# Irreducible polynomial in field extension

I have this problem

Let $$L/K$$ be a finite field extension and $$|L:K|=m$$. Let $$f(x)$$ be an irreducible polynomial of degree $$n$$ in $$K[x]$$ and $$\gcd(m,n)=1$$. Prove that $$f$$ is irreducible over $$L$$.

I'm just a beginner so I'm really confused with the $$\gcd(m,n)=1$$, I don't know how to use it. So help me with this problem. Thank you

• $(m, n) = 1$ means $\gcd(m, n) = 1$. This notation is usually used in number theory's text. IMO, $(a, b)$ is one of the most overloaded notation. It can means coordinate, row vector, open interval, gcd, ... – Alex Vong Mar 6 '17 at 15:49
• Thank you, I know what it is, just don't know how to use it in this situation – chí trung châu Mar 6 '17 at 15:51

$\newcommand{\Size}{\left\lvert #1 \right\rvert}$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 $L[x]$.