Can the factorisation of (p) in a number field K be described by the minimal polynomial of a primitive element? Let $K$ be a number field, with ring of integers $O_K$, and let $\alpha\in O_K$ be a primitive element for the extension $K/Q$, with minimal monic polynomial $f(x)\in Z[x]$. If $p$ is a prime number, then the factorisation of the principal ideal $(p)$ in the Dedekind ring $O_K$ can be described in terms of the reduction mod $p$ of $f(x)$ as long as $p$ does not divide the (finite) index $[O_K:Z[\alpha]]$. This is well known and rather elementary, as one can say.
On the other hand, I have never seen in the literature a more general statement relating the factorisation of $(p)$ with the reduction mod $p$ of $f(x)$ without the assumption that $p$ did not divide $[O_K:Z[\alpha]]$. In this more general setting, the sole mod $p$ reduction of $f(x)$ does not give enough information about the splitting of $(p)$, one does need some extra "characteristic zero" information. Would the $p$-adic valuation of the discriminant of $f(x)$ be enough? At least if one knows that $p$ is tamely ramified in $K$. Have you ever seen this discussed or have you ever thought about it? Many thanks.
 A: The discriminant of the polynomial is equal to the discriminant of the field times the square of the index $[\mathcal{O}_K: \mathbf{Z}[\alpha]]$. Hence, if the $p$-adic valuation of the discriminant of $f(x)$ is $0$ or $1$, then one already has an integral basis $p$-adically.
 Yet already the examples $x^2 - p^2$ and $x^2 - a p^2$ where $a$ is a quadratic non-residue modulo $p$ (and $p$ is odd) show that the situation is hopeless in general, even if one knows that $p$ is unramified.
Since, I imagine, you are trying to determine the behavior of Hecke algebras from computing characteristic polynomials of small Hecke operators, you might be able to get mileage out of the following observation, which says that "small" discriminants must come from ramification rather than index.
Suppose, for example, that $f(x)$ modulo $p$ is exactly divisible by $a(x)^2$ for some irreducible 
polynomial $a(x)$ over $\mathbf{F}_p$.
 By Hensel's lemma, $f(x)$ is divisible by a lift $A(x)$ of $a(x)^2$.
There are the following possibilities:


*

*$A(x)$ is irreducible, and corresponds to an unramified prime of residue degree $2d$.

*$A(x)$ has two irreducible factors, corresponding to a pair of unramified primes of
residue degree $d$.

*$A(x)$ is irreducible, and corresponds to a ramified prime of residue degree $d$ and ramification index $2$.


In the first two cases, the index of $\mathbf{Z}[\alpha]$ in $\mathcal{O}_K$ is divisible
by $p^d$, and so the discriminant is divisible by $p^{2d}$. Hence, if you know that
the valuation of the discriminant is less than $2d$, you can deduce that $A(x)$ corresponds to a ramified prime.
