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Chapter 39: More on Groupoid Schemes > Section 39.14: Finite groupoids

Lemma 39.14.10. Let $(U, R, s, t, c)$ be a groupoid scheme. Let $u \in U$. Assume

  1. $s, t$ are finite morphisms,
  2. $U$ is separated and locally Noetherian,
  3. $\dim(\mathcal{O}_{U, u'}) \leq 1$ for every point $u'$ in the orbit of $u$.

Then $u$ is contained in an $R$-invariant affine open of $U$.

Proof. The $R$-orbit of $u$ is finite. By conditions (2) and (3) it is contained in an affine open $U'$ of $U$, see Varieties, Proposition 32.41.7. Then $t(s^{-1}(U \setminus U'))$ is an $R$-invariant closed subset of $U$ which does not contain $u$. Thus $U \setminus t(s^{-1}(U \setminus U'))$ is an $R$-invariant open of $U'$ containing $u$. Replacing $U$ by this open we may assume $U$ is quasi-affine.

By Lemma 39.14.6 we may replace $U$ by its reduction and assume $U$ is reduced. This means $R$-invariant subschemes $W' \subset W \subset U$ of Lemma 39.14.2 are equal $W' = W$. As $U = t(s^{-1}(\overline{W}))$ some point $u'$ of the $R$-orbit of $u$ is contained in $\overline{W}$ and by Lemma 39.14.6 we may replace $U$ by $\overline{W}$ and $u$ by $u'$. Hence we may assume there is a dense open $R$-invariant subscheme $W \subset U$ such that the morphisms $s_W, t_W$ of the restriction $(W, R_W, s_W, t_W, c_W)$ are finite locally free.

If $u \in W$ then we are done by Groupoids, Lemma 38.24.1 (because $W$ is quasi-affine so any finite set of points of $W$ is contained in an affine open, see Properties, Lemma 27.29.5). Thus we assume $u \not \in W$ and hence none of the points of the orbit of $u$ is in $W$. Let $\xi \in U$ be a point with a nontrivial specialization to a point $u'$ in the orbit of $u$. Since there are no specializations among the points in the orbit of $u$ (Lemma 39.14.8) we see that $\xi$ is not in the orbit. By assumption (3) we see that $\xi$ is a generic point of $U$ and hence $\xi \in W$. As $U$ is Noetherian there are finitely many of these points $\xi_1, \ldots, \xi_m \in W$. Because $s_W, t_W$ are flat the orbit of each $\xi_j$ consists of generic points of irreducible components of $W$ (and hence $U$).

Let $j : U \to \mathop{\rm Spec}(A)$ be an immersion of $U$ into an affine scheme (this is possible as $U$ is quasi-affine). Let $J \subset A$ be an ideal such that $V(J) \cap j(W) = \emptyset$ and $V(J) \cup j(W)$ is closed. Apply Lemma 39.14.7 to the groupoid scheme $(W, R_W, s_W, t_W, c_W)$, the morphism $j|_W : W \to \mathop{\rm Spec}(A)$, the points $\xi_j$, and the ideal $J$ to find an $f \in J$ such that $(j|_W)^{-1}D(f)$ is an $R_W$-invariant affine open containing $\xi_j$ for all $j$. Since $f \in J$ we see that $j^{-1}D(f) \subset W$, i.e., $j^{-1}D(f)$ is an $R$-invariant affine open of $U$ contained in $W$ containing all $\xi_j$.

Let $Z$ be the reduced induced closed subscheme structure on $$ U \setminus j^{-1}D(f) = j^{-1}V(f). $$ Then $Z$ is set theoretically $R$-invariant (but it may not be scheme theoretically $R$-invariant). Let $(Z, R_Z, s_Z, t_Z, c_Z)$ be the restriction of $R$ to $Z$. Since $Z \to U$ is finite, it follows that $s_Z$ and $t_Z$ are finite. Since $u \in Z$ the orbit of $u$ is in $Z$ and agrees with the $R_Z$-orbit of $u$ viewed as a point of $Z$. Since $\dim(\mathcal{O}_{U, u'}) \leq 1$ and since $\xi_j \not \in Z$ for all $j$, we see that $\dim(\mathcal{O}_{Z, u'}) \leq 0$ for all $u'$ in the orbit of $u$. In other words, the $R_Z$-orbit of $u$ consists of generic points of irreducible components of $Z$.

