition. Namely, let \(f : \mathcal{X} \to \mathcal{Y}\) and \(g : \mathcal{Y} \to \mathcal{Z}\) be DM morphisms of algebraic stacks having property \(\mathcal{P}\). By Lemma 050K the composition \(g \circ f\) is DM. Choose an algebraic space \(W\) and a surjective smooth morphism \(W \to \mathcal{Z}\). Choose an algebraic space \(V\) and a surjective étale morphism \(V \to \mathcal{Y} \times_\mathcal{Z} W\) (Lemma 0CIA). Choose an algebraic space \(U\) and a surjective étale morphism \(U \to \mathcal{X} \times_\mathcal{Y} V\). Then the morphisms \(V \to W\) and \(U \to V\) have property \(\mathcal{P}\) by definition. Whence \(U \to W\) has property \(\mathcal{P}\) as we assumed that \(\mathcal{P}\) is stable under composition. Thus, by definition again, we see that \(g \circ f : \mathcal{X} \to \mathcal{Z}\) has property \(\mathcal{P}\).

Remark

Let \(\mathcal{P}\) be a property of morphisms of algebraic spaces which is étale-smooth local on the source-and-target and stable under base change. Then the property of DM morphisms of algebraic stacks defined in Definition 0CIG is stable under arbitrary base change. Namely, let \(f : \mathcal{X} \to \mathcal{Y}\) be a DM morphism of algebraic stacks and \(g : \mathcal{Y}' \to \mathcal{Y}\) be a morphism of algebraic stacks and assume \(f\) has property \(\mathcal{P}\). Then the base change \(\mathcal{Y}' \times_\mathcal{Y} \mathcal{X} \to \mathcal{Y}'\) is a DM morphism by Lemma 050F. Choose an algebraic space \(V\) and a surjective smooth morphism \(V \to \mathcal{Y}\). Choose an algebraic space \(U\) and a surjective étale morphism \(U \to \mathcal{X} \times_\mathcal{Y} V\) (Lemma 0CIA). Finally, choose an algebraic space \(V'\) and a surjective and smooth morphism \(V' \to \mathcal{Y}' \times_\mathcal{Y} V\). Then the morphism \(U \to V\) has property \(\mathcal{P}\) by definition. Whence \(V' \times_V U \to V'\) has property \(\mathcal{P}\) as we assumed that \(\mathcal{P}\) is stable under base change. Considering the diagram \[\xymatrix{ V' \times_V U \ar[r] \ar[d] & \mathcal{Y}' \times_\mathcal{Y} \mathcal{X} \ar[r] \ar[d] & \mathcal{X} \ar[d] \\ V' \ar[r] & \mathcal{Y}' \ar[r] & \mathcal{Y} }\] we see that the left top horizontal arrow is surjective and \[V' \times_V U \to V' \times_\mathcal{Y} (\mathcal{Y}' \times_{\mathcal{Y}'} \mathcal{X}) = V' \times_V (\mathcal{X} \times_\mathcal{Y} V)\] is étale as a base change of \(U \to \mathcal{X} \times_\mathcal{Y} V\), whence by definition we see that the projection \(\mathcal{Y}' \times_\mathcal{Y} \mathcal{X} \to \mathcal{Y}'\) has property \(\mathcal{P}\).

Remark

Let \(\mathcal{P}, \mathcal{P}'\) be properties of morphisms of algebraic spaces which are étale-smooth local on the source-and-target. Suppose that we have \(\mathcal{P} \Rightarrow \mathcal{P}'\) for morphisms of algebraic spaces. Then we also have \(\mathcal{P} \Rightarrow \mathcal{P}'\) for the properties of morphisms of algebraic stacks defined in Definition 0CIG using \(\mathcal{P}\) and \(\mathcal{P}'\). This is clear from the definition.

Étale morphisms

An étale morphism of algebraic stacks should not be defined as a smooth morphism of relative dimension \(0\). Namely, the morphism \[[\mathbf{A}^1_k/\mathbf{G}_{m, k}] \longrightarrow \Spec(k)\] is smooth of relative dimension \(0\) for any choice of action of the group scheme \(\mathbf{G}_{m, k}\) on \(\mathbf{A}^1_k\). This does not correspond to our usual idea that étale morphisms should identify tangent spaces. The example above isn’t quasi-finite, but the morphism \[\mathcal{X} = [\Spec(k)/\mu_{p, k}] \longrightarrow \Spec(k)\] is smooth and quasi-finite (Section 06PT). However, if the characteristic of \(k\) is \(p > 0\), then we see that the representable morphism \(\Spec(k) \to \mathcal{X}\) isn’t étale as the base change \(\mu_{p, k} = \Spec(k) \times_\mathcal{X} \Spec(k) \to \Spec(k)\) is a morphism from a nonreduced scheme to the spectrum of a field. Thus if we define an étale morphism as smooth and locally quasi-finite, then the analogue of Morphisms of Spaces, Lemma 05W3 would fail.

Instead, our approach will be to start with the requirements that “étaleness” should be a property preserved under base change and that if \(\mathcal{X} \to X\) is an étale morphism from an algebraic stack to a scheme, then \(\mathcal{X}\) should be Deligne-Mumford. In other words, we will require étale morphisms to be DM and we will use the material in Section 0CIE to define étale morphisms of algebraic stacks.

In Lemma 0CJ1 we will characterize étale morphisms of algebraic stacks as morphisms which are (a) locally of finite presentation, (b) flat, and (c) have étale diagonal.

The property “étale” of morphisms of algebraic spaces is étale-smooth local on the source-and-target, see Descent on Spaces, Remark 0CG3. It is also stable under base change and fpqc local on the target, see Morphisms of Spaces, Lemma 0466 and Descent on Spaces, Lemma 042B. Hence, by Lemma 0CIF above, we may define what it means for a morphism of algebraic spaces to be étale as follows and it agrees with the already existing notion defined in Properties of Stacks, Section 04XB when the morphism is representable by algebraic spaces because such a morphism is automatically DM by Lemma 050E.

Definition

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. We say \(f\) is étale if \(f\) is DM and the equivalent conditions of Lemma 0CIF hold with \(\mathcal{P} = \etale\).

We will use without further mention that this agrees with the already existing notion of étale morphisms in case \(f\) is representable by algebraic spaces or if \(\mathcal{X}\) and \(\mathcal{Y}\) are representable by algebraic spaces.

