Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 3 / 18I (Galois Theory)

Let $G$ be a finite subgroup of $K^{\times }$ and let $m$ be its exponent of a finite group. A finite abelian group contains an element whose order is its exponent: combine elements of maximal prime-power order across the primary components. Hence $m\le |G|$.

Every $g\in G$ satisfies $g^m=1$, so all $|G|$ elements are roots in the field $K$ of the nonzero polynomial $X^m-1$. The Lagrange root bound over a field gives $|G|\le m$. Therefore $|G|=m$, and the element of order $m$ generates $G$. This proves that a finite subgroup of a field multiplicative group is cyclic.

A primitive root of unity is an $n$th root of unity of multiplicative order exactly $n$.

Let ${\zeta }_m,{\zeta }_n\in K$ be primitive and let $H$ be the finite subgroup of $K^{\times }$ that they generate. By part a, $H$ is cyclic. Its order is divisible by both $m$ and $n$, hence by $N=\mathrm{lcm}(m,n)$. On the other hand every element of $H$ has $N$th power one, so the order of the cyclic group $H$ divides $N$. Its order is therefore exactly $N$, and a generator of $H$ is a primitive $N$th root of unity in $K$.

The cyclotomic polynomial iswhere ${\zeta }_n$ is primitive. Partitioning all $n$th roots by exact order gives the cyclotomic factorizationInduct on $n$. The quotient of the monic integer polynomial $X^n-1$ by the product of the already constructed monic ${\Phi }_d$, $d<n$, lies in $\mathbb{Q}[X]$ and is monic. Gauss lemma then shows that this quotient ${\Phi }_n$ lies in $\mathbb{Z}[X]$.

If the prime $p$ does not divide $n$, then over ${\mathbb{F}}_p$,Thus $X^n-1$ has no repeated root. Every factor, including the reduction of ${\Phi }_n$, is square-free, proving the Separability of a cyclotomic polynomial modulo p.

Choose a primitive root ${\zeta }_n$. Since every root of ${\Phi }_n$ is a power of ${\zeta }_n$, its splitting field is $L=K({\zeta }_n)$. Each $\sigma \in G$ sends ${\zeta }_n$ to another primitive root, uniquely of the formThe assignment $\sigma \mapsto a_{\sigma }$ is a homomorphism. It is injective because an automorphism fixing ${\zeta }_n$ fixes $K({\zeta }_n)=L$.

If ${\Phi }_n$ is irreducible over $K$, the orbit of ${\zeta }_n$ has all $\phi (n)$ primitive roots, so the injection has image of size $\phi (n)$ and is surjective. Conversely, surjectivity makes every primitive root a conjugate of ${\zeta }_n$. Its minimal polynomial over $K$ then has at least $\phi (n)$ roots and divides the degree-$\phi (n)$ polynomial ${\Phi }_n$, so it equals ${\Phi }_n$. This proves the Galois embedding for a cyclotomic polynomial and the claimed equivalence.

Let $L=\mathbb{Q}({\zeta }_n)$. We use the standard cyclotomic containment lemma: every root of unity in $\mathbb{Q}({\zeta }_n)$ has order dividing $\mathrm{lcm}(2,n)$. One proof uses the conductor of a rational cyclotomic field. The conductor attached to a primitive $m$th root is $m$, except that it is $m/2$ when $m\equiv 2(\mathrm{mod}4)$; containment in $\mathbb{Q}({\zeta }_n)$ forces this conductor to divide $n$, which gives $m\mid \mathrm{lcm}(2,n)$.

If $n$ is even, $\mathrm{lcm}(2,n)=n$, and the powers of ${\zeta }_n$ already give all $n$ possible roots of unity in $L$. If $n$ is odd, $-{\zeta }_n$ has order $2n$, so $L$ contains all $2n$th roots of unity, and the containment lemma shows there are no others. Thus the roots of unity in a rational cyclotomic field number