Exercise 5.13

Question

Fixing a prime ideal $P \subseteq A^G$, let $\mathcal{Q}$ be the set of primes in $A$ whose contraction is $P$. Show that $G$ acts transitively on $\mathcal{Q}$.

Amit Awasthi · Accepted Answer

extit{Fixing a prime ideal $P \subseteq A^G$, let $\mathcal{Q}$ be the set of primes in $A$ whose contraction is $P$. Show that $G$ acts transitively on $\mathcal{Q}$.}
\begin{proof}
  Consider $Q_1,Q_2 \in \mathcal{Q}$. If $x \in Q_1$, then $\prod_{\sigma \in G}\sigma(x) \in Q_1 \cap A^G = P \subseteq Q_2$, for $e \in G \implies x \mid \prod\sigma(x)$ and $	au(\prod\sigma(x)) = \prod(\sigma(x))$ for any $	au \in G$ since $G$ is a group. It follows $\sigma(x) \in Q_2$ for some $\sigma$ since $Q_2$ is prime, i.e., $x \in \sigma^{-1}(Q_2)$. Thus $Q_1 \subseteq \bigcup_{\sigma \in G}\sigma^{-1}(Q_2)$. $\sigma^{-1}(Q_2) = Q_2^c$ and so they are prime; by prime avoidance $Q_1 \subseteq \sigma^{-1}(Q_2)$ for some $\sigma \in G$. But since $Q_1 \cap A^G = \sigma^{-1}(Q_2) \cap A^G = P$, by Incomparability we have $Q_1 = \sigma^{-1}(Q_2)$, i.e., $G$ acts transitively on $\mathcal{Q}$.
\end{proof}

Exercise 5.13

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Exercise 5.13

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