Lemma

The composition of étale morphisms is étale.

Proof

Combine Remark 0CIH with Morphisms of Spaces, Lemma 0465.

Lemma

A base change of an étale morphism is étale.

Proof

Combine Remark 0CII with Morphisms of Spaces, Lemma 0466.

Lemma

An open immersion is étale.

Proof

Let \(j : \mathcal{U} \to \mathcal{X}\) be an open immersion of algebraic stacks. Since \(j\) is representable, it is DM by Lemma 050E. On the other hand, if \(X \to \mathcal{X}\) is a smooth and surjective morphism where \(X\) is a scheme, then \(U = \mathcal{U} \times_\mathcal{X} X\) is an open subscheme of \(X\). Hence \(U \to X\) is étale (Morphisms, Lemma 02GP) and we conclude that \(j\) is étale from the definition.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. The following are equivalent

1. \(f\) is étale,

2. \(f\) is DM and for any morphism \(V \to \mathcal{Y}\) where \(V\) is an algebraic space and any étale morphism \(U \to V \times_\mathcal{Y} \mathcal{X}\) where \(U\) is an algebraic space, the morphism \(U \to V\) is étale,

3. there exists some surjective, locally of finite presentation, and flat morphism \(W \to \mathcal{Y}\) where \(W\) is an algebraic space and some surjective étale morphism \(T \to W \times_\mathcal{Y} \mathcal{X}\) where \(T\) is an algebraic space such that the morphism \(T \to W\) is étale.

Proof

Assume (1). Then \(f\) is DM and since being étale is preserved by base change, we see that (2) holds.

Assume (2). Choose a scheme \(V\) and a surjective étale morphism \(V \to \mathcal{Y}\). Choose a scheme \(U\) and a surjective étale morphism \(U \to V \times_\mathcal{Y} \mathcal{X}\) (Lemma 0CIA). Thus we see that (3) holds.

Assume \(W \to \mathcal{Y}\) and \(T \to W \times_\mathcal{Y} \mathcal{X}\) are as in (3). We first check \(f\) is DM. Namely, it suffices to check \(W \times_\mathcal{Y} \mathcal{X} \to W\) is DM, see Lemma 06TZ. By Lemma 050M it suffices to check \(W \times_\mathcal{Y} \mathcal{X}\) is DM. This follows from the existence of \(T \to W \times_\mathcal{Y} \mathcal{X}\) by (the easy direction of) Theorem 06N3.

Assume \(f\) is DM and \(W \to \mathcal{Y}\) and \(T \to W \times_\mathcal{Y} \mathcal{X}\) are as in (3). Let \(V\) be an algebraic space, let \(V \to \mathcal{Y}\) be surjective smooth, let \(U\) be an algebraic space, and let \(U \to V \times_\mathcal{Y} \mathcal{X}\) is surjective and étale (Lemma 0CIA). We have to check that \(U \to V\) is étale. It suffices to prove \(U \times_\mathcal{Y} W \to V \times_\mathcal{Y} W\) is étale by Descent on Spaces, Lemma 042B. We may replace \(\mathcal{X}, \mathcal{Y}, W, T, U, V\) by \(\mathcal{X} \times_\mathcal{Y} W, W, W, T, U \times_\mathcal{Y} W, V \times_\mathcal{Y} W\) (small detail omitted). Thus we may assume that \(Y = \mathcal{Y}\) is an algebraic space, there exists an algebraic space \(T\) and a surjective étale morphism \(T \to \mathcal{X}\) such that \(T \to Y\) is étale, and \(U\) and \(V\) are as before. In this case we know that \[U \to V\text{ is \'etale} \Leftrightarrow \mathcal{X} \to Y\text{ is \'etale} \Leftrightarrow T \to Y\text{ is \'etale}\] by the equivalence of properties (1) and (2) of Lemma 0CIF and Definition 0CIL. This finishes the proof.

Lemma

Let \(\mathcal{X}, \mathcal{Y}\) be algebraic stacks étale over an algebraic stack \(\mathcal{Z}\). Any morphism \(\mathcal{X} \to \mathcal{Y}\) over \(\mathcal{Z}\) is étale.

Proof

The morphism \(\mathcal{X} \to \mathcal{Y}\) is DM by Lemma 050M. Let \(W \to \mathcal{Z}\) be a surjective smooth morphism whose source is an algebraic space. Let \(V \to \mathcal{Y} \times_\mathcal{Z} W\) be a surjective étale morphism whose source is an algebraic space (Lemma 0CIA). Let \(U \to \mathcal{X} \times_\mathcal{Y} V\) be a surjective étale morphism whose source is an algebraic space (Lemma 0CIA). Then \[U \longrightarrow \mathcal{X} \times_\mathcal{Z} W\] is surjective étale as the composition of \(U \to \mathcal{X} \times_\mathcal{Y} V\) and the base change of \(V \to \mathcal{Y} \times_\mathcal{Z} W\) by \(\mathcal{X} \times_\mathcal{Z} W \to \mathcal{Y} \times_\mathcal{Z} W\). Hence it suffices to show that \(U \to W\) is étale. Since \(U \to W\) and \(V \to W\) are étale because \(\mathcal{X} \to \mathcal{Z}\) and \(\mathcal{Y} \to \mathcal{Z}\) are étale, this follows from the version of the lemma for algebraic spaces, namely Morphisms of Spaces, Lemma 05W3.

Unramified morphisms

For a justification of our choice of definition of unramified morphisms we refer the reader to the discussion in the section on étale morphisms Section 0CIK.

In Lemma 0CJ0 we will characterize unramified morphisms of algebraic stacks as morphisms which are locally of finite type and have étale diagonal.

The property “unramified” of morphisms of algebraic spaces is étale-smooth local on the source-and-target, see Descent on Spaces, Remark 0CG3. It is also stable under base change and fpqc local on the target, see Morphisms of Spaces, Lemma 03ZJ and Descent on Spaces, Lemma 042A. Hence, by Lemma 0CIF above, we may define what it means for a morphism of algebraic spaces to be unramified as follows and it agrees with the already existing notion defined in Properties of Stacks, Section 04XB when the morphism is representable by algebraic spaces because such a morphism is automatically DM by Lemma 050E.

Definition

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. We say \(f\) is unramified if \(f\) is DM and the equivalent conditions of Lemma 0CIF hold with \(\mathcal{P} =\)“unramified”.

