Finding the inertia group Set $h(x) = x^5+x^4+x^3+x^2+x-1$, let $L$ be the splitting field of $h$ over $\mathbb{Q}$, and let $p$ be a prime of $L$ lying over $2$.

What is the isomorphism class of the inertia group $I_p$, and how do I find it?

I would be more happy with having some procedure for finding inertia groups, rather than only the isomorphism class of this specific group.
I would also like to find the field $O_L/p$, and the decomposition group $D_p$ if possible.
Some useful information is: $h$ is irreducible over the rational numbers, and $h(x)(x-1) = x^6 - 2x + 1$ which allows for an easy irreducible factorization mod $2$: $h(x) = (x + 1)(x^2 + x + 1)^2$. 
 A: This answer produces $I$ somewhat indirectly. So it actually does not what the OP asked for.
The decomposition group $D$, which is the Galois group of $h(x)$ over $\mathbb Q_2$, can be computed as follows: Using resultants, one sees that the minimal polynomial over $\mathbb Q$ of the difference of two distinct roots of $h(x)$ is \begin{equation}
H(y)=y^{20}+6y^{18}+21y^{16}+46y^{14}-116y^{12}+694y^{10}+1837y^8-1810y^6-1527y^4+8560y^2+9584.
\end{equation}
From the factorization of $h(x)$ over $\mathbb Q_2$ we know that $D$ is a transitive subgroup of $S_4$. Over $\mathbb Q_2$ the polynomial $H(y)$ factors into irreducibles of degrees $4,4,4,8$. The only transitive subgroup of $S_4$ which has these orbit lengths on the $20$ pairs of distinct elements of $\{1,2,3,4,5\}$ is the dihedral group of order $8$.
As Will Sawin already remarked, the inertia group $I$ is a subgroup of $S_2\times S_2$. On the other hand, $D/I$, as a Galois group of an extension of a finite field, is cyclic. This forces $|I|\ge4$, because a dihedral group of order $8$ modulo a normal subgroup of order $\le2$ isn't cyclic. Thus $I=S_2\times S_2$.
There is a paper by Sybilla Beckmann which describes a method to compute inertia groups under certain restrictions. Her theorem does not apply here. The paper finishes with some remarks about how to possibly extend the methods.
A: The inertia group acts on the roots and preserves the reduction modulo $2$. So it is a subgroup of the group of permutations of the roots preserving the reduction mod $2$, which is $S_2 \times S_2 = \mathbb Z/2 \times \mathbb Z/2$.
To compute which one it is, you need to use Hensel lifting to actually factor the polynomial into one linear and two quadratic factors over $W(\mathbb F_4)$ and see what you need to adjoin the square roots of. It's easy to see if two square roots correspond to nontrivial extensions and whether they correspond to the same extension.
The point is that you have a factorization mod $2$, and you can use it to get a factorization mod $4$, and mod $8$, etc., until the roots separate, and then you can compute it. Probably you will not have to go very far.
