## 100.27 Morphisms of finite presentation

The property “locally of finite presentation” of morphisms of algebraic spaces is smooth local on the source-and-target, see Descent on Spaces, Remark 73.20.5. It is also stable under base change and fpqc local on the target, see Morphisms of Spaces, Lemma 66.28.3 and Descent on Spaces, Lemma 73.11.10. Hence, by Lemma 100.16.1 above, we may define what it means for a morphism of algebraic stacks to be locally of finite presentation as follows and it agrees with the already existing notion defined in Properties of Stacks, Section 99.3 when the morphism is representable by algebraic spaces.

Definition 100.27.1. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks.

1. We say $f$ locally of finite presentation if the equivalent conditions of Lemma 100.16.1 hold with $\mathcal{P} = \text{locally of finite presentation}$.

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

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

Lemma 100.27.2. The composition of finitely presented morphisms is of finite presentation. The same holds for morphisms which are locally of finite presentation.

Proof. Combine Remark 100.16.3 with Morphisms of Spaces, Lemma 66.28.2. $\square$

Lemma 100.27.3. A base change of a finitely presented morphism is of finite presentation. The same holds for morphisms which are locally of finite presentation.

Proof. Combine Remark 100.16.4 with Morphisms of Spaces, Lemma 66.28.3. $\square$

Lemma 100.27.4. A morphism which is locally of finite presentation is locally of finite type. A morphism of finite presentation is of finite type.

Proof. Combine Remark 100.16.5 with Morphisms of Spaces, Lemma 66.28.5. $\square$

Lemma 100.27.5. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks.

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

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

Proof. Assume $f : \mathcal{X} \to \mathcal{Y}$ locally of finite type and $\mathcal{Y}$ locally Noetherian. This means there exists a diagram as in Lemma 100.16.1 with $h$ locally of finite type and surjective vertical arrow $a$. By Morphisms of Spaces, Lemma 66.28.7 $h$ is locally of finite presentation. Hence $\mathcal{X} \to \mathcal{Y}$ is locally of finite presentation by definition. This proves (1). If $f$ is of finite type and quasi-separated then it is also quasi-compact and quasi-separated and (2) follows immediately. $\square$

Lemma 100.27.6. Let $f : \mathcal{X} \to \mathcal{Y}$ and $g : \mathcal{Y} \to \mathcal{Z}$ be morphisms of algebraic stacks If $g \circ f$ is locally of finite presentation and $g$ is locally of finite type, then $f$ is locally of finite presentation.

Proof. Choose an algebraic space $W$ and a surjective smooth morphism $W \to \mathcal{Z}$. Choose an algebraic space $V$ and a surjective smooth morphism $V \to \mathcal{Y} \times _\mathcal {Z} W$. Choose an algebraic space $U$ and a surjective smooth morphism $U \to \mathcal{X} \times _\mathcal {Y} V$. The lemma follows upon applying Morphisms of Spaces, Lemma 66.28.9 to the morphisms $U \to V \to W$. $\square$

Lemma 100.27.7. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks with diagonal $\Delta : \mathcal{X} \to \mathcal{X} \times _\mathcal {Y} \mathcal{X}$. If $f$ is locally of finite type then $\Delta$ is locally of finite presentation. If $f$ is quasi-separated and locally of finite type, then $\Delta$ is of finite presentation.

Proof. Note that $\Delta$ is a morphism over $\mathcal{X}$ (via the second projection). If $f$ is locally of finite type, then $\mathcal{X}$ is of finite presentation over $\mathcal{X}$ and $\text{pr}_2 : \mathcal{X} \times _\mathcal {Y} \mathcal{X} \to \mathcal{X}$ is locally of finite type by Lemma 100.17.3. Thus the first statement holds by Lemma 100.27.6. The second statement follows from the first and the definitions (because $f$ being quasi-separated means by definition that $\Delta _ f$ is quasi-compact and quasi-separated). $\square$

Proof. In view of Properties of Stacks, Definition 99.9.1 this follows from Morphisms of Spaces, Lemma 66.28.11. $\square$

Lemma 100.27.9. Let $P$ be a property of morphisms of algebraic spaces which is fppf local on the target and preserved by arbitrary base change. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks representable by algebraic spaces. Let $\mathcal{Z} \to \mathcal{Y}$ be a morphism of algebraic stacks which is surjective, flat, and locally of finite presentation. Set $\mathcal{W} = \mathcal{Z} \times _\mathcal {Y} \mathcal{X}$. Then

$(f\text{ has }P) \Leftrightarrow (\text{the projection }\mathcal{W} \to \mathcal{Z}\text{ has }P).$

For the meaning of this statement see Properties of Stacks, Section 99.3.