We will use without further mention that this agrees with the already existing notion of unramified morphisms in case \(f\) is representable by algebraic spaces or if \(\mathcal{X}\) and \(\mathcal{Y}\) are representable by algebraic spaces.

Lemma

The composition of unramified morphisms is unramified.

Proof

Combine Remark 0CIH with Morphisms of Spaces, Lemma 03ZI.

Lemma

A base change of an unramified morphism is unramified.

Proof

Combine Remark 0CII with Morphisms of Spaces, Lemma 03ZJ.

Lemma

An étale morphism is unramified.

Proof

Follows from Remark 0CIJ and Morphisms of Spaces, Lemma 06CR.

Lemma

An immersion is unramified.

Proof

Let \(j : \mathcal{Z} \to \mathcal{X}\) be an immersion of algebraic stacks. Since \(j\) is representable, it is DM by Lemma 050E. On the other hand, if \(X \to \mathcal{X}\) is a smooth and surjective morphism where \(X\) is a scheme, then \(Z = \mathcal{Z} \times_\mathcal{X} X\) is a locally closed subscheme of \(X\). Hence \(Z \to X\) is unramified (Morphisms, Lemmas 02GB and 02GC) and we conclude that \(j\) is unramified from the definition.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. The following are equivalent

1. \(f\) is unramified,

2. \(f\) is DM and for any morphism \(V \to \mathcal{Y}\) where \(V\) is an algebraic space and any étale morphism \(U \to V \times_\mathcal{Y} \mathcal{X}\) where \(U\) is an algebraic space, the morphism \(U \to V\) is unramified,

3. there exists some surjective, locally of finite presentation, and flat morphism \(W \to \mathcal{Y}\) where \(W\) is an algebraic space and some surjective étale morphism \(T \to W \times_\mathcal{Y} \mathcal{X}\) where \(T\) is an algebraic space such that the morphism \(T \to W\) is unramified.

Proof

Assume (1). Then \(f\) is DM and since being unramified is preserved by base change, we see that (2) holds.

Assume (2). Choose a scheme \(V\) and a surjective étale morphism \(V \to \mathcal{Y}\). Choose a scheme \(U\) and a surjective étale morphism \(U \to V \times_\mathcal{Y} \mathcal{X}\) (Lemma 0CIA). Thus we see that (3) holds.

Assume \(W \to \mathcal{Y}\) and \(T \to W \times_\mathcal{Y} \mathcal{X}\) are as in (3). We first check \(f\) is DM. Namely, it suffices to check \(W \times_\mathcal{Y} \mathcal{X} \to W\) is DM, see Lemma 06TZ. By Lemma 050M it suffices to check \(W \times_\mathcal{Y} \mathcal{X}\) is DM. This follows from the existence of \(T \to W \times_\mathcal{Y} \mathcal{X}\) by (the easy direction of) Theorem 06N3.

Assume \(f\) is DM and \(W \to \mathcal{Y}\) and \(T \to W \times_\mathcal{Y} \mathcal{X}\) are as in (3). Let \(V\) be an algebraic space, let \(V \to \mathcal{Y}\) be surjective smooth, let \(U\) be an algebraic space, and let \(U \to V \times_\mathcal{Y} \mathcal{X}\) is surjective and étale (Lemma 0CIA). We have to check that \(U \to V\) is unramified. It suffices to prove \(U \times_\mathcal{Y} W \to V \times_\mathcal{Y} W\) is unramified by Descent on Spaces, Lemma 042A. We may replace \(\mathcal{X}, \mathcal{Y}, W, T, U, V\) by \(\mathcal{X} \times_\mathcal{Y} W, W, W, T, U \times_\mathcal{Y} W, V \times_\mathcal{Y} W\) (small detail omitted). Thus we may assume that \(Y = \mathcal{Y}\) is an algebraic space, there exists an algebraic space \(T\) and a surjective étale morphism \(T \to \mathcal{X}\) such that \(T \to Y\) is unramified, and \(U\) and \(V\) are as before. In this case we know that \[U \to V\text{ is unramified} \Leftrightarrow \mathcal{X} \to Y\text{ is unramified} \Leftrightarrow T \to Y\text{ is unramified}\] by the equivalence of properties (1) and (2) of Lemma 0CIF and Definition 0CIT. This finishes the proof.

Lemma

An unramified morphism of algebraic stacks is locally quasi-finite.

Proof

This follows from Lemma 0CIY (characterizing unramified morphisms), Lemma 06UF (characterizing locally quasi-finite morphisms), and Morphisms of Spaces, Lemma 05W0 (the corresponding result for algebraic spaces).

Lemma

Let \(\mathcal{X} \to \mathcal{Y} \to \mathcal{Z}\) be morphisms of algebraic stacks. If \(\mathcal{X} \to \mathcal{Z}\) is unramified and \(\mathcal{Y} \to \mathcal{Z}\) is DM, then \(\mathcal{X} \to \mathcal{Y}\) is unramified.

Proof

Assume \(\mathcal{X} \to \mathcal{Z}\) is unramified. By Lemma 050M the morphism \(\mathcal{X} \to \mathcal{Y}\) is DM. Choose a commutative diagram \[\xymatrix{ U \ar[d] \ar[r] & V \ar[d] \ar[r] & W \ar[d] \\ \mathcal{X} \ar[r] & \mathcal{Y} \ar[r] & \mathcal{Z} }\] with \(U, V, W\) algebraic spaces, with \(W \to \mathcal{Z}\) surjective smooth, \(V \to \mathcal{Y} \times_\mathcal{Z} W\) surjective étale, and \(U \to \mathcal{X} \times_\mathcal{Y} V\) surjective étale (see Lemma 0CIA). Then also \(U \to \mathcal{X} \times_\mathcal{Z} W\) is surjective and étale. Hence we know that \(U \to W\) is unramified and we have to show that \(U \to V\) is unramified. This follows from Morphisms of Spaces, Lemma 06G6.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. The following are equivalent

1. \(f\) is unramified, and

2. \(f\) is locally of finite type and its diagonal is étale.

