Introduction
Please see Algebraic Stacks, Section 026L for a brief introduction to algebraic stacks, and please read some of that chapter for our foundations of algebraic stacks. The intent is that in that chapter we are careful to distinguish between schemes, algebraic spaces, algebraic stacks, and starting with this chapter we employ the customary abuse of language where all of these concepts are used interchangeably.
The goal of this chapter is to introduce some basic notions and properties of algebraic stacks. A fundamental reference for the case of quasi-separated algebraic stacks with representable diagonal is [LM-B].
Conventions and abuse of language
We choose a big fppf site \(\Sch_{fppf}\). All schemes are contained in \(\Sch_{fppf}\). And all rings \(A\) considered have the property that \(\Spec(A)\) is (isomorphic) to an object of this big site.
We also fix a base scheme \(S\), by the conventions above an element of \(\Sch_{fppf}\). The reader who is only interested in the absolute case can take \(S = \Spec(\mathbf{Z})\).
Here are our conventions regarding algebraic stacks:
When we say algebraic stack we will mean an algebraic stacks over \(S\), i.e., a category fibred in groupoids \(p : \mathcal{X} \to (\Sch/S)_{fppf}\) which satisfies the conditions of Algebraic Stacks, Definition 026O.
We will say \(f : \mathcal{X} \to \mathcal{Y}\) is a morphism of algebraic stacks to indicate a \(1\)-morphism of algebraic stacks over \(S\), i.e., a \(1\)-morphism of categories fibred in groupoids over \((\Sch/S)_{fppf}\), see Algebraic Stacks, Definition 03YP.
A \(2\)-morphism \(\alpha : f \to g\) will indicate a \(2\)-morphism in the \(2\)-category of algebraic stacks over \(S\), see Algebraic Stacks, Definition 03YP.
Given morphisms \(\mathcal{X} \to \mathcal{Z}\) and \(\mathcal{Y} \to \mathcal{Z}\) of algebraic stacks we abusively call the \(2\)-fibre product \(\mathcal{X} \times_\mathcal{Z} \mathcal{Y}\) the fibre product.
We will write \(\mathcal{X} \times_S \mathcal{Y}\) for the product of the algebraic stacks \(\mathcal{X}\), \(\mathcal{Y}\).
We will often abuse notation and say two algebraic stacks \(\mathcal{X}\) and \(\mathcal{Y}\) are isomorphic if they are equivalent in this \(2\)-category.
Here are our conventions regarding algebraic spaces.
If we say \(X\) is an algebraic space then we mean that \(X\) is an algebraic space over \(S\), i.e., \(X\) is a presheaf on \((\Sch/S)_{fppf}\) which satisfies the conditions of Spaces, Definition 025Y.
A morphism of algebraic spaces \(f :X \to Y\) is a morphism of algebraic spaces over \(S\) as defined in Spaces, Definition 0260.
We will not distinguish between an algebraic space \(X\) and the algebraic stack \(\mathcal{S}_X \to (\Sch/S)_{fppf}\) it gives rise to, see Algebraic Stacks, Lemma 03YS.
In particular, a morphism \(f : X \to \mathcal{Y}\) from \(X\) to an algebraic stack \(\mathcal{Y}\) means a morphism \(f : \mathcal{S}_X \to \mathcal{Y}\) of algebraic stacks. Similarly for morphisms \(\mathcal{Y} \to X\).
Moreover, given an algebraic stack \(\mathcal{X}\) we say \(\mathcal{X}\) is an algebraic space to indicate that \(\mathcal{X}\) is representable by an algebraic space, see Algebraic Stacks, Definition 04SV.
We will use the following notational convention: If we indicate an algebraic stack by a roman capital (such as \(X, Y, Z, A, B, \ldots\)) then it will be the case that its inertia stack is trivial, and hence it is an algebraic space, see Algebraic Stacks, Proposition 04SZ.
Here are our conventions regarding schemes.
If we say \(X\) is a scheme then we mean that \(X\) is a scheme over \(S\), i.e., \(X\) is an object of \((\Sch/S)_{fppf}\).
By a morphism of schemes we mean a morphism of schemes over \(S\).
We will not distinguish between a scheme \(X\) and the algebraic stack \(\mathcal{S}_X \to (\Sch/S)_{fppf}\) it gives rise to, see Algebraic Stacks, Lemma 03YS.
In particular, a morphism \(f : X \to \mathcal{Y}\) from a scheme \(X\) to an algebraic stack \(\mathcal{Y}\) means a morphism \(f : \mathcal{S}_X \to \mathcal{Y}\) of algebraic stacks. Similarly for morphisms \(\mathcal{Y} \to X\).
Moreover, given an algebraic stack \(\mathcal{X}\) we say \(\mathcal{X}\) is a scheme to indicate that \(\mathcal{X}\) is representable, see Algebraic Stacks, Section 02ZQ.
Here are our conventions regarding morphisms of algebraic stacks:
A morphism \(f : \mathcal{X} \to \mathcal{Y}\) of algebraic stacks is representable, or representable by schemes if for every scheme \(T\) and morphism \(T \to \mathcal{Y}\) the fibre product \(T \times_\mathcal{Y} \mathcal{X}\) is a scheme. See Algebraic Stacks, Section 04ST.
A morphism \(f : \mathcal{X} \to \mathcal{Y}\) of algebraic stacks is representable by algebraic spaces if for every scheme \(T\) and morphism \(T \to \mathcal{Y}\) the fibre product \(T \times_\mathcal{Y} \mathcal{X}\) is an algebraic space. See Algebraic Stacks, Definition 02ZW. In this case \(Z \times_\mathcal{Y} \mathcal{X}\) is an algebraic space whenever \(Z \to \mathcal{Y}\) is a morphism whose source is an algebraic space, see Algebraic Stacks, Lemma 0300.
We may abuse notation and say that a diagram of algebraic stacks commutes if the diagram is \(2\)-commutative in the \(2\)-category of algebraic stacks.
Note that every morphism \(X \to \mathcal{Y}\) from an algebraic space to an algebraic stack is representable by algebraic spaces, see Algebraic Stacks, Lemma 045G. We will use this basic result without further mention.
Properties of morphisms representable by algebraic spaces
We will study properties of (arbitrary) morphisms of algebraic stacks in its own chapter. For morphisms representable by algebraic spaces we know what it means to be surjective, smooth, or étale, etc. This applies in particular to morphisms \(X \to \mathcal{Y}\) from algebraic spaces to algebraic stacks. In this section, we recall how this works, we list the properties to which this applies, and we prove a few easy lemmas.
Our first lemma says a morphism is representable by algebraic spaces if it is so after a base change by a flat, locally finitely presented, surjective morphism.
Lemma
Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. Let \(W\) be an algebraic space and let \(W \to \mathcal{Y}\) be surjective, locally of finite presentation, and flat. The following are equivalent
\(f\) is representable by algebraic spaces, and
\(W \times_\mathcal{Y} \mathcal{X}\) is an algebraic space.
Proof
The implication (1) \(\Rightarrow\) (2) is Algebraic Stacks, Lemma 0300. Conversely, let \(W \to \mathcal{Y}\) be as in (2). To prove (1) it suffices to show that \(f\) is faithful on fibre categories, see Algebraic Stacks, Lemma 04Y5. Assumption (2) implies in particular that \(W \times_\mathcal{Y} \mathcal{X} \to W\) is faithful. Hence the faithfulness of \(f\) follows from Stacks, Lemma 06DV.
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. Then we say \(f\) has property \(P\) if and only if for every scheme \(T\) and morphism \(T \to \mathcal{Y}\) the morphism of algebraic spaces \(T \times_\mathcal{Y} \mathcal{X} \to T\) has property \(P\), see Algebraic Stacks, Definition 03YK.
It turns out that if \(f : \mathcal{X} \to \mathcal{Y}\) is representable by algebraic spaces and has property \(P\), then for any morphism of algebraic stacks \(\mathcal{Y}' \to \mathcal{Y}\) the base change \(\mathcal{Y}' \times_\mathcal{Y} \mathcal{X} \to \mathcal{Y}'\) has property \(P\), see Algebraic Stacks, Lemmas 0302 and 045C. If the property \(P\) is preserved under compositions, then this holds also in the setting of morphisms of algebraic stacks representable by algebraic spaces, see Algebraic Stacks, Lemmas 0301 and 045B. Moreover, in this case products \(\mathcal{X}_1 \times \mathcal{X}_2 \to \mathcal{Y}_1 \times \mathcal{Y}_2\) of morphisms representable by algebraic spaces having property \(\mathcal{P}\) have property \(\mathcal{P}\), see Algebraic Stacks, Lemma 045E.
Finally, if we have two properties \(P, P'\) of morphisms of algebraic spaces which are fppf local on the target and preserved by arbitrary base change and if \(P(f) \Rightarrow P'(f)\) for every morphism \(f\), then the same implication holds for the corresponding property of morphisms of algebraic stacks representable by algebraic spaces, see Algebraic Stacks, Lemma 045F. We will use this without further mention in the following and in the following chapters.
The discussion above applies to each of the following properties of morphisms of algebraic spaces
quasi-compact, see Morphisms of Spaces, Lemma 03HF and Descent on Spaces, Lemma 041L,
quasi-separated, see Morphisms of Spaces, Lemma 03KL and Descent on Spaces, Lemma 041N,
universally closed, see Morphisms of Spaces, Lemma 03IS and Descent on Spaces, Lemma 041O,
universally open, see Morphisms of Spaces, Lemma 03Z3 and Descent on Spaces, Lemma 041P,
universally submersive, see Morphisms of Spaces, Lemma 0CFR and Descent on Spaces, Lemma 0CFW,
universal homeomorphism, see Morphisms of Spaces, Lemma 0CFT and Descent on Spaces, Lemma 0CFX,
surjective, see Morphisms of Spaces, Lemma 03MH and Descent on Spaces, Lemma 041Q,
universally injective, see Morphisms of Spaces, Lemma 03MW and Descent on Spaces, Lemma 041R,
locally of finite type, see Morphisms of Spaces, Lemma 03XH and Descent on Spaces, Lemma 041S,
locally of finite presentation, see Morphisms of Spaces, Lemma 03XR and Descent on Spaces, Lemma 041T,
finite type, see Morphisms of Spaces, Lemma 03XH and Descent on Spaces, Lemma 041U,
finite presentation, see Morphisms of Spaces, Lemma 03XR and Descent on Spaces, Lemma 041V,
flat, see Morphisms of Spaces, Lemma 03MO and Descent on Spaces, Lemma 041W,
open immersion, see Morphisms of Spaces, Section 03HB and Descent on Spaces, Lemma 041X,
isomorphism, see Descent on Spaces, Lemma 041Y,
affine, see Morphisms of Spaces, Lemma 03WI and Descent on Spaces, Lemma 041Z,
closed immersion, see Morphisms of Spaces, Section 03HB and Descent on Spaces, Lemma 0420,
separated, see Morphisms of Spaces, Lemma 03KL and Descent on Spaces, Lemma 0421,
proper, see Morphisms of Spaces, Lemma 04WP and Descent on Spaces, Lemma 0422,
quasi-affine, see Morphisms of Spaces, Lemma 03WO and Descent on Spaces, Lemma 0423,
integral, see Morphisms of Spaces, Lemma 03ZS and Descent on Spaces, Lemma 0425,
finite, see Morphisms of Spaces, Lemma 03ZS and Descent on Spaces, Lemma 0426,
(locally) quasi-finite, see Morphisms of Spaces, Lemma 03XL and Descent on Spaces, Lemma 0427,
syntomic, see Morphisms of Spaces, Lemma 03Z9 and Descent on Spaces, Lemma 0428,
smooth, see Morphisms of Spaces, Lemma 03ZE and Descent on Spaces, Lemma 0429,
unramified, see Morphisms of Spaces, Lemma 03ZJ and Descent on Spaces, Lemma 042A,
étale, see Morphisms of Spaces, Lemma 0466 and Descent on Spaces, Lemma 042B,
finite locally free, see Morphisms of Spaces, Lemma 03ZY and Descent on Spaces, Lemma 042C,
monomorphism, see Morphisms of Spaces, Lemma 042P and Descent on Spaces, Lemma 042D,
immersion, see Morphisms of Spaces, Section 03HB and Descent on Spaces, Lemma 042U,
locally separated, see Morphisms of Spaces, Lemma 03KL and Descent on Spaces, Lemma 042F,
Lemma
Let \(P\) be a property of morphisms of algebraic spaces as above. Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks representable by algebraic spaces. The following are equivalent:
\(f\) has \(P\),
for every algebraic space \(Z\) and morphism \(Z \to \mathcal{Y}\) the morphism \(Z \times_\mathcal{Y} \mathcal{X} \to Z\) has \(P\).
Proof
The implication (2) \(\Rightarrow\) (1) is immediate. Assume (1). Let \(Z \to \mathcal{Y}\) be as in (2). Choose a scheme \(U\) and a surjective étale morphism \(U \to Z\). By assumption the morphism \(U \times_\mathcal{Y} \mathcal{X} \to U\) has \(P\). But the diagram \[\xymatrix{ U \times_\mathcal{Y} \mathcal{X} \ar[d] \ar[r] & Z \times_\mathcal{Y} \mathcal{X} \ar[d] \\ U \ar[r] & Z }\] is cartesian, hence the right vertical arrow has \(P\) as \(\{U \to Z\}\) is an fppf covering.
The following lemma tells us it suffices to check \(P\) after a base change by a surjective, flat, locally finitely presented morphism.
Lemma
Let \(P\) be a property of morphisms of algebraic spaces as above. Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks representable by algebraic spaces. Let \(W\) be an algebraic space and let \(W \to \mathcal{Y}\) be surjective, locally of finite presentation, and flat. Set \(V = W \times_\mathcal{Y} \mathcal{X}\). Then \[(f\text{ has }P) \Leftrightarrow (\text{the projection }V \to W\text{ has }P).\]
Proof
The implication from left to right follows from Lemma 04XC. Assume \(V \to W\) has \(P\). Let \(T\) be a scheme, and let \(T \to \mathcal{Y}\) be a morphism. Consider the commutative diagram \[\xymatrix{ T \times_\mathcal{Y} \mathcal{X} \ar[d] & T \times_\mathcal{Y} V \ar[d] \ar[l] \ar[r] & V \ar[d] \\ T & T \times_\mathcal{Y} W \ar[l] \ar[r] & W }\] of algebraic spaces. The squares are cartesian. The bottom left morphism is a surjective, flat morphism which is locally of finite presentation, hence \(\{T \times_\mathcal{Y} V \to T\}\) is an fppf covering. Hence the fact that the right vertical arrow has property \(P\) implies that the left vertical arrow has property \(P\).