Proof. Choose an algebraic space $W$ and a morphism $W \to \mathcal{Z}$ which is surjective, flat, and locally of finite presentation. By Properties of Stacks, Lemma 99.5.2 and Lemmas 100.25.2 and 100.27.2 the composition $W \to \mathcal{Y}$ is also surjective, flat, and locally of finite presentation. Denote $V = W \times _\mathcal {Z} \mathcal{W} = V \times _\mathcal {Y} \mathcal{X}$. By Properties of Stacks, Lemma 99.3.3 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. $\square$

Lemma 100.27.10. Let $\mathcal{P}$ be a property of morphisms of algebraic spaces which is smooth local on the source-and-target and fppf local on the target. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $\mathcal{Z} \to \mathcal{Y}$ be a surjective, flat, locally finitely presented morphism of algebraic stacks. If the base change $\mathcal{Z} \times _\mathcal {Y} \mathcal{X} \to \mathcal{Z}$ has $\mathcal{P}$, then $f$ has $\mathcal{P}$.

Proof. Assume $\mathcal{Z} \times _\mathcal {Y} \mathcal{X} \to \mathcal{Z}$ has $\mathcal{P}$. Choose an algebraic space $W$ and a surjective smooth morphism $W \to \mathcal{Z}$. Observe that $W \times _\mathcal {Z} \mathcal{Z} \times _\mathcal {Y} \mathcal{X} = W \times _\mathcal {Y} \mathcal{X}$. Thus by the very definition of what it means for $\mathcal{Z} \times _\mathcal {Y} \mathcal{X} \to \mathcal{Z}$ to have $\mathcal{P}$ (see Definition 100.16.2 and Lemma 100.16.1) we see that $W \times _\mathcal {Y} \mathcal{X} \to W$ has $\mathcal{P}$. On the other hand, $W \to \mathcal{Z}$ is surjective, flat, and locally of finite presentation (Morphisms of Spaces, Lemmas 66.37.7 and 66.37.5) hence $W \to \mathcal{Y}$ is surjective, flat, and locally of finite presentation (by Properties of Stacks, Lemma 99.5.2 and Lemmas 100.25.2 and 100.27.2). Thus we may replace $\mathcal{Z}$ by $W$.

Choose an algebraic space $V$ and a surjective smooth morphism $V \to \mathcal{Y}$. Choose an algebraic space $U$ and a surjective smooth morphism $U \to V \times _\mathcal {Y} \mathcal{X}$. We have to show that $U \to V$ has $\mathcal{P}$. Now we base change everything by $W \to \mathcal{Y}$: Set $U' = W \times _\mathcal {Y} U$, $V' = W \times _\mathcal {Y} V$, $\mathcal{X}' = W \times _\mathcal {Y} \mathcal{X}$, and $\mathcal{Y}' = W \times _\mathcal {Y} \mathcal{Y} = W$. Then it is still true that $U' \to V' \times _{\mathcal{Y}'} \mathcal{X}'$ is smooth by base change. Hence by Lemma 100.16.1 used in the definition of $\mathcal{X}' \to \mathcal{Y}' = W$ having $\mathcal{P}$ we see that $U' \to V'$ has $\mathcal{P}$. Then, since $V' \to V$ is surjective, flat, and locally of finite presentation as a base change of $W \to \mathcal{Y}$ we see that $U \to V$ has $\mathcal{P}$ as $\mathcal{P}$ is local in the fppf topology on the target. $\square$

Lemma 100.27.11. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $\mathcal{Z} \to \mathcal{Y}$ be a surjective, flat, locally finitely presented morphism of algebraic stacks. If the base change $\mathcal{Z} \times _\mathcal {Y} \mathcal{X} \to \mathcal{Z}$ is locally of finite presentation, then $f$ is locally of finite presentation.

Proof. The property “locally of finite presentation” satisfies the conditions of Lemma 100.27.10. Smooth local on the source-and-target we have seen in the introduction to this section and fppf local on the target is Descent on Spaces, Lemma 73.11.10. $\square$

Lemma 100.27.12. Let $\mathcal{X} \to \mathcal{Y} \to \mathcal{Z}$ be morphisms of algebraic stacks. If $\mathcal{X} \to \mathcal{Z}$ is locally of finite presentation and $\mathcal{X} \to \mathcal{Y}$ is surjective, flat, and locally of finite presentation, then $\mathcal{Y} \to \mathcal{Z}$ is locally of finite presentation.

