## Tag `06DW`

## 70.9. Properties of groups over fields and groupoids on fields

The reader is advised to first look at the corresponding sections for groupoid schemes, see Groupoids, Section 38.7 and More on Groupoids, Section 39.10.

Situation 70.9.1. Here $S$ is a scheme, $k$ is a field over $S$, and $(G, m)$ is a group algebraic spaces over $\mathop{\rm Spec}(k)$.

Situation 70.9.2. Here $S$ is a scheme, $B$ is an algebraic space, and $(U, R, s, t, c)$ is a groupoid in algebraic spaces over $B$ with $U = \mathop{\rm Spec}(k)$ for some field $k$.

Note that in Situation 70.9.1 we obtain a groupoid in algebraic spaces \begin{equation} \tag{70.9.2.1} (\mathop{\rm Spec}(k), G, p, p, m) \end{equation} where $p : G \to \mathop{\rm Spec}(k)$ is the structure morphism of $G$, see Groupoids in Spaces, Lemma 69.14.1. This is a situation as in Situation 70.9.2. We will use this without further mention in the rest of this section.

Lemma 70.9.3. In Situation 70.9.2 the composition morphism $c : R \times_{s, U, t} R \to R$ is flat and universally open. In Situation 70.9.1 the group law $m : G \times_k G \to G$ is flat and universally open.

Proof.The composition is isomorphic to the projection map $\text{pr}_1 : R \times_{t, U, t} R \to R$ by Diagram (70.3.0.2). The projection is flat as a base change of the flat morphism $t$ and open by Morphisms of Spaces, Lemma 58.6.6. The second assertion follows immediately from the first because $m$ matches $c$ in (70.9.2.1). $\square$Note that the following lemma applies in particular when working with either quasi-separated or locally separated algebraic spaces (Decent Spaces, Lemma 59.14.2).

Lemma 70.9.4. In Situation 70.9.2 assume $R$ is a decent space. Then $R$ is a separated algebraic space. In Situation 70.9.1 assume that $G$ is a decent algebraic space. Then $G$ is separated algebraic space.

Proof.We first prove the second assertion. By Groupoids in Spaces, Lemma 69.6.1 we have to show that $e : S \to G$ is a closed immersion. This follows from Decent Spaces, Lemma 59.13.5.Next, we prove the second assertion. To do this we may replace $B$ by $S$. By the paragraph above the stabilizer group scheme $G \to U$ is separated. By Groupoids in Spaces, Lemma 69.28.2 the morphism $j = (t, s) : R \to U \times_S U$ is separated. As $U$ is the spectrum of a field the scheme $U \times_S U$ is affine (by the construction of fibre products in Schemes, Section 25.17). Hence $R$ is separated, see Morphisms of Spaces, Lemma 58.4.9. $\square$

Lemma 70.9.5. In Situation 70.9.2. Let $k \subset k'$ be a field extension, $U' = \mathop{\rm Spec}(k')$ and let $(U', R', s', t', c')$ be the restriction of $(U, R, s, t, c)$ via $U' \to U$. In the defining diagram $$ \xymatrix{ R' \ar[d] \ar[r] \ar@/_3pc/[dd]_{t'} \ar@/^1pc/[rr]^{s'} \ar@{..>}[rd] & R \times_{s, U} U' \ar[r] \ar[d] & U' \ar[d] \\ U' \times_{U, t} R \ar[d] \ar[r] & R \ar[r]^s \ar[d]_t & U \\ U' \ar[r] & U } $$ all the morphisms are surjective, flat, and universally open. The dotted arrow $R' \to R$ is in addition affine.

