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groupoids.tex
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groupoids.tex
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\input{preamble}
% OK, start here.
%
\begin{document}
\title{Groupoid Schemes}
\maketitle
\phantomsection
\label{section-phantom}
\tableofcontents
\section{Introduction}
\label{section-introduction}
\noindent
This chapter is devoted to generalities concerning groupoid schemes.
See for example the beautiful paper \cite{K-M} by Keel and Mori.
\section{Notation}
\label{section-notation}
\noindent
Let $S$ be a scheme. If $U$, $T$ are schemes over $S$ we denote
$U(T)$ for the set of $T$-valued points of $U$ {\it over} $S$. In a formula:
$U(T) = \Mor_S(T, U)$. We try to reserve the letter $T$ to denote
a ``test scheme'' over $S$, as in the discussion that follows.
Suppose we are given schemes $X$, $Y$ over
$S$ and a morphism of schemes $f : X \to Y$ over $S$.
For any scheme $T$ over $S$ we get an induced map of sets
$$
f : X(T) \longrightarrow Y(T)
$$
which as indicated we denote by $f$ also. In fact this construction
is functorial in the scheme $T/S$. Yoneda's Lemma, see Categories,
Lemma \ref{categories-lemma-yoneda}, says that $f$ determines and is
determined by this transformation of functors $f : h_X \to h_Y$.
More generally, we use the same notation for maps between fibre
products. For example, if
$X$, $Y$, $Z$ are schemes over $S$, and if
$m : X \times_S Y \to Z \times_S Z$ is
a morphism of schemes over $S$, then we think of $m$ as corresponding
to a collection of maps between $T$-valued points
$$
X(T) \times Y(T) \longrightarrow Z(T) \times Z(T).
$$
And so on and so forth.
\medskip\noindent
We continue our convention to label projection maps starting with
index $0$, so we have $\text{pr}_0 : X \times_S Y \to X$ and
$\text{pr}_1 : X \times_S Y \to Y$.
\section{Equivalence relations}
\label{section-equivalence-relations}
\noindent
Recall that a {\it relation} $R$ on a set $A$ is just a subset
of $R \subset A \times A$. We usually write $a R b$ to indicate
$(a, b) \in R$. We say the relation is {\it transitive} if
$a R b, b R c \Rightarrow a R c$. We say the relation is
{\it reflexive} if $a R a$ for all $a \in A$. We say the relation is
{\it symmetric} if $a R b \Rightarrow b R a$.
A relation is called an {\it equivalence relation} if
it is transitive, reflexive and symmetric.
\medskip\noindent
In the setting of schemes we are going to relax the notion of a
relation a little bit and just require $R \to A \times A$ to
be a map. Here is the definition.
\begin{definition}
\label{definition-equivalence-relation}
Let $S$ be a scheme. Let $U$ be a scheme over $S$.
\begin{enumerate}
\item A {\it pre-relation} on $U$ over $S$ is any morphism
$j : R \to U \times_S U$. In this case we set
$t = \text{pr}_0 \circ j$ and $s = \text{pr}_1 \circ j$, so
that $j = (t, s)$.
\item A {\it relation} on $U$ over $S$ is a monomorphism
$j : R \to U \times_S U$.
\item A {\it pre-equivalence relation} is a pre-relation
$j : R \to U \times_S U$ such that the image of
$j : R(T) \to U(T) \times U(T)$ is an equivalence relation for
all $T/S$.
\item We say a morphism $R \to U \times_S U$ is
an {\it equivalence relation on $U$ over $S$}
if and only if for every $T/S$ the $T$-valued
points of $R$ define an equivalence relation
on the set of $T$-valued points of $U$.
\end{enumerate}
\end{definition}
\noindent
In other words, an equivalence relation is a pre-equivalence relation
such that $j$ is a relation.
\begin{lemma}
\label{lemma-restrict-relation}
Let $S$ be a scheme.
Let $U$ be a scheme over $S$.
Let $j : R \to U \times_S U$ be a pre-relation.
Let $g : U' \to U$ be a morphism of schemes.
Finally, set
$$
R' = (U' \times_S U')\times_{U \times_S U} R
\xrightarrow{j'}
U' \times_S U'
$$
Then $j'$ is a pre-relation on $U'$ over $S$.
If $j$ is a relation, then $j'$ is a relation.
If $j$ is a pre-equivalence relation, then $j'$ is a pre-equivalence relation.
If $j$ is an equivalence relation, then $j'$ is an equivalence relation.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{definition}
\label{definition-restrict-relation}
Let $S$ be a scheme.
Let $U$ be a scheme over $S$.
Let $j : R \to U \times_S U$ be a pre-relation.
Let $g : U' \to U$ be a morphism of schemes.
The pre-relation $j' : R' \to U' \times_S U'$ is called
the {\it restriction}, or {\it pullback} of the pre-relation $j$ to $U'$.
In this situation we sometimes write $R' = R|_{U'}$.
\end{definition}
\begin{lemma}
\label{lemma-pre-equivalence-equivalence-relation-points}
Let $j : R \to U \times_S U$ be a pre-relation.
