Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 2 / 18G (Galois Theory)

Every $\tau \in \mathrm{Aut}(L({\xi }_N)/K({\xi }_N))$ fixes $K$ and permutes the roots of $f$, so it preserves their splitting field $L$. Restriction therefore defines a homomorphismIf $\rho (\tau )$ is the identity, then $\tau$ fixes both $L$ and ${\xi }_N$, which generate $L({\xi }_N)$; hence $\tau$ is the identity. Thus $\rho$ is injective, and its image is a subgroup:

Let $\sigma (\alpha )={\xi }_d\alpha$ with $\alpha \ne 0$. Thenand these $d$ elements are distinct because ${\xi }_d$ is a primitive root of unity. Thus the stabilizer of $\alpha$ in the cyclic Galois group is trivial. The orbit-stabilizer form of the Fundamental theorem of Galois theory givessoMoreover,The fixed field of the whole Galois group is $K$, and thereforeThis is the basic eigenvector mechanism behind a cyclic Kummer extension.

A group is solvable when its derived serieseventually reaches the trivial group; equivalently, it has a subnormal series with abelian quotients.

The symmetric group $S_4$ is solvable becausehas abelian quotients $V_4$, $C_3$, and $C_2$, where $V_4$ is the Klein four-group. Thus

For $n\ge 2$, the commutator subgroup of $S_n$ is $A_n$. The group $A_5$ is nonabelian and simple, so its commutator subgroup is a nontrivial normal subgroup and must equal $A_5$ itself. Hence the derived series of $S_5$ isand never reaches the identity. Therefore

Let $x_1,\dots ,x_5$ be algebraically independent indeterminates and set $a_i=e_i(x_1,\dots ,x_5)$, the elementary symmetric polynomials. ThenThe Fundamental theorem of symmetric polynomials identifies the fixed field of the natural $S_5$ action on $\mathbb{Q}(x_1,\dots ,x_5)$ with $\mathbb{Q}(a_1,\dots ,a_5)=K$. Every permutation of the roots therefore gives a $K$-automorphism, and the Galois group of a polynomial isA characteristic-zero polynomial is solvable by radicals exactly when its Galois group is a solvable group. Part (c) shows that $S_5$ is not solvable, so