Showing that degree of polynomial reduced modulo $p$ divides degree of splitting field 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!
 A: Cool problem! Here is a solution, which I preface with a hint/sketch, in case anyone reading wants to try to finish it for themself.
First fix some notation: let $\overline{f}$ be the reduction of $f$ mod $p$, $K/\mathbb{Q}$ the splitting field of $f$, $\mathcal{O}_K$ its ring of integers, $\mathfrak{p}$ any prime of $\mathcal{O}_K$ lying over $p$ and $e$ its ramification index.
The hint/sketch: since $\overline{f}$ is irreducible, its splitting field $\kappa$ has degree $k$ over $\mathbb{F}_p$. We'd like to exhibit $\operatorname{Gal}(\kappa/\mathbb{F}_p)$ as a subquotient of the Galois group $\operatorname{Gal}(K/\mathbb{Q})$ via a map from the decomposition group $D(\mathfrak p)$, so then $k\mid |\operatorname{Gal}(K/\mathbb{Q})|=[K:F]$. The trouble is that $f$ is not monic, so it may not have roots in $\mathcal{O}_K$ (or even in the localization $(\mathcal{O}_K)_\mathfrak{p}$), and so it's not immediate that the factorization of $f$ in $K$ gives rise to a factorization of $\overline{f}$ in $\mathcal{O}_K/\mathfrak{p}$. The idea will be to use Vieta's formulas and an extension of the $p$-adic valuation to show that there are exactly $k$ roots of $f$ giving rise to the roots of $\overline{f}$ (namely, roots of $f$ with denominators not in $\mathfrak{p}$).
The solution: we shift attention to the localization $(\mathcal{O}_K)_{\mathfrak{p}}$, and let $\nu:=\nu_{\mathfrak p}$ be the $\mathfrak{p}$-adic valuation, so $\nu(x)=e\nu_p(x)$ for $x\in \mathbb{Q}$. Write $$f(x)=\sum_{i=0}^na_ix^i.$$
Then the hypotheses imply: (1) $\nu(a_i)>0$ for $i>k$; (2) $\nu(a_k)=0$; (3) $\nu(a_0)=0$. Here (1) and (2) are immediate and (3) follows by noting that $\overline{f}$ has no constant term otherwise, which would imply $\overline{f}$ has a root $x=0$ over $\mathbb{F}_p$ (so it is either reducible or has degree $1$). Let $r_i=c_i/d_i$, $1\leq i\leq n$ be the roots of $f$ in $K$, with $c_i,d_i\in (\mathcal{O}_K)_{\mathfrak{p}}$ not both in $\mathfrak{p}$. Let $R=\{r_1,\ldots,r_n\}$. For $0\leq \ell \leq n-1$, Vieta's formulas can be written
$$
a_\ell = (-1)^{n-\ell}\cdot a_n\cdot \sum_{\sigma\subset R\atop |\sigma|=n-\ell}\prod_{r_i\in \sigma}r_i.
$$
Furthermore, we have $a_n=\prod_{i=1}^nd_i$. In particular, $a_0=\prod_{i=1}^nd_i\cdot \prod_{i=1}^nr_i$ is the product of the numerators $c_i$, and (3) above then gives $\nu(c_i)=0$ for all $i$, so $\nu(r_i)=-\nu(d_i)$. Writing this out for $a_k$ and applying $\nu$ gives
$$
0=\nu(a_k) = \nu(a_n)+\nu\left(\sum_{\sigma\subset R\atop |\sigma|=n-k}\prod_{r_i\in \sigma}r_i\right).
$$
Recall the Strong Triangle Inequality: $\nu(\sum_{finite} x_i)\geq \min_i(\nu(x_i))$ with equality unless $\min_i(\nu(x_i))=\nu(x_i)=\nu(x_j)$ for some $i\neq j$.
Applied above, this gives
$$
0=\nu(a_n)+\nu\left(\sum_{\sigma\subset R\atop |\sigma|=n-k}\prod_{r_i\in \sigma}r_i\right)\geq \nu(a_n)+\min_{\sigma\subset R\atop |\sigma|=n-k}\left(\sum_{r_i\in \sigma}\nu(r_i)\right).
$$
Rearranging and applying $\nu(r_i)=-\nu(d_i)$ gives
$$
\nu(a_n)\leq-\min_{\sigma\subset R\atop |\sigma|=n-k}\left(\sum_{r_i\in \sigma}-\nu(d_i)\right)=\max_{\sigma\subset R\atop |\sigma|=n-k}\left(\sum_{r_i\in \sigma}\nu(d_i)\right).
$$
But $\nu(a_n)=\sum_{i=1}^n\nu(d_i)$, so we have
$$\sum_{i=1}^n\nu(d_i)\leq\max_{\sigma\subset R\atop |\sigma|=n-k}\left(\sum_{r_i\in \sigma}\nu(d_i)\right),$$
and we see that there can be at most $n-k$ values of $i$ for which $\nu(d_i)>0$, otherwise the left hand side will be larger than the right.