Consider the relation on points of the scheme $U$ defined by
the rule
$$
x \sim y
\Leftrightarrow
\exists\ r \in R :
t(r) = x,
s(r) = y.
$$
If $j$ is a pre-equivalence relation then this is an
equivalence relation.
\end{lemma}
\begin{proof}
Suppose that $x \sim y$ and $y \sim z$.
Pick $r \in R$ with $t(r) = x$, $s(r) = y$ and
pick $r' \in R$ with $t(r') = y$, $s(r') = z$.
Pick a field $K$ fitting into the following commutative
diagram
$$
\xymatrix{
\kappa(r) \ar[r] & K \\
\kappa(y) \ar[u] \ar[r] & \kappa(r') \ar[u]
}
$$
Denote $x_K, y_K, z_K : \Spec(K) \to U$
the morphisms
$$
\begin{matrix}
\Spec(K) \to \Spec(\kappa(r))
\to
\Spec(\kappa(x)) \to U \\
\Spec(K) \to \Spec(\kappa(r))
\to
\Spec(\kappa(y)) \to U \\
\Spec(K) \to \Spec(\kappa(r'))
\to
\Spec(\kappa(z)) \to U
\end{matrix}
$$
By construction $(x_K, y_K) \in j(R(K))$ and
$(y_K, z_K) \in j(R(K))$. Since $j$ is a pre-equivalence relation
we see that also $(x_K, z_K) \in j(R(K))$.
This clearly implies that $x \sim z$.
\medskip\noindent
The proof that $\sim$ is reflexive and symmetric is omitted.
\end{proof}
\section{Group schemes}
\label{section-group-schemes}
\noindent
Let us recall that a {\it group} is a pair
$(G, m)$ where $G$ is a set, and $m : G \times G \to G$ is
a map of sets with the following properties:
\begin{enumerate}
\item (associativity) $m(g, m(g', g'')) = m(m(g, g'), g'')$
for all $g, g', g'' \in G$,
\item (identity) there exists a unique element $e \in G$
(called the {\it identity}, {\it unit}, or $1$ of $G$) such that
$m(g, e) = m(e, g) = g$ for all $g \in G$, and
\item (inverse) for all $g \in G$ there exists a $i(g) \in G$
such that $m(g, i(g)) = m(i(g), g) = e$, where $e$ is the
identity.
\end{enumerate}
Thus we obtain a map $e : \{*\} \to G$ and a map
$i : G \to G$ so that the quadruple $(G, m, e, i)$
satisfies the axioms listed above.
\medskip\noindent
A {\it homomorphism of groups} $\psi : (G, m) \to (G', m')$
is a map of sets $\psi : G \to G'$ such that
$m'(\psi(g), \psi(g')) = \psi(m(g, g'))$. This automatically
insures that $\psi(e) = e'$ and $i'(\psi(g)) = \psi(i(g))$.
(Obvious notation.) We will use this below.
\begin{definition}
\label{definition-group-scheme}
Let $S$ be a scheme.
\begin{enumerate}
\item A {\it group scheme over $S$} is a pair $(G, m)$, where
$G$ is a scheme over $S$ and $m : G \times_S G \to G$ is
a morphism of schemes over $S$ with the following property:
For every scheme $T$ over $S$ the pair $(G(T), m)$
is a group.
\item A {\it morphism $\psi : (G, m) \to (G', m')$ of group schemes over $S$}
is a morphism $\psi : G \to G'$ of schemes over $S$ such that for
every $T/S$ the induced map $\psi : G(T) \to G'(T)$ is a homomorphism
of groups.
\end{enumerate}
\end{definition}
\noindent
Let $(G, m)$ be a group scheme over the scheme $S$.
By the discussion above (and the discussion in Section \ref{section-notation})
we obtain morphisms of schemes over $S$:
(identity) $e : S \to G$ and (inverse) $i : G \to G$ such that
for every $T$ the quadruple $(G(T), m, e, i)$ satisfies the
axioms of a group listed above.
\medskip\noindent
Let $(G, m)$, $(G', m')$ be group schemes over $S$.
Let $f : G \to G'$ be a morphism of schemes over $S$.
It follows from the definition that $f$ is a morphism
of group schemes over $S$ if and only if the following diagram
is commutative:
$$
\xymatrix{
G \times_S G \ar[r]_-{f \times f} \ar[d]_m &
G' \times_S G' \ar[d]^m \\
G \ar[r]^f & G'
}
$$
\begin{lemma}
\label{lemma-base-change-group-scheme}
Let $(G, m)$ be a group scheme over $S$.
Let $S' \to S$ be a morphism of schemes.
The pullback $(G_{S'}, m_{S'})$ is a group scheme over $S'$.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{definition}
\label{definition-closed-subgroup-scheme}
Let $S$ be a scheme. Let $(G, m)$ be a group scheme over $S$.
\begin{enumerate}
\item A {\it closed subgroup scheme} of $G$ is a closed subscheme
$H \subset G$ such that $m|_{H \times_S H}$ factors through $H$ and induces a
group scheme structure on $H$ over $S$.