Let $I \subset A$ be an ideal such that $V(I) \cap j(U) =\emptyset$ and $V(I) \cup j(U)$ is closed. Apply Lemma 39.14.7 to the groupoid scheme $(Z, R_Z, s_Z, t_Z, c_Z)$, the restriction $j|_Z$, the ideal $I$, and the point $u \in Z$ to obtain $h \in I$ such that $j^{-1}D(h) \cap Z$ is an $R_Z$-invariant open affine containing $u$.

Consider the $R_W$-invariant (Groupoids, Lemma 38.23.2) function $$ g = \text{Norm}_{s_W}(t_W^\sharp(j^\sharp(h)|_W)) \in \Gamma(W, \mathcal{O}_W) $$ (In the following we only need the restriction of $g$ to $j^{-1}D(f)$ and in this case the norm is along a finite locally free morphism of affines.) We claim that $$ V = (W_g \cap j^{-1}D(f)) \cup (j^{-1}D(h) \cap Z) $$ is an $R$-invariant affine open of $U$ which finishes the proof of the lemma. It is set theoretically $R$-invariant by construction. As $V$ is a constuctible set, to see that it is open it suffices to show it is closed under generalization in $U$ (Topology, Lemma 5.19.9 or the more general Topology, Lemma 5.23.5). Since $W_g \cap j^{-1}D(f)$ is open in $U$, it suffices to consider a specialization $u_1 \leadsto u_2$ of $U$ with $u_2 \in j^{-1}D(h) \cap Z$. This means that $h$ is nonzero in $j(u_2)$ and $u_2 \in Z$. If $u_1 \in Z$, then $j(u_1) \leadsto j(u_2)$ and since $h$ is nonzero in $j(u_2)$ it is nonzero in $j(u_1)$ which implies $u_1 \in V$. If $u_1 \not \in Z$ and also not in $W_g \cap j^{-1}D(f)$, then $u_1 \in W$, $u_1 \not \in W_g$ because the complement of $Z = j^{-1}V(f)$ is contained in $W \cap j^{-1}D(f)$. Hence there exists a point $r_1 \in R$ with $s(r_1) = u_1$ such that $h$ is zero in $t(r_1)$. Since $s$ is finite we can find a specialization $r_1 \leadsto r_2$ with $s(r_2) = u_2$. However, then we conclude that $f$ is zero in $u'_2 = t(r_2)$ which contradicts the fact that $j^{-1}D(h) \cap Z$ is $R$-invariant and $u_2$ is in it. Thus $V$ is open.

Observe that $V \subset j^{-1}D(h)$ for our function $h \in I$. Thus we obtain an immersion $$ j' : V \longrightarrow \mathop{\rm Spec}(A_h) $$ Let $f' \in A_h$ be the image of $f$. Then $(j')^{-1}D(f')$ is the principal open determined by $g$ in the affine open $j^{-1}D(f)$ of $U$. Hence $(j')^{-1}D(f)$ is affine. Finally, $j'(V) \cap V(f') = j'(j^{-1}D(h) \cap Z)$ is closed in $\mathop{\rm Spec}(A_h/(f')) = \mathop{\rm Spec}((A/f)_h) = D(h) \cap V(f)$ by our choice of $h \in I$ and the ideal $I$. Hence we can apply Lemma 39.14.9 to conclude that $V$ is affine as claimed above. $\square$

    The code snippet corresponding to this tag is a part of the file more-groupoids.tex and is located in lines 2805–2815 (see updates for more information).

