The Stacks project

108.49 Stack with quasi-compact flat covering which is not algebraic

In this section we briefly describe an example due to Brian Conrad. You can find the example online at this location. Our example is slightly different.

Let $k$ be an algebraically closed field. All schemes and stacks are over $k$ in what follows. Let $G \to \mathop{\mathrm{Spec}}(k)$ be an affine group scheme. In Examples of Stacks, Lemma 93.15.4 we have given several different equivalent ways to view $\mathcal{X} = [\mathop{\mathrm{Spec}}(k)/G]$ as a stack in groupoids over $(\mathit{Sch}/\mathop{\mathrm{Spec}}(k))_{fppf}$. In particular $\mathcal{X}$ classifies fppf $G$-torsors. More precisely, a $1$-morphism $T \to \mathcal{X}$ corresponds to an fppf $G_ T$-torsor $P$ over $T$ and $2$-arrows correspond to isomorphisms of torsors. It follows that the diagonal $1$-morphism

\[ \Delta : \mathcal{X} \longrightarrow \mathcal{X} \times _{\mathop{\mathrm{Spec}}(k)} \mathcal{X} \]

is representable and affine. Namely, given any pair of fppf $G_ T$-torsors $P_1, P_2$ over a scheme $T/k$ the scheme $\mathit{Isom}(P_1, P_2)$ is affine over $T$. The trivial $G$-torsor over $\mathop{\mathrm{Spec}}(k)$ defines a $1$-morphism

\[ f : \mathop{\mathrm{Spec}}(k) \longrightarrow \mathcal{X}. \]

We claim that this is a surjective $1$-morphism. The reason is simply that by definition for any $1$-morphism $T \to \mathcal{X}$ there exists a fppf covering $\{ T_ i \to T\} $ such that $P_{T_ i}$ is isomorphic to the trivial $G_{T_ i}$-torsor. Hence the compositions $T_ i \to T \to \mathcal{X}$ factor through $f$. Thus it is clear that the projection $T \times _\mathcal {X} \mathop{\mathrm{Spec}}(k) \to T$ is surjective (which is how we define the property that $f$ is surjective, see Algebraic Stacks, Definition 92.10.1). In a similar way you show that $f$ is quasi-compact and flat (details omitted). We also record here the observation that

\[ \mathop{\mathrm{Spec}}(k) \times _\mathcal {X} \mathop{\mathrm{Spec}}(k) \cong G \]

as schemes over $k$.

Suppose there exists a surjective smooth morphism $p : U \to \mathcal{X}$ where $U$ is a scheme. Consider the fibre product

\[ \xymatrix{ W \ar[d] \ar[r] & U \ar[d] \\ \mathop{\mathrm{Spec}}(k) \ar[r] & \mathcal{X} } \]

Then we see that $W$ is a nonempty smooth scheme over $k$ which hence has a $k$-point. This means that we can factor $f$ through $U$. Hence we obtain

\[ G \cong \mathop{\mathrm{Spec}}(k) \times _\mathcal {X} \mathop{\mathrm{Spec}}(k) \cong (\mathop{\mathrm{Spec}}(k) \times _ k \mathop{\mathrm{Spec}}(k)) \times _{(U \times _ k U)} (U \times _\mathcal {X} U) \]

and since the projections $U \times _\mathcal {X} U \to U$ were assumed smooth we conclude that $U \times _\mathcal {X} U \to U \times _ k U$ is locally of finite type, see Morphisms, Lemma 29.15.8. It follows that in this case $G$ is locally of finite type over $k$. Altogether we have proved the following lemma (which can be significantly generalized).

Lemma 108.49.1. Let $k$ be a field. Let $G$ be an affine group scheme over $k$. If the stack $[\mathop{\mathrm{Spec}}(k)/G]$ has a smooth covering by a scheme, then $G$ is of finite type over $k$.

Proof. See discussion above. $\square$

To get an explicit example as in the title of this section, take for example $G = (\mu _{2, k})^\infty $ the group scheme of Section 108.48, which is not locally of finite type over $k$. By the discussion above we see that $\mathcal{X} = [\mathop{\mathrm{Spec}}(k)/G]$ has properties (1) and (2) of Algebraic Stacks, Definition 92.12.1, but not property (3). Hence $\mathcal{X}$ is not an algebraic stack. On the other hand, there does exist a scheme $U$ and a surjective, flat, quasi-compact morphism $U \to \mathcal{X}$, namely the morphism $f : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ we studied above.


Comments (2)

Comment #1866 by Ariyan on

Typos: there does exist(s) a scheme an(d) a surjective,...


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 04AG. Beware of the difference between the letter 'O' and the digit '0'.