Now we claim that there are exactly $n-k$ values of $i$ for which $\nu(d_i)>0$. Suppose otherwise, so that there are exactly $\ell>k$ values of $i$ with $\nu(d_i)=0$. Then observe that, in the formula
$$
\nu(a_\ell) = \nu(a_n)+\nu\left(\sum_{\sigma\subset R\atop |\sigma|=n-\ell}\prod_{r_i\in \sigma}r_i\right)
$$
there is a unique term in the sum on the right with minimal valuation, namely $\prod_{r_i:\nu(r_i)\neq 0}r_i$, so the strong triangle inequality will give an equality, and we have
$$
\nu(a_\ell) = \nu(a_n)+\min_{\sigma\subset R\atop |\sigma|=n-\ell}\left(\sum_{r_i\in \sigma}\nu(r_i)\right)=\nu(a_n)+\min_{\sigma\subset R\atop |\sigma|=n-\ell}\left(\sum_{r_i\in \sigma}-\nu(d_i)\right)=\nu(a_n)-\max_{\sigma\subset R\atop |\sigma|=n-\ell}\left(\sum_{r_i\in \sigma}\nu(d_i)\right)=\sum_{r_i:\nu(r_i)\neq 0}\nu(d_i)-\sum_{r_i:\nu(r_i)\neq 0}\nu(d_i)=0
$$
contradicting (1) above, since $\ell>k$. Thus there are exactly $n-k$ values of $i$ with $\nu(d_i)>0$, and so after reindexing the roots $r_i$, we can write the (non-monic) factorization of $f$ over $(\mathcal{O}_K)_{\mathfrak{p}}$ as
$$
f(x)=\prod_{i=1}^{k}(x-c_i)\cdot \prod_{i=k+1}^{n}(d_ix-c_i).
$$
Let $g(x):=\prod_{i=1}^{k}(x-c_i)\in (\mathcal{O}_K)_{\mathfrak{p}}[x]$, and $\overline{g}$ its image in $((\mathcal{O}_K)_{\mathfrak{p}}/\mathfrak{p})[x]$. Notice that $\overline{f}(x)=\prod_{i=k+1}^nc_i\cdot \overline{g}(x)$. In particular, $\overline{f}$ splits over $(\mathcal{O}_K)_{\mathfrak{p}}/\mathfrak{p}$!
Now we're in the home stretch: $(\mathcal{O}_K)_{\mathfrak{p}}/\mathfrak{p}$ contains the splitting field $\kappa$ of $\overline{f}$, so letting $I(\mathfrak p)$ denote the inertia group at $\mathfrak{p}$ we conclude via $$k=[\kappa:\mathbb{F}_p]\mid [((\mathcal{O}_K)_{\mathfrak{p}}/\mathfrak{p}):\kappa][\kappa:\mathbb{F}_p]=[((\mathcal{O}_K)_{\mathfrak{p}}/\mathfrak{p}):\mathbb{F}_p]=\#\operatorname{Gal}(((\mathcal{O}_K)_{\mathfrak{p}}/\mathfrak{p})/\mathbb{F}_p)=\#D(\mathfrak{p})/\#I(\mathfrak{p})\mid \#D(\mathfrak{p})\mid \#\operatorname{Gal}(K/\mathbb{Q})=[K:\mathbb{Q}].$$
A: I'll start with a hint/sketch as Quinn does in his answer, if someone wants to try this before reading the details: factor $f(x)$ in $\mathbf Z_p[x]$  (not just in $\mathbf Q_p[x]$) using Hensel's lemma or Newton polygons to get a factor of degree $k$ in $\mathbf Z_p[x]$ that has an irreducible reduction mod $p$ of degree $k$. Then think of the Galois group of the splitting field of $f(x)$ over $\mathbf Q_p$ as a decomposition group at a prime ideal over $p$ in the integers of the splitting field of $f(x)$ over $\mathbf Q$.
We will use the decomposition group at a prime ideal, as Quinn does in the other solution posted here, but we'll interpret decomposition groups as Galois groups of $\mathfrak p$-adic completions.  This has the advantage that a decomposition group is genuinely a Galois group over $\mathbf Q_p$ even in the case of ramified primes, whereas if you work with localizations then you can say $D(\mathfrak p|p)/I(\mathfrak p|p)$ is a Galois group over $\mathbf F_p$ but you can't say that about $D(\mathfrak p|p)$ itself for all $p$.
