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 $\sigma|_{L^{dec}} = \tau|_{L^{dec}}$?

• I suppose $B$ is the integral closure of $A$ in $L$? – M Turgeon Aug 3 '12 at 15:26
• 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? – M Turgeon Aug 3 '12 at 15:29
• Correct, i will fix this! – Manos Aug 3 '12 at 15:29

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.

• 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 math.stackexchange.com/questions/178108/… I would appreciate it! – Manos Aug 3 '12 at 17:11
• @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, – Matt E Aug 3 '12 at 17:18
• @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, – Matt E Aug 3 '12 at 17:19
• Dear Matt, pass granted, your opinion is respected :-) – Manos Aug 4 '12 at 16:10
• Dear Matt, i am still a little bit confused. Why am i asking that $\sigma G_{\beta} \sigma^{-1} \cap G_{\beta} = G_{\beta}$? – Manos Aug 4 '12 at 16:53