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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