Proof.The morphism $U' \to U$ equals $\mathop{\rm Spec}(k') \to \mathop{\rm Spec}(k)$, hence is affine, surjective and flat. The morphisms $s, t : R \to U$ and the morphism $U' \to U$ are universally open by Morphisms, Lemma 28.22.4. Since $R$ is not empty and $U$ is the spectrum of a field the morphisms $s, t : R \to U$ are surjective and flat. Then you conclude by using Morphisms of Spaces, Lemmas 58.5.5, 58.5.4, 58.6.4, 58.20.5, 58.20.4, 58.29.4, and 58.29.3. $\square$Lemma 70.9.6. In Situation 70.9.2. For any point $r \in |R|$ there exist

- a field extension $k \subset k'$ with $k'$ algebraically closed,
- a point $r' : \mathop{\rm Spec}(k') \to R'$ where $(U', R', s', t', c')$ is the restriction of $(U, R, s, t, c)$ via $\mathop{\rm Spec}(k') \to \mathop{\rm Spec}(k)$
such that

- the point $r'$ maps to $r$ under the morphism $R' \to R$, and
- the maps $s' \circ r', t' \circ r' : \mathop{\rm Spec}(k') \to \mathop{\rm Spec}(k')$ are automorphisms.

Proof.Let's represent $r$ by a morphism $r : \mathop{\rm Spec}(K) \to R$ for some field $K$. To prove the lemma we have to find an algebraically closed field $k'$ and a commutative diagram $$ \xymatrix{ k' & k' \ar[l]^1 & \\ k' \ar[u]^\tau & K \ar[lu]^\sigma & k \ar[l]^-s \ar[lu]_i \\ & k \ar[lu]^i \ar[u]_t } $$ where $s, t : k \to K$ are the field maps coming from $s \circ r$ and $t \circ r$. In the proof of More on Groupoids, Lemma 39.10.5 it is shown how to construct such a diagram. $\square$Lemma 70.9.7. In Situation 70.9.2. If $r : \mathop{\rm Spec}(k) \to R$ is a morphism such that $s \circ r, t \circ r$ are automorphisms of $\mathop{\rm Spec}(k)$, then the map $$ R \longrightarrow R, \quad x \longmapsto c(r, x) $$ is an automorphism $R \to R$ which maps $e$ to $r$.

Proof.Proof is identical to the proof of More on Groupoids, Lemma 39.10.6. $\square$Lemma 70.9.8. In Situation 70.9.2 the algebraic space $R$ is geometrically unibranch. In Situation 70.9.1 the algebraic space $G$ is geometrically unibranch.

Proof.Let $r \in |R|$. We have to show that $R$ is geometrically unibranch at $r$. Combining Lemma 70.9.5 with Descent on Spaces, Lemma 65.8.1 we see that it suffices to prove this in case $k$ is algebraically closed and $r$ comes from a morphism $r : \mathop{\rm Spec}(k) \to R$ such that $s \circ r$ and $t \circ r$ are automorphisms of $\mathop{\rm Spec}(k)$. By Lemma 70.9.7 we reduce to the case that $r = e$ is the identity of $R$ and $k$ is algebraically closed.Assume $r = e$ and $k$ is algebraically closed. Let $A = \mathcal{O}_{R, e}$ be the étale local ring of $R$ at $e$ and let $C = \mathcal{O}_{R \times_{s, U, t} R, (e, e)}$ be the étale local ring of $R \times_{s, U, t} R$ at $(e, e)$. By More on Algebra, Lemma 15.89.9 the minimal prime ideals $\mathfrak q$ of $C$ correspond $1$-to-$1$ to pairs of minimal primes $\mathfrak p, \mathfrak p' \subset A$. On the other hand, the composition law induces a flat ring map $$ \xymatrix{ A \ar[r]_{c^\sharp} & C & \mathfrak q \\ & A \otimes_{s^\sharp, k, t^\sharp} A \ar[u] & \mathfrak p \otimes A + A \otimes \mathfrak p' \ar@{|}[u] } $$ Note that $(c^\sharp)^{-1}(\mathfrak q)$ contains both $\mathfrak p$ and $\mathfrak p'$ as the diagrams $$ \xymatrix{ A \ar[r]_{c^\sharp} & C \\ A \otimes_{s^\sharp, k} k \ar[u] & A \otimes_{s^\sharp, k, t^\sharp} A \ar[l]_{1 \otimes e^\sharp} \ar[u] } \quad\quad \xymatrix{ A \ar[r]_{c^\sharp} & C \\ k \otimes_{k, t^\sharp} A \ar[u] & A \otimes_{s^\sharp, k, t^\sharp} A \ar[l]_{e^\sharp \otimes 1} \ar[u] } $$ commute by (70.3.0.1). Since $c^\sharp$ is flat (as $c$ is a flat morphism by Lemma 70.9.3), we see that $(c^\sharp)^{-1}(\mathfrak q)$ is a minimal prime of $A$. Hence $\mathfrak p = (c^\sharp)^{-1}(\mathfrak q) = \mathfrak p'$. $\square$