Proof

Assume \(f\) is unramified. Then \(f\) is DM hence we can choose algebraic spaces \(U\), \(V\), a smooth surjective morphism \(V \to \mathcal{Y}\) and a surjective étale morphism \(U \to \mathcal{X} \times_\mathcal{Y} V\) (Lemma 0CIA). Since \(f\) is unramified the induced morphism \(U \to V\) is unramified. Thus \(U \to V\) is locally of finite type (Morphisms of Spaces, Lemma 05VZ) and we conclude that \(f\) is locally of finite type. The diagonal \(\Delta : \mathcal{X} \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}\) is a morphism of algebraic stacks over \(\mathcal{Y}\). The base change of \(\Delta\) by the surjective smooth morphism \(V \to \mathcal{Y}\) is the diagonal of the base change of \(f\), i.e., of \(\mathcal{X}_V = \mathcal{X} \times_\mathcal{Y} V \to V\). In other words, the diagram \[\xymatrix{ \mathcal{X}_V \ar[r] \ar[d] & \mathcal{X}_V \times_V \mathcal{X}_V \ar[d] \\ \mathcal{X} \ar[r] & \mathcal{X} \times_\mathcal{Y} \mathcal{X} }\] is cartesian. Since the right vertical arrow is surjective and smooth it suffices to show that the top horizontal arrow is étale by Properties of Stacks, Lemma 06TY. Consider the commutative diagram \[\xymatrix{ U \ar[d] \ar[r] & U \times_V U \ar[d] \\ \mathcal{X}_V \ar[r] & \mathcal{X}_V \times_V \mathcal{X}_V }\] All arrows are representable by algebraic spaces, the vertical arrows are étale, the left one is surjective, and the top horizontal arrow is an open immersion by Morphisms of Spaces, Lemma 05W1. This implies what we want: first we see that \(U \to \mathcal{X}_V \times_V \mathcal{X}_V\) is étale as a composition of étale morphisms, and then we can use Properties of Stacks, Lemma 06M2 to see that \(\mathcal{X}_V \to \mathcal{X}_V \times_V \mathcal{X}_V\) is étale because being étale (for morphisms of algebraic spaces) is local on the source in the étale topology (Descent on Spaces, Lemma 06F4).

Assume \(f\) is locally of finite type and that its diagonal is étale. Then \(f\) is DM by definition (as étale morphisms of algebraic spaces are unramified). As above this means we can choose algebraic spaces \(U\), \(V\), a smooth surjective morphism \(V \to \mathcal{Y}\) and a surjective étale morphism \(U \to \mathcal{X} \times_\mathcal{Y} V\) (Lemma 0CIA). To finish the proof we have to show that \(U \to V\) is unramified. We already know that \(U \to V\) is locally of finite type. Arguing as above we find a commutative diagram \[\xymatrix{ U \ar[d] \ar[r] & U \times_V U \ar[d] \\ \mathcal{X}_V \ar[r] & \mathcal{X}_V \times_V \mathcal{X}_V }\] where all arrows are representable by algebraic spaces, the vertical arrows are étale, and the lower horizontal one is étale as a base change of \(\Delta\). It follows that \(U \to U \times_V U\) is étale for example by Lemma 0CIR⁷. Thus \(U \to U \times_V U\) is an étale monomorphism hence an open immersion (Morphisms of Spaces, Lemma 05W5). Then \(U \to V\) is unramified by Morphisms of Spaces, Lemma 05W1.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. The following are equivalent

1. \(f\) is étale, and

2. \(f\) is locally of finite presentation, flat, and unramified,

3. \(f\) is locally of finite presentation, flat, and its diagonal is étale.

Proof

The equivalence of (2) and (3) follows immediately from Lemma 0CJ0. Thus in each case the morphism \(f\) is DM. Then we can choose Then we can choose algebraic spaces \(U\), \(V\), a smooth surjective morphism \(V \to \mathcal{Y}\) and a surjective étale morphism \(U \to \mathcal{X} \times_\mathcal{Y} V\) (Lemma 0CIA). To finish the proof we have to show that \(U \to V\) is étale if and only if it is locally of finite presentation, flat, and unramified. This follows from Morphisms of Spaces, Lemma 06LU (and the more trivial Morphisms of Spaces, Lemmas 06CR, 0468, and 0467).

Proper morphisms

The notion of a proper morphism plays an important role in algebraic geometry. Here is the definition of a proper morphism of algebraic stacks.

Definition

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. We say \(f\) is proper if \(f\) is separated, finite type, and universally closed.

This does not conflict with the already existing notion of a proper morphism of algebraic spaces: a morphism of algebraic spaces is proper if and only if it is separated, finite type, and universally closed (Morphisms of Spaces, Definition 03ZM) and we’ve already checked the compatibility of these notions in Lemma 04YS, Section 06FR, and Lemmas 0512. Similarly, if \(f : \mathcal{X} \to \mathcal{Y}\) is a morphism of algebraic stacks which is representable by algebraic spaces then we have defined what it means for \(f\) to be proper in Properties of Stacks, Section 04XB. However, the discussion in that section shows that this is equivalent to requiring \(f\) to be separated, finite type, and universally closed and the same references as above give the compatibility.

Lemma

A base change of a proper morphism is proper.

Proof

See Lemmas 050F, 06FU, and 0514.

Lemma

A composition of proper morphisms is proper.

Proof

See Lemmas 050K, 06FT, and 0515.

Lemma

A closed immersion of algebraic stacks is a proper morphism of algebraic stacks.

Proof

A closed immersion is by definition representable (Properties of Stacks, Definition 04YL). Hence this follows from the discussion in Properties of Stacks, Section 04XB and the corresponding result for morphisms of algebraic spaces, see Morphisms of Spaces, Lemma 04XZ.

Lemma

Consider a commutative diagram \[\xymatrix{ \mathcal{X} \ar[rr] \ar[rd] & & \mathcal{Y} \ar[ld] \\ & \mathcal{Z} & }\] of algebraic stacks.

1. If \(\mathcal{X} \to \mathcal{Z}\) is universally closed and \(\mathcal{Y} \to \mathcal{Z}\) is separated, then the morphism \(\mathcal{X} \to \mathcal{Y}\) is universally closed. In particular, the image of \(|\mathcal{X}|\) in \(|\mathcal{Y}|\) is closed.

2. If \(\mathcal{X} \to \mathcal{Z}\) is proper and \(\mathcal{Y} \to \mathcal{Z}\) is separated, then the morphism \(\mathcal{X} \to \mathcal{Y}\) is proper.

