Describing the Inertia group of a number field Let $ K \subseteq L$ be number fields and $\pi$ be a prime ideal of $L$. $G = \operatorname{Gal}\left(L/K\right)$
Let $D = \{\sigma \in G\:|\:\sigma(\pi) = \pi\} $ be the decomposition group for $\pi$ and $I = \{\sigma \in G\:|\:\sigma(\alpha)\equiv\alpha\pmod{\pi}\: \forall \alpha\in\mathcal{O}_{L}\}\subseteq D$ be the inertia group for $\pi$. 
We can calculate the number of elements in $D = ef$ easily where e is the ramification index. However, the only proof I have seen of the size of $I$ is by showing that it is the kernel of $D$ in some homomorphism.
Is there a way to describe the $e$ elements of $I$ (or $D$) directly or barring that, some way to show its size directly (only for $I$)?
Edit:I am assuming that $L/K$ is Galois. Thinking a bit more:
Can one define $I$ by the maximal power of $\pi$ that an automorphism $\sigma$ in it fixes as a set? Clearly, every automorphism in $I$ has to fix upto $\pi^{k}$, $1 \leq k \leq e$. Can one show that there is exactly one such automorphism for each $k$? That would directly show the order, correct?
 A: I assume from your notation and the nature of your question that your extension is Galois. Not particularly easy. First of all, the original proof of the size of $D$ comes from the orbit-stabilizer theorem and the fact that $G$ acts transitively on the primes above the primes of a base field. Ordinarily the constants $e,f$ are easier to compute than $r=|D|$ because of your ability to reduce to a localization or completion and compute there. Computing the order of the inertial subgroup is going to similarly be complicated. You usually use the completion to give yourself a residue field, and then use the surjectivity onto its Galois group to give you the inertial subgroup--which is ultimately a local object anyways when you think about how it's defined.
I would argue that the definition of $I$ is quite direct, and makes sense in terms of the localization, but if that's not to your sensibilities, there's always the so-called "higher ramification theory" approach wherein you pick some $\mathfrak{P}|\mathfrak{p}$ in $L$ and pass to $\mathfrak{P}$-adic completions and then define
$$I=\{\sigma\in G : \min_{x\in\mathcal{O}_{L,\mathfrak{P}}} v_{\mathfrak{P}}(\sigma(x)-x)\ge 0\}.$$
But this is ultimately just a restatement of the definition (albeit a useful one).

Addendum
Because the comments are getting unwieldy, I'll just add this as a lemma rather than try and continue there.
Lemma Let $\sigma$ be a field automorphism and $I,J$ be ideals of a subring of the field. Then $\sigma(IJ)=\sigma(I)\sigma(J)$.
Proof:  By definition

$$IJ=\left\{\sum_{i,j}x_iy_j : x_i\in I, y_j\in J\right\}$$

Then
$$\sigma(IJ)=\left\{\sigma\left(\sum_{i,j}x_iy_j\right): x_i\in I, y_j\in J\right\}$$
$$=\left\{\sum_{i,j}\sigma(x_i)\sigma(y_j): x_i\in I, y_j\in J\right\}$$
because $\sigma$ is a field automorphism
$$=\left\{\sum_{i,j}\sigma(x_i)\sigma(y_j) : \sigma(x_i)\in \sigma(I), \sigma(y_j)\in \sigma(J)\right\}=\sigma(I)\sigma(J)$$
Both of the last two qualities by definition. Setting $I=J=\pi$ and choosing $\sigma\in D$ we get the specific result.
