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My question refers to the proof of Proposition 2.4, p. 341 in Lang's Algebra. Here is the context:

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 $p$ be a maximal ideal of $A$ and let $\beta$ be a maximal ideal of $B$ that lies above $p$, i.e. $A \cap \beta=p$. 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 $L^{dec}$. We know that $\beta$ is the only prime ideal of $B$ lying over $Q=\beta \cap B^{dec}$ in $B^{dec}$.

In the proof of Proposition 2.4, it is argued that for any $\sigma$ not inside $G_{\beta}$ we have that $Q_{\sigma} = \sigma^{-1} \beta \cap B^{dec} \neq Q$. This is true because if that was not the case, $\sigma^{-1} \beta$ would also lie above $Q$ and this is a contradiction because it would imply that $\beta = \sigma \beta$. Then the chinese remainder theorem is invoked to show the existence of a $y \in B^{dec}$ so that given $x \in B^{dec}$ we have that $y = x (mod Q)$ and $y = 1 (mod Q_{\sigma})$ for all $\sigma$ not inside $G_{\beta}$. To use the chinese remainder theorem, we need at least the ideals to be distinct. In particular we want $Q_{\sigma} \neq Q_{\tau}$ for any $\sigma, \tau$ not inside $G_{\beta}$. Why is that true?

My concern: consider $\sigma, \tau$ be distinct elements of $G$, not inside $G_{\beta}$ but inside the same coset of $G_{\beta}$ in $G$. Then $\sigma \beta = \tau \beta$ and consequently $Q_{\sigma} = Q_{\tau}$.

Added:

This is how his proof continues. There exists such a $y$ as above and so $y=x (mod \, \beta), y=1 (mod \, \sigma^{-1} \beta)$ for each $\sigma$ not in $G_{\beta}$. The second congruence gives $\sigma y=1 (mod \, \beta)$ for all $\sigma$ not in $G_{\beta}$. Taking the norm of $y$ from $L^{dec}$ to $K$ gives $N^{L^{dec}}_K=x (mod \, \beta)$. This later congruence is also true $mod \, B^{dec}$ and so $mod \, Q$, from which he concludes that $A/p = B^{dec} / Q$ (what we want to prove), since the norm is in $A$.

Any insights? Thanks.

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Your last point seems to hold-what you've shared so far doesn't obviously require the full strength of the claim in the last sentence of your last long paragraph, but only that $Q \neq Q_\sigma$ for $\sigma$ not in $G_\beta$. Then we'd be golden. But he explicitly wants to apply the crt to all the $Q_\tau$, not just these two? –  Kevin Carlson Aug 2 '12 at 18:04
    
I think he wants to apply the CRT to all of the $Q_{\sigma}$, because he then takes the norm of $y$ which is the product of the $\sigma y$. –  Manos Aug 2 '12 at 18:07
    
Well, I won't claim I've gotten the proof completely down, but in taking the norm of $y$ in $L^{dec}$ over $K$ he should only be taking the product of $\sigma y$ for $\sigma$ in $G/G_\beta$ the Galois group of $L$ over $K$. More specifically, we can get the condition on $y$'s congruences just from considering the conjugates of $Q$-since the congruences are certainly constant on conjugacy classes of $\sigma$. –  Kevin Carlson Aug 2 '12 at 18:33
    
Can you explain this a little bit better, i am not following. By definition, the norm is the product of the $\sigma y$ for those $\sigma$ that are distinct embeddings of $L^{dec}$ over $K$. Maybe you are in the right path... –  Manos Aug 2 '12 at 18:38
    
Ok, i see it now, the distinct embeddings of $L^{dec}$ over $K$ are precisely coset representatives of $G_{\beta}$, elements of $G/G_{\beta}$ as you say. So the question then boils down to the following: let $\sigma, \tau$ be distinct coset representatives of $G_{\beta}$. Why is it true that $Q_{\sigma} \neq Q_{\tau}$? If $Q_{\sigma}=Q_{\tau}$ then $\sigma \beta, \tau \beta$ must be conjugates by some element of $G_{\beta}$, say $\sigma \beta = \phi \tau \beta, \phi \in G_{\beta}$. Then... –  Manos Aug 2 '12 at 18:46

1 Answer 1

up vote 1 down vote accepted

If $\tau ^{-1} \sigma \in G_\beta$ then we certainly have that $Q_\sigma = Q_\tau$ and it may be that $\tau, \sigma \notin G_\beta$. This is irrelevant to the proof, however. Say $Q_\sigma = Q_\tau$ and $y \equiv 1 \pmod{Q_\sigma}$. You don't need to use Chinese remainder theorem to say that $y \equiv 1 \pmod{Q_\tau}$. This is in fact, the same statement as the previous one.

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