Past exam of the options II course of the University of Oxford / 2007 / B (Introduction to Fields)

An element is algebraic over $F$ if it is a zero of some nonzero polynomial in $F[x]$. An extension is algebraic if every element is algebraic. If $[K:F]=d<\infty$, the $d+1$ elements $1,\alpha ,\dots ,{\alpha }^d$ are linearly dependent over $F$, proving every $\alpha \in K$ is algebraic.

The converse fails for the algebraic closure $\overline{\mathbb{Q}}$ inside $\mathbb{C}$. All its elements are algebraic by definition, but it contains $2^{1/n}$ of degree $n$ for every positive integer $n$, since $x^n-2$ is irreducible by Eisenstein's criterion at two. A finite extension could not contain elements of unbounded degree, so this algebraic extension is infinite.

All sixth roots of unity lie in $\mathbb{Q}({\zeta }_6)$, with ${\zeta }_6=(1+\sqrt{-3})/2$. Its quadratic polynomial $x^2-x+1$ has no rational root, so

A primitive seventh root generates all seventh roots, so the splitting field is $\mathbb{Q}({\zeta }_7)$. Its minimal polynomial is the given irreducible cyclotomic polynomial $1+x+\cdots +x^6$. Therefore $\boxed{[K:\mathbb{Q}]=6}$.

A primitive eighth root satisfies ${\zeta }_8=(1+i)/\sqrt{2}$, while $i={\zeta }_8^2$ and $\sqrt{2}={\zeta }_8+{\zeta }_8^{-1}$. Thus $K=\mathbb{Q}(i,\sqrt{2})$. The real elements of $\mathbb{Q}(i)$ are rational, so it does not contain $\sqrt{2}$. The tower law gives

The cubic $x^3+x+1$ has no zero in ${\mathbb{F}}_2$, so is irreducible. Its quotient field has eight elements. Every nonzero element of that field satisfies $x^7=1$ by the order of its multiplicative group, so the polynomial splits there. It does not already split in the base field; a root of that irreducible cubic, which divides $x^7-1$, is needed. Hence

In characteristic two, repeated squaring gives $x^8-1=(x-1{)}^8$. There is only the root one, already in the base field. Thus $\boxed{K={\mathbb{F}}_2,[K:{\mathbb{F}}_2]=1}$. Repeated roots do not require a larger splitting field.