$K/k$ field extension, $[K:k]=n,f(x)\in k[x]$ irreducible over $k$ with degree $m$, and $f$ has a root in $K\implies m$ divides $n$

Question

Let $K/k$ a field extension, with $[K:k]=n\in\mathbb{N}$. Let $f(x)\in k[x]$, with $\deg (f)=m$. I need to show that if $f$ has a root $\alpha\in K$, so $m|n$.

I know that in $K[x]$ we can write $f(x)=(x-\alpha)g(x)$, where $g(x)\in k[x]$, but this implies $\deg(g)=m-1$. What can I do?

Answer 1

If $\alpha\in K$ is this root, then $k(\alpha)$ is an intermediate field between $k$ and $K$. This implies that $$ [K:k]=[K:k(\alpha)]\cdot [k(\alpha):k]\\ n=[K:k(\alpha)]\cdot m $$