Proof. Choose an algebraic space $W$ and a surjective smooth morphism $W \to \mathcal{Z}$. Choose an algebraic space $V$ and a surjective smooth morphism $V \to W \times _\mathcal {Z} \mathcal{Y}$. Choose an algebraic space $U$ and a surjective smooth morphism $U \to V \times _\mathcal {Y} \mathcal{X}$. We know that $U \to V$ is flat and locally of finite presentation and that $U \to W$ is locally of finite presentation. Also, as $\mathcal{X} \to \mathcal{Y}$ is surjective we see that $U \to V$ is surjective (as a composition of surjective morphisms). Hence the lemma reduces to the case of morphisms of algebraic spaces. The case of morphisms of algebraic spaces is Descent on Spaces, Lemma 73.16.1. $\square$

Lemma 100.27.13. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks which is surjective, flat, and locally of finite presentation. Then for every scheme $U$ and object $y$ of $\mathcal{Y}$ over $U$ there exists an fppf covering $\{ U_ i \to U\}$ and objects $x_ i$ of $\mathcal{X}$ over $U_ i$ such that $f(x_ i) \cong y|_{U_ i}$ in $\mathcal{Y}_{U_ i}$.

Proof. We may think of $y$ as a morphism $U \to \mathcal{Y}$. By Properties of Stacks, Lemma 99.5.3 and Lemmas 100.27.3 and 100.25.3 we see that $\mathcal{X} \times _\mathcal {Y} U \to U$ is surjective, flat, and locally of finite presentation. Let $V$ be a scheme and let $V \to \mathcal{X} \times _\mathcal {Y} U$ smooth and surjective. Then $V \to \mathcal{X} \times _\mathcal {Y} U$ is also surjective, flat, and locally of finite presentation (see Morphisms of Spaces, Lemmas 66.37.7 and 66.37.5). Hence also $V \to U$ is surjective, flat, and locally of finite presentation, see Properties of Stacks, Lemma 99.5.2 and Lemmas 100.27.2, and 100.25.2. Hence $\{ V \to U\}$ is the desired fppf covering and $x : V \to \mathcal{X}$ is the desired object. $\square$

Lemma 100.27.14. Let $f_ j : \mathcal{X}_ j \to \mathcal{X}$, $j \in J$ be a family of morphisms of algebraic stacks which are each flat and locally of finite presentation and which are jointly surjective, i.e., $|\mathcal{X}| = \bigcup |f_ j|(|\mathcal{X}_ j|)$. Then for every scheme $U$ and object $x$ of $\mathcal{X}$ over $U$ there exists an fppf covering $\{ U_ i \to U\} _{i \in I}$, a map $a : I \to J$, and objects $x_ i$ of $\mathcal{X}_{a(i)}$ over $U_ i$ such that $f_{a(i)}(x_ i) \cong y|_{U_ i}$ in $\mathcal{X}_{U_ i}$.

Proof. Apply Lemma 100.27.13 to the morphism $\coprod _{j \in J} \mathcal{X}_ j \to \mathcal{X}$. (There is a slight set theoretic issue here – due to our setup of things – which we ignore.) To finish, note that a morphism $x_ i : U_ i \to \coprod _{j \in J} \mathcal{X}_ j$ is given by a disjoint union decomposition $U_ i = \coprod U_{i, j}$ and morphisms $U_{i, j} \to \mathcal{X}_ j$. Then the fppf covering $\{ U_{i, j} \to U\}$ and the morphisms $U_{i, j} \to \mathcal{X}_ j$ do the job. $\square$

Lemma 100.27.15. Let $f : \mathcal{X} \to \mathcal{Y}$ be flat and locally of finite presentation. Then $|f| : |\mathcal{X}| \to |\mathcal{Y}|$ is open.