\item An {\it open subgroup scheme} of $G$ is an open subscheme
$G' \subset G$ such that $m|_{G' \times_S G'}$ factors through $G'$
and induces a group scheme structure on $G'$ over $S$.
\end{enumerate}
\end{definition}
\noindent
Alternatively, we could say that $H$ is a closed subgroup scheme of $G$
if it is a group scheme over $S$ endowed with a morphism of group schemes
$i : H \to G$ over $S$ which identifies $H$ with a closed subscheme of $G$.
\begin{definition}
\label{definition-smooth-group-scheme}
Let $S$ be a scheme. Let $(G, m)$ be a group scheme over $S$.
\begin{enumerate}
\item We say $G$ is a {\it smooth group scheme} if the structure
morphism $G \to S$ is smooth.
\item We say $G$ is a {\it flat group scheme} if the structure
morphism $G \to S$ is flat.
\item We say $G$ is a {\it separated group scheme} if the structure
morphism $G \to S$ is separated.
\end{enumerate}
Add more as needed.
\end{definition}
\section{Examples of group schemes}
\label{section-examples-group-schemes}
\begin{example}[Multiplicative group scheme]
\label{example-multiplicative-group}
Consider the functor which associates
to any scheme $T$ the group $\Gamma(T, \mathcal{O}_T^*)$
of units in the global sections of the structure sheaf.
This is representable by the scheme
$$
\mathbf{G}_m = \Spec(\mathbf{Z}[x, x^{-1}])
$$
The morphism giving the group structure is the morphism
\begin{eqnarray*}
\mathbf{G}_m \times \mathbf{G}_m & \to & \mathbf{G}_m \\
\Spec(\mathbf{Z}[x, x^{-1}] \otimes_{\mathbf{Z}} \mathbf{Z}[x, x^{-1}])
& \to &
\Spec(\mathbf{Z}[x, x^{-1}]) \\
\mathbf{Z}[x, x^{-1}] \otimes_{\mathbf{Z}} \mathbf{Z}[x, x^{-1}]
& \leftarrow &
\mathbf{Z}[x, x^{-1}] \\
x \otimes x & \leftarrow & x
\end{eqnarray*}
Hence we see that $\mathbf{G}_m$ is a group scheme over $\mathbf{Z}$.
For any scheme $S$ the base change $\mathbf{G}_{m, S}$ is a
group scheme over $S$ whose functor of points is
$$
T/S
\longmapsto
\mathbf{G}_{m, S}(T) = \mathbf{G}_m(T) = \Gamma(T, \mathcal{O}_T^*)
$$
as before.
\end{example}
\begin{example}[Roots of unity]
\label{example-roots-of-unity}
Let $n \in \mathbf{N}$.
Consider the functor which associates
to any scheme $T$ the subgroup of $\Gamma(T, \mathcal{O}_T^*)$
consisting of $n$th roots of unity.
This is representable by the scheme
$$
\mu_n = \Spec(\mathbf{Z}[x]/(x^n - 1)).
$$
The morphism giving the group structure is the morphism
\begin{eqnarray*}
\mu_n \times \mu_n & \to & \mu_n \\
\Spec(
\mathbf{Z}[x]/(x^n - 1)
\otimes_{\mathbf{Z}}
\mathbf{Z}[x]/(x^n - 1))
& \to &
\Spec(\mathbf{Z}[x]/(x^n - 1)) \\
\mathbf{Z}[x]/(x^n - 1) \otimes_{\mathbf{Z}} \mathbf{Z}[x]/(x^n - 1)
& \leftarrow &
\mathbf{Z}[x]/(x^n - 1) \\
x \otimes x & \leftarrow & x
\end{eqnarray*}
Hence we see that $\mu_n$ is a group scheme over $\mathbf{Z}$.
For any scheme $S$ the base change $\mu_{n, S}$ is a
group scheme over $S$ whose functor of points is
$$
T/S
\longmapsto
\mu_{n, S}(T) = \mu_n(T) = \{f \in \Gamma(T, \mathcal{O}_T^*) \mid f^n = 1\}
$$
as before.
\end{example}
\begin{example}[Additive group scheme]
\label{example-additive-group}
Consider the functor which associates
to any scheme $T$ the group $\Gamma(T, \mathcal{O}_T)$
of global sections of the structure sheaf.
This is representable by the scheme
$$
\mathbf{G}_a = \Spec(\mathbf{Z}[x])
$$
The morphism giving the group structure is the morphism
\begin{eqnarray*}
\mathbf{G}_a \times \mathbf{G}_a & \to & \mathbf{G}_a \\
\Spec(\mathbf{Z}[x] \otimes_{\mathbf{Z}} \mathbf{Z}[x])
& \to &
\Spec(\mathbf{Z}[x]) \\
\mathbf{Z}[x] \otimes_{\mathbf{Z}} \mathbf{Z}[x]
& \leftarrow &
\mathbf{Z}[x] \\
x \otimes 1 + 1 \otimes x & \leftarrow & x
\end{eqnarray*}
Hence we see that $\mathbf{G}_a$ is a group scheme over $\mathbf{Z}$.