Proof

Assume \(\mathcal{X} \to \mathcal{Z}\) is universally closed and \(\mathcal{Y} \to \mathcal{Z}\) is separated. We factor the morphism as \(\mathcal{X} \to \mathcal{X} \times_\mathcal{Z} \mathcal{Y} \to \mathcal{Y}\). The first morphism is proper (Lemma 050I) hence universally closed. The projection \(\mathcal{X} \times_\mathcal{Z} \mathcal{Y} \to \mathcal{Y}\) is the base change of a universally closed morphism and hence universally closed, see Lemma 0514. Thus \(\mathcal{X} \to \mathcal{Y}\) is universally closed as the composition of universally closed morphisms, see Lemma 0515. This proves (1). To deduce (2) combine (1) with Lemmas 050M, 050Y, and 06U9.

Lemma

Let \(\mathcal{Z}\) be an algebraic stack. Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks over \(\mathcal{Z}\). If \(\mathcal{X}\) is universally closed over \(\mathcal{Z}\) and \(f\) is surjective then \(\mathcal{Y}\) is universally closed over \(\mathcal{Z}\). In particular, if also \(\mathcal{Y}\) is separated and of finite type over \(\mathcal{Z}\), then \(\mathcal{Y}\) is proper over \(\mathcal{Z}\).

Proof

Assume \(\mathcal{X}\) is universally closed and \(f\) surjective. Denote \(p : \mathcal{X} \to \mathcal{Z}\), \(q : \mathcal{Y} \to \mathcal{Z}\) the structure morphisms. Let \(\mathcal{Z}' \to \mathcal{Z}\) be a morphism of algebraic stacks. The base change \(f' : \mathcal{X}' \to \mathcal{Y}'\) of \(f\) by \(\mathcal{Z}' \to \mathcal{Z}\) is surjective (Properties of Stacks, Lemma 04ZU) and the base change \(p' : \mathcal{X}' \to \mathcal{Z}'\) of \(p\) is closed. If \(T \subset |\mathcal{Y}'|\) is closed, then \((f')^{-1}(T) \subset |\mathcal{X}'|\) is closed, hence \(p'((f')^{-1}(T)) = q'(T)\) is closed. So \(q'\) is closed.

Scheme theoretic image

Here is the definition.

Definition

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. The scheme theoretic image of \(f\) is the smallest closed substack \(\mathcal{Z} \subset \mathcal{Y}\) through which \(f\) factors⁸.

We often denote \(f : \mathcal{X} \to \mathcal{Z}\) the factorization of \(f\). If the morphism \(f\) is not quasi-compact, then (in general) the construction of the scheme theoretic image does not commute with restriction to open substacks of \(\mathcal{Y}\). However, if \(f\) is quasi-compact then the scheme theoretic image commutes with flat base change (Lemma 0CMK).

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. Let \(g : \mathcal{W} \to \mathcal{X}\) be a morphism of algebraic stacks which is surjective, flat, and locally of finite presentation. Then the scheme theoretic image of \(f\) exists if and only if the scheme theoretic image of \(f \circ g\) exists and if so then these scheme theoretic images are the same.

Proof

Assume \(\mathcal{Z} \subset \mathcal{Y}\) is a closed substack and \(f \circ g\) factors through \(\mathcal{Z}\). To prove the lemma it suffices to show that \(f\) factors through \(\mathcal{Z}\). Consider a scheme \(T\) and a morphism \(T \to \mathcal{X}\) given by an object \(x\) of the fibre category of \(\mathcal{X}\) over \(T\). We will show that \(f(x)\) is in fact in the fibre category of \(\mathcal{Z}\) over \(T\). Namely, the projection \(T \times_\mathcal{X} \mathcal{W} \to T\) is a surjective, flat, locally finitely presented morphism. Hence there is an fppf covering \(\{T_i \to T\}\) such that \(T_i \to T\) factors through \(T \times_\mathcal{X} \mathcal{W} \to T\) for all \(i\). Then \(T_i \to \mathcal{X}\) factors through \(\mathcal{W}\) and hence \(T_i \to \mathcal{Y}\) factors through \(\mathcal{Z}\). Thus \(f(x)|_{T_i}\) is an object of \(\mathcal{Z}\). Since \(\mathcal{Z}\) is a strictly full substack, we conclude that \(f(x)\) is an object of \(\mathcal{Z}\) as desired.

Lemma

Let \(f : \mathcal{Y} \to \mathcal{X}\) be a morphism of algebraic stacks. Then the scheme theoretic image of \(f\) exists.

Proof

Choose a scheme \(V\) and a surjective smooth morphism \(V \to \mathcal{Y}\). By Lemma 0CMJ we may replace \(\mathcal{Y}\) by \(V\). Thus it suffices to show that if \(X \to \mathcal{X}\) is a morphism from a scheme to an algebraic stack, then the scheme theoretic image exists. Choose a scheme \(U\) and a surjective smooth morphism \(U \to \mathcal{X}\). Set \(R = U \times_\mathcal{X} U\). We have \(\mathcal{X} = [U/R]\) by Algebraic Stacks, Lemma 04T5. By Properties of Stacks, Lemma 0507 the closed substacks \(\mathcal{Z}\) of \(\mathcal{X}\) are in \(1\)-to-\(1\) correspondence with \(R\)-invariant closed subschemes \(Z \subset U\). Let \(Z_1 \subset U\) be the scheme theoretic image of \(X \times_\mathcal{X} U \to U\). Observe that \(X \to \mathcal{X}\) factors through \(\mathcal{Z}\) if and only if \(X \times_\mathcal{X} U \to U\) factors through the corresponding \(R\)-invariant closed subscheme \(Z\) (details omitted; hint: this follows because \(X \times_\mathcal{X} U \to X\) is surjective and smooth). Thus we have to show that there exists a smallest \(R\)-invariant closed subscheme \(Z \subset U\) containing \(Z_1\).

