A problem related to localization

Question

Let $A$ be integrally closed (not necessary be Dedekind), $K=\operatorname{Frac}(A),L/K$ Galois, $B$ is the integral closure of $A$ in $L,p$ is a maximal ideal in $A$. We know that the galois group $G$ acts transitively on maximal ideals in $B$ lying over $p$. If we weaken $p$ maximal to prime, let’s consider localization. Then we know that $G$ acts transitively on prime ideals in $B_{p}$ lying over $A_{p}$’s unique maximal ideal $m_p$. But now does $G$ acts transitively on prime ideal in $B$ lying over $p$?

Answer 1

Atiyah–Macdonald Exercise 5.13:

In the situation of Exercise 12 [Let $G$ be a finite group of automorphisms of a ring $A$, and let $A^G$ denote the subring of $G$-invariants, that is of all $x \in A$ such that $\sigma(x) = x$ for all $\sigma \in G$.], let $\mathfrak p$ be a prime ideal of $A^G$, and let $P$ be the set of prime ideals of $A$ whose contraction is $\mathfrak p$. Show that $G$ acts transitive on $P$. In particular, $P$ is finite.

So it suffices to show that $B^G = A$. The $\supseteq$ direction is clear. The $\subseteq$ direction follows from $B^G \subseteq B \cap K = A$ since $A$ is integrally closed in $K$.

So the answer is yes.

Answer 2

Thanks for Kneey Lau's illuminating proof, and I try to explain the original proof more detailedlly.

First, supposing that $p$ is maximal in A and $\beta_1$, $\beta_2$ are different prime ideals in B lying over p ( using $\beta|p$ as a symbol to express that), if for any $\sigma$ in G we have $\sigma(\beta_1)\not=\beta_2$, using CRT we can find an element x suffices $$x=0\ (module\ \beta_1) $$ $$x=1\ (module\ \sigma(\beta_2)$$ thus $$\sigma(x)=1\ (module\ \beta_2$$ thus $N_{L/K}(x)=\prod\sigma(x)\in A\cap\beta_1\ but\notin\beta_2$, causing a contridiction.

Second, for $p$ is a prime ideal, considering localize A at p, let $m_p$ be the unique maximal ideal in $A_p$, and in the former case, (omitting some verification here), we can let $q_1$, $q_2$ be maximal ideal in $B_p$ and $B_p|m_p$, let $\beta_i=q_i\cap B$.
Then the problem is already solved , the prolem I missed before is that $\sigma(\beta_1) \subseteq q_2\cap B=\beta_2$ can conclude $\sigma(\beta_1)=\beta_2$