Situation 77.9.1. Here $S$ is a scheme, $k$ is a field over $S$, and $(G, m)$ is a group algebraic space over $\mathop{\mathrm{Spec}}(k)$.

## 77.9 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 39.7 and More on Groupoids, Section 40.10.

Situation 77.9.2. 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 $U = \mathop{\mathrm{Spec}}(k)$ for some field $k$.

Note that in Situation 77.9.1 we obtain a groupoid in algebraic spaces

where $p : G \to \mathop{\mathrm{Spec}}(k)$ is the structure morphism of $G$, see Groupoids in Spaces, Lemma 76.14.1. This is a situation as in Situation 77.9.2. We will use this without further mention in the rest of this section.

Lemma 77.9.3. In Situation 77.9.2 the composition morphism $c : R \times _{s, U, t} R \to R$ is flat and universally open. In Situation 77.9.1 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 (77.3.0.2). The projection is flat as a base change of the flat morphism $t$ and open by Morphisms of Spaces, Lemma 65.6.6. The second assertion follows immediately from the first because $m$ matches $c$ in (77.9.2.1).
$\square$

Note that the following lemma applies in particular when working with either quasi-separated or locally separated algebraic spaces (Decent Spaces, Lemma 66.15.2).

Lemma 77.9.4. In Situation 77.9.2 assume $R$ is a decent space. Then $R$ is a separated algebraic space. In Situation 77.9.1 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 76.6.1 we have to show that $e : S \to G$ is a closed immersion. This follows from Decent Spaces, Lemma 66.14.5.

Next, we prove the first 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 76.28.2 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 26.17). Hence $R$ is separated, see Morphisms of Spaces, Lemma 65.4.9. $\square$

Lemma 77.9.5. In Situation 77.9.2. Let $k \subset k'$ be a field extension, $U' = \mathop{\mathrm{Spec}}(k')$ and let $(U', R', s', t', c')$ be the restriction of $(U, R, s, t, c)$ via $U' \to U$. In the defining diagram

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 $\mathop{\mathrm{Spec}}(k') \to \mathop{\mathrm{Spec}}(k)$, 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 29.23.4. 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 65.5.5, 65.5.4, 65.6.4, 65.20.5, 65.20.4, 65.30.4, and 65.30.3.
$\square$

Lemma 77.9.6. In Situation 77.9.2. For any point $r \in |R|$ there exist

a field extension $k \subset k'$ with $k'$ algebraically closed,

a point $r' : \mathop{\mathrm{Spec}}(k') \to R'$ where $(U', R', s', t', c')$ is the restriction of $(U, R, s, t, c)$ via $\mathop{\mathrm{Spec}}(k') \to \mathop{\mathrm{Spec}}(k)$

such that

the point $r'$ maps to $r$ under the morphism $R' \to R$, and

the maps $s' \circ r', t' \circ r' : \mathop{\mathrm{Spec}}(k') \to \mathop{\mathrm{Spec}}(k')$ are automorphisms.

**Proof.**
Let's represent $r$ by a morphism $r : \mathop{\mathrm{Spec}}(K) \to R$ for some field $K$. To prove the lemma we have to find an algebraically closed field $k'$ and a commutative diagram

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 40.10.5 it is shown how to construct such a diagram. $\square$

Lemma 77.9.7. In Situation 77.9.2. If $r : \mathop{\mathrm{Spec}}(k) \to R$ is a morphism such that $s \circ r, t \circ r$ are automorphisms of $\mathop{\mathrm{Spec}}(k)$, then the map

is an automorphism $R \to R$ which maps $e$ to $r$.

**Proof.**
Proof is identical to the proof of More on Groupoids, Lemma 40.10.6.
$\square$

Lemma 77.9.8. In Situation 77.9.2 the algebraic space $R$ is geometrically unibranch. In Situation 77.9.1 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 77.9.5 with Descent on Spaces, Lemma 72.8.1 we see that it suffices to prove this in case $k$ is algebraically closed and $r$ comes from a morphism $r : \mathop{\mathrm{Spec}}(k) \to R$ such that $s \circ r$ and $t \circ r$ are automorphisms of $\mathop{\mathrm{Spec}}(k)$. By Lemma 77.9.7 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 15.104.9 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

Note that $(c^\sharp )^{-1}(\mathfrak q)$ contains both $\mathfrak p$ and $\mathfrak p'$ as the diagrams

commute by (77.3.0.1). Since $c^\sharp $ is flat (as $c$ is a flat morphism by Lemma 77.9.3), 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 64.9. We also use the dimension of the local ring defined in Properties of Spaces, Section 64.10 and transcendence degree of points, see Morphisms of Spaces, Section 65.33.

Lemma 77.9.9. In Situation 77.9.2 assume $s, t$ are locally of finite type. For all $r \in |R|$

$\dim (R) = \dim _ r(R)$,

the transcendence degree of $r$ over $\mathop{\mathrm{Spec}}(k)$ via $s$ equals the transcendence degree of $r$ over $\mathop{\mathrm{Spec}}(k)$ 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 $\mathop{\mathrm{Spec}}(k)$ 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

and similarly for $t$ (the second equality by Morphisms, Lemma 29.28.1). 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 $\mathop{\mathrm{Spec}}(k')$ (see Lemma 77.9.5) is obtained from $R$ by two base changes by morphisms of fields. Thus Morphisms of Spaces, Lemma 65.34.3 shows the dimension of $R$ at a point is unchanged by this operation. Hence in order to prove (1) we may assume, by Lemma 77.9.6, that $r$ is represented by a morphism $r : \mathop{\mathrm{Spec}}(k) \to R$ such that both $s \circ r$ and $t \circ r$ are automorphisms of $\mathop{\mathrm{Spec}}(k)$. In this case there exists an automorphism $R \to R$ which maps $r$ to $e$ (Lemma 77.9.7). 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 77.9.10. In Situation 77.9.1 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 77.9.9 via (77.9.2.1).
$\square$

Lemma 77.9.11. In Situation 77.9.2 assume $s, t$ are locally of finite type. Let $G = \mathop{\mathrm{Spec}}(k) \times _{\Delta , \mathop{\mathrm{Spec}}(k) \times _ B \mathop{\mathrm{Spec}}(k), t \times s} R$ be the stabilizer group algebraic space. Then we have $\dim (R) = \dim (G)$.

**Proof.**
Since $G$ and $R$ are equidimensional (see Lemmas 77.9.9 and 77.9.10) 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 65.23.2, 65.39.8, and 65.28.5 and Morphisms, Lemma 29.15.6). Hence $V$ is locally connected (see Properties, Lemma 28.5.5 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 $a, b : T \to \mathop{\mathrm{Spec}}(k)$ 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 33.31.1 we see that $k_ a = k_ b$. Thus we obtain the following commutative diagram

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 $\dim _ e(R) = \dim _ e(G)$. $\square$

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