Lemma 65.14.6. Notation and assumptions as in Lemma 65.14.3. If \mathop{\mathrm{Spec}}(k) \to U/G is a morphism, then there exist
a finite Galois extension k'/k,
a finite subgroup H \subset G,
an isomorphism H \to \text{Gal}(k'/k), and
an H-equivariant morphism \mathop{\mathrm{Spec}}(k') \to U.
Conversely, such data determine a morphism \mathop{\mathrm{Spec}}(k) \to U/G.
Proof.
Consider the fibre product V = \mathop{\mathrm{Spec}}(k) \times _{U/G} U. Here is a diagram
\xymatrix{ V \ar[r] \ar[d] & U \ar[d] \\ \mathop{\mathrm{Spec}}(k) \ar[r] & U/G }
Then V is a nonempty scheme étale over \mathop{\mathrm{Spec}}(k) and hence is a disjoint union V = \coprod _{i \in I} \mathop{\mathrm{Spec}}(k_ i) of spectra of fields k_ i finite separable over k (Morphisms, Lemma 29.36.7). We have
\begin{align*} V \times _{\mathop{\mathrm{Spec}}(k)} V & = (\mathop{\mathrm{Spec}}(k) \times _{U/G} U) \times _{\mathop{\mathrm{Spec}}(k)}(\mathop{\mathrm{Spec}}(k) \times _{U/G} U) \\ & = \mathop{\mathrm{Spec}}(k) \times _{U/G} U \times _{U/G} U \\ & = \mathop{\mathrm{Spec}}(k) \times _{U/G} U \times G \\ & = V \times G \end{align*}
The action of G on U induces an action of a : G \times V \to V. The displayed equality means that G \times V \to V \times _{\mathop{\mathrm{Spec}}(k)} V, (g, v) \mapsto (a(g, v), v) is an isomorphism. In particular we see that for every i we have an isomorphism H_ i \times \mathop{\mathrm{Spec}}(k_ i) \to \mathop{\mathrm{Spec}}(k_ i \otimes _ k k_ i) where H_ i \subset G is the subgroup of elements fixing i \in I. Thus H_ i is finite and is the Galois group of k_ i/k. We omit the converse construction.
\square
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