    \begin{lemma}
    \label{lemma-find-affine-codimension-1}
    Let $(U, R, s, t, c)$ be a groupoid scheme. Let $u \in U$. Assume
    \begin{enumerate}
    \item $s, t$ are finite morphisms,
    \item $U$ is separated and locally Noetherian,
    \item $\dim(\mathcal{O}_{U, u'}) \leq 1$ for every point $u'$
    in the orbit of $u$.
    \end{enumerate}
    Then $u$ is contained in an $R$-invariant affine open of $U$.
    \end{lemma}
    
    \begin{proof}
    The $R$-orbit of $u$ is finite. By conditions (2) and (3) it is contained
    in an affine open $U'$ of $U$, see
    Varieties, Proposition
    \ref{varieties-proposition-finite-set-of-points-of-codim-1-in-affine}.
    Then $t(s^{-1}(U \setminus U'))$ is an $R$-invariant
    closed subset of $U$ which does not contain $u$. Thus
    $U \setminus t(s^{-1}(U \setminus U'))$ is an $R$-invariant open
    of $U'$ containing $u$.
    Replacing $U$ by this open we may assume $U$ is quasi-affine.
    
    \medskip\noindent
    By Lemma \ref{lemma-find-affine-integral} we may replace $U$ by its reduction
    and assume $U$ is reduced. This means $R$-invariant subschemes
    $W' \subset W \subset U$ of
    Lemma \ref{lemma-finite-flat-over-almost-dense-subscheme}
    are equal $W' = W$. As $U = t(s^{-1}(\overline{W}))$ some point
    $u'$ of the $R$-orbit of $u$ is contained in $\overline{W}$
    and by Lemma \ref{lemma-find-affine-integral}
    we may replace $U$ by $\overline{W}$ and $u$ by $u'$.
    Hence we may assume there is
    a dense open $R$-invariant subscheme $W \subset U$ such that
    the morphisms $s_W, t_W$ of the restriction $(W, R_W, s_W, t_W, c_W)$ are
    finite locally free.
    
    \medskip\noindent
    If $u \in W$ then we are done by
    Groupoids, Lemma \ref{groupoids-lemma-find-invariant-affine}
    (because $W$ is quasi-affine so any finite set of points
    of $W$ is contained in an affine open, see
    Properties, Lemma \ref{properties-lemma-ample-finite-set-in-affine}).
    Thus we assume $u \not \in W$ and hence none of the points of the
    orbit of $u$ is in $W$. Let $\xi \in U$
    be a point with a nontrivial specialization to a point $u'$ in the orbit
    of $u$. Since there are no specializations among the points in the
    orbit of $u$ (Lemma \ref{lemma-no-specializations-map-to-same-point})
    we see that $\xi$ is not in the orbit.
    By assumption (3) we see that $\xi$ is a generic point of $U$
    and hence $\xi \in W$.
    As $U$ is Noetherian there are finitely many of these
    points $\xi_1, \ldots, \xi_m \in W$. Because $s_W, t_W$ are flat the orbit
    of each $\xi_j$ consists of generic points of irreducible components
    of $W$ (and hence $U$).
    
    \medskip\noindent
    Let $j : U \to \Spec(A)$ be an immersion of $U$ into an affine scheme
    (this is possible as $U$ is quasi-affine). Let $J \subset A$
    be an ideal such that $V(J) \cap j(W) = \emptyset$ and $V(J) \cup j(W)$
    is closed. Apply Lemma \ref{lemma-find-almost-invariant-function}
    to the groupoid scheme $(W, R_W, s_W, t_W, c_W)$, the morphism
    $j|_W : W \to \Spec(A)$, the points $\xi_j$, and the ideal $J$
    to find an $f \in J$ such that $(j|_W)^{-1}D(f)$ is an $R_W$-invariant
    affine open containing $\xi_j$ for all $j$. Since $f \in J$
    we see that $j^{-1}D(f) \subset W$, i.e., $j^{-1}D(f)$ is
    an $R$-invariant affine open of $U$ contained in $W$
    containing all $\xi_j$.
    
    \medskip\noindent
    Let $Z$ be the reduced induced closed subscheme structure on
    $$
    U \setminus j^{-1}D(f) = j^{-1}V(f).
    $$
    Then $Z$ is set theoretically
    $R$-invariant (but it may not be scheme theoretically $R$-invariant).
    Let $(Z, R_Z, s_Z, t_Z, c_Z)$ be the restriction of $R$ to $Z$.
    Since $Z \to U$ is finite, it follows that $s_Z$ and $t_Z$ are finite.
    Since $u \in Z$ the orbit of $u$ is in $Z$ and agrees with the
    $R_Z$-orbit of $u$ viewed as a point of $Z$. Since
    $\dim(\mathcal{O}_{U, u'}) \leq 1$ and since $\xi_j \not \in Z$
    for all $j$, we see that $\dim(\mathcal{O}_{Z, u'}) \leq 0$ for
    all $u'$ in the orbit of $u$. In other words, the $R_Z$-orbit of $u$
    consists of generic points of irreducible components of $Z$.
    
