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Tag 06MQ

Chapter 90: Properties of Algebraic Stacks > Section 90.11: Residual gerbes

Lemma 90.11.4. 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 $\mathop{\rm Spec}(k) \to \mathcal{Z}$. 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 $T = \mathcal{Z}' \times_\mathcal{Z} \mathop{\rm Spec}(k)$ is nonempty, see Lemma 90.4.3. Now $T$ is an algebraic space and the projection $T \to \mathop{\rm Spec}(k)$ is a monomorphism. Hence $T = \mathop{\rm Spec}(k)$, see Morphisms of Spaces, Lemma 58.10.8. We conclude that $\mathop{\rm Spec}(k) \to \mathcal{Z}$ factors through $\mathcal{Z}'$. Suppose the morphism $z : \mathop{\rm Spec}(k) \to \mathcal{Z}$ is given by the object $\xi$ over $\mathop{\rm Spec}(k)$. We have just seen that $\xi$ is isomorphic to an object $\xi'$ of $\mathcal{Z}'$ over $\mathop{\rm Spec}(k)$. Since $z$ is 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 90.8.4). Hence we win. $\square$

    The code snippet corresponding to this tag is a part of the file stacks-properties.tex and is located in lines 2519–2525 (see updates for more information).

    \begin{lemma}
    \label{lemma-monomorphism-into-point}
    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 $\Spec(k) \to \mathcal{Z}$. Then either $\mathcal{Z}'$
    is empty or $\mathcal{Z}' \to \mathcal{Z}$ is an equivalence.
    \end{lemma}
    
    \begin{proof}
    We may assume that $\mathcal{Z}'$ is nonempty. In this case the
    fibre product $T = \mathcal{Z}' \times_\mathcal{Z} \Spec(k)$
    is nonempty, see
    Lemma \ref{lemma-points-cartesian}.
    Now $T$ is an algebraic space and the projection $T \to \Spec(k)$
    is a monomorphism. Hence $T = \Spec(k)$, see
    Morphisms of Spaces, Lemma
    \ref{spaces-morphisms-lemma-monomorphism-toward-field}.
    We conclude that $\Spec(k) \to \mathcal{Z}$ factors through
    $\mathcal{Z}'$. Suppose the morphism $z : \Spec(k) \to \mathcal{Z}$
    is given by the object $\xi$ over $\Spec(k)$. We have just seen that
    $\xi$ is isomorphic to an object $\xi'$ of $\mathcal{Z}'$ over
    $\Spec(k)$. Since $z$ is
    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 \ref{lemma-monomorphism}).
    Hence we win.
    \end{proof}

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