The seventh cyclotomic polynomial is
It is irreducible over , and all its roots , , lie in . Hence is a degree-six Finite Galois extension, with
The 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, put
The cyclotomic relation gives , while direct multiplication gives . Hence
Its discriminant is , so
For the real cubic subfield of the seventh cyclotomic field, divide
by . With ,
so
This cubic has no rational root, and therefore
The requested primitive elements, minimal polynomials, and automorphism groups are consequently
More explicitly, the nontrivial automorphism of sends
and the three automorphisms of cyclically permute
Finally, is abelian, so every subgroup is normal. The subextensions of an abelian Galois extension theorem shows that
For , the answer is yes. A root of is or , so
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 make
Galois. This is false: has two nonreal roots absent from the real field . Hence is a proper subfield of .
For , the answer is yes. Given , take
which is coprime to . Every root of in characteristic has exact multiplicative order . The degree
is the least positive such that . The value works, and no can work because
Therefore
Every element algebraic over lies in some finite field , so
These 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.