In the following lemma we use dimension of algebraic spaces (at a point) as defined in Properties of Spaces, Section 57.8. We also use the dimension of the local ring defined in Properties of Spaces, Section 57.9 and transcendence degree of points, see Morphisms of Spaces, Section 58.32.

Lemma 70.9.9. In Situation 70.9.2 assume $s, t$ are locally of finite type. For all $r \in |R|$

- $\dim(R) = \dim_r(R)$,
- the transcendence degree of $r$ over $\mathop{\rm Spec}(k)$ via $s$ equals the transcendence degree of $r$ over $\mathop{\rm Spec}(k)$ via $t$, and
- if the transcendence degree mentioned in (2) is $0$, then $\dim(R) = \dim(\mathcal{O}_{R, \overline{r}})$.

Proof.Let $r \in |R|$. Denote $\text{trdeg}(r/_{\!\! s}k)$ the transcendence degree of $r$ over $\mathop{\rm Spec}(k)$ via $s$. Choose an étale morphism $\varphi : V \to R$ where $V$ is a scheme and $v \in V$ mapping to $r$. Using the definitions mentioned above the lemma we see that $$ \dim_r(R) = \dim_v(V) = \dim(\mathcal{O}_{V, v}) + \text{trdeg}_{s(k)}(\kappa(v)) = \dim(\mathcal{O}_{R, \overline{r}}) + \text{trdeg}(r/_{\!\! s}k) $$ and similarly for $t$ (the second equality by Morphisms, Lemma 28.27.1). Hence we see that $\text{trdeg}(r/_{\!\! s}k) = \text{trdeg}(r/_{\!\! t}k)$, i.e., (2) holds.Let $k \subset k'$ be a field extension. Note that the restriction $R'$ of $R$ to $\mathop{\rm Spec}(k')$ (see Lemma 70.9.5) is obtained from $R$ by two base changes by morphisms of fields. Thus Morphisms of Spaces, Lemma 58.33.3 shows the dimension of $R$ at a point is unchanged by this operation. Hence in order to prove (1) we may assume, by Lemma 70.9.6, that $r$ is represented by a morphism $r : \mathop{\rm Spec}(k) \to R$ such that both $s \circ r$ and $t \circ r$ are automorphisms of $\mathop{\rm Spec}(k)$. In this case there exists an automorphism $R \to R$ which maps $r$ to $e$ (Lemma 70.9.7). Hence we see that $\dim_r(R) = \dim_e(R)$ for any $r$. By definition this means that $\dim_r(R) = \dim(R)$.

Part (3) is a formal consequence of the results obtained in the discussion above. $\square$

Lemma 70.9.10. In Situation 70.9.1 assume $G$ locally of finite type. For all $g \in |G|$

- $\dim(G) = \dim_g(G)$,
- if the transcendence degree of $g$ over $k$ is $0$, then $\dim(G) = \dim(\mathcal{O}_{G, \overline{g}})$.

Proof.Immediate from Lemma 70.9.9 via (70.9.2.1). $\square$Lemma 70.9.11. In Situation 70.9.2 assume $s, t$ are locally of finite type. Let $G = \mathop{\rm Spec}(k) \times_{\Delta, \mathop{\rm Spec}(k) \times_B \mathop{\rm Spec}(k), t \times s} R$ be the stabilizer group algebraic space. Then we have $\dim(R) = \dim(G)$.

