Let $f(x)\in F[x]$ be an irreducible polynomial of degree $n$ over the field $F$. For any $K$ is any Galois extension of $F$, show that $f(x)$ factors in $K[x]$ see this (whether or not $K$ is contained in the Galois closure $L$ of $f(x)$).
I have proved:
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 $d=[K(\alpha):K]$?
The steps to prove this are : 1) Show that the factorisation of $f(x)$ over $K$ is the same as its factorisation over $L \cap K$.
2) Then show that the factors of $f(x)\in F[x]$ over $L \cap K$ correspond to the orbits of $H=Gal(L/L \cap K)$ on the roots of $f(x)$ and then proceed.
What's a good way to prove this, or at least proceed? Thank you for any insight.
Edit: Let me clarify some of the doubts raised in the comment suppose $f$ is irreducible over $K$ then let it be. The question is not that $f(x)$ factors or not! The question is show that $f(x)$ factors in $K[x]$(whether or not $K$ is contained in the Galois closure $L$ of $f(x)$).
Observe my highlighted part: there is a possibility of $m$ being $1$ then $F(\alpha) \cap K=F$.
