TITLE: Showing that degree of polynomial reduced modulo $p$ divides degree of splitting field
QUESTION [8 upvotes]: Let $f$ be a polynomial of degree $n$ with integer coefficients and $p$ a prime for which $f$, considered modulo $p$, is a degree-$k$ irreducible polynomial over $\mathbb{F}_p$. Show that $k$ divides the degree of the splitting field of $f$ over $\mathbb{Q}$.
(this is Miklos Schweitzer 2020, Problem 10)
My miscellaneous thoughts: We know that for a polynomial of degree $n$, the degree of its splitting field is at most $n!$ from induction on the Ring Tower Theorem. Additionally, since $f(x)$ is irreducible over $\mathbb{F}_p$, then the field $\mathbb{F}_p[x]/(f(x))$ has a root of $f(x)$. Now suppose $k = 2$. Then $f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 \in \mathbb{Z}[x]$ reduces to $f(x) \equiv b_2 x^2 + b_1x + b_0 \in \mathbb{F}_p [x]$ where $b_i \equiv a_i \pmod{p}$ where applicable. I'm really not sure if these things are particularly useful or how to combine them.
This is really hard!

REPLY [2 votes]: If $p\mid f(0)$ then $\overline f(x)=cx$ (using the notation from the previous posts) for some $c\in\mathbb F_p^\times$, and thus $k=1$. So let's assume $p\nmid f(0)$ from now on. Let $x_1,\ldots,x_n$ be the roots of $f(x)$ in $K$ and let * denotes the reciprocal polynomial. Then $f^*(x)$ has degree $n$ (since $f(0)\ne0$) and roots $1/x_1,\ldots,1/x_n$. Furthermore, we have $f(0)/x_1,\ldots,f(0)/x_n\in\mathcal O_K$ since they are roots of the monic polynomial $g(x):=f(0)^{n-1}\cdot f^*(x/f(0))\in\mathbb Z[x]$. Let $\mathfrak P$ be a prime ideal of $\mathcal O_K$ above $p$, then since $g(x)$ splits in $\mathcal O_K$, $\overline g(x)$ splits in $\mathcal O_K/\mathfrak P$. We have $\overline{f^*}(x)=x^{n-k}\cdot\overline f^*(x)$, and since $p\nmid f(0)$, $\overline f^*(x)$ has degree $k$ and is irreducible over $\mathbb F_p$, and $\overline g(x)=\overline{f(0)}^{k-1}x^{n-k}\cdot\overline f^*\big(x/\overline{f(0)}\big)$. The polynomial $\overline f^*\big(x/\overline{f(0)}\big)$ is also irreducible over $\mathbb F_p$ of degree $k$, and thus the splitting field of $\overline g(x)$ is just $\mathbb F_p[x]\big/\big(\overline f^*\big(x/\overline{f(0)}\big)\big)$ and
$$k=\big[\mathbb F_p[x]\big/\big(\overline f^*\big(x/\overline{f(0)}\big)\big):\mathbb F_p\big]\mid\big[\mathcal O_K/\mathfrak P:\mathbb F_p\big]\mid\big[K:\mathbb Q\big].$$