[Math] Decomposition group and inertia group

algebraic-number-theoryramification

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 $P$ with ramification index e$(Q|P)=e$ and inertia degree f$(Q|P)=f$.

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 decomposition field (the fixed field of $D(Q|P)$). Define $Q_D=Q\cap L_D$.

The inertia group is $$E(Q|P)=\lbrace \sigma\in D(Q|P):\sigma(x)\equiv x\text{ mod } Q\text{ for all } x\in O_L\rbrace$$
Let $L_E$ be the inertia field (the fixed field of $E(Q|P)$). Define $Q_E=Q\cap L_E$.

From algebraic number theory we know the following

  1. e$(Q_D|P)=1$ and f$(Q_D|P)=1$
  2. e$(Q_E|P)=1$ and f$(Q_E|P)=f$

I want to find a prime $Q_D'$ of $L_D$ lying over $P$ such that e$(Q_D'|P)\neq1$ and f$(Q_D'|P)\neq1$.

Similarly, is there any prime $Q_E'$ of $L_E$ lying over $P$ such that e$(Q_E'|P)\neq1$ and f$(Q_E'|P)\neq f$ ?

What we know for sure is that to find such $Q_D'$ (resp. $Q_E'$), we must choose $P$ and $Q$ such that $D(Q|P)$ (resp. $E(Q|P)$) is not normal in $G$. (Corollary 2 of Theorem 28, Number fields, Daniel A Marcus)

Best Answer

We are looking for a collection of ramification and splitting behaviours over a single prime $P$, all of which are possible, but rarely combine in a single easily-calculated example. I have produced a case where each of the things you ask for does indeed occur, in order to illustrate the general case. But I should point out that it is easy to produce individual lower-degree examples demonstrating separately each of the requirements you have placed on the primes above $P$.

Recall the well-known formula describing the splitting of a prime $P$ in an extension $L/K$ in terms of its ramification and residue field extension indices: \begin{equation*} \Sigma_{Q\mid P}\ e_{Q}f_{Q} = [L:K]. \end{equation*} As you have pointed out, when the extension is Galois all of the $e_Q$ are equal to some fixed $e$, and all of the $f_Q$ are equal to a fixed value $f$; so this reduces to \begin{equation} efg = [L:K]. \end{equation} where we have used the standard notation $g$ for the number of primes of $L$ above the prime $P$ of $K$ in a Galois extension.

We would like to see different types of behaviour at each of the levels $L$, $L_D$ and $L_E$ for some prime $Q$ of $L$ above $P$, which implies that we need $e\geq2$ and $f\geq2$. Moreover in order to have any chance of seeing different outcomes above the same prime $P$, we need that $P$ split into at least 2 distinct primes at the level $L_D$, otherwise $L_D=K$. Hence we also need $g\geq2$. So the degree of the extension $L/K$ needs to be at least 8.

Finally, as you also point out, we need that the Galois group Gal($L/K$) contain non-normal subgroups in order that any prime have a chance that the fixed field of its decomposition and/or inertia groups be non-Galois; otherwise we are just in a lower-degree version of the above formula. It is not sufficient, by the way, that the extension simply be non-Abelian, since for example the quaternion group Q8 is non-Abelian but has no non-trivial non-normal subgroups.

All of this forces the extension to have degree at least 16 (the non-normal subgroups of the dihedral group D4 of order 8 do not allow enough varied behaviour). We also would like an extension whose Galois group is furnished with lots of non-normal subgroups. So the easiest place to start would be something with Galois group S4. In order to simplify things let us assume that $K=\mathbb{Q}$, and take some "general" quartic extension, the simplest interesting one of which might be the splitting field $L$ of the polynomial $q(x) = x^4+x+1$.

I used MAGMA for the following calculations. This extension $L/\mathbb{Q}$ is ramified only above $P=229$, splitting into six primes with residue field extension (="inertia") degree $f=2$ and ramification degree $e=2$. Choosing any of these primes $Q$ say we have an inertia group $E(Q\mid P)$ which is cyclic of order 2 ($\cong C_2$) and a decomposition group $D(Q\mid P)$ which is isomorphic to $C_2^2$.

We then calculate that there is always a prime $Q_D'$ of $L_D$ over $P$ with $e_{Q_D'}=f_{Q_D'}=2$.

Similarly there is always a prime $Q_E'$ of $L_E$ over $P$ with $e_{Q_E'}=2$ and $f_{Q_E'}=1$.

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