The Stacks project

Lemma 105.13.6. Let the algebraic stack $\mathcal{X}$ be well-nigh affine. The morphism

\[ f : \mathcal{X} \longrightarrow M \]

of Lemma 105.13.4 is a uniform categorical moduli space.

Proof. We already know that $M$ is a uniform categorical moduli space in the category of affine schemes. By Lemma 105.12.4 it suffices to show that the base change $f' : M' \times _ M \mathcal{X} \to M'$ is a categorical moduli space for any flat morphism $M' \to M$ of affine schemes. Observe that $\mathcal{X}' = M' \times _ M \mathcal{X}$ is well-nigh affine by Lemma 105.13.3. This after replacing $\mathcal{X}$ by $\mathcal{X}'$ and $M$ by $M'$, we reduce to proving $f$ is a categorical moduli space.

Let $g : \mathcal{X} \to Y$ be a morphism where $Y$ is an algebraic space. We have to show that $g = h \circ f$ for a unique morphism $h : M \to Y$.

Uniqueness. Suppose we have two morphisms $h_ i : M \to Y$ with $g = h_1 \circ f = h_2 \circ f$. Let $M' \subset M$ be the equalizer of $h_1$ and $h_2$. Then $M' \to M$ is a monomorphism and $f : \mathcal{X} \to M$ factors through $M'$. Thus $M' \to M$ is a universal homeomorphism. We conclude $M'$ is affine (Morphisms, Lemma 29.45.5). But then since $f : \mathcal{X} \to M$ is a categorical moduli space in the category of affine schemes, we see $M' = M$.

Existence. Below we will show that for every $p \in M$ there exists a cartesian square

\[ \xymatrix{ \mathcal{X}' \ar[r] \ar[d] & \mathcal{X} \ar[d] \\ M' \ar[r] & M } \]

with $M' \to M$ an étale morphism of affines and $p$ in the image such that the composition $\mathcal{X}' \to \mathcal{X} \to Y$ factors through $M'$. This means we can construct the map $h : M \to Y$ étale locally on $M$. Since $Y$ is a sheaf for the étale topology and by the uniqueness shown above, this is enough (small detail omitted).

Let $y \in |Y|$ be the image of $p$. Let $(V, v) \to (Y, y)$ be an étale morphism with $V$ affine. Consider $\mathcal{X}' = V \times _ Y \mathcal{X}$. Observe that $\mathcal{X}' \to \mathcal{X}$ is separated and étale as the base change of $V \to Y$. Moreover, $\mathcal{X}' \to \mathcal{X}$ induces isomorphisms on automorphism groups (Morphisms of Stacks, Remark 100.19.5) as this is true for $V \to Y$, see Morphisms of Stacks, Lemma 100.45.5. Choose a presentation $\mathcal{X} = [U/R]$ as in Lemma 105.13.2. Set $U' = \mathcal{X}' \times _\mathcal {X} U = V \times _ Y U$ and choose $u' \in U'$ mapping to $p$ and $v$ (possible by Properties of Spaces, Lemma 65.4.3). Since $U' \to U$ is separated and étale we see that every finite set of points of $U'$ is contained in an affine open, see More on Morphisms, Lemma 37.45.1. On the other hand, the morphism $U' \to \mathcal{X}'$ is surjective, finite, flat, and locally of finite presentation. Setting $R' = U' \times _{\mathcal{X}'} U'$ we see that $s', t' : R' \to U'$ are finite locally free. By Groupoids, Lemma 39.24.1 there exists an $R'$-invariant affine open subscheme $U'' \subset U'$ containing $u'$. Let $\mathcal{X}'' \subset \mathcal{X}'$ be the corresponding open substack. Then $\mathcal{X}''$ is well-nigh affine. By Lemma 105.13.5 we obtain a cartesian square

\[ \xymatrix{ \mathcal{X}'' \ar[r] \ar[d] & \mathcal{X} \ar[d] \\ M'' \ar[r] & M } \]

with $M'' \to M$ étale. Since $\mathcal{X}'' \to M''$ is a categorical moduli space in the category of affine schemes we obtain a morphism $M'' \to V$ such that the composition $\mathcal{X}'' \to \mathcal{X}' \to V$ is equal to the composition $\mathcal{X}'' \to M'' \to V$. This proves our claim and finishes the proof. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0DUR. Beware of the difference between the letter 'O' and the digit '0'.