Lemma 79.9.6. In Situation 79.9.2. For any point $r \in |R|$ there exist
a field extension $k'/k$ 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. $\square$
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)