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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.

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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.

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  • $\begingroup$ Could you please give an example of this? Many thanks $\endgroup$ – Josh F. Jun 21 '13 at 19:27
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    $\begingroup$ I did, namely $K=\mathbb{Q}$, $L=\mathbb{Q}(i)$, $\mathfrak{P}=(3)\subset\mathbb{Z}[i]$. This prime ideal $\mathfrak{P}$ is sent to itself by both elements of $\mathrm{Gal}(L/K)$, namely the identity and complex conjugation. $\endgroup$ – Zev Chonoles Jun 21 '13 at 19:29
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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).

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  • $\begingroup$ So suppose we have two unequal non-zero prime ideals $\frak{P}$ and $\frak{P}^{\prime}$ and their intersection $\frak{Q}$. Since $\frak{Q}$ is a prime ideal does it mean that $\frak{P}=\frak{Q}=\frak{P}^{\prime}$ if $\frak{Q}$ is non-zero by unique factorisation of ideals? Therefore we must have $\frak{Q}=(0)$? And I take it that this in turn implies that $\frak{P}+{\frak{P}^{\prime}}$= $B$ ? $\endgroup$ – Josh F. Jun 21 '13 at 19:47

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