Past exam of the Probability, statistics and options II course of the University of Oxford / 2007 / B / 2

An element $\alpha$ is an algebraic element over $F$ if some nonzero polynomial in $F[X]$ vanishes at $\alpha$. A field extension is an algebraic extension if each of its elements is an algebraic element over $F$. If $[K:F]=d<\infty$, the $d+1$ vectors $1,\alpha ,\dots ,{\alpha }^d$ are linearly dependent, producing such a polynomial. Thus every finite field extension is an algebraic extension.

The converse fails for the field of algebraic numbers $\overline{\mathbb{Q}}$. It is an algebraic extension of $\mathbb{Q}$ by definition, but it contains primitive $p$th roots of unity of degree $p-1$ for arbitrarily large primes $p$, by the permitted irreducibility of the cyclotomic polynomial. The tower law would make every such degree at most $[\overline{\mathbb{Q}}:\mathbb{Q}]$ if the latter were finite, a contradiction.

The sixth roots of unity all belong to $\mathbb{Q}({\zeta }_6)$, where ${\zeta }_6=(1+\sqrt{-3})/2$. Its minimal polynomial of an algebraic element is $X^2-X+1$, which has no rational root. Thus the splitting field is $\mathbb{Q}(\sqrt{-3})$ and

A primitive seventh root of unity ${\zeta }_7$ generates all roots. Its minimal polynomial of an algebraic element is the permitted irreducible cyclotomic polynomial $1+X+\cdots +X^6$. Hence the splitting field has

A primitive eighth root of unity is ${\zeta }_8=(1+i)/\sqrt{2}$, and $\mathbb{Q}({\zeta }_8)=\mathbb{Q}(i,\sqrt{2})$. Indeed $i={\zeta }_8^2$ and $\sqrt{2}={\zeta }_8+{\zeta }_8^{-1}$, while the displayed expression gives the reverse containment. Since $\mathbb{Q}(\sqrt{2})\subset \mathbb{R}$, adjoining $i$ has degree two; adjoining $\sqrt{2}$ to $\mathbb{Q}$ also has degree two. By the tower law,

The polynomial $X^3+X+1$ has no root in ${\mathbb{F}}_2$, so it is an irreducible polynomial. The quotient ring of ${\mathbb{F}}_2[X]$ by the ideal it generates is therefore a field with eight elements. The seven nonzero elements form a group of order seven, so each satisfies $x^7=1$ by Lagrange's theorem. Thus this field splits $X^7-1$. A root different from one has order seven; it cannot lie in ${\mathbb{F}}_2$ and generates this degree-three field extension, since three is prime and the tower law applies. Therefore

In characteristic two, the binomial theorem gives $X^8-1=(X-1{)}^8$. The only root is already in ${\mathbb{F}}_2$, so the splitting field is the original field, despite the repeated root: