Technical question on integral ring extensions 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 $\sigma|_{L^{dec}} = \tau|_{L^{dec}}$?
 A: 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.