The first step is to factor $f(x)$ in $\mathbf Z_p[x]$ into a factor of degree $k$ and a complementary factor, where the first factor reduces mod $p$ to an irreducible polynomial in $\mathbf F_p[x]$.  I will show two ways to do this: by Hensel's lemma and by Newton polygons.
Method 1: Factoring $f(x)$ in $\mathbf Z_p[x]$ by Hensel's lemma.
We are going to use a version of Hensel's lemma in $\mathbf Z_p[x]$ that allows for polynomials whose leading coefficients are not $p$-adic units. Here is the version I have in mind.
Theorem:  Let $f(x) \in \mathbf Z_p[x]$ with at least one coefficient being in $\mathbf Z_p^\times$, and assume the reduced polynomial $\overline{f}(x)$ in $\mathbf F_p[x]$ can be written as $\alpha(x)\beta(x)$ where $\alpha(x)$ and $\beta(x)$ are relatively prime polynomials in $\mathbf F_p[x]$. Then in $\mathbf Z_p[x]$ we can write $f(x) = a(x)b(x)$ where (i) $\overline{a}(x) = \alpha(x)$ and $\overline{b}(x) = \beta(x)$ in $\mathbf F_p[x]$ and (ii) $\deg a(x) = \deg \alpha(x)$.
Proof: See Gouvea, "$p$-Adic Numbers: an Introduction" (2nd ed.) Theorem 3.4.6, p. 72 or Borevich and Shafarevich, "Number Theory," Theorem 2, p. 275 for a proof.
Remark. I am using the hypotheses as they appear in Borevich and Shafarevich. Gouvea instead essentially assumes $\alpha(x)$ is monic and says in the conclusion that $a(x)$ is monic, but that is morally the same kind of condition as $\deg a(x) = \deg \alpha(x)$ in the theorem above.
Note $\overline{f}(x)$ is not $0$ in $\mathbf F_p[x]$ since some coefficient of $f(x)$ (not necessarily the leading coefficient!) is in $\mathbf Z_p^\times$.
We are not requiring $\deg b(x) = \deg \beta(x)$, and the following example illustrates that.
Example. In $\mathbf Z_2[x]$ let $f(x) = 6x^5 + 4x^3 + x^2 + x + 3$. Then in $\mathbf F_2[x]$, $\overline{f}(x) = x^2+x+1$, which is irreducible and the degree has dropped from $5$ down to $2$. We can write $\overline{f}(x)$ in $\mathbf F_2[x]$ as a product of relatively prime factors $\alpha(x)\beta(x)$ in two ways, and each lifts to a corresponding factorization of $f(x)$ as $a(x)b(x)$ in $\mathbf Z_2[x]$ where $\deg a(x) = \deg \alpha(x)$.
First way: $\alpha(x) = 1$ and $\beta(x) = x^2+x+1$. Then we can use $a(x) = 1$ and $b(x) = f(x)$. Indeed, $f(x) = 1 \cdot f(x)$, $\overline{1}(x) = \alpha(x)$, $\overline{f}(x) = \beta(x)$, and $\deg a(x) = \deg \alpha(x)$ (but $\deg b(x) = 6 \not= \deg \beta(x))$. This is how the theorem looks when we use $\alpha(x) = 1$. It is silly, but it is valid.
Second way: $\alpha(x) = x^2+x+1$ and $\beta(x) = 1$. Then the theorem says $f(x) = a(x)b(x)$ in $\mathbf Z_2[x]$ where $\overline{a}(x) = x^2+x+1$, $\overline{b}(x) = 1$, and $\deg a(x) = 2$ (but $\deg b(x) = 3 \not= \deg \beta(x)$). Using PARI,
$$
a(x) = x^2 + (1 + 2 + 2^5 + 2^{15} + 2^{16} + \cdots)x + (1 + 2^2 + 2^5 + 2^6 + 2^{10} + 2^{13} + 2^{16} + \cdots)
$$
and
$$
b(x) = 6x^3 + (2 + 2^2 + 2^3 + 2^5 + 2^8 + 2^9 + 2^{10} + 2^{11} + 2^{12} + 2^{13} + 2^{14} + 2^{15} + \cdots)x^2 + (2^2 + 2^3 + 2^4 + 2^6 + 2^9 + 2^{14} + 2^{16} + \cdots)x + (1 + 2 + 2^2 + 2^6 + 2^{11} + 2^{14} + 2^{15} + \cdots).
$$
Notice $\overline{a}(x) = x^2+x+1$ and $\overline{b}(x) = 1$ in $\mathbf F_2[x]$.  This concludes the example.