Lemma
Let \(P\) be a property of morphisms of algebraic spaces as above. 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 representable by algebraic spaces, surjective, flat, and locally of finite presentation. Set \(\mathcal{W} = \mathcal{Z} \times_\mathcal{Y} \mathcal{X}\). Then \[(f\text{ has }P) \Leftrightarrow (\text{the projection }\mathcal{W} \to \mathcal{Z}\text{ has }P).\]
Proof
Choose an algebraic space \(W\) and a morphism \(W \to \mathcal{Z}\) which is surjective, flat, and locally of finite presentation. By the discussion above 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 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 \(P\) be a property of morphisms of algebraic spaces as above. Let \(\tau \in \{\etale, smooth, syntomic, fppf\}\). Let \(\mathcal{X} \to \mathcal{Y}\) and \(\mathcal{Y} \to \mathcal{Z}\) be morphisms of algebraic stacks representable by algebraic spaces. Assume
\(\mathcal{X} \to \mathcal{Y}\) is surjective and étale, smooth, syntomic, or flat and locally of finite presentation,
the composition has \(P\), and
\(P\) is local on the source in the \(\tau\) topology.
Then \(\mathcal{Y} \to \mathcal{Z}\) has property \(P\).
Proof
Let \(Z\) be a scheme and let \(Z \to \mathcal{Z}\) be a morphism. Set \(X = \mathcal{X} \times_\mathcal{Z} Z\), \(Y = \mathcal{Y} \times_\mathcal{Z} Z\). By (1) \(\{X \to Y\}\) is a \(\tau\) covering of algebraic spaces and by (2) \(X \to Z\) has property \(P\). By (3) this implies that \(Y \to Z\) has property \(P\) and we win.
Lemma
Let \(g : \mathcal{X}' \to \mathcal{X}\) be a morphism of algebraic stacks which is representable by algebraic spaces. Let \([U/R] \to \mathcal{X}\) be a presentation. Set \(U' = U \times_\mathcal{X} \mathcal{X}'\), and \(R' = R \times_\mathcal{X} \mathcal{X}'\). Then there exists a groupoid in algebraic spaces of the form \((U', R', s', t', c')\), a presentation \([U'/R'] \to \mathcal{X}'\), and the diagram \[\xymatrix{ [U'/R'] \ar[d]_{[\text{pr}]} \ar[r] & \mathcal{X}' \ar[d]^g \\ [U/R] \ar[r] & \mathcal{X} }\] is \(2\)-commutative where the morphism \([\text{pr}]\) comes from a morphism of groupoids \(\text{pr} : (U', R', s', t', c') \to (U, R, s, t, c)\).
Proof
Since \(U \to \mathcal{Y}\) is surjective and smooth, see Algebraic Stacks, Lemma 04X0 the base change \(U' \to \mathcal{X}'\) is also surjective and smooth. Hence, by Algebraic Stacks, Lemma 04T5 it suffices to show that \(R' = U' \times_{\mathcal{X}'} U'\) in order to get a smooth groupoid \((U', R', s', t', c')\) and a presentation \([U'/R'] \to \mathcal{X}'\). Using that \(R = V \times_\mathcal{Y} V\) (see Groupoids in Spaces, Lemma 04M9) this follows from \[R' = U \times_\mathcal{X} U \times_\mathcal{X} \mathcal{X}' = (U \times_\mathcal{X} \mathcal{X}') \times_{\mathcal{X}'} (U \times_\mathcal{X} \mathcal{X}')\] see Categories, Lemmas 02XC and 02XD. Clearly the projection morphisms \(U' \to U\) and \(R' \to R\) give the desired morphism of groupoids \(\text{pr} : (U', R', s', t', c') \to (U, R, s, t, c)\). Hence the morphism \([\text{pr}]\) of quotient stacks by Groupoids in Spaces, Lemma 046Q.
We still have to show that the diagram \(2\)-commutes. It is clear that the diagram \[\xymatrix{ U' \ar[d]_{\text{pr}_U} \ar[r]_{f'} & \mathcal{X}' \ar[d]^g \\ U \ar[r]^f & \mathcal{X} }\] \(2\)-commutes where \(\text{pr}_U : U' \to U\) is the projection. There is a canonical \(2\)-arrow \(\tau : f \circ t \to f \circ s\) in \(\Mor(R, \mathcal{X})\) coming from \(R = U \times_\mathcal{X} U\), \(t = \text{pr}_0\), and \(s = \text{pr}_1\). Using the isomorphism \(R' \to U' \times_{\mathcal{X}'} U'\) we get similarly an isomorphism \(\tau' : f' \circ t' \to f' \circ s'\). Note that \(g \circ f' \circ t' = f \circ t \circ \text{pr}_R\) and \(g \circ f' \circ s' = f \circ s \circ \text{pr}_R\), where \(\text{pr}_R : R' \to R\) is the projection. Thus it makes sense to ask if [04Y7]\[\begin{equation} \tau \star \text{id}_{\text{pr}_R} = \text{id}_g \star \tau'. \end{equation}\] Now we make two claims: (1) if Equation (04Y7) holds, then the diagram \(2\)-commutes, and (2) Equation (04Y7) holds. We omit the proof of both claims. Hints: part (1) follows from the construction of \(f = f_{can}\) and \(f' = f'_{can}\) in Algebraic Stacks, Lemma 04T4. Part (2) follows by carefully working through the definitions.
Remark
Let \(\mathcal{Y}\) be an algebraic stack. Consider the following \(2\)-category:
An object is a morphism \(f : \mathcal{X} \to \mathcal{Y}\) which is representable by algebraic spaces,
a \(1\)-morphism \((g, \beta) : (f_1 : \mathcal{X}_1 \to \mathcal{Y}) \to (f_2 : \mathcal{X}_2 \to \mathcal{Y})\) consists of a morphism \(g : \mathcal{X}_1 \to \mathcal{X}_2\) and a \(2\)-morphism \(\beta : f_1 \to f_2 \circ g\), and
a \(2\)-morphism between \((g, \beta), (g', \beta') : (f_1 : \mathcal{X}_1 \to \mathcal{Y}) \to (f_2 : \mathcal{X}_2 \to \mathcal{Y})\) is a \(2\)-morphism \(\alpha : g \to g'\) such that \((\text{id}_{f_2} \star \alpha) \circ \beta = \beta'\).
Let us denote this \(2\)-category \(\textit{Spaces}/\mathcal{Y}\) by analogy with the notation of Topologies on Spaces, Section 03Y6. Now we claim that in this \(2\)-category the morphism categories \[\Mor_{\textit{Spaces}/\mathcal{Y}}( (f_1 : \mathcal{X}_1 \to \mathcal{Y}), (f_2 : \mathcal{X}_2 \to \mathcal{Y}))\] are all setoids. Namely, a \(2\)-morphism \(\alpha\) is a rule which to each object \(x_1\) of \(\mathcal{X}_1\) assigns an isomorphism \(\alpha_{x_1} : g(x_1) \longrightarrow g'(x_1)\) in the relevant fibre category of \(\mathcal{X}_2\) such that the diagram \[\xymatrix{ & f_2(x_1) \ar[ld]_{\beta_{x_1}} \ar[rd]^{\beta'_{x_1}} \\ f_2(g(x_1)) \ar[rr]^{f_2(\alpha_{x_1})} & & f_2(g'(x_1)) }\] commutes. But since \(f_2\) is faithful (see Algebraic Stacks, Lemma 04Y5) this means that if \(\alpha_{x_1}\) exists, then it is unique! In other words the \(2\)-category \(\textit{Spaces}/\mathcal{Y}\) is very close to being a category. Namely, if we replace \(1\)-morphisms by isomorphism classes of \(1\)-morphisms we obtain a category. We will often perform this replacement without further mention.
Points of algebraic stacks
Let \(\mathcal{X}\) be an algebraic stack. Let \(K, L\) be two fields and let \(p : \Spec(K) \to \mathcal{X}\) and \(q : \Spec(L) \to \mathcal{X}\) be morphisms. We say that \(p\) and \(q\) are equivalent if there exists a field \(\Omega\) and a \(2\)-commutative diagram \[\xymatrix{ \Spec(\Omega) \ar[r] \ar[d] & \Spec(L) \ar[d]^q \\ \Spec(K) \ar[r]^p & \mathcal{X}. }\]
Lemma
The notion above does indeed define an equivalence relation on morphisms from spectra of fields into the algebraic stack \(\mathcal{X}\).
Proof
It is clear that the relation is reflexive and symmetric. Hence we have to prove that it is transitive. This comes down to the following: Given a diagram \[\xymatrix{ \Spec(\Omega) \ar[r]_b \ar[d]_a & \Spec(L) \ar[d]^q & \Spec(\Omega') \ar[l]^{b'} \ar[d]^{a'} \\ \Spec(K) \ar[r]^p & \mathcal{X} & \Spec(K') \ar[l]_{p'} }\] with both squares \(2\)-commutative we have to show that \(p\) is equivalent to \(p'\). By the \(2\)-Yoneda lemma (see Algebraic Stacks, Section 04SS) the morphisms \(p\), \(p'\), and \(q\) are given by objects \(x\), \(x'\), and \(y\) in the fibre categories of \(\mathcal{X}\) over \(\Spec(K)\), \(\Spec(K')\), and \(\Spec(L)\). The \(2\)-commutativity of the squares means that there are isomorphisms \(\alpha : a^*x \to b^*y\) and \(\alpha' : (a')^*x' \to (b')^*y\) in the fibre categories of \(\mathcal{X}\) over \(\Spec(\Omega)\) and \(\Spec(\Omega')\). Choose any field \(\Omega''\) and embeddings \(\Omega \to \Omega''\) and \(\Omega' \to \Omega''\) agreeing on \(L\). Then we can extend the diagram above to \[\xymatrix{ & \Spec(\Omega'') \ar[ld]_c \ar[d]^{q'} \ar[rd]^{c'} \\ \Spec(\Omega) \ar[r]_b \ar[d]_a & \Spec(L) \ar[d]^q & \Spec(\Omega') \ar[l]^{b'} \ar[d]^{a'} \\ \Spec(K) \ar[r]^p & \mathcal{X} & \Spec(K') \ar[l]_{p'} }\] with commutative triangles and \[(q')^*(\alpha')^{-1} \circ (q')^*\alpha : (a \circ c)^*x \longrightarrow (a' \circ c')^*x'\] is an isomorphism in the fibre category over \(\Spec(\Omega'')\). Hence \(p\) is equivalent to \(p'\) as desired.
Definition
Let \(\mathcal{X}\) be an algebraic stack. A point of \(\mathcal{X}\) is an equivalence class of morphisms from spectra of fields into \(\mathcal{X}\). The set of points of \(\mathcal{X}\) is denoted \(|\mathcal{X}|\).
This agrees with our definition of points of algebraic spaces, see Properties of Spaces, Definition 03BU. Moreover, for a scheme we recover the usual notion of points, see Properties of Spaces, Lemma 03BV. If \(f : \mathcal{X} \to \mathcal{Y}\) is a morphism of algebraic stacks then there is an induced map \(|f| : |\mathcal{X}| \to |\mathcal{Y}|\) which maps a representative \(x : \Spec(K) \to \mathcal{X}\) to the representative \(f \circ x : \Spec(K) \to \mathcal{Y}\). This is well defined: namely \(2\)-isomorphic \(1\)-morphisms remain \(2\)-isomorphic after pre- or post-composing by a \(1\)-morphism because you can horizontally pre- or post-compose by the identity of the given \(1\)-morphism. This holds in any (strict) \((2, 1)\)-category. If \[\xymatrix{ \mathcal{X} \ar[d] \ar[r] & \mathcal{Y} \ar[d] \\ \mathcal{W} \ar[r] & \mathcal{Z} }\] is a \(2\)-commutative diagram of algebraic stacks, then the diagram of sets \[\xymatrix{ |\mathcal{X}| \ar[d] \ar[r] & |\mathcal{Y}| \ar[d] \\ |\mathcal{W}| \ar[r] & |\mathcal{Z}| }\] is commutative. In particular, if \(\mathcal{X} \to \mathcal{Y}\) is an equivalence then \(|\mathcal{X}| \to |\mathcal{Y}|\) is a bijection.
Lemma
Let \[\xymatrix{ \mathcal{Z} \times_\mathcal{Y} \mathcal{X} \ar[r] \ar[d] & \mathcal{X} \ar[d] \\ \mathcal{Z} \ar[r] & \mathcal{Y} }\] be a fibre product of algebraic stacks. Then the map of sets of points \[|\mathcal{Z} \times_\mathcal{Y} \mathcal{X}| \longrightarrow |\mathcal{Z}| \times_{|\mathcal{Y}|} |\mathcal{X}|\] is surjective.
Proof
Namely, suppose given fields \(K\), \(L\) and morphisms \(\Spec(K) \to \mathcal{X}\), \(\Spec(L) \to \mathcal{Z}\), then the assumption that they agree as elements of \(|\mathcal{Y}|\) means that there is a common extension \(M/K\) and \(M/L\) such that \(\Spec(M) \to \Spec(K) \to \mathcal{X} \to \mathcal{Y}\) and \(\Spec(M) \to \Spec(L) \to \mathcal{Z} \to \mathcal{Y}\) are \(2\)-isomorphic. And this is exactly the condition which says you get a morphism \(\Spec(M) \to \mathcal{Z} \times_\mathcal{Y} \mathcal{X}\).
Lemma
Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks which is representable by algebraic spaces. The following are equivalent:
\(|f| : |\mathcal{X}| \to |\mathcal{Y}|\) is surjective, and
\(f\) is surjective (in the sense of Section 04XB).
Proof
Assume (1). Let \(T \to \mathcal{Y}\) be a morphism whose source is a scheme. To prove (2) we have to show that the morphism of algebraic spaces \(T \times_\mathcal{Y} \mathcal{X} \to T\) is surjective. By Morphisms of Spaces, Definition 03ME this means we have to show that \(|T \times_\mathcal{Y} \mathcal{X}| \to |T|\) is surjective. Applying Lemma 04XH we see that this follows from (1).
