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

An element is algebraic over $F$ when some nonzero polynomial in $F[x]$ vanishes at it. Its minimal polynomial is the monic polynomial of least positive degree with this property; scaling a least-degree nonzero vanishing polynomial makes it monic.

If $m_1,m_2$ are two such polynomials, they have the same least degree. Their difference has smaller degree because both are monic, and it also vanishes at $a$. Minimality forces the difference to be zero. Hence the monic minimal polynomial is unique.

Divide $f$ by $m$ in the polynomial ring over the field: $f=qm+r$, where $r=0$ or $\mathrm{deg}r<\mathrm{deg}m$. If $f(a)=0$, then $r(a)=0$, and minimality forces $r=0$. Conversely a multiple of $m$ vanishes at $a$. ThereforeThis includes the zero polynomial.

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