# Prime ideals in galois extensions

This is with reference to proposition 1 in Robert Ash's notes

I don't think the Dedekind assumption is necessary. Explicitly, if $A$ is an integral domain with fraction field $K$ and $L/K$ is galois, let $B$ be the integral closure of $A$ in $L$. Then, I claim that any element of the galois group $G$ of $L/K$ acts on the set of prime ideals lying over a prime $P\subseteq A$.

Let $Q$ be a prime ideal in $B$ lying over $P$. Then, if $\sigma \in G$ and $\sigma(x) \sigma(y) \in \sigma(Q)$ then, $\sigma(xy)\in \sigma(Q)$. Applying, $\sigma ^{-1}$, we get $xy\in Q$ and hence either $x$ or $y$ is in $Q$ and thus, either $\sigma(x)$ or $\sigma(y)$ is in $\sigma(Q)$.

Finally, $\sigma$ is an automorphism and hence preserves intersections and also fixes $A$ and hence $P$, so $\sigma(Q) \cap A= \sigma(Q) \cap \sigma(A) = \sigma(Q\cap A)= \sigma(P)=P$.

Are there any holes in this argument?

I am doubting since I have seen three sets of number theory notes which use the Dedekind assumption.

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I don't think there is any problem with this bit; the issues really start showing up later; e.g., you use the fact that primes are maximal in Theorem 1 (transitivity of the action of the Galois group on the set of primes lying over $P$). – Arturo Magidin Oct 9 '11 at 22:48
@Arturo: Thanks for the confirmation. – BMI Oct 10 '11 at 1:48