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Tag 0DTI

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

Lemma 90.11.13. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $x \in |\mathcal{X}|$ with image $y \in |\mathcal{Y}|$. Assume the residual gerbes $\mathcal{Z}_x \subset \mathcal{X}$ and $\mathcal{Z}_y \subset \mathcal{Y}$ of $x$ and $y$ exist and that there exists a morphism $\mathop{\rm Spec}(k) \to \mathcal{X}$ in the equivalence class of $x$ such that $$ \mathop{\rm Spec}(k) \times_\mathcal{X} \mathop{\rm Spec}(k) \longrightarrow \mathop{\rm Spec}(k) \times_\mathcal{Y} \mathop{\rm Spec}(k) $$ is an isomorphism. Then $\mathcal{Z}_x \to \mathcal{Z}_y$ is an isomorphism.

Proof. Let $k'/k$ be an extension of fields. Then $$ \mathop{\rm Spec}(k') \times_\mathcal{X} \mathop{\rm Spec}(k') \longrightarrow \mathop{\rm Spec}(k') \times_\mathcal{Y} \mathop{\rm Spec}(k') $$ is the base change of the morphism in the lemma by the faithfully flat morphism $\mathop{\rm Spec}(k' \otimes k') \to \mathop{\rm Spec}(k \otimes k)$. Thus the property described in the lemma is independent of the choice of the morphism $\mathop{\rm Spec}(k) \to \mathcal{X}$ in the equivalence class of $x$. Thus we may assume that $\mathop{\rm Spec}(k) \to \mathcal{Z}_x$ is surjective, flat, and locally of finite presentation. In this situation we have $$ \mathcal{Z}_x = [\mathop{\rm Spec}(k)/R] $$ with $R = \mathop{\rm Spec}(k) \times_\mathcal{X} \mathop{\rm Spec}(k)$. See proof of Lemma 90.11.5. Since also $R = \mathop{\rm Spec}(k) \times_\mathcal{Y} \mathop{\rm Spec}(k)$ we conclude that the morphism $\mathcal{Z}_x \to \mathcal{Z}_y$ of Lemma 90.11.12 is fully faithful by Algebraic Stacks, Lemma 84.16.1. We conclude for example by Lemma 90.11.11. $\square$

    The code snippet corresponding to this tag is a part of the file stacks-properties.tex and is located in lines 2874–2889 (see updates for more information).

    \begin{lemma}
    \label{lemma-residual-gerbe-isomorphic}
    Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks.
    Let $x \in |\mathcal{X}|$ with image $y \in |\mathcal{Y}|$.
    Assume the residual gerbes $\mathcal{Z}_x \subset \mathcal{X}$
    and $\mathcal{Z}_y \subset \mathcal{Y}$ of $x$ and $y$ exist
    and that there exists a morphism $\Spec(k) \to \mathcal{X}$
    in the equivalence class of $x$ such that
    $$
    \Spec(k) \times_\mathcal{X} \Spec(k)
    \longrightarrow
    \Spec(k) \times_\mathcal{Y} \Spec(k)
    $$
    is an isomorphism. Then $\mathcal{Z}_x \to \mathcal{Z}_y$
    is an isomorphism.
    \end{lemma}
    
    \begin{proof}
    Let $k'/k$ be an extension of fields. Then
    $$
    \Spec(k') \times_\mathcal{X} \Spec(k')
    \longrightarrow
    \Spec(k') \times_\mathcal{Y} \Spec(k')
    $$
    is the base change of the morphism in the lemma by the
    faithfully flat morphism $\Spec(k' \otimes k') \to \Spec(k \otimes k)$.
    Thus the property described in the lemma is independent of the
    choice of the morphism $\Spec(k) \to \mathcal{X}$ in the
    equivalence class of $x$. Thus we may assume that
    $\Spec(k) \to \mathcal{Z}_x$ is surjective, flat, and
    locally of finite presentation. In this situation we have
    $$
    \mathcal{Z}_x = [\Spec(k)/R]
    $$
    with $R = \Spec(k) \times_\mathcal{X} \Spec(k)$. See
    proof of Lemma \ref{lemma-improve-unique-point}.
    Since also $R = \Spec(k) \times_\mathcal{Y} \Spec(k)$
    we conclude that the morphism $\mathcal{Z}_x \to \mathcal{Z}_y$
    of Lemma \ref{lemma-residual-gerbe-functorial}
    is fully faithful by
    Algebraic Stacks, Lemma \ref{algebraic-lemma-map-space-into-stack}.
    We conclude for example by Lemma \ref{lemma-residual-gerbe-unique}.
    \end{proof}

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