The action of a Galois group on a prime ideal of a Dedekind domain

Question

This is a slight variant of a question I asked earlier.

Let $A$ be a commutative Dedekind domain and $K$ its field of fractions. Let $L/K$ be a finite Galois extension with Galois group $G$ and let $B$ be the integral closure of $A$ in $L$.

If $\frak{P}$ is a non-zero prime (maximal) ideal of $B$ is it true that as $\sigma$ runs through $G$ the prime ideals $\sigma\frak{P}$ of $B$ are all COPRIME? If so, why?

Help would again be very much appreciated.

Answer 1

This is already answered by Lubin's comment on your earlier question - given that it is sometimes the case that $\sigma(\mathfrak{P})$ equals $\mathfrak{P}$ for all $\sigma\in\mathrm{Gal}(L/K)$, e.g. $K=\mathbb{Q}$, $L=\mathbb{Q}(i)$, $\mathfrak{P}=(3)\subset\mathbb{Z}[i]$, then certainly there can be no guarantee that they will be coprime.

Answer 2

They are always either coprime or equal -- proof: their intersection is a prime ideal (by general ring theory) contained in both, so is 0 (since you have a Dedekind domain).