Irreducible polynomial in field extension

Solution 1:

$\newcommand{\Size}[1]{\left\lvert #1 \right\rvert}$Let $g$ be an irreducible factor of $f$ in $L[x]$, of degree $q \le n$. Let $\alpha$ be a root of $g$ (and thus of $f$) in some extension of $K$, so that $\Size{K[\alpha] : K} = n$, and $\Size{L[\alpha] : L} = q$

We have $$\Size{L[\alpha] : K} = \Size{L[\alpha] : L} \cdot \Size{L : K} = q m.$$

Now $K[\alpha] \subseteq L[\alpha]$, so $$ q m = \Size{L[\alpha] : K} = \Size{L[\alpha] : K[\alpha]} \cdot \Size{K[\alpha] : K} $$ implies $\Size{K[\alpha] : K} = n \mid q m$.

Since $(n, m) = 1$, we have that $n \mid q$, so $n = q$ and $f$ is irreducible in $L[x]$.