Why does an irreducible polynomial split into irreducible factors of equal degree over a Galois extension? I've been struggling to prove this fact over the past day or so.

Suppose $f(x)\in F[X]$ is irreducible over a field $F$ with $\deg(f)=n$, and let $L$ be the splitting field of $f(x)$ over $F$ with $\alpha$ a root of $f(x)$ in $L$. Let $K/F$ be a Galois extension, with $K\subset L$. Then why does $f(x)$ split into a product of $m$ irreducible polynomials of degree $d$ over $K$, with $m=[F(\alpha)\cap K:F]$ and $d=[K(\alpha):K]$?

If $G$ is the Galois group of $L/F$, then I let $H$ be the corresponding subgroup of $K$, which is normal since $K/F$ is Galois. I know $H$ permutes the roots of $f(x)$, specifically fixing the roots of $f(x)$ in $K$. I know some basic equalities like $[F(\alpha):F]=n$, and $[K:F]=|G:H|$, but I'm not seeing how to start a good argument. 
What's a good way to prove this, or at least proceed? Thank you for any insight.
 A: This may also be viewed in the context of factorization of prime ideals in integral extensions of Dedekind domains (possibly a sledgehammer for a small task).
The prime (ideal)  $(h(x)) \subset F[x]$ splits into products
of powers of prime ideals  $\prod_{i=1}^g (h_i(x))^{e_i}$ in $K[x]$.
 Then a standard theorem states
$\sum_{i=1}^g e_if_i= n=[K:F]$ where $e_i$ are ramification indices, $f_i$ residual degrees, and $g$ the number of primes lying above $f(x)$. In the special case $K$ is Galois over $F$, all $e_i$'s are equal  and all $f_i$'s are equal.
And so we have isomorphisms of residual fields by the transitive action of $Gal\,(K/F)$ on the set of $h_j$'s:  $K[x]/(h_i(x))\cong K[x]/(h_j(x))$. So degree of $h_i(x) = $ degree of $h_j$.  (As transitive action of Galois group needs just Chinese Remainder Theorem, this answer, I hope, is  not a circular argument.)
A: Factor $f(x)$ into irreducibles in $K[x]$,
$$f(x) = q_1(x)\cdots q_m(x).$$
If $\alpha$ is a root of $q_1(x)$ and $\beta$ is a root of $q_i(x)$, then $q_1(x)$ is the irreducible polynomial of $\alpha$ over $K$, and $q_i(x)$ is the irreducible polynomial of $\beta$ over $K$. 
Since the action of $G$ is transitive on the roots of $f$ there exists $\sigma\in G$ such that $\sigma(\alpha)=\beta$. Since $K$ is Galois, $\sigma(K)=K$, so $\sigma(q_1(x))\in K[x]$ is a polynomial that has $\sigma(\alpha)=\beta$ as a root. Therefore, $q_i(x)|\sigma(q_1(x))$. Since both are irreducible, it follows that $\deg(q_i) = \deg(\sigma(q_1))=\deg(q_1)$. So all irreducible factors of $f(x)$ in $K$ have the same degree. 
The degree is equal to the degree of the extension $K(\alpha)/K$ (which is the degree of $q_1(x)$). 
A for $m$, the number of factors, we have:
$n=[F(\alpha):F] = [F(\alpha):F(\alpha)\cap K][F(\alpha)\cap K:F]$. 
I claim that $[F(\alpha):F(\alpha)\cap K] = [K(\alpha):K]$.
Indeed, since $K$ is Galois over $F$, then $K(\alpha)$ is Galois over $F(\alpha)$ ($K$ is the splitting field of some polynomial over $F$, and this same polynomial works for $K(\alpha)$), and $K$ is Galois over $K\cap F(\alpha)$. If $\sigma\in\mathrm{Gal}(K(\alpha)/F(\alpha))$, then restricting $\sigma$ to $K$ gives a homomorphism $\mathrm{Gal}(K(\alpha)/F(\alpha))\to \mathrm{Gal}(K/K\cap F(\alpha))$. If $\sigma$ restricts to the identity on $K$, then it must be the identity on $K(\alpha)$ (it fixes $\alpha$ since it fixes $F(\alpha)$ pointwise), so the map $\mathrm{Gal}(K(\alpha)/F(\alpha))\to \mathrm{Gal}(K/K\cap F(\alpha))$ is one-to-one.
Let $H'$ be the image of this map. Then $H'$ fixes $K\cap F(\alpha)$ pointwise, and if $k\in K$ is fixed by all elements of $H'$, then $k$ must be fixed by all elements of $\mathrm{Gal}(K(\alpha)/F(\alpha))$, hence $k\in F(\alpha)\cap K$. So $F(\alpha)\cap K$ is the fixed field of $H'$, hence $H'=\mathrm{Gal}(K/K\cap F(\alpha))$. 
Thus, $[K(\alpha):F(\alpha)] = [K:K\cap F(\alpha)]$. 
Now, 
$$\begin{align*}
[K(\alpha):K\cap F(\alpha)] &= [K(\alpha):K][K:K\cap F(\alpha)]\\
\text{ and }[K(\alpha):K\cap F(\alpha)] &= [K(\alpha):F(\alpha)][F(\alpha):K\cap F(\alpha)].\end{align*}$$ 
Therefore, $[K(\alpha):K] = [F(\alpha):K\cap F(\alpha)]$, as claimed. 
So we have that
$$\begin{align*}
n &= [F(\alpha):F]= [F(\alpha):F(\alpha)\cap K][F(\alpha)\cap K:F]\\
 &= [K(\alpha):K][F(\alpha)\cap K:F] \\
&= d[F(\alpha)\cap K:F].
\end{align*}$$ 
Since $n=dm$, then
$$m = [F(\alpha)\cap K:F],$$
as claimed. 
To see the action of $G$ is transitive, note that $F(\alpha)$ is isomorphic to $F(\beta)$ over $F$, since $\alpha$ and $\beta$ have the same irreducible over $F$; hence there is an isomorphism $\sigma\colon F(\alpha)\to F(\beta)$ that restricts to the identity on $F$ and sends $\alpha$ to $\beta$. Since $L$ is Galois over $F$, and $F(\alpha),F(\beta)\subseteq L$, $\sigma$ extends to an element of $G$.