Proof.Since $G$ and $R$ are equidimensional (see Lemmas 70.9.9 and 70.9.10) it suffices to prove that $\dim_e(R) = \dim_e(G)$. Let $V$ be an affine scheme, $v \in V$, and let $\varphi : V \to R$ be an étale morphism of schemes such that $\varphi(v) = e$. Note that $V$ is a Noetherian scheme as $s \circ \varphi$ is locally of finite type as a composition of morphisms locally of finite type and as $V$ is quasi-compact (use Morphisms of Spaces, Lemmas 58.23.2, 58.38.8, and 58.28.5 and Morphisms, Lemma 28.14.6). Hence $V$ is locally connected (see Properties, Lemma 27.5.5 and Topology, Lemma 5.9.6). Thus we may replace $V$ by the connected component containing $v$ (it is still affine as it is an open and closed subscheme of $V$). Set $T = V_{red}$ equal to the reduction of $V$. Consider the two morphisms $a, b : T \to \mathop{\rm Spec}(k)$ given by $a = s \circ \varphi|_T$ and $b = t \circ \varphi|_T$. Note that $a, b$ induce the same field map $k \to \kappa(v)$ because $\varphi(v) = e$! Let $k_a \subset \Gamma(T, \mathcal{O}_T)$ be the integral closure of $a^\sharp(k) \subset \Gamma(T, \mathcal{O}_T)$. Similarly, let $k_b \subset \Gamma(T, \mathcal{O}_T)$ be the integral closure of $b^\sharp(k) \subset \Gamma(T, \mathcal{O}_T)$. By Varieties, Proposition 32.31.1 we see that $k_a = k_b$. Thus we obtain the following commutative diagram $$ \xymatrix{ k \ar[rd]^a \ar[rrrd] \\ & k_a = k_b \ar[r] & \Gamma(T, \mathcal{O}_T) \ar[r] & \kappa(v) \\ k \ar[ru]_b \ar[rrru] } $$ As discussed above the long arrows are equal. Since $k_a = k_b \to \kappa(v)$ is injective we conclude that the two morphisms $a$ and $b$ agree. Hence $T \to R$ factors through $G$. It follows that $R_{red} = G_{red}$ in an open neighbourhood of $e$ which certainly implies that $\dim_e(R) = \dim_e(G)$. $\square$

