1. ## Galois Theory Question

Let E = Q(α,ω), α = 21/3, ω = e2πi/3, (Q is the field of rationals)
Is E a simple extension of Q?

My attempt:
Let β = α + ω, find a monic irreducible polynomial, p in Q[x] with root β.
p should have degree 6, implying E = Q(β)

β3 = (α + ω)3
= α3 + 3α2ω + 3αω2 + ω3
= 2 + 3α2ω + 3αω2 + 1
= 3αω(α + ω) + 3
= 3αωβ + 3
β3  3 = 3αωβ
3  3)3 = (3αωβ)3
3)3 + 3(β3) 2(-3) + 3(β3) (-3)2 + (-3)3 = 33α3ω3β3
β9 - 9β6 + 27β3  27 = 9(2)(1)β3
= 18 β3
β9 - 9β6 + 9β3  27 = 0

However, if I define p(x) = x9  9x6 + 9x3  27 = 0, my calculator tells me p(21/3 + e2πi/3) ≠ 0.
Also I cant see how to reduce p such that p = fg with f,g in Q[x], deg(f) = 6, and f(β) = 0.

2. ## Re: Galois theory question

Try $\beta=\alpha\omega.$ Then $\beta$ is a root of $x^6-4.$

3. ## Re: Galois theory question

$x^6 - 4 = (x^3 - 2)(x^3 + 2)$
In fact, $E$ is a splitting field for $x^3 - 2 = (x - \alpha)(x - \alpha\omega)(x - \alpha\omega^2)$

I need an irreducible polynomial in $Q$

4. ## Re: Galois Theory Question

i have a different idea, prove Q(α+ω) = Q(α,ω). obviously Q(α,ω) contains Q(α+ω). so if we can show α,ω are in Q(α+ω), we're done. in fact all we NEED to do is show ω is in Q(α+ω), because then: α = α+ω - ω is also in Q(α+ω).

note that: 0 = α3 - 2 = (α+ω - ω)3 - 2 = (α+ω)3 - 3(α+ω)2ω + 3(α+ω)ω2 - ω3 - 2

= (α+ω)3 - 3(α+ω)2ω + 3(α+ω)(-ω-1) - 3 (since ω2+ω+1 = 0)

= (α+ω)3 - 3(α+ω)2ω - 3(α+ω)ω - 3(α+ω) - 3, so that:

ω(3(α+ω)2+3(α+ω)) = (α+ω)3 - 3(α+ω) - 3, thus:

ω = [(α+ω)3 - 3(α+ω) - 3]/(3(α+ω)2+3(α+ω)), which is in Q(α+ω).

by the way, there is an error in your p(x), it should be:

p(x) = x9-9x6-27x3-27 (unless i've also made an error. could be.)