Past exam of the number theory course of the University of Oxford / 2010 / A / 2

The Galois group $\mathrm{Gal}(L/K)$ is the group of all field automorphisms of $L$ fixing every element of $K$, with composition as its operation. A finite extension is a Galois extension if it is both normal and separable; equivalently, it is a splitting extension of separable polynomials, or its group has order $[L:K]$. Normality means that every irreducible polynomial over $K$ with a root in $L$ splits there.

The Fundamental theorem of Galois theory gives an inclusion-reversing bijection for a finite Galois extension $L/K$ with group $G$:It satisfies $[L:L^H]=|H|$ and $[L^H:K]=[G:H]$. An intermediate extension $M/K$ is Galois precisely when its corresponding subgroup $H$ is normal, in which case restriction induces $\mathrm{Gal}(M/K)\cong G/H$.

The cyclotomic polynomial isBy the allowed irreducibility, this is the minimal polynomial of $\xi$ and its degree is $\phi (n)$. The field $\mathbb{Q}(\xi )$ contains every $n$th root of unity and is the splitting field of $X^n-1$. This polynomial has no repeated roots in characteristic zero, so the extension is Galois.

An automorphism sends $\xi$ to another primitive root ${\xi }^a$, where $a$ is a unit modulo $n$. Conversely each such conjugate defines an embedding, and it is an automorphism of $\mathbb{Q}(\xi )$ because $a$ has an inverse modulo $n$. Composition corresponds to multiplication of exponents. Therefore

Let $a=\sqrt[n]{2}$ and $F=\mathbb{Q}(\xi )$. The roots of $X^n-2$ are $a{\xi }^j$, so $L=F(a)$. This splitting extension is separable in characteristic zero, hence Galois over $\mathbb{Q}$. The subextension $F/\mathbb{Q}$ is itself Galois by part (iii), so restriction gives a surjectionwith normal kernel $H=\mathrm{Gal}(L/F)$. Surjectivity follows from the extension theorem for embeddings and normality of $L/\mathbb{Q}$.

For $\sigma \in H$, write $\sigma (a)=a{\xi }^{j(\sigma )}$. This defines an injective homomorphism $H\to ⟨\xi ⟩$: injectivity holds because $L=F(a)$, and multiplicativity holds because $H$ fixes all roots of unity in $F$. A subgroup of a cyclic group is cyclic. The first isomorphism theorem therefore givesThere is no need to assume that $|H|=n$; the intersection with the cyclotomic field can reduce its order.