Let be a positive integer. The subfield of generated over by an -th root of unity is called the -th cyclotomic field, denoted by .
Proposition
Let be a prime, and let . The Galois group of over is a cyclic group of order . It is isomorphic to the multiplicative group . Futhermore, for any subfield of , the Galois group of over is a cyclic group.
Proof Let be the Galois group of over . Clearly, is the splitting field of the irreducible cyclotomic polynomial , so is a Galois extension, and . Note that , the elements in are precisely -isomorphisms , which are maps that maps to . In , will act as an automorphism and send to . They satisfy thus can be naturally identified with elements in . Similarly, the element in the Galois group of over are of the form, so this Galois group must be a subgroup of , which has to be cyclic.
e.g. Let , we have the following diagram of Galois extensions hold:
One can see that is the only subgroup of order 4 in (see proposition). is the fixed field of , because elements in cyclicly permutes , , , , and , the only symmetric expression that is fixed by elements in is . As is a normal subgroup of , is also a Galois extension over , and .
In fact, has minimal polynomial , whose resolvent polynomial is with non-square discriminant. This means has to be either or according to the table. is cyclic, so must be , which verifies our previous arguments.