Separable polynomial with splitting field an unramified extension?

Question

I am trying to prove a theorem and it seems that I need that an irreducible polynomial $f(x)$ that is separable over $\mathfrak{p}$ has its splitting field an unramified extension of $\mathbb{Q}_\mathfrak{p}$, the original field.

Is this true? And if so, any hints on how to approach it?

Answer 1

I think what you are asking is how to prove the following:

Let $f \in \mathbb{Z}[X]$ be a monic irreducible polynomial. Let $L$ be the splitting field of $f$ over $\mathbb{Q}$. If the reduction of $f$ modulo $p$ is squarefree, then $L/\mathbb{Q}$ is unramified at $p$.

If $\mathbb{Q} \subseteq K \subseteq L$ are fields, $p$ a rational prime, then $L$ unramified at $p$ implies $K$ is unramified in $p$. On the other hand, a compositum of unramified extensions at a given prime remains unramified. So your question is equivalent to asking the following question:

Let $f \in \mathbb{Z}[X]$ be a monic irreducible polynomial. Let $\alpha$ be any root of $f$, and let $K = \mathbb{Q}(\alpha)$. If the reduction of $f$ modulo $p$ is squarefree, then $K/\mathbb{Q}$ is unramified at $p$.

Lemma: let $F$ be a perfect field, let $g \in F[X]$ be monic and irreducible. If $\alpha_1, ... , \alpha_t$ are all the roots of $g$, counting multiplicity, in some algebraic closure of $F$, then the discriminant $\textrm{disc}(g)$ of $g$, is the element $$\prod\limits_{i\neq j} (\alpha_i - \alpha_j)$$ It lies in $F$ and does not depend on the choice of algebraic closure. In fact, it is a polynomial function of the coefficients of $g$. The polynomial $g$ has a repeated root if and only if $g$ is squarefree as a product of irreducibles, if and only if the discriminant of $F$ is zero.

Now let $d$ be the discriminant of $f$. There is a notion of a discriminant of a number field, denoted $\delta = \delta(K/\mathbb{Q})$. It is an integer, well defined up to sign. The important result is that $K$ is unramified at a given prime number $q$ if and only if $q$ divides $\delta$.

Now, since $f$ is monic, $\mathbb{Z}[\alpha] \subseteq \mathcal O_K$, the ring of integers of $K$. To these two rings we may associate integers, which we call their discriminants. As a result of that subset inclusion, one can show the the discriminant of $\mathcal O_K$ (which turns out to be $\delta$) divides the discriminant of $\mathbb{Z}[\alpha]$ (which turns out to be $d = \textrm{disc } f$).

We claim that $K$ is unramified at $p$. If not, then $p$ divides $d$, whence the reduction of $d$ modulo $p$ is zero. But one can show that the reduction of $d$ modulo $p$ is exactly the discriminant of the reduction of $f$ modulo $p$. Thus $\overline{f} \in \mathbb{F}_p[X]$ is not squarefree, contradiction.