Lower bounds on the index of $\mathbf Z[X]/(P)$ in the ring of integers of a number field Let $P$ in $\mathbf Z[X]$ be an irreducible polynomial. Let $\mathcal O$ be the ring of integers of the number field $K:=\mathbf Q[X]/(P)$ and $i$ be the index of $\mathbf Z[X]/(P)$ in $\mathcal O$.
Are there known lower bound  (or even better divisibility conditions) for $i$?
What if $P$ is assumed to be totally real?
To be more specific, can someone explain to me the following numerical phenomenon?
Let $H$ be the following symmetric matrix
$$H:= \begin{bmatrix}2&1&1&2\cr
 1&2&1&2\cr
 1&1&2&2\cr
 2&2&2&4\cr
\end{bmatrix}
$$
($H$ is the Gram matrix of the root lattice $D_4$. It is positive definite and has determinant $4$.) 
Let $S$ be any integral symmetric matrix with determinant $1$, let $M=H \cdot S$ and $P=\chi_M$ be the characteristic polynomial of $M$
(note that $P$ is still totally real since $M$ is autoadjoint for a scalar product).
When $P$ is irreducible, pari/GP told me on thousands of examples that $i$ is always divisible by 4 (and I found not a single exception to this).
More strangely, I tested also in other dimensions (that is with $H$ the Gram matrix of $D_n$) and then found several examples with $i=1$ or $i=2$ when $n\neq 4$.
I can understand that the discriminant of $P$ is divisible by a (not so big) power of $2$ wich leaves some possibility for $i$ to be even, but I don't understand why dimension $n=4$ seems to be specific.
 A: $\def\F{\mathbf{F}}$
$\def\Z{\mathbf{Z}}$
$\def\OL{\mathcal{O}}$
Let's start only with the assumption that $\det(H)$ is divisible by $4$, and that $H$ and $S$ are integral matrices. Let $P(x)$ be the characteristic polynomial of $M = HS$, and assume that $P(x)$ is irreducible
and $\OL = \Z[x]/P(x)$ is the ring of integers of $K$, at least at the prime $2$ (So really, we are assuming that $\OL_K \otimes \Z_2 = \OL \otimes \Z_2 = \Z_2[x]/P(x))$. Note that $P(0) = \det(M)$ is also divisible by $4$.
The first claim is that $\mathfrak{p} = (2,x)$ is an unramified prime of $\OL$ of norm $2$. It certainly has norm $2$, since $\OL/\mathfrak{p} = \Z[x]/(2,x,P(x)) = \F_2$.
Now let $I = (4,x)$. Since $I \subset \mathfrak{p}^2$, it must be the case that either $I = \mathfrak{p}$ or $I = \mathfrak{p}^2$. Yet
$$\OL/I = \Z[x]/(4,x,P(x)) = \Z/(4,P(0)) = \Z/4,$$
since $P(0) = \det(M)$ is divisible by $4$. Hence $I = \mathfrak{p}^2$, and $\mathfrak{p}$ is unramfied (since otherwise $\OL/\mathfrak{p}^2$ would have characteristic two).
I next claim that $\mathfrak{p}$ is the only prime above $2$ dividing $x$. For this, we compute that
$$\OL \otimes \Z_2/x = \Z_2[x]/(P(x),x) = \Z_2/P(0).$$
The latter is a local ring, and so $x$ can be divisible by at most one prime above $2$.
Moreover, the valuation of $x$ at the localization of $\OL$ at $\mathfrak{p}$ is the valuation of $P(0)$. In particular, all the other roots of $P(x)$ must have valuation zero, and hence,
from the Newton Polygon, we must have
$$P(x) \equiv 0 + x + O(x^2) \mod 2.$$
We have deduced the following criterion: if $P(x) \equiv 0 \mod (2,x^2)$, then we deduce that $\OL =\Z[x]/P(x)$ cannot have index prime to $2$. 
However, we can compute $P(x) \mod 2$ in terms of $H$ and $S$ modulo $2$. It thus remains to show:
Lemma: If $S \in M_4(\F_2)$, then the coefficient of $x$ in the characteristic polynomial of $M = HS$ is even.
Proof: Writing $S = [s_{ij}]$, one can explicitly compute the coefficient of $x$ in $\mathrm{det}(HS - x)$, and it is $2$ times a (complicated) integral polynomial in the $s_{ij}$.
Hence, one obtains the stronger claim that the result holds for all integral matrices $S$ (that is, no
symmetric assumption is needed, nor any assumption on the determinant, although if $\det(S) = 0$ then $P(x)$ is obviously not irreducible).
Now why this particular $H$ has this property I am not sure, but the result depends only on $H \mod 2$ plus the assumption that $\det(H)$ is divisible by $4$.
