Consider a Galois extension $L/K$ of number fields and denote by $B,A$ the corresponding integer rings. Let $Q$ be a prime in $B$ lying over $P$ (which is in $A$). Let $D,I$ denote the decomposition and inertia groups of $Q$ respectively. Consider the following statement:
The following are equivalent:
- $I$ is trivial,
- $P$ is unramified in $B$.
I am looking for a proof of this statement which would be as direct as possible. I have found two different proofs of this fact, one of them contained in in Marcus's "Number Fields", and one in these notes. Both make essential use of the $efg=n$ identity ($g$ - number of primes in $B$ lying over $P$, $n$ - degree of $L/K$) to find that $|I|=e$ after working out different numbers, e.g. $|D|$ and $|D/I|$. However, I feel like we should be able to prove this without having to do such computations. Does anyone know of such a proof?
Thanks in advance
PS. I am mostly interested in the implication "$P$ unramified" $\Rightarrow$ "$I$ trivial", so I am willing to accept an answer which provides only this single implication.
To me, the crux behind the definition of the decomposition and inertia groups is as follows:
Consider the extension of fields $L/L^I/L^D/K$, where $L^D$ and $L^I$ are the fixed fields of $D$ and $I$ respectively. Then
In other words these groups allow us to decompose $L/K$ into totally split, totally ramified and inert extensions, allowing us to reduce our efforts to each of these three types alone.
We will show that $P$ is ramified in $L/L^I$ (in fact it is totally ramified). In particular, if $P$ is unramified in $L$, then $L/L^I$ must be trivial, so $I = 1$. This can be shown without assuming $efg = n$ as follows:
Let $P_I = Q\cap L^I$ be the prime of $L^I$ under $Q$. Let $D_I$ be the decomposition group of $Q/P_I$. Then our key observation is that $$D_I=D\cap I = I = \mathrm{Gal}(L/L^I).$$
Since $D_I=\mathrm{Gal}(L/L^I)$ acts transitively on the primes above $P_I$, and it fixes $Q$, $Q$ must be the only prime of $L$ above $P_I$.
In a similar way, the inertia group of $Q/P_I$ is $I$. Since by definition, $I$ is the kernel of the map $$D_I\to \mathrm{Gal}(\mathbb F_Q/\mathbb F_{P_I}),$$ it follows that the residue degree of $Q/P_I$ is $1$.
So $f=g=1$. At this point, if we could assume that $efg =n$, we could deduce that $Q/P_I$ is totally ramified. However, to deduce the $L/L^I$ is ramified, we only need to assume that $efg>1$, which is a much weaker fact.
For the converse, I think that using the fact that $efg = n$ is necessary, although I'd be interested if someone can bypass this fact.