TITLE: Why does an irreducible polynomial split into irreducible factors of equal degree over a Galois extension?
QUESTION [19 upvotes]: 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.

REPLY [5 votes]: 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.)