Conversely, assume (2). Let \(y : \Spec(K) \to \mathcal{Y}\) be a morphism from the spectrum of a field into \(\mathcal{Y}\). By assumption the morphism \(\Spec(K) \times_{y, \mathcal{Y}} \mathcal{X} \to \Spec(K)\) of algebraic spaces is surjective. By Morphisms of Spaces, Definition 03ME this means there exists a field extension \(K'/K\) and a morphism \(\Spec(K') \to \Spec(K) \times_{y, \mathcal{Y}} \mathcal{X}\) such that the left square of the diagram \[\xymatrix{ \Spec(K') \ar[r] \ar[d] & \Spec(K) \times_{y, \mathcal{Y}} \mathcal{X} \ar[d] \ar[r] & \mathcal{X} \ar[d] \\ \Spec(K) \ar@{=}[r] & \Spec(K) \ar[r]^-y & \mathcal{Y} }\] is commutative. This shows that \(|X| \to |\mathcal{Y}|\) is surjective.
Here is a lemma explaining how to compute the set of points in terms of a presentation.
Lemma
Let \(\mathcal{X}\) be an algebraic stack. Let \(\mathcal{X} = [U/R]\) be a presentation of \(\mathcal{X}\), see Algebraic Stacks, Definition 04TI. 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 assumption means that we have a smooth groupoid \((U, R, s, t, c)\) in algebraic spaces, and an equivalence \(f : [U/R] \to \mathcal{X}\). We may assume \(\mathcal{X} = [U/R]\). The induced morphism \(p : U \to \mathcal{X}\) is smooth and surjective, see Algebraic Stacks, Lemma 04X0. Hence \(|U| \to |\mathcal{X}|\) is surjective by Lemma 04XI. Note that \(R = U \times_\mathcal{X} U\), see Groupoids in Spaces, Lemma 04M9. Hence 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.
Remark
The result of Lemma 04XJ can be generalized as follows. Let \(\mathcal{X}\) be an algebraic stack. Let \(U\) be an algebraic space and let \(f : U \to \mathcal{X}\) be a surjective morphism (which makes sense by Section 04XB). Let \(R = U \times_\mathcal{X} U\), let \((U, R, s, t, c)\) be the groupoid in algebraic spaces, and let \(f_{can} : [U/R] \to \mathcal{X}\) be the canonical morphism as constructed in Algebraic Stacks, Lemma 04T4. Then the image of \(|R| \to |U| \times |U|\) is an equivalence relation and \(|\mathcal{X}| = |U|/|R|\). The proof of Lemma 04XJ works without change. (Of course in general \([U/R]\) is not an algebraic stack, and in general \(f_{can}\) is not an isomorphism.)
Lemma
There exists a unique topology on the sets of points of algebraic stacks with the following properties:
for every morphism of algebraic stacks \(\mathcal{X} \to \mathcal{Y}\) the map \(|\mathcal{X}| \to |\mathcal{Y}|\) is continuous, and
for every morphism \(U \to \mathcal{X}\) which is flat and locally of finite presentation with \(U\) an algebraic space the map of topological spaces \(|U| \to |\mathcal{X}|\) is continuous and open.
Proof
Choose a morphism \(p : U \to \mathcal{X}\) which is surjective, flat, and locally of finite presentation with \(U\) an algebraic space. Such exist by the definition of an algebraic stack, as a smooth morphism is flat and locally of finite presentation (see Morphisms of Spaces, Lemmas 04AJ and 04TA). We define a topology on \(|\mathcal{X}|\) by the rule: \(W \subset |\mathcal{X}|\) is open if and only if \(|p|^{-1}(W)\) is open in \(|U|\). To show that this is independent of the choice of \(p\), let \(p' : U' \to \mathcal{X}\) be another morphism which is surjective, flat, locally of finite presentation from an algebraic space to \(\mathcal{X}\). Set \(U'' = U \times_\mathcal{X} U'\) so that we have a \(2\)-commutative diagram \[\xymatrix{ U'' \ar[r] \ar[d] & U' \ar[d] \\ U \ar[r] & \mathcal{X} }\] As \(U \to \mathcal{X}\) and \(U' \to \mathcal{X}\) are surjective, flat, locally of finite presentation we see that \(U'' \to U'\) and \(U'' \to U\) are surjective, flat and locally of finite presentation, see Lemma 04XC. Hence the maps \(|U''| \to |U'|\) and \(|U''| \to |U|\) are continuous, open and surjective, see Morphisms of Spaces, Definition 03ME and Lemma 042S. This clearly implies that our definition is independent of the choice of \(p : U \to \mathcal{X}\).
Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. By Algebraic Stacks, Lemma 04T1 we can find a \(2\)-commutative diagram \[\xymatrix{ U \ar[d]_x \ar[r]_a & V \ar[d]^y \\ \mathcal{X} \ar[r]^f & \mathcal{Y} }\] with surjective smooth vertical arrows. Consider the associated commutative diagram \[\xymatrix{ |U| \ar[d]_{|x|} \ar[r]_{|a|} & |V| \ar[d]^{|y|} \\ |\mathcal{X}| \ar[r]^{|f|} & |\mathcal{Y}| }\] of sets. If \(W \subset |\mathcal{Y}|\) is open, then by the definition above this means exactly that \(|y|^{-1}(W)\) is open in \(|V|\). Since \(|a|\) is continuous we conclude that \(|a|^{-1}|y|^{-1}(W) = |x|^{-1}|f|^{-1}(W)\) is open in \(|W|\) which means by definition that \(|f|^{-1}(W)\) is open in \(|\mathcal{X}|\). Thus \(|f|\) is continuous.
Finally, we have to show that if \(U\) is an algebraic space, and \(U \to \mathcal{X}\) is flat and locally of finite presentation, then \(|U| \to |\mathcal{X}|\) is open. Let \(V \to \mathcal{X}\) be surjective, flat, and locally of finite presentation with \(V\) an algebraic space. Consider the commutative diagram \[\xymatrix{ |U \times_\mathcal{X} V| \ar[r]_e \ar[rd]_f & |U| \times_{|\mathcal{X}|} |V| \ar[d]_c \ar[r]_d & |V| \ar[d]^b \\ & |U| \ar[r]^a & |\mathcal{X}| }\] Now the morphism \(U \times_\mathcal{X} V \to U\) is surjective, i.e, \(f : |U \times_\mathcal{X} V| \to |U|\) is surjective. The left top horizontal arrow is surjective, see Lemma 04XH. The morphism \(U \times_\mathcal{X} V \to V\) is flat and locally of finite presentation, hence \(d \circ e : |U \times_\mathcal{X} V| \to |V|\) is open, see Morphisms of Spaces, Lemma 042S. Pick \(W \subset |U|\) open. The properties above imply that \(b^{-1}(a(W)) = (d \circ e)(f^{-1}(W))\) is open, which by construction means that \(a(W)\) is open as desired.
Definition
Let \(\mathcal{X}\) be an algebraic stack. The underlying topological space of \(\mathcal{X}\) is the set of points \(|\mathcal{X}|\) endowed with the topology constructed in Lemma 04XL.
This definition does not conflict with the already existing topology on \(|\mathcal{X}|\) if \(\mathcal{X}\) is an algebraic space.
Lemma
Let \(\mathcal{X}\) be an algebraic stack. Every point of \(|\mathcal{X}|\) has a fundamental system of quasi-compact open neighbourhoods. In particular \(|\mathcal{X}|\) is locally quasi-compact in the sense of Topology, Definition 0068.
Proof
This follows formally from the fact that there exists a scheme \(U\) and a surjective, open, continuous map \(U \to |\mathcal{X}|\) of topological spaces. Namely, if \(U \to \mathcal{X}\) is surjective and smooth, then Lemma 04XL guarantees that \(|U| \to |\mathcal{X}|\) is continuous, surjective, and open.
Surjective morphisms
Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks which is representable by algebraic spaces. In Section 04XB we have already defined what it means for \(f\) to be surjective. In Lemma 04XI we have seen that this is equivalent to requiring \(|f| : |\mathcal{X}| \to |\mathcal{Y}|\) to be surjective. This clears the way for the following definition.
Definition
Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. We say \(f\) is surjective if the map \(|f| : |\mathcal{X}| \to |\mathcal{Y}|\) of associated topological spaces is surjective.
Here are some lemmas.
Lemma
The composition of surjective morphisms is surjective.
Proof
Omitted.
Lemma
The base change of a surjective morphism is surjective.
Proof
Omitted. Hint: Use Lemma 04XH.
Lemma
Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. Let \(\mathcal{Y}' \to \mathcal{Y}\) be a surjective morphism of algebraic stacks. If the base change \(f' : \mathcal{Y}' \times_\mathcal{Y} \mathcal{X} \to \mathcal{Y}'\) of \(f\) is surjective, then \(f\) is surjective.
Proof
Immediate from Lemma 04XH.
Lemma
Let \(\mathcal{X} \to \mathcal{Y} \to \mathcal{Z}\) be morphisms of algebraic stacks. If \(\mathcal{X} \to \mathcal{Z}\) is surjective so is \(\mathcal{Y} \to \mathcal{Z}\).
Proof
Immediate.
Quasi-compact algebraic stacks
The following definition is equivalent with the definition for algebraic spaces by Properties of Spaces, Lemma 03E4.
Definition
Let \(\mathcal{X}\) be an algebraic stack. We say \(\mathcal{X}\) is quasi-compact if and only if \(|\mathcal{X}|\) is quasi-compact.
Lemma
Let \(\mathcal{X}\) be an algebraic stack. The following are equivalent:
\(\mathcal{X}\) is quasi-compact,
there exists a surjective smooth morphism \(U \to \mathcal{X}\) with \(U\) an affine scheme,
there exists a surjective smooth morphism \(U \to \mathcal{X}\) with \(U\) a quasi-compact scheme,
there exists a surjective smooth morphism \(U \to \mathcal{X}\) with \(U\) a quasi-compact algebraic space, and
there exists a surjective morphism \(\mathcal{U} \to \mathcal{X}\) of algebraic stacks such that \(\mathcal{U}\) is quasi-compact.
Proof
We will use Lemma 04XI. Suppose \(\mathcal{U}\) and \(\mathcal{U} \to \mathcal{X}\) are as in (5). Then since \(|\mathcal{U}| \to |\mathcal{X}|\) is surjective and continuous we conclude that \(|\mathcal{X}|\) is quasi-compact. Thus (5) implies (1). The implications (2) \(\Rightarrow\) (3) \(\Rightarrow\) (4) \(\Rightarrow\) (5) are immediate. Assume (1), i.e., \(\mathcal{X}\) is quasi-compact, i.e., that \(|\mathcal{X}|\) is quasi-compact. Choose a scheme \(U\) and a surjective smooth morphism \(U \to \mathcal{X}\). Then since \(|U| \to |\mathcal{X}|\) is open we see that there exists a quasi-compact open \(U' \subset U\) such that \(|U'| \to |X|\) is surjective (and still smooth). Choose a finite affine open covering \(U' = U_1 \cup \ldots \cup U_n\). Then \(U_1 \amalg \ldots \amalg U_n \to \mathcal{X}\) is a surjective smooth morphism whose source is an affine scheme (Schemes, Lemma 01I5). Hence (2) holds.
Lemma
A finite disjoint union of quasi-compact algebraic stacks is a quasi-compact algebraic stack.
Proof
This is clear from the corresponding topological fact.
Properties of algebraic stacks defined by properties of schemes
Any smooth local property of schemes gives rise to a corresponding property of algebraic stacks via the following lemma. Note that a property of schemes which is smooth local is also étale local as any étale covering is also a smooth covering. Hence for a smooth local property \(P\) of schemes we know what it means to say that an algebraic space has \(P\), see Properties of Spaces, Section 03E5.
Lemma
Let \(\mathcal{P}\) be a property of schemes which is local in the smooth topology, see Descent, Definition 0348. Let \(\mathcal{X}\) be an algebraic stack. The following are equivalent
for some scheme \(U\) and some surjective smooth morphism \(U \to \mathcal{X}\) the scheme \(U\) has property \(\mathcal{P}\),
for every scheme \(U\) and every smooth morphism \(U \to \mathcal{X}\) the scheme \(U\) has property \(\mathcal{P}\),
for some algebraic space \(U\) and some surjective smooth morphism \(U \to \mathcal{X}\) the algebraic space \(U\) has property \(\mathcal{P}\), and
for every algebraic space \(U\) and every smooth morphism \(U \to \mathcal{X}\) the algebraic space \(U\) has property \(\mathcal{P}\).
If \(\mathcal{X}\) is a scheme \(U\) this is equivalent to \(\mathcal{P}(U)\). If \(\mathcal{X}\) is an algebraic space \(X\) this is equivalent to \(X\) having property \(\mathcal{P}\).
Proof
Let \(U \to \mathcal{X}\) surjective and smooth with \(U\) an algebraic space. Let \(V \to \mathcal{X}\) be a smooth morphism with \(V\) an algebraic space. Choose schemes \(U'\) and \(V'\) and surjective étale morphisms \(U' \to U\) and \(V' \to V\). Finally, choose a scheme \(W\) and a surjective étale morphism \(W \to V' \times_\mathcal{X} U'\). Then \(W \to V'\) and \(W \to U'\) are smooth morphisms of schemes as compositions of étale and smooth morphisms of algebraic spaces, see Morphisms of Spaces, Lemmas 04XX and 03ZD. Moreover, \(W \to V'\) is surjective as \(U' \to \mathcal{X}\) is surjective. Hence, we have \[\mathcal{P}(U) \Leftrightarrow \mathcal{P}(U') \Rightarrow \mathcal{P}(W) \Rightarrow \mathcal{P}(V') \Leftrightarrow \mathcal{P}(V)\] where the equivalences are by definition of property \(\mathcal{P}\) for algebraic spaces, and the two implications come from Descent, Definition 0348. This proves (3) \(\Rightarrow\) (4).
The implications (2) \(\Rightarrow\) (1), (1) \(\Rightarrow\) (3), and (4) \(\Rightarrow\) (2) are immediate.
Definition
Let \(\mathcal{X}\) be an algebraic stack. Let \(\mathcal{P}\) be a property of schemes which is local in the smooth topology. We say \(\mathcal{X}\) has property \(\mathcal{P}\) if any of the equivalent conditions of Lemma 04YF hold.
Remark
Here is a list of properties which are local for the smooth topology (keep in mind that the fpqc, fppf, and syntomic topologies are stronger than the smooth topology):
locally Noetherian, see Descent, Lemma 034C,
Jacobson, see Descent, Lemma 0368,
locally Noetherian and \((S_k)\), see Descent, Lemma 036A,
Cohen-Macaulay, see Descent, Lemma 036B,
reduced, see Descent, Lemma 034E,
normal, see Descent, Lemma 034F,
locally Noetherian and \((R_k)\), see Descent, Lemma 036C,
regular, see Descent, Lemma 036D,
Nagata, see Descent, Lemma 036E.
