Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 1 / 18I / b / Solution

The seventh cyclotomic polynomial isIt is irreducible over $\mathbb{Q}$, and all its roots ${\zeta }_7^a$, $a=1,\dots ,6$, lie in $L=\mathbb{Q}({\zeta }_7)$. Hence $L/\mathbb{Q}$ is a degree-six Finite Galois extension, withThe group $(\mathbb{Z}/7\mathbb{Z}{)}^{\times }$ is cyclic of order six; for example,

By the Galois correspondence, the subfields are the fixed fields of the four subgroups of $C_6$:Thus these are all the subfields.

For the Quadratic Gaussian period in the seventh cyclotomic field, putThe cyclotomic relation gives $\eta +\bar{\eta }=-1$, while direct multiplication gives $\eta \bar{\eta }=2$. HenceIts discriminant is $-7$, so

For the real cubic subfield of the seventh cyclotomic field, divideby ${\zeta }_7^3$. With $\theta ={\zeta }_7+{\zeta }_7^{-1}$,soThis cubic has no rational root, and therefore

The requested primitive elements, minimal polynomials, and automorphism groups are consequentlyMore explicitly, the nontrivial automorphism of $\mathbb{Q}(\eta )$ sendsand the three automorphisms of $\mathbb{Q}(\theta )$ cyclically permute

Finally, $G$ is abelian, so every subgroup is normal. The subextensions of an abelian Galois extension theorem shows that

Solved by gpt-5.6-sol high.