For any scheme $S$ the base change $\mathbf{G}_{a, S}$ is a
group scheme over $S$ whose functor of points is
$$
T/S
\longmapsto
\mathbf{G}_{a, S}(T) = \mathbf{G}_a(T) = \Gamma(T, \mathcal{O}_T)
$$
as before.
\end{example}
\begin{example}[General linear group scheme]
\label{example-general-linear-group}
Let $n \geq 1$.
Consider the functor which associates
to any scheme $T$ the group
$$
\text{GL}_n(\Gamma(T, \mathcal{O}_T))
$$
of invertible $n \times n$ matrices over
the global sections of the structure sheaf.
This is representable by the scheme
$$
\text{GL}_n = \Spec(\mathbf{Z}[\{x_{ij}\}_{1 \leq i, j \leq n}][1/d])
$$
where $d = \det((x_{ij}))$ with $(x_{ij})$ the $n \times n$ matrix
with entry $x_{ij}$ in the $(i, j)$-spot.
The morphism giving the group structure is the morphism
\begin{eqnarray*}
\text{GL}_n \times \text{GL}_n & \to & \text{GL}_n \\
\Spec(\mathbf{Z}[x_{ij}, 1/d] \otimes_{\mathbf{Z}}
\mathbf{Z}[x_{ij}, 1/d])
& \to &
\Spec(\mathbf{Z}[x_{ij}, 1/d]) \\
\mathbf{Z}[x_{ij}, 1/d] \otimes_{\mathbf{Z}} \mathbf{Z}[x_{ij}, 1/d]
& \leftarrow &
\mathbf{Z}[x_{ij}, 1/d] \\
\sum x_{ik} \otimes x_{kj} & \leftarrow & x_{ij}
\end{eqnarray*}
Hence we see that $\text{GL}_n$ is a group scheme over $\mathbf{Z}$.
For any scheme $S$ the base change $\text{GL}_{n, S}$ is a
group scheme over $S$ whose functor of points is
$$
T/S
\longmapsto
\text{GL}_{n, S}(T) = \text{GL}_n(T) =\text{GL}_n(\Gamma(T, \mathcal{O}_T))
$$
as before.
\end{example}
\begin{example}
\label{example-determinant}
The determinant defines a morphisms of group schemes
$$
\det : \text{GL}_n \longrightarrow \mathbf{G}_m
$$
over $\mathbf{Z}$. By base change it gives a morphism
of group schemes $\text{GL}_{n, S} \to \mathbf{G}_{m, S}$
over any base scheme $S$.
\end{example}
\begin{example}[Constant group]
\label{example-constant-group}
Let $G$ be an abstract group. Consider the functor
which associates to any scheme $T$ the group
of locally constant maps $T \to G$ (where $T$ has the Zariski topology
and $G$ the discrete topology). This is representable by the scheme
$$
G_{\Spec(\mathbf{Z})} =
\coprod\nolimits_{g \in G} \Spec(\mathbf{Z}).
$$
The morphism giving the group structure is the morphism
$$
G_{\Spec(\mathbf{Z})}
\times_{\Spec(\mathbf{Z})}
G_{\Spec(\mathbf{Z})}
\longrightarrow
G_{\Spec(\mathbf{Z})}
$$
which maps the component corresponding to the pair $(g, g')$ to the
component corresponding to $gg'$. For any scheme $S$ the base change
$G_S$ is a group scheme over $S$ whose functor of points is
$$
T/S
\longmapsto
G_S(T) = \{f : T \to G \text{ locally constant}\}
$$
as before.
\end{example}
\section{Properties of group schemes}
\label{section-properties-group-schemes}
\noindent
In this section we collect some simple properties of group schemes which
hold over any base.
\begin{lemma}
\label{lemma-group-scheme-separated}
Let $S$ be a scheme.
Let $G$ be a group scheme over $S$.
Then $G \to S$ is separated (resp.\ quasi-separated) if and only if
the identity morphism $e : S \to G$ is a closed immersion
(resp.\ quasi-compact).
\end{lemma}
\begin{proof}
We recall that by
Schemes, Lemma \ref{schemes-lemma-section-immersion}
we have that $e$ is an immersion which is a closed immersion
(resp.\ quasi-compact) if $G \to S$ is separated (resp.\ quasi-separated).
For the converse, consider the diagram
$$
\xymatrix{
G \ar[r]_-{\Delta_{G/S}} \ar[d] &
G \times_S G \ar[d]^{(g, g') \mapsto m(i(g), g')} \\
S \ar[r]^e & G
}
$$
It is an exercise in the functorial point of view in algebraic geometry
to show that this diagram is cartesian. In other words, we see that
$\Delta_{G/S}$ is a base change of $e$. Hence if $e$ is a
closed immersion (resp.\ quasi-compact) so is $\Delta_{G/S}$, see
Schemes, Lemma \ref{schemes-lemma-base-change-immersion}
(resp.\ Schemes, Lemma
\ref{schemes-lemma-quasi-compact-preserved-base-change}).