We'll now apply Hensel's lemma to the contest problem. We are told $f(x) \in \mathbf Z[x]$ has degree $n$ and $\overline{f}(x) = \pi(x)$ in $\mathbf F_p[x]$ is irreducible of degree $k$, so $1 \leq k \leq n$. Use Hensel's lemma with $\alpha(x) = \pi(x)$ and $\beta(x) = 1$, as in the "second way" of the example above.  We obtain $f(x) = a(x)b(x)$ in $\mathbf Z_p[x]$ where $\overline{a}(x) = \pi(x)$, $\overline{b}(x) = 1$, and $\deg a(x) = \deg \pi(x) = k$. That means $a(x)$ has a leading coefficient in $\mathbf Z_p^\times$, since its degree did not drop under reduction mod $p$. So $a(x)$ is a polynomial of degree $k$ in $\mathbf Z_p[x]$ whose reduction mod $p$ is irreducible of degree $k$.
Method 2:  Factoring $f(x)$ in $\mathbf Z_p[x]$ by Newton polygons.
By the hypotheses of the problem, the constant term $f(0)$ is not divisible by $p$, the coefficient of $x^k$ in $f(x)$ is not divisible by $p$, coefficients of $f(x)$ of degree strictly between $0$ and $k$ may or may not be divisible by $p$ (but at least they are integral) and the coefficients of $f(x)$ of degree greater than $k$ are all divisible by $p$. Therefore the first segment of the $p$-adic Newton polygon of $f(x)$ is the segment on the $x$-axis from $(0,0)$ to $(k,0)$. Further segments of the $p$-adic Newton polygon of $f(x)$ have positive slope, so $f(x)$ has $k$ roots in $\overline{\mathbf Q_p}$ with $p$-adic absolute value $1$ ($p$-adic valuation $0$) and its other roots in $\overline{\mathbf Q_p}$ have $p$-adic absolute value greater than $1$.
Numbers algebraic over $\mathbf Q_p$ that are roots of the same irreducible polynomial in $\mathbf Q_p[x]$ have the same $p$-adic absolute value, so the $k$ roots of $p$-adic absolute value $1$ are $\mathbf Q_p$-conjugate only to other roots of $p$-adic absolute value $1$. Therefore if we let $a(x)$ be the product of the monic irreducible factors of $f(x)$ in $\mathbf Q_p[x]$  whose roots have $p$-adic absolute value $1$,  then $\deg a(x) = k$, $a(x)$ is in $\mathbf Z_p[x]$ (not just $\mathbf Q_p[x]$!) and $f(x) = a(x)b(x)$ where $b(x) \in \mathbf Q_p[x]$.  We want to show $b(x) \in \mathbf Z_p[x]$. Since $a(x)$ is monic in $\mathbf Z_p[x]$ and $f(x) \in \mathbf Z_p[x]$, we can apply the division algorithm in $\mathbf Z_p[x]$ to write $f(x) = a(x)q(x) + r(x)$ in $\mathbf Z_p[x]$ where $r(x) = 0$ or $\deg r(x) < k$.
Viewing this in $\mathbf Q_p[x]$ and comparing it to $f(x) = a(x)b(x)$ in $\mathbf Q_p[x]$, the uniqueness of the division algorithm  in $\mathbf Q_p[x]$  tells us that $q(x) = b(x)$ and $r(x) = 0$. Therefore $b(x) \in \mathbf Z_p[x]$, so the factorization $f(x) = a(x)b(x)$ is in $\mathbf Z_p[x]$. Why is $a(x) \bmod p$ irreducible in $\mathbf F_p[x]$?
Reduce the equation $f(x) = a(x)b(x)$ modulo $p$: in $\mathbf F_p[x]$, $\overline{f}(x) = \overline{a}(x)\overline{b}(x)$. Since $a(x)$ is monic of degree $k$, $\overline{a}(x)$ has degree $k$ in $\mathbf F_p[x]$. By the hypothesis of the problem, $\overline{f}(x)$ is irreducible of degree $k$.    Therefore by unique factorization in $\mathbf F_p[x]$, $\overline{a}(x)$ is irreducible and $\overline{b}(x)$ is a nonzero constant.
Now we continue with the rest of the argument, knowing (by two methods) that $f(x)$ factors as $a(x)b(x)$ in $\mathbf Z_p[x]$ where $a(x) \bmod p$ is irreducible of degree $k$.  By the reduction mod $p$ test, $a(x)$ is irreducible in $\mathbf Q_p[x]$.
