Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 18J (Galois Theory)

Fix a primitive $n$th root of unity ${\zeta }_n$. The $n$th cyclotomic polynomial isIts roots are exactly the roots of unity of order $n$.

Partitioning all $n$th roots of unity by exact order gives the cyclotomic factorizationWe prove ${\Phi }_n\in \mathbb{Z}[X]$ by induction on $n$. The assertion is clear for $n=1$. If it holds for every proper divisor of $n$, thenThe denominator is monic and belongs to $\mathbb{Z}[X]$. Exact polynomial division shows that the quotient lies in $\mathbb{Q}[X]$, and Gauss's lemma makes this monic quotient an element of $\mathbb{Z}[X]$.

Now let $K\subseteq \mathbb{C}$. The minimal polynomial of ${\zeta }_n$ over $K$ divides ${\Phi }_n$, so every $K$-conjugate of ${\zeta }_n$ is another primitive root ${\zeta }_n^a$ and lies in $K({\zeta }_n)$. The extension is therefore normal. It is separable because the characteristic is zero, and hence it is Galois.

Every $\sigma \in \mathrm{Gal}(K({\zeta }_n)/K)$ hasfor some $a\in (\mathbb{Z}/n\mathbb{Z}{)}^{\times }$. Since ${\zeta }_n$ generates the extension, this gives an injective homomorphismThe group on the right is abelian, so the Galois group is abelian, as summarized by the cyclotomic field construction.

A real subfield is constructible when all its elements are straightedge-and-compass constructible; algebraically, it is contained in a tower of quadratic extensions of $\mathbb{Q}$.

Let $\zeta =e^{2\pi i/17}$ andThe cyclotomic extension $\mathbb{Q}(\zeta )/\mathbb{Q}$ has Galois groupwhich is cyclic of order $16$. Complex conjugation corresponds to $-1$, so the maximal real subfield hasa cyclic group of order $8$.

This group has a chainin which every index is two. By the Galois correspondence, the fixed fields formwith every successive degree equal to two. Thus $F$, and in particular $\cos (2\pi /17)$, is constructible. This is the constructibility of the real seventeenth cyclotomic field.

For $K=\mathbb{R}$, the answer is yes. A root of ${\Phi }_4(X)=X^2+1$ is $i$ or $-i$, so

For $K=\mathbb{Q}$, the answer is no. Every $\mathbb{Q}({\zeta }_n)/\mathbb{Q}$ is finite abelian Galois. If $\sqrt[3]{2}$ belonged to ${\mathbb{Q}}^{\mathrm{cyc}}$, then it would belong to some $\mathbb{Q}({\zeta }_n)$. The subextensions of an abelian Galois extension result would makeGalois. This is false: $X^3-2$ has two nonreal roots absent from the real field $\mathbb{Q}(\sqrt[3]{2})$. Hence ${\mathbb{Q}}^{\mathrm{cyc}}$ is a proper subfield of $\overline{\mathbb{Q}}$.

For $K={\mathbb{F}}_p$, the answer is yes. Given $m\ge 1$, takewhich is coprime to $p$. Every root of ${\Phi }_n$ in characteristic $p$ has exact multiplicative order $n$. The degreeis the least positive $d$ such that $n\mid p^d-1$. The value $d=m$ works, and no $d<m$ can work becauseThereforeEvery element algebraic over ${\mathbb{F}}_p$ lies in some finite field ${\mathbb{F}}_{p^m}$, soThese are respectively the maximal cyclotomic extension of the real numbers, the maximal cyclotomic extension of the rational numbers, and the maximal cyclotomic extension of a finite field.