Lemma 58.12.3 (0BSS)—The Stacks project

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Table of contents
Part 3: Topics in Scheme Theory
Chapter 58: Fundamental Groups of Schemes
Section 58.12: Group actions and integral closure
Lemma 58.12.3 (cite)

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Lemma 58.12.3. Let $G$ be a finite group acting on a ring $R$ . Let $R^ G \to A$ be a ring map. Let $\mathfrak q' \subset A \otimes _{R^ G} R$ be a prime lying over the prime $\mathfrak q \subset R$ . Then

$I_\mathfrak q = \{ \sigma \in G \mid \sigma (\mathfrak q) = \mathfrak q\text{ and } \sigma \bmod \mathfrak q = \text{id}_{\kappa (\mathfrak q)}\}$

is equal to

$I_{\mathfrak q'} = \{ \sigma \in G \mid \sigma (\mathfrak q') = \mathfrak q'\text{ and } \sigma \bmod \mathfrak q' = \text{id}_{\kappa (\mathfrak q')}\}$

Proof. Since $\mathfrak q$ is the inverse image of $\mathfrak q'$ and since $\kappa (\mathfrak q) \subset \kappa (\mathfrak q')$ , we get $I_{\mathfrak q'} \subset I_\mathfrak q$ . Conversely, if $\sigma \in I_\mathfrak q$ , the $\sigma$ acts trivially on the fibre ring $A \otimes _{R^ G} \kappa (\mathfrak q)$ . Thus $\sigma$ fixes all the primes lying over $\mathfrak q$ and induces the identity on their residue fields. $\square$

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