\end{proof}
\begin{lemma}
\label{lemma-flat-action-on-group-scheme}
Let $S$ be a scheme.
Let $G$ be a group scheme over $S$.
Let $T$ be a scheme over $S$ and let $\psi : T \to G$ be a morphism over $S$.
If $T$ is flat over $S$, then the morphism
$$
T \times_S G \longrightarrow G, \quad
(t, g) \longmapsto m(\psi(t), g)
$$
is flat. In particular, if $G$ is flat over $S$, then
$m : G \times_S G \to G$ is flat.
\end{lemma}
\begin{proof}
Consider the diagram
$$
\xymatrix{
T \times_S G \ar[rrr]_{(t, g) \mapsto (t, m(\psi(t), g))} & & &
T \times_S G \ar[r]_{\text{pr}} \ar[d] &
G \ar[d] \\
& & &
T \ar[r] &
S
}
$$
The left top horizontal arrow is an isomorphism and the
square is cartesian. Hence the lemma follows from
Morphisms, Lemma \ref{morphisms-lemma-base-change-flat}.
\end{proof}
\begin{lemma}
\label{lemma-group-scheme-module-differentials}
Let $(G, m, e, i)$ be a group scheme over the scheme $S$.
Denote $f : G \to S$ the structure morphism. Assume $f$ is flat.
Then there exist canonical isomorphisms
$$
\Omega_{G/S} \cong f^*\mathcal{C}_{S/G} \cong f^*e^*\Omega_{G/S}
$$
where $\mathcal{C}_{S/G}$ denotes the conormal sheaf of the
immersion $e$. In particular, if $S$ is the spectrum of a field, then
$\Omega_{G/S}$ is a free $\mathcal{O}_G$-module.
\end{lemma}
\begin{proof}
In
Morphisms, Lemma \ref{morphisms-lemma-differentials-affine}
we identified $\Omega_{G/S}$ with the conormal sheaf of the
diagonal morphism $\Delta_{G/S}$. In the proof of
Lemma \ref{lemma-group-scheme-separated}
we showed that $\Delta_{G/S}$ is a base change of the immersion $e$
by the morphism $(g, g') \mapsto m(i(g), g')$. This morphism
is isomorphic to the morphism $(g, g') \mapsto m(g, g')$
hence is flat by
Lemma \ref{lemma-flat-action-on-group-scheme}.
Hence we get the first isomorphism by
Morphisms, Lemma \ref{morphisms-lemma-conormal-functorial-flat}.
By
Morphisms, Lemma \ref{morphisms-lemma-differentials-relative-immersion-section}
we have $\mathcal{C}_{S/G} \cong e^*\Omega_{G/S}$.
\medskip\noindent
If $S$ is the spectrum of a field, then $G \to S$ is flat, and
any $\mathcal{O}_S$-module on $S$ is free.
\end{proof}
\section{Properties of group schemes over a field}
\label{section-properties-group-schemes-field}
\noindent
In this section we collect some simple properties of group schemes over a
field.
\begin{lemma}
\label{lemma-group-scheme-over-field-open-multiplication}
If $(G, m)$ is a group scheme over a field $k$, then the
multiplication map $m : G \times_k G \to G$ is open.
\end{lemma}
\begin{proof}
The multiplication map is isomorphic to the projection map
$\text{pr}_0 : G \times_k G \to G$
because the diagram
$$
\xymatrix{
G \times_k G \ar[d]^m \ar[rrr]_{(g, g') \mapsto (m(g, g'), g')} & & &
G \times_k G \ar[d]^{(g, g') \mapsto g} \\
G \ar[rrr]^{\text{id}} & & & G
}
$$
is commutative with isomorphisms as horizontal arrows. The projection
is open by
Morphisms, Lemma \ref{morphisms-lemma-scheme-over-field-universally-open}.
\end{proof}
\begin{lemma}
\label{lemma-group-scheme-over-field-separated}
Let $G$ be a group scheme over a field.
Then $G$ is a separated scheme.
\end{lemma}
\begin{proof}
Say $S = \Spec(k)$ with $k$ a field, and let $G$ be a group scheme
over $S$. By
Lemma \ref{lemma-group-scheme-separated}
we have to show that $e : S \to G$ is a closed immersion. By
Morphisms, Lemma
\ref{morphisms-lemma-algebraic-residue-field-extension-closed-point-fibre}
the image of $e : S \to G$ is a closed point of $G$.
It is clear that $\mathcal{O}_G \to e_*\mathcal{O}_S$ is surjective,
since $e_*\mathcal{O}_S$ is a skyscraper sheaf supported at the neutral
element of $G$ with value $k$. We conclude that $e$ is a closed immersion by
Schemes, Lemma \ref{schemes-lemma-characterize-closed-immersions}.
\end{proof}
\begin{lemma}
\label{lemma-group-scheme-field-geometrically-irreducible}
Let $G$ be a group scheme over a field $k$.
