## 105.3 Thickenings

The following terminology may not be completely standard, but it is convenient. If $\mathcal{Y}$ is a closed substack of an algebraic stack $\mathcal{X}$, then the morphism $\mathcal{Y} \to \mathcal{X}$ is representable.

Definition 105.3.1. Thickenings.

1. We say an algebraic stack $\mathcal{X}'$ is a thickening of an algebraic stack $\mathcal{X}$ if $\mathcal{X}$ is a closed substack of $\mathcal{X}'$ and the associated topological spaces are equal.

2. Given two thickenings $\mathcal{X} \subset \mathcal{X}'$ and $\mathcal{Y} \subset \mathcal{Y}'$ a morphism of thickenings is a morphism $f' : \mathcal{X}' \to \mathcal{Y}'$ of algebraic stacks such that $f'|_\mathcal {X}$ factors through the closed substack $\mathcal{Y}$. In this situation we set $f = f'|_\mathcal {X} : \mathcal{X} \to \mathcal{Y}$ and we say that $(f, f') : (\mathcal{X} \subset \mathcal{X}') \to (\mathcal{Y} \subset \mathcal{Y}')$ is a morphism of thickenings.

3. Let $\mathcal{Z}$ be an algebraic stack. We similarly define thickenings over $\mathcal{Z}$ and morphisms of thickenings over $\mathcal{Z}$. This means that the algebraic stacks $\mathcal{X}'$ and $\mathcal{Y}'$ are endowed with a structure morphism to $\mathcal{Z}$ and that $f'$ fits into a suitable $2$-commutative diagram of algebraic stacks.

Let $\mathcal{X} \subset \mathcal{X}'$ be a thickening of algebraic stacks. Let $U'$ be a scheme and let $U' \to \mathcal{X}'$ be a surjective smooth morphism. Setting $U = \mathcal{X} \times _{\mathcal{X}'} U'$ we obtain a morphism of thickenings

$(U \subset U') \longrightarrow (\mathcal{X} \subset \mathcal{X}')$

and $U \to \mathcal{X}$ is a surjective smooth morphism. We can often deduce properties of the thickening $\mathcal{X} \subset \mathcal{X}'$ from the corresponding properties of the thickening $U \subset U'$. Sometimes, by abuse of language, we say that a morphism $\mathcal{X} \to \mathcal{X}'$ is a thickening if it is a closed immersion inducing a bijection $|\mathcal{X}| \to |\mathcal{X}'|$.

Lemma 105.3.2. Let $i : \mathcal{X} \to \mathcal{X}'$ be a morphism of algebraic stacks. The following are equivalent

1. $i$ is a thickening of algebraic stacks (abuse of language as above), and

2. $i$ is representable by algebraic spaces and is a thickening in the sense of Properties of Stacks, Section 99.3.

In this case $i$ is a closed immersion and a universal homeomorphism.

Proof. By More on Morphisms of Spaces, Lemmas 75.9.10 and 75.9.8 the property $P$ that a morphism of algebraic spaces is a (first order) thickening is fpqc local on the base and stable under base change. Thus the discussion in Properties of Stacks, Section 99.3 indeed applies. Having said this the equivalence of (1) and (2) follows from the fact that $P = P_1 + P_2$ where $P_1$ is the property of being a closed immersion and $P_2$ is the property of being surjective. (Strictly speaking, the reader should also consult More on Morphisms of Spaces, Definition 75.9.1, Properties of Stacks, Definition 99.9.1 and the discussion following, Morphisms of Spaces, Lemma 66.5.1, Properties of Stacks, Section 99.5 to see that all the concepts all match up.) The final assertion is clear from the foregoing. $\square$

We will use the lemma without further mention. Using the same references More on Morphisms of Spaces, Lemmas 75.9.10 and 75.9.8 as used in the lemma, allows us to define a first order thickening as follows.

Definition 105.3.3. We say an algebraic stack $\mathcal{X}'$ is a first order thickening of an algebraic stack $\mathcal{X}$ if $\mathcal{X}$ is a closed substack of $\mathcal{X}'$ and $\mathcal{X} \to \mathcal{X}'$ is a first order thickening in the sense of Properties of Stacks, Section 99.3.