The code snippet corresponding to this tag is a part of the file `spaces-more-groupoids.tex` and is located in lines 624–1022 (see updates for more information).

```
\section{Properties of groups over fields and groupoids on fields}
\label{section-properties-groupoids-on-fields}
\noindent
The reader is advised to first look at the corresponding sections for
groupoid schemes, see
Groupoids, Section \ref{groupoids-section-properties-group-schemes-field}
and
More on Groupoids,
Section \ref{more-groupoids-section-properties-groupoids-on-fields}.
\begin{situation}
\label{situation-group-over-field}
Here $S$ is a scheme, $k$ is a field over $S$, and
$(G, m)$ is a group algebraic spaces over $\Spec(k)$.
\end{situation}
\begin{situation}
\label{situation-groupoid-on-field}
Here $S$ is a scheme, $B$ is an algebraic space, and
$(U, R, s, t, c)$ is a groupoid in algebraic spaces over $B$
with $U = \Spec(k)$ for some field $k$.
\end{situation}
\noindent
Note that in
Situation \ref{situation-group-over-field}
we obtain a groupoid in algebraic spaces
\begin{equation}
\label{equation-groupoid-from-group}
(\Spec(k), G, p, p, m)
\end{equation}
where $p : G \to \Spec(k)$ is the structure morphism of $G$, see
Groupoids in Spaces, Lemma \ref{spaces-groupoids-lemma-groupoid-from-action}.
This is a situation as in
Situation \ref{situation-groupoid-on-field}.
We will use this without further mention in the rest of this section.
\begin{lemma}
\label{lemma-groupoid-on-field-open-multiplication}
In
Situation \ref{situation-groupoid-on-field}
the composition morphism $c : R \times_{s, U, t} R \to R$ is flat and
universally open.
In
Situation \ref{situation-group-over-field}
the group law $m : G \times_k G \to G$ is flat and
universally open.
\end{lemma}
\begin{proof}
The composition is isomorphic to the projection map
$\text{pr}_1 : R \times_{t, U, t} R \to R$ by
Diagram (\ref{equation-pull}).
The projection is flat as a base change of the flat morphism $t$
and open by
Morphisms of Spaces,
Lemma \ref{spaces-morphisms-lemma-space-over-field-universally-open}.
The second assertion follows immediately from the first because
$m$ matches $c$ in (\ref{equation-groupoid-from-group}).
\end{proof}
\noindent
Note that the following lemma applies in particular when working
with either quasi-separated or locally separated algebraic spaces
(Decent Spaces, Lemma \ref{decent-spaces-lemma-locally-separated-decent}).
\begin{lemma}
\label{lemma-group-scheme-over-field-separated}
In Situation \ref{situation-groupoid-on-field}
assume $R$ is a decent space. Then $R$ is a separated algebraic space.
In Situation \ref{situation-group-over-field} assume that
$G$ is a decent algebraic space. Then $G$ is separated algebraic space.
\end{lemma}
\begin{proof}
We first prove the second assertion. By Groupoids in Spaces,
Lemma \ref{spaces-groupoids-lemma-group-scheme-separated}
we have to show that $e : S \to G$ is a closed immersion.
This follows from Decent Spaces, Lemma
\ref{decent-spaces-lemma-finite-residue-field-extension-finite}.
\medskip\noindent
Next, we prove the second assertion. To do this we may replace $B$ by $S$.
By the paragraph above the stabilizer group scheme $G \to U$ is separated. By
Groupoids in Spaces, Lemma \ref{spaces-groupoids-lemma-diagonal}
the morphism $j = (t, s) : R \to U \times_S U$ is separated.
As $U$ is the spectrum of a field the scheme
$U \times_S U$ is affine (by the construction of fibre products in
Schemes, Section \ref{schemes-section-fibre-products}).
Hence $R$ is separated, see
Morphisms of Spaces, Lemma
\ref{spaces-morphisms-lemma-separated-over-separated}.
\end{proof}
\begin{lemma}
