Let ω be a normal complex number such that ω^5=1. Abstract Algebra: Galois Theory

Let ω be a normal complex number such that ω^5=1. if ϕ is the automorphism of Q(ω) that carries ω to ω^4, find the fixed field of <ϕ>.

Note that since ϕ(ω) = ω^4, we have
ϕ(ω^2) = ϕ(ω)^2 = ω^8 = ω^3
ϕ(ω^3) = ϕ(ω)^3 = ω^12 = ω^2
ϕ(ω^4) = ϕ(ω)^4 = ω^16 = ω.

Next, note that Q(ω) = {a + bω + cω^2 + dω^3 : a, b, c, d in Q}, since 1 + ω + ... + ω^4 = 0.
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By definition, Fix(<ϕ>) = {x in Q(ω) : ϕ(x) = x}

So, we solve a + bω^4 + cω^3 + dω^2 = a + bω + cω^2 + dω^3
==> bω^3 + cω^2 + dω = b + cω + dω^2
==> bω^3 + (c - d)ω^2 + (d - c)ω - b = 0
==> b = 0, c - d = 0, since {1, ω, ω^2, ω^3} is a Q-basis for Q(ω)

So, Fix(<ϕ>) = {a + c(ω^2 + ω^3) : a, c in Q}.


Note that (ω^2 + ω^3)^2 = ω^4 + 2 + ω = (-ω^3 - ω^2 - ω - 1) + 2 + ω = 1 - ω^3 - ω^2.
So, we can say that Fix(<ϕ>) = Q(ω^2 + ω^3).

I hope this helps!