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Tag 02Z5

Lemma 56.14.6. Notation and assumptions as in Lemma 56.14.3. If $\mathop{\rm Spec}(k) \to U/G$ 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 $\mathop{\rm Spec}(k') \to U$.

Conversely, such data determine a morphism $\mathop{\rm Spec}(k) \to U/G$.

Proof. Consider the fibre product $V = \mathop{\rm Spec}(k) \times_{U/G} U$. Here is a diagram $$\xymatrix{ V \ar[r] \ar[d] & U \ar[d] \\ \mathop{\rm Spec}(k) \ar[r] & U/G }$$ Then $V$ is a nonempty scheme étale over $\mathop{\rm Spec}(k)$ and hence is a disjoint union $V = \coprod_{i \in I} \mathop{\rm Spec}(k_i)$ of spectra of fields $k_i$ finite separable over $k$ (Morphisms, Lemma 28.34.7). We have \begin{align*} V \times_{\mathop{\rm Spec}(k)} V & = (\mathop{\rm Spec}(k) \times_{U/G} U) \times_{\mathop{\rm Spec}(k)}(\mathop{\rm Spec}(k) \times_{U/G} U) \\ & = \mathop{\rm Spec}(k) \times_{U/G} U \times_{U/G} U \\ & = \mathop{\rm 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 $G \times V \to V \times_{\mathop{\rm Spec}(k)} V$, $(g, v) \mapsto (a(g, v), v)$ is an isomorphism. In particular we see that for every $i$ we have an isomorphism $H_i \times \mathop{\rm Spec}(k_i) \to \mathop{\rm Spec}(k_i \otimes_k k_i)$ 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. $\square$

The code snippet corresponding to this tag is a part of the file spaces.tex and is located in lines 2324–2335 (see updates for more information).

\begin{lemma}
\label{lemma-quotient-field-map}
Notation and assumptions as in Lemma \ref{lemma-quotient}.
If $\Spec(k) \to U/G$ is a morphism, then there exist
\begin{enumerate}
\item a finite Galois extension $k'/k$,
\item a finite subgroup $H \subset G$,
\item an isomorphism $H \to \text{Gal}(k'/k)$, and
\item an $H$-equivariant morphism $\Spec(k') \to U$.
\end{enumerate}
Conversely, such data determine a morphism $\Spec(k) \to U/G$.
\end{lemma}

\begin{proof}
Consider the fibre product $V = \Spec(k) \times_{U/G} U$.
Here is a diagram
$$\xymatrix{ V \ar[r] \ar[d] & U \ar[d] \\ \Spec(k) \ar[r] & U/G }$$
Then $V$ is a nonempty scheme \'etale over $\Spec(k)$ and hence is a
disjoint union $V = \coprod_{i \in I} \Spec(k_i)$
of spectra of fields $k_i$ finite separable over $k$
(Morphisms, Lemma \ref{morphisms-lemma-etale-over-field}).
We have
\begin{align*}
V \times_{\Spec(k)} V
& =
(\Spec(k) \times_{U/G} U) \times_{\Spec(k)}(\Spec(k) \times_{U/G} U) \\
& =
\Spec(k) \times_{U/G} U \times_{U/G} U \\
& =
\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
$G \times V \to V \times_{\Spec(k)} V$, $(g, v) \mapsto (a(g, v), v)$
is an isomorphism. In particular we see that for every $i$ we have
an isomorphism $H_i \times \Spec(k_i) \to \Spec(k_i \otimes_k k_i)$
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.
\end{proof}

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