The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

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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