## 101.48 Decent algebraic stacks

This section is the analogue of Decent Spaces, Section 68.6. In particular, the following definition is compatible with the notion of a decent algebraic space defined there.

Definition 101.48.1. Let $\mathcal{X}$ be an algebraic stack. We say $\mathcal{X}$ is decent if for every $x \in |\mathcal{X}|$ the equivalent conditions of Properties of Stacks, Lemma 100.14.1 are satisfied.

Some people would rephrase this definition by saying that every point of $\mathcal{X}$ is quasi-compact. A slightly stronger condition would be to ask that any morphism $\mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ in the equivalence class of $x$ is quasi-separated as well as quasi-compact.

Lemma 101.48.2. A quasi-separated algebraic stack $\mathcal{X}$ is decent. More generally, if $\Delta : \mathcal{X} \to \mathcal{X} \times \mathcal{X}$ is quasi-compact, then $\mathcal{X}$ is decent.

Proof. Namely, if $\mathcal{X}$ is quasi-separated, then any morphism $f : T \to \mathcal{X}$ whose source is a quasi-compact scheme $T$, is quasi-compact, see Lemma 101.7.7. If $\Delta$ is on known to be quasi-compact, then one uses the description

$T \times _{f, \mathcal{X}, f'} T' = (T \times T') \times _{(f, f'), \mathcal{X} \times \mathcal{X}, \Delta } \mathcal{X}$

to prove this. Details omitted. $\square$

Lemma 101.48.3. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Assume $Y$ is decent and $f$ is representable (by schemes) or $f$ is representable by algebraic spaces and quasi-separated. Then $\mathcal{X}$ is decent.

Proof. Let $x \in |\mathcal{X}|$ with image $y \in |\mathcal{Y}|$. Choose a morphism $y : \mathop{\mathrm{Spec}}(k) \to \mathcal{Y}$ in the equivalence class defining $y$. Set $\mathcal{X}_ y = \mathop{\mathrm{Spec}}(k) \times _{y, \mathcal{Y}} \mathcal{X}$. Choose a point $x' \in |\mathcal{X}_ y|$ mapping to $x$, see Properties of Stacks, Lemma 100.4.3. Choose a morphism $x' : \mathop{\mathrm{Spec}}(k') \to \mathcal{X}_ y$ in the equivalence class of $x'$. Diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(k') \ar[r]_{x'} & \mathcal{X}_ y \ar[r] \ar[d] & \mathcal{X} \ar[d] \\ & \mathop{\mathrm{Spec}}(k) \ar[r]^ y & \mathcal{Y} }$

The morphism $y$ is quasi-compact if $\mathcal{Y}$ is decent. Hence $\mathcal{X}_ y \to \mathcal{X}$ is quasi-compact as a base change (Lemma 101.7.3). Thus to conclude it suffices to prove that $x'$ is quasi-compact (Lemma 101.7.4). If $f$ is representable, then $\mathcal{X}_ y$ is a scheme and $x'$ is quasi-compact. If $f$ is representable by algebraic spaces and quasi-separated, then $\mathcal{X}_ y$ is a quasi-separated algebraic space and hence decent (Decent Spaces, Lemma 68.17.2). $\square$

Lemma 101.48.4. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. If $f$ is quasi-compact and surjective and $\mathcal{X}$ is decent, then $\mathcal{Y}$ is decent.

Proof. Let $x : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ be a morphism where $k$ is a field and denote $y = f \circ x$. Since $f$ is surjective, every point of $|\mathcal{Y}|$ arises in this manner, see Properties of Stacks, Lemma 100.4.4. Consider an affine scheme $T$ and morphism $T \to \mathcal{Y}$. It suffices to show that $T \times _{\mathcal{Y}, y} \mathop{\mathrm{Spec}}(k)$ is quasi-compact, see Lemma 101.7.10. We have

$(T \times _{\mathcal{Y}} \mathcal{X}) \times _{\mathcal{X}, x} \mathop{\mathrm{Spec}}(k) = T \times _{\mathcal{Y}, y} \mathop{\mathrm{Spec}}(k)$

The morphism $T \times _{\mathcal{Y}} \mathcal{X} \to T$ is quasi-compact hence $T \times _\mathcal {Y} \mathcal{X}$ is quasi-compact. Since $x$ is a quasi-compact morphism as $\mathcal{X}$ is decent we see that the displayed fibre product is quasi-compact. $\square$

Lemma 101.48.5. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. If $\mathcal{X}$ is a gerbe over $\mathcal{Y}$ and $\mathcal{X}$ is decent, then $\mathcal{Y}$ is decent.

Proof. Assume $\mathcal{X}$ is a gerbe over $\mathcal{Y}$ and $\mathcal{X}$ is decent. Note that $f$ is a universal homeomorphism by Lemma 101.28.13. Thus the lemma follows from Lemma 101.48.4. $\square$

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