Past exam of the options I course of the University of Oxford / 2005 / B / 1 / iii / Solution

If $m=uv$ with both factors of positive degree, evaluation gives $u(a)v(a)=0$. Since $K$ is a field, one factor vanishes at $a$, contrary to the minimal degree of $m$. Thus $m$ is an irreducible polynomial over $F$.

For the final unnumbered request, $\zeta =e^{\pi i/3}=\frac{1}{2}+i\frac{\sqrt{3}}{2}$ satisfies ${\zeta }^2-\zeta +1=0$. The monic quadratic $x^2-x+1$ has no rational root, since its discriminant is $-3$. It is therefore an irreducible polynomial over $\mathbb{Q}$, and