Then
\begin{enumerate}
\item every local ring $\mathcal{O}_{G, g}$ of $G$ has a unique
minimal prime ideal,
\item there is exactly one irreducible component $Z$ of $G$
passing through $e$, and
\item $Z$ is geometrically irreducible over $k$.
\end{enumerate}
\end{lemma}
\begin{proof}
For any point $g \in G$ there exists a field extension
$k \subset K$ and a $K$-valued point $g' \in G(K)$ mapping to
$g$. If we think of $g'$ as a $K$-rational point of the
group scheme $G_K$, then we see that
$\mathcal{O}_{G, g} \to \mathcal{O}_{G_K, g'}$ is a faithfully flat
local ring map (as $G_K \to G$ is flat, and a local flat ring map
is faithfully flat, see
Algebra, Lemma \ref{algebra-lemma-local-flat-ff}).
The result for $\mathcal{O}_{G_K, g'}$ implies the
result for $\mathcal{O}_{G, g}$, see
Algebra, Lemma \ref{algebra-lemma-injective-minimal-primes-in-image}.
Hence in order to prove (1) it suffices to
prove it for $k$-rational points $g$ of $G$. In this case
translation by $g$ defines an automorphism $G \to G$
which maps $e$ to $g$. Hence $\mathcal{O}_{G, g} \cong \mathcal{O}_{G, e}$.
In this way we see that (2) implies (1), since irreducible components
passing through $e$ correspond one to one with minimal prime ideals
of $\mathcal{O}_{G, e}$.
\medskip\noindent
In order to prove (2) and (3) it suffices to prove (2) when $k$
is algebraically closed. In this case, let $Z_1$, $Z_2$ be two
irreducible components of $G$ passing through $e$.
Since $k$ is algebraically closed the closed subscheme
$Z_1 \times_k Z_2 \subset G \times_k G$ is irreducible too, see
Varieties, Lemma \ref{varieties-lemma-bijection-irreducible-components}.
Hence $m(Z_1 \times_k Z_2)$ is contained in an irreducible
component of $G$. On the other hand it contains
$Z_1$ and $Z_2$ since $m|_{e \times G} = \text{id}_G$ and
$m|_{G \times e} = \text{id}_G$. We conclude $Z_1 = Z_2$ as desired.
\end{proof}
\begin{remark}
\label{remark-warning-group-scheme-geometrically-irreducible}
Warning: The result of
Lemma \ref{lemma-group-scheme-field-geometrically-irreducible}
does not mean that every irreducible component of $G/k$ is
geometrically irreducible. For example the group scheme
$\mu_{3, \mathbf{Q}} = \Spec(\mathbf{Q}[x]/(x^3 - 1))$
over $\mathbf{Q}$ has two irreducible components corresponding
to the factorization $x^3 - 1 = (x - 1)(x^2 + x + 1)$.
The first factor corresponds to the irreducible component
passing through the identity, and the second irreducible component
is not geometrically irreducible over $\Spec(\mathbf{Q})$.
\end{remark}
\begin{lemma}
\label{lemma-group-scheme-finite-type-field}
Let $G$ be a group scheme which is locally of finite type over a
field $k$. Then $G$ is equidimensional and
$\dim(G) = \dim_g(G)$ for all $g \in G$.
For any closed point $g \in G$ we have $\dim(G) = \dim(\mathcal{O}_{G, g})$.
\end{lemma}
\begin{proof}
Let us first prove that $\dim_g(G) = \dim_{g'}(G)$ for any
pair of points $g, g' \in G$. By
Morphisms, Lemma \ref{morphisms-lemma-dimension-fibre-after-base-change}
we may extend the ground field at will. Hence we may assume that
both $g$ and $g'$ are defined over $k$. Hence there exists an
automorphism of $G$ mapping $g$ to $g'$, whence the equality.
By
Morphisms, Lemma \ref{morphisms-lemma-dimension-fibre-at-a-point}
we have
$\dim_g(G) = \dim(\mathcal{O}_{G, g}) +
\text{trdeg}_k(\kappa(g))$.
On the other hand, the dimension of $G$ (or any open subset of $G$)
is the supremum of the dimensions of the local rings of of $G$, see
Properties, Lemma \ref{properties-lemma-codimension-local-ring}.
Clearly this is maximal for closed points $g$ in which case
$\text{trdeg}_k(\kappa(g)) = 0$ (by the Hilbert Nullstellensatz, see
Morphisms, Section \ref{morphisms-section-points-finite-type}).
Hence the lemma follows.
\end{proof}
\noindent
The following result is sometimes referred to as Cartier's theorem.
\begin{lemma}
\label{lemma-group-scheme-characteristic-zero-smooth}
Let $G$ be a group scheme which is locally of finite type
over a field $k$ of characteristic zero. Then the structure
morphism $G \to \Spec(k)$ is smooth, i.e., $G$ is a smooth
group scheme.