Any smooth local property of germs of schemes gives rise to a corresponding property of algebraic stacks. Note that a property of germs which is smooth local is also étale local. Hence for a smooth local property of germs of schemes \(P\) we know what it means to say that an algebraic space \(X\) has property \(P\) at \(x \in |X|\), see Properties of Spaces, Section 04YE.
Lemma
Let \(\mathcal{X}\) be an algebraic stack. Let \(x \in |\mathcal{X}|\) be a point of \(\mathcal{X}\). Let \(\mathcal{P}\) be a property of germs of schemes which is smooth local, see Descent, Definition 04N1. The following are equivalent
for any smooth morphism \(U \to \mathcal{X}\) with \(U\) a scheme and \(u \in U\) with \(a(u) = x\) we have \(\mathcal{P}(U, u)\),
for some smooth morphism \(U \to \mathcal{X}\) with \(U\) a scheme and some \(u \in U\) with \(a(u) = x\) we have \(\mathcal{P}(U, u)\),
for any smooth morphism \(U \to \mathcal{X}\) with \(U\) an algebraic space and \(u \in |U|\) with \(a(u) = x\) the algebraic space \(U\) has property \(\mathcal{P}\) at \(u\), and
for some smooth morphism \(U \to \mathcal{X}\) with \(U\) an algebraic space and some \(u \in |U|\) with \(a(u) = x\) the algebraic space \(U\) has property \(\mathcal{P}\) at \(u\).
If \(\mathcal{X}\) is representable, then this is equivalent to \(\mathcal{P}(\mathcal{X}, x)\). If \(\mathcal{X}\) is an algebraic space then this is equivalent to \(\mathcal{X}\) having property \(\mathcal{P}\) at \(x\).
Proof
Let \(a : U \to \mathcal{X}\) and \(u \in |U|\) as in (3). Let \(b : V \to \mathcal{X}\) be another smooth morphism with \(V\) an algebraic space and \(v \in |V|\) with \(b(v) = x\) also. Choose a scheme \(U'\), an étale morphism \(U' \to U\) and \(u' \in U'\) mapping to \(u\). Choose a scheme \(V'\), an étale morphism \(V' \to V\) and \(v' \in V'\) mapping to \(v\). By Lemma 04XH there exists a point \(\overline{w} \in |V' \times_\mathcal{X} U'|\) mapping to \(u'\) and \(v'\). Choose a scheme \(W\) and a surjective étale morphism \(W \to V' \times_\mathcal{X} U'\). We may choose a \(w \in |W|\) mapping to \(\overline{w}\) (see Properties of Spaces, Lemma 03H5). Then \(W \to V'\) and \(W \to U'\) are smooth morphisms of schemes as compositions of étale and smooth morphisms of algebraic spaces, see Morphisms of Spaces, Lemmas 04XX and 03ZD. Hence \[\mathcal{P}(U, u) \Leftrightarrow \mathcal{P}(U', u') \Leftrightarrow \mathcal{P}(W, w) \Leftrightarrow \mathcal{P}(V', v') \Leftrightarrow \mathcal{P}(V, v)\] The outer two equivalences by Properties of Spaces, Definition 04RC and the other two by what it means to be a smooth local property of germs of schemes. This proves (4) \(\Rightarrow\) (3).
The implications (1) \(\Rightarrow\) (2), (2) \(\Rightarrow\) (4), and (3) \(\Rightarrow\) (1) are immediate.
Definition
Let \(\mathcal{P}\) be a property of germs of schemes which is smooth local. Let \(\mathcal{X}\) be an algebraic stack. Let \(x \in |\mathcal{X}|\). We say \(\mathcal{X}\) has property \(\mathcal{P}\) at \(x\) if any of the equivalent conditions of Lemma 04YI holds.
Monomorphisms of algebraic stacks
We define a monomorphism of algebraic stacks in the following way. We will see in Lemma 04ZZ that this is compatible with the corresponding \(2\)-category theoretic notion.
Definition
Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. We say \(f\) is a monomorphism if it is representable by algebraic spaces and a monomorphism in the sense of Section 04XB.
First some basic lemmas.
Lemma
Let \(\mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. Let \(\mathcal{Z} \to \mathcal{Y}\) be a monomorphism. Then \(\mathcal{Z} \times_\mathcal{Y} \mathcal{X} \to \mathcal{X}\) is a monomorphism.
Proof
This follows from the general discussion in Section 04XB.
Lemma
Compositions of monomorphisms of algebraic stacks are monomorphisms.
Proof
This follows from the general discussion in Section 04XB and Morphisms of Spaces, Lemma 042O.
Lemma
Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. The following are equivalent:
\(f\) is a monomorphism,
\(f\) is fully faithful,
the diagonal \(\Delta_f : \mathcal{X} \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}\) is an equivalence, and
there exists an algebraic space \(W\) and a surjective, flat morphism \(W \to \mathcal{Y}\) which is locally of finite presentation such that \(V = \mathcal{X} \times_\mathcal{Y} W\) is an algebraic space, and the morphism \(V \to W\) is a monomorphism of algebraic spaces.
Proof
The equivalence of (1) and (4) follows from the general discussion in Section 04XB and in particular Lemmas 04ZP and 04XD.
The equivalence of (2) and (3) is Categories, Lemma 04Z7.
Assume the equivalent conditions (2) and (3). Then \(f\) is representable by algebraic spaces according to Algebraic Stacks, Lemma 04Y5. Moreover, the \(2\)-Yoneda lemma combined with the fully faithfulness implies that for every scheme \(T\) the functor \[\Mor(T, \mathcal{X}) \longrightarrow \Mor(T, \mathcal{Y})\] is fully faithful. Hence given a morphism \(y : T \to \mathcal{Y}\) there exists up to unique \(2\)-isomorphism at most one morphism \(x : T \to \mathcal{X}\) such that \(y \cong f \circ x\). In particular, given a morphism of schemes \(h : T' \to T\) there exists at most one lift \(\tilde h : T' \to T \times_\mathcal{Y} \mathcal{X}\) of \(h\). Thus \(T \times_\mathcal{Y} \mathcal{X} \to T\) is a monomorphism of algebraic spaces, which proves that (1) holds.
Finally, assume that (1) holds. Then for any scheme \(T\) and morphism \(y : T \to \mathcal{Y}\) the fibre product \(T \times_\mathcal{Y} \mathcal{X}\) is an algebraic space, and \(T \times_\mathcal{Y} \mathcal{X} \to T\) is a monomorphism. Hence there exists up to unique isomorphism exactly one pair \((x, \alpha)\) where \(x : T \to \mathcal{X}\) is a morphism and \(\alpha : f \circ x \to y\) is a \(2\)-morphism. Applying the \(2\)-Yoneda lemma this says exactly that \(f\) is fully faithful, i.e., that (2) holds.
Lemma
A monomorphism of algebraic stacks induces an injective map of sets of points.
Proof
Let \(f : \mathcal{X} \to \mathcal{Y}\) be a monomorphism of algebraic stacks. Suppose that \(x_i : \Spec(K_i) \to \mathcal{X}\) be morphisms such that \(f \circ x_1\) and \(f \circ x_2\) define the same element of \(|\mathcal{Y}|\). Applying the definition we find a common extension \(\Omega\) with corresponding morphisms \(c_i : \Spec(\Omega) \to \Spec(K_i)\) and a \(2\)-isomorphism \(\beta : f \circ x_1 \circ c_1 \to f \circ x_1 \circ c_2\). As \(f\) is fully faithful, see Lemma 04ZZ, we can lift \(\beta\) to an isomorphism \(\alpha : x_1 \circ c_1 \to x_1 \circ c_2\). Hence \(x_1\) and \(x_2\) define the same point of \(|\mathcal{X}|\) as desired.
Lemma
Let \(\mathcal{X} \to \mathcal{X}' \to \mathcal{Y}\) be morphisms of algebraic stacks. If \(\mathcal{X} \to \mathcal{X}'\) is a monomorphism then the canonical diagram \[\xymatrix{ \mathcal{X} \ar[r] \ar[d] & \mathcal{X} \times_\mathcal{Y} \mathcal{X} \ar[d] \\ \mathcal{X}' \ar[r] & \mathcal{X}' \times_\mathcal{Y} \mathcal{X}' }\] is a fibre product square.
Proof
We have \(\mathcal{X} = \mathcal{X} \times_{\mathcal{X}'} \mathcal{X}\) by Lemma 04ZZ. Thus the result by applying Categories, Lemma 04Z1.
Immersions of algebraic stacks
Immersions of algebraic stacks are defined as follows.
Definition
Immersions.
A morphism of algebraic stacks is called an open immersion if it is representable, and an open immersion in the sense of Section 04XB.
A morphism of algebraic stacks is called a closed immersion if it is representable, and a closed immersion in the sense of Section 04XB.
A morphism of algebraic stacks is called an immersion if it is representable, and an immersion in the sense of Section 04XB.
This is not the most convenient way to think about immersions for us. For us it is a little bit more convenient to think of an immersion as a morphism of algebraic stacks which is representable by algebraic spaces and is an immersion in the sense of Section 04XB. Similarly for closed and open immersions. 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 a (closed, resp. open) immersion. Then \(\mathcal{Z} \times_\mathcal{Y} \mathcal{X} \to \mathcal{X}\) is a (closed, resp. open) immersion.
Proof
This follows from the general discussion in Section 04XB.
Lemma
Compositions of immersions of algebraic stacks are immersions. Similarly for closed immersions and open immersions.
Proof
This follows from the general discussion in Section 04XB and Spaces, Lemma 02YV.
Lemma
Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. Let \(W\) be an algebraic space and let \(W \to \mathcal{Y}\) be a surjective, flat morphism which is locally of finite presentation. The following are equivalent:
\(f\) is an (open, resp. closed) immersion, and
\(V = W \times_\mathcal{Y} \mathcal{X}\) is an algebraic space, and \(V \to W\) is an (open, resp. closed) immersion.
Proof
This follows from the general discussion in Section 04XB and in particular Lemmas 04ZP and 04XD.
Lemma
An immersion is a monomorphism.
Proof
See Morphisms of Spaces, Lemma 042R.
Lemma
If \(f : \mathcal{X} \to \mathcal{Y}\) is an immersion, then \(|f| : |\mathcal{X}| \to |\mathcal{Y}|\) is a homeomorphism onto a locally closed subset. If \(f\) is a closed, resp. open immersion, then \(|f|\) is closed, resp. open.
Proof
Omitted.
The following two lemmas explain how to think about immersions in terms of presentations.
Lemma
Let \((U, R, s, t, c)\) be a smooth groupoid in algebraic spaces. Let \(i : \mathcal{Z} \to [U/R]\) be an immersion. Then there exists an \(R\)-invariant locally closed subspace \(Z \subset U\) and a presentation \([Z/R_Z] \to \mathcal{Z}\) where \(R_Z\) is the restriction of \(R\) to \(Z\) such that \[\xymatrix{ [Z/R_Z] \ar[dr] \ar[rr] & & \mathcal{Z} \ar[ld]^i \\ & [U/R] }\] is \(2\)-commutative. If \(i\) is a closed (resp. open) immersion then \(Z\) is a closed (resp. open) subspace of \(U\).
Proof
By Lemma 04Y6 we get a commutative diagram \[\xymatrix{ [U'/R'] \ar[dr] \ar[rr] & & \mathcal{Z} \ar[ld] \\ & [U/R] }\] where \(U' = \mathcal{Z} \times_{[U/R]} U\) and \(R' = \mathcal{Z} \times_{[U/R]} R\). Since \(\mathcal{Z} \to [U/R]\) is an immersion we see that \(U' \to U\) is an immersion of algebraic spaces. Let \(Z \subset U\) be the locally closed subspace such that \(U' \to U\) factors through \(Z\) and induces an isomorphism \(U' \to Z\). It is clear from the construction of \(R'\) that \(R' = U' \times_{U, t} R = R \times_{s, U} U'\). This implies that \(Z \cong U'\) is \(R\)-invariant and that the image of \(R' \to R\) identifies \(R'\) with the restriction \(R_Z = s^{-1}(Z) = t^{-1}(Z)\) of \(R\) to \(Z\). Hence the lemma holds.
Lemma
Let \((U, R, s, t, c)\) be a smooth groupoid in algebraic spaces. Let \(\mathcal{X} = [U/R]\) be the associated algebraic stack, see Algebraic Stacks, Theorem 04TK. Let \(Z \subset U\) be an \(R\)-invariant locally closed subspace. Then \[[Z/R_Z] \longrightarrow [U/R]\] is an immersion of algebraic stacks, where \(R_Z\) is the restriction of \(R\) to \(Z\). If \(Z \subset U\) is open (resp. closed) then the morphism is an open (resp. closed) immersion of algebraic stacks.
Proof
Recall that by Groupoids in Spaces, Definition 044F (see also discussion following the definition) we have \(R_Z = s^{-1}(Z) = t^{-1}(Z)\) as locally closed subspaces of \(R\). Hence the two morphisms \(R_Z \to Z\) are smooth as base changes of \(s\) and \(t\). Hence \((Z, R_Z, s|_{R_Z}, t|_{R_Z}, c|_{R_Z \times_{s, Z, t} R_Z})\) is a smooth groupoid in algebraic spaces, and we see that \([Z/R_Z]\) is an algebraic stack, see Algebraic Stacks, Theorem 04TK. The assumptions of Groupoids in Spaces, Lemma 04ZN are all satisfied and it follows that we have a \(2\)-fibre square \[\xymatrix{ Z \ar[d] \ar[r] & [Z/R_Z] \ar[d] \\ U \ar[r] & [U/R] }\] It follows from this and Lemma 04ZP that \([Z/R_Z] \to [U/R]\) is representable by algebraic spaces, whereupon it follows from Lemma 04XD that the right vertical arrow is an immersion (resp. closed immersion, resp. open immersion) if and only if the left vertical arrow is.
We can define open, closed, and locally closed substacks as follows.
Definition
Let \(\mathcal{X}\) be an algebraic stack.
An open substack of \(\mathcal{X}\) is a strictly full subcategory \(\mathcal{X}' \subset \mathcal{X}\) such that \(\mathcal{X}'\) is an algebraic stack and \(\mathcal{X}' \to \mathcal{X}\) is an open immersion.