\label{lemma-restrict-groupoid-on-field}
In
Situation \ref{situation-groupoid-on-field}.
Let $k \subset k'$ be a field extension, $U' = \Spec(k')$
and let $(U', R', s', t', c')$ be the restriction of
$(U, R, s, t, c)$ via $U' \to U$. In the defining diagram
$$
\xymatrix{
R' \ar[d] \ar[r] \ar@/_3pc/[dd]_{t'} \ar@/^1pc/[rr]^{s'} \ar@{..>}[rd] &
R \times_{s, U} U' \ar[r] \ar[d] &
U' \ar[d] \\
U' \times_{U, t} R \ar[d] \ar[r] &
R \ar[r]^s \ar[d]_t &
U \\
U' \ar[r] &
U
}
$$
all the morphisms are surjective, flat, and universally open.
The dotted arrow $R' \to R$ is in addition affine.
\end{lemma}
\begin{proof}
The morphism $U' \to U$ equals $\Spec(k') \to \Spec(k)$,
hence is affine, surjective and flat. The morphisms $s, t : R \to U$
and the morphism $U' \to U$ are universally open by
Morphisms, Lemma \ref{morphisms-lemma-scheme-over-field-universally-open}.
Since $R$ is not empty and $U$ is the spectrum of a field the morphisms
$s, t : R \to U$ are surjective and flat. Then you conclude by using
Morphisms of Spaces, Lemmas
\ref{spaces-morphisms-lemma-base-change-surjective},
\ref{spaces-morphisms-lemma-composition-surjective},
\ref{spaces-morphisms-lemma-composition-open},
\ref{spaces-morphisms-lemma-base-change-affine},
\ref{spaces-morphisms-lemma-composition-affine},
\ref{spaces-morphisms-lemma-base-change-flat}, and
\ref{spaces-morphisms-lemma-composition-flat}.
\end{proof}
\begin{lemma}
\label{lemma-groupoid-on-field-explain-points}
In
Situation \ref{situation-groupoid-on-field}.
For any point $r \in |R|$ there exist
\begin{enumerate}
\item a field extension $k \subset k'$ with $k'$ algebraically closed,
\item a point $r' : \Spec(k') \to R'$ where
$(U', R', s', t', c')$ is the restriction of $(U, R, s, t, c)$
via $\Spec(k') \to \Spec(k)$
\end{enumerate}
such that
\begin{enumerate}
\item the point $r'$ maps to $r$ under the morphism $R' \to R$, and
\item the maps
$s' \circ r', t' \circ r' : \Spec(k') \to \Spec(k')$
are automorphisms.
\end{enumerate}
\end{lemma}
\begin{proof}
Let's represent $r$ by a morphism $r : \Spec(K) \to R$ for some
field $K$. To prove the lemma we have to find an algebraically closed
field $k'$ and a commutative diagram
$$
\xymatrix{
k' & k' \ar[l]^1 & \\
k' \ar[u]^\tau & K \ar[lu]^\sigma & k \ar[l]^-s \ar[lu]_i \\
& k \ar[lu]^i \ar[u]_t
}
$$
where $s, t : k \to K$ are the field maps coming from
$s \circ r$ and $t \circ r$. In the proof of
More on Groupoids,
Lemma \ref{more-groupoids-lemma-groupoid-on-field-explain-points}
it is shown how to construct such a diagram.
\end{proof}
\begin{lemma}
\label{lemma-groupoid-on-field-move-point}
In
Situation \ref{situation-groupoid-on-field}.
If $r : \Spec(k) \to R$ is a morphism such that
$s \circ r, t \circ r$ are automorphisms of $\Spec(k)$, then the map
$$
R \longrightarrow R, \quad
x \longmapsto c(r, x)
$$
is an automorphism $R \to R$ which maps $e$ to $r$.
\end{lemma}
\begin{proof}
Proof is identical to the proof of
More on Groupoids,
Lemma \ref{more-groupoids-lemma-groupoid-on-field-move-point}.
\end{proof}
\begin{lemma}
\label{lemma-groupoid-on-field-geometrically-irreducible}
In
Situation \ref{situation-groupoid-on-field}
the algebraic space $R$ is geometrically unibranch. In
Situation \ref{situation-group-over-field}
the algebraic space $G$ is geometrically unibranch.
\end{lemma}
\begin{proof}
Let $r \in |R|$. We have to show that $R$ is geometrically unibranch
at $r$. Combining
Lemma \ref{lemma-restrict-groupoid-on-field}
with
Descent on Spaces, Lemma \ref{spaces-descent-lemma-descend-unibranch}
we see that it suffices to prove this in case $k$ is algebraically closed
and $r$ comes from a morphism $r : \Spec(k) \to R$ such that
