= 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 \b[$m$ is an <irreducible polynomial> over $F$].
For the final unnumbered request, $\zeta=e^{\pi i/3}=\frac12+i\frac{\sqrt3}2$ 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
$$
\boxed{m_\zeta(x)=x^2-x+1.}
$$
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