Lemma 40.10.5. Let $S$ be a scheme. Let $(U, R, s, t, c)$ be a groupoid scheme over $S$. Assume $U = \mathop{\mathrm{Spec}}(k)$ with $k$ a field. For any point $r \in R$ there exist

a field extension $k'/k$ with $k'$ algebraically closed,

a point $r' \in R'$ 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 $s', t' : R' \to \mathop{\mathrm{Spec}}(k')$ induce isomorphisms $k' \to \kappa (r')$.

**Proof.**
Translating the geometric statement into a statement on fields, this means that we can find a diagram

\[ \xymatrix{ k' & k' \ar[l]^1 & \\ k' \ar[u]^\tau & \kappa (r) \ar[lu]^\sigma & k \ar[l]^-s \ar[lu]_ i \\ & k \ar[lu]^ i \ar[u]_ t } \]

where $i : k \to k'$ is the embedding of $k$ into $k'$, the maps $s, t : k \to \kappa (r)$ are induced by $s, t : R \to U$, and the map $\tau : k' \to k'$ is an automorphism. To produce such a diagram we may proceed in the following way:

Pick $i : k \to k'$ a field map with $k'$ algebraically closed of very large transcendence degree over $k$.

Pick an embedding $\sigma : \kappa (r) \to k'$ such that $\sigma \circ s = i$. Such a $\sigma $ exists because we can just choose a transcendence basis $\{ x_\alpha \} _{\alpha \in A}$ of $\kappa (r)$ over $k$ and find $y_\alpha \in k'$, $\alpha \in A$ which are algebraically independent over $i(k)$, and map $s(k)(\{ x_\alpha \} )$ into $k'$ by the rules $s(\lambda ) \mapsto i(\lambda )$ for $\lambda \in k$ and $x_\alpha \mapsto y_\alpha $ for $\alpha \in A$. Then extend to $\tau : \kappa (\alpha ) \to k'$ using that $k'$ is algebraically closed.

Pick an automorphism $\tau : k' \to k'$ such that $\tau \circ i = \sigma \circ t$. To do this pick a transcendence basis $\{ x_\alpha \} _{\alpha \in A}$ of $k$ over its prime field. On the one hand, extend $\{ i(x_\alpha )\} $ to a transcendence basis of $k'$ by adding $\{ y_\beta \} _{\beta \in B}$ and extend $\{ \sigma (t(x_\alpha ))\} $ to a transcendence basis of $k'$ by adding $\{ z_\gamma \} _{\gamma \in C}$. As $k'$ is algebraically closed we can extend the isomorphism $\sigma \circ t \circ i^{-1} : i(k) \to \sigma (t(k))$ to an isomorphism $\tau ' : \overline{i(k)} \to \overline{\sigma (t(k))}$ of their algebraic closures in $k'$. As $k'$ has large transcendence degree we see that the sets $B$ and $C$ have the same cardinality. Thus we can use a bijection $B \to C$ to extend $\tau '$ to an isomorphism

\[ \overline{i(k)}(\{ y_\beta \} ) \longrightarrow \overline{\sigma (t(k))}(\{ z_\gamma \} ) \]

and then since $k'$ is the algebraic closure of both sides we see that this extends to an automorphism $\tau : k' \to k'$ as desired.

This proves the lemma.
$\square$

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