$s \circ r$ and $t \circ r$
are automorphisms of $\Spec(k)$. By
Lemma \ref{lemma-groupoid-on-field-move-point}
we reduce to the case that $r = e$ is the identity of $R$ and $k$ is
algebraically closed.
\medskip\noindent
Assume $r = e$ and $k$ is algebraically closed. Let
$A = \mathcal{O}_{R, e}$ be the \'etale local ring of
$R$ at $e$ and let
$C = \mathcal{O}_{R \times_{s, U, t} R, (e, e)}$
be the \'etale local ring of $R \times_{s, U, t} R$ at $(e, e)$.
By More on Algebra, Lemma
\ref{more-algebra-lemma-minimal-primes-tensor-strictly-henselian}
the minimal prime ideals $\mathfrak q$ of $C$ correspond $1$-to-$1$
to pairs of minimal primes $\mathfrak p, \mathfrak p' \subset A$.
On the other hand, the composition law induces a flat ring map
$$
\xymatrix{
A \ar[r]_{c^\sharp} & C & \mathfrak q \\
& A \otimes_{s^\sharp, k, t^\sharp} A \ar[u] &
\mathfrak p \otimes A + A \otimes \mathfrak p' \ar@{|}[u]
}
$$
Note that $(c^\sharp)^{-1}(\mathfrak q)$ contains both $\mathfrak p$ and
$\mathfrak p'$ as the diagrams
$$
\xymatrix{
A \ar[r]_{c^\sharp} & C \\
A \otimes_{s^\sharp, k} k \ar[u] &
A \otimes_{s^\sharp, k, t^\sharp} A \ar[l]_{1 \otimes e^\sharp} \ar[u]
}
\quad\quad
\xymatrix{
A \ar[r]_{c^\sharp} & C \\
k \otimes_{k, t^\sharp} A \ar[u] &
A \otimes_{s^\sharp, k, t^\sharp} A \ar[l]_{e^\sharp \otimes 1} \ar[u]
}
$$
commute by (\ref{equation-diagram}).
Since $c^\sharp$ is flat (as $c$ is a flat morphism by
Lemma \ref{lemma-groupoid-on-field-open-multiplication}),
we see that $(c^\sharp)^{-1}(\mathfrak q)$ is a minimal prime
of $A$. Hence $\mathfrak p = (c^\sharp)^{-1}(\mathfrak q) = \mathfrak p'$.
\end{proof}
\noindent
In the following lemma we use dimension of algebraic spaces (at a point)
as defined in
Properties of Spaces, Section \ref{spaces-properties-section-dimension}.
We also use the dimension of the local ring defined in
Properties of Spaces, Section
\ref{spaces-properties-section-dimension-local-ring}
and transcendence degree of points, see
Morphisms of Spaces, Section \ref{spaces-morphisms-section-relative-dimension}.
\begin{lemma}
\label{lemma-groupoid-on-field-locally-finite-type-dimension}
In
Situation \ref{situation-groupoid-on-field}
assume $s, t$ are locally of finite type.
For all $r \in |R|$
\begin{enumerate}
\item $\dim(R) = \dim_r(R)$,
\item the transcendence degree of $r$ over $\Spec(k)$
via $s$ equals the transcendence degree of $r$ over $\Spec(k)$
via $t$, and
\item if the transcendence degree mentioned in (2) is $0$, then
$\dim(R) = \dim(\mathcal{O}_{R, \overline{r}})$.
\end{enumerate}
\end{lemma}
\begin{proof}
Let $r \in |R|$. Denote $\text{trdeg}(r/_{\!\! s}k)$ the transcendence
degree of $r$ over $\Spec(k)$ via $s$. Choose an \'etale morphism
$\varphi : V \to R$ where $V$ is a scheme and $v \in V$ mapping to $r$.
Using the definitions mentioned above the lemma we see that
$$
\dim_r(R) = \dim_v(V) =
\dim(\mathcal{O}_{V, v}) + \text{trdeg}_{s(k)}(\kappa(v)) =
\dim(\mathcal{O}_{R, \overline{r}}) + \text{trdeg}(r/_{\!\! s}k)
$$
and similarly for $t$ (the second equality by
Morphisms, Lemma \ref{morphisms-lemma-dimension-fibre-at-a-point}).
Hence we see that $\text{trdeg}(r/_{\!\! s}k) = \text{trdeg}(r/_{\!\! t}k)$,
i.e., (2) holds.
\medskip\noindent
Let $k \subset k'$ be a field extension. Note that the restriction $R'$
of $R$ to $\Spec(k')$ (see
Lemma \ref{lemma-restrict-groupoid-on-field})
is obtained from $R$ by two base changes by morphisms of fields. Thus
Morphisms of Spaces,
Lemma \ref{spaces-morphisms-lemma-dimension-fibre-after-base-change}
shows the dimension of $R$ at a point is unchanged by this operation.
Hence in order to prove (1) we may assume, by
