Galois Theory: A Study of Cyclotomic Field ... - Scripps College

18 Cyclotomic Field Extensions
group G(Q(ζ)/Q) is p − 1.
4.2 Structure of the Galois Group
Theorem 16. Let p ∈ Z be prime and ζ = ζ p . Then the Galois group of Q(ζ) over
Q is isomorphic to the multiplicative group (Z/pZ) × of nonzero elements of the
ring Z/pZ and is a cyclic group of order p − 1.
Proof. Let G be the Galois group of Q(ζ) over Q. We will need a map
v : G → (Z/pZ) × as follows: Let σ be an automorphism in G. Then σ will
carry ζ to another root of the polynomial x p + · · · + x + 1, call it ζ i . Since
ζ has multiplicative order p, our exponent i will be an integer modulo p.
We set v(σ) = i. We need to verify that v is multiplicative. If τ is another
element of G such that v(τ) = j, that is, τ(ζ) = ζ j , then
στ(ζ) = σ(ζ j ) = σ(ζ) j = ζ ij = ζ ij .
The identity automorphism will send ζ to ζ, so v(id) = 1. We know that
v is a homomorphism to (Z/pZ) × since v is multiplicative and v(σ) does
not equal 0 for any σ. The homomorphism is injective since ζ generates
Q(ζ) and the action of an automorphism is determined when we know
its action on ζ. Then we know G is isomorphic to its image in (Z/pZ) × .
This means that G is cyclic. The order of G is equal to that of (Z/pZ) × , or
|G| = |(Z/pZ) × |, which equals p − 1. These two groups are isomorphic. □
Since (Z/pZ) × is a cyclic group, so is every subgroup. The Galois group
G of K = Q(ζ p ) over Q is a cyclic group of order p − 1, we know G has one