A closed substack of \(\mathcal{X}\) is a strictly full subcategory \(\mathcal{X}' \subset \mathcal{X}\) such that \(\mathcal{X}'\) is an algebraic stack and \(\mathcal{X}' \to \mathcal{X}\) is a closed immersion.
A locally closed substack of \(\mathcal{X}\) is a strictly full subcategory \(\mathcal{X}' \subset \mathcal{X}\) such that \(\mathcal{X}'\) is an algebraic stack and \(\mathcal{X}' \to \mathcal{X}\) is an immersion.
This definition should be used with caution. Namely, if \(f : \mathcal{X} \to \mathcal{Y}\) is an equivalence of algebraic stacks and \(\mathcal{X}' \subset \mathcal{X}\) is an open substack, then it is not necessarily the case that the subcategory \(f(\mathcal{X}')\) is an open substack of \(\mathcal{Y}\). The problem is that it may not be a strictly full subcategory; but this is also the only problem. Here is a formal statement.
Lemma
For any immersion \(i : \mathcal{Z} \to \mathcal{X}\) there exists a unique locally closed substack \(\mathcal{X}' \subset \mathcal{X}\) such that \(i\) factors as the composition of an equivalence \(i' : \mathcal{Z} \to \mathcal{X}'\) followed by the inclusion morphism \(\mathcal{X}' \to \mathcal{X}\). If \(i\) is a closed (resp. open) immersion, then \(\mathcal{X}'\) is a closed (resp. open) substack of \(\mathcal{X}\).
Proof
Omitted.
Lemma
Let \([U/R] \to \mathcal{X}\) be a presentation of an algebraic stack. There is a canonical bijection \[\text{locally closed substacks }\mathcal{Z}\text{ of }\mathcal{X} \longrightarrow R\text{-invariant locally closed subspaces }Z\text{ of }U\] which sends \(\mathcal{Z}\) to \(U \times_\mathcal{X} \mathcal{Z}\). Moreover, a morphism of algebraic stacks \(f : \mathcal{Y} \to \mathcal{X}\) factors through \(\mathcal{Z}\) if and only if \(\mathcal{Y} \times_\mathcal{X} U \to U\) factors through \(Z\). Similarly for closed substacks and open substacks.
Proof
By Lemmas 0505 and 04YN we find that the map is a bijection. If \(\mathcal{Y} \to \mathcal{X}\) factors through \(\mathcal{Z}\) then of course the base change \(\mathcal{Y} \times_\mathcal{X} U \to U\) factors through \(Z\). Converse, suppose that \(\mathcal{Y} \to \mathcal{X}\) is a morphism such that \(\mathcal{Y} \times_\mathcal{X} U \to U\) factors through \(Z\). We will show that for every scheme \(T\) and morphism \(T \to \mathcal{Y}\), given by an object \(y\) of the fibre category of \(\mathcal{Y}\) over \(T\), the object \(y\) is in fact in the fibre category of \(\mathcal{Z}\) over \(T\). Namely, the fibre product \(T \times_\mathcal{X} U\) is an algebraic space and \(T \times_\mathcal{X} U \to T\) is a surjective smooth morphism. Hence there is an fppf covering \(\{T_i \to T\}\) such that \(T_i \to T\) factors through \(T \times_\mathcal{X} U \to T\) for all \(i\). Then \(T_i \to \mathcal{X}\) factors through \(\mathcal{Y} \times_\mathcal{X} U\) and hence through \(Z \subset U\). Thus \(y|_{T_i}\) is an object of \(\mathcal{Z}\) (as \(Z\) is the fibre product of \(U\) with \(\mathcal{Z}\) over \(\mathcal{X}\)). Since \(\mathcal{Z}\) is a strictly full substack, we conclude that \(y\) is an object of \(\mathcal{Z}\) as desired.
Lemma
Let \(\mathcal{X}\) be an algebraic stack. The rule \(\mathcal{U} \mapsto |\mathcal{U}|\) defines an inclusion preserving bijection between open substacks of \(\mathcal{X}\) and open subsets of \(|\mathcal{X}|\).
Proof
Choose a presentation \([U/R] \to \mathcal{X}\), see Algebraic Stacks, Lemma 04T5. By Lemma 0507 we see that open substacks correspond to \(R\)-invariant open subschemes of \(U\). On the other hand Lemmas 04XJ and 04XL guarantee these correspond bijectively to open subsets of \(|\mathcal{X}|\).
Lemma
Let \(\mathcal X\) be an algebraic stack. Let \(U\) be an algebraic space and \(U \to \mathcal X\) a surjective smooth morphism. For an open immersion \(V \hookrightarrow U\), there exists an algebraic stack \(\mathcal Y\), an open immersion \(\mathcal Y \to \mathcal X\), and a surjective smooth morphism \(V \to \mathcal Y\).
Proof
We define a category fibred in groupoids \(\mathcal Y\) by letting the fiber category \(\mathcal{Y}_T\) over an object \(T\) of \((\Sch/S)_{fppf}\) be the full subcategory of \(\mathcal{X}_T\) consisting of all \(y \in \Ob(\mathcal{X}_T)\) such that the projection morphism \(V \times_{\mathcal X, y} T \to T\) surjective. Now for any morphism \(x : T \to \mathcal X\), the \(2\)-fibred product \(T \times_{x, \mathcal X} \mathcal Y\) has fiber category over \(T'\) consisting of triples \((f : T' \to T, y \in \mathcal{X}_{T'}, f^*x \simeq y)\) such that \(V \times_{\mathcal X, y} T' \to T'\) is surjective. Note that \(T \times_{x, \mathcal X} \mathcal Y\) is fibered in setoids since \(\mathcal Y \to \mathcal X\) is faithful (see Stacks, Lemma 05UI). Now the isomorphism \(f^*x \simeq y\) gives the diagram \[\xymatrix{ V \times_{\mathcal X, y} T' \ar[d] \ar[r] & V \times_{\mathcal X, x} T \ar[r] \ar[d] & V \ar[d] \\ T' \ar[r]^f & T \ar[r]^x & \mathcal X }\] where both squares are cartesian. The morphism \(V \times_{\mathcal X, x} T \to T\) is smooth by base change, and hence open. Let \(T_0 \subset T\) be its image. From the cartesian squares we deduce that \(V \times_{\mathcal X, y} T' \to T'\) is surjective if and only if \(f\) lands in \(T_0\). Therefore \(T \times_{x, \mathcal X} \mathcal Y\) is representable by \(T_0\), so the inclusion \(\mathcal Y \to \mathcal X\) is an open immersion. By Algebraic Stacks, Lemma 05UN we conclude that \(\mathcal{Y}\) is an algebraic stack. Lastly if we denote the morphism \(V \to \mathcal X\) by \(g\), we have \(V \times_{\mathcal X} V \to V\) is surjective (the diagonal gives a section). Hence \(g\) is in the image of \(\mathcal{Y}_V \to \mathcal{X}_V\), i.e., we obtain a morphism \(g' : V \to \mathcal{Y}\) fitting into the commutative diagram \[\xymatrix{ V \ar[r] \ar[d]^{g'} & U \ar[d] \\ \mathcal{Y} \ar[r] & \mathcal{X} }\] Since \(V \times_{g, \mathcal X} \mathcal Y \to V\) is a monomorphism, it is in fact an isomorphism since \((1, g')\) defines a section. Therefore \(g' : V \to \mathcal Y\) is a smooth morphism, as it is the base change of the smooth morphism \(g : V \to \mathcal{X}\). It is surjective by our construction of \(\mathcal{Y}\) which finishes the proof of the lemma.
Lemma
Let \(\mathcal X\) be an algebraic stack and \(\mathcal{X}_i \subset \mathcal X\) a collection of open substacks indexed by \(i \in I\). Then there exists an open substack, which we denote \(\bigcup_{i\in I} \mathcal{X}_i \subset \mathcal X\), such that the \(\mathcal{X}_i\) are open substacks covering it.
Proof
We define a fibred subcategory \(\mathcal{X}' = \bigcup_{i \in I} \mathcal{X}_i\) by letting the fiber category over an object \(T\) of \((\Sch/S)_{fppf}\) be the full subcategory of \(\mathcal{X}_T\) consisting of all \(x \in \Ob(\mathcal{X}_T)\) such that the morphism \(\coprod_{i \in I} (\mathcal{X}_i \times_{\mathcal X} T) \to T\) is surjective. Let \(x_i \in \Ob((\mathcal{X}_i)_T)\). Then \((x_i, 1)\) gives a section of \(\mathcal{X}_i \times_{\mathcal X} T \to T\), so we have an isomorphism. Thus \(\mathcal{X}_i \subset \mathcal{X}'\) is a full subcategory. Now let \(x \in \Ob(\mathcal{X}_T)\). Then \(\mathcal{X}_i \times_{\mathcal X} T\) is representable by an open subscheme \(T_i \subset T\). The \(2\)-fibred product \(\mathcal{X}' \times_{\mathcal X} T\) has fiber over \(T'\) consisting of \((y \in \mathcal{X}_{T'}, f : T' \to T, f^*x \simeq y)\) such that \(\coprod (\mathcal{X}_i \times_{\mathcal X, y} T') \to T'\) is surjective. The isomorphism \(f^*x \simeq y\) induces an isomorphism \(\mathcal{X}_i \times_{\mathcal X, y} T' \simeq T_i \times_T T'\). Then the \(T_i \times_T T'\) cover \(T'\) if and only if \(f\) lands in \(\bigcup T_i\). Therefore we have a diagram \[\xymatrix{ T_i \ar[r] \ar[d] & \bigcup T_i \ar[r] \ar[d] & T \ar[d] \\ \mathcal{X}_i \ar[r] & \mathcal{X}' \ar[r] & \mathcal{X} }\] with both squares cartesian. By Algebraic Stacks, Lemma 05UN we conclude that \(\mathcal{X'} \subset \mathcal{X}\) is algebraic and an open substack. It is also clear from the cartesian squares above that the morphism \(\coprod_{i \in I} \mathcal{X}_i \to \mathcal{X}'\) which finishes the proof of the lemma.
Lemma
Let \(\mathcal X\) be an algebraic stack and \(\mathcal X' \subset \mathcal X\) a quasi-compact open substack. Suppose that we have a collection of open substacks \(\mathcal{X}_i \subset \mathcal X\) indexed by \(i \in I\) such that \(\mathcal{X}' \subset \bigcup_{i \in I} \mathcal{X}_i\), where we define the union as in Lemma 05UQ. Then there exists a finite subset \(I' \subset I\) such that \(\mathcal{X}' \subset \bigcup_{i \in I'} \mathcal{X}_i\).
Proof
Since \(\mathcal X\) is algebraic, there exists a scheme \(U\) with a surjective smooth morphism \(U \to \mathcal X\). Let \(U_i \subset U\) be the open subscheme representing \(\mathcal{X}_i \times_{\mathcal X} U\) and \(U' \subset U\) the open subscheme representing \(\mathcal{X}' \times_{\mathcal X} U\). By hypothesis, \(U'\subset \bigcup_{i\in I} U_i\). From the proof of Lemma 04YC, there is a quasi-compact open \(V \subset U'\) such that \(V \to \mathcal{X}'\) is a surjective smooth morphism. Therefore there exists a finite subset \(I' \subset I\) such that \(V \subset \bigcup_{i \in I'} U_i\). We claim that \(\mathcal{X}' \subset \bigcup_{i \in I'} \mathcal{X}_i\). Take \(x \in \Ob(\mathcal{X}'_T)\) for \(T \in \Ob((\Sch/S)_{fppf})\). Since \(\mathcal{X}' \to \mathcal{X}\) is a monomorphism, we have cartesian squares \[\xymatrix{ V \times_\mathcal{X} T \ar[r] \ar[d] & T \ar[d]^x \ar@{=}[r] & T \ar[d]^x \\ V \ar[r] & \mathcal{X}' \ar[r] & \mathcal X }\] By base change, \(V \times_{\mathcal X} T \to T\) is surjective. Therefore \(\bigcup_{i \in I'} U_i \times_{\mathcal X} T \to T\) is also surjective. Let \(T_i \subset T\) be the open subscheme representing \(\mathcal{X}_i \times_{\mathcal X} T\). By a formal argument, we have a Cartesian square \[\xymatrix{ U_i \times_{\mathcal{X}_i} T_i \ar[r] \ar[d] & U \times_{\mathcal X} T \ar[d] \\ T_i \ar[r] & T }\] where the vertical arrows are surjective by base change. Since \(U_i \times_{\mathcal{X}_i} T_i \simeq U_i \times_{\mathcal X} T\), we find that \(\bigcup_{i \in I'} T_i = T\). Hence \(x\) is an object of \((\bigcup_{i\in I'} \mathcal{X}_i)_T\) by definition of the union. Observe that the inclusion \(\mathcal{X}' \subset \bigcup_{i \in I'} \mathcal{X}_i\) is automatically an open substack.
Lemma
Let \(\mathcal X\) be an algebraic stack. Let \(\mathcal{X}_i\), \(i \in I\) be a set of open substacks of \(\mathcal{X}\). Assume
\(\mathcal{X} = \bigcup_{i \in I} \mathcal{X}_i\), and
each \(\mathcal{X}_i\) is an algebraic space.
Then \(\mathcal{X}\) is an algebraic space.
Proof
Apply Stacks, Lemma 05W9 to the morphism \(\coprod_{i \in I} \mathcal{X}_i \to \mathcal{X}\) and the morphism \(\text{id} : \mathcal{X} \to \mathcal{X}\) to see that \(\mathcal{X}\) is a stack in setoids. Hence \(\mathcal{X}\) is an algebraic space, see Algebraic Stacks, Proposition 04SZ.
Lemma
Let \(\mathcal X\) be an algebraic stack. Let \(\mathcal{X}_i\), \(i \in I\) be a set of open substacks of \(\mathcal{X}\). Assume
\(\mathcal{X} = \bigcup_{i \in I} \mathcal{X}_i\), and
each \(\mathcal{X}_i\) is a scheme
Then \(\mathcal{X}\) is a scheme.
Proof
By Lemma 05WE we see that \(\mathcal{X}\) is an algebraic space. Since any algebraic space has a largest open subspace which is a scheme, see Properties of Spaces, Lemma 03JH we see that \(\mathcal{X}\) is a scheme.
The following lemma is the analogue of More on Groupoids, Lemma 04LH.