Lemma \ref{lemma-groupoid-on-field-explain-points},
that $r$ is represented by a morphism $r : \Spec(k) \to R$ such
that both $s \circ r$ and $t \circ r$ are automorphisms of $\Spec(k)$.
In this case there exists an automorphism $R \to R$ which maps $r$ to $e$
(Lemma \ref{lemma-groupoid-on-field-move-point}).
Hence we see that $\dim_r(R) = \dim_e(R)$ for any $r$. By definition this
means that $\dim_r(R) = \dim(R)$.
\medskip\noindent
Part (3) is a formal consequence of the results obtained in the discussion
above.
\end{proof}
\begin{lemma}
\label{lemma-group-over-field-locally-finite-type-dimension}
In
Situation \ref{situation-group-over-field}
assume $G$ locally of finite type.
For all $g \in |G|$
\begin{enumerate}
\item $\dim(G) = \dim_g(G)$,
\item if the transcendence degree of $g$ over $k$ is $0$, then
$\dim(G) = \dim(\mathcal{O}_{G, \overline{g}})$.
\end{enumerate}
\end{lemma}
\begin{proof}
Immediate from
Lemma \ref{lemma-groupoid-on-field-locally-finite-type-dimension}
via (\ref{equation-groupoid-from-group}).
\end{proof}
\begin{lemma}
\label{lemma-groupoid-on-field-dimension-equal-stabilizer}
In
Situation \ref{situation-groupoid-on-field}
assume $s, t$ are locally of finite type.
Let
$G = \Spec(k)
\times_{\Delta, \Spec(k) \times_B \Spec(k), t \times s} R$
be the stabilizer group algebraic space.
Then we have $\dim(R) = \dim(G)$.
\end{lemma}
\begin{proof}
Since $G$ and $R$ are equidimensional (see
Lemmas \ref{lemma-groupoid-on-field-locally-finite-type-dimension} and
\ref{lemma-group-over-field-locally-finite-type-dimension})
it suffices to prove that $\dim_e(R) = \dim_e(G)$. Let $V$ be an affine scheme,
$v \in V$, and let $\varphi : V \to R$ be an \'etale morphism of schemes
such that $\varphi(v) = e$. Note that $V$ is a Noetherian scheme as
$s \circ \varphi$ is locally of finite type as a composition of morphisms
locally of finite type and as $V$ is quasi-compact (use
Morphisms of Spaces, Lemmas
\ref{spaces-morphisms-lemma-composition-finite-type},
\ref{spaces-morphisms-lemma-etale-locally-finite-presentation}, and
\ref{spaces-morphisms-lemma-finite-presentation-finite-type}
and
Morphisms, Lemma \ref{morphisms-lemma-finite-type-noetherian}).
Hence $V$ is locally connected (see
Properties, Lemma \ref{properties-lemma-Noetherian-topology}
and
Topology, Lemma \ref{topology-lemma-locally-Noetherian-locally-connected}).
Thus we may replace $V$ by the connected component containing $v$ (it
is still affine as it is an open and closed subscheme of $V$).
Set $T = V_{red}$ equal to the reduction of $V$. Consider the two
morphisms $a, b : T \to \Spec(k)$ given by
$a = s \circ \varphi|_T$ and $b = t \circ \varphi|_T$. Note that
$a, b$ induce the same field map $k \to \kappa(v)$ because $\varphi(v) = e$!
Let $k_a \subset \Gamma(T, \mathcal{O}_T)$ be the integral closure of
$a^\sharp(k) \subset \Gamma(T, \mathcal{O}_T)$. Similarly, let
$k_b \subset \Gamma(T, \mathcal{O}_T)$ be the integral closure of
$b^\sharp(k) \subset \Gamma(T, \mathcal{O}_T)$. By
Varieties, Proposition \ref{varieties-proposition-unique-base-field}
we see that $k_a = k_b$. Thus we obtain the following commutative diagram
$$
\xymatrix{
k \ar[rd]^a \ar[rrrd] \\
& k_a = k_b \ar[r] & \Gamma(T, \mathcal{O}_T) \ar[r] & \kappa(v) \\
k \ar[ru]_b \ar[rrru]
}
$$
As discussed above the long arrows are equal.
Since $k_a = k_b \to \kappa(v)$ is injective we conclude that
the two morphisms $a$ and $b$ agree. Hence $T \to R$ factors through $G$.
It follows that $R_{red} = G_{red}$ in an open neighbourhood of $e$
which certainly implies that $\dim_e(R) = \dim_e(G)$.
\end{proof}
```

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