An irreducible polynomial of degree n over field is irreducible over extension of degree m if m and n are coprime

Question

Here is the same question:
An irreducible polynomial of degree coprime to the degree of an extension is irreducible over this extension
But I can't understand the first step.
Does "monic irreducible of $u$ over $K$" mean the irreducible polynomial over $K$ with $u$ as its root? Why the monic polynomial of u must divide f?

Answer 1

The "monic irreducible of $u$ over $K$" is understood to be the minimal polynomial of $u$ over $K$, let's call it $m_{u,K}$. That $m_{u,K}$ divides $f$ follows by a simple argument: By the Euclidean algorithm in $K[x]$ we can write $$f(x)=q(x)m_{u,K}(x)+r(x),$$ with $\deg(r(x))<\deg(m_{u,K}(x))$. Notice that $$0=f(u)=q(u)m_{u,K}(u)+r(u),$$ which implies that $r(u)=0$ but that contradicts the minimality of $m_{u,K}$ unless $r(x)=0$. Therefore you conclude that $m_{u,K}$ divides $f$.