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

1. a finite Galois extension $k'/k$,

2. a finite subgroup $H \subset G$,

3. an isomorphism $H \to \text{Gal}(k'/k)$, and

4. 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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