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.
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.
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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