Lemma
Let \(\mathcal{P}, \mathcal{Q}, \mathcal{R}\) be properties of morphisms of algebraic spaces. Assume
\(\mathcal{P}, \mathcal{Q}, \mathcal{R}\) are fppf local on the target and stable under arbitrary base change,
\(\text{smooth} \Rightarrow \mathcal{R}\),
for any morphism \(f : X \to Y\) which has \(\mathcal{Q}\) there exists a largest open subspace \(W(\mathcal{P}, f) \subset X\) such that \(f|_{W(\mathcal{P}, f)}\) has \(\mathcal{P}\), and
for any morphism \(f : X \to Y\) which has \(\mathcal{Q}\), and any morphism \(Y' \to Y\) which has \(\mathcal{R}\) we have \(Y' \times_Y W(\mathcal{P}, f) = W(\mathcal{P}, f')\), where \(f' : X_{Y'} \to Y'\) is the base change of \(f\).
Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks representable by algebraic spaces. Assume \(f\) has \(\mathcal{Q}\). Then
there exists a largest open substack \(\mathcal{X}' \subset \mathcal{X}\) such that \(f|_{\mathcal{X}'}\) has \(\mathcal{P}\), and
if \(\mathcal{Z} \to \mathcal{Y}\) is a morphism of algebraic stacks representable by algebraic spaces which has \(\mathcal{R}\) then \(\mathcal{Z} \times_\mathcal{Y} \mathcal{X}'\) is the largest open substack of \(\mathcal{Z} \times_\mathcal{Y} \mathcal{X}\) over which the base change \(\text{id}_\mathcal{Z} \times f\) has property \(\mathcal{P}\).
Proof
Choose a scheme \(V\) and a surjective smooth morphism \(V \to \mathcal{Y}\). Set \(U = V \times_\mathcal{Y} \mathcal{X}\) and let \(f' : U \to V\) be the base change of \(f\). The morphism of algebraic spaces \(f' : U \to V\) has property \(\mathcal{Q}\). Thus we obtain the open \(W(\mathcal{P}, f') \subset U\) by assumption (3). Note that \(U \times_\mathcal{X} U = (V \times_\mathcal{Y} V) \times_\mathcal{Y} \mathcal{X}\) hence the morphism \(f'' : U \times_\mathcal{X} U \to V \times_\mathcal{Y} V\) is the base change of \(f\) via either projection \(V \times_\mathcal{Y} V \to V\). By our choice of \(V\) these projections are smooth, hence have property \(\mathcal{R}\) by (2). Thus by (4) we see that the inverse images of \(W(\mathcal{P}, f')\) under the two projections \(\text{pr}_i : U \times_\mathcal{X} U \to U\) agree. In other words, \(W(\mathcal{P}, f')\) is an \(R\)-invariant subspace of \(U\) (where \(R = U \times_\mathcal{X} U\)). Let \(\mathcal{X}'\) be the open substack of \(\mathcal{X}\) corresponding to \(W(\mathcal{P}, f)\) via Lemma 0505. By construction \(W(\mathcal{P}, f') = \mathcal{X}' \times_\mathcal{Y} V\) hence \(f|_{\mathcal{X}'}\) has property \(\mathcal{P}\) by Lemma 04XD. Also, \(\mathcal{X}'\) is the largest open substack such that \(f|_{\mathcal{X}'}\) has \(\mathcal{P}\) as the same maximality holds for \(W(\mathcal{P}, f)\). This proves (A).
Finally, if \(\mathcal{Z} \to \mathcal{Y}\) is a morphism of algebraic stacks representable by algebraic spaces which has \(\mathcal{R}\) then we set \(T = V \times_\mathcal{Y} \mathcal{Z}\) and we see that \(T \to V\) is a morphism of algebraic spaces having property \(\mathcal{R}\). Set \(f'_T : T \times_V U \to T\) the base change of \(f'\). By (4) again we see that \(W(\mathcal{P}, f'_T)\) is the inverse image of \(W(\mathcal{P}, f)\) in \(T \times_V U\). This implies (B); some details omitted.
Remark
Warning: Lemma 06M3 should be used with care. For example, it applies to \(\mathcal{P}=\)“flat”, \(\mathcal{Q}=\)“empty”, and \(\mathcal{R}=\)“flat and locally of finite presentation”. But given a morphism of algebraic spaces \(f : X \to Y\) the largest open subspace \(W \subset X\) such that \(f|_W\) is flat is not the set of points where \(f\) is flat!
Remark
Notwithstanding the warning in Remark 06M4 there are some cases where Lemma 06M3 can be used without causing ambiguity. We give a list. In each case we omit the verification of assumptions (1) and (2) and we give references which imply (3) and (4). Here is the list:
\(\mathcal{Q} =\)“locally of finite type”, \(\mathcal{R} = \emptyset\), and \(\mathcal{P} =\)“relative dimension \(\leq d\)”. See Morphisms of Spaces, Definition 06LR and Morphisms of Spaces, Lemmas 04NT and 04NS.
\(\mathcal{Q} =\)“locally of finite type”, \(\mathcal{R} = \emptyset\), and \(\mathcal{P} =\)“locally quasi-finite”. This is the case \(d = 0\) of the previous item, see Morphisms of Spaces, Lemma 04NV. On the other hand, properties (3) and (4) are spelled out in Morphisms of Spaces, Lemma 04NW.
\(\mathcal{Q} =\)“locally of finite type”, \(\mathcal{R} = \emptyset\), and \(\mathcal{P} =\)“unramified”. This is Morphisms of Spaces, Lemma 05W2.
\(\mathcal{Q} =\)“locally of finite presentation”, \(\mathcal{R} =\)“flat and locally of finite presentation”, and \(\mathcal{P} =\)“flat”. See More on Morphisms of Spaces, Theorem 05WV and Lemma 05WW. Note that here \(W(\mathcal{P}, f)\) is always exactly the set of points where the morphism \(f\) is flat because we only consider this open when \(f\) has \(\mathcal{Q}\) (see loc.cit.).
\(\mathcal{Q} =\)“locally of finite presentation”, \(\mathcal{R} =\)“flat and locally of finite presentation”, and \(\mathcal{P}=\)“étale”. This follows on combining (06M8) and (06M9) because an unramified morphism which is flat and locally of finite presentation is étale, see Morphisms of Spaces, Lemma 06LU.
Add more here as needed (compare with the longer list at More on Groupoids, Remark 047W).
Reduced algebraic stacks
We have already defined reduced algebraic stacks in Section 04YE.
Lemma
Let \(\mathcal{X}\) be an algebraic stack. Let \(T \subset |\mathcal{X}|\) be a closed subset. There exists a unique closed substack \(\mathcal{Z} \subset \mathcal{X}\) with the following properties: (a) we have \(|\mathcal{Z}| = T\), and (b) \(\mathcal{Z}\) is reduced.
Proof
Let \(U \to \mathcal{X}\) be a surjective smooth morphism, where \(U\) is an algebraic space. Set \(R = U \times_\mathcal{X} U\), so that there is a presentation \([U/R] \to \mathcal{X}\), see Algebraic Stacks, Lemma 04T5. As usual we denote \(s, t : R \to U\) the two smooth projection morphisms. By Lemma 04XJ we see that \(T\) corresponds to a closed subset \(T' \subset |U|\) such that \(|s|^{-1}(T') = |t|^{-1}(T')\). Let \(Z \subset U\) be the reduced induced algebraic space structure on \(T'\), see Properties of Spaces, Definition 047X. The fibre products \(Z \times_{U, t} R\) and \(R \times_{s, U} Z\) are closed subspaces of \(R\) (Spaces, Lemma 02YW). The projections \(Z \times_{U, t} R \to Z\) and \(R \times_{s, U} Z \to Z\) are smooth by Morphisms of Spaces, Lemma 03ZE. Thus as \(Z\) is reduced, it follows that \(Z \times_{U, t} R\) and \(R \times_{s, U} Z\) are reduced, see Remark 04YH. Since \[|Z \times_{U, t} R| = |t|^{-1}(T') = |s|^{-1}(T') = R \times_{s, U} Z\] we conclude from the uniqueness in Properties of Spaces, Lemma 03IQ that \(Z \times_{U, t} R = R \times_{s, U} Z\). Hence \(Z\) is an \(R\)-invariant closed subspace of \(U\). By the correspondence of Lemma 0507 we obtain a closed substack \(\mathcal{Z} \subset \mathcal{X}\) with \(Z = \mathcal{Z} \times_\mathcal{X} U\). Then \([Z/R_Z] \to \mathcal{Z}\) is a presentation (Lemma 0505). Then \(|\mathcal{Z}| = |Z|/|R_Z| = |T'|/\sim\) is the given closed subset \(T\). We omit the proof of unicity.
Lemma
Let \(\mathcal{X}\) be an algebraic stack. If \(\mathcal{X}' \subset \mathcal{X}\) is a closed substack, \(\mathcal{X}\) is reduced and \(|\mathcal{X}'| = |\mathcal{X}|\), then \(\mathcal{X}' = \mathcal{X}\).
Proof
Choose a presentation \([U/R] \to \mathcal{X}\) with \(U\) a scheme. As \(\mathcal{X}\) is reduced, we see that \(U\) is reduced (by definition of reduced algebraic stacks). By Lemma 0507 \(\mathcal{X}'\) corresponds to an \(R\)-invariant closed subscheme \(Z \subset U\). But now \(|Z| \subset |U|\) is the inverse image of \(|\mathcal{X}'|\), and hence \(|Z| = |U|\). Hence \(Z\) is a closed subscheme of \(U\) whose underlying sets of points agree. By Schemes, Lemma 0356 the map \(\text{id}_U : U \to U\) factors through \(Z \to U\), and hence \(Z = U\), i.e., \(\mathcal{X}' = \mathcal{X}\).
Lemma
Let \(\mathcal{X}\), \(\mathcal{Y}\) be algebraic stacks. Let \(\mathcal{Z} \subset \mathcal{X}\) be a closed substack Assume \(\mathcal{Y}\) is reduced. A morphism \(f : \mathcal{Y} \to \mathcal{X}\) factors through \(\mathcal{Z}\) if and only if \(f(|\mathcal{Y}|) \subset |\mathcal{Z}|\).
Proof
Assume \(f(|\mathcal{Y}|) \subset |\mathcal{Z}|\). Consider \(\mathcal{Y} \times_\mathcal{X} \mathcal{Z} \to \mathcal{Y}\). There is an equivalence \(\mathcal{Y} \times_\mathcal{X} \mathcal{Z} \to \mathcal{Y}'\) where \(\mathcal{Y}'\) is a closed substack of \(\mathcal{Y}\), see Lemmas 0501 and 0506. Using Lemmas 04XH, 0500, and 0504 we see that \(|\mathcal{Y}'| = |\mathcal{Y}|\). Hence we have reduced the lemma to Lemma 050A.
Definition
Let \(\mathcal{X}\) be an algebraic stack. Let \(Z \subset |\mathcal{X}|\) be a closed subset. An algebraic stack structure on \(Z\) is given by a closed substack \(\mathcal{Z}\) of \(\mathcal{X}\) with \(|\mathcal{Z}|\) equal to \(Z\). The reduced induced algebraic stack structure on \(Z\) is the one constructed in Lemma 0509. The reduction \(\mathcal{X}_{red}\) of \(\mathcal{X}\) is the reduced induced algebraic stack structure on \(|\mathcal{X}|\).
In fact we can use this to define the reduced induced algebraic stack structure on a locally closed subset.
Remark
Let \(X\) be an algebraic stack. Let \(T \subset |\mathcal{X}|\) be a locally closed subset. Let \(\partial T\) be the boundary of \(T\) in the topological space \(|\mathcal{X}|\). In a formula \[\partial T = \overline{T} \setminus T.\] Let \(\mathcal{U} \subset \mathcal{X}\) be the open substack of \(X\) with \(|\mathcal{U}| = |\mathcal{X}| \setminus \partial T\), see Lemma 06FJ. Let \(\mathcal{Z}\) be the reduced closed substack of \(\mathcal{U}\) with \(|\mathcal{Z}| = T\) obtained by taking the reduced induced closed subspace structure, see Definition 050C. By construction \(\mathcal{Z} \to \mathcal{U}\) is a closed immersion of algebraic stacks and \(\mathcal{U} \to \mathcal{X}\) is an open immersion, hence \(\mathcal{Z} \to \mathcal{X}\) is an immersion of algebraic stacks by Lemma 0502. Note that \(\mathcal{Z}\) is a reduced algebraic stack and that \(|\mathcal{Z}| = T\) as subsets of \(|X|\). We sometimes say \(\mathcal{Z}\) is the reduced induced substack structure on \(T\).
Residual gerbes
In the Stacks project we would like to define the residual gerbe of an algebraic stack \(\mathcal{X}\) at a point \(x \in |\mathcal{X}|\) to be a monomorphism of algebraic stacks \(m_x : \mathcal{Z}_x \to \mathcal{X}\) where \(\mathcal{Z}_x\) is a reduced algebraic stack having a unique point which is mapped by \(m_x\) to \(x\). It turns out that there are many issues with this notion; existence is not clear in general and neither is uniqueness. We resolve the uniqueness issue by imposing a slightly stronger condition on the algebraic stacks \(\mathcal{Z}_x\). We discuss this in more detail by working through a few simple lemmas regarding reduced algebraic stacks having a unique point.
Lemma
Let \(\mathcal{Z}\) be an algebraic stack. Let \(k\) be a field and let \(\Spec(k) \to \mathcal{Z}\) be surjective and flat. Then any morphism \(\Spec(k') \to \mathcal{Z}\) where \(k'\) is a field is surjective and flat.
Proof
Consider the fibre square \[\xymatrix{ T \ar[d] \ar[r] & \Spec(k) \ar[d] \\ \Spec(k') \ar[r] & \mathcal{Z} }\] Note that \(T \to \Spec(k')\) is flat and surjective hence \(T\) is not empty. On the other hand \(T \to \Spec(k)\) is flat as \(k\) is a field. Hence \(T \to \mathcal{Z}\) is flat and surjective. It follows from Morphisms of Spaces, Lemma 05VY (via the discussion in Section 04XB) that \(\Spec(k') \to \mathcal{Z}\) is flat. It is clear that it is surjective as by assumption \(|\mathcal{Z}|\) is a singleton.
Lemma
Let \(\mathcal{Z}\) be an algebraic stack. The following are equivalent
\(\mathcal{Z}\) is reduced and \(|\mathcal{Z}|\) is a singleton,
there exists a surjective flat morphism \(\Spec(k) \to \mathcal{Z}\) where \(k\) is a field, and
there exists a locally of finite type, surjective, flat morphism \(\Spec(k) \to \mathcal{Z}\) where \(k\) is a field.
