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Let $A$ be an integral domain, integrally closed in its field of quotients $K$ and let $L$ be a finite Galois extension of $K$ with group $G$. Let $B$ be the integral closure of $A$ in $L$. Let $p$ be a maximal ideal of $A$ and let $\beta$ be a maximal ideal of $B$ such that $A \cap \beta=p$. Let $G_{\beta}$ be the subgroup of $G$ consisting of those automorphisms $\sigma$ such that $\sigma \beta = \beta$. Let $L^{dec}$ be the fixed field of $G_{\beta}$ in $L$ and let $B^{dec}$ be the integral closure of $A$ in $L^{dec}$.

Let $\sigma, \tau \in G$ be such that $(\sigma \beta) \cap B^{dec} = (\tau \beta) \cap B^{dec}$. Does this then imply that $\sigma|_{L^{dec}} = \tau|_{L^{dec}}$?

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    I suppose $B$ is the integral closure of $A$ in $L$?2012-08-03
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    Also, on the last line, do you take the intersection of the Galois conjugate of $\beta$ with $B^{dec}$, or do you apply your Galois element to the intersection?2012-08-03
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    Correct, i will fix this!2012-08-03

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Let me suppose that $A$ is in fact a Dedekind domain, so that I don't have to think through various foundational questions.

Then $G$ acts transitively on the set of $\beta$ lying over $p$, and $G_{\beta}$ is the stabilizer of $\beta$. So we see that $\sigma \beta = \tau \beta$ if and only if $\sigma G_{\beta} = \tau G_{\beta}$, which holds if and only if $\sigma_{| L^{dec}} = \tau_{| L^{dec}}$.

So your question amounts to asking if $\sigma\beta \cap B^{dec} = \tau\beta\cap B^{dec}$ implies that $\sigma \beta = \tau\beta$. So you are asking if a prime above $p$ is determined by its restriction to $B^{dec}$, or equivalently, you are asking if for every prime $\beta'$ above $p$ in $B^{dec}$, there is a unique prime above $\beta'$ in $B$.

Continuing to translate, since the decomposition group at $\sigma\beta$ for the extension $L/L^{dec}$ is $\sigma G_{\beta} \sigma^{-1} \cap G_{\beta}$, you are asking if $\sigma G_{\beta} \sigma^{-1} \cap G_{\beta} = G_{\beta}$ for each $\sigma \in G$, or equivalently, if $G_{\beta}$ is normal in $G$.

So the answer to your question will be yes if $G_{\beta}$ is normal in $G$, and no otherwise.

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    I lost you after the third paragraph. In particular, the decomposition group at $\sigma \beta$ is $\sigma G_{\beta} \sigma^{-1}$. Why is it also $\sigma G_{\beta} \sigma^{-1} \cap G_{\beta}$? Also, you seem to be able to answer this related question as well http://math.stackexchange.com/questions/178108/integral-galois-extensions-proposition-2-4-lang/178142#178142 I would appreciate it!2012-08-03
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    @Manos: Dear Manos, The decomp. gp. at $\sigma\beta$ for $L$ over $K$ is $\sigma G_{\beta} \sigma^{-1}$. To compute the decomposition group for the extension $L/L^{dec}$, we have to intersect the decomp. group for $L/K$ with the Galois group for $L/L^{dec}$; the latter group is $G_{\beta}$. Regards,2012-08-03
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    @manos: Dear Manos, I saw that other question, but I'll take a pass, if that's okay. (I don't really enjoy Lang's treatment; it has a certain technical aspect that I find a bit unpleasant.) Regards,2012-08-03
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    Dear Matt, pass granted, your opinion is respected :-)2012-08-04
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    Dear Matt, i am still a little bit confused. Why am i asking that $\sigma G_{\beta} \sigma^{-1} \cap G_{\beta} = G_{\beta}$?2012-08-04
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    @Manos: Dear Manos, Because you are asking that there only be one prime in $L$ above $\sigma\beta \cap B^{dec}$, so you are asking that the index of the decomposition group at $\sigma\beta$ in $Gal(L/L^{dec})$ (i.e. in $G_{\beta}$) be equal to $1$. Regards,2012-08-04
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    Thanks Matt, i see it now :-)2012-08-06