Let \(\mathcal{I}_1 \subset \mathcal{O}_U\) be the quasi-coherent ideal sheaf corresponding to the closed subscheme \(Z_1 \subset U\). Let \(Z_\alpha\), \(\alpha \in A\) be the set of all \(R\)-invariant closed subschemes of \(U\) containing \(Z_1\). For \(\alpha \in A\), let \(\mathcal{I}_\alpha \subset \mathcal{O}_U\) be the quasi-coherent ideal sheaf corresponding to the closed subscheme \(Z_\alpha \subset U\). The containment \(Z_1 \subset Z_\alpha\) means \(\mathcal{I}_\alpha \subset \mathcal{I}_1\). The \(R\)-invariance of \(Z_\alpha\) means that \[s^{-1}\mathcal{I}_\alpha \cdot \mathcal{O}_R = t^{-1}\mathcal{I}_\alpha \cdot \mathcal{O}_R\] as (quasi-coherent) ideal sheaves on (the algebraic space) \(R\). Consider the image \[\mathcal{I} = \Im\left( \bigoplus\nolimits_{\alpha \in A} \mathcal{I}_\alpha \to \mathcal{I}_1 \right) = \Im\left( \bigoplus\nolimits_{\alpha \in A} \mathcal{I}_\alpha \to \mathcal{O}_X \right)\] Since direct sums of quasi-coherent sheaves are quasi-coherent and since images of maps between quasi-coherent sheaves are quasi-coherent, we find that \(\mathcal{I}\) is quasi-coherent. Since pull back is exact and commutes with direct sums we find \[s^{-1}\mathcal{I} \cdot \mathcal{O}_R = t^{-1}\mathcal{I} \cdot \mathcal{O}_R\] Hence \(\mathcal{I}\) defines an \(R\)-invariant closed subscheme \(Z \subset U\) which is contained in every \(Z_\alpha\) and contains \(Z_1\) as desired.

Lemma

Let \[\xymatrix{ \mathcal{X}_1 \ar[d] \ar[r]_{f_1} & \mathcal{Y}_1 \ar[d] \\ \mathcal{X}_2 \ar[r]^{f_2} & \mathcal{Y}_2 }\] be a commutative diagram of algebraic stacks. Let \(\mathcal{Z}_i \subset \mathcal{Y}_i\), \(i = 1, 2\) be the scheme theoretic image of \(f_i\). Then the morphism \(\mathcal{Y}_1 \to \mathcal{Y}_2\) induces a morphism \(\mathcal{Z}_1 \to \mathcal{Z}_2\) and a commutative diagram \[\xymatrix{ \mathcal{X}_1 \ar[r] \ar[d] & \mathcal{Z}_1 \ar[d] \ar[r] & \mathcal{Y}_1 \ar[d] \\ \mathcal{X}_2 \ar[r] & \mathcal{Z}_2 \ar[r] & \mathcal{Y}_2 }\]

Proof

The scheme theoretic inverse image of \(\mathcal{Z}_2\) in \(\mathcal{Y}_1\) is a closed substack of \(\mathcal{Y}_1\) through which \(f_1\) factors. Hence \(\mathcal{Z}_1\) is contained in this. This proves the lemma.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a quasi-compact morphism of algebraic stacks. Then formation of the scheme theoretic image commutes with flat base change.

Proof

Let \(\mathcal{Y}' \to \mathcal{Y}\) be a flat morphism of algebraic stacks. Choose a scheme \(V\) and a surjective smooth morphism \(V \to \mathcal{Y}\). Choose a scheme \(V'\) and a surjective smooth morphism \(V' \to \mathcal{Y}' \times_\mathcal{Y} V\). We may and do assume that \(V = \coprod_{i \in I} V_i\) is a disjoint union of affine schemes and that \(V' = \coprod_{i \in I} \coprod_{j \in J_i} V_{i, j}\) is a disjoint union of affine schemes with each \(V_{i, j}\) mapping into \(V_i\). Let

1. \(\mathcal{Z} \subset \mathcal{Y}\) be the scheme theoretic image of \(f\),

2. \(\mathcal{Z}' \subset \mathcal{Y}'\) be the scheme theoretic image of the base change of \(f\) by \(\mathcal{Y}' \to \mathcal{Y}\),

3. \(Z \subset V\) be the scheme theoretic image of the base change of \(f\) by \(V \to \mathcal{Y}\),

4. \(Z' \subset V'\) be the scheme theoretic image of the base change of \(f\) by \(V' \to \mathcal{Y}\).

If we can show that (a) \(Z = V \times_\mathcal{Y} \mathcal{Z}\), (b) \(Z' = V' \times_{\mathcal{Y}'} \mathcal{Z}'\), and (c) \(Z' = V' \times_V Z\) then the lemma follows: the inclusion \(\mathcal{Z}' \to \mathcal{Z} \times_\mathcal{Y} \mathcal{Y}'\) (Lemma 0CPV) has to be an isomorphism because after base change by the surjective smooth morphism \(V' \to \mathcal{Y}'\) it is.

Proof of (a). Set \(R = V \times_\mathcal{Y} V\). By Properties of Stacks, Lemma 0507 the rule \(\mathcal{Z} \mapsto \mathcal{Z} \times_\mathcal{Y} V\) defines a \(1\)-to-\(1\) correspondence between closed substacks of \(\mathcal{Y}\) and \(R\)-invariant closed subspaces of \(V\). Moreover, \(f : \mathcal{X} \to \mathcal{Y}\) factors through \(\mathcal{Z}\) if and only if the base change \(g : \mathcal{X} \times_\mathcal{Y} V \to V\) factors through \(\mathcal{Z} \times_\mathcal{Y} V\). We claim: the scheme theoretic image \(Z \subset V\) of \(g\) is \(R\)-invariant. The claim implies (a) by what we just said.

For each \(i\) the morphism \(\mathcal{X} \times_\mathcal{Y} V_i \to V_i\) is quasi-compact and hence \(\mathcal{X} \times_\mathcal{Y} V_i\) is quasi-compact. Thus we can choose an affine scheme \(W_i\) and a surjective smooth morphism \(W_i \to \mathcal{X} \times_\mathcal{Y} V_i\). Observe that \(W = \coprod W_i\) is a scheme endowed with a smooth and surjective morphism \(W \to \mathcal{X} \times_\mathcal{Y} V\) such that the composition \(W \to V\) with \(g\) is quasi-compact. Let \(Z \to V\) be the scheme theoretic image of \(W \to V\), see Morphisms, Section 01R5 and Morphisms of Spaces, Section 082W. It follows from Lemma 0CMJ that \(Z \subset V\) is the scheme theoretic image of \(g\). To show that \(Z\) is \(R\)-invariant we claim that both \[\text{pr}_0^{-1}(Z), \text{pr}_1^{-1}(Z) \subset R = V \times_\mathcal{Y} V\] are the scheme theoretic image of \(\mathcal{X} \times_\mathcal{Y} R \to R\). Namely, we first use Morphisms of Spaces, Lemma 089E to see that \(\text{pr}_0^{-1}(Z)\) is the scheme theoretic image of the composition \[W \times_{V, \text{pr}_0} R = W \times_\mathcal{Y} V \to \mathcal{X} \times_\mathcal{Y} R \to R\] Since the first arrow here is surjective and smooth we see that \(\text{pr}_0^{-1}(Z)\) is the scheme theoretic image of \(\mathcal{X} \times_\mathcal{Y} R \to R\). The same argument applies that \(\text{pr}_1^{-1}(Z)\). Hence \(Z\) is \(R\)-invariant.

