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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 distinct? If so, why?

Any help would be very much appreciated.

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    $\begingroup$ That may happen and it may fail to happen. This is what the arithmetic of $A$, or if you prefer, of the extension $B\supset A$, is all about. Take $A=\mathbb Z$ and $B=\mathbb Z[i]$. Then the primes $\mathfrak P$ above rational primes $\equiv1\pmod4$ occur in conjugate pairs, while all other primes (including $(1+i)$) are selfconjugate. $\endgroup$ – Lubin Jun 21 '13 at 18:29
  • $\begingroup$ @Lubin: Your comment is a perfectly good answer to the question. For various site-mechanical reasons it is better that questions actually get answered in the formal sense. So could you please leave your comment as an actual answer? $\endgroup$ – Pete L. Clark Jun 21 '13 at 18:47
  • $\begingroup$ @Lubin: many thanks. $\endgroup$ – Josh F. Jun 21 '13 at 18:49
  • $\begingroup$ @Lubin: I asked the question because I couldn't quite understand the proof in Neukirch's ANT of the result that (in our set-up) $G$ acts transitively on the set of all prime ideals $\frak{P}$ of $B$ lying above a given prime ideal $\frak{p}$ of $A$. The proof uses the Chinese Remainder Theorem applied to the prime ideals $\frak{P}^{\prime}$ and $\sigma\frak{P}$ for $\sigma \in G$ lying above $\frak{p}$. So let me ask you another question: are the ideals $\sigma\frak{P}$ coprime to each other? $\endgroup$ – Josh F. Jun 21 '13 at 19:03
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    $\begingroup$ @user68418 Coprime and unequal are the same for non-zero primes in a Dedekind domain. Neukirch probably takes the distinct conjugates of $\mathfrak{P}$. I think you could also prove this via "prime avoidance". $\endgroup$ – TTS Jun 21 '13 at 19:24

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