Decomposition group Daniel A Marcus Theorem 28

Question

I am reading the book Number Fields by Daniel A Marcus. I have the following doubt.

Let $L/K$ be a Galois extension with Galois group $G$. Let $O_K$ and $O_L$ be the ring of algebraic integers of $K$ and $L$ respectively. Let $P\subseteq O_K$ be a prime. Let $Q\subseteq O_L$ be a prime lying over $P$. Let $D(Q|P)$ be the decomposition group of $Q$. In other words $$D(Q|P)=\lbrace\sigma\in G\text{ }|\text{ }\sigma(Q)=Q\rbrace$$ Let $L_D$ be the deocompostion field (the fixed field of $D(Q|P)$). Let $Q_D$ be a prime lying over $P$ in $L_D$.

Is it true that the prime $Q$ will lie over the prime $Q_D$ ? If so, then why ?

This question came to my mind because of the following reason. Say, we have a Galois extension $L/F$. Let $P\subseteq O_F$ be a prime. Let $Q\subseteq O_L$ be a prime lying over $P$. Say, $K$ is an intermediate field between $F$ and $L$. Let $Q'$ be a prime of $O_K$ lying over $P$. I don't have any example, but I have a feeling that it is not necessarily true that $Q$ will lie over $Q'$. (Can you find some example of this ?)

Answer 1

No this is almost always false. For a simple counterexample, take any prime which splits completely in the extension. The decomposition group will be trivial, so $L_D=L$. However, there are plenty of primes of $L$ lying over $P$ which are not equal to $Q$.

On the other hand, if you let $P_D=Q\cap L_D$, then $P_D$ will be a prime of $L_D$ lying over $P$ and under $Q$. This is, by the definition of lying over, the only prime of $L_D$ for which this works.

Answer 2

If you mean $Q_D= Q\cap \mathcal{O}_{L_D}$, then yes: In the decomposition field you have all the primes that appear in the eventual decomposition in the full extension, $L$. This is basically because you know that the Galois group acts transitively on the set of primes above a given prime, then applying the orbit-stabilizer theorem, since $G$ acts on $S = \{Q|P: Q\text{ is a prime of L}\}$. Then as $[L_D:K] = |G|/|\text{stab}_G\;{Q}|$ for any $Q|P$ (all stabilizers are conjugate, so they have the same size), we see the primes of $L_D$ are exactly the primes of $L$ in the sense that $Q'\mathcal{O}_L=Q$ is a prime-power ideal of $L$ for every $Q'$ a prime of $L_D$ above $P$.

But then the lying over clause is obvious, just using the Galois theory.

If you really just mean $Q$ is arbitrary (which I doubt) then the answer is clearly "almost assuredly not" as there is no clear connection between any two given primes, and each primes of $\mathcal{O}_L$ can lie above only one prime of any subfield (in this case $L_D$), and in fact we know that $Q\cap\mathcal{O}_{L_D}$ is the required prime.