    \medskip\noindent
    Let $I \subset A$ be an ideal such that $V(I) \cap j(U) =\emptyset$
    and $V(I) \cup j(U)$ is closed. Apply
    Lemma \ref{lemma-find-almost-invariant-function} to
    the groupoid scheme $(Z, R_Z, s_Z, t_Z, c_Z)$, the restriction $j|_Z$,
    the ideal $I$, and the point $u \in Z$ to obtain $h \in I$ such that
    $j^{-1}D(h) \cap Z$ is an $R_Z$-invariant open affine containing $u$.
    
    \medskip\noindent
    Consider the $R_W$-invariant (Groupoids, Lemma
    \ref{groupoids-lemma-determinant-trick}) function
    $$
    g = 
    \text{Norm}_{s_W}(t_W^\sharp(j^\sharp(h)|_W)) \in \Gamma(W, \mathcal{O}_W)
    $$
    (In the following we only need the restriction of $g$ to $j^{-1}D(f)$ and
    in this case the norm is along a finite locally free morphism of affines.)
    We claim that
    $$
    V = (W_g \cap j^{-1}D(f)) \cup (j^{-1}D(h) \cap Z)
    $$
    is an $R$-invariant affine open of $U$ which finishes the proof of the lemma.
    It is set theoretically $R$-invariant by construction. As $V$ is a
    constuctible set, to see that it is open it suffices to show it is
    closed under generalization in $U$ (Topology, Lemma
    \ref{topology-lemma-characterize-closed-Noetherian}
    or the more general
    Topology, Lemma
    \ref{topology-lemma-constructible-stable-specialization-closed}).
    Since $W_g \cap j^{-1}D(f)$ is open in $U$, it suffices to consider
    a specialization $u_1 \leadsto u_2$ of $U$ with
    $u_2 \in j^{-1}D(h) \cap Z$.
    This means that $h$ is nonzero in $j(u_2)$ and $u_2 \in Z$.
    If $u_1 \in Z$, then $j(u_1) \leadsto j(u_2)$ and since
    $h$ is nonzero in $j(u_2)$ it is nonzero in $j(u_1)$ which
    implies $u_1 \in V$. If $u_1 \not \in Z$ and
    also not in $W_g \cap j^{-1}D(f)$, then $u_1 \in W$, $u_1 \not \in W_g$
    because the complement of $Z = j^{-1}V(f)$ is contained in $W \cap j^{-1}D(f)$.
    Hence there exists a point $r_1 \in R$ with $s(r_1) = u_1$
    such that $h$ is zero in $t(r_1)$. Since $s$ is finite we
    can find a specialization $r_1 \leadsto r_2$ with $s(r_2) = u_2$.
    However, then we conclude that $f$ is zero in $u'_2 = t(r_2)$
    which contradicts the fact that $j^{-1}D(h) \cap Z$
    is $R$-invariant and $u_2$ is in it. Thus $V$ is open.
    
    \medskip\noindent
    Observe that $V \subset j^{-1}D(h)$ for our function $h \in I$.
    Thus we obtain an immersion
    $$
    j' : V \longrightarrow \Spec(A_h)
    $$
    Let $f' \in A_h$ be the image of $f$. Then $(j')^{-1}D(f')$
    is the principal open determined by $g$ in the affine
    open $j^{-1}D(f)$ of $U$.
    Hence $(j')^{-1}D(f)$ is affine. Finally,
    $j'(V) \cap V(f') = j'(j^{-1}D(h) \cap Z)$
    is closed in $\Spec(A_h/(f')) = \Spec((A/f)_h) = D(h) \cap V(f)$
    by our choice of $h \in I$ and the ideal $I$. Hence we can apply
    Lemma \ref{lemma-get-affine}
    to conclude that $V$ is affine as claimed above.
    \end{proof}

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