I have two questions about the proof of the following

Proposition: Let $A$ be a ring, integrally closed in its quotient field $K$. Let $L$ be a finite Galois extension of $K$ with group $G$. Let $P$ be a maximal ideal of $A$, and let $I$, $J$ be prime ideals of the integral closure $B$ of $A$ in $L$ lying above $p$. Then there exists $\sigma\in G$ such that $I=\sigma J$.

Proof: Suppose that $I\neq\sigma J$ for any $\sigma\in G$. There exists an element $x\in B$ such that $x\equiv 0\pmod I$, $x\equiv 1\pmod{\sigma J}$, all $\sigma\in G$ (use the Chinese remainder theorem). The norm $\prod_{\sigma\in G}\sigma x$ lies in $B\cap K = A$...

Question 1: Why can we use the Chinese remainder theorem here?

Question 2: Why does the norm lies in $K$ on the last line?

  • 1
    $\begingroup$ The norm is a map $L\to K$. (The expression is $G$-invariant) $\endgroup$ – Hagen von Eitzen Jan 15 '14 at 10:56

Since $I$ and $J$ are prime ideals of $B$ lying over the maximal ideal $P$ of $A$ and $B$ is integral over $A$, $I$ and $J$ are also maximal, and for any $\sigma\in G$, $\sigma(J)$ is maximal as well. So, if $I\neq\sigma(J)$, $I$ and $\sigma(J)$ are distinct maximal ideals, hence comaximal, and the Chinese remainder theorem can be applied to the collection of ideals $\{I\}\cup\{\sigma(J):\sigma\in G\}$, assuming these sets are disjoint (as Lang does in his proof).

Applying any $g\in G$ to the norm $\prod_{\sigma\in G}\sigma(x)$ just gives the same product in a possibly different order. So each $g\in G$ fixes the norm of $x$, and since $L/K$ is Galois with group $G$, this norm must lie in $K$. It is also integral over $A$, so it lies in $A$.


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