Statement (b) is proved in exactly the same way as one proves (a).

Proof of (c). Let \(Z_i \subset V_i\) be the scheme theoretic image of \(\mathcal{X} \times_\mathcal{Y} V_i \to V_i\) and let \(Z_{i, j} \subset V_{i, j}\) be the scheme theoretic image of \(\mathcal{X} \times_\mathcal{Y} V_{i, j} \to V_{i, j}\). Clearly it suffices to show that the inverse image of \(Z_i\) in \(V_{i, j}\) is \(Z_{i, j}\). Above we’ve seen that \(Z_i\) is the scheme theoretic image of \(W_i \to V_i\) and by the same token \(Z_{i, j}\) is the scheme theoretic image of \(W_i \times_{V_i} V_{i, j} \to V_{i, j}\). Hence the equality follows from the case of schemes (Morphisms, Lemma 081I) and the fact that \(V_{i, j} \to V_i\) is flat.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a quasi-compact morphism of algebraic stacks. Let \(\mathcal{Z} \subset \mathcal{Y}\) be the scheme theoretic image of \(f\). Then \(|\mathcal{Z}|\) is the closure of the image of \(|f|\).

Proof

Let \(z \in |\mathcal{Z}|\) be a point. Choose an affine scheme \(V\), a point \(v \in V\), and a smooth morphism \(V \to \mathcal{Y}\) mapping \(v\) to \(z\). Then \(\mathcal{X} \times_\mathcal{Y} V\) is a quasi-compact algebraic stack. Hence we can find an affine scheme \(W\) and a surjective smooth morphism \(W \to \mathcal{X} \times_\mathcal{Y} V\). By Lemma 0CMK the scheme theoretic image of \(\mathcal{X} \times_\mathcal{Y} V \to V\) is \(Z = \mathcal{Z} \times_\mathcal{Y} V\). Hence the inverse image of \(|\mathcal{Z}|\) in \(|V|\) is \(|Z|\) by Properties of Stacks, Lemma 04XH. By Lemma 0CMJ \(Z\) is the scheme theoretic image of \(W \to V\). By Morphisms of Spaces, Lemma 082Z we see that the image of \(|W| \to |Z|\) is dense. Hence the image of \(|\mathcal{X} \times_\mathcal{Y} V| \to |Z|\) is dense. Observe that \(v \in Z\). Since \(|V| \to |\mathcal{Y}|\) is open, a topology argument tells us that \(z\) is in the closure of the image of \(|f|\) as desired.

Lemma

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks which is representable by algebraic spaces and separated. Let \(\mathcal{V} \subset \mathcal{Y}\) be an open substack such that \(\mathcal{V} \to \mathcal{Y}\) is quasi-compact. Let \(s : \mathcal{V} \to \mathcal{X}\) be a morphism such that \(f \circ s = \text{id}_\mathcal{V}\). Let \(\mathcal{Y}'\) be the scheme theoretic image of \(s\). Then \(\mathcal{Y}' \to \mathcal{Y}\) is an isomorphism over \(\mathcal{V}\).

Proof

By Lemma 050Y the morphism \(s : \mathcal{V} \to \mathcal{Y}\) is quasi-compact. Hence the construction of the scheme theoretic image \(\mathcal{Y}'\) of \(s\) commutes with flat base change by Lemma 0CMK. Thus to prove the lemma we may assume \(\mathcal{Y}\) is representable by an algebraic space and we reduce to the case of algebraic spaces which is Morphisms of Spaces, Lemma 0CP3.

Valuative criteria

We need to be careful when stating the valuative criterion. Namely, in the formulation we need to speak about commutative diagrams but we are working in a \(2\)-category and we need to make sure the \(2\)-morphisms compose correctly as well!

Definition

Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. Consider a \(2\)-commutative solid diagram [0CLB]\[\begin{equation} \vcenter{ \xymatrix{ \Spec(K) \ar[r]_-x \ar[d]_j & \mathcal{X} \ar[d]^f \\ \Spec(A) \ar[r]^-y \ar@{..>}[ru] & \mathcal{Y} } } \end{equation}\] where \(A\) is a valuation ring with field of fractions \(K\). Let \[\gamma : y \circ j \longrightarrow f \circ x\] be a \(2\)-morphism witnessing the \(2\)-commutativity of the diagram. (Notation as in Categories, Sections 003D and 003G.) Given (0CLB) and \(\gamma\) a dotted arrow is a triple \((a, \alpha, \beta)\) consisting of a morphism \(a : \Spec(A) \to \mathcal{X}\) and \(2\)-arrows \(\alpha : a \circ j \to x\), \(\beta : y \to f \circ a\) such that \(\gamma = (\text{id}_f \star \alpha) \circ (\beta \star \text{id}_j)\), in other words such that \[\xymatrix{ & f \circ a \circ j \ar[rd]^{\text{id}_f \star \alpha} \\ y \circ j \ar[ru]^{\beta \star \text{id}_j} \ar[rr]^\gamma & & f \circ x }\] is commutative. A morphism of dotted arrows \((a, \alpha, \beta) \to (a', \alpha', \beta')\) is a \(2\)-arrow \(\theta : a \to a'\) such that \(\alpha = \alpha' \circ (\theta \star \text{id}_j)\) and \(\beta' = (\text{id}_f \star \theta) \circ \beta\).

The preceding definition is a special case of Categories, Definition 0H18. The category of dotted arrows depends on \(\gamma\) in general. If \(\mathcal{Y}\) is representable by an algebraic space (or if automorphism groups of objects over fields are trivial), then of course there is at most one \(\gamma\) and one does not need to check the commutativity of the triangle. More generally, we have Lemma 0CLD. The commutativity of the triangle is important in the proof of compatibility with base change, see proof of Lemma 0CLE.

