
Lemma 15.96.5. Let $G$ be a finite group acting on a ring $R$. For any two primes $\mathfrak q, \mathfrak q' \subset R$ lying over the same prime in $R^ G$ there exists a $\sigma \in G$ with $\sigma (\mathfrak q) = \mathfrak q'$.

Proof. The extension $R^ G \subset R$ is integral because every $x \in R$ is a root of the monic polynomial $\prod _{\sigma \in G}(T - \sigma (x))$ in $R^ G[T]$. Thus there are no inclusion relations among the primes lying over a given prime $\mathfrak p$ (Algebra, Lemma 10.35.20). If the lemma is wrong, then we can choose $x \in \mathfrak q'$, $x \not\in \sigma (\mathfrak q)$ for all $\sigma \in G$. See Algebra, Lemma 10.14.2. Then $y = \prod _{\sigma \in G} \sigma (x)$ is in $R^ G$ and in $\mathfrak p = R^ G \cap \mathfrak q'$. On the other hand, $x \not\in \sigma (\mathfrak q)$ for all $\sigma$ means $\sigma (x) \not\in \mathfrak q$ for all $\sigma$. Hence $y \not\in \mathfrak q$ as $\mathfrak q$ is a prime ideal. This is impossible as $y \in \mathfrak p \subset \mathfrak q$. $\square$

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