Proof. Choose a scheme $V$ and a smooth surjective morphism $V \to \mathcal{Y}$. Choose a scheme $U$ and a smooth surjective morphism $U \to V \times _\mathcal {Y} \mathcal{X}$. By assumption the morphism of schemes $U \to V$ is flat and locally of finite presentation. Hence $U \to V$ is open by Morphisms, Lemma 29.25.10. By construction of the topology on $|\mathcal{Y}|$ the map $|V| \to |\mathcal{Y}|$ is open. The map $|U| \to |\mathcal{X}|$ is surjective. The result follows from these facts by elementary topology. $\square$

Lemma 100.27.16. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $\mathcal{Z} \to \mathcal{Y}$ be a surjective, flat, locally finitely presented morphism of algebraic stacks. If the base change $\mathcal{Z} \times _\mathcal {Y} \mathcal{X} \to \mathcal{Z}$ is quasi-compact, then $f$ is quasi-compact.

Proof. We have to show that given $\mathcal{Y}' \to \mathcal{Y}$ with $\mathcal{Y}'$ quasi-compact, we have $\mathcal{Y}' \times _\mathcal {Y} \mathcal{X}$ is quasi-compact. Denote $\mathcal{Z}' = \mathcal{Z} \times _\mathcal {Y} \mathcal{Y}'$. Then $|\mathcal{Z}'| \to |\mathcal{Y}'|$ is open, see Lemma 100.27.15. Hence we can find a quasi-compact open substack $\mathcal{W} \subset \mathcal{Z}'$ mapping onto $\mathcal{Y}'$. Because $\mathcal{Z} \times _\mathcal {Y} \mathcal{X} \to \mathcal{Z}$ is quasi-compact, we know that

$\mathcal{W} \times _\mathcal {Z} \mathcal{Z} \times _\mathcal {Y} \mathcal{X} = \mathcal{W} \times _\mathcal {Y} \mathcal{X}$

is quasi-compact. And the map $\mathcal{W} \times _\mathcal {Y} \mathcal{X} \to \mathcal{Y}' \times _\mathcal {Y} \mathcal{X}$ is surjective, hence we win. Some details omitted. $\square$

Lemma 100.27.17. Let $f : \mathcal{X} \to \mathcal{Y}$, $g : \mathcal{Y} \to \mathcal{Z}$ be composable morphisms of algebraic stacks with composition $h = g \circ f : \mathcal{X} \to \mathcal{Z}$. If $f$ is surjective, flat, locally of finite presentation, and universally injective and if $h$ is separated, then $g$ is separated.

Proof. Consider the diagram

$\xymatrix{ \mathcal{X} \ar[r]_\Delta \ar[rd] & \mathcal{X} \times _\mathcal {Y} \mathcal{X} \ar[r] \ar[d] & \mathcal{X} \times _\mathcal {Z} \mathcal{X} \ar[d] \\ & \mathcal{Y} \ar[r] & \mathcal{Y} \times _\mathcal {Z} \mathcal{Y} }$

The square is cartesian. We have to show the bottom horizontal arrow is proper. We already know that it is representable by algebraic spaces and locally of finite type (Lemma 100.3.3). Since the right vertical arrow is surjective, flat, and locally of finite presentation it suffices to show the top right horizontal arrow is proper (Lemma 100.27.9). Since $h$ is separated, the composition of the top horizontal arrows is proper.

Since $f$ is universally injective $\Delta$ is surjective (Lemma 100.14.5). Since the composition of $\Delta$ with the projection $\mathcal{X} \times _\mathcal {Y} \mathcal{X} \to \mathcal{X}$ is the identity, we see that $\Delta$ is universally closed. By Morphisms of Spaces, Lemma 66.9.8 we conclude that $\mathcal{X} \times _\mathcal {Y} \mathcal{X} \to \mathcal{X} \times _\mathcal {Z} \mathcal{X}$ is separated as $\mathcal{X} \to \mathcal{X} \times _\mathcal {Z} \mathcal{X}$ is separated. Here we use that implications between properties of morphisms of algebraic spaces can be transferred to the same implications between properties of morphisms of algebraic stacks representable by algebraic spaces; this is discussed in Properties of Stacks, Section 99.3. Finally, we use the same principle to conlude that $\mathcal{X} \times _\mathcal {Y} \mathcal{X} \to \mathcal{X} \times _\mathcal {Z} \mathcal{X}$ is proper from Morphisms of Spaces, Lemma 66.40.7. $\square$

Comment #5126 by Matthieu Romagny on

Typo in the section introduction : replace "spaces" by stacks" in the sentence "we may define what it means for a morphism of algebraic spaces..."

Comment #5328 by on

No idea how you ever found this, but you did and thanks! Fixed here.

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