Lemma 98.11.4. Let $\mathcal{Z}' \to \mathcal{Z}$ be a monomorphism of algebraic stacks. Assume there exists a field $k$ and a locally finitely presented, surjective, flat morphism $\mathop{\mathrm{Spec}}(k) \to \mathcal{Z}$. Then either $\mathcal{Z}'$ is empty or $\mathcal{Z}' \to \mathcal{Z}$ is an equivalence.

Proof. We may assume that $\mathcal{Z}'$ is nonempty. In this case the fibre product $T = \mathcal{Z}' \times _\mathcal {Z} \mathop{\mathrm{Spec}}(k)$ is nonempty, see Lemma 98.4.3. Now $T$ is an algebraic space and the projection $T \to \mathop{\mathrm{Spec}}(k)$ is a monomorphism. Hence $T = \mathop{\mathrm{Spec}}(k)$, see Morphisms of Spaces, Lemma 65.10.8. We conclude that $\mathop{\mathrm{Spec}}(k) \to \mathcal{Z}$ factors through $\mathcal{Z}'$. Suppose the morphism $z : \mathop{\mathrm{Spec}}(k) \to \mathcal{Z}$ is given by the object $\xi$ over $\mathop{\mathrm{Spec}}(k)$. We have just seen that $\xi$ is isomorphic to an object $\xi '$ of $\mathcal{Z}'$ over $\mathop{\mathrm{Spec}}(k)$. Since $z$ is surjective, flat, and locally of finite presentation we see that every object of $\mathcal{Z}$ over any scheme is fppf locally isomorphic to a pullback of $\xi$, hence also to a pullback of $\xi '$. By descent of objects for stacks in groupoids this implies that $\mathcal{Z}' \to \mathcal{Z}$ is essentially surjective (as well as fully faithful, see Lemma 98.8.4). Hence we win. $\square$

There are also:

• 2 comment(s) on Section 98.11: Residual gerbes

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).