Lemma 15.110.6. Let $G$ be a finite group acting on a ring $R$. Let $\mathfrak q \subset R$ be a prime lying over $\mathfrak p \subset R^ G$. Then $\kappa (\mathfrak q)/\kappa (\mathfrak p)$ is an algebraic normal extension and the map

\[ D = \{ \sigma \in G \mid \sigma (\mathfrak q) = \mathfrak q\} \longrightarrow \text{Aut}(\kappa (\mathfrak q)/\kappa (\mathfrak p)) \]

is surjective^{1}.

**Proof.**
With $A = (R^ G)_\mathfrak p$ and $B = A \otimes _{R^ G} R$ we see that $A = B^ G$ as localization is flat, see Lemma 15.110.4. Observe that $\mathfrak pA$ and $\mathfrak qB$ are prime ideals, $D$ is the stabilizer of $\mathfrak qB$, and $\kappa (\mathfrak p) = \kappa (\mathfrak pA)$ and $\kappa (\mathfrak q) = \kappa (\mathfrak qB)$. Thus we may replace $R$ by $B$ and assume that $\mathfrak p$ is a maximal ideal. Since $R^ G \subset R$ is an integral ring extension, we find that the maximal ideals of $R$ are exactly the primes lying over $\mathfrak p$ (follows from Algebra, Lemmas 10.36.20 and 10.36.22). By Lemma 15.110.5 there are finitely many of them $\mathfrak q = \mathfrak q_1, \mathfrak q_2, \ldots , \mathfrak q_ m$ and they form a single orbit for $G$. By the Chinese remainder theorem (Algebra, Lemma 10.15.4) the map $R \to \prod _{j = 1, \ldots , m} R/\sigma (\mathfrak q_ j)$ is surjective.

First we prove that the extension is normal. Pick an element $\alpha \in \kappa (\mathfrak q)$. We have to show that the minimal polynomial $P$ of $\alpha $ over $\kappa (\mathfrak p)$ splits completely. By the above we can choose $a \in \mathfrak q_2 \cap \ldots \cap \mathfrak q_ m$ mapping to $\alpha $ in $\kappa (\mathfrak q)$. Consider the polynomial $Q = \prod _{\sigma \in G} (T - \sigma (a))$ in $R^ G[T]$. The image of $Q$ in $R[T]$ splits completely into linear factors, hence the same is true for its image in $\kappa (\mathfrak q)[T]$. Since $P$ divides the image of $Q$ in $\kappa (\mathfrak p)[T]$ we conclude that $P$ splits completely into linear factors over $\kappa (\mathfrak q)$ as desired.

Since $\kappa (\mathfrak q)/\kappa (\mathfrak p)$ is normal we may assume $\kappa (\mathfrak q) = \kappa _1 \otimes _{\kappa (\mathfrak p)} \kappa _2$ with $\kappa _1/\kappa (\mathfrak p)$ purely inseparable and $\kappa _2/\kappa (\mathfrak p)$ Galois, see Fields, Lemma 9.27.3. $\alpha \in \kappa _2$ which generates $\kappa _2$ over $\kappa (\mathfrak p)$ if it is finite and a subfield of degree $> |G|$ if it is infinite (to get a contradiction). This is possible by Fields, Lemma 9.19.1. Pick $a$, $P$, and $Q$ as in the previous paragraph. If $\alpha ' \in \kappa _2$ is a Galois conjugate of $\alpha $, then the above shows there exists a $\sigma \in G$ such that $\sigma (a)$ maps to $\alpha '$. By our choice of $a$ (vanishing at other maximal ideals) this implies $\sigma \in D$ and that the image of $\sigma $ in $\text{Aut}(\kappa (\mathfrak q)/\kappa (\mathfrak p))$ maps $\alpha $ to $\alpha '$. Hence the surjectivity or the desired absurdity in case $\alpha $ has degree $> |G|$ over $\kappa (\mathfrak p)$.
$\square$

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