## 101.33 Smooth morphisms

The property “being smooth” 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 67.37.3 and Descent on Spaces, Lemma 74.11.26. Hence, by Lemma 101.16.1 above, we may define what it means for a morphism of algebraic spaces to be smooth as follows and it agrees with the already existing notion defined in Properties of Stacks, Section 100.3 when the morphism is representable by algebraic spaces.

Definition 101.33.1. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. We say $f$ is smooth if the equivalent conditions of Lemma 101.16.1 hold with $\mathcal{P} = \text{smooth}$.

Proof. Combine Remark 101.16.3 with Morphisms of Spaces, Lemma 67.37.2. $\square$

Proof. Combine Remark 101.16.4 with Morphisms of Spaces, Lemma 67.37.3. $\square$

Lemma 101.33.4. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $\mathcal{Z} \to \mathcal{Y}$ be a surjective, flat, locally finitely presented morphism of algebraic stacks. If the base change $\mathcal{Z} \times _\mathcal {Y} \mathcal{X} \to \mathcal{Z}$ is smooth, then $f$ is smooth.

Proof. The property “smooth” satisfies the conditions of Lemma 101.27.10. 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 74.11.26. $\square$

Lemma 101.33.5. A smooth morphism of algebraic stacks is locally of finite presentation.

Proof. Omitted. $\square$

Lemma 101.33.6. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. There is a maximal open substack $\mathcal{U} \subset \mathcal{X}$ such that $f|_\mathcal {U} : \mathcal{U} \to \mathcal{Y}$ is smooth. Moreover, formation of this open commutes with

1. precomposing by smooth morphisms,

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

3. base change by flat morphisms provided $f$ is locally of finite presentation.

Proof. Choose a commutative diagram

$\xymatrix{ U \ar[d]_ a \ar[r]_ h & V \ar[d]^ b \\ \mathcal{X} \ar[r]^ f & \mathcal{Y} }$

where $U$ and $V$ are algebraic spaces, the vertical arrows are smooth, and $a : U \to \mathcal{X}$ surjective. There is a maximal open subspace $U' \subset U$ such that $h_{U'} : U' \to V$ is smooth, see Morphisms of Spaces, Lemma 67.37.9. Let $\mathcal{U} \subset \mathcal{X}$ be the open substack corresponding to the image of $|U'| \to |\mathcal{X}|$ (Properties of Stacks, Lemmas 100.4.7 and 100.9.12). By the equivalence in Lemma 101.16.1 we find that $f|_\mathcal {U} : \mathcal{U} \to \mathcal{Y}$ is smooth and that $\mathcal{U}$ is the largest open substack with this property.

Assertion (1) follows from the fact that being smooth is smooth local on the source (this property was used to even define smooth morphisms of algebraic stacks). Assertions (2) and (3) follow from the case of algebraic spaces, see Morphisms of Spaces, Lemma 67.37.9. $\square$

Lemma 101.33.7. Let $X \to Y$ be a smooth morphism of algebraic spaces. Let $G$ be a group algebraic space over $Y$ which is flat and locally of finite presentation over $Y$. Let $G$ act on $X$ over $Y$. Then the quotient stack $[X/G]$ is smooth over $Y$.

This holds even if $G$ is not smooth over $Y$!

Proof. The quotient $[X/G]$ is an algebraic stack by Criteria for Representability, Theorem 97.17.2. The smoothness of $[X/G]$ over $Y$ follows from the fact that smoothness descends under fppf coverings: Choose a surjective smooth morphism $U \to [X/G]$ where $U$ is a scheme. Smoothness of $[X/G]$ over $Y$ is equivalent to smoothness of $U$ over $Y$. Observe that $U \times _{[X/G]} X$ is smooth over $X$ and hence smooth over $Y$ (because compositions of smooth morphisms are smooth). On the other hand, $U \times _{[X/G]} X \to U$ is locally of finite presentation, flat, and surjective (because it is the base change of $X \to [X/G]$ which has those properties for example by Criteria for Representability, Lemma 97.17.1). Therefore we may apply Descent on Spaces, Lemma 74.8.4. $\square$

Lemma 101.33.8. Let $\pi : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. If $\mathcal{X}$ is a gerbe over $\mathcal{Y}$, then $\pi$ is surjective and smooth.

Proof. We have seen surjectivity in Lemma 101.28.8. By Lemma 101.33.4 it suffices to prove to the lemma after replacing $\pi$ by a base change with a surjective, flat, locally finitely presented morphism $\mathcal{Y}' \to \mathcal{Y}$. By Lemma 101.28.7 we may assume $\mathcal{Y} = U$ is an algebraic space and $\mathcal{X} = [U/G]$ over $U$ with $G \to U$ flat and locally of finite presentation. Then we win by Lemma 101.33.7. $\square$

Comment #8835 by yujitomo on

Is this a typo?? (the sentence after the statement of Lemma 0DLS)

This holds even if $G$ is not smooth over $S$! (←$Y$??)

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