\end{lemma}
\begin{proof}
By
Lemma \ref{lemma-group-scheme-module-differentials}
the module of differentials of $G$ over $k$ is free.
Hence smoothness follows from
Varieties, Lemma \ref{varieties-lemma-char-zero-differentials-free-smooth}.
\end{proof}
\begin{remark}
\label{remark-when-reduced}
Any group scheme over a field of characteristic $0$ is reduced, see
\cite[I, Theorem 1.1 and I, Corollary 3.9, and II, Theorem 2.4]{Perrin-thesis}
and also
\cite[Proposition 4.2.8]{Perrin}.
This was a question raised in
\cite[page 80]{Oort}.
We have seen in
Lemma \ref{lemma-group-scheme-characteristic-zero-smooth}
that this holds when the group scheme is locally of finite type.
\end{remark}
\begin{lemma}
\label{lemma-reduced-group-scheme-prefect-field-characteristic-p-smooth}
Let $G$ be a group scheme which is locally of finite type
over a perfect field $k$ of characteristic $p > 0$ (see
Lemma \ref{lemma-group-scheme-characteristic-zero-smooth}
for the characteristic zero case). If $G$ is reduced then the structure
morphism $G \to \Spec(k)$ is smooth, i.e., $G$ is a smooth
group scheme.
\end{lemma}
\begin{proof}
By
Lemma \ref{lemma-group-scheme-module-differentials}
the sheaf $\Omega_{G/k}$ is free. Hence the lemma follows from
Varieties, Lemma \ref{varieties-lemma-char-p-differentials-free-smooth}.
\end{proof}
\begin{remark}
\label{remark-reduced-smooth-not-true-general}
Let $k$ be a field of characteristic $p > 0$.
Let $\alpha \in k$ be an element which is not a $p$th power.
The closed subgroup scheme
$$
G = V(x^p + \alpha y^p) \subset \mathbf{G}_{a, k}^2
$$
is reduced and irreducible but not smooth (not even normal).
\end{remark}
\begin{lemma}
\label{lemma-reduced-subgroup-scheme-perfect}
Let $G$ be a group scheme over a perfect field $k$.
Then the reduction $G_{red}$ of $G$ is a closed subgroup scheme of $G$.
\end{lemma}
\begin{proof}
Omitted. Hint: Use that $G_{red} \times_k G_{red}$ is reduced by
Varieties, Lemmas \ref{varieties-lemma-perfect-reduced} and
\ref{varieties-lemma-geometrically-reduced-any-base-change}.
\end{proof}
\noindent
The next lemma will be generalized slightly in
More on Groupoids, Lemma
\ref{more-groupoids-lemma-open-image-is-closed}.
Namely, if $G' \to G$ is a morphism of group schemes over a field
whose image is open, then its image is closed.
\begin{lemma}
\label{lemma-open-subgroup-closed-over-field}
Let $G$ be group scheme over a field $k$.
Let $G' \subset G$ be an open subgroup scheme.
Then $G'$ is open and closed in $G$.
\end{lemma}
\begin{proof}
Suppose that $k \subset K$ is a field extension such that
$G'_K \subset G_K$ is closed. Then it follows from
Morphisms, Lemma \ref{morphisms-lemma-fpqc-quotient-topology}
that $G'$ is closed (as $G_K \to G$ is flat, quasi-compact and surjective).
Hence it suffices to prove the lemma after replacing $k$ by some
extension. Choose $K$ to be an algebraically closed field extension of
very large cardinality. Then by
Varieties, Lemma \ref{varieties-lemma-make-Jacobson},
we see that $G_K$ is a Jacobson scheme all of whose closed points have residue
field equal to $K$. In other words we may assume $G$ is a Jacobson
scheme all of whose closed points have residue field $k$.
\medskip\noindent
Let $Z = G \setminus G'$. We have to show that $Z$ is open.
Because $G$ is Jacobson and $Z$ is closed
the closed points of $Z$ are dense in $Z$.
Moreover any closed point $z \in Z$ is a $k$-rational point and hence
we translation by $z$ defines an automorphism $L_z : G \to G$,
$g \mapsto m(z, g)$ with $e \mapsto z$. As $G'$ is a subgroup scheme we
conclude that $L_z(G') \subset Z$. Altogether we see that
$$
Z = \bigcup\nolimits_{z \in Z(k)} L_z(G')
$$
is a union of open subsets, and hence open as desired.
\end{proof}
\begin{lemma}
\label{lemma-immersion-group-schemes-closed-over-field}
Let $i : G' \to G$ be an immersion of group schemes over a field $k$.
Then $i$ is a closed immersion, i.e., $i(G')$ is a closed subgroup scheme
of $G$.
\end{lemma}
\begin{proof}
To show that $i$ is a closed immersion it suffices to show that
$i(G')$ is a closed subset of $G$. Let $k \subset k'$ be a perfect
extension of $k$. If $i(G'_{k'}) \subset G_{k'}$ is closed, then
$i(G') \subset G$ is closed by
Morphisms, Lemma \ref{morphisms-lemma-fpqc-quotient-topology}
(as $G_{k'} \to G$ is flat, quasi-compact and surjective).