Lemma

In the situation of Definition 0CLA the category of dotted arrows is a groupoid. If \(\Delta_f\) is separated, then it is a setoid.

Proof

Since \(2\)-arrows are invertible it is clear that the category of dotted arrows is a groupoid. Given a dotted arrow \((a, \alpha, \beta)\) an automorphism of \((a, \alpha, \beta)\) is a \(2\)-morphism \(\theta : a \to a\) satisfying two conditions. The first condition \(\beta = (\text{id}_f \star \theta) \circ \beta\) signifies that \(\theta\) defines a morphism \((a, \theta) : \Spec(A) \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}\). The second condition \(\alpha = \alpha \circ (\theta \star \text{id}_j)\) implies that the restriction of \((a, \theta)\) to \(\Spec(K)\) is the identity. Picture \[\xymatrix{ \mathcal{I}_{\mathcal{X}/\mathcal{Y}} \ar[d] & & \Spec(K) \ar[d]^j \ar[ll]_{(a \circ j, \text{id})} \\ \mathcal{X} & & \Spec(A) \ar[ll]_a \ar[llu]_{(a, \theta)} }\] In other words, if \(G \to \Spec(A)\) is the group algebraic space we get by pulling back the relative inertia by \(a\), then \(\theta\) defines a point \(\theta \in G(A)\) whose image in \(G(K)\) is trivial. Certainly, if the identity \(e : \Spec(A) \to G\) is a closed immersion, then this can happen only if \(\theta\) is the identity. Looking at Lemma 0CL0 we obtain the result we want.

Lemma

In Definition 0CLA assume \(\mathcal{I}_\mathcal{Y} \to \mathcal{Y}\) is proper (for example if \(\mathcal{Y}\) is separated or if \(\mathcal{Y}\) is separated over an algebraic space). Then the category of dotted arrows is independent (up to noncanonical equivalence) of the choice of \(\gamma\) and the existence of a dotted arrow (for some and hence equivalently all \(\gamma\)) is equivalent to the existence of a diagram \[\xymatrix{ \Spec(K) \ar[r]_-x \ar[d]_j & \mathcal{X} \ar[d]^f \\ \Spec(A) \ar[r]^-y \ar[ru]_a & \mathcal{Y} }\] with \(2\)-commutative triangles (without checking the \(2\)-morphisms compose correctly).

Proof

Let \(\gamma, \gamma' : y \circ j \longrightarrow f \circ x\) be two \(2\)-morphisms. Then \(\gamma^{-1} \circ \gamma'\) is an automorphism of \(y\) over \(\Spec(K)\). Hence if \(\mathit{Isom}_\mathcal{Y}(y, y) \to \Spec(A)\) is proper, then by the valuative criterion of properness (Morphisms of Spaces, Lemma 0A40) we can find \(\delta : y \to y\) whose restriction to \(\Spec(K)\) is \(\gamma^{-1} \circ \gamma'\). Then we can use \(\delta\) to define an equivalence between the category of dotted arrows for \(\gamma\) to the category of dotted arrows for \(\gamma'\) by sending \((a, \alpha, \beta)\) to \((a, \alpha, \beta \circ \delta)\). The final statement is clear.

Lemma

Assume given a \(2\)-commutative diagram \[\xymatrix{ \Spec(K) \ar[r]_-{x'} \ar[d]_j & \mathcal{X}' \ar[d]^p \ar[r]_q & \mathcal{X} \ar[d]^f \\ \Spec(A) \ar[r]^-{y'} & \mathcal{Y}' \ar[r]^g & \mathcal{Y} }\] with the right square \(2\)-cartesian. Choose a \(2\)-arrow \(\gamma' : y' \circ j \to p \circ x'\). Set \(x = q \circ x'\), \(y = g \circ y'\) and let \(\gamma : y \circ j \to f \circ x\) be the composition of \(\gamma'\) with the \(2\)-arrow implicit in the \(2\)-commutativity of the right square. Then the category of dotted arrows for the left square and \(\gamma'\) is equivalent to the category of dotted arrows for the outer rectangle and \(\gamma\).

Proof

(We do not know how to prove the analogue of this lemma if instead of the category of dotted arrows we look at the set of isomorphism classes of morphisms producing two \(2\)-commutative triangles as in Lemma 0CLD; in fact this analogue may very well be wrong.) First proof: this lemma is a special case of Categories, Lemma 0H1A. Second proof: we are allowed to replace \(\mathcal{X}'\) by the \(2\)-fibre product \(\mathcal{Y}' \times_\mathcal{Y} \mathcal{X}\) as described in Categories, Lemma 0040. Then the object \(x'\) becomes the triple \((y' \circ j, x, \gamma)\). Then we can go from a dotted arrow \((a, \alpha, \beta)\) for the outer rectangle to a dotted arrow \((a', \alpha', \beta')\) for the left square by taking \(a' = (y', a, \beta)\) and \(\alpha' = (\text{id}_{y' \circ j}, \alpha)\) and \(\beta' = \text{id}_{y'}\). Details omitted.

Lemma

Assume given a \(2\)-commutative diagram \[\xymatrix{ \Spec(K) \ar[r]_-x \ar[dd]_j & \mathcal{X} \ar[d]^f \\ & \mathcal{Y} \ar[d]^g \\ \Spec(A) \ar[r]^-z & \mathcal{Z} }\] Choose a \(2\)-arrow \(\gamma : z \circ j \to g \circ f \circ x\). Let \(\mathcal{C}\) be the category of dotted arrows for the outer rectangle and \(\gamma\). Let \(\mathcal{C}'\) be the category of dotted arrows for the square \[\xymatrix{ \Spec(K) \ar[r]_-{f \circ x} \ar[d]_j & \mathcal{Y} \ar[d]^g \\ \Spec(A) \ar[r]^-z & \mathcal{Z} }\] and \(\gamma\). Then \(\mathcal{C}\) is equivalent to a category \(\mathcal{C}''\) which has the following property: there is a functor \(\mathcal{C}'' \to \mathcal{C}'\) which turns \(\mathcal{C}''\) into a category fibred in groupoids over \(\mathcal{C}'\) and whose fibre categories are categories of dotted arrows for certain squares of the form \[\xymatrix{ \Spec(K) \ar[r]_-x \ar[d]_j & \mathcal{X} \ar[d]^f \\ \Spec(A) \ar[r]^-y & \mathcal{Y} }\] and some choices of \(y \circ j \to f \circ x\).