Proof
Assume (1). Let \(W\) be a scheme and let \(W \to \mathcal{Z}\) be a surjective smooth morphism. Then \(W\) is a reduced scheme. Let \(\eta \in W\) be a generic point of an irreducible component of \(W\). Since \(W\) is reduced we have \(\mathcal{O}_{W, \eta} = \kappa(\eta)\). It follows that the canonical morphism \(\eta = \Spec(\kappa(\eta)) \to W\) is flat. We see that the composition \(\eta \to \mathcal{Z}\) is flat (see Morphisms of Spaces, Lemma 03MN). It is also surjective as \(|\mathcal{Z}|\) is a singleton. In other words (2) holds.
Assume (2). Let \(W\) be a scheme and let \(W \to \mathcal{Z}\) be a surjective smooth morphism. Choose a field \(k\) and a surjective flat morphism \(\Spec(k) \to \mathcal{Z}\). Then \(W \times_\mathcal{Z} \Spec(k)\) is an algebraic space smooth over \(k\), hence regular (see Spaces over Fields, Lemma 06M1) and in particular reduced. Since \(W \times_\mathcal{Z} \Spec(k) \to W\) is surjective and flat we conclude that \(W\) is reduced (Descent on Spaces, Lemma 06MI). In other words (1) holds.
It is clear that (3) implies (2). Finally, assume (2). Pick a nonempty affine scheme \(W\) and a smooth morphism \(W \to \mathcal{Z}\). Pick a closed point \(w \in W\) and set \(k = \kappa(w)\). The composition \[\Spec(k) \xrightarrow{w} W \longrightarrow \mathcal{Z}\] is locally of finite type by Morphisms of Spaces, Lemmas 03XG and 06MH. It is also flat and surjective by Lemma 06MM. Hence (3) holds.
The following lemma singles out a slightly better class of singleton algebraic stacks than the preceding lemma.
Lemma
Let \(\mathcal{Z}\) be an algebraic stack. The following are equivalent
\(\mathcal{Z}\) is reduced, locally Noetherian, and \(|\mathcal{Z}|\) is a singleton, and
there exists a locally finitely presented, surjective, flat morphism \(\Spec(k) \to \mathcal{Z}\) where \(k\) is a field.
Proof
Assume (2) holds. By Lemma 06MN we see that \(\mathcal{Z}\) is reduced and \(|\mathcal{Z}|\) is a singleton. Let \(W\) be a scheme and let \(W \to \mathcal{Z}\) be a surjective smooth morphism. Choose a field \(k\) and a locally finitely presented, surjective, flat morphism \(\Spec(k) \to \mathcal{Z}\). Then \(W \times_\mathcal{Z} \Spec(k)\) is an algebraic space smooth over \(k\), hence locally Noetherian (see Morphisms of Spaces, Lemma 04ZK). Since \(W \times_\mathcal{Z} \Spec(k) \to W\) is flat, surjective, and locally of finite presentation, we see that \(\{W \times_\mathcal{Z} \Spec(k) \to W\}\) is an fppf covering and we conclude that \(W\) is locally Noetherian (Descent on Spaces, Lemma 06MJ). In other words (1) holds.
Assume (1). Pick a nonempty affine scheme \(W\) and a smooth morphism \(W \to \mathcal{Z}\). Pick a closed point \(w \in W\) and set \(k = \kappa(w)\). Because \(W\) is locally Noetherian the morphism \(w : \Spec(k) \to W\) is of finite presentation, see Morphisms, Lemma 01TV. Hence the composition \[\Spec(k) \xrightarrow{w} W \longrightarrow \mathcal{Z}\] is locally of finite presentation by Morphisms of Spaces, Lemmas 03XQ and 04AJ. It is also flat and surjective by Lemma 06MM. Hence (2) holds.
Lemma
Let \(\mathcal{Z}' \to \mathcal{Z}\) be a monomorphism of algebraic stacks. Assume there exists a field \(k\) and a locally finitely presented, surjective, flat morphism \(\Spec(k) \to \mathcal{Z}\). Then either \(\mathcal{Z}'\) is empty or \(\mathcal{Z}' \to \mathcal{Z}\) is an equivalence.
Proof
We may assume that \(\mathcal{Z}'\) is nonempty. In this case the fibre product \(T = \mathcal{Z}' \times_\mathcal{Z} \Spec(k)\) is nonempty, see Lemma 04XH. Now \(T\) is an algebraic space and the projection \(T \to \Spec(k)\) is a monomorphism. Hence \(T = \Spec(k)\), see Morphisms of Spaces, Lemma 06MG. We conclude that \(\Spec(k) \to \mathcal{Z}\) factors through \(\mathcal{Z}'\). Suppose the morphism \(z : \Spec(k) \to \mathcal{Z}\) is given by the object \(\xi\) over \(\Spec(k)\). We have just seen that \(\xi\) is isomorphic to an object \(\xi'\) of \(\mathcal{Z}'\) over \(\Spec(k)\). Since \(z\) is surjective, flat, and locally of finite presentation we see that every object of \(\mathcal{Z}\) over any scheme is fppf locally isomorphic to a pullback of \(\xi\), hence also to a pullback of \(\xi'\). By descent of objects for stacks in groupoids this implies that \(\mathcal{Z}' \to \mathcal{Z}\) is essentially surjective (as well as fully faithful, see Lemma 04ZZ). Hence we win.
Lemma
Let \(\mathcal{Z}\) be an algebraic stack. Assume \(\mathcal{Z}\) satisfies the equivalent conditions of Lemma 06MN. Then there exists a unique strictly full subcategory \(\mathcal{Z}' \subset \mathcal{Z}\) such that \(\mathcal{Z}'\) is an algebraic stack which satisfies the equivalent conditions of Lemma 06MP. The inclusion morphism \(\mathcal{Z}' \to \mathcal{Z}\) is a monomorphism of algebraic stacks.
Proof
The last part is immediate from the first part and Lemma 04ZZ. Pick a field \(k\) and a morphism \(\Spec(k) \to \mathcal{Z}\) which is surjective, flat, and locally of finite type. Set \(U = \Spec(k)\) and \(R = U \times_\mathcal{Z} U\). The projections \(s, t : R \to U\) are locally of finite type. 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 \[f : \mathcal{Z}' \longrightarrow \mathcal{Z}\] which is fully faithful. Hence this morphism is representable by algebraic spaces, see Algebraic Stacks, Lemma 04Y5 and a monomorphism, see Lemma 04ZZ. By Criteria for Representability, Lemma 06FH the morphism \(U \to \mathcal{Z}'\) is surjective, flat, and locally of finite presentation. Hence \(\mathcal{Z}'\) is an algebraic stack which satisfies the equivalent conditions of Lemma 06MP. By Algebraic Stacks, Lemma 03YQ we may replace \(\mathcal{Z}'\) by its essential image in \(\mathcal{Z}\). Hence we have proved all the assertions of the lemma except for the uniqueness of \(\mathcal{Z}' \subset \mathcal{Z}\). Suppose that \(\mathcal{Z}'' \subset \mathcal{Z}\) is a second such algebraic stack. Then the projections \[\mathcal{Z}' \longleftarrow \mathcal{Z}' \times_\mathcal{Z} \mathcal{Z}'' \longrightarrow \mathcal{Z}''\] are monomorphisms. The algebraic stack in the middle is nonempty by Lemma 04XH. Hence the two projections are isomorphisms by Lemma 06MQ and we win.
Example
Here is an example where the morphism constructed in Lemma 06MR isn’t an isomorphism. This example shows that imposing that residual gerbes are locally Noetherian is necessary in Definition 06MU. In fact, the example is even an algebraic space! Let \(\text{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})\) be the absolute Galois group of \(\mathbf{Q}\) with the pro-finite topology. Let \[U = \Spec(\overline{\mathbf{Q}}) \times_{\Spec(\mathbf{Q})} \Spec(\overline{\mathbf{Q}}) = \text{Gal}(\overline{\mathbf{Q}}/\mathbf{Q}) \times \Spec(\overline{\mathbf{Q}})\] (we omit a precise explanation of the meaning of the last equal sign). Let \(G\) denote the absolute Galois group \(\text{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})\) with the discrete topology viewed as a constant group scheme over \(\Spec(\overline{\mathbf{Q}})\), see Groupoids, Example 03YW. Then \(G\) acts freely and transitively on \(U\). Let \(X = U/G\), see Spaces, Definition 02Z3. Then \(X\) is a non-noetherian reduced algebraic space with exactly one point. Furthermore, \(X\) has a (locally) finite type point: \[x : \Spec(\overline{\mathbf{Q}}) \longrightarrow U \longrightarrow X\] Indeed, every point of \(U\) is actually closed! As \(X\) is an algebraic space over \(\overline{\mathbf{Q}}\) it follows that \(x\) is a monomorphism. So \(x\) is the morphism constructed in Lemma 06MR but \(x\) is not an isomorphism. In fact \(\Spec(\overline{\mathbf{Q}}) \to X\) is the residual gerbe of \(X\) at \(x\).
It will turn out later that under mild assumptions on the algebraic stack \(\mathcal{X}\) the equivalent conditions of the following lemma are satisfied for every point \(x \in |\mathcal{X}|\) (see Morphisms of Stacks, Section 06UH).
Lemma
Let \(\mathcal{X}\) be an algebraic stack. Let \(x \in |\mathcal{X}|\) be a point. The following are equivalent
there exists an algebraic stack \(\mathcal{Z}\) and a monomorphism \(\mathcal{Z} \to \mathcal{X}\) such that \(|\mathcal{Z}|\) is a singleton and such that the image of \(|\mathcal{Z}|\) in \(|\mathcal{X}|\) is \(x\),
there exists a reduced algebraic stack \(\mathcal{Z}\) and a monomorphism \(\mathcal{Z} \to \mathcal{X}\) such that \(|\mathcal{Z}|\) is a singleton and such that the image of \(|\mathcal{Z}|\) in \(|\mathcal{X}|\) is \(x\),
there exists an algebraic stack \(\mathcal{Z}\), a monomorphism \(f : \mathcal{Z} \to \mathcal{X}\), and a surjective flat morphism \(z : \Spec(k) \to \mathcal{Z}\) where \(k\) is a field such that \(x = f(z)\).
Moreover, if these conditions hold, then there exists a unique strictly full subcategory \(\mathcal{Z}_x \subset \mathcal{X}\) such that \(\mathcal{Z}_x\) is a reduced, locally Noetherian algebraic stack and \(|\mathcal{Z}_x|\) is a singleton which maps to \(x\) via the map \(|\mathcal{Z}_x| \to |\mathcal{X}|\).
Proof
If \(\mathcal{Z} \to \mathcal{X}\) is as in (1), then \(\mathcal{Z}_{red} \to \mathcal{X}\) is as in (2). (See Section 0508 for the notion of the reduction of an algebraic stack.) Hence (1) implies (2). It is immediate that (2) implies (1). The equivalence of (2) and (3) is immediate from Lemma 06MN.
At this point we’ve seen the equivalence of (1) – (3). Pick a monomorphism \(f : \mathcal{Z} \to \mathcal{X}\) as in (2). Note that this implies that \(f\) is fully faithful, see Lemma 04ZZ. Denote \(\mathcal{Z}' \subset \mathcal{X}\) the essential image of the functor \(f\). Then \(f : \mathcal{Z} \to \mathcal{Z}'\) is an equivalence and hence \(\mathcal{Z}'\) is an algebraic stack, see Algebraic Stacks, Lemma 03YQ. Apply Lemma 06MR to get a strictly full subcategory \(\mathcal{Z}_x \subset \mathcal{Z}'\) as in the statement of the lemma. This proves all the statements of the lemma except for uniqueness.
In order to prove the uniqueness suppose that \(\mathcal{Z}_x \subset \mathcal{X}\) and \(\mathcal{Z}'_x \subset \mathcal{X}\) are two strictly full subcategories as in the statement of the lemma. Then the projections \[\mathcal{Z}'_x \longleftarrow \mathcal{Z}'_x \times_\mathcal{X} \mathcal{Z}_x \longrightarrow \mathcal{Z}_x\] are monomorphisms. The algebraic stack in the middle is nonempty by Lemma 04XH. Hence the two projections are isomorphisms by Lemma 06MQ and we win.
Having explained the above we can now make the following definition.
Definition
Let \(\mathcal{X}\) be an algebraic stack. Let \(x \in |\mathcal{X}|\).
We say the residual gerbe of \(\mathcal{X}\) at \(x\) exists if the equivalent conditions (1), (2), and (3) of Lemma 06MT hold.
If the residual gerbe of \(\mathcal{X}\) at \(x\) exists, then the residual gerbe of \(\mathcal{X}\) at \(x\)1 is the strictly full subcategory \(\mathcal{Z}_x \subset \mathcal{X}\) constructed in Lemma 06MT.
In particular we know that \(\mathcal{Z}_x\) (if it exists) is a locally Noetherian, reduced algebraic stack and that there exists a field and a surjective, flat, locally finitely presented morphism \[\Spec(k) \longrightarrow \mathcal{Z}_x.\] We will see in Morphisms of Stacks, Lemma 06QK that \(\mathcal{Z}_x\) is a gerbe. Existence of residual gerbes is discussed in Morphisms of Stacks, Section 06UH.
Example
Let \(X\) be a scheme and let \(x \in X\) be a point. Then the monomorphism \(x \to X\) is the residual gerbe of \(X\) at \(x\) where we, as usual, identify \(x\) with the scheme \(x = \Spec(\kappa(x))\). If \(X\) is an algebraic space and \(x \in |X|\), then the residual gerbe at \(x\) (which is called the residual space) always exists, see Decent Spaces, Section 06QU.
The residual gerbe, if it exists, is a regular algebraic stack by the following lemma.
Lemma
A reduced, locally Noetherian algebraic stack \(\mathcal{Z}\) such that \(|\mathcal{Z}|\) is a singleton is regular.
Proof
Let \(W \to \mathcal{Z}\) be a surjective smooth morphism where \(W\) is a scheme. Let \(k\) be a field and let \(\Spec(k) \to \mathcal{Z}\) be surjective, flat, and locally of finite presentation (see Lemma 06MP). The algebraic space \(T = W \times_\mathcal{Z} \Spec(k)\) is smooth over \(k\) in particular regular, see Spaces over Fields, Lemma 06M1. Since \(T \to W\) is locally of finite presentation, flat, and surjective it follows that \(W\) is regular, see Descent on Spaces, Lemma 06MK. By definition this means that \(\mathcal{Z}\) is regular.