Hence we may and do assume $k$ is perfect. We will use without further
mention that products of reduced schemes over $k$ are reduced.
We may replace $G'$ and $G$ by their reductions, see
Lemma \ref{lemma-reduced-subgroup-scheme-perfect}.
Let $\overline{G'} \subset G$ be the closure of $i(G')$ viewed
as a reduced closed subscheme. By
Varieties, Lemma \ref{varieties-lemma-closure-of-product}
we conclude that $\overline{G'} \times_k \overline{G'}$
is the closure of the image of $G' \times_k G' \to G \times_k G$. Hence
$$
m\Big(\overline{G'} \times_k \overline{G'}\Big)
\subset \overline{G'}
$$
as $m$ is continuous. It follows that $\overline{G'} \subset G$
is a (reduced) closed subgroup scheme. By
Lemma \ref{lemma-open-subgroup-closed-over-field}
we see that $i(G') \subset \overline{G'}$ is also closed
which implies that $i(G') = \overline{G'}$ as desired.
\end{proof}
\begin{lemma}
\label{lemma-group-scheme-field-countable-affine}
Let $G$ be a group scheme over a field.
There exists an open and closed subscheme $G' \subset G$
which is a countable union of affines.
\end{lemma}
\begin{proof}
Let $e \in U(k)$ be a quasi-compact open neighbourhood of the identity
element. By replacing $U$ by $U \cap i(U)$ we may assume that
$U$ is invariant under the inverse map. As $G$ is separated this is
still a quasi-compact set. Set
$$
G' = \bigcup\nolimits_{n \geq 1} m_n(U \times_k \ldots \times_k U)
$$
where $m_n : G \times_k \ldots \times_k G \to G$ is the $n$-slot
multiplication map
$(g_1, \ldots, g_n) \mapsto m(m(\ldots (m(g_1, g_2), g_3), \ldots ), g_n)$.
Each of these maps are open (see
Lemma \ref{lemma-group-scheme-over-field-open-multiplication})
hence $G'$ is an open subgroup scheme. By
Lemma \ref{lemma-open-subgroup-closed-over-field}
it is also a closed subgroup scheme.
\end{proof}
\begin{remark}
\label{remark-easy}
If $G$ is a group scheme over a field, is there always a quasi-compact
open and closed subgroup scheme? Or is there a counter example?
\end{remark}
\section{Actions of group schemes}
\label{section-action-group-scheme}
\noindent
Let $(G, m)$ be a group and let $V$ be a set.
Recall that a {\it (left) action} of $G$ on $V$ is given
by a map $a : G \times V \to V$ such that
\begin{enumerate}
\item (associativity) $a(m(g, g'), v) = a(g, a(g', v))$ for all
$g, g' \in G$ and $v \in V$, and
\item (identity) $a(e, v) = v$ for all $v \in V$.
\end{enumerate}
We also say that $V$ is a {\it $G$-set} (this usually means we
drop the $a$ from the notation -- which is abuse of notation).
A {\it map of $G$-sets} $\psi : V \to V'$ is any set map
such that $\psi(a(g, v)) = a(g, \psi(v))$ for all $v \in V$.
\begin{definition}
\label{definition-action-group-scheme}
Let $S$ be a scheme. Let $(G, m)$ be a group scheme over $S$.
\begin{enumerate}
\item An {\it action of $G$ on the scheme $X/S$} is
a morphism $a : G \times_S X \to X$ over $S$ such that
for every $T/S$ the map $a : G(T) \times X(T) \to X(T)$
defines the structure of a $G(T)$-set on $X(T)$.
\item Suppose that $X$, $Y$ are schemes over $S$ each endowed
with an action of $G$. An {\it equivariant} or more precisely
a {\it $G$-equivariant} morphism $\psi : X \to Y$
is a morphism of schemes over $S$ such
that for every $T/S$ the map $\psi : X(T) \to Y(T)$ is
a morphism of $G(T)$-sets.
\end{enumerate}
\end{definition}
\noindent
In situation (1) this means that the diagrams
\begin{equation}
\label{equation-action}
\xymatrix{
G \times_S G \times_S X \ar[r]_-{1_G \times a} \ar[d]_{m \times 1_X} &
G \times_S X \ar[d]^a \\
G \times_S X \ar[r]^a & X
}
\quad
\xymatrix{
G \times_S X \ar[r]_-a & X \\
X\ar[u]^{e \times 1_X} \ar[ru]_{1_X}
}
\end{equation}
are commutative. In situation (2) this just means that the diagram
$$
\xymatrix{
G \times_S X \ar[r]_-{\text{id} \times f} \ar[d]_a &
G \times_S Y \ar[d]^a \\
X \ar[r]^f & Y
}
$$
commutes.
\begin{definition}
\label{definition-free-action}
Let $S$, $G \to S$, and $X \to S$ as in
Definition \ref{definition-action-group-scheme}.
Let $a : G \times_S X \to X$ be an action of $G$ on $X/S$.