Let $r_1, \ldots, r_n$ be a full set of roots of $f(x)$ in an extension of $\mathbf Q_p$.  Then $\mathbf Q_p(r_1, \ldots, r_n)$ is a splitting field of $f(x)$ over $\mathbf Q_p$ and $K := \mathbf Q(r_1, \ldots, r_n)$ is a splitting field of $f(x)$ over $\mathbf Q$.  We want to show $k \mid [K:\mathbf Q]$.
Some $r_i$ is a root of $a(x)$, so $[\mathbf Q_p(r_i):\mathbf Q_p] = \deg a(x) = k$ on account of irreducibility of $a(x)$ over $\mathbf Q_p$. Therefore $k$ divides $[\mathbf Q_p(r_1,\ldots,r_n):\mathbf Q_p]$.
Let $\mathfrak p$ be a prime ideal lying over $p$ in $\mathcal O_K$. Then the $\mathfrak p$-adic completion $K_{\mathfrak p}$ is isomorphic to $\mathbf Q_p(r_1,\ldots, r_n)$. (Since $K/\mathbf Q$ is Galois, its completions at different prime ideals lying over $p$ are isomorphic as extensions of $\mathbf Q_p$, so the specific choice of $\mathfrak p$ lying over $p$ is not important; any one is as good as any other.) The extension $K_{\mathfrak p}/\mathbf Q_p$ is Galois and ${\rm Gal}(K_{\mathfrak p}/\mathbf Q_p)$ is isomorphic to the decomposition group $D(\mathfrak p|p)$ in ${\rm Gal}(K/\mathbf Q)$. From the "$efg = n$" formula for Galois extensions of number fields, $|{\rm Gal}(K/\mathbf Q)|$ is divisible by $|D(\mathfrak p|p)|$, so $[K:\mathbf Q]$ is divisible by $|{\rm Gal}(K_{\mathfrak p}/\mathbf Q_p)|$, which equals $[K_{\mathfrak p}:\mathbf Q_p]$, and that is divisible by $k$. Thus $k \mid [K:\mathbf Q]$.
In fact, since $a(x)$ has an irreducible reduction mod $p$, each root of $r$ of $a(x)$ generates an unramified extension of $\mathbf Q_p$, so $[\mathbf Q_p(r):\mathbf Q_p]$ divides the degree of the residue field extension for $K_{\mathfrak p}/\mathbf Q_p$. Residue field degrees do not change under completion, so $k$ divides $f(\mathfrak p|p)$, which is also what Quinn found at the end of his answer ($k \mid [(\mathcal O_K)_{\mathfrak p}/\mathfrak p:\mathbf F_p]$).
Боже мой, какая замечательная задача!
A: 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
$$k=\big[\mathbb F_p[x]\big/\big(\overline f^*\big(x/\overline{f(0)}\big)\big):\mathbb F_p\big]\mid\big[\mathcal O_K/\mathfrak P:\mathbb F_p\big]\mid\big[K:\mathbb Q\big].$$
A: $f\in \Bbb{Z}[x]$ is such that $f\bmod p$ is irreducible and non-constant. Factorize $f=\prod_j f_j\in \Bbb{Z}[x]$, then one of the $f_j$ is such that $f_j=c f\bmod p$ with $c\in \Bbb{F}_p^\times$.
Let $L$ be the splitting field of $f$ so that $f_j$ has a root $a/b\in Frac(O_L)$.
Note that $p$ is not a unit of $\Bbb{Z}[a/b]\cong\Bbb{Z}[x]/(f_j)$. Since $O_L[a/b]$ is integral over $\Bbb{Z}[a/b]$ it means that $O_L[a/b]$ contains a maximal ideal $P$ above $p$.
Being finite fields $O_L[a/b]/P$ contains and is integral over $O_L/(P\cap O_L)$, thus $a/b\in (O_L)_{(P\cap O_L)}$ and $O_L/(P\cap O_L)\cong O_L[a/b]/P$.
So $O_L/(P\cap O_L)$ contains $\Bbb{Z}[a/b]/(p) \cong \Bbb{F}_p[x]/(f)$.
Thus $\deg(f\bmod p)$ divides $[O_L/(P\cap O_L):\Bbb{F}_p]$.
The factorization $pO_L = \prod_j Q_j^e$ with $Q_1=P\cap O_L$ and $N(Q_j)=N(Q_1)=p^{[O_L/(P\cap O_L):\Bbb{F}_p]}, N(pO_L)=p^{[L:\Bbb{Q}]}$ gives that $\deg(f\bmod p)$ divides $[L:\Bbb{Q}]$.