Lemma
Let \(\mathcal{X}\) be an algebraic stack. Let \(x \in |\mathcal{X}|\). Assume that the residual gerbe \(\mathcal{Z}_x\) of \(\mathcal{X}\) exists. Let \(f : \Spec(K) \to \mathcal{X}\) be a morphism where \(K\) is a field in the equivalence class of \(x\). Then \(f\) factors through the inclusion morphism \(\mathcal{Z}_x \to \mathcal{X}\).
Proof
Choose a field \(k\) and a surjective flat locally finite presentation morphism \(\Spec(k) \to \mathcal{Z}_x\). Set \(T = \Spec(K) \times_\mathcal{X} \mathcal{Z}_x\). By Lemma 04XH we see that \(T\) is nonempty. As \(\mathcal{Z}_x \to \mathcal{X}\) is a monomorphism we see that \(T \to \Spec(K)\) is a monomorphism. Hence by Morphisms of Spaces, Lemma 06MG we see that \(T = \Spec(K)\) which proves the lemma.
Lemma
Let \(\mathcal{X}\) be an algebraic stack. Let \(x \in |\mathcal{X}|\). Let \(\mathcal{Z}\) be an algebraic stack satisfying the equivalent conditions of Lemma 06MP and let \(\mathcal{Z} \to \mathcal{X}\) be a monomorphism such that the image of \(|\mathcal{Z}| \to |\mathcal{X}|\) is \(x\). Then the residual gerbe \(\mathcal{Z}_x\) of \(\mathcal{X}\) at \(x\) exists and \(\mathcal{Z} \to \mathcal{X}\) factors as \(\mathcal{Z} \to \mathcal{Z}_x \to \mathcal{X}\) where the first arrow is an equivalence.
Proof
Let \(\mathcal{Z}_x \subset \mathcal{X}\) be the full subcategory corresponding to the essential image of the functor \(\mathcal{Z} \to \mathcal{X}\). Then \(\mathcal{Z} \to \mathcal{Z}_x\) is an equivalence, hence \(\mathcal{Z}_x\) is an algebraic stack, see Algebraic Stacks, Lemma 03YQ. Since \(\mathcal{Z}_x\) inherits all the properties of \(\mathcal{Z}\) from this equivalence it is clear from the uniqueness in Lemma 06MT that \(\mathcal{Z}_x\) is the residual gerbe of \(\mathcal{X}\) at \(x\).
Lemma
Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. Let \(x \in |\mathcal{X}|\) with image \(y \in |\mathcal{Y}|\). If the residual gerbes \(\mathcal{Z}_x \subset \mathcal{X}\) and \(\mathcal{Z}_y \subset \mathcal{Y}\) of \(x\) and \(y\) exist, then \(f\) induces a commutative diagram \[\xymatrix{ \mathcal{X} \ar[d]_f & \mathcal{Z}_x \ar[l] \ar[d] \\ \mathcal{Y} & \mathcal{Z}_y \ar[l] }\]
Proof
Choose a field \(k\) and a surjective, flat, locally finitely presented morphism \(\Spec(k) \to \mathcal{Z}_x\). The morphism \(\Spec(k) \to \mathcal{Y}\) factors through \(\mathcal{Z}_y\) by Lemma 06MW. Thus \(\mathcal{Z}_x \times_\mathcal{Y} \mathcal{Z}_y\) is a nonempty substack of \(\mathcal{Z}_x\) hence equal to \(\mathcal{Z}_x\) by Lemma 06MQ.
Lemma
Let \(f : \mathcal{X} \to \mathcal{Y}\) be a morphism of algebraic stacks. Let \(x \in |\mathcal{X}|\) with image \(y \in |\mathcal{Y}|\). Assume the residual gerbes \(\mathcal{Z}_x \subset \mathcal{X}\) and \(\mathcal{Z}_y \subset \mathcal{Y}\) of \(x\) and \(y\) exist and that there exists a morphism \(\Spec(k) \to \mathcal{X}\) in the equivalence class of \(x\) such that \[\Spec(k) \times_\mathcal{X} \Spec(k) \longrightarrow \Spec(k) \times_\mathcal{Y} \Spec(k)\] is an isomorphism. Then \(\mathcal{Z}_x \to \mathcal{Z}_y\) is an isomorphism.
Proof
Let \(k'/k\) be an extension of fields. Then \[\Spec(k') \times_\mathcal{X} \Spec(k') \longrightarrow \Spec(k') \times_\mathcal{Y} \Spec(k')\] is the base change of the morphism in the lemma by the faithfully flat morphism \(\Spec(k' \otimes k') \to \Spec(k \otimes k)\). Thus the property described in the lemma is independent of the choice of the morphism \(\Spec(k) \to \mathcal{X}\) in the equivalence class of \(x\). Thus we may assume that \(\Spec(k) \to \mathcal{Z}_x\) is surjective, flat, and locally of finite presentation. In this situation we have \[\mathcal{Z}_x = [\Spec(k)/R]\] with \(R = \Spec(k) \times_\mathcal{X} \Spec(k)\). See proof of Lemma 06MR. Since also \(R = \Spec(k) \times_\mathcal{Y} \Spec(k)\) we conclude that the morphism \(\mathcal{Z}_x \to \mathcal{Z}_y\) of Lemma 0DTH is fully faithful by Algebraic Stacks, Lemma 04T4. We conclude for example by Lemma 06MX.
Dimension of a stack
We can define the dimension of an algebraic stack \(\mathcal{X}\) at a point \(x\), using the notion of dimension of an algebraic space at a point (Properties of Spaces, Definition 04N5). In the following lemma the output may be \(\infty\) either because \(\mathcal{X}\) is not quasi-compact or because we run into the phenomenon described in Examples, Section 02JC.
Lemma
Let \(\mathcal{X}\) be a locally Noetherian algebraic stack over a scheme \(S\). Let \(x \in |\mathcal{X}|\) be a point of \(\mathcal{X}\). Let \([U/R] \to \mathcal{X}\) be a presentation (Algebraic Stacks, Definition 04TI) where \(U\) is a scheme. Let \(u \in U\) be a point that maps to \(x\). Let \(e : U \to R\) be the “identity” map and let \(s : R \to U\) be the “source” map, which is a smooth morphism of algebraic spaces. Let \(R_u\) be the fiber of \(s : R \to U\) over \(u\). The element \[\dim_x(\mathcal{X}) = \dim_u(U) - \dim_{e(u)}(R_u) \in \mathbf{Z} \cup \infty\] is independent of the choice of presentation and the point \(u\) over \(x\).
Proof
Since \(R \to U\) is smooth, the scheme \(R_u\) is smooth over \(\kappa(u)\) and hence has finite dimension. On the other hand, the scheme \(U\) is locally Noetherian, but this does not guarantee that \(\dim_u(U)\) is finite. Thus the difference is an element of \(\mathbf{Z} \cup \{\infty\}\).
Let \([U'/R'] \to \mathcal{X}\) and \(u' \in U'\) be a second presentation where \(U'\) is a scheme and \(u'\) maps to \(x\). Consider the algebraic space \(P = U \times_\mathcal{X} U'\). By Lemma 04XH there exists a \(p \in |P|\) mapping to \(u\) and \(u'\). Since \(P \to U\) and \(P \to U'\) are smooth we see that \(\dim_p(P) = \dim_u(U) + \dim_p(P_u)\) and \(\dim_p(P) = \dim_{u'}(U') + \dim_p(P_{u'})\), see Morphisms of Spaces, Lemma 0AFI. Note that \[R'_{u'} = \Spec(\kappa(u')) \times_\mathcal{X} U' \quad\text{and}\quad P_u = \Spec(\kappa(u)) \times_\mathcal{X} U'\] Let us represent \(p \in |P|\) by a morphism \(\Spec(\Omega) \to P\). Since \(p\) maps to both \(u\) and \(u'\) it induces a \(2\)-morphism between the compositions \(\Spec(\Omega) \to \Spec(\kappa(u')) \to \mathcal{X}\) and \(\Spec(\Omega) \to \Spec(\kappa(u)) \to \mathcal{X}\) which in turn defines an isomorphism \[\Spec(\Omega) \times_{\Spec(\kappa(u'))} R'_{u'} \cong \Spec(\Omega) \times_{\Spec(\kappa(u))} P_u\] as algebraic spaces over \(\Spec(\Omega)\) mapping the \(\Omega\)-rational point \((1, e'(u'))\) to \((1, p)\) (some details omitted). We conclude that \[\dim_{e'(u')}(R'_{u'}) = \dim_p(P_u)\] by Morphisms of Spaces, Lemma 04NS. By symmetry we have \(\dim_{e(u)}(R_u) = \dim_p(P_{u'})\). Putting everything together we obtain the independence of choices.
We can use the lemma above to make the following definition.
Definition
Let \(\mathcal{X}\) be a locally Noetherian algebraic stack over a scheme \(S\). Let \(x \in |\mathcal{X}|\) be a point of \(\mathcal{X}\). Let \([U/R] \to \mathcal{X}\) be a presentation (Algebraic Stacks, Definition 04TI) where \(U\) is a scheme and let \(u \in U\) be a point that maps to \(x\). We define the dimension of \(\mathcal{X}\) at \(x\) to be the element \(\dim_x(\mathcal{X}) \in \mathbf{Z} \cup \infty\) such that \[\dim_x(\mathcal{X}) = \dim_u(U)-\dim_{e(u)}(R_u).\] with notation as in Lemma 0AFM.
The dimension of a stack at a point agrees with the usual notion when \(\mathcal{X}\) is a scheme (Topology, Definition 0055), or more generally when \(\mathcal{X}\) is a locally Noetherian algebraic space (Properties of Spaces, Definition 04N5).
Definition
Let \(S\) be a scheme. Let \(\mathcal{X}\) be a locally Noetherian algebraic stack over \(S\). The dimension \(\dim(\mathcal{X})\) of \(\mathcal{X}\) is defined to be \[\dim(\mathcal{X}) = \sup\nolimits_{x \in |\mathcal{X}|} \dim_x(\mathcal{X})\]
This definition of dimension agrees with the usual notion if \(\mathcal{X}\) is a scheme (Properties, Lemma 04MU) or an algebraic space (Properties of Spaces, Definition 04N6).
Remark
If \(\mathcal{X}\) is a nonempty stack of finite type over a field, then \(\dim(\mathcal{X})\) is an integer. For an arbitrary locally Noetherian algebraic stack \(\mathcal{X}\), \(\dim(\mathcal{X})\) is in \(Z\cup \{\pm \infty\}\), and \(\dim(\mathcal{X}) = -\infty\) if and only if \(\mathcal{X}\) is empty.
Example
Let \(X\) be a scheme of finite type over a field \(k\), and let \(G\) be a group scheme of finite type over \(k\) which acts on \(X\). Then the dimension of the quotient stack \([X/G]\) is equal to \(\dim(X)-\dim(G)\). In particular, the dimension of the classifying stack \(BG=[\Spec(k)/G]\) is \(-\dim(G)\). Thus the dimension of an algebraic stack can be a negative integer, in contrast to what happens for schemes or algebraic spaces.
Local irreducibility
We have defined the geometric number of branches of a scheme at a point in Properties, Section 0BQ1 and for an algebraic space at a point in Properties of Spaces, Section 06DJ. Let \(n \in \mathbf{N}\). For a local ring \(A\) set \[P_n(A) = \text{the number of geometric branches of }A\text{ is }n\] For a smooth ring map \(A \to B\) and a prime ideal \(\mathfrak q\) of \(B\) lying over \(\mathfrak p\) of \(A\) we have \[P_n(A_\mathfrak p) \Leftrightarrow P_n(B_\mathfrak q)\] by More on Algebra, Lemma 0DQ1. As in Properties of Spaces, Remark 0BBL we may use \(P_n\) to define an étale local property \(\mathcal{P}_n\) of germs \((U, u)\) of schemes by setting \(\mathcal{P}_n(U, u) = P_n(\mathcal{O}_{U, u})\). The corresponding property \(\mathcal{P}_n\) of an algebraic space \(X\) at a point \(x\) (see Properties of Spaces, Definition 04RC) is just the property “the number of geometric branches of \(X\) at \(x\) is \(n\)”, see Properties of Spaces, Definition 0DQ4. Moreover, the property \(\mathcal{P}_n\) is smooth local, see Descent, Definition 04N1. This follows either from the equivalence displayed above or More on Morphisms, Lemma 0DQ2. Thus Definition 04YJ applies and we obtain a notion for algebraic stacks at a point.
Definition
Let \(\mathcal{X}\) be an algebraic stack. Let \(x \in |\mathcal{X}|\).
The number of geometric branches of \(\mathcal{X}\) at \(x\) is either \(n \in \mathbf{N}\) if the equivalent conditions of Lemma 04YI hold for \(\mathcal{P}_n\) defined above, or else \(\infty\).
We say \(\mathcal{X}\) is geometrically unibranch at \(x\) if the number of geometric branches of \(\mathcal{X}\) at \(x\) is \(1\).
Finiteness conditions and points
This section is the analogue of Decent Spaces, Section 03JR for points of algebraic stacks.
Lemma
Let \(\mathcal{X}\) be an algebraic stack. Let \(x \in |\mathcal{X}|\) be a point. The following are equivalent
some morphism \(\Spec(k) \to \mathcal{X}\) in the equivalence class of \(x\) is quasi-compact, and
any morphism \(\Spec(k) \to \mathcal{X}\) in the equivalence class of \(x\) is quasi-compact.
Proof
Let \(\Spec(k) \to \mathcal{X}\) be in the equivalence class of \(x\). Let \(k'/k\) be a field extension. Then we have to show that \(\Spec(k) \to \mathcal{X}\) is quasi-compact if and only if \(\Spec(k') \to \mathcal{X}\) is quasi-compact. This follows from Morphisms of Spaces, Lemma 040W and the principle of Algebraic Stacks, Lemma 045F.
Sometimes people say that a point \(x \in |\mathcal{X}|\) satisfying the equivalent conditions of Lemma 0DTK is a “quasi-compact point”.
This clashes with [LM-B] in spirit, but not in fact. Namely, in Chapter 11 they associate to any point on any quasi-separated algebraic stack a gerbe (not necessarily algebraic) which they call the residual gerbe. We will see in Morphisms of Stacks, Lemma 06RD that on a quasi-separated algebraic stack every point has a residual gerbe in our sense which is then equivalent to theirs. For more information on this topic see [rydh_etale_devissage, Appendix B].↩︎