## 101.17 Morphisms of finite type

The property “locally of finite type” of morphisms of algebraic spaces is smooth local on the source-and-target, see Descent on Spaces, Remark 74.20.5. It is also stable under base change and fpqc local on the target, see Morphisms of Spaces, Lemma 67.23.3 and Descent on Spaces, Lemma 74.11.9. Hence, by Lemma 101.16.1 above, we may define what it means for a morphism of algebraic spaces to be locally of finite type as follows and it agrees with the already existing notion defined in Properties of Stacks, Section 100.3 when the morphism is representable by algebraic spaces.

Definition 101.17.1. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks.

1. We say $f$ locally of finite type if the equivalent conditions of Lemma 101.16.1 hold with $\mathcal{P} = \text{locally of finite type}$.

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

Lemma 101.17.2. The composition of finite type morphisms is of finite type. The same holds for locally of finite type.

Proof. Combine Remark 101.16.3 with Morphisms of Spaces, Lemma 67.23.2. $\square$

Lemma 101.17.3. A base change of a finite type morphism is finite type. The same holds for locally of finite type.

Proof. Combine Remark 101.16.4 with Morphisms of Spaces, Lemma 67.23.3. $\square$

Proof. Combine Remark 101.16.5 with Morphisms of Spaces, Lemma 67.23.7. $\square$

Lemma 101.17.5. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. If $f$ is locally of finite type and $\mathcal{Y}$ is locally Noetherian, then $\mathcal{X}$ is locally Noetherian.

Proof. Let

$\xymatrix{ U \ar[d] \ar[r] & V \ar[d] \\ \mathcal{X} \ar[r] & \mathcal{Y} }$

be a commutative diagram where $U$, $V$ are schemes, $V \to \mathcal{Y}$ is surjective and smooth, and $U \to V \times _\mathcal {Y} \mathcal{X}$ is surjective and smooth. Then $U \to V$ is locally of finite type. If $\mathcal{Y}$ is locally Noetherian, then $V$ is locally Noetherian. By Morphisms, Lemma 29.15.6 we see that $U$ is locally Noetherian, which means that $\mathcal{X}$ is locally Noetherian. $\square$

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

Lemma 101.17.6. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $W \to \mathcal{Y}$ be a surjective, flat, and locally of finite presentation where $W$ is an algebraic space. If the base change $W \times _\mathcal {Y} \mathcal{X} \to W$ is locally of finite type, then $f$ is locally of finite type.

Proof. Choose an algebraic space $V$ and a surjective smooth morphism $V \to \mathcal{Y}$. Choose an algebraic space $U$ and a surjective smooth morphism $U \to V \times _\mathcal {Y} \mathcal{X}$. We have to show that $U \to V$ is locally of finite presentation. Now we base change everything by $W \to \mathcal{Y}$: Set $U' = W \times _\mathcal {Y} U$, $V' = W \times _\mathcal {Y} V$, $\mathcal{X}' = W \times _\mathcal {Y} \mathcal{X}$, and $\mathcal{Y}' = W \times _\mathcal {Y} \mathcal{Y} = W$. Then it is still true that $U' \to V' \times _{\mathcal{Y}'} \mathcal{X}'$ is smooth by base change. Hence by our definition of locally finite type morphisms of algebraic stacks and the assumption that $\mathcal{X}' \to \mathcal{Y}'$ is locally of finite type, we see that $U' \to V'$ is locally of finite type. Then, since $V' \to V$ is surjective, flat, and locally of finite presentation as a base change of $W \to \mathcal{Y}$ we see that $U \to V$ is locally of finite type by Descent on Spaces, Lemma 74.11.9 and we win. $\square$

Lemma 101.17.7. Let $\mathcal{X} \to \mathcal{Y} \to \mathcal{Z}$ be morphisms of algebraic stacks. Assume $\mathcal{X} \to \mathcal{Z}$ is locally of finite type and that $\mathcal{X} \to \mathcal{Y}$ is representable by algebraic spaces, surjective, flat, and locally of finite presentation. Then $\mathcal{Y} \to \mathcal{Z}$ is locally of finite type.

Proof. Choose an algebraic space $W$ and a surjective smooth morphism $W \to \mathcal{Z}$. Choose an algebraic space $V$ and a surjective smooth morphism $V \to W \times _\mathcal {Z} \mathcal{Y}$. Set $U = V \times _\mathcal {Y} \mathcal{X}$ which is an algebraic space. We know that $U \to V$ is surjective, flat, and locally of finite presentation and that $U \to W$ is locally of finite type. Hence the lemma reduces to the case of morphisms of algebraic spaces. The case of morphisms of algebraic spaces is Descent on Spaces, Lemma 74.16.2. $\square$

Lemma 101.17.8. Let $f : \mathcal{X} \to \mathcal{Y}$, $g : \mathcal{Y} \to \mathcal{Z}$ be morphisms of algebraic stacks. If $g \circ f : \mathcal{X} \to \mathcal{Z}$ is locally of finite type, then $f : \mathcal{X} \to \mathcal{Y}$ is locally of finite type.

Proof. We can find a diagram

$\xymatrix{ U \ar[r] \ar[d] & V \ar[r] \ar[d] & W \ar[d] \\ \mathcal{X} \ar[r] & \mathcal{Y} \ar[r] & \mathcal{Z} }$

where $U$, $V$, $W$ are schemes, the vertical arrow $W \to \mathcal{Z}$ is surjective and smooth, the arrow $V \to \mathcal{Y} \times _\mathcal {Z} W$ is surjective and smooth, and the arrow $U \to \mathcal{X} \times _\mathcal {Y} V$ is surjective and smooth. Then also $U \to \mathcal{X} \times _\mathcal {Z} V$ is surjective and smooth (as a composition of a surjective and smooth morphism with a base change of such). By definition we see that $U \to W$ is locally of finite type. Hence $U \to V$ is locally of finite type by Morphisms, Lemma 29.15.8 which in turn means (by definition) that $\mathcal{X} \to \mathcal{Y}$ is locally of finite type. $\square$

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