If $(U \subset U') \to (\mathcal{X} \subset \mathcal{X}')$ is a smooth cover by a scheme as above, then this simply means that $U \subset U'$ is a first order thickening. Next we formulate the obligatory lemmas.

Lemma 105.3.4. Let $\mathcal{Y} \subset \mathcal{Y}'$ be a thickening of algebraic stacks. Let $\mathcal{X}' \to \mathcal{Y}'$ be a morphism of algebraic stacks and set $\mathcal{X} = \mathcal{Y} \times _{\mathcal{Y}'} \mathcal{X}'$. Then $(\mathcal{X} \subset \mathcal{X}') \to (\mathcal{Y} \subset \mathcal{Y}')$ is a morphism of thickenings. If $\mathcal{Y} \subset \mathcal{Y}'$ is a first order thickening, then $\mathcal{X} \subset \mathcal{X}'$ is a first order thickening.

Proof. See discussion above, Properties of Stacks, Section 99.3, and More on Morphisms of Spaces, Lemma 75.9.8. $\square$

Lemma 105.3.5. If $\mathcal{X} \subset \mathcal{X}'$ and $\mathcal{X}' \subset \mathcal{X}''$ are thickenings of algebraic stacks, then so is $\mathcal{X} \subset \mathcal{X}''$.

Proof. See discussion above, Properties of Stacks, Section 99.3, and More on Morphisms of Spaces, Lemma 75.9.9 $\square$

Example 105.3.6. Let $\mathcal{X}'$ be an algebraic stack. Then $\mathcal{X}'$ is a thickening of the reduction $\mathcal{X}'_{red}$, see Properties of Stacks, Definition 99.10.4. Moreover, if $\mathcal{X} \subset \mathcal{X}'$ is a thickening of algebraic stacks, then $\mathcal{X}'_{red} = \mathcal{X}_{red} \subset \mathcal{X}$. In other words, $\mathcal{X} = \mathcal{X}'_{red}$ if and only if $\mathcal{X}$ is a reduced algebraic stack.

Lemma 105.3.7. Let $(f, f') : (\mathcal{X} \subset \mathcal{X}') \to (\mathcal{Y} \subset \mathcal{Y}')$ be a morphism of thickenings of algebraic stacks. Then $\mathcal{X} \times _\mathcal {Y} \mathcal{X} \to \mathcal{X}' \times _{\mathcal{Y}'} \mathcal{X}'$ is a thickening and the canonical diagram

$\xymatrix{ \mathcal{X} \ar[r]_-\Delta \ar[d] & \mathcal{X} \times _\mathcal {Y} \mathcal{X} \ar[d] \\ \mathcal{X}' \ar[r]^-{\Delta '} & \mathcal{X}' \times _{\mathcal{Y}'} \mathcal{X}' }$

is cartesian.

Proof. Since $\mathcal{X} \to \mathcal{Y}'$ factors through the closed substack $\mathcal{Y}$ we see that $\mathcal{X} \times _\mathcal {Y} \mathcal{X} = \mathcal{X} \times _{\mathcal{Y}'} \mathcal{X}$. Hence $\mathcal{X} \times _\mathcal {Y} \mathcal{X} \to \mathcal{X}' \times _{\mathcal{Y}'} \mathcal{X}'$ is isomorphic to the composition

$\mathcal{X} \times _{\mathcal{Y}'} \mathcal{X} \to \mathcal{X} \times _{\mathcal{Y}'} \mathcal{X}' \to \mathcal{X}' \times _{\mathcal{Y}'} \mathcal{X}'$

both of which are thickenings as base changes of thickenings (Lemma 105.3.4). Hence so is the composition (Lemma 105.3.5). Since $\mathcal{X} \to \mathcal{X}'$ is a monomorphism, the final statement of the lemma follows from Properties of Stacks, Lemma 99.8.6 applied to $\mathcal{X} \to \mathcal{X}' \to \mathcal{Y}'$. $\square$

