Directly showing that a prime is unramified iff the inertia group is trivial 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.
 A: 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


*

*$P$ splits completely in $L^D$.

*Any prime $P_D$ of $L^D$ above $P$ is inert in $L^I$.

*Let $P_I$ be the prime of $L^I$ above $P_D$. Then $P_I$ is totally ramified in $L$.


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.
A: Recall inertia is the kernel $\mathrm{Gal}(O_K/O_k)\to \mathrm{Gal}(\kappa’/\kappa)$, inertia being trivial means this map is injective. Given homomorphisms $f_1,f_2\colon O_K\to O_K$ which have same reduction modulo $m$. Then $\overline{g}=\overline{f_1}-\overline{f_2}$ is a derivation $O_K\to m/m^2$. Unramifiedness implies this morphism is zero since $Der_{O_k}(O_k, m/m^2)=Hom(\Omega_{O_K/O_k}, m/m^2)=Hom(0,m/m^2)$ is zero. Thus $g$ maps into $m^2$ keep doing show image of $g$ has arbitraryly large valuation. Thus $g=0,f_1=f_2$.
