Past exam of the mathematics course of the University of Cambridge 2021 ii Paper 1 18I b Solution Created 2026-09-24 Updated 2026-09-25
The seventh cyclotomic polynomial isIt is irreducible over , and all its roots , , lie in . Hence is a degree-six Finite Galois extension, withThe group is cyclic of order six; for example,
By the Galois correspondence, the subfields are the fixed fields of the four subgroups of :Thus these are all the subfields.
For the Quadratic Gaussian period in the seventh cyclotomic field, putThe cyclotomic relation gives , while direct multiplication gives . HenceIts discriminant is , so
For the real cubic subfield of the seventh cyclotomic field, divideby . With ,soThis cubic has no rational root, and therefore
The requested primitive elements, minimal polynomials, and automorphism groups are consequentlyMore explicitly, the nontrivial automorphism of sendsand the three automorphisms of cyclically permute
Finally, is abelian, so every subgroup is normal. The subextensions of an abelian Galois extension theorem shows that
Past exam of the mathematics course of the University of Cambridge 2025 ii Paper 4 18J c Solution Created 2026-09-24 Updated 2026-09-25
For , the answer is no. Every is finite abelian Galois. If belonged to , then it would belong to some . The subextensions of an abelian Galois extension result would makeGalois. This is false: has two nonreal roots absent from the real field . Hence is a proper subfield of .
For , the answer is yes. Given , takewhich is coprime to . Every root of in characteristic has exact multiplicative order . The degreeis the least positive such that . The value works, and no can work becauseThereforeEvery element algebraic over lies in some finite field , 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.