Lemma 105.3.8. Let $(f, f') : (\mathcal{X} \subset \mathcal{X}') \to (\mathcal{Y} \subset \mathcal{Y}')$ be a morphism of thickenings of algebraic stacks. Let $\Delta : \mathcal{X} \to \mathcal{X} \times _\mathcal {Y} \mathcal{X}$ and $\Delta ' : \mathcal{X}' \to \mathcal{X}' \times _{\mathcal{Y}'} \mathcal{X}'$ be the corresponding diagonal morphisms. Then each property from the following list is satisfied by $\Delta$ if and only if it is satisfied by $\Delta '$: (a) representable by schemes, (b) affine, (c) surjective, (d) quasi-compact, (e) universally closed, (f) integral, (g) quasi-separated, (h) separated, (i) universally injective, (j) universally open, (k) locally quasi-finite, (l) finite, (m) unramified, (n) monomorphism, (o) immersion, (p) closed immersion, and (q) proper.

Proof. Observe that

$(\Delta , \Delta ') : (\mathcal{X} \subset \mathcal{X}') \longrightarrow (\mathcal{X} \times _\mathcal {Y} \mathcal{X} \subset \mathcal{X}' \times _{\mathcal{Y}'} \mathcal{X}')$

is a morphism of thickenings (Lemma 105.3.7). Moreover $\Delta$ and $\Delta '$ are representable by algebraic spaces by Morphisms of Stacks, Lemma 100.3.3. Hence, via the discussion in Properties of Stacks, Section 99.3 the lemma follows for cases (a), (b), (c), (d), (e), (f), (g), (h), (i), and (j) by using More on Morphisms of Spaces, Lemma 75.10.1.

Lemma 105.3.7 tells us that $\mathcal{X} = (\mathcal{X} \times _\mathcal {Y} \mathcal{X}) \times _{(\mathcal{X}' \times _{\mathcal{Y}'} \mathcal{X}')} \mathcal{X}'$. Moreover, $\Delta$ and $\Delta '$ are locally of finite type by the aforementioned Morphisms of Stacks, Lemma 100.3.3. Hence the result for cases (k), (l), (m), (n), (o), (p), and (q) by using More on Morphisms of Spaces, Lemma 75.10.3. $\square$

As a consequence we obtain the following pleasing result.

Lemma 105.3.9. Let $\mathcal{X} \subset \mathcal{X}'$ be a thickening of algebraic stacks. Then

1. $\mathcal{X}$ is an algebraic space if and only if $\mathcal{X}'$ is an algebraic space,

2. $\mathcal{X}$ is a scheme if and only if $\mathcal{X}'$ is a scheme,

3. $\mathcal{X}$ is DM if and only if $\mathcal{X}'$ is DM,

4. $\mathcal{X}$ is quasi-DM if and only if $\mathcal{X}'$ is quasi-DM,

5. $\mathcal{X}$ is separated if and only if $\mathcal{X}'$ is separated,

6. $\mathcal{X}$ is quasi-separated if and only if $\mathcal{X}'$ is quasi-separated, and

Proof. In each case we reduce to a question about the diagonal and then we use Lemma 105.3.8 applied to the morphism of thickenings

$(\mathcal{X} \subset \mathcal{X}') \to \left(\mathop{\mathrm{Spec}}(\mathbf{Z}) \subset \mathop{\mathrm{Spec}}(\mathbf{Z})\right)$

We do this after viewing $\mathcal{X} \subset \mathcal{X}'$ as a thickening of algebraic stacks over $\mathop{\mathrm{Spec}}(\mathbf{Z})$ via Algebraic Stacks, Definition 93.19.2.

Case (1). An algebraic stack is an algebraic space if and only if its diagonal is a monomorphism, see Morphisms of Stacks, Lemma 100.6.3 (this also follows immediately from Algebraic Stacks, Proposition 93.13.3).

Case (2). By (1) we may assume that $\mathcal{X}$ and $\mathcal{X}'$ are algebraic spaces and then we can use More on Morphisms of Spaces, Lemma 75.9.5.

Case (3) – (6). Each of these cases corresponds to a condition on the diagonal, see Morphisms of Stacks, Definitions 100.4.1 and 100.4.2. $\square$

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